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// Exa 6.8 // To calculate the minimum number of PN chips that are required for each frequency word. clc; clear all; Bss=600; //Hopping bandwidth in MHz stepsize=400; // in Hz //solution No_of_Tones=Bss*10^6/stepsize; Min_chips_required=log2(No_of_Tones); printf('Minimum number of chips required are %d ...
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off echo; write "Premier exemple: utilisation interactive sur une equation simple"; write "desir(); %appel de DESIR"; desir(); 3; %ordre de l'equation 1;x;x;x**6; %coefficients non; %correction ? non; ...
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clc clear printf("Example 9.13 | Page number 288 \n\n"); //Find magnitude and direction of heat transfer //Given data V_He = 0.3 //m^3 //volume of Helium p_He = 20e5 //Pa //pressure of Helium T_He = 273+30 //K //Temperature of Helium V_O2 = 0.7 //m^3 //volume of O2 p_O2 = 6e5 //Pa //pressure of O2 T_O2 = 27...
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function cels=casc(x,z) //cels=casc(x,z) //Creates cascade realization of filter //from a matrix of coefficients // x :(4xN)-Matrix where each column is a cascade // :element, the first two column entries being // :the numerator coefficients and the second two // :column entries being the denomina...
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clc d_r = 13.6e03 // Density of mercury in kg/m^3 g = 9.81 // Acceleration due to gravity in m/s^2 z = 710e-03 // Steam flow pressure in m z0 = 772e-03 // Reading of barometer in m P = 1.4e06 // Gauge pressure of applied steam in Pa P0 = d_r*g*z0 // Atmospheric pressure in Pa Pi = P+P0 // Inlet steam pressure i...
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// find minimum trigger voltage,maximum capacitor voltage ,width of output pulse // Electronic Principles // By Albert Malvino , David Bates // Seventh Edition // The McGraw-Hill Companies // Example 23-6, page 918 clear;clc; close; // Given data C=0.47*10^-6;// capacitance in faraday R=33*10^3;// resista...
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clear //Given M=0.075 //kg /mol m=1.2*10**-6 //kg A=6.0*10**23 ///mol t=9.6*10**18 N=170 //Calculation n=(A*m)/M l=N/t T=0.693/l //Result printf("\n Half life of K-40 is %0.3f *10**9 years",T/(24.0*3600.0*365)*10**-9)
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clear clc disp('the five digits can be arranged in 5! ways =') factorial(5) disp('of which 4! will begin with 0=') factorial(4) disp('so,total no. of five digit numbers=5!-4!=') factorial(5)-factorial(4) disp('the numbers ending in 04,12,20,24,32,40 will be divisible by 4') disp('numbers ending in 04=3!') fa...
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clc; funcprot(0); //Example 8.5 Flying Level at Altitude // Initialisation of variables W = 2000; S = 350; V = 100*1.467; rho = 0.002378; // Calculations Cl = 2*(W/S)/(rho*V^2); alpha = -1.75; // From fig 8.15 Cd = 0.014; // From fig 8.15 D = Cd*(rho/2)*S*V^2; HP = D*V/550; // At 10000 f...
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//chapter 18 Ex 9 clc; clear; close; sTrain=54; tTrainP=20; tTrainM=12; sMan=6; sRelative=(sTrain-sMan)*5/18; //Difference since opposite in direction lTrain=sRelative*tTrainM; lTotal=tTrainP*sTrain*5/18; lPlatform=lTotal-lTrain; printf("The length of train is %d m and length of platform is %d m",lTrain,lP...
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<?xml version="1.0" encoding="utf-8"?> <test> <description>Linear stability with Velocity Correction Scheme and a half mode Ev = (2.4868e-03,1.5835e-01i) which corresponds to a multiplier of (1.00249,0.158351). Note without the restart file it need 1000's iterations</description> <executable>IncNavierStokesSol...
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function [pt]=smooth(ptd,pas) // [lhs,rhs]=argn(0) [m,n]=size(ptd) d=splin(ptd(1,:),ptd(2,:)) if rhs=1 then l=abs(ptd(1,n)-ptd(1,1));pas=l/100;end pt=[ptd(1,1)+pas:pas:ptd(1,n)] pt=[ptd(:,1) [pt;interp(pt,ptd(1,:),ptd(2,:),d)] ptd(:,n)]
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//<-- NO CHECK REF --> exec('eqm/aerodata_f16.sci'); angle_list = -8:4:12; [S, K, DA, L] = angle_interp(angle_list, -8); assert_checktrue(abs(S-1)<=1e-7); assert_checktrue(abs(K-2)<=1e-7); assert_checktrue(abs(DA+1)<=1e-7); assert_checktrue(abs(L-1)<=1e-7); [S, K, DA, L] = angle_interp(angle_list, -9); assert_checktrue...
