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//Desenvolver um programa em Matlab que leia 20 //valores correspondentes as notas de //PCI. As notas variam de 0 a 10, somente valores //inteiros. Calcular e escrever a //Frequência Absoluta e a Frequência Relativa //das notas lidas. //Obs: //a)Frequência Absoluta é a quantidade de vezes que uma nota ocorreu no conjunto. //b)Frequência Relativa é a Frequência Absoluta //dividida pela quantidade de notas lidas. //c)Escrever a nota, a sua Frequência Absoluta e a sua Frequência Relativa.
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Ch04Ex11.sce
// Scilab Code Ex4.11:: Page-4.23 (2009) clc; clear; mu_o = 1.5442; // Refractive index of ordinary wave mu_e = 1.5533; // Refractive index of extraordinary wave lambda = 5000e-008; // Wavelength of light used, m // As (mu_o - mu_e)*t = lambda/4, solving for t t = lambda/(4*(mu_e - mu_o)); // Least thickness of plate for which emergent beam is plane polarised, cm printf("\nThe least thickness of plate for which emergent beam is plane polarised = %4.2e cm", t); // Result // The least thickness of plate for which emergent beam is plane polarised = 1.37e-003 cm
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clear // // // //Variable declaration n1=1.5; //core refractive index n2=1.48; //cladding refractive index n=1; //Calculations NA=sqrt(n1^2-n2^2); //numerical aperture i0=asin(NA/n); //maximum entrance angle(radian) i0=i0*180/%pi ; //maximum entrance angle(degrees) //Result printf("\n numerical aperture is %0.5f ",NA) printf("\n maximum entrance angle is %0.2f degrees",i0)
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ENV_ENCRYPT.tst
-- VectorCAST 21.sp6 (01/11/22) -- Test Case Script -- -- Environment : ENV_ENCRYPT -- Unit(s) Under Test: encrypt -- -- Script Features TEST.SCRIPT_FEATURE:C_DIRECT_ARRAY_INDEXING TEST.SCRIPT_FEATURE:CPP_CLASS_OBJECT_REVISION TEST.SCRIPT_FEATURE:MULTIPLE_UUT_SUPPORT TEST.SCRIPT_FEATURE:REMOVED_CL_PREFIX TEST.SCRIPT_FEATURE:MIXED_CASE_NAMES TEST.SCRIPT_FEATURE:STANDARD_SPACING_R2 TEST.SCRIPT_FEATURE:OVERLOADED_CONST_SUPPORT TEST.SCRIPT_FEATURE:UNDERSCORE_NULLPTR TEST.SCRIPT_FEATURE:FULL_PARAMETER_TYPES TEST.SCRIPT_FEATURE:STRUCT_DTOR_ADDS_POINTER TEST.SCRIPT_FEATURE:STRUCT_FIELD_CTOR_ADDS_POINTER TEST.SCRIPT_FEATURE:STATIC_HEADER_FUNCS_IN_UUTS TEST.SCRIPT_FEATURE:VCAST_MAIN_NOT_RENAMED -- -- Unit: encrypt -- Subprogram: transmit_Info -- Test Case: encrypt.transmit_Info.failure TEST.UNIT:encrypt TEST.SUBPROGRAM:transmit_Info TEST.NEW TEST.NAME:encrypt.transmit_Info.failure TEST.STUB:encrypt.generate_private_key TEST.VALUE:encrypt.transmit_Info.name:<<malloc 14>> TEST.VALUE:encrypt.transmit_Info.name:"Tim Schneider" TEST.VALUE:encrypt.transmit_Info.number:<<malloc 17>> TEST.VALUE:encrypt.transmit_Info.number:"0000111122223333" TEST.VALUE:encrypt.transmit_Info.secCode:<<malloc 4>> TEST.VALUE:encrypt.transmit_Info.secCode:"012" TEST.VALUE:encrypt.transmit_Info.Info:12.34 TEST.VALUE:uut_prototype_stubs.matrix_multiply.return:-1 TEST.EXPECTED:encrypt.transmit_Info.return:MACRO=FAILURE TEST.END -- Test Case: encrypt.transmit_Info.good TEST.UNIT:encrypt TEST.SUBPROGRAM:transmit_Info TEST.NEW TEST.NAME:encrypt.transmit_Info.good TEST.VALUE:encrypt.transmit_Info.name:<<malloc 14>> TEST.VALUE:encrypt.transmit_Info.name:"Tim Schneider" TEST.VALUE:encrypt.transmit_Info.number:<<malloc 17>> TEST.VALUE:encrypt.transmit_Info.number:"0000111122223333" TEST.VALUE:encrypt.transmit_Info.secCode:<<malloc 4>> TEST.VALUE:encrypt.transmit_Info.secCode:"012" TEST.VALUE:encrypt.transmit_Info.Info:12.34 TEST.EXPECTED:encrypt.transmit_Info.return:MACRO=SUCCESS TEST.END
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//Variable declaration t=1 //thickness(mil) e=1.6*10**-19 //charge on electron(C) Pp=10**17 //concentration of phosphorous(atoms/cm^3) Bn=5*10**16 //boron concentration(atoms/cm^3) un=.135 //mobility(m^2/Vs) //Calculations n=(Pp-Bn)*10**6 //net concentration(atoms/cm^3) g=e*un*n //conductivity() rho=10**6/(g*25) //resistivity(ohm mil) Rs=rho/t //sheet resistance(ohm mil^2) //Results printf ("Sheet resistance is %.f ohm(mil**2)",Rs)
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//Problem 15.02: //initializing the variables: //calculation: //The first step is to convert the equipment, installation, and operating costs to total\ncosts by multiplying each by the total gas flow, 100,000 acfm. Hence, for the finned exchanger, //the total costs are Equipmentcost = 100000*3.1 // in $ Installationcost = 100000*0.80 // in $ Operatingcost = 100000*0.06 // in $ //Note that the operating costs are on an annualized basis. The equipment cost and the installation\ncost must then be converted to an annual basis using the CRF. From Equation (15.3) CRF = (0.1)*(1+0.1)^20/[(1+0.1)^20 - 1] //The annual costs for the equipment and the installation is given by the product of the CRF and\nthe total costs of each: Equipmentannualcost = 0.11746*Equipmentcost Installationannualcost = 0.11746*Installationcost //The calculations for the 4-pass and the 2-pass exchangers are performed in the same manner.\nThe three preheaters can be compared after all the annual costs are added. The tabulated results\nare provided in Table 15.5. // total annual costs CF = 65000 C4 = 77385 C2 = 60111 printf("\n\nResult\n\n") printf("\n According to the analysis, Total Annual Costs for Finned exchanger = $%.0f, for 4-Pass Exchanger = $%.0f and for 2-Pass Exchanger = $%.0f.\nTherefore 2-pass exchanger is the most economically attractive device since the annual cost is the lowest.",CF,C4,C2)
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// Example 16.2 format('v',6) clc; clear; close; // given data A=20000; B= 0.02; Vin= 1;// in mV Vin= Vin*10^-3;// in V // The closed loop voltage gain, A_CL= A/(1+A*B); // The output voltage, Vout= Vin*A_CL;// in V // The error voltage, Verror= Vout/A;// in V Vout= Vout*10^3;// in mV Verror= Verror*10^6;// in µV disp(A_CL,"The value of A_CL is : "); disp(Vout,"The value of Vout in mV is : ") disp(Verror,"The value of Verror in µV is : ")
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clc //solution // initialization of variables Cp=1.0 // specific heat at constant pressure k=1.4 // polytropic index for air T1=25+273 // temperature at compressor inlet T3=850+273 // maximum temperature in kelvin r=5 // pressure ratio=P2/P1 & P4/P3 T2=T1*(r)^((k-1)/k) // temperature after compression T4=T3*(1/r)^((k-1)/k) // final temperature Wcomp=Cp*(T2-T1) // compressor work Wturb=Cp*(T3-T4) // turbine work BWR=Wcomp/Wturb // back work ratio printf("The BWR is %0.1f %%\n",BWR*100) Effi=1-r^((1-k)/k) // thermal efficiency printf(" The thermal efficiency is %0.1f %% \n",Effi*100)
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// Exa 6.15 clc; clear; close; // Given data I_DSS = 30;// in mA V_GS = -5;// in V V_GS_off = -8;// in V I_D = I_DSS*(1-(V_GS/V_GS_off))^2;// in mA disp(I_D,"The drain current in mA is");
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//pagenumber 228 example 46 clear vb=0.8;//volt beta1=100; vce=0.2;//volt vcc=10;//volt rb=200*10^3;//ohm //collector resistance ib=(5-0.7)/rb; colres=(vcc-vce)/(beta1*ib); disp("min collector resistance = "+string((colres))+"ohm");
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//Electric Drives:concepts and applications by V.subrahmanyam //Publisher:Tata McGraw-Hill //Edition:Second //Ex3_10 clc; clear; Vs=400;//Supply voltage in V f=50;//Frequency in Hz Rd=15;//Resistance in ohm pf=0.2588;//Powerfactor Vdia=1.35*Vs*pf; disp(Vdia,"Average value of load voltage in V is:") Id=Vdia/Rd; disp(Id,"Average value of load current in A is:") P=Vdia*Id; disp(P,"Power dissipation in W is:")
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clc; clear; format('v',11); r0=[1,1,1]; //Vector ro r1=[1,2,3]; //Vector r1 //Displaying the given points in vectorial form disp(r0,'The given two points are: ro='); disp(r1,'r1='); R=r1-r0; modR=sqrt(R(1)^2+R(2)^2+R(3)^2); //Distance between the two given points unit_R=R/modR; //Unit vector along the vector from ro towards r1 p1=5*unit_R+r0; //Point at 5cm from ro towards r1 disp(p1,'The point at distance of 5cm away from r0 and towards r1 is:p1= '); p2=-5*unit_R+r0; disp(p2,'The point at distance of 5cm away from r0 and away from r1 is:p2='); //Point at 5cm from ro and away from r1 disp('Equation of the line passing through the given points:r=t(r1-r0)+r0'); disp('to find the intersection of this line with X-Y plane:z=0'); t=-1*sqrt(5)/2; disp(t,'The value of the parameter t='); //Displaying the equation of the line //Computing the location of the point of intersection x=t*unit_R(1)+r0(1); y=t*unit_R(2)+r0(2); p1=[x,y,0]; // Point of intersaction with X-Y plane disp(p1,'The point of intersection with X-Y plane:p1='); disp('to find the intersection with x-z plane:y=0'); t=-1*sqrt(5); //The value of the parameter t disp(t,'The value of the parameter t='); x=t*unit_R(1)+r0(1); z=t*unit_R(3)+r0(3); p2=[x 0 z]; //Point of intersection with X-Z plane disp(p2,'The point of intersection with X-Z plane:p2='); disp('to find the intersection with y-z plane:x=0'); disp('as we are getting 0=1,we can say that the line does not intersect with the Y-Z plane');
