pierreqi/scilab_code · Datasets at Hugging Face

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/2240/CH32/EX31.3/EX31_3.sce

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EX31_3.sce

// Grob's Basic Electronics 11e // Chapter No. 31 // Example No. 31_3 clear; clc; //Calculate the following quantities: Pl, Pcc, Pdmax & percent efficiency // Given data Vin = 20; // Input Voltage=20 Volts(p-p) Vopp = 20; // Output Voltage(p-p)=20 Volts(p-p) Vcc = 24; // Supply Voltage(Collector)=24 Volts Vop = 10; // Output Voltage(peak)=10 Volts Rl = 8; // Load Resistor=8 Ohms Vopp1 = Vopp*Vopp; Pl = (Vopp1/(8*Rl)); disp (Pl,'The Load Power in Watts'); Icc = ((Vop/Rl)*0.318); Pcc = Vcc*Icc disp (Pcc,'The DC Input Power in Watts'); eff = ((Pl/Pcc)*100); disp (eff,'The Efficiency in % is'); Pd = (Vcc*Vcc)/(40*Rl); disp (Pd,'The Maximum Power Dissipation in Watts');

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/165/CH4/EX4.22/ex4_22.sce

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ex4_22.sce

//Example 4.22 close; clc; E=3; //in volts Rm=100; //In ohms Im=1*10^-3; //in amperes //Rs value that will give FSD current Rs=E/Im-Rm; printf('\nValue of Rs that will limit Current to FSD = %.2f k ohm\n',Rs/1000) function [Rx]=deflection(x,Rs,Rm) Rx=(Rs+Rm)/x-(Rs+Rm); endfunction //For 20% deflection funcprot(0); x=20/100; //x=I/Im [Rx]=deflection(x,Rs,Rm); printf('\nValue of Rx that will 20 percent deflection = %.2f k ohm\n',Rx/1000) //For 40% deflection funcprot(0); x=40/100; //x=I/Im [Rx]=deflection(x,Rs,Rm); printf('\nValue of Rx that will 40 percent deflection = %.2f k ohm\n',Rx/1000) //For 50% deflection funcprot(0); x=50/100; //x=I/Im [Rx]=deflection(x,Rs,Rm); printf('\nValue of Rx that will 50 percent deflection = %.2f k ohm\n',Rx/1000) //For 75% deflection funcprot(0); x=75/100; //x=I/Im [Rx]=deflection(x,Rs,Rm); printf('\nValue of Rx that will 75 percent deflection = %.2f k ohm\n',Rx/1000) //For 90% deflection funcprot(0); x=90/100; //x=I/Im [Rx]=deflection(x,Rs,Rm); printf('\nValue of Rx that will 90 percent deflection = %.2f k ohm\n',Rx/1000) //For 100% deflection funcprot(0); x=100/100; //x=I/Im [Rx]=deflection(x,Rs,Rm); printf('\nValue of Rx that will 100 percent deflection = %.2f k ohm\n',Rx/1000)

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/479/CH2/EX2.1/Example_2_1.sce

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Example_2_1.sce

//Chemical Engineering Thermodynamics //Chapter 2 //P-V-T Relations //Example 2.1 clear; clc; //Given m = 140;//m is the mass of N2 in Kg P = 4.052*(10^5);//P is the pressure of the system in Pa V = 30;//V is the volume of the system in m^3 R = 8314.4;// R is the gas constant //To determine temperature required T = P*V/((m/28)*R);//T is the temperature of the system in K mprintf('Temperature of the system is %f K',T); //end

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/tests/mutex.tst

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scartill/semu

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mutex.tst

source $SEMU_ROOT/lib/kernel/build.list SRC="hw.sasm $KERNEL_SRC mutex/app.sasm" CMP=mutex/output.log

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/991/CH4/EX4.6/Example4_6.sce

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Example4_6.sce

//Example 4.6. clc ni=1.5*10^10 un=1300 up=500 q=1.6*10^-19 nos=5*10^22 disp("(a) In intrensic condition, n=p=ni") disp("Hence, sigma_i = q*ni*(un+up)") format(8) sigma_i = q*ni*(un+up) disp(sigma_i,"sigma_i(S/cm) = ") disp("(b) Number of silicon atoms/cm^3 = 5*10^22") ND=5*10^22/10^8 disp(ND,"Hence, ND(cm^-3) = ") disp("Further, n = ND") disp("Therefore, p = ni^2/n = ni^2/ND") p=ni^2/ND disp(p,"p(cm^-3) = ") // wrong answer in textbook disp("Thus p << n. Hence p may be neglected while calculating the conductivity.") disp("Hence, sigma = n*q*un = ND*q*un") sigma=ND*q*un disp(sigma,"sigma(S/cm) = ") NA=(5*10^22)/(5*10^7) disp(NA,"(c) NA(cm^-3) = ") disp("Further, p = NA") disp("Hence, n = ni^2/p = ni^2/NA") n=ni^2/NA disp(n,"n(cm^-3)= ") disp("Thus p >> n. Hence n may be neglected while calculating the conductivity.") disp("Hence, sigma = p*q*up = NA*q*up") sigma1=NA*q*up disp(sigma1,"sigma(S/cm) = ") disp("(d) With both types of impurities present simultaneously, the net acceptor impurity density is,") Na=NA-ND disp(Na,"Na(cm^-3) = NA - ND = ") disp("Hence, sigma = Na*q*up") sigma2=Na*q*up disp(sigma2,"sigma(S/cm) = ")

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/620/CH6/EX6.3/example6_3.sce

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example6_3.sce

r1=400; r2=2*10^3; g3=10^(-3); v=10; disp("Part a"); g=1/r1+1/r2+g3; disp("the total conductance (in mS) is"); disp(g*10^3); disp("Part b"); r=1/g; disp("the combined resistance (in Ω) is"); disp(r); disp("Part c"); i=v/r; disp("the total current drawn (in mA) from the source is"); disp(i*10^3);

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/3311/CH2/EX2.12/Ex2_12.sce

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Ex2_12.sce

// chapter 2 // example 2.12 // Calculate one cycle surge current rating // page-54-55 clear; clc; // given I_sub=3000; // in A (half cycle surge current rating) f=50; // in Hz (supply frequency) T=100, t=50; // time ration for one and half cycle respectively // calculate // I^2*T=I_sub^2*t, therefore we get I=sqrt(I_sub^2*(t/T)); // calculation of one cycle surge current rating printf("\nThe one cycle surge current rating \tI= %.2f A",I); rating=(I_sub/(5*sqrt(2)))^2*(1/(2*f)); // calculation of i^2*t rating printf("\n\nThe i^2*t rating is \t %.f A^2.s",rating); // Note: The answer in the book for i^2*t rating is wrong due to calculation mistake

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/2084/CH9/EX9.8w/9_8w.sce

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9_8w.sce

//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 9.8w //calculation of the distance from launching point //given data u=100//speed(in m/s) of the projectile theta=37//angle(in degree) of the projectile above horizontal g=10//gravitational acceleration(in m/s^2) of the earth //calculation xcm=(2*u*u*sind(theta)*cosd(theta))/g//range of original projectile //also xcm=((m1*x1)+(m2*x2))/(m1+m2) //here m1=M/4 and m2=3*M/4 x1=xcm/2//since small part falls from heighest point i.e half of range x2=(4/3)*((xcm*((1/4)+(3/4)))-(x1/4)) printf('the distance of landing of heavier piece from launching point is %d m',x2)

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/3769/CH5/EX5.35/Ex5_35.sce

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Ex5_35.sce

clear //Given R1=5 //ohm R2=5.0 //ohm R=6 //Calculation n=(1/(R-R1)*R2) //Result printf("\n There are %0.3f resistance are in parallel", n)

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/2534/CH5/EX5.11/Ex5_11.sce

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Ex5_11.sce

//Ex5_11 clc Rf = 10 RL = 150 eta_r = 40.6/(1+Rf/RL) disp("Rf = "+string(Rf)+"ohm")//forward resistance disp("RL = "+string(RL)+"ohm")//load resistance disp("eta_r = 40.6/(1+Rf/RL) = "+string(eta_r)+"%")//rectification efficiency

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/278/CH9/EX9.11/ex_9_11.sce

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ex_9_11.sce

//solution //given b=200//mm t=12.5//mm ft=80//N/mm^2 T=65//N/mm^2 fc=160//N/mm^2 pi=3.14 printf("the value of d is,%f mm\n",6*sqrt(t)) //standard value of d=21.5mm d=21.5//mm //let n be number of rivets Pt=(b-d)*t*ft//N Ps=1.75*(pi/4)*d^2*T//N Pc=d*t*fc//N n=Pt/Ps t1=0.75*t//mm Pt1=(b-d)*t*ft//N Pt2=(b-2*d)*t*ft+Ps//N Pt3=(b-2*d)*t*ft+(3*Ps)//N Ps5=5*Ps//N//for 5 rivets Pc5=5*Pc//N//for 5 rivets P=b*t*ft//N printf("the value of forces is,%f N\n,%f N\n,%f N\n,%f N\n,%f N\n",Pt1,Pt2,Pt3,Ps5,Pc5) //eff=least(Pt1.Pt2,Pt3,Ps5,Pc5)/P eff=Pt1/P//since Pt1 is least p=3*d +5//mm//pitch m=1.5*d//mm d1=2.5*d//mm//dis btw rows of rivets printf("the diameter is,%f mm\n",d) printf("the nuber of rivets is,%f\n",n) printf("the thickness of strap is,%f mm\n",t1) printf("the eff is,%f\n",eff) printf("the pitch is,%f mm\n",p) printf("the marginl pitch is,%f mm\n",m) printf("the dis btw row is,%f mm",d1)

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/1223/CH10/EX10.12/Ex10_12.sce

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Ex10_12.sce

clear; clc; //Example 10.12 //lambda=y yn=0.01; yp=0.01; Vtn=1; Kn=1; Iref=0.5; gm=2*sqrt(Kn*Iref); printf('\ntransconductance =%.2fmA/V\n',gm) go=yn*Iref; printf('\nsmall signal transistor conductance=%.4f mA/V\n',go) go2=go; //for Rl=infinity Av=-gm/(go+go2); printf('\nvoltage gain=%.2f \n',Av) Rl=100;//Kohm gl=0.01; Av=-gm/(go+gl+go2); printf('\nvoltage gain=%.2f \n',Av)

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/788/CH14/EX14.8.b/14_8_soln.sce

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14_8_soln.sce

clc; pathname=get_absolute_file_path('14_8_soln.sce') filename=pathname+filesep()+'14_8_data.sci' exec(filename) // Solutions: // inlet pressure, p_in=101; //kPa // actual power, act_kW=((p_in*Q)/(17.1*(eff_o/100)))*(((p_out+101)/p_in)^0.286-1); //kW // electric power required to drive electric motor, elect_kW=act_kW/(eff_mot/100); //kW // rounding off the above answer elect_kW=fix(elect_kW)+(fix(round((elect_kW-fix(elect_kW))*10))/10); //kW // cost of electricity, yearly_cost=elect_kW*t*cost_per_wat; //$/yr // Results: printf("\n Results: ") printf("\n The cost of electricity per year is %.0f $.",yearly_cost) printf("\n The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)")

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/2345/CH2/EX2.15/Ex2_15.sce

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Ex2_15.sce

//Finding resistance //Example 2.15(pg. 28) clc clear n=100//no of slots c=12//conductors per slot Lm=300// mean length of turn in cm a=1.5*0.2//cross section of each conductor in cm^2 s=1.72*(10^-6)//specific resistance of copper at 20 degreeC p=4// poles t=20,T=75//temp in degreeC k0=0.00427//temp coefficient of resistivity for copper L=n*c*Lm//total length of conductors Ls=L/p//length of conductors in each parallel path s0=s*(1-(k0*t)) printf('Thus specific resistance at 0 degree C is %e ohm-cm \n',s0) RT=(s0*Ls)/a printf('Thus resistance at working temp of 75 degree C is %3.4f ohm',RT)

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/app/views/listings/listings.sce

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TheBrenny/Web-Dev-and-Security

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2023-06-17T08:33:35.176024

2021-06-15T05:07:20

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listings.sce

[[i= partials/header ]] [[i= partials/navbar ]] <div class="container center"> <div class="search"> <input placeholder="Search" type="text" name="search" value="[[search]]"> <div class="btn">Search</div> </div> <hr style="margin:2em;height:2px;"> [[?= listings.length == 0 ]] <p>It seems that you're looking for something that's already been bought, or is not yet up for sale. Check back later, or try a different search term.</p> [[3=]] <div id="listingTable" class="table"> <div class="row head"> <div class="cell center">Preview</div> <div class="cell center">Name</div> <div class="cell center">Description</div> <div class="cell center">Cost</div> <div class="cell center">Seller</div> <div class="cell center" style="width:2em;">Add</div> </div> [[e= item in listings ]] [[c= components/listingrow || item=item link="true" showAddButton="true" ]] [[?==]] </div> [[?==]] </div> <link rel="stylesheet" href="/assets/css/listings.css"> <script src="/assets/js/listings.js"></script> [[i= partials/footer ]]

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/Positive_Negative_test/Netezza-Base-DataMining/SP_VarCluster-NZ-01.tst

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SP_VarCluster-NZ-01.tst

-- Fuzzy Logix, LLC: Functional Testing Script for DB Lytix functions on Netezza -- -- Copyright (c): 2015 Fuzzy Logix, LLC -- -- NOTICE: All information contained herein is, and remains the property of Fuzzy Logix, LLC. -- The intellectual and technical concepts contained herein are proprietary to Fuzzy Logix, LLC. -- and may be covered by U.S. and Foreign Patents, patents in process, and are protected by trade -- secret or copyright law. Dissemination of this information or reproduction of this material is -- strictly forbidden unless prior written permission is obtained from Fuzzy Logix, LLC. -- -- -- Functional Test Specifications: -- -- Test Category: Data Mining Functions -- -- Test Unit Number: FLAggClustering-TD-01 -- -- Name(s): FLAggClustering -- -- Description: Uses a principal component analysis for dimensionality reduction in order to cluster a given set of input variables into a smaller representative set. -- The number of output clusters depend on the contribution level specified. -- Applications: -- Signature: SP_VarCluster(IN TableName VARCHAR(256), -- IN MatrixType VARCHAR(10), -- IN Contrib DOUBLE PRECISION, -- IN TableOutput BYTEINT, -- OUT ResultTable VARCHAR(256) -- -- Parameters: See Documentation -- -- Return value: VARCHAR(256) -- -- Last Updated: 04-02-2014 -- -- Author: <gandhari.sen@fuzzyl.com> -- BEGIN: TEST SCRIPT -- .run file=../PulsarLogOn.sql ---- Drop and recreate the test table DROP TABLE tblVarClusterTest; CREATE TABLE tblVarClusterTest ( ObsID BIGINT, VarID INTEGER, Num_Val DOUBLE PRECISION) DISTRIBUTE ON (ObsID); ---- BEGIN: NEGATIVE TEST(s) -- Case 1a: ---- Incorrect table name CALL SP_VarCluster('tblVarClusterTestNotExist','COVAR', 0.75, 'OutPutTable'); CALL SP_VarCluster(NULL,'COVAR', 0.75, 'OutPutTable'); CALL SP_VarCluster('','COVAR', 0.75, 'OutPutTable'); -- Result: Fuzzy Logix specific error message -- Case1b : ---- No data in table CALL SP_VarCluster('tblVarClusterTest', 'COVAR', 0.75, 'OutPutTable'); --goes through and returns empty set --Populate the table INSERT INTO tblVarClusterTest SELECT * FROM tblLogRegr; -- Case 2: CALL SP_VarCluster('tblVarClusterTest','CLOVAR', 0.75, 'OutPutTable'); --Case 3 CALL SP_VarCluster('tblVarClusterTest','COVAR', -10.75, 'OutPutTable'); CALL SP_VarCluster('tblVarClusterTest','COVAR', 1.75, 'OutPutTable'); ---Case 4 CALL SP_VarCluster('tblVarClusterTest','COVAR', 0.9, ''); CALL SP_VarCluster('tblVarClusterTest','COVAR', 0.9, NULL); DROP TABLE OutPutTable; ---- END: NEGATIVE TEST(s) ---- BEGIN: POSITIVE TEST(s) -- Case a: DELETE FROM tblVarClusterTest; INSERT INTO tblVarClusterTest SELECT * FROM tblLinRegr WHERE VarID > 0; ---- Perform Clustering with non-sparse data and small number of VarIDs CALL SP_VarCluster('tblVarClusterTest', 'COVAR', 0.75, 'ResultTable'); -- Result: standard outputs Drop table ResultTable; DELETE FROM tblVarClusterTest; INSERT INTO tblVarClusterTest SELECT * FROM tblLinRegr WHERE VarID > 0 AND NUM_VAL>0; ---- Perform Clustering with non-sparse data and small number of VarIDs CALL SP_VarCluster('tblVarClusterTest', 'COVAR', 0.75, 'ResultTable'); -- Result: standard outputs Drop table ResultTable; ---- Perform Clustering with sparse data CALL FLVarCluster('tbllogregr', 'CORREL', 0.75, 'ResultTable'); -- Result: standard outputs Drop table ResultTable; ----DROP the test table DROP TABLE tblVarClusterTest; -- END: POSITIIVE TEST(s) -- END: TEST SCRIPT