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clear; clc; disp("--------------Example 3.30---------------") dBpkm= -0.3 ; // dB/km p1=2; // 2 mW - initial power distance = 5; // 5km dB= dBpkm*distance; // loss in the cable in decibel ratio=10^(dB/10); // dB=10*log10(ratio) p2 = p1*ratio; // ratio = p2/p1 printf("The power of the signal at 5 km is %2.1f mW...
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//Chapter 5, Example 5.5, page 196 clc //Initialisation f1=2.5 //frequency in MHz f2=6.3 //frequency in MHz K=1.1 // K factor //Calculation fse=1.05*f1*2 //frequency in MHz ...
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clc; //page 263 n=3; // no of component r=60;//mm, radius l=100;//mm length of cylinder V=[0.5*4/3*%pi*(r)^3,%pi*r*r*l,-%pi/3*r*r*l];//mm^3, Volumes of Hemisphere, cylinder and cone respectively x=[-3/8*r,l/2,3/4*l];//mm, x components of centroids of Hemisphere, cylinder and cone respectively sumV=0; sumxV=0;...
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function [y] = mtlb_isreal(x) y=(x==real(x))
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Xlam = 3.2; i =2; prob = cdfpoi("PQ", i, Xlam); disp(prob)
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clc horsepower=2.5 //rating of induction motor in horsepower at half load Vl=230 //terminal voltage of motor in volts Il=7 //load current of motor in amperes pf=0.8 //power factor of the machine Pin=sqrt(3)*Vl*Il*pf //input power in watts mprintf("Pin=%f W\n",Pin)//The answer may vary due to roundoff error ...
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// Modified by Minho Kim (9 Oct 2014) load Buffer.hdl, output-file Buffer.out, compare-to Buffer.cmp, output-list in%B3.1.3 out%B3.1.3; set in 0, eval, output; set in 1, eval, output;
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//Example_a_4_5 page no:177 clc; Vrms=sqrt(5^2+(5^2/2));//the values are taken by comparing the given equation with rms equation disp(Vrms,"the rms value of the waveform is");
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Example12_12.sce
//Example 12.12 //Program to determine the average incident optical power required to //maintain given SNR clear; clc ; close ; //Given data Lambda=1*10^(-6); //metre - WAVELENGTH h= 6.626*10^(-34); //J/K - PLANK's CONSTANT c=2.998*10^8; //m/s - VELOCITY OF...
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//EX13_17 PG-13.10 clc clear printf("conversion of hexadecimal no 9B2.1A to its decimal equivalent =") N=(9*16^2)+(11*16^1)+(2*16^0)+(1*16^(-1))+(10*16^(-2)); printf(" %.1f",N)
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turkish.high.tst
peynirli_pasta peynirli_pastalarınızı N;ACC;PL;PSS2P bisküvi bisküvilerinizde N;LOC;PL;PSS2P ön önlerine N;DAT;PL;PSS3P sosluk sosluğunun N;GEN;SG;PSS3S sürtük sürtüklerinde N;LOC;SG;PSS3P muz muzlarımıza N;DAT;PL;PSS1P seçmen seçmenlerinde N;LOC;PL;PSS3S düldül düldüllerimi N;ACC;PL;PSS1S güdüm güdüme N;DAT;SG işkence...
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clc // initialization of variables T1=300+273 // initial temperature in kelvin P1=600 // initial pressure in kPa P2=40 // final pressure in kPa //solution //please refer to steam table for values v1=0.4344 // specific volume from steam table @ 573k and 600 kPa v2=v1 // rigid container u1=2801 // specific int...
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Ch05Ex3.sce
// Scilab code Ex5.3: Pg 158 (2005) clc; clear; h = 6.63e-34; // Plank's constant, J-s lamda = 1e-10; // de Broglie wavelength of neutron, m p = h/lamda; // Momentum associated with neutron, kg-m/s m_n = 1.66e-27; // Mass of neutron, kg e = 1.6e-19; // Energy equivalent of 1 eV, J/eV K = p^2/(2...