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getElementaryMatrix.m
function [E] = getElementaryRowMatrix(i, j, value, numberOfRows) // create an identity matrix E = eye(numberOfRows); E(i, j) = value; endfunction
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res_log=read('../data/gps_kalman_log.m',-1,16); [r_n,c_n]=size(res_log); idx=[1:r_n]; // First plot KF state //xset('color',3); //xset('background', 1); xbasc(); xset('window',0); // plot pos_x result subplot(511); plot2d(res_log(:,2)); xtitle("ECEF_x",'100ms','m'); xset('background', 4); // plot pos_y result subplot(512); plot2d(res_log(:,3)); xtitle("ECEF_y",'100ms','m'); xset('background', 4); // plot pos_z result subplot(513); plot2d(res_log(:,4)); xtitle("ECEF_z", '100ms','m'); xset('background', 4); // plot clk_bias result subplot(514); plot2d(res_log(:,5)); xtitle("clk_bias", '100ms','m'); xset('background', 4); // plot clk_drift result subplot(515); plot2d(res_log(:,6)); xtitle("clk_drift", '100ms','m/s'); xset('background', 4); // Then plot corrections xset('window',1); // plot corr_pos_x result subplot(511); plot2d(res_log(:,7)); xtitle("ECEF_x correction",'100ms','m'); xset('background', 4); // plot corr_pos_y result subplot(512); plot2d(res_log(:,8)); xtitle("ECEF_y_correction",'100ms','m'); xset('background', 4); // plot corr_pos_z result subplot(513); plot2d(res_log(:,9)); xtitle("ECEF_z_correction", '100ms','m'); xset('background', 4); // plot corr_bias result subplot(514); plot2d(res_log(:,10)); xtitle("clk_bias_correction", '100ms','m'); xset('background', 4); // plot corr_drift result subplot(515); plot2d(res_log(:,11)); xtitle("clk_drift_correction", '100ms','m/s'); xset('background', 4); // Then plot diag of P_matrix xset('window',2); // plot p_pos_x result subplot(511); plot2d(res_log(:,12)); xtitle("ECEF_x_cov",'100ms','m^2'); xset('background', 4); // plot p_pos_y result subplot(512); plot2d(res_log(:,13)); xtitle("ECEF_y_cov",'100ms','m^2'); xset('background', 4); // plot p_pos_z result subplot(513); plot2d(res_log(:,14)); xtitle("ECEF_z_cov", '100ms','m^2'); xset('background', 4); // plot p_bias result subplot(514); plot2d(res_log(:,15)); xtitle("clk_bias_cov", '100ms','m^2'); xset('background', 4); // plot p_drift result subplot(515); plot2d(res_log(:,16)); xtitle("clk_drift_cov", '100ms','(m/s)^2'); xset('background', 4);
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//Householder Matrix clc; clear; close(); format('v',7); e = [1;0;0]; x = [-1;1;4]; disp(e , 'e = '); disp(x , 'x = '); //considering the positive k according to sign convention k = sqrt(x'*x); disp(k,'k = '); u = x - k*e; disp(u,'u = '); Q = eye(3,3) - 2*u*u'/(u'*u); disp(Q,'Householder Matrix : ')
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// //readings ir=8.652 fr=6.798 c=20 m=100//natural scale n=1 sc=100 a2=m*(fr-ir-10*(n)+c) a2=a2*sc printf("\n A= %0.3f ",a2) printf("\n required area is %0.3f square meters',a2)
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// Example 1.3 clc; clear; close; // Given data format('v',6); VCC= 12;//in V VEE= -12;// in V RC= 10;//in kΩ RE= 10;// in kΩ RB= 20;// in kΩ VBE= 0.7;// in V // Part (a) beta_dc= 75; // Tail current, IT= 2*IE= VEE/RE (ignoring VBE), hence IT= abs(VEE)/RE;// in mA IC= IT/2;//collector current in mA // output voltage, Vout1= VCC-IC*RC;// in V IT= (abs(VEE)-VBE)/RE;// tail current in mA (on considering VBE) IC= IT/2;//collector current in mA Vout2= VCC-IC*RC;// in V // Tail current, IT= (abs(VEE)-VBE)/(RE+RB/(2*beta_dc));// in mA IC= IT/2;//collector current in mA // output voltage, Vout3= VCC-IC*RC;// in V disp("Part (a) : There are three different values of output voltage in volts"); disp(Vout1); disp(Vout2); disp(Vout3); // Part (b) IT= abs(VEE)/RE;// in mA IC= IT/2;//collector current in mA IB= IC/(beta_dc);// base current in mA IB= IB*10^3;// in µA VB= -IB*RB;//base voltage in mV VB= VB*10^-3;// in V disp("Part (b) : "); disp(IB,"The value of base current in µA is : "); disp(VB,"The value of base voltage in volts is : "); // Part (c) beta_dc1= 60; beta_dc2= 80; IB1= IC/beta_dc1;//base current for transistor Q1, in mA IB1= IB1*10^3;// in µA disp("Part (c)") disp(IB1,"The value of base current for transistor Q1 in µA is : "); VB1= -IB1*RB;// in mV VB1= VB1*10^-3;// in V disp(VB1,"The value of base voltage for transistor Q1 in volts is : "); IB2= IC/beta_dc2;//base current for transistor Q2, in mA IB2= IB2*10^3;// in µA disp(IB2,"The value of base current for transistor Q2 in µA is : "); VB2= -IB2*RB;// in mV VB2= VB2*10^-3;// in V disp(VB2,"The value of base voltage for transistor Q2 in volts is : "); // Note : In the part (c), the unit of base current for transistor Q2 in the book is wrong it will be µA
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//chapter5,Example5_3,pg 98 V=1450 A1=112*0.03//absorption due to plastered wall A2=130*0.06//absorption due to wooden floor A3=170*0.04//absorption due to plastd. celing A4=20*0.06//absorption due to wooden door A5=100*1//absorption due to cushioned chairs sum_as=A1+A2+A3+A4+A5 T1=(0.161*V)/sum_as//reverberation time case-1 T2=(0.161*V)/(sum_as+(60*4.7))//persons=60,A=4.7 case-2 T3=(0.161*V)/(sum_as+(100*4.7))//seat cushioned=100 rev. case-3 printf("rev. time for case-1\n") printf("T1=%.3f sec",T1) printf("\nrev. time for case-2\n") printf("T2=%.3f sec",T2) printf("\nrev. time for case-3\n") printf("T3=%.3f sec",T3)
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//Calculations on diesel engine fuel pump clc,clear //Given: V_b=7 //Volume of fuel in the barrel in cc D_l=3,L_l=700 //Diameter and length of fuel delivery line in mm V_iv=3 //Volume of fuel in the injection valve in cc P2=200 //Delivery pressure in bar P1=1 //Sump pressure in bar V_d=0.15 //Volume to be delivered in cc C=78.8D-6 //Coefficient of compressibility d=8 //Diameter of the plunger in mm //Solution: V_l=%pi/4*D_l^2*L_l*10^-3 //Volume of fuel in delivery line in cc V1=V_b+V_l+V_iv //Total initial fuel volume in cc deltaV=C*(P2-P1)*V1 //Change in volume due to compression in cc V_p=deltaV+V_d //Displaced volume by plunger in cc A_p=%pi/4*d^2*10^-2 //Area of the plunger in cm^2 l=V_p/A_p //Effective stroke of plunger in cm //Results: printf("\n The plunger displacement = %.3f cc",V_p) printf("\n The effective stroke of the plunger, l = %.2f mm\n\n",l*10)
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w=10;//intensity of distributed load, in N/m L=10;//leangth in m E=200000;//in N/mm^2 I=0.5;// moment of Inertia of cross section, in m^4
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clc clear //Input data l=80;//Latent heat of fusion of ice in cal/g L=l*4.2*10^7;//Latent heat of fusion in ergs/g V=0.091;//The change in specific volume when 1 g of water freezes into ice in cc t1=0;//The actual freezing point of ice in degree centigrade t2=-1;//The given temperature at which ice must freeze in degree centigrade p=1;//The atmospheric pressure in atmospheres //Calculations T1=t1+273;//The actual freezing point of ice in K T2=t2+273;//The given temperature at which ice must freeze in K T=T1-T2;//The change in temperature in K P=(L*T)/(V*T1);//The pressure in dynes/cm^2 P1=P/10^6;//The pressure in atmospheres P2=P1+p;//The pressure under which ice would freeze in atmospheres //Output printf('The pressure under which ice would freeze at -1 degree centigrade is %3.1f atmospheres ',P2)
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//Ex13_22 PG-13.12 clc clear printf("conversion of decimal number 3509 to its hexadecimal equivalent =") a=[3509]; x=dec2hex(a) printf(" %s",x)
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clear // // // //Variable declaration h=6.6*10^-34; //planck's constant(J-sec) m=9.1*10^-31; //mass of electron(kg) c=3*10^8; //velocity of light(m/sec) e=1.6*10^-19; //charge of electron(c) E=1000; //energy of electron(eV) //Calculations lamda_p=h*c*10^10/(E*e); //wavelength of photon(angstrom) lamda_e=h*10^10/sqrt(2*m*E*e); //wavelength of electron(angstrom) //Result printf("\n wavelength of photon is %0.1f angstrom",lamda_p) printf("\n wavelength of electron is %0.2f angstrom",lamda_e)
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clear; clc; g=15; p=10; o=8; d=1; c=3; y=o+d+c; oc=g*p/y; mprintf("the overcurrent factor=%f",oc)