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clc(); clear; //To measure magnetisation of paramagnetic salt //(M1*T1)=(M2*T2).Therefore M2=(M1*T1)/T2 M1=2; //Initial magnetisation in A/m T1=305; //Initial temperature in K T2=321; M2=(M1*T1)/T2 //M2 is magnetisation at 321K printf("Magnetisation at 321 K is %f A/m",M2);

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//===================================================================================== //Chapter 12 example 27 clc;clear all; //variable declaration f = 450*10^3; //resistance inHz C = 250*10^-12; //capcaitance in F Rsh = 0.75; //resistance in Ω Q = 105; //Q-factor //calculations w = 2*(%pi)*f; L = 1/(((w)^2)*(C)); //inductance in uH R = ((w*L)/(Q))-Rsh; //resistance of the coil in Ω //result mprintf("inductance = %3.2f uH",(L*10^6)); mprintf("\n resistance of the coil = %3.2f Ω",R);

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//Example 6.2 // force clc; clear; close; I=70;// in amperes B=0.4;//flus density in Wb/m^2 n=1;//turns F=B*n*I;// in newton disp(F,"force in newtons is")

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// Chapter 3 example 10 //------------------------------------------------------------------------------ clc; clear; // Given data C = 30; // per unit capacitance in pF/m Vp = 260; // velocity of propagation in m/us f = 500*10^6 // freq in Hz Zl = 50; // terminating load impedance in ohm // calculations v = Vp/10^-6 // conversion from m/us to m/s C1 = C*10^-12 // conversion from pF/m to F/m // 1/sqrt(LC) = Vp L = 1/(v^2 * C1); // per unit inductance Zo = sqrt(L/C1); // charecteristic impedance in ohm lamda = v/f // wavelength b = (2*%pi)/lamda // phase shift constant p = (Zl - Zo)/(Zl + Zo); // Reflection coefficient // Output mprintf('Per Unit inductance = %d nH/m\n Charecteristic Impedance = %d ohm\n Phase shift Constant = %d rad/m\n Reflection co-efficient = %3.3f',L*10^9,Zo,b,abs(p)); //------------------------------------------------------------------------------

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clc; //page no 518 //prob no. 15.2 //Dielelectric strength of air=3MV/m e=3*10^6;//electric field strength Z=377;//impedance of air Pd=(e^2)/Z;//Determination of power density disp('GW/m2',Pd/10^9,'The max power density is');

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// calculate the power loss and relative error clc; I=64*10^-3; R=3200; P=(I^2)*R; disp(P,'Power(W)=') Re=2*0.75+0.2; disp(Re,'Relative error (%)=')

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//ques11 //Enthalpy of Compressed Liquid clear clc //the water exists as a compressed liquid at the specified state //(a) using compressed liquid table P=15000;//pressure in kPa T=100;//temperature in C h=430.39;//heat of water in kJ/kg from Table A-7 printf('(a) Heat of water using compressed liquid table = %.2f kJ/kg \n',h); //(b) Approximating the compressed liquid as a saturated liquid at 100°C h=419.17;//heat of water at liquid state in kJ/Kg ie hf @ 100C printf(' (b) Heat of Water by approximating compressed liquid as saturated = %.2f kJ/kg \n',h); //(c) Using correction method vf=0.001//specific volume of water in saturated liquid state @100C Psat=101.42//saturated pressure in kPa from Table h=h+vf*(P-Psat)//corrected value of heat of water at given state in kJ/kg printf(' (c) Heat of water using correction method = %.2f kJ/kg ',h);

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// Copyright (c) 2015, Embedded Solutions // All rights reserved. // This file is released under the 3-clause BSD license. See COPYING-BSD. mode(-1); lines(0); function main_builder() TOOLBOX_NAME = "microdaq"; TOOLBOX_TITLE = "MicroDAQ toolbox"; toolbox_dir = get_absolute_file_path("builder.sce"); // Check Scilab's version // ============================================================================= // check minimal version (xcosPal required) if ~isdef('xcosPal') then // and xcos features required error(gettext('Scilab 5.3.2 or more is required.')); end // Check modules_manager module availability // ============================================================================= if ~isdef('tbx_build_loader') then error(msprintf(gettext('%s module not installed.'), 'modules_manager')); end if ~isdir(toolbox_dir+filesep()+"images"+filesep()+"h5") [status, msg] = mkdir(toolbox_dir+filesep()+"images"+filesep()+"h5"); if and(status <> [1 2]) error(msg); end end // Action // ============================================================================= sci_ver_str = getversion('scilab', 'string_info'); tbx_builder_macros(toolbox_dir); // tbx_builder_help(toolbox_dir); if sci_ver_str == 'scilab-5.5.2' then tbx_build_loader(TOOLBOX_NAME, toolbox_dir); else tbx_build_loader(toolbox_dir); end // help_path = pathconvert(toolbox_dir+'jar/'+sci_ver_str); // if isdir(toolbox_dir+'/jar/'+sci_ver_str) == %F then // mkdir(help_path); // end // movefile(pathconvert(toolbox_dir+'jar/scilab_en_US_help.jar', %F), pathconvert(help_path+'scilab_en_US_help.jar', %F)); //delete mblockstmpdir file to force loader to rebuild mdaq palette deletefile(toolbox_dir+filesep()+"etc"+filesep()+"mblockstmpdir"); endfunction if with_module('xcos') then main_builder(); clear main_builder; // remove main_builder on stack end

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clc //solution // initialization of variables r=18 // compression ratio k=1.4 // polytropic index for air R=0.287 // specific gas constant for air T1=200+273 // lower temperaure in kelvin P1=200 // low pressure in kPa T3=2000 // higher temperature of cycle in kelvin v1=R*T1/P1 // specific volume at state 1 in m^3 //using table E.1 u1=340 // specific internal energy in kJ/kg vr1=198.1 // in m^3/kg vr2=vr1*(1/r) // as r=v1/v2 // now finding corresponding values from table E.1 T2=1310 // temperature in kelvin Pr2=34 // pressure in kPa h2=1408 // specific entropy in kJ/kg v2=v1/18 // volume at state 2 P2=R*T2/v2 // pressure at state 2 h3=2252.1 // specific enthalpy in kJ/kg from table E.1 vr3=2.776 P3=P2 // diesel cycle v3=R*T3/P3 // after compression volume v4=v1 // isochoric process vr4=vr3*v4/v3 // isentropic process // now using Vr4 we read corresponding value from table E.1 T4=915 // final temperature in kelvin u4=687.5 // specific internal energy at state 4 qin=h3-h2 // using first law qout=u4-u1 // heat rejected Wnet=qin-qout // net work effi=100*Wnet/qin // thermal efficiency printf("The thermal efficiency is %0.2f %% \n",effi) MEP=Wnet/(v1-v2) // expression of mean effective pressure in terms of work printf(" The MEP is %0.2f kPa \n",MEP) erroreffi=(66.6-effi)*100/effi // error in efficiency errorMEP=(515-MEP)*100/MEP // error in MEP printf(" The %% error in efficiency is %0.1f %% \n",erroreffi) printf(" The %% error in MEP is %0.1f %% \n",errorMEP) // the answers are slight different due to approximation in textbook ... here answers are precise

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temp.sce

clear mite_473_977_10uA; mite_473_977_10uA=csvRead("~/rasp30/prog_assembly/libs/scilab_code/characterization/char_miteADC/data_miteADC473_977_chip"+chip_num+brdtype); //polyfit [p_mite_977_10uA,S_mite_977_10uA]=polyfit(mite_473_977_10uA(:,1), mite_473_977_10uA(:,2),7); size_a=size(mite_473_977_10uA); //MITE_range_977 = mite_473_977_10uA(size_a(1,1),1):1:mite_473_977_10uA(1,1); MITE_range_977 = mite_473_977_10uA(size_a(1,1),1):1:9600; MITE_fit_977 = polyval(p_mite_977_10uA,MITE_range_977,S_mite_977_10uA); scf(7);clf(7); plot2d("nn", mite_473_977_10uA(:,1), mite_473_977_10uA(:,2));p = get("hdl"); p.children.mark_style = 9; p.children.thickness = 3; p.children.line_mode="off";p.children.mark_foreground=1; plot2d("nn", MITE_range_977, MITE_fit_977, style=1);p = get("hdl"); p.children.line_style = 1; p.children.thickness = 3; p.children.thickness = 3;p.children.line_mode="on"; a=gca();//a.data_bounds=[0 0; 150 2.6]; legend("mite_473_977","mite_473_978","mite_473_979","in_upper_right"); xtitle("","ADC code","Vg (V)"); // Vg=0.005 !!!!! to make 0x0078 clear onchip_dac02_char_data; onchip_dac02_char_data=csvRead("~/rasp30/prog_assembly/libs/scilab_code/characterization/char_onchipDAC/data_onchipDAC02_chip"+chip_num+brdtype); [p_onchip_dac02,S_onchip_dac02]=polyfit(onchip_dac02_char_data(:,1), onchip_dac02_char_data(:,2),3); DAC02_HEX_range = hex2dec('0000'):1:hex2dec('007F'); DAC02_fit = polyval(p_onchip_dac02,DAC02_HEX_range,S_onchip_dac02); DAC02_char_table = [DAC02_HEX_range(:) DAC02_fit(:)]; // Plot the data scf(6); clf(6); plot2d("nn", onchip_dac02_char_data(:,1), onchip_dac02_char_data(:,2));p = get("hdl"); p.children.mark_style = 9; p.children.thickness = 3; p.children.line_mode="off";p.children.mark_foreground=3; plot2d("nn", DAC02_HEX_range, DAC02_fit, style=3);p = get("hdl"); p.children.line_style = 1; p.children.thickness = 3; p.children.thickness = 3;p.children.line_mode="on"; a=gca();a.data_bounds=[0 0; 150 2.6]; legend("DAC00","DAC01","DAC02","DAC03","DAC04","DAC05","DAC06","DAC07","DAC08","DAC09","in_upper_right"); xtitle("","ADC code","Vg (V)");

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Ex9_1.sce

//chapter9 //example9.1 //page142 printf("in fig. (i), the conventional current coming out of battery flows in the \nbranch circuits. In diode D1,the conventional current flows in the \ndirection of arrowhead and hence this diode is forward biased. \nHowever in diode D2, the conventional current flows opposite \nto arrowhead and hence this diode is reverse biased.\n \n") printf("in fig. (ii), During the positive half cycle of input ac voltage, the \nconventional current flows in the direction of arrowhead and hence diode \nis forward biased. However, during the negative half cycle \nof input ac voltage, the diode is reverse biased.\n \n") printf("in fig. (iii), During the positive half cycle of input ac voltage, the \nconventional current flows in the direction of arrowhead in D1 but it flows \nopposite to arrowhead in D2. So during positive half cycle, \ndiode D1 is forward biased and diode D2 is reverse biased. \nHowever in the negative half cycle of the input ac voltage, diode D2 \nis forward biased and diode D1 is reverse biased.\n \n") printf("in fig. (iv), During the positive half cycle of input ac voltage, \nboth diodes are reverse biased. However in the negative half cycle of the \ninput ac voltage, both diodes are forward biased.\n \n")

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example20_sce.sce

//chapter 5 //example 5.20 //page 2 printf("\n") printf("given") Vcc=10;Rc=1*10^3;Rb=6.8*10^3;Vs=5; disp(" hFE calculation") Ic=Vcc/Rc Ib=(Vs-Vbe)/Rb hFE=Ic/Ib disp("when hFE=10") hFE=10 Ic=hFE*Ib Vce=Vcc-(Ic*Rc)

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ex1_11.sce

//Page Number: 1.14 //Example 1.11 clc; //(a) //Given //Signal is x(t) = rect(t) //rect(t) = 1 for -a< |t| < a and 0 elsewhere //Therefore //We find out fourier transform of x(t)= 1 for -a< |t| < a thus, x=1; a= 200; //Assume t= -a : 1 : a; //range for fourier transform y=fft(x); disp(y,'Fourier transform of x(t)='); //(b) //Given //Signal is x(t) = rect(t) //rect(t) = 1 for -a/4< |t| < a/4 and 0 elsewhere //Therefore //We find out fourier transform of x(t)= 1 for -a/4< |t| < a/4 thus, x=1; a= 200; //Assume t= -a/4 : 1 : a/4;//range for fourer transform y=fft(x); disp(y,'Fourier transform of x(t)='); //(c) //Given //Signal is x(t) = rect(t) //rect(t) = 1 for b < |t| < b + a/2 and 0 elsewhere //Therefore //We find out fourier transform of x(t)= 1 for b < |t| < b+ a/2 thus, x=1; a= 200; //Assume b=100; //Assume t= b : 1 : (b+(a/2));//range for fourer transform y=fft(x); disp(y,'Fourier transform of x(t)=');

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4_10.sce

//All the quantities are expressed in SI units c = 1.5; //airfoil chord length Rex_cr = 1e6; //critical Reynold's number Re_c = 3.1e6; //Reynold's number at the trailing edge //the point of transition is given by x1 = Rex_cr/Re_c*c; //the various skin friction coefficients are given as Cf1_laminar = 1.328/sqrt(Rex_cr); //this is a mistake in the book in calulation of this quantity thus the answer in book is wrong Cfc_turbulent = 0.074/(Re_c^0.2); Cf1_turbulent = 0.074/(Rex_cr^0.2); //thus the total skin friction coefficient is given by Cf = x1/c*Cf1_laminar + Cfc_turbulent - x1/c*Cf1_turbulent; //taking both sides of plate into account Net_Cf = 2*Cf; printf("\nRESULTS\n--------\nThe net skin friction coefficient is\n Net Cf = %1.5f",Net_Cf)