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example8_23.sce
//Chapter 8 //Example 8_23 //Page 193 clear;clc; l=500; w=1.5; t=1600; h2=90; h1=30; h=h2-h1; printf("x1+x2=500 \n"); d=h*2*t/w/l; printf("x2-x1=%.0f \n\n", d); A=[1 1; -1 1]; b=[l; d]; X=A\b; x1=X(1); x2=X(2); s1=w*x1^2/2/t; cl=h1-s1; x=l/2-x1; smid=w*x^2/2/t; clmp=cl+smid; printf("x1 = %.0f m \n", x1); printf("...
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//Exa 6.7 clc; clear; close; //given data R1=2.1;//in Mohm R2=270;//in Kohm RD=4.7;//in Kohm RS=1.5;//in Kohm VDD=20;//in Volt VP=-4;//in Volt IDSS=8;//in mA //step 1 : Find VGS : VG=R2*10^3*VDD/(R1*10^6+R2*10^3);//in Volt disp("VS=ID*RS-VGS Volt"); disp("VGS=VG-VS=2.28-1.5*ID") //step 2 : Find ID : di...
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ex7_7.sce
//Communication Techniqu;es : example 7-7 : (pg 314) NF=20; df=10^6; x=10*log10(df); S=-174+NF+x; a=5;//input intercept dr=2/3*(a-S); printf("\nS = -174dBm + NF + 10log10df + S/N = %.f dB",S);//sensitivity printf("\ndynamic range = 2/3.(input intercept-noise floor) = %.d dB",dr);
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//Ex10_4 Pg-518 clc R1=1.5*10^(3) //resistor R1 in ohm Rf=75*10^(3) //feedback resistor in ohm Vin=10*10^(-3) //input voltage in V funi=1*10^(6) //unity frequency in Hz Acl=(-1)*Rf/R1 //closed loop gain printf("Magnitude of Closed loop gain = %.0f \n",abs(Acl)) fcl=funi/abs(Acl) //closed loop frequency p...
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clc // initialization of variables // The reaction equation is //C3H8 + 5O2---> 8CO2 + 4H2O // All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol hfCO2=-393520 // enthalpy associated with CO2 hfH2O=-241820 // enthalpy associated with gaseous H2O hfC3H8=103850// enthalpy of fo...
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Example2_8.sci
//Chapter 2_Thick Film and Thin Film Hybrid ICs //Caption : Thickness //Example2.8: The bulk resistivity of nichrom is 120uohm-cm. Calculate the thickness T in angstroms of a film with sheet resistivity of 100ohm/square. // Solution: function T=thickness(Ps,p)// Ps: sheet resistivity of nichrom=100ohm/square, p:bu...
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// Scilab code Exa4.7.1: To calculate the energy and power released during fission of U-235 : Page 189 (2011) m = 0.001; // Mass of U-235 lost during fission, Kg c = 3e+08; // Velocity of light, m/s E = m*c^2; // Energy released during fission, J E_t = E/(4e+09*1000); // Energy requires TNT, Kt printf("\n Energy...
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clc(); clear; //Given : v = 343; // velocity of sound in m/s lambda = 1; // wavelength in cm // 1 cm = 1.0*10^-2 m f = v/(lambda*10^-2); //frequency in Hz printf("Frequency is %.1f kHz",f*10^-3);
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// Exa 4.10 clc; clear; close; format('v',8) // Given data V1 = 10;// in V V2 = 5;// in V I1 = 5.8;// in mA I2 = 5;// in mA delV_C = V1-V2;// in V delI_C = I1-I2;// in mA r_out = delV_C/delI_C;// in k ohm disp(r_out,"The dynamic output resistance in k ohm is");
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// Example 6_1 clc;funcprot(0); // Given data V=300;// ft/s D=6/12;// ft R=D/2;// ft Z=15;// ft g=32.174;// ft/s^2 g_c=32.174;// lbm.ft/lbf.s^2 // Calculation // From the superheated steam table, Table C.3a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that, at 100. psia an...
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// The code was developed under Horizon2020 Framework Programme // Project: 748767 — SIMFREE function [zX,zY] = SSSeIQcohrec2pol(X,Phet) // Coherent receiver of X and Y polarized optical signals // // Calling Sequence // [zX,zY] = SSSeIQcohrec2pol(X,Phet) // // Parameters // X : Optical...