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funcprot(0); C = [1 1 0 0; 0 1 1 0; 0 0 1 1; 1 0 0 1; 1 0 1 0; 0 1 0 1]; K = 6; L = 4; N = 2; chain = [0 0 0 0]; chain_old = chain; p = 0.5; q = 0.5; function z=get_state_number(chain) [n,m]=size(chain); if n>m then chain=chain'; end z=0; for i=1:K if and(chain==C(i,:)) then z=i; return; end end endfunction function [z,c,zo]=bsu_step(i,j,p,q,chain,chain_old) z=chain; zo=chain_old; c=zeros(1,L); pr = p; if zo(i)==1 then pr = q; end if (chain(j)==0) & (chain(i)==1) & (rand()<pr) then z(j)=1; z(i)=0; c(j)=1; zo(i)=1; zo(j)=1; else c(i)=1; zo(i)=1; end endfunction function [z,c]=bsu_update(p,q,chain) zo=chain; c=zeros(1,L); ic=c; z=chain; // (4,1),(3,4),(2,3),(1,2) [z,ic,zo]=bsu_step(L,1,p,q,z,zo); c=c+ic; for i=L:-1:2 [z,ic,zo]=bsu_step(i-1,i,p,q,z,zo); c=c+ic; end endfunction function pm=bsu_probability_matrix(p,q,num_of_iterations) pm=zeros(K,K); for k=1:K // for each possible initial configuration for iter=1:num_of_iterations chain=C(k,:); [z,c]=bsu_update(p,q,chain); kk=get_state_number(z); pm(k,kk)=pm(k,kk)+1; end end for k=1:K pm(:,k)=pm(:,k)/sum(pm(:,k)); end endfunction
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exec('diffuse_utils.m'); exec('diffuse.m'); exec('diffuse_drv.m'); exec('diffuse_analysis.m'); chdir('results'); sirout=diffuse_analysis('diffuse_datlist.lis',10);
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//Pressure diference(in mm of mecury): p=30; //Density of water(in kg/m^3): dw=1000; //Aceleration due to gravity(in m/sec^2): g=9.81; //Density of air(in kg/m^3): da=1.23; //Specific gravity of mercury: SG=13.6;
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// Copyright (c) 2017, Embedded Solutions // All rights reserved. // This file is released under the 3-clause BSD license. See COPYING-BSD. // FFT, AI scan function - DEMO figH = figure("Figure_name","MicroDAQ FFT demo"); notInitialized = 1; divFac = 10; axisH = []; // AI scan parameters scanFreq = 100000; channel = 1; AIRange = [-5 5]; isDifferential = %F; numOfSmaples = scanFreq/divFac; duration = -1; // Initialize analog input scanning mdaqAIScanInit(channel, AIRange, isDifferential, scanFreq, duration); while %T // Acquire data [data result] = mdaqAIScanRead(numOfSmaples, 1); // Calculate FFT data = data - mean(data); y = fft(data'); f = scanFreq*(0:(numOfSmaples/10))/numOfSmaples; n = size(f,'*'); // Update plot if is_handle_valid(figH) then if notInitialized == 1 then clf(); plot(f,abs(y(1:n))); title("FFT", "fontsize", 3); xlabel("Frequency [Hz]","fontsize", 3); notInitialized = 0; axisH = gca(); else axisH.children.children.data(:,2) = abs(y(1:n))'; end else // Stop scanning mdaqAIScanStop(); break; end end // Close plot mprintf("\nFFT demo has been stopped."); if is_handle_valid(figH) then close(figH); end
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load EnhancedALU.hdl, output-file EnhancedALU.out, compare-to EnhancedALU.cmp, output-list D%B1.16.1 AM%B1.16.1 cBits%B1.6.1 outM%B1.16.1 zr%B2.1.1 ng%B2.1.1; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B101010, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B111111, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B111010, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B001100, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B110000, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B001101, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B110001, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B001111, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B110011, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B011111, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B110111, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B001110, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B110010, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B000010, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B010011, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B000111, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B000000, eval, output; set D %B0000000000000010, set AM %B0000000000000100, set cBits %B010101, eval, output;
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clc p1=1.05*10^5; //N/m^2 V1=0.04; //m^3 T1=288; //K p2=4.8*10^5; T2=T1; R0=8314; M=28; disp("(i) The change of entropy =") R=R0/M; m=p1*V1/R/T1; dS=m*R*log(p1/p2) disp("Decrease in entropy =") disp(-dS) disp("J/K") disp("(ii)Heat rejected = ") Q=T1*(-dS); disp("Q=") disp(Q) disp("J") W=Q; disp("Work done = ") disp(W) disp("J") V2=p1*V1/p2; v1=V1/m; //specific volume v2=V2/m; //specific volume v=v2:0.01:v1; function p=f(v) p=p1*v1/v endfunction plot(v,f) p=p1 plot(v,p,'--') p=[0 p2] v=[v2 v2] plot(v,p,'--') p=[0 p1] v=[v1 v1] plot(v,p,'--') xtitle("p-v diagram", "v(m^3/kg)", "p(N/m^2)") xset('window', 1) T=[288 288] s=[10 (10-dS)] plot(s,T) s=[10 10] T=[0 288] plot(s,T,'--') s=[(10-dS) (10-dS)] T=[0 288] plot(s,T,'--') xtitle("T-s diagram", "s(kJ/kg K)", "T(K)")
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//OptoElectronics and Fibre Optics Communication, by C.K Sarkar and B.C Sarkar //Example 9.3 //OS=Windows 10 ////Scilab version Scilab 6.0.0-beta-2(64 bit) clc; clear; //given L=1.2//link length in Km Gama_o=12.7;//optical output pulse of 3dB width in nanoseconds Gama_i=0.4;//optical input pulse of 3dB width in nanosseconds q=(Gama_o)^2; w=(Gama_i)^2; e=q-w; u=sqrt(e); v=1.2; Gama_3dB=u/v;//3dB pulse dispersion for the fibre in ns/Km mprintf("\n The 3dB pulse dispersion for the fibre is=%.2f ns/Km",Gama_3dB); Bopt=0.44/(Gama_3dB*1e-9);//fibre bandwidth length productmultiplication by 1e-9 as gama is in nsKm mprintf("\n The fibre bandwidth length product is=%.2f MHzKm",Bopt/1e6); //multiplication by 1e6 to convert unit from Hz to MHz //the answer vary due to rounding
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errcatch(-1,"stop");mode(2);// 10.5 function b = flip(a) b = a(length(a):-1:1); endfunction; exit();
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@relation unknow @attribute at1 real[0.0,100.0] @attribute at2 real[0.0,100.0] @attribute at3 real[0.0,100.0] @attribute at4 real[0.0,100.0] @attribute at5 real[0.0,100.0] @attribute at6 real[0.0,100.0] @attribute at7 real[0.0,100.0] @attribute at8 real[0.0,100.0] @attribute at9 real[0.0,100.0] @attribute at10 real[0.0,100.0] @attribute at11 real[0.0,100.0] @attribute at12 real[0.0,100.0] @attribute at13 real[0.0,100.0] @attribute at14 real[0.0,100.0] @attribute at15 real[0.0,100.0] @attribute at16 real[0.0,100.0] @attribute class{0,1,2,3,4,5,6,7,8,9} @inputs at1,at2,at3,at4,at5,at6,at7,at8,at9,at10,at11,at12,at13,at14,at15,at16 @outputs class @data 5 5 0 0 4 4 5 5 1 1 3 3 5 5 2 2 8 8 8 0 8 8 8 8 5 5 1 1 0 0 5 5 7 7 9 9 6 6 9 9 9 9 1 1 9 4 9 5 6 6 0 0 6 6 5 5 3 3 7 7 3 3 0 0 3 3 1 1 1 1 3 2 5 5 5 9 5 5 8 8 7 7 6 6 2 2 3 3 0 0 9 9 8 8 7 7 5 5 1 1 0 0 5 3 0 0 3 3 5 5 4 4 9 9 4 4 3 3 9 9 0 0 4 4 4 4 4 4 6 6 3 3 8 8 1 1 4 4 8 8 9 9 0 0 4 4 9 9 8 8 0 0 4 4 2 2 3 3 2 2 6 6 9 9 0 0 5 5 9 9 7 7 6 6 3 3 3 3 1 1 1 1 2 2 7 7 2 2 6 6 9 9 7 7 3 3 7 7 0 0 3 3 7 7 8 8 8 8 2 2 1 1 0 0 7 7 0 0 7 7 1 1 6 6 7 7 7 7 7 7 1 1 8 8 8 8 4 4 1 1 2 2 1 1 9 9 4 4 6 6 2 2 2 2 2 2 7 7 2 2 1 1 2 2 1 1 6 6 8 8 8 8 3 3 4 4 3 3 7 7 6 6 9 9 2 2 1 1 6 6 7 7 4 4 0 0 8 8 6 6 6 6 0 0 8 8 5 5 6 6 9 9 7 7 9 9 0 0 1 2 1 1 5 5 8 8 1 1 5 5 4 4 8 8 0 0 5 5 8 8 0 0 0 8 4 4 9 9 9 9 8 8 0 0 4 4 2 2 7 7 2 2 7 7 9 9 2 2 0 0 7 7 1 1 7 7 7 7 1 1 3 3 3 3 2 2 4 4 6 6 2 2 6 6 1 1 2 2 9 9 2 2 3 3 5 5 0 0 5 5 4 4 5 5 6 6 5 3 3 3 6 6 4 4 4 4 4 4 3 3 2 2 6 6 2 2 4 4 4 4
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s=poly(0,'s') h=syslin('c',((20*s^2+2000*s)/(s^2+12*s+20))) clf();bode(h,0.1,100);
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function V = novaMatriz(n, m, f) V = zeros(n,m) for i = 1:n for j = 1:m V(i,j) = f(i,j) end end endfunction function y = senoDeINaJ(i,j) y = sin(i)^j endfunction function y = apenasI(i,j) y = i endfunction function y = iVezesENaJ(i, j) y = i*exp(j) endfunction //criar matriz com M = novaMatriz(5, 6, iVezesENaJ)
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// To plot various types of vertical bar charts in a graphics window x=3:5; y=1:3; x1= [1,4,5]; y1=5*rand (3,3); y2= [1, -2,3]; subplot (2,3,1), bar(y); subplot (2,3,2), bar(x, y); subplot (2,3,3), bar (x, y1); subplot (2,3,4), bar (x, y1,"stacked"); subplot (2,3,5), bar (x, y2); subplot (2,3,6), bar (x, y1,.2,"green");
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function y=f(x) y=sin(x)/x endfunction x=[-2*%pi:0.02:2*%pi]';size(x) // x coordinate y=f(x);size(y) // y is not the correct size y=sin(x)./x; size(y) // y is the correct size y=feval(x,f); size(y) // y is the correct size