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gmv_incremental.sce

// Exemplo: regulador de variância mínima clear; xdel(winsid()); clc; // Condições Iniciais z = %z; Ts = 0.1; tfinal = 50; N = round( tfinal/Ts ); t = 0:Ts:N*Ts-Ts; yr(1:50)=0; yr(51:N+1)=1; y(1:3)=0; u(1:3)=0; e(1:3)=0; // Parametros fornecidos Bz = [0.0234015 0.5489666 0.0751607] b0 = Bz(1); b1 = Bz(2); b2 = Bz(3); Az = [1 -0.4641429 0.077525 0.0429111]; a1 = Az(2); a2 = Az(3); a3 = Az(4); // Ruido branco Variancia = 0.0007362; xi = grand(N, "mn", 0, Variancia); // Projeto do GMV Incremental du(1:3)=0; Delta = [1 -1]; // Delta = 1-z^-1 Abarz = conv(Delta,Az); // Delta*Az abar1 = Abarz(2); abar2 = Abarz(3); abar3 = Abarz(4); abar4 = Abarz(5); f0 = -abar1; f1 = -abar2; f2 = -abar3; f3 = -abar4; q0 = 1; for k = 4:N // Simula modelo do processo y(k) = -a1*y(k-1) -a2*y(k-2) -a3*y(k-3)+b0*u(k-1) +b1*u(k-2) +b2*u(k-3)+ xi(k); // Parcela do regulador de variância mínima du(k) = (1/(b0+q0))*( -b1*du(k-1) -b2*du(k-2)+yr(k+1) -f0*y(k) -f1*y(k-1) -f2*y(k-2) -f3*y(k-3)); u(k) = u(k-1) + du(k); e(k) = yr(k) - y(k); end subplot(3,1,1); plot(t,yr(1:N),'k') ylabel('yr(t) [Unid.]'); xlabel('Tempo [s]'); legend({'Referência'}, 'Location','northwestoutside') subplot(3,1,2); plot(t,u,'b') ylabel('u(t) [Unid.]'); xlabel('Tempo [s]'); legend({'Controle'},'Location','northwestoutside') subplot(3,1,3); plot(t, y, 'r') ylabel('y(t) [Unid.]'); xlabel('Tempo [s]'); legend({'Saída'}, 'Location','northwestoutside')

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Ex3_12.sce

clc // Fundamental of Electric Circuit // Charles K. Alexander and Matthew N.O Sadiku // Mc Graw Hill of New York // 5th Edition // Part 1 : DC Circuits // Chapter 3: Methods of Analysis // Example 3 - 12 clear; clc; close; // Given data Vs = 4.00 VBE = 0.70 Rb = 20000.00 beta = 50.00 // // Calculations // Calculations Ib Ib = ((Vs - VBE)/Rb)*10^6; // Calculations Ic Ic = beta * Ib; // Calculations Vo Vo = 6.00 - 100*Ic; // // Display the result disp("Example 3-12 Solution : "); printf(" \n Ib = Current basis = %.3f A",Ib) printf(" \n Ic = Current collector = %.3f A",Ic) printf(" \n Vo = Voltage collector - emitter = %.3f A",Vo)

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Example3_4.sci

clear; clc(); // To calculate heat loss from pipe d1=10.75/12; // outer diameter of pipe in ft x1=1.5/12; // thickness of insulation 1 in ft x2=2/12; // thickness of insulation 2 in ft d2= d1+2*x1; // diameter of insulation 1 in ft d3=d2+2*x2; // diameter of insulation 1 in ft t1=700; // inner surface temperature of composite insulation in degF t2=110; // outer surface temperature of composite insulation in degF k1=0.05; //thermal conductivity of material 1 in Btu/hr-ft-degF k2=0.039; // thermal conductivity of material 2 in Btu/hr-ft-degF q=2*%pi*(t1-t2)/(log(d2/d1)/k1+log(d3/d2)/k2); // heat loss per linear foot in Btu/hr printf("\n The heat loss is found to be %d Btu/hr-ft", q);

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luu.sci

function[L,U,X]=LU_Parker(A,B) % Now use a vector y to solve 'Ly=b' y=zeros(m,1); y(1)=B(1)/L(1,1); for i=2:m y(i)=-L(i,1)*y(1); for k=2:i-1 y(i)=y(i)-L(i,k)*y(k); y(i)=(B(i)+y(i))/L(i,i); end; end; % Now we use this y to solve Ux = y x=zeros(m,1); x(1)=y(1)/U(1,1); for i=2:m x(i)=-U(i,1)*x(1); for k=i:m x(i)=x(i)-U(i,k)*x(k); x(i)=(y(i)+x(i))/U(i,i); end; end A = [12,-1,3; 4,-4,9;1,7,-4]; b = [8;62;-51]; [B,C,D]=lu(A)

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chapter3_ex1.sce

clc clear //input n1=420;//number of conductors in armature of a d.c. machine phi=0.024;//flux produced by each pole in weber e=250;//desired e.m.f in volts n2=4;//number of poles of the d.c. machine //calculations N=n1/2;//number of conductors per path and there are two parallel paths //e1= e.m.f induced per conductor=(4*0.024*w)/(2*%pi) where w is the required angular velocity in rad/s w=e/((n1*(48*10^-3))/(2*%pi));//required angular velocity in rad/s //output mprintf('the armature of hte machine must have an angular velocity of %3.0f rad/s',w)

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// Ex3_5 clc; // Given: A=87; // Solution: z=(40*A)/(0.6*(A^(2/3))+80); printf("The stable nuclied of the isobaric series is Sr atomic no. = %f",z) // nereast integer is 38 printf("\n Hence the nuclides of z<38 fall on the left of the limb of B vs Z parabola while the nuclides of z>38 fall on the right limb of the parabola.")

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// Example 5.5, page no-312 clear clc Cd=0.6 D=150*10^-3 d=75*10^-3 p=250 g=9.8 rho=1000 s=75*10^-3 //(a) Q=Cd*3.14*s^2*sqrt(2*g*p/rho)/(4*sqrt(1-(d/D)^4)) printf("(a) For orifice plate\nQ=%f m^3/sec = %.3f litres/sec",Q,Q*1000) Cd1=0.99 Q2=Cd1*3.14*s^2*sqrt(2*g*p/rho)/(4*sqrt(1-(d/D)^4)) printf("\n\n(b)For venturi tube\nQ=%f m^3/sec = %.2f litres/sec",Q2,Q2*1000)

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arma_rnd.sci

<<<<<<< HEAD //Return a simulation of the ARMA model. //Calling Sequence //arma_rnd (a, b, v, t, n) //arma_rnd (a, b, v, t) ======= function res = arma_rnd (a, b, v, t, n) //Return a simulation of the ARMA model. //Calling Sequence //arma_rnd (a, b, v, t, n) //arma_rnd (a, b, v, t) >>>>>>> 6bbb00d0f0128381ee95194cf7d008fb6504de7d //Parameters //a: vector //b: vector //v: Variance //t: Length of output vector //n: Number of dummy x(i) used for initialization <<<<<<< HEAD ======= >>>>>>> 6bbb00d0f0128381ee95194cf7d008fb6504de7d //Description //This is an Octave function. //The ARMA model is defined by // //x(n) = a(1) * x(n-1) + … + a(k) * x(n-k) // + e(n) + b(1) * e(n-1) + … + b(l) * e(n-l) //in which k is the length of vector a, l is the length of vector b and e is Gaussian white noise with variance v. The function returns a vector of length t. // //The optional parameter n gives the number of dummy x(i) used for initialization, i.e., a sequence of length t+n is generated and x(n+1:t+n) is returned. If n is omitted, n = 100 is used. <<<<<<< HEAD //Examples //a = [1 2 3 4 5]; //b = [7 8 9 10 11]; ======= //Examples //a = [1 2 3 4 5]; //b = [7; 8; 9; 10; 11]; >>>>>>> 6bbb00d0f0128381ee95194cf7d008fb6504de7d //v = 10; //t = 5; //n = 100; //arma_rnd (a, b, v, t, n) <<<<<<< HEAD //Output : // ans = // // 61400.907 // 158177.11 // 407440.29 // 1049604. // 2703841.3 //function res = arma_rnd (a, b, v, t, n) //funcprot(0); //lhs = argn(1) //rhs = argn(2) //if (rhs < 5 | rhs > 6) //error("Wrong number of input arguments.") //end // //select(rhs) // // case 5 then // res = callOctave("arma_rnd",a, b, v, t) // // case 6 then // res = callOctave("arma_rnd",a, b, v, t, n) // // end //endfunction function x = arma_rnd (a, b, v, t, n) funcprot(); [nargout,nargin] = argn() ; if (nargin == 4) n = 100; elseif (nargin == 5) if (~ isscalar (n)) error ("arma_rnd: N must be a scalar"); end else error("arma_rnd: invalid input"); end if ((min (size (a)) > 1) | (min (size (b)) > 1)) error ("arma_rnd: A and B must not be matrices"); end if (~ isscalar (t)) error ("arma_rnd: T must be a scalar"); end ar = length (a); br = length (b); a = matrix (a, ar, 1); b = matrix (b, br, 1); // Apply our notational convention. a = [1; -a]; b = [1; b]; n = min (n, ar + br); e = sqrt (v) * rand(t + n, 1); x = filter (b, a, e); x = x(n + 1 : t + n); ======= //ans = // // -1.6176e+05 // -4.1663e+05 // -1.0732e+06 // -2.7648e+06 // -7.1221e+06 funcprot(0); lhs = argn(1) rhs = argn(2) if (rhs < 5 | rhs > 6) error("Wrong number of input arguments.") end select(rhs) case 5 then res = callOctave("arma_rnd",a, b, v, t) case 6 then res = callOctave("arma_rnd",a, b, v, t, n) end >>>>>>> 6bbb00d0f0128381ee95194cf7d008fb6504de7d endfunction

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//Book - Power System: Analysis & Design 5th Edition //Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J. Overbye //Chapter - 10 ; Example 10.3 //Scilab Version - 6.0.0 ; OS - Windows clc; clear; Crnttap=6; //Current tap setting in Amperes TDsetting=1; //Time dial setting CTratio=100/5; //CT ratio IZB=[5 0.5; 8 0.8; 15 1.5]; //Secondary output current in Amperes and burden resistance in Ohm RC_multiple_Crntap=IZB(1,1)/Crnttap; //Relay current in the multiple of the current tap setting printf('\nCase: a'); if (RC_multiple_Crntap<1) then printf('\nFor the relay current in the multiple of the current tap setting %0.4f \nThe relay will not operate',RC_multiple_Crntap); else printf('\nFor the relay current in the multiple of the current tap setting %0.4f \nThe relay will operate after %0.2f Seconds',RC_multiple_Crntap,time); end RC_multiple_Crntap=IZB(2,1)/Crnttap; //Relay current in the multiple of the current tap setting time=6 //Relay operating time from figure 10.12 in Seconds printf('\n\nCase: b'); if (RC_multiple_Crntap<1) then printf('\nFor the relay current in the multiple of the current tap setting %0.4f \nThe relay will not operate',RC_multiple_Crntap); else printf('\nFor the relay current in the multiple of the current tap setting %0.4f \nThe relay will operate after %d Seconds',RC_multiple_Crntap,time); end RC_multiple_Crntap=IZB(3,1)/Crnttap; //Relay current in the multiple of the current tap setting time=1.2 //Relay operating time from figure 10.12 in Seconds printf('\n\nCase: c'); if (RC_multiple_Crntap<1) then printf('\nFor the relay current in the multiple of the current tap setting %0.4f \nThe relay will not operate',RC_multiple_Crntap); else printf('\nFor the relay current in the multiple of the current tap setting %0.4f \nThe relay will operate after %0.2f Seconds',RC_multiple_Crntap,time); end

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//Example 1_34 clc(); clear; //To calculate the refractive index of the liquid D1=1.40 //units in centimeters D1=1.40*10^-2 //units in meters D2=1.27 //units in centimeters D2=1.27*10^-2 //units in meters u=(D1/D2)^2 printf("Refractive index of the liquid is %.3f",u)

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//Problem 21.10: A shunt generator supplies a 20 kW load at 200 V through cables of resistance, R = 100 mohm. If the field winding resistance, Rf=D 50ohm and the armature resistance, Ra = 40 mohm, determine (a) the terminal voltage, and (b) the e.m.f. generated in the armature. //initializing the variables: Ps = 20000; // in Watts Vs = 200; // in Volts Rs = 0.1; // in ohms Rf = 50; // in ohms Ra = 0.04; // in ohms //calculation: //Load current, I Is = Ps/Vs //Volt drop in the cables to the load Vd = Is*Rs //Hence terminal voltage, V = Vs + Vd //Field current, If If = V/Rf //Armature current Ia = If + Is //Generated e.m.f. E E = V + Ia*Ra printf("\n\n Result \n\n") printf("\n (a)terminal voltage is %.0f V ",V) printf("\n (b)generated e.m.f. is %.2f V ",E)

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.TH factors 8 "April 1993" "Scilab Group" "Scilab Function" .so man1/sci.an .SH NAME factors - numeric real factorization .SH CALLING SEQUENCE .nf [lnum,g]=factors(pol [,'flag']) [lnum,lden,g]=factors(rat [,'flag']) .fi .SH PARAMETERS .TP pol : real polynomial .TP rat : real rational polynomial (\fVrat=pol1/pol2\fR) .TP lnum : list of polynomials (of degrees 1 or 2) .TP lden : list of polynomials (of degrees 1 or 2) .TP g : real number .TP flag : character string \fV'c'\fR or \fV'd'\fR .SH DESCRIPTION returns the factors of polynomial \fVpol\fR in the list \fVlnum\fR and the "gain" g. .LP One has pol= g times product of entries of the list lnum. If argument of \fVfactors\fR is a 1x1 rational \fVrat=pol1/pol2\fR, the factors of the numerator \fVpol1\fR and the denominator \fVpol2\fR are returned in the lists \fVlnum\fR and \fVlden\fR respectively. .LP The "gain" is returned as \fVg\fR. .LP If \fVflag\fR is \fV'c'\fR (resp. \fV'd'\fR), the roots of \fVpol\fR are refected wrt the imaginary axis (resp. the unit circle), i.e. the factors in \fVlnum\fR are stable polynomials. .LP Same thing if \fVfactors\fR is invoked with a rational arguments: the entries in \fVlnum\fR and \fVlden\fR are stable polynomials.