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//Caption: Information capacity //Example 9.56 //page no 444 //Find Information capacity of telephone clear; clc; B=3.4*10^3; SNR=30 SN=10^(SNR/10); C=B*log2(1+SN)//Information capacity printf("Information capacity of telephone is \n\n \tC = %.2f kbps",C/1000);
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clc; h1=3442.6; s1=7.066; s2=s1; sf2=0.391; sfg2=8.13; x2=(s2-sf2)/sfg2 hf2=112; hfg2=2438; h2=hf2+x2*hfg2; h3=112; W12_=h1-h2; Q=h1-h3; Ceff=(h1-h2)/(h1-h3); disp(Ceff,"cycle efficiency is:"); ssc=1/(h1-h2); disp("kg/kW h",ssc,"specific steam consumption is:"); disp("cycle efficiency has increased due to superhe...
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// find ac output voltage // Electronic Principles // By Albert Malvino , David Bates // Seventh Edition // The McGraw-Hill Companies // Example 18-12, page 687 clear; clc; close; // Given data Rf=100*10^3;// in ohms from the given figure R1=20*10^3;// in ohms from the given figure R2=10*10^3;// in ohms f...
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// 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/03/a/Reg16Bit.tst load Reg16Bit.hdl, output-file Reg16Bit.out, compare-to Reg16Bit.cmp, output-list time%S1.4.1 ip1%D1.6.1 load%B2.1.2 out%D1.6.1; set ip1 0, set ...
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clear all; clc; disp("Ex 4_5") disp("u = 0.3841i - 0.5121j + 0.7682k") a1=acos(0.3841) a=a1*180/%pi printf('\n\nalpha = %.1f degrees',a) b1=acos(-0.5121) b=b1*180/%pi printf('\n\nbeta = %.0f degrees',b) c1=acos(0.7682) c=c1*180/%pi printf('\n\ngamma = %.1f degrees',c)
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//Example 3.24.C clc; Syms s t; x=laplace((1+0.5*exp(-6*t)+0.2*exp(-3*t)),t,s); disp(x);
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//Example 10.1 //System of Non Linear Equations //Page no. 311 clc;clear;close; deff('y=f(x)','y=x^2-exp(2*x)-4') deff('y=f1(x)','y=2*x-2*exp(2*x)') x0=0;e=0.00001 for i=1:10 x1=x0-f(x0)/f1(x0) e1=abs(x0-x1) x0=x1; if abs(x0)<e then break; end end printf('\n\nThe solution of...
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clear; clc r1=.02 r2=.05 r3=.03 Ic1=100 Id1=180 Va=255; Vb=250 dV=abs(Va-Vb) Ia=(dV+(r1*0)+(r2*Ic1)+(r3*(Id1+Ic1)))/(r1+r2+r3) Ib=-(Ia-(Ic1+Id1)) Vc=Va-Ia*r1 Vd=Vc-((Ia-Ic1)*r2) mprintf("IA= %.0f A, IB=%.0fA, Vc=%.2f V, Vd=%.2f V", Ia, Ib, Vc,Vd)
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function [ECmInv,EeV] = HCl(var,J) //Written by Aditi and O.S.K.S. Sastri //var is vector consisting of De, b and Re values. //Typical input for var for HCl is [5,1,1.27455] //J is rotational quantum number. //J = 0 gives pure vibrational levels //J = 1 gives energy eigen values corresponding to //first excit...
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//calculating resistance of the shunt i=20D-3 //current throught the coil r=4 //resistance of coil V=i*r I=2 //total current to be measured Is=I-i //current through shunt R=V/Is //Ohm's law mprintf("Resistance of the shunt=%f ohm\n", R) //solving part (ii) V=30 //voltage to be measured R=V/i-r mprint...
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clear clc M1=152.2;//molar mass of carbon in gm T1=451.55;//melting point temp in K T2=433.85;//melting point temp in K(for unknown compound) w2=0.0386;//mass of unknown compound in gm w1=0.522;//mass of camphor in solution in gm R=8.314;//J/Kmol DelHm_f=6.844;//in KJ Kf=(((R*T1^2)/(DelHm_f*10^3))*(M1/1000)) ...
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function Cam_Init(vp) { vp.custom.cam_rotating=false; vp.custom.cam_rotatingpos=false; } function Cam_Rotate(vp,newdir,durat) { vp.custom.cam_rotating=true; vp.custom.cam_rotate_dir1=deref(vp.CameraDir); vp.custom.cam_rotate_angle=vecangle(vp.CameraDir,newdir); vp.custom.cam_rotdir=vecnorm(vp.CameraD...