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n = int(input("ENTER order n :")); m=int(input("m:? ")) A = zeros(n,m); disp("Enter elements of A :\n") for i=1:n for j=1:m A(i,j) = int(input("enter element "+string(i)+","+string(j)+" : ")); end end f = 0; if(~(n == m)) then disp("Matrix is singular\n"); f = 1; end c = 1; if(f == 0) then t = n; for i=1:n if(A(i,i) == 0) then for k = i+1:n if(~(A(k,i) == 0)) for w = 1:m t = A(i,w) A(i,w) = A(k,w) A(k,w) = t end end if(k == n) then f = 1; end end end if(f == 1) then break; end end end if(f == 1) then disp("Matrix is singular\n"); end L = eye(n,m); rem = 0; if(f == 0) then for i = 1:n for k = i+1:n rem = A(k,i)/A(i,i); L(k,i) = rem; for j = 1:m A(k,j) = A(k,j) - A(i,j)*rem; end end end end disp(A,"matrix U :"); disp(L,"matrix L :")
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MEF.sce
clear all; getf("fGauss.sci"); getf("fEF.sci");getf("fMLS.sci"); //=============================================================== // Cas 1D : // // Resolution de l'equation : u,xx + f = 0 // // ========================== -k0*u(0)= -u,x(0) // u(0) = 0 // // u,x(1)= T // // // // f= 6*d*x // // // u = f0 + e*x - d*x^3 // //=============================================================== // Parametres de l'equation // ======================== T = -2; d = 5; k0 = 10; k00=1; if (k0==0) k00=0; k0=1; end // Nombre de Particules : Discretisation N=19; h=1/N; xp = [0.0:h:1.0]; nnodes = length(xp); ncells = nnodes-1; // Choix de la Methode: Methode Elements Finis // ==================== MEF=0; MLSType='spline quadratique'; mp=1; dm=1.1; // Points de Gauss // =============== [gg,weight,jac] = fGauss(h,ncells); hhg=gg(2)-gg(1); // ========================== // Initialistion des Matrices : Mise a Zero // ========================== k = zeros(nnodes) ; f = zeros(nnodes,1); GG = zeros(nnodes,1); // Boucle sur les points de Gauss // ============================== for j = 1:length(gg) xg = gg(j); weight1=weight(j); // Calcul Phi(xg), dPhi(xg) if (MEF==1) [phi,dphi] = fEF(xg,xp,hhg); end; if (MEF==0) [phi,dphi] = fMLS(xg,xp,h,mp,dm,MLSType); end; // Calcul Matrice de rigidite : k et Second Membre : f if j == 1 GG(1:3,1) = -phi(1:3)'; k = k+k00*k0*phi'*phi; else if j<length(gg) k = k+(weight1*jac)*(dphi'*dphi); fbody=6*d*xg ; f = f+(weight1*fbody*jac)*phi'; end if j==length(gg) f= f+T*phi'; end end end // ========== // Resolution // ========== q=[0]; if (k00*k0==0) Encastrement = k0*k00; mat = [k GG; GG' zeros(1)]; depl = mat\[f' q]'; nnodesT=length(depl)-1 else Ressort = k0*k00; mat = [k]; depl = mat\[f]; nnodesT=length(depl) end; u = depl(1:nnodesT); uxp = depl(1:nnodes); // ===================== // Tracer de la solution // ===================== clear xe; clear sol; he=h/10; xe = [0.0:he:1.0]; // Fonction de Formes for j = 1:length(xe) xg = xe(j); [phi,dphi] = fEF(xg,xp,he); for i=1:nnodesT Forme(j,i)=phi(i); end; end // Construction de la solution u=Sum_i u_i Forme_i sol=zeros(length(xe)); for j=1:length(xe) sol(j)=0.; for i=1:nnodesT sol(j)=sol(j)+u(i)*Forme(j,i); end end // plot2d(xe,sol,style=1); plot2d(xp,uxp,style=-1);
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pathname=get_absolute_file_path('3_17.sce') filename=pathname+filesep()+'3_17_data.sci' exec(filename) //Temperature at the end of compression stroke(in K) T2=r^((Cp/Cv)-1)*T1 //Temperature at he start of expansion stroke(in K) T3=CV/(AF*Cp)+T2 //Cutoff ratio rc=T3/T2 //Efficiency of diesel cycle n=1-(1/(y*r^(y-1))*(((rc^y)-1)/(rc-1))) printf("\n\nRESULTS\n\n") printf("\nEfficiency of diesel cycle:%f\n",n*100)
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// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India. //Chapter-5,Example 14,Page 182 //Title: Power output of turbine //================================================================================================================ clear clc //INPUT P=3;//pressure of superheated steam in MPa T_enter=300;//entrance temperature of superheated steam in degree celsius T_exit=45;//final temperature at which the steam leaves in degree celsisus m=1;//mass flow rate of steam in kg/s //CALCULATION //From steam tables corresponding to P and T_enter si=6.5422;//entropy of steam at the entrance in kJ/kgK hi=2995.1;//entahlpy of steam at the entrance in kJ/kg //From steam tables corresponding to T_exit sf=0.6383;//entropy of saturated liquid in kJ/kgK hf=188.35;//enthalpy of saturated liquid in kJ/kg sg=8.1661;//entropy of saturated vapour in kJ/kgK hg=2583.3;//entahlpy of saturayed vapour in kJ/kg Xe=(si-sf)/(sg-sf);//calculation of quality of steam at the exit (no unit) he=((1-Xe)*hf)+(Xe*hg);//calculation of enthalpy of steam at the exit in kJ/kg Ws=-m*(he-hi);//calculation of power output from turbine using the first law of thermodynamics on the control-volume in kW //OUTPUT mprintf("\n The power output from the turbine=%0.1f kW\n",Ws); //===============================================END OF PROGRAM===================================================
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//ques1 //obtaining formulas for from unit considerations clear clc d=850;//density m^3/kg V=2;//volume m^3 m=d*V;//mass Kg printf("Mass of the sample m =%.0f Kg",m);
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Ch4_4_49.sce
clc disp("Example 4.49") printf("\n") disp("Determine the minimum & maximum triggering voltage for a UJT") printf("Given\n") Vbb=20 //intrinsic ratios nmin=0.6 nmax=0.8 V=0.7 //minimum triggering voltage is Vpmini=nmin*Vbb+Vd //maximum triggering voltage is Vpmax=nmax*Vbb+Vd printf("Minimum triggering Voltage \n%f volt\n",Vpmini) printf("Maximum triggering Voltage \n%f volt\n",Vpmax)
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18_01.sce
//pathname=get_absolute_file_path('18.01.sce') //filename=pathname+filesep()+'18.01-data.sci' //exec(filename) //Heat extracted by carnot cycle(in kJ/min): Q1=500 //Temperature of refrigerated space(in K): T1=-16+273 //Atmospheric temperature(in K): T2=27+273 //Heat rejected(in kJ/min): Q2=Q1*(T2/T1) //Work input required(in kJ/min): W=Q2-Q1 printf("\n RESULT \n") printf("\nWork input = %f kJ/min",W)
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example19_6.sce
clc // Given that T1 = 2.18 // temperature in first case in K lambda1 = 16 // penetration depth at 2.18 K in nm T2 = 8.1 // temperature in second case in K lambda2 = 96 // penetration depth at 8.1 K in nm // Sample Problem 6 on page no. 19.15 printf("\n # PROBLEM 6 # \n") printf("Standard formula used \n ") printf(" lambda = lambda_0 * (1 - (T / T_c)^4)^(-1/2) \n") Tc = (((lambda2^2 * T2^4) - (T1^4 * lambda1^2)) / (lambda2^2 - lambda1^2))^(1/4) printf("\n Critical temperature is %f K.",Tc)
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Ex7_3.sci
//Example 7.3 //Find the convolution of the sequences. clear all x1=[4 1 -3 -1]; x2=[1 6 -2 3]; y=convol(x1,x2); disp(y,'The convolution of the above sequences is:');
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SolEx2_26.sce
//Solved Example Ex2.26 page no 61 clear clc R1=6//kΩ R2=3//kΩ V1=5//v V2=10//v Rth=(R1*R2/(R1+R2)) printf("Rth = %0.3f",Rth) R2=(R1*Rth/(R1-Rth)) printf("\nR2 = %0.3f",R2)
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Ex2_3.sce
// Example 2.3 // Rp=(4+4)||(8+4) Rp=(8*12)/(8+12); // By Voltage divider rule disp(' voltage Across Foue resisrance = '+string(Rp)+' Ohm'); // p 20 2.3
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Ex14_8.sce
//===================================================================================== //Chapter 14 example 8 clc; clear all; //variable declaration n =3; //number of full digits v1 = 10; //voltage in V v2 = 100; //voltage in V //calculations R = 1/(10^n); //resolution R1 = R*v1; //resolution on 1V range in V R2 = R*v2; //resolution on 10V range in V //result mprintf("R = %3.4f V",R); mprintf("\nthe meter cannot distinguish the values that differ from each by less than 0.001 of full scale"); mprintf("\nR1 = %3.4f V",R1); mprintf("\nany decimal upto second decimal can be displayed "); mprintf("\nhence 15.45 can be dislayed as 15.45") mprintf("\n R2 = %3.4f V",R2); mprintf("\nany deccimal upto one decimal can be displayed "); mprintf("\nhence 25.65 can be dislayed as 025.6 instead of 25.65");
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Ex8_4.sce
// Initilization of variables W=1000 //N // Load to be lifted n=5 // no. of pulleys E=75 //% // Efficiency // Calculations // Velocity Ratio is given as, V.R=n // Mechanical Advantage (M.A) is, M.A=(E/100)*V.R // from formulae, Efficiency=E=M.A/V.R P=W/M.A //N // Effort required // Results clc printf('The effort required to lift the load of 1000 N is %f N \n',P)
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14_5.sce
clear; clc; fc=1000;Rk=600; L=Rk/(4*%pi*fc); C=1/(4*%pi*fc*Rk); printf("Thus,the series elements are two capacitors of value %f microfarad each and shunt inductance of value %f mH.",round(C*(10^3)*10^6)/10^5,fix(L*(10^3)*100)/100);
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Ex7_3.sce
clear //given //find out brake mean effective pressure of a 6-cylinder D=5**0.75 bhp=120. l=8. m=6. n=1000. bmep=1008000*((bhp/(D**2)*l*m*n)) printf("\n \n brake mean effective pressure %.2f psi",bmep/2.95)