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// Bode plots for Example 5.7 on page 141 // 5.2 exec('label.sci',-1); omega = linspace(0,%pi); g1 = 0.5 ./ (cos(omega)-0.5+%i*sin(omega)); mag1 = abs(g1); angle1 = phasemag(g1); g2 = (0.5+0.5*cos(omega)-1.5*%i*sin(omega)) ... * 0.25 ./ (1.25-cos(omega)); mag2 = abs(g2); angle2 = phasemag(g2); subplot(2,1,1) plot(omega,mag1,omega,mag2,'--'); label('',4,' ','Magnitude',4); subplot(2,1,2); plot(omega,angle1,omega,angle2,'--'); label('',4,'w (rad/s)','Phase',4);

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clc; //Example 1.4 //Page no 8 //Solution //(a) p1=(10^-2.7)*(10^-3); disp('W', p1,"-27dBm in absolute power is, "); //(b) p2=(10^1.3)*(10^-3); disp('W', p2,"13dBm in absolute power is, "); //(c) p3=(10^4)*(10^-3); disp('W', p3,"40dBm in absolute power is, "); //(d) p4=(10^-5.3)*(10^-3); disp('W', p4,"-53dBm in absolute power is, ");

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P4_6.sce

//EXAMPLE 4.6 //Order of LP filter clc; clear; ap = 1 //Peak passband ripple in dB as = 40 //min. stopband atteuation in dB wp = 1000 //Hz ws = 5000 //Hz k = wp/ws; disp(1/k,'1/k = '); k1 = 1/(sqrt((10^(0.1* as)-1)/(10^(0.1*ap)-1))); disp(1/k1,'1/k1 = '); N=ceil(log10(sqrt((10^(0.1* as)-1)/(10^(0.1*ap)-1)))/log10(1/k)); disp(N,'order of the filter is :');

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// Example 4_16 clc;funcprot(0); // Given data P_1=7;// The inlet pressure in kPa T_1=420;// The inlet temperature in °C d_1=200;// The inlet diameter in mm V_1=400;// The inlet velocity in m/s V_2=700;// The exit velocity in m/s c_p=1000;// J/kg.K R=287;// J/kg.K k=1.4;// The specific heat ratio // Calculation // (a) T_2=((V_1^2-V_2^2)/(2*c_p))+T_1;// The exit temperature in °C // (b) rho_1=(P_1*10^3)/(R*(T_1+273));// kg/m^3 A_1=(%pi*(d_1/1000)^2)/4;// m^2 m=rho_1*A_1*V_1;// The mass flux in kg/s // (c) rho_2=rho_1*((T_2+273)/(T_1+273))^(1/(k-1));// The density at the exit in kg/m^3 d_2=sqrt((rho_1*(d_1/1000)^2*V_1)/(rho_2*V_2));// The exit diameter in m printf("\n(a)The exit temperature,T_2=%3.0f°C \n(b)The mass flux,m=%0.4f kg/s \n(c)The exit diameter,d_2=%0.3f m or %3.0f mm",T_2,m,d_2,d_2*10^3);

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//scilab 5.4.1 //windows 7 operating system //chapter 4:Metal-Semiconductor Contacts clc clear //given e=1.6*10^-19//e=charge of an electron in C Fa=7*10^6//Fa=reverse bias field in V/m Es=13.1*8.854*10^-12//(Es/Eo)=13.1;Eo=8.854*10^-12 dQ=((e*Fa)/(4*%pi*Es))^(1/2)//dQ=barrier lowering in V disp("V",dQ,"dQ=") Xm=(dQ)/(2*Fa)//Xm=position of the maximum barrier height disp("m",Xm,"Xm=")

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// Example no. 2.10 // To design single mode fiber such that absolute accumulated dispersion should not exceed 1100ps/nm // Page no. 77 clc; clear; // Given data lambda1=1530; // Left edge of wavelength range in nm lambda2=1560; // Rigth edge of wavelength range in nm lambda0=1545; // Center of the band in nm L=80; // Fiber length in km disp('We choose center of band (lambda_0) for large maximum allowable dispersion slope.'); Dlambda2=1100/L; // Dispersion at rigth edge of band in ps/nm/km S=Dlambda2/(lambda2-lambda0); // Dispersion slope in ps/nm^2/km // Displaying the result in command window printf('\n Dispersion slope = %0.3f ps/nm^2/km',S);

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clc //initialisation of variables clear t= 75 //sec h= 10.5 //in h1= 13.5 //in //CALCULATIONS r= t*%pi*sqrt(2*h^2)/log((sqrt(2*h1^2)+h1)/(sqrt(2*h^2)-h)) t= -r*((1/h1)-(1/h)) //RESULTS printf ('A/K = %.f ',r) printf ('\n Time taken = %.1f sec',t)

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//Calculate frequency of oscillations clear; clc; //soltion //given R=22*10^3;//ohm C=100*10^-12;//F fo=1/(2*%pi*R*C); printf("The frequency of oscillations= %.2f KHz\n",fo/1000);

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we we wy wy wy x1 x2 klasa1 klasa2 klasa3 -6,449 6,677 1 0 0 -5,973 6,435 1 0 0 -5,476 6,725 1 0 0 -4,731 7,16 1 0 0 -3,882 7,184 1 0 0 -2,847 7,16 1 0 0 -1,915 7,136 1 0 0 -0,756 7,184 1 0 0 0,155 7,063 1 0 0 1,439 6,967 1 0 0 1,832 6,604 1 0 0 2,598 6,701 1 0 0 3,095 6,122 1 0 0 4,813 2,958 1 0 0 4,896 1,195 1 0 0 5,041 -0,351 1 0 0 5,041 -1,824 1 0 0 3,985 -5,639 1 0 0 -1,584 -6,388 1 0 0 -3,923 -6,34 1 0 0 -6,946 -4,77 1 0 0 -7,194 -2,886 1 0 0 -7,774 -2,138 1 0 0 -7,878 -0,326 1 0 0 -7,94 0,64 1 0 0 -7,981 2,572 1 0 0 -7,629 4,334 1 0 0 -5,083 -5,929 1 0 0 1,335 -6,485 1 0 0 4,834 -4,408 1 0 0 -4,731 3,151 0 1 0 -4,151 3,537 0 1 0 -1,315 3,682 0 1 0 -0,342 3,707 0 1 0 1,087 3,586 0 1 0 1,646 2,596 0 1 0 2,205 1,485 0 1 0 2,433 1,026 0 1 0 2,329 0,108 0 1 0 2,308 -1,22 0 1 0 2,163 -1,703 0 1 0 1,894 -2,596 0 1 0 1,046 -2,935 0 1 0 0,176 -3,176 0 1 0 -0,466 -3,128 0 1 0 -0,569 -3,683 0 1 0 -1,19 -3,393 0 1 0 -1,894 -3,587 0 1 0 -3,778 -2,621 0 1 0 -4,648 -2,379 0 1 0 -4,71 -1,172 0 1 0 -5,372 -0,254 0 1 0 -5,124 0,229 0 1 0 -5,186 1,509 0 1 0 -2,764 3,707 0 1 0 -3,302 -3,538 0 1 0 -2,412 -2,91 0 1 0 1,812 1,847 0 1 0 -5,435 -1,606 0 1 0 -5,124 2,161 0 1 0 -1,812 0,784 0 0 1 -1,108 0,833 0 0 1 -0,942 0,471 0 0 1 -1,315 0,229 0 0 1 -1,19 -0,037 0 0 1 -1,48 -0,713 0 0 1 -1,956 -0,568 0 0 1 -1,894 -0,109 0 0 1 -2,743 0,132 0 0 1 -3,095 0,132 0 0 1 -3,074 -0,182 0 0 1 -2,805 -0,447 0 0 1 -0,031 -0,495 0 0 1 0,238 -0,351 0 0 1 0,259 -0,012 0 0 1 0,114 0,374 0 0 1 -0,362 0,06 0 0 1 0,652 -0,109 0 0 1 -0,88 -0,737 0 0 1 -0,611 -0,254 0 0 1 -0,714 -0,012 0 0 1 -0,445 0,35 0 0 1 -0,424 0,736 0 0 1 -2,681 0,567 0 0 1 -2,205 0,374 0 0 1 -2,578 -0,423 0 0 1 -2,267 -0,085 0 0 1 -1,522 -0,23 0 0 1 -1,542 0,277 0 0 1 -2,184 0,784 0 0 1

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10.sce

//Variable declaration EF=5.5*1.602*10**-19; //fermi energy of silver(J) tow=3.97*10**-14; //relaxation time(s) m=9.11*10**-31; //mass(kg) //Calculation vf=sqrt(2*EF/m); //fermi velocity(m/s) lamda=vf*tow; //mean free path(m) //Result printf('fermi velocity is %0.3f *10**6 m/s \n',(vf/10**6)) printf('mean free path is %0.3f *10**-8 m \n',(lamda*10**8))

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ex7_5.sce

//Example 7.5, page no-440 clear clc //(a) A=0.226 B=195 t=60 v=A*t-B/t printf("(a) Fluid X\n v = %.2f centipoises",v) A1=0.220 B1=135 t1=140 v1=A1*t1-B1/t1 printf("\n(b)Fluid Y\n v = %.1f centipoises",v1)

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v1=3300 v2=600 f=50 Ac=25/10000 l=1.2 Bmax=1.2 fluxmax1=Bmax*Ac N1=v1/4.44/f/fluxmax1 N2=v2/4.44/f/fluxmax1 disp(N1,N2) I2=20 pf=0.8 I1=N2/N1*I2 I1=I1*(pf-%i*sin(acos(0.8))) disp(I1) Hmax=250 ATmax=Hmax*l immax=ATmax/N1/sqrt(2) Ii=0 I1=-%i*immax+I1 disp("lagging", real(I1)/norm(I1), norm(I1))

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Example10_4.sce

exec('collision.sci', -1) //Given that m1 = 30*10^-3 //in kg h1 = 8*10^-2 //in m m2 = 75*10^-3 //in kg g = 9.8 //in m/s^2 e = 1 //Sample Problem 10-4 printf("**Sample Problem 10-4**\n") //velocity just before collision Vi = zeros(1,2) Vi(1) = sqrt(2*g*h1) Vi(2) = 0 Vf= fsolve([0,0], collision) printf("The velocity of m1 after collision is %fm/s", abs(Vf(1)))

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Ex8_7.sce

clc; N=10000; // Speed in rpm Q=600; // Flow rate m^2/min rp=4; // Pressure ratio eff_c=0.82; // Compressor efficiency T01=293; // Inlet temperature in kelvin p01=1.0; // Inlet pressure in bar Cp=1.005;// Specific heat at constant pressure in kJ/kg K Cv=0.717;// Specific heat at constant volume in kJ/kg K r=1.4; // Specific heat ratio R=287; // Characteristic gas constant in J/kg K ca=60; // Axial velocity in m/s D2_D1=2 ;// Diameter ratio T_03=T01*rp^((r-1)/r); T03=T01+(T_03-T01)/eff_c; u2=sqrt (Cp*10^3*(T03-T01)); Wc=u2^2; // Work of compression D2=(u2*60/(3.14*N)); D1=D2/D2_D1; T1=T01-(ca^2/(2-Cp*10^3)); p1=p01*(T1/T01)^(r/(r-1)); row1=p1*10^5/(R*T1); Wroot=(Q/60)*(1/(ca*3.14*D1)); u1=3.14*N*D1/60; alpha_root=atand (ca/u1); alpha_tip= atand (ca/u2); disp ("(i).Power input "); disp ("kW/kg/s",Wc/1000,"Wc = "); disp ("(ii).Impeller Diameters"); disp ("m",D2,"D2 = ","m",D1,"D1 = "); disp ("(iii).Impeller and diffuser blade angles at inlet"); disp ("degree",alpha_tip,"alpha_tip = ","degree",alpha_root,"alpha_root = ");

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example3_4.sce

// To find the Shut resistance of the ammeter // example 3-4 in paage 43 clc; //Given data A=['b' 'a']; Im=100*10^-6;// FSD(Im) in ampere Rm=1000;// Coil resistance is 1 K-ohm // calculation I=10;// FSD initialisation m=0; while I>0.1, I=I/10; Vm=Im*Rm;//voltage across the meter in volts Is=I-Im;//current through shunt resistance in ampere Rs=Vm/Is;//shunt resistance in ohm m=m+1; printf("(%c) shunt resistance value for %.1f A FSD is %f ohm\n ",A(m),I,Rs); end

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LA_Ass3.sce

disp("Enter the Matrix B") B = [input("value"), input("value"); input("value"), input("value"); input("value"), input("value")]; disp("the matrix is:") disp(B) disp("enter the matrix b") b = [input("value"); input("value"); input("value")]; disp("the matrix is:") disp(b) function least_squares(B,b) x = (B'*B)\(B'*b); disp(x, 'x='); C = x(1,1); D = x(2,1); disp(C, 'C='); disp(D, 'D='); disp('The line of the Best Fit is b=C + Dt'); endfunction

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Ex6_6.sce

// SAMPLE PROBLEM 6/6 clc;funcprot(0); // Given data alpha_0=3;// rad/s^2 m=70;// kg k=0.250;// The radius of gyration in m mu_s=0.25;// The coefficient of static friction g=9.81;// The acceleration due to gravity in m/s^2 DCbar=0.30;// m r_A=0.250;// m r_Bi=0.150;// m r_Bo=0.450;// m // Calculation a_t=r_A*alpha_0;// m/s^2 alpha=a_t/DCbar;// rad/s^2 abar=r_Bo*alpha;// m/s^2 function[X]=force(y) // SigmaF_x=m*abar_x X(1)=(y(1)-y(2))-(m*-abar); N=(m*g);// N // SigmaM_G=Ibar*alpha X(2)=((r_Bo*y(1))-(r_Bi*y(2)))-(m*k^2*alpha); endfunction y=[10 100]; z=fsolve(y,force); F=z(1);// N T=z(2);// N printf("\nThe tension in the cable,T=%3.1f N \nThe friction force exerted by the horizontal surface on the spool,F=%2.1f N",T,F); N=(m*g);// N F_max=mu_s*N;// N // If the coefficient of static friction had been 0.1 mu_s=0.1;// The coefficient of static friction F=mu_s*(m*g);// N // SigmaM_C=Ibar*alpha + m*abar*r T=((m*(r_A^2)*alpha)+(m*abar*r_Bo))/DCbar;// N printf("\nThe tension in the cable,T=%3.1f N",T);

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6_3.sce

clc; clear; //Exmaple 6.3 ic=0.05 //Initial concentration (5%) fc=0.2 //Final concentration (20%) T_dash=373 //B.P of water in [K] bpe=5 //Boiling point elevation[K] mf_dot=5000 //[Basis] feed to evaporator in [kg/h] //Material balance of solute mdash_dot=ic*mf_dot/fc //[kg/h] //Overall material balance mv_dot=mf_dot-mdash_dot //Water evaporated [kg/h] lambda_s=2185 //Latent heat of condensation of steam[kJ/kg] lambda_v=2257 //Latent heat of vaporisation of water [kJ/kg] lambda=lambda_v //[kJ/kg] T=T_dash+bpe //Temperature of thick liquor[K] Tf=298 //Temperature of feed [K] Cpf=4.187 //Sp. heat of feed in [kJ/kg.K] //Heat balance over evaporator=ms_dot ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s //Steam consumption [kg/h] Eco=mv_dot/ms_dot //Economy of evaporator Ts=399 //Saturation temperature of steam in [K] dT=Ts-T //Temperature driving force [K] U=2350 //[W/sq m.K] Q=ms_dot*lambda_s //Rate of heat transfer in [kJ/kg] Q=Q*1000/3600 //[J/s]=[W] A=Q/(U*dT) //Heat transfer area in [sq m] printf("\nANSWER Economoy pf evaporator is %f \n",Eco); printf("\nHeat tarnsfer area to be provided = %f sq m\n",A);

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Example13_3.sce

// Electric Machinery and Transformers // Irving L kosow // Prentice Hall of India // 2nd editiom // Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY // Example 13-3 clear; clc; close; // Clear the work space and console. // Given data // Class A insulation T_A = 105 ; // Temperature in degree celsius recorded by the embedded detectors life_orig = 5 ; // Life in years of the motor (standard) // Class B insulation T_B = 130 ; // Temperature in degree celsius recorded by the embedded detectors // Calculations delta_T = T_B - T_A ; // Positive temperature difference betw the given // max hottest spot temperature of its insulation and the ambient temperature recorded. // T_A and T_B are chosen from table 13-1 E = 2 ^ (delta_T/10); // Life extension factor Life_calc = life_orig * E ; // Increased life expectancy of the motor in years // Display the results disp("Example 13-3 Solution : "); printf(" \n Life extension factor : E = %.2f \n ",E ); printf(" \n Increased life expectancy of the motor : Life_calc = %.1f years ",Life_calc);