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//Example 4.2 : resistance of each coil clc; clear; close; //given data : V=300; // voltage in volts W=360; // power lost in one coil in watt I=6; // current in A R1=V/I; R=V^2/W; a=(1/R1)-(1/R); r2=1/a; disp(R,"resistance of 360W coil1,R(ohm) = ") disp(r2,"resistance of second coil2,r2(ohm) = ")
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// Scilab Code Ex 1.19 Miller indices of the crystal plane: Page-25 (2010) m = 2; n = 3; p = 6; // Coefficients of intercepts along three axes m_inv = 1/m; // Reciprocate the first coefficient n_inv = 1/n; // Reciprocate the second coefficient p_inv = 1/p; // Reciprocate the third coefficient ...
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clc //ex6.3 R=1000/(2*%pi); //resistance C=10*10^-6; //capacitance f_B=1/(2*%pi*R*C); //half-power frequency //the three parts of V_in are V_1=5*cos(20*%pi*t)+5*cos(200*%pi*t)+5*cos(2000*%pi*t) //first component V_in_1 V_in_1=5*complex(cos(0),sin(0)); //V_in_1 phasor w_1=20*%pi; //ome...
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// x=[]; dim=2; [x,perm,nshifts] = shiftdata(x,dim); disp(x); disp(perm); disp(nshifts); //output
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function [menor,medio,maior]=teste(x,y,z) if x>y then if y>z then menor=z medio=y maior=x elseif x>z then menor=y medio=z maior=x else menor=y medio=x maior=z end elseif y>...
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//Chapter-4, Illustration 17, Page 206 //Title: Steam Nozzles and Steam Turbines //============================================================================= clc clear //INPUT DATA m=2;//Mass flow rate of steam in kg/s W=130;//Turbine power in kW U=175;//Blade velocity in m/s C1=400;//Steam velocity in m/...
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// Scilab Code Ex2.35:: Page-2.27 (2009) clc; clear; t = 3.8e-05; // Thickness of the transparent film, cm mu = 1.5; // Refractive index of the transparent film i = 45; // Angle of incidence of the light ray on the transparent film, degrees lambda = 5700e-008; // Wavelength of light, cm // As mu = sind(...
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function r=%b_a_s(a,b) // Copyright INRIA A=zeros(a) A(a)=1 r=A+b
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// This file is adapted from part of www.nand2tetris.org // and the book "The Elements of Computing Systems" // by Nisan and Schocken, MIT Press. load Mux8.hdl, output-file Mux8.out, compare-to Mux8.cmp, output-list a%B1.8.1 b%B1.8.1 sel%D2.1.2 out%B1.8.1; set a 0, set b 0, set sel 0, eval, output; set sel 1, eval, ...
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clc; v=6600; // rated voltage of motor xs=20 ; // per phase synchronous reactance p=500000; // VA rating of motor il=p/(sqrt(3)*v); // rated armature current vt=v/sqrt(3); // per phase rated voltage disp('case a'); de=10; // load angle c1=1; c2=-2*vt*cosd(de); c3=vt^2-(il*xs)^2; // coefficients of quadratic ...
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function txt=get_block_info(scs_m,k) // Copyright INRIA txt=[] o=scs_m(k) select o(1) case 'Block' then txt = standard_document(o,k) txt=[txt;' '] if o(3)(1)=='super'|o(3)(1)=='csuper' then objet=o(3)(8) infos = objet(1) if size(infos(2),'*')==2 then txt = [txt;'Super Block Documentation: '+...
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//CHAPTER 5_ Force,Torque and Shaft Power Measurement //Caption : Load cell // Example 2// Page 295 disp("b=.2") disp("h=.05") disp("Sg=2") disp("Rg=120") disp("sig_f=150*10^6") b=.2 //('enter the width of load cell=:') h=.05 //('enter the thickness of load cell=:') Sg=2; Rg=120; sig_f=150*10^6 //('en...
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exp2_8cpp.sce
clc disp("example 2.8") disp("the chronological load curve is plotted in fig 1") a=[0 5 9 18 20 22 24] //time in matrix format b=[50 50 100 100 150 80 50]//load in matrix format for x=1:6 z(1,x)=((b(1,x)+b(1,x+1))/2)*(a(1,(x+1))-a(1,x)) end e=sum(z); printf("energy required required by the system in 24 hrs ...
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EX8_35.sce
//EXAMPLE 8-35 PG NO-552 R1=200; R2=200; R3=100; Zoc=[R1*(R2+R3)/(R1+R2+R3)]; disp('i) IMPEDANCE (Zoc) is = '+string (Zoc) +' ohm '); Zsc=(R1*R3)/(R1+R3); disp('ii) IMPEDANCE (Zsc) is = '+string (Zsc) +' ohm '); Zo=[Zoc*Zsc]^0.5; disp('iii) IMPEDANCE (Zo) is = '+string (Zo) +'...