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Resultat de prince hertz.tst
5860000;931 @1 114;4448;75.91;37.54;30.00;25016;426.9;5.62;3.7 20;15;27;38;47;55;62;67;72;76;80;83;85;87;89;91;92;93;94;95;96 @2 0;1411;24.09; 1;842;14.38;75.91 2;678;11.57;61.53 3;523;8.93;49.95 4;407;6.95;41.03 5;326;5.57;34.07 6;270;4.62;28.50 7;230;3.93;23.88 8;197;3.36;19.95 9;167;2.85;16.59 10;139;2.38;13.74 11;115;1.98;11.36 12;96;1.64;9.38 13;80;1.37;7.74 14;66;1.14;6.37 15;55;0.94;5.23 16;45;0.77;4.29 17;36;0.63;3.52 18;30;0.52;2.89 19;25;0.43;2.37 20;113;1.94;1.94 @3 P(33%);10.3;2.4;1.8 M(33%);60.7;14.2;41.5 G(33%);355.9;83.4;13.4 @4 04/03/2002;1480;33.3;25.3 05/03/2002;459;10.3;33.1 06/03/2002;295;6.6;38.1 07/03/2002;633;14.2;48.9 08/03/2002;291;6.5;53.9 09/03/2002;206;4.6;57.4 10/03/2002;13;0.3;57.6 11/03/2002;144;3.2;60.1 12/03/2002;68;1.5;61.2 13/03/2002;69;1.6;62.4 14/03/2002;128;2.9;64.6 15/03/2002;242;5.4;68.7 16/03/2002;78;1.8;70.1 18/03/2002;164;3.7;72.9 19/03/2002;69;1.5;74.0 20/03/2002;25;0.6;74.5 21/03/2002;36;0.8;75.1 22/03/2002;36;0.8;75.7 23/03/2002;12;0.3;75.9 @5 du 04/03/02 au 10/03/02;3376;75.9;57.6 du 11/03/02 au 17/03/02;730;16.4;70.1 du 18/03/02 au 24/03/02;342;7.7;75.9 @6 du 04/03/02 au 10/03/02;3376;75.9;57.6 du 11/03/02 au 17/03/02;730;16.4;70.1 du 18/03/02 au 24/03/02;342;7.7;75.9 @7 @8
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13_4.sce
clc //initialisation of variables clear k1= -9130 //cal k2= 7.46 //cal K^-1 k3= -3.69*10^-3 //K^-2 k4= 0.235*10^-6 //K^-3 k5= -12.07 T= 298 //K R= 1.987 //cal deg^-1 mole^-1 //CALCULATIONS dF= k1+k2*T*log(T)+k3*T^2+k4*T^3+k5*R*T //RESULTS printf ('Free energy = %.f cal',dF)
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Ex3_10.sce
clc //given that E = 100 // Energy of X ray beam in KeV theta = 30 // Scattering angle in degree m = 9.1e-31 // mass of electron in kg c = 3e8 // Speed of light in m/s printf("Example 3.10") E_rest = m*c^2/(1.6e-19*1e3) // Rest mass energy in KeV k = 1/E + (1-cos(theta*%pi/180))/(E_rest) del_e = E - 1/k // Energy of recoiled electron printf("\n Energy of recoiled electron is %f KeV\n\n\n",del_e)
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ex8_14.sce
//Page Number: 8.14 //Example 8.14 clc; //Given GG1=20;//dB G1=(10^(GG1/10)); FF1=6;//dB F1=(10^(FF1/10)); GG2=60;//dB G2=(10^(GG2/10)); FF2=16;//dB F2=(10^(FF2/10)); LF=3; //dB FC=(10^(LF/10)); GC=1/FC; //(a)Overall Noise Figure //Usinng F=(F1+((F2-1)/G1)+((F3-1)(G1*G2))); Fa=(F1+((FC-1)/G1)+((F2-1)/(G1*GC))); FadB=(10*(log10(Fa))); disp('db',FadB,'Overall Noise Figure:'); //(b)Noise figure, if pre-amplifier is removed and gain increased by 20dB Fb=FC+((F2-1)/GC); FbdB=(10*(log10(Fb))); disp('db',FbdB,'Overall Noise Figure:'); //(c)Change in noise figure //Again usinng F=(F1+((F2-1)/G1)+((F3-1)(G1*G2))); Fc=(FC+((F1-1)/GC)+((F2-1)/(G1*GC))); FcdB=(10*(log10(Fc))); disp('db',FcdB,'Overall Noise Figure:');
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read_tar_pgm_result.sce
function [tar_pgm_result]=read_tar_pgm_result(result_file_name,m_graph,time_scale) fd = mopen(result_file_name,'r'); j=1; str_temp = mgetstr(7,fd); while str_temp ~= "0xffff ", tar_pgm_result(j,1) = msscanf(str_temp,"%x"); for i=2:m_graph tar_pgm_result(j,i) = msscanf(mgetstr(7,fd),"%x"); end str_temp = mgetstr(7,fd); j=j+1; end mclose(fd); tar_pgm_result(:,2)=tar_pgm_result(:,2)*time_scale; endfunction
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Ex7_20.sce
//Exa 7.20 clc; clear; close; // given : eta_0=377 //intrinsic impedance in ohm disp("when Zd=73+%i*42.5") Zd=73+%i*42.5 // dipole impedance // formula : zs*zd=(eta_0)^2/4 Zs=eta_0^2/(4*Zd) // slot impedance in ohm disp(Zs,"complementary slot impedance in ohm:") disp("when Zd=67+%i*0") Zd=67+%i*0 // dipole impedance // formula : zs*zd=(eta_0)^2/4 Zs=eta_0^2/(4*Zd) // slot impedance in ohm disp(Zs,"complementary slot impedance in ohm:") disp("when Zd=710+%i*0") Zd=710+%i*0 // dipole impedance // formula : zs*zd=(eta_0)^2/4 Zs=eta_0^2/(4*Zd) // slot impedance in ohm disp(Zs,"complementary slot impedance in ohm:") disp("when Zd=500+%i*0") Zd=500+%i*0 // dipole impedance // formula : zs*zd=(eta_0)^2/4 Zs=eta_0^2/(4*Zd) // slot impedance in ohm disp(Zs,"complementary slot impedance in ohm:") disp("when Zd=50+%i*20") Zd=50+%i*20 // dipole impedance // formula : zs*zd=(eta_0)^2/4 Zs=eta_0^2/(4*Zd) // slot impedance in ohm disp(Zs,"complementary slot impedance in ohm:") disp("when Zd=50-%i*25") Zd=50-%i*25 // dipole impedance // formula : zs*zd=(eta_0)^2/4 Zs=eta_0^2/(4*Zd) // slot impedance in ohm disp(Zs,"complementary slot impedance in ohm:") disp("when Zd=300+%i*0") Zd=300+%i*0 // dipole impedance // formula : zs*zd=(eta_0)^2/4 Zs=eta_0^2/(4*Zd) // slot impedance in ohm disp(Zs,"complementary slot impedance in ohm:")
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clear //Dada a taxa de amostragem de 2500 Hz, e usando como exemplo f_1=9 e f_2=400 f_1 = 9; f_2 = 400; //especificamos w_1, w_2 de acordo w_1 = 2*%pi*f_1; w_2 = 2*%pi*f_2; //Usaremos agora um laço para simular o comportamento em tempo real do sistema: for k=1:7500 //nosso filtro é representado por uma equação diferença, assim, devemos escolher //polos e zeros tal que o comportamento do filtro seja como desejamos //para um filtro ressonador digital, temos H(z) = (1-z^{-2})/1-2rcos(w_0)z^{-1}+r^2z^{-2} //o valor de r determina quão amplificada será a frequência w_0 em relação às demais //escolhemos w_0 como sendo a frequência correspondente a f_1, cos(2*%pi*9/1250) //para realçar o efeito do filtro, utilizamos 3 filtros iguais em cascata x_1(k) = sin(w_1*k/2500)+sin(w_2*k/2500); t(k) = k/2500; if k==1 y_1(k) = x_1(k); y_2(k) = y_1(k); y_3(k) = y_2(k); end if k==2 y_1(k) = x_1(k)+1.8980561*y_1(k-1); y_2(k) = y_1(k)+1.8980561*y_2(k-1); y_3(k) = y_2(k)+1.8980561*y_3(k-1); end if k>=3 y_1(k) = x_1(k)-x_1(k-2)+1.8980561*y_1(k-1)-0.9025*y_1(k-2); y_2(k) = y_1(k)-y_1(k-2)+1.8980561*y_2(k-1)-0.9025*y_2(k-2); y_3(k) = y_2(k)-y_2(k-2)+1.8980561*y_3(k-1)-0.9025*y_3(k-2); end // para f_2 = 400, temos cos(w_0) = cos(2*%pi*400/1250), e utilizamos apenas um dispositivo //if k==1 //y_1(k) = x_1(k); //end //if k==2 //y_1(k) = x_1(k)- 0.8089807*y_1(k-1); //end //if k>=3 //y_1(k) = x_1(k)-x_1(k-2)- 0.8089807*y_1(k-1)-0.9025*y_1(k-2); //end end //plot(t,y_1) plot(t,y_3)
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//Example 1-5, Page No - 18 clear clc f1=902000000 f2=928000000 bandwidth=f2-f1 printf('Width of the band is %d Megahertz',bandwidth/1000000)
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clear; clc; printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 12.9 Page 766 \n')// Example 12.9 // Total hemispherical emissivity of fire brick wall // Total emissive power of brick wall // Absorptivity of the wall to irradiation from coals Ts = 500 ;//[K] temperature of brick surface Tc = 2000 ;//[K] Temperature of coal exposed stfncnstt = 5.67*10^-8; //[W/m^2.K^4] Stefan-Boltzmann constant // From the given graph of emissivities e1 = .1; //between wavelength 0 micro-m- 1.5 micro-m e2 = .5; //between wavelength 1.5 micro-m- 10 micro-m e3 = .8; //greater than wavelength 10 micro-m //From Table 12.1 //For wl1 = 1.5 micro-m and T = 500 K, At wl1*T = 750 micro-m.K F0wl1 = 0; //For wl2 = 10 micro-m and T = 500 K, At wl2*T = 5000 micro-m.K F0wl2 = .634; //From equation 12.36 e = e1*F0wl1 + e2*F0wl2 + e3*(1-F0wl1-F0wl2); //Equation 12.26 and 12.35 E = e*stfncnstt*Ts^4; //From Table 12.1 //For wl1 = 1.5 micro-m and T = 2000 K, At wl1*T = 3000 micro-m.K F0wl1c = 0.273; //For wl2 = 10 micro-m and T = 2000 K, At wl2*T = 20000 micro-m.K F0wl2c = .986; ac = e1*F0wl1c + e2*[F0wl2c-F0wl1c] + e3*(1-F0wl2c); printf('\n Total hemispherical emissivity of fire brick wall = %.3f \n Total emissive power of brick wall = %i W/m^2.\n Absorptivity of the wall to irradiation from coals = %.3f',e,E,ac);
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//EX13_48 Pg-26 clc clear printf("subtraction of 111001 from 101011 using 2''s complement method") printf("\n\n we know that 101011<111001\n\n") printf(" Therefore 101011-111001 =") x=['101011']; y=['111001']; //binary to decimal conversion// x=bin2dec(x) y=bin2dec(y) y1=bitcmp(y,6)//one's complement of the larger number y2=y1+1;//2's complement of the larger number //subtraction of larger number from smaller number a=x+y2;//result is in two complement a1=bitcmp(a,6)//one's complement of the result a2=a1+1;//final answer s=dec2bin(a2) printf(" -00%s",s)//final answer is -ve