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Mux4Way.tst

// This file is part of www.nand2tetris.org // and the book "The Elements of Computing Systems" // by Nisan and Schocken, MIT Press. // File name: projects/01/Mux4Way.tst load Mux4Way.hdl, output-file Mux4Way.out, compare-to Mux4Way.cmp, output-list a%B2.1.2 b%B2.1.2 c%B2.1.2 d%B2.1.2 sel%B2.2.2 out%B2.1.2 ; set a 0, set b 0, set c 0, set d 0, set sel 0, eval, output; set sel 1, eval, output; set sel 2, eval, output; set sel 3, eval, output; set a %B1, set sel 0, eval, output; set a 0, set b %B1, set sel 1, eval, output; set b 0, set c %B1, set sel 2, eval, output; set c 0, set d %B1, set sel 3, eval, output; set a 0, set b 1, set c 1, set d 1, set sel 0, eval, output; set a 1, set b 0, set sel 1, eval, output; set b 1, set c 0, set sel 2, eval, output; set c 1, set d 0, set sel 3, eval, output;

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example14_45.sce

clc // Given that t = 300 // temperature in K k = 1.376e-23 // Boltzmann's constant in J/K c = 3e8 // velocity of light in m/sec h = 6.62e-34 // Planck constant in J-sec e = 1.6e-19 // charge on an electron in C m_ = 4 * 1.67e-27 // mass of helium atom in kg m = 1.67e-27 // mass of hydrogen atom in kg // Sample Problem 45 on page no. 14.39 printf("\n # PROBLEM 45 # \n") printf("Standard formula used \n ") printf(" lambda = h/(3*m*k*T)^1/2\n") lambda1 = h / sqrt(3 * m * k * t) lambda2 = h / sqrt(3 * m_ * k * t) r = lambda1 / lambda2 printf("\n Ratio of de-Broglie wavelengths is %d .",r)

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ex4_20.sce

//Chapter-4, Example 4.20, Page 147 //============================================================================= clc clear //CALCULATIONS //given voltage eqn is v=100+(100*sqrt(2))*sin(314*t) volts W=314;//freq in rad/sec R=5;//resistance in ohms X=12;//reactance in ohms Z=R+((%i)*(X));//impedance in ohms Idc=100/R;//dc current in A Iac=(100)/(sqrt((R)^2+(X)^2));//rms value of ac component of current Pt=(R*(Idc^2))+(R*(Iac^2));//total power in Watts V1=sqrt((100)^2+(100)^2);//supplied voltage in Rms in volts I1=sqrt((20)^2+(7.69)^2);//current in Rms in Amps Z1=V1/I1;//circuit impedance in ohms Pf=Pt/(V1*I1);//Power factor mprintf("thus circuit impedance,Power expended and Power factors are %1.1f Ohms ,%1.0f W and %1.3f respectively",Z1,Pt,Pf); //=================================END OF PROGRAM==============================

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clc //Initialization of variables T1=70+460 //R P1=14.7 //psia g=1.4 r=15 rc=2 cp=0.24 cp2=0.1715 //calculations T2=T1*(r)^(g-1) T3=rc*T2 T4=T3*(rc/r)^(g-1) Qh=cp*(T3-T2) Ql=cp2*(T4-T1) W=Qh-Ql eta=W/Qh //results printf("Work output = %d B/lbm",W) printf("\n Efficiency = %.1f percent",eta*100)

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EX3_19.sce

clc;funcprot(0);//Example 3.19 //Initilisation of Variables L1=0.2;.....//Length of slab in m L2=0.1;......//Breath of slab in m Ti=225;...//Initial temparature of brick in degrees celcius h=60;...//heat transfer coefficient on surface of the brick in W/m*K Ta=25;...//Ambient air temparature in degrees celcius t=3600;....//time for measuring temparature in s K=0.7;.....//thermal conductivity of brick in W/m K al=0.5*10^-6;.....//thermal diffucivity of brick in m^2/s //calculations invBi1=(2*K)/(h*L1);....//inverse of biot number for Length of slab Fo1=(al*t)/(L1/2)^2;.....//Fourier number for length of slab teta1=0.8;.....//Value got from hesler chart for slab from 1/Bi=0.12 and Fo=0.18 invBi2=(2*K)/(h*L2);....//inverse of biot number for Length of slab Fo2=(al*t)/(L2/2)^2;.....//Fourier number for length of slab teta2=0.3;.....//Value got from hesler chart for slab from 1/Bi=0.28 and Fo=0.72 teta3=teta1*teta2;....//Realtion for dimentionless temparature for column To=((Ti-Ta)*teta3)+Ta;.....//Center temparature of the column in degrees celcius disp(To,"Center temparature of the column degrees celcius:")

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/767/CH2/EX2.3.1/CH02Exa2_3_1.sci

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CH02Exa2_3_1.sci

// Scilab code Exa2.3.1 To calculate the mass of decayed radioactive material: Page 126 (2011) t_prime = 1600; // Half life of radioactive material, years t = 2000; // Total time, years lambda = 0.6931/t_prime; // Decay constant, years^(-1) m0 = 1; // The mass of radioactive substance at t0, mg m = m0* %e^(-(lambda*t)); // Ratio of total number of atoms and number of atoms disintegrat, mg a = 1-m; // The amount of radioactive substance decayed, mg printf("\nThe amount of radioactive substance decayed : %6.4f mg", a) // Result // The amount of radioactive substance decayed : 0.5795 mg

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example13_8.sce

clear clc //Example 13.8 FLOW RATE FOR A TRIANGULAR WEIR H=0.43; //head on weir[m] g=9.81; //[m/s^2] //Discharge Q=0.179*sqrt(2*g*(H^5)) //[m^3/s] printf("\n The flow of water over the weir = %.3f m^3/s.\n",Q)

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/2594/CH6/EX6.11/Ex6_11.sce

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clc Nd=1*10^18 disp("Nd = "+string(Nd)+" cm^-3") //initializing value of donor concentration. Na=-1*10^18 disp("Na = "+string(Na)+" cm^3") //initializing value of acceptor concentration. Er=11.9 disp("Er = "+string(Er)) //initializing value of relative dielectric permittivity constant . Eo=8.854*10^-14 disp("Eo = "+string(Eo)+" F/cm") //initializing value of dielectric constant of free space. e=1.6*10^-19 disp("e = "+string(e)+" columns") //initializing value of charge of electrons. Vt=0.0259 disp("Vt = "+string(Vt)+" eV") //initializing value of thermal voltage. Vbd=15 disp("Vbd = "+string(Vbd)+" eV") //initializing value of break down voltage. W=2*10^-4 disp("W = "+string(W)+" cm") //initializing value of the distance over which doping profile varies. E=Eo*Er disp("total permittivity,E=Eo*Er)="+string(E)+" F/cm")//calculation a=((Nd-Na)/(W)) disp("slope of doping profile curve,a=((Nd-Na)/(W))="+string(a)+" cm^-4")//calculation Emax=(((Vbd)^2)*9*e*a/(32*E))^(1/3) disp("Emax=(((Vbd)^2)*9*e*a/(32*E))^(1/3)="+string(Emax)+" V/cm")//calculation

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/1949/CH1/EX1.7/1_7.sce

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1_7.sce

//Chapter-1,Example 1_7,Page 1-19 clc() //Given Data: i=30*%pi/180 //angle of incidence u=1.43 //Refractive index of a soap film lam=6*10^-7 //wavelength of light n=1 //For minimum thickness //Calculations: //u=sin i/sin r //Snell's law .So, r=asin(sin(i)/u) //angle of reflection //Now, condition of minima in transmitted system is //2ut*cos r=(2n-1)lam/2 t=lam/(2*2*u*cos(r)) //minimum thickness of film printf('Minimum thickness of film is =%.9f m',t)

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clc; clear; W=2.45*(10^(-2));//N D=3.8*(10^(-2));//m U=12;//m/s //W=L d=1.23;//kg/(m^3) CL=2*W/(d*(U^2)*%pi*(D^2)/4); W=0.5*d*(U^2)*(D^2)*%pi*CL/4; //using this value of CL, omega*D/(2*U)=x is found as x=0.9; omega=2*U*x/D;//rad/sec angvel=omega*60/(2*%pi);//rpm; where angvel is angular velocity disp("rpm",angvel,"The angular velocity=")