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//Ex18_2 Pg-945 clc T=2000 //temperature in Kelvin f=5*10^(14) // frequency in Hz h=6.6*10^(-34) //planck constant k=1.38*10^(-23) //Boltzmann constant R=exp((h*f)/(k*T)) //ratio of spontaneous and stimulated emisson printf(" R = %.2f*1e5",R*1e-5)
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function [M1,M2, sing] = QRhausholder(M, n) sing=%f; eta = M(1,1); for k=1:n//n for i = k:n if eta <= abs(M(i,k)) then eta = abs(M(i,k)) end end sum=0; sumM=0; if eta == 0 then M1(k)=0; M2...
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(get-info :authors) (get-info :name) (get-info :version) (get-info :error-behavior)
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example_5_2.sce
//Chapter 5 //Example 5.2 //page 134 //To find voltage at the bus at the power station end clc;clear; base_MVA=5; base_kV=33; pf=0.85; cable_impedance=(8+%i*2.5); cable_impedance=cable_impedance*base_MVA/(base_kV^2); transf_imp_star=(0.06+%i*0.36)/3; //equivalent star impedance of winding of the transformer Zt=(trans...
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example12_19.sce
clc // Given that m = 0.0001 // mass of Sr(90) in gm t = 28 // half life of Sr(90) in year t_ = 9 // time in sec // Sample Problem 19 on page no. 12.38 printf("\n # PROBLEM 19 # \n") printf("Standard formula used \n") printf(" lambda = 0.693 / t_1/2 (Decay constant) \n del_N = N_0*lambda*t (disintegration of samp...
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//0 denotes False and 1 denotes true b=[0,1]; //binary operation + on the set of bits for i=1:2 for j=1:2 k = b(i)& b(j); disp(k) end end //binary operation * on the set of bits for i=1:2 for j=1:2 k = b(i)| b(j); disp(k) end end //unary operation ' on the set of bits k=~b clear; D=[1,2,5,7,10,14,3...
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13_23.sce
//To find speed, direction and torque clc //Given: TP=144, TQ=120, TR=120, TX=36, TY=24, TZ=30 NI=1500 //rpm P=7.5*1000 //W eta=0.8 //Solution: //Refer Fig. 13.30 and Table 13.25 //Calculating the values of x and y //From the fourth row of the table, x+y = -1500 .....(i) //Also, y-x*(TZ/TR) = 0, or -x...
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example18_11.sce
clc // Given that n = 200 // no. of turns l = 0.5 // the mean length of iron wire in m phi = 4e-4 // magnetic flux in Weber a = 4e-4 // area of cross section in m^2 mu = 6.5e-4 // permeability of iron in wb/Am mu_ = 4 * %pi * 1e-7 // magnetic permeability of space // Sample Problem 11 on page no. 18.25 printf("\n # ...
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Ex6_5.sce
errcatch(-1,"stop");mode(2);//caption:Find resolution and voltage //Ex6.5 N=3//bit of D/A convertor V=5//full scale voltage(in V) A=0.001//magnitude of accuracy R=1/2^N disp(R,'resolution(in V)=') Ac=A*V disp(Ac,'accuracy(in V)=') exit();
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exa_5_2.sce
// Exa 5.2 clc; clear; close; // Given data C_T1= 15; // in pF Vb1=8;// in V Vb2= 12; // in V // As C_T proportional to 1/sqrt(Vb), and // C_T1/C_T2= sqrt(Vb2/Vb1), so C_T2= C_T1*sqrt(Vb1/Vb2);// in pF disp(C_T2,"The value of C_T2 in pF is : ")
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Example15_7.sce
//Example 15.7. clc format(6) fo1=540*10^3 fo2=1650*10^3 L=1*10^-3 disp("Given L = 1 mH") disp("fo ranges from 540-1650 kHz") disp("The resonant frequency is given by") disp(" fo = 1 / 2pi*sqrt(L*C)") disp("Therefore, C = 1 / 4*pi^2*fo^2*L") Cmax = 1 / (4*%pi^2*fo1^2*L) x1=Cmax*10^12 disp(x1,"Wh...
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Example18_1.sce
close(); clear; clc; //meter resistance 'rm', meter current 'Ifs', Maximum current to be measured 'I', Maximum voltage measured 'V' rm = 5; //ohm Ifs = 0.015; //A //(i) I = 2; //A //resistance of shunt required 'Rsh' Rsh = Ifs*rm/(I-Ifs); //ohm mprintf("(i) Resistance of shunt, Rsh = %0.6f ohm\n\n",Rsh); ...