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//To find work done clc //Given: p=12,d=40 //mm mu=0.16 W=2500 //N //Solutiom: //Work done in drawing the wagons together agianst a steady load of 2500 N: //Calculating the helix angle alpha=atan(p/(%pi*d)) //radians //Calculating the effort required at the circumference of the screw phi=atan(mu) //Limiting angle of friction, radians P=W*tan(alpha+phi) //N //Calculating the torque required to overcome friction between the screw and nut T=P*d/(2*1000) //N-m //Calculating the number of turns required N=240/(2*p) //Calculating the work done W1=T*2*%pi*N //Work done, N-m //Work done in drawing the wagons together when the load increases from 2500 N to 6000 N: W2=W1*(6000-2500)/2500 //Work done, N-m //Results: printf("\n\n Work done in drawing the wagons together agianst a steady load of 2500 N = %.1f N-m.\n",W1) printf(" Work done in drawing the wagons together when the load increases from 2500 N to 6000 N = %.1f N-m.\n\n",W2)
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// Chapter 5 additional Example 7 //============================================================================== clc; clear; // input data // given crystal has BCC structure r = 1.2*10^-10; // atomic radius in m // Calculations a = (4*r)/sqrt(3); // lattice constant V = a^3; // volume of cell //Output mprintf('Volume of the cell = %3.3e m^3',V); //==============================================================================
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//Problem 36.14: A voltage waveform having a fundamental of maximum value 400 V and a third harmonic of maximum value 10 V is applied to the circuit shown in Figure 36.18. Determine (a) the fundamental frequency for resonance with the third harmonic, and (b) the maximum value of the fundamental and third harmonic components of current. //initializing the variables: V1m = 400; // in volts V3m = 10; // in volts C = 0.2E-6; // in farads R = 2; // in ohms L = 0.5; // in Henry //calculation: //Resonance with the third harmonic means that w = (1/(9*L*C))^0.5 //fundamental frequency, f f = w/(2*%pi) //At the fundamental frequency, //impedance Z1 Z1 = R + %i*(w*L - 1/(w*C)) Z1mag = (real(Z1)^2 + imag(Z1)^2)^0.5 phiZ1 = atan(imag(Z1)/real(Z1)) //Maximum value of current at the fundamental frequency, I1m = V1m/Z1mag //At the third harmonic frequency, Z3 = R + %i*(3*w*L - 1/(3*w*C)) Z3mag = (real(Z3)^2 + imag(Z3)^2)^0.5 phiZ3 = atan(imag(Z3)/real(Z3)) //Maximum value of current at the third harmonic frequency, I3m = V3m/Z3 printf("\n\n Result \n\n") printf("\n(a)fundamental frequency for resonance with the third harmonic is %.2f Hz",f) printf("\n(b)Maximum value of current at the fundamental frequency is %.3f A and at the third harmonic frequency %.2f A",I1m, I3m)
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//calculates// s=%s; sys1=syslin('c',9/(s*(s+1.8))); syms Td ; sys2=1+(s*Td); sys3=sys1*sys2; H=1; CL=sys3/.H; //Calculates closed-loop transfer function disp(CL,"C(s)/R(s)") // compare CL with Wn^2/(s^2+2*zeta*Wn+Wn^2) [num,den]=numden(CL) //extracts num & den of symbolic function CL den=den/5; cof_a_0 = coeffs(den,'s',0) // coeff of den of symbolic function CL cof_a_1 = coeffs(den,'s',1) //Wn^2= cof_a_0,comparing the coefficients Wn=sqrt(cof_a_0) disp(Wn,"natural frequency Wn") // Wn=natural frequency //cof_a_1=2*zeta*Wn zeta=cof_a_1/(2*Wn) zeta=1;disp(zeta,"for criticaly damped function zeta") Td=((2*Wn)-1.8)/9 Ts=4/(zeta*Wn); Ts=dbl(Ts); disp(Ts,"settling time Ts")
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clc //initialisation of variables d=200//mm p=8//in n=0.8//in s=100//m //CALCULATIONS R=(d*n)/p*p//mm //RESULTS printf('the numerical value is =% f mm',R)
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1 Jan 2 Feb 3 Mär 4 Apr 5 Mai 6 Jun 7 Jul 8 Aug 9 Sep 10 Okt 11 Nov 12 Dez
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clear //Given A=60 //Degree //Calculation // a=sqrt(2)*sin(30*3.14/180.0) b=asin(a)*180/3.14 c=(b*2)-A i=(A+c)/2.0 r=A/2.0 //Result printf("\n (i) Angle of minimum deviation is %0.0f Degree",c) printf("\n (ii) Angle of incidence is %0.0f Degree",i) printf("\n (iii) The angle of refraction is %0.3f Degree", r)
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S=0.80//specific gravity of oil h=60//cm
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//=========================================================================== //chapter 7 example 19 clc; clear all; //variable declaration W1 = 3000; //wattmeter reading in W W2 = 1000; //wattmeter reading in W f = 50; //frequency in HZ V = 400; //voltage in V //calculations VP = V/sqrt(3); //voltage in V P = W1+W2; //input power in kW phi = atan(((W1-W2)/(W1+W2))*sqrt(3)); //phase angle in radians phi1 = phi*180/%pi; //phase angle in degrees pf =cos(phi1*%pi/180); //power factor lagging IL = P/((sqrt(3))*V*pf); //line current in A ZP =VP/IL; //impedance of the circuit per phase in Ω R = ZP*pf; //resistance per phase Ω XL = sqrt((ZP^2 )-(R^2)); //reactance per phase in Ω L = XL/(2*%pi*f); //inducatance per phase in H //result mprintf("resistance per phase = %3.2f Ω",R); mprintf("\ninducatance per phase in = %3.3f H",L);
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I=../images/tilesets/tileset.png; 0,0,0,1,0,1,2,1; 0,0,3,0,0,0,2,2; 3,1,1,1,2,0,0,4; 1,3,1,0,3,1,2,0; 1,1,1,3,2,1,1,0; 2,0,0,17,2,2,2,3; 0,0,0,1,2,0,1,1; 0,2,2,2,0,0,0,1; 1,1,0,2,0,0,0,4; 0,0,1,2,0,0,1,2; 0,2,1,0,0,0,0,3; 1,0,0,0,0,0,0,1; 3,2,0,2,0,0,0,2; 2,0,0,0,1,2,2,3; 1,1,1,0,0,0,0,1;
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clc; //Example 5.4 //page no. 45 printf("Example 5.4 page no. 45\n\n"); //refer to example no 5.3 //determine dynamic viscosity and kinematic viscosity omega=26.2//angular rotation speed D=0.25//diameter of fixed inner cylinder of viscometer v=omega*D/2 printf("\n omega=%f rad/s\n diameter D =%f ft\n linear velocity =%2f ft/s",omega,D,v); d=0.001//clearance betwween two cylinder of visometer vel. gradient =v/(d/12)//velocity gradient gc=32.14//gravitational constant printf("\n clearance d=%5f ft\n vel. gradient=%f 1/s\n gravitational constant gc=%3f ft/s*S",d,vel. gradient,gc); tou=311.7//shear stress tou meu=gc*tou/vel. gradient printf("\n tou=%f psf\n meu=%f lb/ft*s",tou,meu); rho=60.528//density of oil neu=meu/rho//kinamatic viscosity printf("\n kinematic viscosity=%5f (ft*ft)/s",neu);
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//Example 9.6 //Calculate the heat transfer area required //(assuming equal area for the three effects) //Rate of steam consumption, Steam economy //given fc=9.5 //%,feed concentration pc=50 //%, product conc. ft=40 // C,feed temp. er=2000 //kg NaOH/h, evaporation rate vp=714 //mm Hg, vaccum pr. in last effect //heat transfer coefficients, W/m^2 C h1=6000 //for first effect h2=3500 //for second effect h3=2500 //for third effect //calculatiin Wf=er/(fc/100) //kg/h, 2 tons NaOH per hour, feed rate Wp=er/(pc/100) //kg/h, product rate ter=Wf-Wp //kg/h, total evaporation rate //steam p=3.3 //bar,assumed saturated //from steam table Ts=137 //C,temp. l_s=2153 //kj/kg, latent heat pl=760-vp //mm Hg,pressure in the last effect bp=37 //C,boiling point of water //refer to fig. 9.24 attd=Ts-bp //C,apparent total temp. drop //let assume the following evaporation rate for three effects in kg/h ev1=5600 ev2=5680 ev3=5773 //conc. in three effects c1=er/(Wf-ev1) c2=er/(Wf-ev1-ev2) c3=0.5 //given //boiling point elevations in three effects in C bpe1=3.5 bpe2=8 bpe3=39 attda=attd-(bpe1+bpe2+bpe3) //actual total temp. drop available //temp. drop in three effects //from eq. 9.23 dt1=attda*((1/h1)/((1/h1)+(1/h2)+(1/h3))) dt2=attda*((1/h2)/((1/h1)+(1/h2)+(1/h3))) dt3=attda*((1/h3)/((1/h1)+(1/h2)+(1/h3))) //from table 9.4 //enthalpy of solution in three effects in kj/kg i1=486 i2=385 i3=460 //enthalpy of vapour generated for three effects in kj/kg is1=2729 is2=2691 is3=2646 //Enthalpy of condensate over effect 1,2,3 in kj/kg il1=0 il2=519 il3=418 //Enthalpy balance over effect 1 ef=145 //kj/kg,enthalpy of feed //from energy balance eq. //Ws1=0.96Ws-3200......(1) //enthalpy balanc over effect 2 //Ws2=0.9146Ws1+922...........(2) //enthalpy balanc over effet 3 //Ws3=1.073Ws2+0.0343Ws1-722........(3) //ter=Ws1+Ws2+Ws3=17053..........(4) //Solving above four eqns by matrix A=[0.96,-1,0,0;0,0.9146,-1,0;0,0.0343,1.073,-1;0,1,1,1] B=[3200;-922;722;17053] X=inv(A)*B Ws=X([1]) Ws1=X([2]) Ws2=X([3]) Ws3=X([4]) //calculation of heat transfer areas iver effect 1, 2 ,3 A1=Ws*l_s*10^3/(h1*dt1*3600) A2=Ws1*(is1-il2)*10^3/(h2*dt2*3600) A3=Ws2*(is2-il3)*10^3/(h3*dt3*3600) //Revised dt avar=(A1+A2+A3)/3 dt1_=(A1/avar)*dt1 dt2_=(A2/avar)*dt2 dt3_=attda-dt1_-dt2_ //from table 9.5 //enthalpy of vapour generated over effect 1,2,3 in kj/kg is1_=2720 is2_=2685 is3_=2646 //enthalpy of soln on 1,2,3 in kj/kg i1_=470 i2_=380 i3_=460 //enthalpy of condensate over effect 1 ,2,3 in kj/kg il1_=0 il2_=513 il3_=412 //enthalpy balance ove effect 1,2,3 gives Ws_=8854 Ws1_=5432 Ws2_=5812 Ws3_=5809 //revised heat transfer areas for effect 1 ,2,3 in m^2 A1_=Ws_*l_s*1000/(h1*dt1_*3600) A2_=Ws1_*(is1_-il2_)*10^3/(h2*dt2_*3600) A3_=Ws2_*(is2_-il3_)*10^3/(h3*22.5*3600) avar_=(A1_+A2_+A3_)/3 SE=ter/Ws_ printf("The areas are now reasonably close \n") printf("Steam Rate is % f Kg/h \n",Ws_) printf("Steam economy is %f",SE)