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2020-09-04T20:21:44.756895

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SYSTEM.TST

************************************************************************* * * SYSTEM.S * -------- * * MODPlayer low level routines: Hardware accesses etc. * * * * last modified: 13-Jan-1993 * 14-sep-1994 (tas -> bset) ************************************************************************* include "pdefs.inc" SOFTDMA_COOKIE equ 'SSND' SOFTDMA_MAGIC equ 'CLPF' ; ----------------------------------------------------------------------- ; Globals: ; ----------------------------------------------------------------------- ; ; exported subroutines: ; xdef p_global_init xdef acc_malloc ; special malloc for ACC's xdef alloc_fast_ram xdef free_memory xdef init_DMA_sound, exit_DMA_sound xdef check_sound_running xdef supexec xdef read_cookie ; ; exported variables: ; xdef has_DMA_sound ;charflag xdef has_68020 ;charflag xdef softdma_interface ;Pointer! xdef sdmabufadr1,sdmabufadr2 ; ; imported subroutines: ; xref mt_music ;vbl seq. xref Paula ;chip emulator xref SoftPaula ;softdma interface xref calc_voltab ; ; imported variables: ; xref AudioFrameSize ;from synth.s xref voltabP xdef LeftBuffer, RightBuffer xref p_interpolate ;from player.s xref p_overload xref p_cpuload xref p_state xref p_boost xref p_samplefreq xref paula_frequencies ; ----------------------------------------------------------------------- ; internally used defines: ; ----------------------------------------------------------------------- INTBIT equ 7 ;Port 7 interrupt PCM_VECTOR equ $013C ;associated MFP vector HBL_VECTOR equ $0068 ;Autovector IRQ. IERA equ $FFFFFA07 ;MFP Registers IERB equ $FFFFFA09 IPRA equ $FFFFFA0B IPRB equ $FFFFFA0D ISRA equ $FFFFFA0F ISRB equ $FFFFFA11 IMRA equ $FFFFFA13 IMRB equ $FFFFFA15 * STE/TT DMA SOUNDCHIP: SDMACNTR equ $FFFF8900 FRMBASEHI equ $FFFF8902 FRMBASEMID equ $FFFF8904 FRMBASELO equ $FFFF8906 FRMCNTHI equ $FFFF8908 FRMCNTMID equ $FFFF890A FRMCNTLO equ $FFFF890C FRMENDHI equ $FFFF890E FRMENDMID equ $FFFF8910 FRMENDLO equ $FFFF8912 SMODCNTR equ $FFFF8920 MWDATA equ $FFFF8922 MWMASK equ $FFFF8924 XB_MAGIC equ -12 XB_ID equ -8 XB_OLDVEC equ -4 XB_CODE equ 0 ; ----------------------------------------------------------------------- ; Code ; ----------------------------------------------------------------------- text super ;MUST use SuperMode! ;STE mit 8.0106 MHz: ; dc.w 6258,12517,25033,50066; (ste) ;TT mit 32.215905 MHz: tt_paula_frequencies: dc.w 6292,12584,25167,50337 ;(tt) ;Falcon030 mit neuem Soundsubsystem 25.175 MHz: falcon_paula_frequencies: dc.w 6146,12292,24585,49170 ;(falcon030) * * p_global_init: initialisiert den Player, alloziert Speicher. * Param: keine * Return: Fehlercode * p_global_init: movem.l d3-d7/a2-a6,-(sp) move.l a0,basepage ;Bpage Pointer merken lea getsys(pc),a0 bsr supexec ;Systemdaten lesen tst.b has_DMA_sound ;STE DMA vorhanden? beq .no_dmasound ;nein -> kein ST ram alloc clr.l softdma_interface ;du nicht! move.w machine_type(pc),d0 ;Welche Maschine? lea tt_paula_frequencies(pc),a0 cmpi.w #2,d0 ;TT: hher! beq.s .setfreq lea falcon_paula_frequencies(pc),a0 cmpi.w #3,d0 ;Falcon: niedriger bne.s .skip .setfreq: lea paula_frequencies(pc),a1 move.l (a0)+,(a1)+ move.l (a0)+,(a1)+ .skip: move.l #BUFFERSIZE,d0 clr.w -(sp) ;Modus 0: ST-Ram only move.l d0,-(sp) ;Amount move.w mxalloc(pc),-(sp) ;Mxalloc trap #1 addq.w #8,sp tst.l d0 spl can_work bmi.s .noplayer ;ohne ST Ram kein Player ! addq.l #3,d0 andi.w #$FFFC,d0 ;LongWord aligned for TT! movea.l d0,a0 lea buffer1(a0),a1 ;Erster Puffer move.l a1,sdmabufadr1 lea buffer2(a0),a1 move.l a1,sdmabufadr2 .softentry: tst.b has_68020 ;Volume Table ntig?? bne.s .novoltab move.l #TOGGLEVOLUME*256,d0 bsr fm_alloc ;returns addr in a0 & d0 move.l d0,voltabP bsr calc_voltab .novoltab: moveq #E_OK,d0 .globlin1: clr.w p_state ;Idle State movem.l (sp)+,d3-d7/a2-a6 rts .noplayer: moveq #E_FRAMEBUFFER,d0 bra.s .globlin1 .no_dmasound: move.l softdma_interface(pc),d0 ;Christians DMA Emulator? sne can_work ;dann drfen wir! beq.s .no_dmasound2 movea.l d0,a1 ; move.l #PaulaHook,SDMA_HOOK(a1) ;Anmelden! lea paula_frequencies(pc),a0 ;our samplefreqs move.l SDMA_SF0(a1),(a0)+ move.l SDMA_SF2(a1),(a0)+ bra.s .softentry ;weiter .no_dmasound2: moveq #E_DMASOUND,d0 ;ansonsten fehler! bra.s .globlin1 supexec: pea (a0) move.w #38,-(sp) trap #14 addq.l #6,sp rts getsys: lea fstacktop(pc),a0 move.l a0,d0 andi.w #$FFFC,d0 ;Stack auf longwords move.l d0,faststack sf has_DMA_sound ;Annahme: kein DMA sound sf has_68020 ;Annahme: 68000 movea.l _sysbase.w,a0 move.l $28(a0),os_act_pd ;systemvektor fr spter cmpi.w #$0205,2(a0) ;OS-Version bge.s .istos3 move.w #Malloc,mxalloc ;Extended Alloc Call OK?! .istos3: movea.l _p_cookies.w,a0 ;Cookie move.l a0,d0 beq.s .get_end ;kein Cookiejar -> Abbruch .readjar: move.l (a0)+,d0 ;Cookie holen beq.s .get_end ;fertig. move.l (a0)+,d1 ;Value holen cmpi.l #'_CPU',d0 ;Prozessortyp? beq.s .getproc cmpi.l #'_SND',d0 ;Soundhardware? beq.s .getsnd cmpi.l #'_MCH',d0 ;Maschinentyp? beq.s .getmachine cmpi.l #SOFTDMA_COOKIE,d0 ;SoftSound?? beq.s .softsound cmpi.l #'MiNT',d0 ;MTOS / MiNT? beq.s .isMiNT bra.s .readjar .get_end: rts .getproc: cmpi.w #20,d1 ;op d'manst n 68020! sge has_68020 ;020/030 Flag bra.s .readjar .getsnd: btst #1,d1 sne has_DMA_sound ;OK! bra.s .readjar ;DMA Soundchip vorhanden .getmachine: swap d1 ;get highword move.w d1,machine_type bra.s .readjar .isMiNT: clr.l basepage ;Kein Basepage gefummels bra.s .readjar .softsound: movea.l d1,a1 cmpi.l #SOFTDMA_MAGIC,(a1) ;gltig? bne.s .readjar move.l d1,softdma_interface ;Pointer zu SoftDma merken. bra.s .readjar * * Mxalloc, FASTRAM preferred (fr Programm und Daten) * Param.: d0.l Anzahl Bytes die verlangt werden * Return: d0.l & a0 Startadresse acc_malloc: ;PureC Interface move.l a2,-(sp) bsr alloc_fast_ram movea.l (sp)+,a2 rts alloc_fast_ram: move.l basepage(pc),d1 ;Verbiegerei NICHT unter MiNT oder beq.s is_applic ;als Programm! move.l a3,-(sp) movea.l os_act_pd(pc),a3 ;Zeiger auf Processpointer move.l (a3),-(sp) ;retten move.l d1,(a3) ;meine Basepage, ich bin der Owner! bsr.s fm_alloc move.l (sp)+,(a3) ;os_act_pd zurcksetzen movea.l (sp)+,a3 rts is_applic: fm_alloc: move.w #3,-(sp) ;Modus 3 move.l d0,-(sp) ;Amount move.w mxalloc(pc),-(sp) ;Mxalloc trap #1 addq.w #8,sp movea.l d0,a0 rts * * Speicherblock freigeben: * * Param.: d0: Blockadresse free_memory: move.l a2,-(sp) tst.l d0 beq.s freem1 move.l d0,-(sp) move.w #Mfree,-(sp) trap #1 addq.w #6,sp freem1: movea.l (sp)+,a2 rts ;--------------------------------- cookie jar ------------------ ; Deklaration in C: ; int read_cookie(long ID, long *ret); ; ; ID: Cookie Jar Identifier. ; *ret: return value ; Funktion gibt 1 (TRUE) oder 0 (FALSE) zurck. ; read_cookie: ;fr'n Linker! movem.l d3/a2-a3,-(sp) move.l d0,d3 ;ID retten movea.l a0,a3 ;Returnpointer lea getpointer(pc),a0 bsr supexec tst.l d0 ;Pointer? beq.s no_cookiejar btst #0,d0 ;odd? bne.s no_cookiejar movea.l d0,a0 ;Pointer seems OK check_next: move.l (a0)+,d1 ;Cookie ID beq.s cookie_not_found move.l (a0)+,d2 ;entry cmp.l d1,d3 ;gesuchtes Objekt? bne.s check_next move.l d2,(a3) ;dem User das Objekt bergeben moveq #TRUE,d0 bra.s cookie_exit cookie_not_found: no_cookiejar: moveq #FALSE,d0 cookie_exit: movem.l (sp)+,d3/a2-a3 rts getpointer: move.l _p_cookies.w,d0 rts ********************************************************************** * XBRA - UTILITIES * * sollten im Supermode aufgerufen werden, ->BUS ERROR * ********************************************************************** ******************************************************************************** * Xb_install: fge neuen XBRA Teilnehmer in eine Vektorliste ein * * Param: a0: Vektor (&root, zeiger auf Liste!) * * a1: Pointer auf zu installierende Routine, Header erforderlich * * Ret: d0: -1 bei Fehler (bsp: Doppelinstallation) * * 0 wenn OK + FLAGS!! * * Reg's: d0-d2,a0-a2 * ******************************************************************************** Xb_install: cmpi.l #'XBRA',XB_MAGIC(a1) ;Routine braucht einen Header! bne.s Xb_error movea.l a0,a2 ;save vector move.l XB_ID(a1),d0 ;Identifier pea (a1) ; &routine bsr.s Xb_find ;Suche ihn movea.l (sp)+,a1 bpl.s Xb_error ;und vermeide Doppelinstallation move.l (a2),XB_OLDVEC(a1) ;Flle Header aus (erst hier!) move.l a1,(a2) ;insert into list moveq #0,d0 ;OK rts ******************************************************************************** * Xb_remove: entferne einen XBRA Teilnehmer aus einer Liste * * Param: a0: vector (&root) * * d0: XB_ID des Todeskandidaten * * Ret: d0: -1 bei Fehler (nicht gefunden) * * >=0 wenn OK (Verschachtelungstiefe) * * Regs: d0-d2/a0-a1 * ******************************************************************************** Xb_remove: bsr.s Xb_find ;suchen bmi.s Xb_error ;nicht vorhanden beq.s Xb_remv ;Root Level, erstes Listenelement move.l XB_OLDVEC(a0),XB_OLDVEC(a1) ; prev->oldvec=xbra->oldvec bra.s Xb_remv1 Xb_remv: move.l XB_OLDVEC(a0),(a1) ; root=xbra->oldvec Xb_remv1: clr.l XB_OLDVEC(a0) rts Xb_error: moveq #-1,d0 rts ******************************************************************************** * Xb_find: suche einen XBRA-Teilnehmer * * Param: a0: vector (&root der Liste) * * d0: XB_ID des Kandidaten * * Ret: d0: -1 : Fehler - nichts gefunden * * >=0 : Verschachtelungstiefe an der Fundstelle * * a0: xbra (Adresse der gefundenen XBRA-Struktur) * * a1: parent (Adresse der vorherigen XBRA-Struktur) oder &root * * &root nur wenn Verschachtelungstiefe==0 * * Regs: d0-d2,a0-a1 * ******************************************************************************** Xb_find: move.l d0,d1 movea.l a0,a1 ; parent=root; movea.l (a0),a0 ; xbra=*root; moveq #0,d2 ; level=0; bra.s Xb_find2 Xb_find1: addq.w #1,d2 ; level++; movea.l a0,a1 ; parent=xbra; movea.l XB_OLDVEC(a0),a0 ; xbra=xbra->oldvec; Xb_find2: bsr.s Xb_check ; while ( (check(xbra)) bne.s Xb_error ; && cmp.l XB_ID(a0),d1 ; (xbra->magic!=mymagic) bne.s Xb_find1 ; ); move.w d2,d0 ;gefunden! rts ******************************************************************************** * Xb_check: teste die in a0 spezifizierte XBRA Struktur * * Param: a0: &xbra (Zeiger auf XBRA Struktur) * * Ret: d0: 0 = XBRA, -1: no xbra * * Flags: EQ = XBRA * * Regs: d0 * ******************************************************************************** Xb_check: move.l a0,d0 ;Teste diese Struktur btst #0,d0 ;Ungerade? bne.s Xb_error cmp.l #$0100,d0 ;NICHT in den Systemvektoren! blt.s Xb_error ;Sicherheitshalber cmpi.l #'XBRA',XB_MAGIC(a0) ;hoffentlich keine Bomben ! bne.s Xb_error moveq #0,d0 rts ;--------------------------------------------------------------- * * softy-routinen: * * * Hauptroutine: * PaulaHook: movem.l d3-d7/a2-a6,-(sp) move.l sp,savestack movea.l faststack(pc),sp ;use own stack (32 bit fastram) lea p_overload(pc),a0 subq.w #1,(a0) bpl.s .domusic clr.w (a0) .domusic: bsr mt_music ; not.w frameflag ;Pufferumschaltung bne.s .setbuf2 ;Nchster puffer ist Nummer 2 .setbuf1: lea LeftBuffer(pc),a0 ;Frame 1 berechnen bsr SoftPaula ;Chip Emulator lea LeftBuffer(pc),a0 ;und Frame 1 setzen! bra.s .setbuf .setbuf2: lea RightBuffer(pc),a0 ;Frame 2 berechnen bsr SoftPaula lea RightBuffer(pc),a0 ;Adresse des 2ten Puffers .setbuf: move.w AudioFrameSize(pc),d1 ;Lnge des Puffers exg d0,d1 ;bitsize kommt korrekt zurck. move.l softdma_interface(pc),a1 move.w p_boost(pc),SDMA_BOOST(a1) move.l SDMA_SETFRAME(a1),a1 jsr (a1) movea.l savestack(pc),sp movem.l (sp)+,d3-d7/a2-a6 ;ab hier wieder reentrant rts * * new init: * softdma_init: movea.l d0,a0 movea.l SDMA_INIT(a0),a1 move.l #PaulaHook,SDMA_HOOK(a0) lea LeftBuffer(pc),a0 ;erster (Mono-)Puffer clr.w frameflag ;current running buffer! move.w AudioFrameSize,d0 ;Lnge der Daten moveq #0,d1 ;Range = 0 bit :-) move.w p_samplefreq(pc),d2 ;Diese Frequenz... jmp (a1) ;jsr/rts * * new exit: * softdma_exit: movea.l d0,a0 movea.l SDMA_EXIT(a0),a1 jmp (a1) ;stop it. * * Testroutine zum evtl. Anschmeissen des Sounds nach Fremd-Abbruch * * check_sound_running: move.l softdma_interface(pc),d0 bne.s .nocheck ;Nicht fr's SoftDMA! btst #0,SDMACNTR+1.w ;DMA running? beq.s relaunch_dma .nocheck: rts * * Routinen zum Starten/Stoppen des Chips * - supervisor mode erforderlich * init_DMA_sound: move.l softdma_interface(pc),d0 bne.s softdma_init andi.w #!%11,SDMACNTR.w ;Stop DMA & Repeat relaunch_dma: btst #INTBIT,gpip.w ;Monitor Type ? bne.s m_high bclr #INTBIT,aer.w bra.s m_cont m_high: bset #INTBIT,aer.w m_cont: lea HBL_VECTOR.w,a0 ; sound synthesizer... lea HBL_interrupt(pc),a1 bsr Xb_install ; ignore errors! lea Sound_interrupt(pc),a0 move.l a0,PCM_VECTOR.w ;Set the Exception Vector bclr #INTBIT,IPRA.w ;MFP_PendA.w bset #INTBIT,IMRA.w ;Mask 'Sound Active' interupt. bset #INTBIT,IERA.w ;Enable 'Sound Active' interupt move.b p_samplefreq+1(pc),SMODCNTR+1.w ;Chip Mode Control movea.l sdmabufadr1(pc),a0 ;Adresse des 1ten Puffers bsr SetFrame clr.w frameflag ;current running buffer! ori.w #%11,SDMACNTR.w ;chipstart! rts exit_DMA_sound: move.l softdma_interface(pc),d0 bne softdma_exit move sr,-(sp) ;Except. #24: spurious interrupt ori #$700,sr ;hope it helps... andi.w #!%10,SDMACNTR.w ;DMA Repeat aus bclr #INTBIT,IERA.w ;weitere Irq's unterdrcken clr.l PCM_VECTOR.w ;Vektor zurcksetzen clr.w p_cpuload ;Indikator lschen clr.w p_overload lea HBL_VECTOR.w,a0 ; sound synthesizer... move.l #'PAUL',d0 ; ID to remove bsr Xb_remove ; ignore errors! move (sp)+,sr rts * * interrupt overlapping... * overlap: bclr #INTBIT,ISRA.w ;End Of Interrupt -> wg. IKBD irq move.w #OVERTICKS,p_overload ;25 irq's = 0.5 sec move.w AudioFrameSize(pc),p_cpuload ;--- call song data interpreter (although overload...) movem.l d0-a6,-(sp) bsr mt_music movem.l (sp)+,d0-a6 rte ********************************************************************** * * INTERRUPT-ROUTINE, ALLE 20 MSEC AUFZURUFEN! * * dc.l 'XBRA','PAUL',0 Sound_interrupt: bset.b #7,interrupt_pending bne.s overlap ;HORROR: berlappende IRQ's!!!! bclr #INTBIT,ISRA.w ;End Of Interrupt -> wg. IKBD irq andi.w #$F8FF,(sp) ;irq level 0 rte ;END of high-priority interrupt ; ; Hooked before HBL irq: ; dc.l 'XBRA','PAUL',0 HBL_interrupt: ori #$300,sr movem.l d0-a6,-(sp) ;**** !!!!! NON REENTRANT CODE !!!!! **** move.l sp,savestack movea.l faststack(pc),sp ;use own stack (32 bit fastram) tst.w p_overload ble.s .peak_load subq.w #1,p_overload ;decrement overload .peak_load: not.w frameflag ;Pufferumschaltung bne.s .setbuf2 ;Nchster puffer ist Nummer 2 .setbuf1: movea.l sdmabufadr1(pc),a0 ;Frame 1 berechnen bsr Paula ;Chip Emulator move.l sdmabufadr2(pc),d2 movea.l sdmabufadr1(pc),a0 ;und Frame 1 setzen! bra.s .setbuf .setbuf2: movea.l sdmabufadr2(pc),a0 ;Frame 2 berechnen bsr Paula move.l sdmabufadr1(pc),d2 movea.l sdmabufadr2(pc),a0 ;Adresse des 2ten Puffers .setbuf: move.l d2,-(sp) ;save sdmabufadr... bsr SetFrame ;--- call song data interpreter --------; bsr mt_music ; ;---------------------------------------; ; Rechenzeitverbrauch messen! lea FRMCNTHI.w,a0 ;Chipregister movep.l -1(a0),d1 ;akt. Adresse auslesen sub.l (sp)+,d1 ;im Lowword Differenz! lea p_cpuload(pc),a6 add.w d1,(a6) lsr.w (a6) ;/2 ;-- ok! -- movea.l savestack(pc),sp movem.l (sp)+,d0-a6 ;ab hier wieder reentrant sf interrupt_pending ; jump to original HBL handler: move.l (HBL_interrupt-4)(pc),-(sp) rts ; rte * ***************************************************************************** * * Setze nchsten Abspielbereich: von a0 bis a1; * Param: a0: FrameStart * Return: void * Global: FrameSize is read from AudioFrameSize! * SetFrame: move sr,d0 ori #$0700,sr pea (a0) move.b 3(sp),FRMBASELO+1.w move.b 2(sp),FRMBASEMID+1.w move.b 1(sp),FRMBASEHI+1.w ;Start des Frames move.w AudioFrameSize(pc),d1 adda.w d1,a0 ;Stereo: L & R ! adda.w d1,a0 move.l a0,(sp) move.b 3(sp),FRMENDLO+1.w move.b 2(sp),FRMENDMID+1.w move.b 1(sp),FRMENDHI+1.w ;Ende des Frames addq.l #4,sp move d0,sr rts ; ----------------------------------------------------------------------- ; initialized variables ; ----------------------------------------------------------------------- data mxalloc: dc.w Mxalloc ; ----------------------------------------------------------------------- ; variables ; ----------------------------------------------------------------------- bss faststack: ds.l 1 ;Eigener Stackbereich! savestack: ds.l 1 sdmabufadr1: ds.l 1 ;Adresse der Framebuffer fr den D/A-Wandler sdmabufadr2: ds.l 1 ;es sind 2 wegen Double buffering technik os_act_pd: ds.l 1 ;Zeiger auf den act_pd vektor basepage: ds.l 1 ;Zeiger auf eigene BasePage last_freqtabcheck: ds.l 1 ;Letzte berechnete Samplingfrequenz softdma_interface: ds.l 1 ;DMA Emulator? frameflag: ds.w 1 ;Welcher Puffer gerade gespielt wird machine_type: ds.w 1 ;Maschinentyp aus Cookie Jar (_MCH) has_DMA_sound: ds.b 1 has_68020: ds.b 1 ;>'020!! can_work: ds.b 1 ;gesetzt wenn Arbeit mglich... interrupt_pending: ds.b 1 ;seeeehhhr wichtig even ds.l FSTACK ;interner Stack fstacktop: ds.l 1 ; ----------------------------------------------------------------------- ; end of file ; -----------------------------------------------------------------------

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//Problem 2.10: An electric kettle has a resistance of 30. What current will flow when it is connected to a 240 V supply? Find also the power rating of the kettle. //initializing the variables: V = 240; // in Volts R = 30; // in ohms //calculation: I = V/R P = V*I printf("\n\nResult\n\n") printf("\nCurrent(I): %.0f Ampere(A)",I) printf("\nPower(P): %.0f Watt(W)\n",P)

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laba9e.sce

//9e clf t=-%pi:0.3:%pi; plot3d1(t,t, sin(t)'*cos(t), 35, 45, 'X@Y@Z', [2,2,4])

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FOSSEE-Internship/FOSSEE-Control-Systems-Toolbox

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testfiltbutter.sce

data=rand(100,3); n=5;band=[0.2 0.8]; u=filtbutter(data,n,band) savematfile("testbutter.mat",'data','n','band') //here for a given input ,if the output of these filters will be same for given same //input the corresponding result of rbs function will also be same as rest all part of //rbs function just adjusts the outputs from the filter with their corresponding sign //and levels.