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//Example 9.1 m1=26.0;//Mass of 1st child (kg) m2=32.0;//Mass of 2nd child (kg) r1=1.60;//Distance of 1st child from pivot (m) r_p=0;//Distance of supporting force of pivot from pivot (m) g=9.80;//Acceleration due to gravity (m/s^2) theta=90;//Angle (deg) //Torque tau=r*(m*g)*sind(theta) //Torque due to support...
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clear //Given N=100 A=0.10 //m**2 f=0.5 //Hz B=0.01 //T //Calculation // w=2*%pi*f E0=N*A*B*w //Result printf("\n Maximum voltage generated in the coil is %0.3f V",E0)
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exec("lib/page_rank_matrice.sce"); M = bool2s(rand(500, 500)<0.5); A = page_rank_matrice(M, 0.85)
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10_2.sce
//To find weight and coefficient of friction clc //Given: P1=1500,P2=1720 //N alpha1=12,alpha2=15 //degrees //Solution: //Refer Fig. 10.10 //Effort applied parallel to the plane, P1=W*(sind(alpha1)+mu*cosd(alpha1)), or P1/W-mu*cosd(alpha1)=sind(alpha1) .....(i) //Effort applied parallel to the plane, P2=W*(s...
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//Chapter-2, Example 2.8, Page 95 //============================================================================= clc; clear; //INPUT DATA l=0.3;//length in m d=1.5*10^-2;//diameter in m N=900;//no of turns ur=1;//relative permeability in free space u0=4*%pi*10^-7;//permeability in free space I=5;//current in...
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// Examle 3.2 w=0.2; // Energy stored i=0.2; // Current L1=(2*w)/i^2; // The value of Inductor Using case-1 disp(' The value of Inductor Using case-1 = '+string(L1)+' H'); v=10; // Voltage di1=0.1; ...
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Ex12_25.sce
//Chapter 12 //page no 493 //given clc; clear all; R=800; //in V/W Pin=1.5; //in mW m=0.04; Voutp=R*Pin*m; printf("\n Vout(peak) = %0.0f mV",Voutp); Vavg=Voutp/sqrt(2); printf("\n Vavg = %0.1f mV",Vavg); //in dB Vavgd=20*log10(Vavg*10^-3); printf("\n Vavg(in dBmV) = %0.1f ",Vavgd);...
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Ex3_6.sce
clc //Initialization of variables P1 = 1456 // lb/ft^2 d = 0.001756 //slugs/ft^3 V = 293 // ft/s // Calculations P11 = P1/144 //psi P2 = P1 + (d*V^2)/2 // lb/ft^2 P22 = d*V^2/(2*144) //psi D = P22 //psi // Results printf ("The static pressure at the altitude is %.2f psi",P11) printf ("\nThe pressure in ter...
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large.tst
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Exa13_4.sce
//Exa 13.4 clc; clear; close; //given data : //initial cash outflows ICO=80000;//in Rs. //Total Present Value calculated in Exa13.3 P=97922;//in Rs disp(P,"Total present value(in Rs) is : ") //Profitability Index at 10% discount rate PI=P/ICO;//unitless disp(PI,"Profitability Index at 10% discount rate is ...
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5_16.sce
clear; clc; R=45;L=1.2*(10^-3);G=5*(10^-6);C=0.065*(10^-6);w=20000;l=22*(10^-3);s=1.1; pf=0.005//pf=power factor //value of pf as taken in solution r=pf*w*L; Rc=R+(r/s); Lc=L+(l/s); P=sqrt((Rc+(%i*w*Lc))*(G+(%i*w*C))); theta=round(atan(imag(P),real(P))*180/%pi); a=abs(P)*cos(theta*%pi/180); printf("Attenuatio...
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Ex15_14.sce
// Example 15_14 clc;funcprot(0); // Given data p=0.100;// MPa T_a=298;// K T_b=2000;// K R=0.0083143;// MJ/kgmole.K // Calculation // (a) gbar0_f_H2O=-228.583;// kJ/kgmole // since H2 and O2 are elements, their molar specific Gibbs function of formation is zero. Then, from Table 15.7, gbar0_f_H2=0;// kJ...