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clc //initialisation p16=80//cm v16=432//cc t=273//k po=76//cm t=16//c t16=273+t//k T=273//k poxy=0.0014 cfe=0.09 t1=15//c t2=184//c m1=2//gm //calculations v0=(p16*v16*T)/(po*t16) m=poxy*v0 h=m1*cfe*(t1+t2) l=h/m //results printf(' latent heat= % 1f cal',l)
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clc //Initialization of variables t=28.4 //min //calculations n=log2(8) time=n*t printf("Time required = %.1f min",time)
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clear // //variable declaration P=(200) //loading,KN E=200*1000 d1=40 //Young's modulus,N/mm^2 A= %pi*(d1**2)/4 //Area of uniform portion**mm^2 L1=1500 //length of uniform portion,mm d2=60 //diameter of tapered section,mm L2=500 //length of tapered section,mm //Extensions of uniform portion and tapering portion are worked out separately and then added to get extension of the given bar. //Extension of uniform portion delta1=(P*1000*L1)/(A*E) printf("\n delta1= %0.3f mm",delta1) delta2=(P*1000*4*L2)/(E*%pi*d1*d2) printf("\n delta2= %0.3f mm",delta2) T=delta1 + delta2 printf("\n Total extension %0.3f mm",T)
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//Chapter-1, Example 1.5, Page 1.17 //============================================================================= clc clear //INPUT DATA n=48;//Number of slots z=16;//Number of conductors per slot q=0.018;//Flux per pole in Wb P=4;//Number of poles N=1000;//Speed of armature in rpm A=2;//Number of parallel paths //CALCULATIONS Z=(n*z);//Number of conductors Eg=(q*Z*N*P)/(60*A);//Generated emf in V //OUTPUT mprintf('Generated emf is %3.1f V',Eg) //=================================END OF PROGRAM==============================
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##### diagonal test, right (ascending) boardsize 19 play b a1 play b b2 00 gogui-rules_final_result #?[unknown] play b c3 play b d4 10 gogui-rules_final_result #?[unknown] play b e5 20 gogui-rules_final_result #?[black] 21 genmove w #?[resign] ##### diagonal test, left (descending) boardsize 18 30 gogui-rules_final_result #?[unknown] play w e1 play w d2 40 gogui-rules_final_result #?[unknown] play w c3 play w b4 50 gogui-rules_final_result #?[unknown] play w a5 60 gogui-rules_final_result #?[white] 61 genmove b #?[resign] ##### horizontal test boardsize 17 70 gogui-rules_final_result #?[unknown] play w d15 play w e15 play w f15 80 gogui-rules_final_result #?[unknown] play w g15 play w h15 90 gogui-rules_final_result #?[white] ##### vertical test boardsize 16 100 gogui-rules_final_result #?[unknown] play b j4 play b j5 play b j6 110 gogui-rules_final_result #?[unknown] play b j7 play b j8 120 gogui-rules_final_result #?[black] boardsize 5 130 gogui-rules_final_result #?[unknown] play b a1 play b a2 play w a3 play w a4 play b a5 play w b1 play w b2 play b b3 play b b4 play w b5 play w c3 play b c2 140 gogui-rules_final_result #?[unknown] genmove w genmove b genmove w 150 gogui-rules_final_result #?[unknown] genmove b genmove w genmove b genmove w 160 gogui-rules_final_result #?[unknown] genmove b genmove w genmove b genmove w 170 gogui-rules_final_result #?[unknown] genmove b 180 gogui-rules_final_result #?[unknown] genmove w 190 gogui-rules_final_result #?[draw] 200 genmove w #?[pass] 210 genmove b #?[pass] boardsize 10 220 gogui-rules_legal_moves #? [A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 B1 B2 B3 B4 B5 B6 B7 B8 B9 B10 C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 D1 D2 D3 D4 D5 D6 D7 D8 D9 D10 E1 E2 E3 E4 E5 E6 E7 E8 E9 E10 F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 G1 G2 G3 G4 G5 G6 G7 G8 G9 G10 H1 H2 H3 H4 H5 H6 H7 H8 H9 H10 J1 J2 J3 J4 J5 J6 J7 J8 J9 J10 K1 K2 K3 K4 K5 K6 K7 K8 K9 K10] ##### test hopeful winners boardsize 10 play w a1 play w b2 play w c3 play w e5 230 gogui-rules_final_result #?[unknown] play w f6 play w g7 play w h8 play w k10 240 gogui-rules_final_result #?[unknown] play b d4 play b j9 250 gogui-rules_final_result #?[unknown] play w d5 play w c5 play w b5 play b a5 260 gogui-rules_final_result #?[unknown] play w b6 play w b7 play w b8 play w b10 270 gogui-rules_final_result #?[unknown] play w j1 play w j2 play w j3 play w j4 play w h1 play w h2 play w h3 play w h4 play w g1 play w g2 play w g3 play w g4 play w k1 play w k2 play w k3 play w k4 280 gogui-rules_final_result #?[unknown] gogui-rules_board
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clc //initialization of variables T1=80+460 //R T2=90+460 //R P=14.5 //lb/in^2 cp=0.24 //calculations disp("From steam tables,") hg2=1096.6 hf3=48.02 Pg2=0.5069 hf2=hf3 Pair=P-Pg2 wg2=0.622*Pg2/Pair hgv1=1100.9 wwv1=(cp*(T1-T2)+wg2*(hg2-hf3))/(hgv1-hf3) Pg=0.6982 xi=wwv1*(P-Pg)/(Pg*0.622) //results printf("Specific humidity = %.4f lbm/lbm",wwv1) printf("\n relative humidity = %.2f",xi)
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//1ère étape - Load path_name = "D:\Users\ADRIEN KEGLER\Documents\Visual Studio 2015\Projects\Exolife\Projet-Exolife\images\Encelade_Surface.pbm"; img_in = readpbm(path_name); y = 0; //Création de l'histogramme histo = histogramme(img_in); // Déterminance du niveau le plus élevé L_max = max_histo(histo); //Application du seuil img_out = Seuil(img_in, L_max); : //Localisation du/des pixel(s) avec le plus haut niveau Coord = zeros(histo(L_max), 2); for i =1:size(img_out, "r"); for j = 1:size(img_out, "c"); if img_out(i,j) == 255 then y = y+1 Coord(y,1)= i; Coord(y, 2)=j; end, end; end; disp(Coord); //Step 3 - Show display_gray(img_out); //Etape N°4 - Save writepbm(img_out, "D:\Users\ADRIEN KEGLER\Documents\Visual Studio 2015\Projects\Exolife\Projet-Exolife\images\A1.pbm");
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s=100 rand("seed",s) n=5; A=rand(n,n); GB = zeros(n,n); for k=1:n GB(k,k) = A(k,k) ; if k<n GB(k+1,k) = A(k+1,k); GB(k,k+1) = A(k+1,k); end end disp(GB) [L,U]= FactoLU(GB); /* U=triu(B) disp(U); L=tril(B) disp(L); */
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clear all; clc; disp("Scilab Code Ex 1.14 : ") //Given: shear_allow = 55; //MPa l_ac = 200; //mm l_cd= 75; //mm l_de = 50; //mm l_ce = l_cd + l_de; load_d =15; //kN load_e = 25; //kN //Internal Shear Force: //summation Mc = 0 f_ab = ((load_d*l_cd +load_e*(3/5)*l_ce)/l_ac); c_x =-load_d + (load_e*(4/5)); //resolving C in x dir c_y = load_d + (load_e*(3/5)); //resolving C in y dir f_c = sqrt(c_x^2 + c_y^2); //kN V = f_c/2; //Required Area A = ((V*10^3)/(shear_allow)); //A = V/Allowable shear in mm^2 d = ((sqrt((4*A)/%pi))) // Area = (%pi\4)d^2 in mm^2 chosen_d = ceil(ceil(d))+1; //Displaying Results: printf("\n\nThe force at AB = %.2f kN",f_ab); printf("\nThe resultant force at C = %.2f kN",f_c); printf("\nThe area of pin = %.2f mm^2",A); printf("\nThe diameter of pin = %.2f mm",chosen_d); //---------------------------------------------------------------END--------------------------------------------------------------------------------------
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// Example 8.8 // From the diagram 8.15 R1=1000; // Resistance of 1 kilo-Ohms R2=10000; // Resistance of 10 kilo-Ohms R3=1000; // Resistance of 1 kilo-Ohms Rth=[(R1+R2)*R3]/(R1+R2+R3); // Equivalent resistance C=10*10^-6; // capacitor t=Rth*C; // Time constant V=30; // Source voltage Vc=V*(R1/(R1+R2)); // Voltage across the capacitor // Apply KVL to outer loop // we get 30-Io*R1-15= 0 Io=15/R1; // Current in the outer loop Iin=V/(R1+R2+R3); // Open=ckt current // We know that ==> it=Iin+[Io-Iin]*e(-t1/t) t1=0.001; // Assume t1=1 mS it=Iin+[Io-Iin]*exp(-t1/t); // Current i(t) disp(' Current i(t) is = '+string(it)+' Amp oR i(t)= 2.5+(15-2.5)*e(-t/9.17ms) mA'); // p 287 8.8
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// Exa 7.15 clc; clear; // Given // An LVDT Vo = 1.25; // Output voltage Dmax = 0.0025;// max. deviation of linearity L = 0.75; // weight of load in kgf // Solution Linearity = (Dmax/Vo)*100; printf(' The linearity at a given load 0.65/kgf = %.1f percent \n',Linearity);
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clc; s1=5.615;//kJ/kg K t1=311;//C t2=300;//C t3=350;//C s2=7.124+(t1-t2)/(t3-t2)*(7.301-7.124); T=t1+273;//K Q=T*(s2-s1); disp("heat supplied is:"); disp("kJ/kg",Q) u1=2545;//kJ/kg u2=2794+(t1-t2)/(t3-t2)*(2875-2794); W=(u2-u1)-Q disp("work done by the steam is:"); disp("kJ/kg",-W)