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/213/CH14/EX14.10/14_10.sce

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14_10.sce

//To find centrifugal and gyroscopic effects clc //Given: m=2500 //kg x=1.5, R=30, dW=0.75, rW=dW/2, h=0.9 //m v=24*1000/3600 //m/s G=5 IW=18, IE=12 //kg-m^2 //Solution: //Calculating the road reaction on each wheel r=m*9.81/4 //Road reaction on each wheel, N //Calculating the angular velocity o the wheels omegaW=v/rW //rad/s //Calculating the angular velocity of precession omegaP=v/R //rad/s //Calculating the gyroscopic couple due to one pair of wheels and axle CW=round(2*IW*omegaW*omegaP) //N-m //Calculating the gyroscopic couple due to the rotating parts of the motor and gears CE=round(2*IE*G*omegaW*omegaP) //N-m //Calculating the net gyroscopic couple C=CW-CE //N-m //Calculating the reaction due to gyroscopic couple at each of the outer or inner wheels P=2*(-C)/(2*x) //N //Calculating the centrifugal force FC=m*v^2/R //N //Calculating the overturning couple CO=FC*h //N-m //Calculating the reaction due to overturning couple at each of the outer and inner wheels Q=2*CO/(2*x) //N //Calculating the vertical force exerted on each outer wheel PO=m*9.81/4-P/2+Q/2 //N //Calculating the vertical force exerted on each inner wheel PI=m*9.81/4+P/2-Q/2 //N //Results: printf("\n\n Vertical force exerted on each outer wheel, PO = %.2f N.\n\n",PO) printf(" Vertical force exerted on each inner wheel, PI = %.2f N.\n\n",PI)

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/tests/test_xrep_1_d.tst

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test_xrep_1_d.tst

// [(start test_spec)] if( !get_obj( ).get_key( ).empty( ) ) { string value; if( !is_null( get_obj( ).First( ) ) ) { if( !value.empty( ) ) value += ","; value += to_string( get_obj( ).First( ) ); } if( !is_null( get_obj( ).Second( ) ) ) { if( !value.empty( ) ) value += ","; value += to_string( get_obj( ).Second( ) ); } if( !is_null( get_obj( ).Third( ) ) ) { if( !value.empty( ) ) value += ","; value += to_string( get_obj( ).Third( ) ); } get_obj( ).Field( value ); } // [(finish test_spec)]

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/3204/CH17/EX17.1/Ex17_1.sce

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Ex17_1.sce

// Initilization of variables m=0.1 // kg // mass of ball // Calculations // Consider the respective F.B.D. // For component eq'n in x-direction delta_t=0.015 // seconds // time for which the ball &the bat are in contact v_x_1=-25 // m/s v_x_2=40*cosd(40) // m/s F_x_average=((m*(v_x_2))-(m*(v_x_1)))/(delta_t) // N // For component eq'n in y-direction delta_t=0.015 // sceonds v_y_1=0 // m/s v_y_2=40*sind(40) // m/s F_y_average=((m*v_y_2)-(m*(v_y_1)))/(delta_t) // N F_average=sqrt(F_x_average^2+F_y_average^2) // N // Results clc printf('The average impules force exerted by the bat on the ball is %f N \n',F_average)

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Ex17_2.sce

clc P1 = 0.18 // Diffuser static pressure in MPa R = 0.287 // Gas constant T1 = 37 // Static temperature P0 = 0.1// Atmospheric pressure in MPa A1 = 0.11 // intake area in m^2 V1 = 267 // Inlet velocity in m/s w = (P1*1e3/(R*(T1+273)))*A1*V1 // mass flow rate g = 1.4 // Heat capacity ratio c1 = sqrt(g*R*(T1+273)*1000) // velocity M1 = V1/c1 // Mach number A1A_ = 1.0570 // A1/A* A* = A_ P1P01 = 0.68207 // pressure ratio T1T01 = 0.89644// Temperature ratio F1F_ = 1.0284// Impulse function ratio A2A1 = 0.44/0.11 // Area ratio A2A_ = A2A1*A1A_// Area ratio M2 = 0.135 // Mach number P2P02 = 0.987 // Pressure ratio T2T02 = 0.996 // Temperature ratio F2F_ = 3.46// Impulse function ratio P2P1 = P2P02/P1P01 // Pressure ratio T2T1 = T2T02/T1T01// Temperature ratio F2F1 = F2F_/F1F_ // Impulse function ratio P2 = P2P1*P1 // Outlet pressure T2 = T2T1*(T1+273) // Outlet temperature A2 = A2A1*A1 // Exit area F1 = P1*A1*(1+g*M1^2) // Impulse function F2 = F2F1*F1 // Impulse function Tint = F2-F1 // Internal thrust Text = P0*(A2-A1) // External thrust NT = Tint - Text // Net thrust printf("\n Example 17.2 \n") printf("\n Mass flow rate of air through diffuser is %f Kg/s",w) printf("\n Mach number of leaving air is %f ",M2) printf("\n Temperature of leaving air is %f degree celcius",T2-273) printf("\n Pressure of leaving air is %f MPa ",P2) printf("\n Net thrust is %f kN",NT*1e3) //The answers vary due to round off error

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sh_asys_2pc.tst

-v G_USER=jnguyen -v G_CONFIG=1.0 -v G_TBTYPE=asys -v G_TST_TITLE="Advanced system config" -v G_PROD_TYPE=MC524WR -v G_HTTP_DIR=test/ -v G_FTP_DIR=/log/autotest -v G_TESTBED=tb1 -v G_FROMRCPT=qaman -v G_FTPUSR=root -v G_FTPPWD=@ctiontec123 -v U_USER=admin -v U_PWD=admin1 -v G_LIBVERSION=1.0 -v G_LOG=$SQAROOT/automation/logs -v U_COMMONLIB=$SQAROOT/lib/$G_LIBVERSION/common -v U_COMMONBIN=$SQAROOT/bin/$G_LIBVERSION/common -v U_TBCFG=$SQAROOT/config/$G_LIBVERSION/testbed -v U_TBPROF=$SQAROOT/config/$G_LIBVERSION/common -v U_VERIWAVE=$SQAROOT/bin/1.0/veriwave/ -v U_MI424=$SQAROOT/bin/1.0/mi424wr/ -v U_TESTPATH=$SQAROOT/platform/1.0/verizon/testcases/asys/json #this value used to setup dut configuration -v U_DEBUG=3 -v U_RUBYBIN=$SQAROOT/bin/$G_LIBVERSION/rbin -v U_VZBIN=$SQAROOT/bin/$G_LIBVERSION/vz_bin -v U_COMMONJSON=$SQAROOT/platform/1.0/verizon2/testcases/common/json -v U_COAX=0 #$G_PFVERSION=1.0 #------------------------------ # Set up the test environment. #------------------------------ #-nc $SQAROOT/config/$G_CONFIG/common/testbedcfg_env.xml -nc $SQAROOT/config/$G_CONFIG/common/testbedcfg.xml; -nc $SQAROOT/platform/1.0/verizon2/testcases/common/tcases/login_logout.xml -nc $SQAROOT/platform/1.0/verizon2/testcases/common/tcases/fw_upgrage_image.xml;pass=init -nc $SQAROOT/platform/1.0/verizon2/testcases/common/tcases/fw_upgrage_image.xml;pass=init -nc $SQAROOT/platform/1.0/verizon2/testcases/common/tcases/fw_upgrage_image.xml;fail=finish -label init -nc $SQAROOT/platform/1.0/verizon2/testcases/common/tcases/reset_dut_to_default.xml -nc $SQAROOT/platform/1.0/verizon2/testcases/common/tcases/tc_init_dut.xml;pass=next -nc $SQAROOT/platform/1.0/verizon2/testcases/common/tcases/tc_init_dut.xml;pass=next -nc $SQAROOT/platform/1.0/verizon2/testcases/common/tcases/tc_init_dut.xml;fail=finish -label next -nc $SQAROOT/platform/1.0/verizon2/testcases/common/tcases/set_default_time.xml -nc $SQAROOT/platform/1.0/verizon2/testcases/common/tcases/tc_init_ping.xml;fail=finish -nc $SQAROOT/platform/1.0/verizon2/testcases/common/tcases/enable_tnet.xml #------------------------------ # Test cases #------------------------------ -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_mailcapacity_06041003900.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_mailcapacity_06041003901.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_mailcapacity_06041003902.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_mailcapacity_06041003903.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_mailcapacity_06041003904.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_mailcapacity_06041003905.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_logbuf_06041003906.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_logbuf_06041003907.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_logbuf_06041003908.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_logbuf_06041003909.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_logbuf_06041003910.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_logbuf_06041003911.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_rmsyslog_06041003912.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_rmsyslog_06041003913.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_rmsyslog_06041003914.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_rmsyslog_06041003915.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_rmsyslog_06041003916.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_rmsyslog_06041003917.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_secmailcapacity_06041003918.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_secmailcapacity_06041003919.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_secmailcapacity_06041003920.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_secmailcapacity_06041003921.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_secmailcapacity_06041003922.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_secmailcapacity_06041003923.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_seclogbuf_06041003924.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_seclogbuf_06041003925.xml #-tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_seclogbuf_06041003926.xml #-tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_seclogbuf_06041003927.xml #-tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_seclogbuf_06041003928.xml #-tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_seclogbuf_06041003929.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_rmsecsyslog_06041003930.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_rmsecsyslog_06041003931.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_rmsecsyslog_06041003933.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_rmsecsyslog_06041003934.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpport_06041003900.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpport_06041003901.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpport_06041003902.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpport_06041003903.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpsport_06041003904.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpsport_06041003905.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpsport_06041003906.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpsport_06041003907.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_telport_06041003849.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_telport_06041003850.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_telport_06041003851.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_telport_06041003852.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_telport_06041003853.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_logdisable_06041003908.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_logenable_06041003907.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_ses60_06041004100.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_ses7200_06041004101.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_ses59_06041004102.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_ses7201_06041004103.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_autorefresh_06041000001.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_autorefresh_06041000002.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_promptpasswd_06041000003.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_promptpasswd_06041000004.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_changewarn_06041000005.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_changewarn_06041000006.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpport_06041003841.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpport_06041003842.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpport_06041003843.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpport_06041003844.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpsport_06041003845.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpsport_06041003846.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpsport_06041003847.xml -tc $SQAROOT/platform/1.0/verizon/testcases/asys/tcases/tc_httpsport_06041003848.xml # #------------------------------ # Checkout #------------------------------ -label finish -nc $SQAROOT/config/$G_CONFIG/common/finalresult.xml -nc $SQAROOT/config/$G_CONFIG/common/uploadlog.xml -nc $SQAROOT/config/$G_CONFIG/common/email.xml

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Example8_9.sce

clc //Given that R_e = 6.4e6 // radius of Earth in km M_e = 6e24 // mass of Earth in kg G = 6.67e-11 // universal gravitational constant u = 6e3 // initial speed of rocket in m/s // sample problem 9 page No. 302 printf("\n\n\n # Problem 9 # \n") printf("Standard formula used U_f - U_i = 1/2 * m *(u^2 - v^2)\n ") h = ((R_e * 1e3)^2 * u^2) / (2 * G * M_e - R_e * u^2) / 1000 // calculation of Height reached by rocket before returning to Earth printf ("\n Height reached by rocket before returning is %e km.",h)

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Ch1_1_48.sce

clc disp("Example 1.48") printf("\n") disp("Find the maximum forward current") T1=25 //to find maximum forward current at this temperature T2=65 //to find maximum forward current at this temperature PT1=600*10^-3 //maximum power dissipation at 25c D=5*10^-3 //derating factor VT1=0.6 //forward voltage drop(constant at all temperature) VT2=VT1 IT1=PT1/VT1 //maximum forward current at T1 PT2=PT1-((T2-T1)*D) IT2=PT2/VT2 //maximum forward current at T2 printf("Forward current at temperature T1=\n%f Ampere\n",IT1) printf("Forward current at temperature T2=\n%f Ampere\n",IT2)

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/1358/CH4/EX4.14/Example414.sce

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Example414.sce

// Display mode mode(0); // Display warning for floating point exception ieee(1); clear; clc; disp("Turbomachinery Design and Theory,Rama S. R. Gorla and Aijaz A. Khan, Chapter 4, Example 14") disp("The absolute Mach number of the air at the impeller tip is:") disp("M2 = C2/(gamma*R*T2)^0.5") disp("where T2 is the static temperature at the impeller tip. Let us first calculate C2 and T2.") U2 = 364; sigma = 0.89; Cw2 = sigma*U2 disp("From the velocity triangle,") Cr2 = 28; C2 = (Cr2^2+Cw2^2)^0.5 disp("With zero whirl at the inlet") disp("W/m = sigam*U2^2 = Cp (T02 - T01)") T01 = 288; Cp = 1005; T02 = T01 + sigma*U2^2 / Cp disp("Static Temperature") T2 = T02 - C2^2 /(2*Cp) gamma = 1.4; R = 287; M2 = (C2^2/(gamma*R*T2))^0.5 disp("Using the isentropic P–T relation:") disp("Ratioa = P02/P01 ") etac = 0.88; Ratioa = (1+etac * (T02/T01 - 1))^3.5 disp("Ratiob = P2/P02") Ratiob = (T2/T02)^3.5 P01 = 1*100; disp("Static Pressure in kPa") P2 = Ratiob*Ratioa*P01 rho2 = P2*1000/(R*T2) disp("Mass flow: in kg/s") A = 0.085;//m2 m = rho2*Cr2*A