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ex4_5.sce
//AC Circuits : example 4.5 :(pg 4.6) kf=1.2; kp=1.5; Vavg=10; disp("kf=1.2"); disp("kp=1.5"); disp("Vavg=10"); disp("form factor kf=(Vrms/Vavg)"); Vrms=(kf*Vavg); printf("\nVrms=%.f V",Vrms); disp("peak factor kp=(Vm/Vrms)"); Vm=(kp*Vrms); printf("\nVm=%.f V",Vm);
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sce
Ch05Ex25.sce
// Scilab code Ex5.25: Pg 183 (2008) clc; clear; N = 400; // Number of turns in a coil l = 0.25; // Effective length of coil, m A = 4.5e-04; // Cross-sectional area, m^2 mew_r = 180; // Relative ...
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Ex18_28.sce
// scilab Code Exa 18.28 centrifugal Air compressor T01=335; // in Kelvin p01=1.02; // Initial Pressure in bar beta1=61.4; // air angle at the inlet of axial inducer blades gamma=1.4; N=7200; // rotor Speed in RPM d1=0.175; // Mean Blade ring diameter at entry d2=0.5; // impeller diameter at exit cp=1005; //...
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2_8.sce
clear all; clc; disp("Ex 2_8") f=100 b1=60//given angle theta in degrees b=b1*%pi/180 g1=45//given angle theta in degrees g=g1*%pi/180 p=sqrt(1-(cos(b)^2)-(cos(g)^2)) p1=p*(-1) printf('\n\ncos(alpha) = + or - %.1f',p) l1=acos(p) lf=l1*180/%pi l2=acos(p1) lf1=l2*180/%pi printf('\n\n alpha = %.0f degrees or %.0f degrees'...
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9_18.sce
//Compute moment of inertia //refer fig. 9.43 Ixx=((125*60^3)/36)+(125*(60/2)*(60+60/3)^2)+((125*60^3)/36)+(125*(60/2)*(2*60/3)^2)+((125*60^3)/36)+(125*(60/2)*(60/3)^2)+((125*60^3)/36)+(125*(60/2)*(60/3)^2) //mm^4 printf("Ixx=%.2d mm^4",Ixx)
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testexo2b.sce
n=6;m=5; L=([1:n]')*ones(1,m); // line number C=ones(n,1)*[1:m]; // column number // N = matrix of element positions N=L+(C-1)*n // method 1 N=zeros(n,m);N(:)=[1:n*m] // method 2
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sce
Ex4_33.sce
clear // //the eqns are formed using the given diagram //the derivations from the eqns are obtained as below using matrices for their construction //the below eqns are in polar form delta=0.3165 delta1=5.95 delta2=6.82 v1=delta1/delta printf("\n v1 at -47.63 is= %0.1f V",v1) v2=delta2/delta printf("\n v2 at -42.30 is=...
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Ex7_5.sce
//Example 7_5 page no:263 clc; Vmag=50; Vang=0; Rmag=9.22; Rang=77.47; Imag=Vmag/Rmag; Iang=Vang-Rang; V1mag=5.42*5.38;//voltage across (2+j5)ohm V1ang=-77.47+68.19; I2mag=(20*4)/9.22; I2ang=120-77.47; V2mag=I2mag*5.38;//voltage across (2+j5)ohm due to current I2 is V2ang=42.53+68.19; V1rel=V1mag*(cosd(V1...
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EX2_105.sce
//EXAMPLE-2.105 PG-NO143 N1=250; //number of turn I1=2; //current Q1=0.3*10^-3; //phi L1=(N1*Q1)/I1; V2=63.75; K=0.85; x=10^3; //x=di/dt M=V2/x; L2=((V2/K)^2)/((37.510^-3)^0.5); Q12=0.255*10^-3; y=1.275*10^-7; //y=dQ12/dt N2=V2/y ; disp('i) L1 = '+st...
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example10_6.sce
vb=13.2; vg=14.4; rb=0.5; rg=0.1; i=(vg-vb)/(rb+rg); vrb=i*rb; vth=vb+vrb; rth=rb*rg/(rb+rg); disp("Part a"); rl1=1; i1=vth/(rth+rl1); disp("current (in A) through the load resistor is"); disp(i1); disp("Part b"); rl2=2; i2=vth/(rth+rl2); disp("current (in A) through the load resistor is"); disp(i2); di...
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Exa3_14.sce
//Exa:3.14 clc; clear; close; V_dc=125;//in volts V_a=200;//average output voltage (in volts) T_on=1*10^-3;//in seconds alpha=V_a/(V_a+V_dc);//duty cycle f=alpha/T_on; disp(f,'Frequency Of Switching pulse (in hertz)=')
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2013-07-31T06:53:59
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bow.18_3.tst
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