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// waveform.sci is a scilab file to read source parameters. It is developed for a scilab based circuit simulator. It is written by Yogesh Dilip Save (yogessave@gmail.com). // Copyright (C) 2012 Yogesh Dilip Save // This program is free software; you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation; either version 2 of the License, or (at your option) any later version. // This program is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details. // You should have received a copy of the GNU General Public License along with this program; if not, write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA. function value=sine(param,t) pi=3.14; value=param(3)*sin(2*pi*param(4)*t); endfunction function value=pulse(param,t) v1=param(2); // Initial value v2=param(3); // Pulsed value td=param(4); // Delay time tr=param(5); // Rise time tf=param(6); // Fall time pw=param(7); // Pulse width per=param(8); // Pulse period while(t>per) t=t-per; end if(v1>v2) tr_back=tr; tr=tf; tf=tr_back; end if(t<td) value=v1; elseif(t<td+tr) va=v1; ta=td; vb=v2; tb=td+tr; value=(vb-va)/(tb-ta)*(t-ta)+va; elseif(t<td+tr+pw) value=v2; elseif(t<td+tr+pw+tf) va=v2; ta=td+tr+pw; vb=v1; tb=td+tr+pw+tf; value=(vb-va)/(tb-ta)*(t-ta)+va; else value=v1; end endfunction
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--> -->H = !--error 2 Facteur invalide. -->H = { -->0 1 0 0 0 0 0 0 0 0 -->1 0 1 0 0 0 0 0 0 0 -->0 0 0 1 1 1 0 0 0 0 -->0 1 0 0 0 0 0 1 1 0 -->0 0 0 0 0 0 0 0 0 0 -->0 0 0 1 0 0 0 0 0 0 -->0 0 0 1 1 0 0 0 0 1 -->0 0 0 0 0 0 1 0 1 0 -->0 0 0 0 1 0 0 0 0 0 -->0 0 0 0 0 0 0 0 0 0} H = 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 1. 1. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 1. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 1. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. -->M=H'*H M = 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 2. 0. 0. 0. 0. 0. 1. 1. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 3. 2. 1. 0. 0. 0. 1. 0. 0. 0. 2. 3. 1. 0. 0. 0. 1. 0. 0. 0. 1. 1. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 1. 0. 0. 0. 0. 0. 1. 1. 0. 0. 1. 0. 0. 0. 0. 1. 1. 2. 0. 0. 0. 0. 1. 1. 0. 0. 0. 0. 1. -->[A,B]=spec(M) B = column 1 to 7 - 8.166D-16 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.1715729 0. 0. 0. 0. 0. 0. 0. 0.4679111 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 1.6527036 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. column 8 to 10 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 2. 0. 0. 0. 3.8793852 0. 0. 0. 5.8284271 A = column 1 to 5 0. - 0.7071068 0. 0. 0. 3.263D-16 0. - 8.913D-17 - 0.5773503 - 5.674D-16 0. 0.7071068 0. 0. 0. - 3.987D-16 0. 0.2705981 - 4.097D-16 0.7071068 - 2.156D-16 0. 0.2705981 2.383D-16 - 0.7071068 7.748D-16 0. - 0.6532815 2.216D-16 2.759D-17 - 0.5773503 0. - 6.978D-16 - 0.4285251 - 5.607D-16 - 0.5773503 0. - 4.151D-16 0.6565385 4.105D-16 0.5773503 0. 5.781D-16 0.2280134 3.672D-16 5.141D-16 0. - 0.6532815 7.507D-17 - 2.759D-17 column 6 to 10 0. 0. - 0.7071068 0. 0. 2.616D-17 - 0.5773503 0. - 0.5773503 - 3.899D-17 0. 0. - 0.7071068 0. 0. 5.092D-17 - 9.769D-17 0. - 6.917D-18 0.6532815 1.061D-16 7.546D-17 0. 3.151D-18 0.6532815 0.7071068 - 1.319D-16 0. - 7.397D-17 0.2705981 3.526D-16 0.6565385 0. - 0.2280134 - 2.636D-17 1.273D-16 - 0.2280134 0. - 0.4285251 - 2.200D-17 - 1.845D-16 0.4285251 0. - 0.6565385 - 1.273D-16 - 0.7071068 1.372D-16 0. 1.474D-17 0.2705981
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Example_28_2.sce
//clear// clear; clc; //Example 28.2 //Given x = 0.14; xavg = 0.10; t = 3; //[min] x =[10.24,9.3,7.94,10.24,11.08,10.03,11.91,9.72,9.20,10.76,10.97,10.55]/100; //Solution mu = xavg; N =12; xbar = mean(x); //Substituing in Eq.(28.20) Ip = sqrt((N-1)*mu*(1-mu)/(sum(x^2)-xbar*sum(x))); //Using Eq.(28.18) s = stdev(x); disp(s,'s =',Ip,'Ip =')
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// example:-6.8,page no.-316. // program to design a three section chebysev transformer. Zl=100;Zo=50;taom=0.05;N=3;A=0.05; thetam=asec(cosh((1/N)*acosh((1/taom)*abs((Zl-Zo)/(Zl+Zo)))))*(180/%pi); x=(cosh((1/N)*acosh((1/taom)*abs((Zl-Zo)/(Zl+Zo))))) tao_o=A*(x^3)/2; tao_1=(3*A*(x^3-x))/2; // from symmetry tao_3=tao_0; Z1=Zo*((1+tao_o)/(1-tao_o)); Z2=Z1*((1+tao_1)/(1-tao_1)); Z3=Zl*((1-tao_o)/(1+tao_o)); disp(Z1,Z2,Z3,'the characteristic impedences are = ')
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clc; clear; E0=100 //energy of the incident photon in keV E=90 //energy of the scattered photon in keV m=9.1*10^-31 //mass in kg c=3*10^8 //velocity of light in m/s //calculation delta_E=E0-E //energy lost in keV mc_square=(m*c^2)/(1.6*10^-19*10^3) //calculating one part of the formula phi=acosd(1-(delta_E/E*mc_square/E0)) mprintf("The scattering angle of the photon is = %2.1f degree",phi)
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//Example 2.3: Reduce a given expression clc; //clears the console clear; //clears all existing variables //the given expression is as follows// disp(' Given Expression- A''B''C''+ A''BC''+ A''BC + AB''C''') disp('A''C''+ A''BC + AB''C''') disp('A''C''+ A''B + AB''C''') disp('A''C''+ A''B + B''C''') disp('The reduced expression is = ') disp('B''C''+ A''B') //final reduced expression is displayed//
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modulate17.sce
//i/p arg x is a vector x=[0 0.5 0.4 0.3 0.9 1]; fc=100; fs=500; y = modulate(x,fc,fs,'ppm'); disp(y); //output // column 1 to 7 // // 0. 0. 0. 0. 0. 0. 0. // // column 8 to 14 // // 0. 0. 0. 0. 0. 0. 0. // // column 15 to 21 // // 0. 0. 0. 0. 0. 0. 0. // // column 22 to 28 // // 0. 0. 0. 0. 0. 0. 0. // // column 29 to 30 // // 0. 0.
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7s3.sce
clear; clc; disp("Example 7.3"); funcprot(0); function[a1]=quick(a); a=gsort(a);//IN BUILT QUICK SORT FUNCTION n=length(a); a1=[]; for i=1:n a1=[a1(:,:) a(n+1-i)]; end disp(a1,"Sorted array is:"); endfunction //Calling Routine: a=[26 5 37 1 61 11 59 15 48 19] disp(a,"Given Array"); a1=quick(a)
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//Find the voltage-gain ratio V2/V1 //Solved Example 1.15 page no 23 clear clc printf("\nFind the voltage-gain ratio V2/V1") //Let V=V2/V1 RL=2000 h11=100 //ohm h12=0.0025 //ohm h21=20 //ohm h22=0.001 //mS V=1/(h12-(h11/h21)*((1/RL)+h22)) printf("\n The value of V2/V1=%0.1f",V)
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Example9_14.sce
// Electric Machinery and Transformers // Irving L kosow // Prentice Hall of India // 2nd editiom // Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS // Example 9-14 clear; clc; close; // Clear the work space and console. // Given data T_max = 17.75 ; // Maximum torque developed in lb-ft s_max = 0.3 ; // Slip for which T_max occurs s_a = 0.0333 ; // slip (case a) s_b = 1.0 ; // slip (case b) // Calculations // Subscript a in T indicates case a T_a = T_max * ( 2 / ((s_max/s_a) + (s_a/s_max)) ); // Full-load torque in lb-ft // Subscript b in T indicates case b T_b = T_max * ( 2 / ((s_max/s_b) + (s_b/s_max)) ); // Starting torque in lb-ft // Display the results disp("Example 9-14 Solution : "); printf(" \n a: Full-load torque at slip = %.4f \n T = %.1f lb-ft\n",s_a,T_a); printf(" \n b: Starting torque at slip = %.1f \n T = %.2f lb-ft\n",s_b,T_b);
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eg7_14.sce
clear; //clc(); d=20; s=0.5; r=20/1000; dab=20; dbc=20; dca=40; dsl=((sqrt(2)*0.7788*r*(s*s*s))^(1/4)); dm=(dab*dbc*dca)^(1/3); lb=2*log([dm/dsl]); xlb=2*(%pi)*lb*50; dsc=(sqrt(2)*r*(s^3))^(1/4); cn=1/(18*(10^(9))*(log([dm/dsc]))); printf("the capacitance is: %.2f*10^(-9) F/km\n",cn*10^(12)); xcb=1/(2*(%pi)*50*cn*1000); printf(" the reactance is: %.2f*10^(5) Ohm/km\n",xcb*.00001);
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anguloProyectil.sci
function fx = anguloProyectil(theta0) v0 = 30; g = 9.81; x = 90; y0 = 1.8; y = 1; fx = tan(theta0)*x-(g./(2*(v0^2)*(cos(theta0))^2))*x^2+y0-y; endfunction
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//clc() funcprot(0) //f(x) = log(x) disp("secant method") for i = 1:4 if i==1 then x(i) = 0.5; else if i==2 then x(i) = 5; else x(i) =x(i-1) - log(x(i-1)) * (x(i-2) - x(i-1))/(log(x(i-2)) - log(x(i-1))); end end end disp(x(1:4),"x =") disp("thus, secant method is divergent") disp("Now, False position method") xl = 0.5; xu = 5; for i = 1:3 m = log(xl); n = log(xu); xr = xu - n*(xl - xu)/(m - n); disp(xr,"xr = ") w = log(xr); if m*w < 0 then xu = xr; else xl = xr; end end disp("thus, false position method is convergent")
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// Example 1.4 Scilab mode(0) function p = Gauss(x,mu,std) // Eq. (1.5) probability density function p = 1/std/sqrt(2*%pi)*exp(-1/2*((x-mu)/std).^2) endfunction // data x = [66 69 58 100 83 42 54 69 49 64 59 30 51 67 64 53 64.. 58 49 81 49 77 70 73 64 58 89 80 82 67 87 74 62 63 51 80]; // calculations mu = mean(x) std = stdev(x) COV = std/mu histplot(7,x); xtitle('','Strength F','probability p(F)') xx = [min(x):1:max(x)]; pdensity = Gauss(xx,mu,std); plot(xx,pdensity)