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instruction_nf.sce

scenario = "instruction for ball movement judgment without feedback"; no_logfile = true; scenario_type = trials; active_buttons = 1; button_codes = 99; screen_width = 1024; screen_height = 768; screen_bit_depth = 16; default_font_size = 20; begin; picture {} default; trial { trial_type = first_response; trial_duration = forever; picture { text { caption = "I N S T R U K T I O N\n\nIn den nächsten zwei Blöcken bearbeiten Sie wieder dieselbe Aufgabe.\n\nBitte antworten Sie erst, wenn Sie dazu aufgefordert werden.\n\nAuch diesmal bekommen Sie wieder die Rückmeldungen wie im vorherigen Block,\n\naber gelegentlich kommt eine weitere Rückmeldung,\n\ndie nichts darüber aussagt, ob Sie richtig oder falsch gedrückt haben:\n\n\n\n\n\n\nWeiter mit Leertaste..."; }; x = 0; y = 0; bitmap { filename = "nf_wob.bmp"; width = 80; height = 80; }; x = 0; y = -150; }; time = 0; };

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/1673/CH4/EX4.8/4_8.sce

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4_8.sce

//curve fitting by sum of exponentials //example 4.8 //page 137 clc;clear;close; x=[1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1. 1.9]; y=[1.54 1.67 1.81 1.97 2.15 2.35 2.58 2.83 3.11]; s1=y(1)+y(5)-2*y(3); h=x(2)-x(1); I1=0; for i=1:3 if i==1|i==3 then I1=I1+y(i) elseif (modulo(i,2))==0 then I1=I1+4*y(i) elseif (modulo(i,2))~=0 then I1=I1+2*y(i) end end I1=(I1*h)/3 I2=0; for i=3:5 if i==3|i==5 then I2=I2+y(i) elseif (modulo(i,2))==0 then I2=I2+4*y(i) elseif (modulo(i,2))~=0 then I2=I2+2*y(i) end end I2=(I2*h)/3; for i=1:5 y1(i)=(1.0-x(i))*y(i); end for i=5:9 y2(i)=(1.4-x(i))*y(i); end I3=0; for i=1:3 if i==1|i==3 then I3=I3+y1(i) elseif (modulo(i,2))==0 then I3=I3+4*y1(i) elseif (modulo(i,2))~=0 then I3=I3+2*y1(i) end end I3=(I3*h)/3 I4=0; for i=3:5 if i==3|i==5 then I4=I4+y2(i) elseif (modulo(i,2))==0 then I4=I4+4*y2(i) elseif (modulo(i,2))~=0 then I4=I4+2*y2(i) end end I4=(I4*h)/3; s2=y(5)+y(9)-2*y(7); I5=0; for i=5:7 if i==5|i==7 then I5=I5+y(i) elseif (modulo(i,2))==0 then I5=I5+4*y(i) elseif (modulo(i,2))~=0 then I5=I5+2*y(i) end end I5=(I5*h)/3; I6=0; for i=7:9 if i==7|i==9 then I6=I6+y(i) elseif (modulo(i,2))==0 then I6=I6+4*y(i) elseif (modulo(i,2))~=0 then I6=I6+2*y(i) end end I6=(I6*h)/3; I7=0; for i=5:7 if i==5|i==7 then I7=I7+y2(i) elseif (modulo(i,2))==0 then I7=I7+4*y2(i) elseif (modulo(i,2))~=0 then I7=I7+2*y2(i) end end I7=(I7*h)/3; I8=0; for i=7:9 if i==7|i==9 then I8=I8+y2(i) elseif (modulo(i,2))==0 then I8=I8+4*y2(i) elseif (modulo(i,2))~=0 then I8=I8+2*y2(i) end end I8=(I8*h)/3; A=[1.81 2.180;2.88 3.104]; C=[2.10;3.00]; Z=A^-1*C X=poly(0,'X'); y=X^2-Z(1,1)*X-Z(2,1); R=roots(y) printf(' the unknown value of equation is %1.0f %1.0f',R(1,1),R(2,1));

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/1943/CH4/EX4.2/Ex4_2.sce

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Ex4_2.sce

clc clear //Input data CO2=11.5;//Percentage of carbondioxide present in combustion in % O2=2.7;//Percentage of oxygen present in the combustion in % CO=0.7;//Percentage of carbonmonoxide present in the combuston in % //Calculations a=85.1/3.76;//Equating moles for nitrogen from the equation x=(CO2+CO)/3;//Equating moles for carbon from the equation b=[a-CO2-(CO/2)-O2]*2;//Equating moles for oxygen from the equation y=a/x;//Moles of oxygen supplied for one mole of propane gas z=5;//Theoretically 5 moles of oxygen are required for reacting E=[(y-z)/z]*100;//The excess of air supplied in % //Output printf('The percentage excess air used is = %3.1f percentage',E)

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/52/CH11/EX11.5/Program11_5.sce

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Program11_5.sce

//Program 11.5 //Program To Calculate the value of the function //Y=A*X1+B*X2+C*X3 clear; clc; close; //Data A=1; B=2; C=3; X1=4; X2=5; X3=6; //Compute the function Y=A*X1+B*X2+C*X3; //Display the result in command window disp(Y,"Y = A*X1+B*X2+C*X3 = ");

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clc// // // //Variable declaration n1=1.563; //refractive index of core n2=1.498; //refractive index of cladding //Calculation NA=sqrt(n1^2-n2^2); //numerical aperture alpha_i=asin(NA); //angle of acceptance(radian) alpha_i=(alpha_i*180/%pi); //angle(degrees) alpha_id=int(alpha_i); alpha_im=60*(alpha_i-alpha_id); //Result printf("\n angle of acceptance is %0.3f degrees %0.1f minutes",alpha_id,alpha_im) printf("\n answer varies due to rounding off errors")

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//check op when the i/p contains imaginary elements A = [4*%i -2 1; 9 5 7]; v=var(A,0,1); v2=var(A,0,2); disp(v); disp(v2); //output //!--error 10000 //stdev: Wrong type for input argument #1: A real matrix expected. //at line 58 of function stdev called by : //at line 64 of function var called by : //v=var(A,0,1); //corresponding MATLAB o/p // 48.5000 24.5000 18.0000 // // 7.6667 // 4.0000 //

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// A Texbook on POWER SYSTEM ENGINEERING // A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar // DHANPAT RAI & Co. // SECOND EDITION // PART I : GENERATION // CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION // EXAMPLE : 7.25 : // Page number 85 clear ; clc ; close ; // Clear the work space and console // Given data w1 = 1.0 // Week 1 Q1 = 200.0 // Discharge during week 1(m^2/sec) w2 = 2.0 // Week 2 Q2 = 300.0 // Discharge during week 2(m^2/sec) w3 = 3.0 // Week 3 Q3 = 1100.0 // Discharge during week 3(m^2/sec) w4 = 4.0 // Week 4 Q4 = 700.0 // Discharge during week 4(m^2/sec) w5 = 5.0 // Week 5 Q5 = 900.0 // Discharge during week 5(m^2/sec) w6 = 6.0 // Week 6 Q6 = 800.0 // Discharge during week 6(m^2/sec) w7 = 7.0 // Week 7 Q7 = 600.0 // Discharge during week 7(m^2/sec) w8 = 8.0 // Week 8 Q8 = 1000.0 // Discharge during week 8(m^2/sec) w9 = 9.0 // Week 9 Q9 = 500.0 // Discharge during week 9(m^2/sec) w10 = 10.0 // Week 10 Q10 = 400.0 // Discharge during week 10(m^2/sec) w11 = 11.0 // Week 11 Q11 = 500.0 // Discharge during week 11(m^2/sec) w12 = 12.0 // Week 12 Q12 = 700.0 // Discharge during week 12(m^2/sec) w13 = 13.0 // Week 13 Q13 = 100.0 // Discharge during week 13(m^2/sec) no_week = 13.0 // Total weeks of discharge // Calculations Q_average = (Q1+Q2+Q3+Q4+Q5+Q6+Q7+Q8+Q9+Q10+Q11+Q12+Q13)/no_week // Average weekly discharge(m^3/sec) // Hydrograph W = [0,w1,w1,w2,w2,w3,w3,w4,w4,w5,w5,w6,w6,w7,w7,w8,w8,w9,w9,w10,w10,w11,w11,w12,w12,w13,w13,w13] Q = [200,Q1,Q2,Q2,Q3,Q3,Q4,Q4,Q5,Q5,Q6,Q6,Q7,Q7,Q8,Q8,Q9,Q9,Q10,Q10,Q11,Q11,Q12,Q12,Q13,Q13,Q13,0] a = gca() a.thickness = 2 // sets thickness of plot plot(W,Q) // Plotting hydrograph q = Q_average w = [0,w1,w2,w3,w4,w5,w6,w7,w8,w9,w10,w11,w12,w13,14] q_dash = [q,q,q,q,q,q,q,q,q,q,q,q,q,q,q] // Plotting average weekly discharge plot(w,q_dash,'r--') a.x_label.text = 'Time(week)' // labels x-axis a.y_label.text = 'Q(m^3/sec)' // labels y-axis xtitle("Fig E7.4 . Plot of Hydrograph") xset('thickness',2) // sets thickness of axes xstring(13,560,'Q_av') xstring(12.02,110,'Q_min') xstring(2.02,1110,'Q_max') // Results disp("PART I - EXAMPLE : 7.25 : SOLUTION :-") printf("\nThe hydrograph is shown in the Figure E7.4") printf("\nAverage discharge available for the whole period = %.f m^3/sec", Q_average)

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clc clear pi=22/7; angle1=input('enter angle of rotation in yaw in degrees : '); a1=pi*angle1/180; angle2=input('enter angle of rotation in pitch in degrees : '); a2=pi*angle2/180; angle3=input('enter angle of rotation in roll in degrees : '); a3=pi*angle3/180; i=[2;3;4]; r1=[1 0 0;0 cos(a1) -sin(a1 );0 sin(a1) cos(a1)]; r2=[cos(a2) 0 sin(a2);0 1 0;-sin(a2) 0 cos(a2)]; r3=[cos(a3) -sin(a3) 0;sin(a3) cos(a3) 0;0 0 1] ; ypr2=r1*i; ypr2=r2*ypr2; ypr2=r3*ypr2; disp('the ypr matrix is : ') disp(ypr2); rpy3=r3*r2; rpy3=rpy3*r1; rpy3=rpy3*i; disp(' The rpy matrix is : ') disp(rpy3); if (isequal(ypr2,rpy3)) then disp('Hence proved'); end;

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clc;clear; //EXample 5.16 //given data Wo=5000;//wavelength in angstrom V=3.1;//stopping potential in V //calcualtion W=1/((V/12400)+(1/Wo)); disp(W,'The unknown wavelength in Angstrom')

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// FUNDAMENTALS OF ELECTICAL MACHINES // M.A.SALAM // NAROSA PUBLISHING HOUSE // SECOND EDITION // Chapter 7 : THREE-PHASE INDUCTION MOTOR // Example : 7.6 clc;clear; // clears the console and command history // Given data V_1 = 150 // supply voltage in V P = 4 // number of poles f = 50 // frequency in Hz Z_1 = 0.12+%i*0.16 // per phase standstill stator impedance in ohm Z_2 = 0.22+%i*0.28 // per phase standstill rotor impedance in ohm R_2 = real(Z_2) // from Z_2 // caclulations Z_eq = Z_1+Z_2 // equivalent impedance in ohm P_mech = 3*V_1^2/(2*(R_2+abs(Z_eq))) // maximum mechanical power developed in W s_mp = R_2/(abs(Z_eq)+R_2) // slip W_s = 2*%pi*2*f/P // since N_s = f/(P/2) and W_s = 2*%pi*N_s W = (1-s_mp)*W_s // speed of rotor in rad/s T_mxm = P_mech/W // miximum torque in N-m // display the result disp("Example 7.6 solution"); printf(" \n maximum mechanical power developed \n P_mech = %.f W \n", P_mech); printf(" \n Maximum torque \n T_mxm = %.3f N-m \n", T_mxm); printf(" \n Maximum slip \n s_mp = %.2f \n", s_mp); printf(" \n NOTE : Error in calculation of P_mech and T_mxm ");

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//Example 12.3, page 445 clc h=6.63*10^-34//in J-s I=(2*%pi)^2*2.66*10^-47//in kg-m2 m_H=1/(6.02*10^26)//in kg E=(h^2)/I printf("\n The energy is %e J",E) s=.59*10^-19//in J k=1.38*10^-23//in j/k T=(s)/k printf("\n The temperature is %f K",T) //Answer diffrence is because of round off

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//data R2=62.5*10^3 //ohm E=1 L1=320*10^(-9) //H L2=20*10^(-9) //H //formula and result printf("\nresult:-") n=sqrt(E*L1/L2) printf("\nn=sqrt(E*L1/L2)=%.0f",n) R1=n^(2)*R2 printf("\nZ1=n^2*Z2\nY1=Y2/n^2=1/n^2*complex(1/R2,W*C2)\n\nR1=n^2*R2=%.0e ohm\n",R1) C1=(6+1/4^2*30.7)*10^-12 printf("\nC1=%.2e F\n",C1) Wo=1/sqrt(L1*C1) printf("\nWo=1/sqrt(L1*C1)=%.4e rad/s\n",Wo) fo=Wo/(2*%pi) printf("\nfo=Wo/(2*pi)=%.2e Hz\n",fo) Q=R1/(Wo*L1) printf("\nQ=R1/(Wo*L1)=%.4f",Q)

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// Example 22_15 clc;funcprot(0); //Given data T_1=400;// °C p_1=40;// bar p_2=2;// bar p_3=0.5;// bar p_4=0.05;// bar n_t1=75/100;// The isentropic efficiency of the first stage of the turbine n_t2=80/100;// The isentropic efficiency of the second stage of the turbine n_t3=85/100;// The isentropic efficiency of the third stage of the turbine m_s=10;// The steam flow in kg/sec // Calculation // From h-s chart: h_1=3210;// kJ/kg h_2a=2562;// kJ/kg h_2=h_1-((n_t1)*(h_1-h_2a));// kJ/kg h_3a=2508;// kJ/kg h_3=h_2-((n_t2)*(h_2-h_3a));// kJ/kg h_4a=2232;// kJ/kg h_4=h_3-((n_t3)*(h_3-h_4a));// kJ/kg // From steam tables h_f8=502;// kJ/kg(2 bar) h_f10=h_f8;// kJ/kg h_f6=339;// kJ/kg(0.5 bar) h_f7=h_f6;// kJ/kg h_f9=h_f6;// kJ/kg h_f5=136;// kJ/kg(0.05 bar) //Assume m_1=y(1);m_2=y(2) function[X]=mass(y) X(1)=(y(1)*(h_2-h_f10))-((1-y(1))*(h_f8-h_f7)); X(2)=(y(2)*(h_3-h_f9))-((1-y(1)-y(2))*(h_f6-h_f5)); endfunction y=[0.01 0.01]; z=fsolve(y,mass); m_1=z(1);// kJ/kg m_2=z(2);// kJ/kg W=(h_1-h_2)+((1-m_1)*(h_2-h_3))+((1-m_1-m_2)*(h_3-h_4));// kJ/kg P=W*m_s;// Power developed by the turbine in kW Q_s=h_1-h_f10;// Heat supplied per kg of steam in kJ/kg n_th=(W/Q_s)*100;// Thermal efficiency of the cycle in % printf('\n(a)Steam bled for regenerative heaters per kg of steam to turbine,m_1=%0.4f kJ/kg & m_2=%0.4f kJ/kg \n(b)Power developed by the turbine=%0.0f kW \n(c)Thermal efficiency of the cycle=%0.2f percentage',m_1,m_2,P,n_th); // The answer provided in the textbook is wrong