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function [str]=pol2str(p) //polynomial to character string // // p : polynomial (real) // str : chracter string //! // Copyright INRIA n=degree(p) var=varn(p) nv=length(var);while part(var,nv)==' ' then nv=nv-1,end;var=part(var,1:nv); string(20) p=coeff(p) string(10) if p(1)<>0 then if p(1)<0 then str=string(p(1)...
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Name=Counter-Striker Peek PlayerCharacters=Counter-Striker BotCharacters=Counter-Striker Bot PEEKER.bot IsChallenge=true Timelimit=120.0 PlayerProfile=Counter-Striker AddedBots=Counter-Striker Bot PEEKER.bot PlayerMaxLives=0 BotMaxLives=0 PlayerTeam=1 BotTeams=2 MapName=794569841.map MapScale=3.0 BlockProjectilePredict...
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//pathname=get_absolute_file_path('4.15.sce') //filename=pathname+filesep()+'4.15-data.sci' //exec(filename) //Maximum temperature(in K): T1=500+273 //Minimum temperature(in K): T2=200+273 //Temperature of the body(in K): T3=450+273 //Efficiency: n=1-T2/T1 //Ratio of W to Q1: r1=n //COP of pump: COP=T3/(T...
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// Ex1_6 Page:15 (2014) clc; clear; c = 1; // For simplicity assume speed of light to be unity, m/s m0 = 1; // For simplicity assume the rest mass, kg d = 1; // Percentage difference of relativistic mass from the rest mass, kg m = m0*(1+20/100); // Effective mass of the body while moving, kg v = poly(0, "v")...
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// Initiization of variables a=1 // m/s^2 // acceleration of car A u_B=36*(1000/3600) // m/s // velocity of car B u=0 // m/s // initial velocity of car A d=32.5 // m // position of car A from north of crossing t=5 // seconds // Calculations // CAR A: Absolute motion using eq'n v=u+at we have, v=u+(a*t) // m/s ...
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clc //initialisation of variables p=1.01325 //pressure in bar w1=0.01468 td=20 //temp in degrees tw=40 //temp in degrees //CALCULATIONS ha=(1.005*td+w1*(2500+1.86*td)) w2=(ha-(1.005*tw))/(2500+1.86*tw) //RESULTS printf('humidity rate is %2fkg/kg of da',ha) printf('\nw2 is %2fkg/kg of da',w2)
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clc //initialisation of variables p=10^5//N/m^2 l=1//m a=0.2//m^2 n=5 //CALCULATIONS power=2*p*l*a*n/746 //results printf(' \n horse power of engine= % 1f H P',power)
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clear //Given Pc=500 //watts //Calculation Ps=(1/2.0)*(Pc) Pt=Pc+Ps //Result printf("\n (i) sideband power is %0.3f W",Ps) printf("\n (ii) power of AM wave is %0.3f W",Pt)
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// DFT – magnitude and phase plot// clc; clear all; close; x1=[1 1 1 0]; X1=fft(x1,-1); disp(X1,"X1(k) ="); subplot(2,3,1);plot2d3(x1);title('Sequence 1') subplot(2,3,2);plot2d3(abs(X1));title('Absolute value of DFT') subplot(2,3,3);plot2d3(atan(imag(X1),real(X1)));title('Phase of DFT') x2=[1 1 1 0 0 0 0 0]; X2=fft(x2...
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//Example6.1 // design an inverting amplifier with a closed loop voltage gain of Av = -5 clc; clear; close; Av = -5 ; Is = 5*10^-6 ; // A Rs = 1*10^3 ; // ohm // input voltage source Vs = sinwt volts // in an inverting amplifier frequency effect is neglected then i/p volt Vin = 1 V and total resistance equal...
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//Example 8.18 clc disp("The fig shows the connections for 74LS191 to get desire operation. We can design the combinational circuit for such counter from the truth table shown below.") disp("") disp("Q3 Q2 Q1 Q0 Y") disp("0 0 0 0 0") disp("0 0 0 1 0") disp("0 0 1 0 0") disp("...
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//chapter 4 //end fire array printf("\n"); n=16; d=0.25; L=(n-1)*d; m=1; //part a HPBW=57.3*sqrt((2*m)/L); printf("the HPBW is %g degree",HPBW); //part b D=4*L; Ddb=10*log10(D); printf("\nthe directivity is %d",D); printf("\nthe directivity in db is %gdb",Ddb); //part c A=4*(%pi)/D; printf("\nthe beam ...
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clc; clear; Fc=500;//carrier frequency in kHz Fm=1;// message signal frequency in kHz //a) USB=Fc+Fm; LSB=Fc-Fm; disp(USB,"USBI(in kHZ)="); disp(LSB,"LSB(in kHz)="); //b) Bandwidth=USB-LSB; disp(Bandwidth,"Bandwidth(in kHZ)=") //c) Fm=1.5;//message signal frequency in kHz USB1=Fc+Fm; LSB1=Fc-Fm...
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sd lkjasd l;kajsd lkasd sdg sdf g dsf gs df gs df g df gsdf g sdf g sd fgs df g spp -- test.lua 文件末尾注释-- test.lua 文件末尾注释-- test.lua 文件末尾注释-- test.lua 文件末尾注释
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//Example 3.13 //Program to Compute circular convolution of following sequences //x1[n]=[1,-1,-2,3,-1] //x2[n]=[1,2,3] clear; clc ; close ; x1=[1,-1,-2,3,-1]; x2=[1,2,3]; //Loop for zero padding the smaller sequence out of the two n1=length(x1); n2=length(x2); n3=n2-n1; if (n3>=0) then x1=[x1,zeros(1,n3)...
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//x+y=19, xy=84 clear; clc; close; x=poly(0,'x'); //substitute y=19-x in xy=84 Y=x*(19-x)-84; x=roots(Y); y=19-x; mprintf('the solutions are: \n') mprintf("(x,y)=(%i,%i) \n",x,y)
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clear clc disp('Data is not provided,only graph is provided')
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//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 8.4w //calculation of the work done by the given force //given data function F=f(x) F=(10+(.50*x)) endfunction x1=0//initial position(in m) of the particle x2=2//final position(in m) of the particle //calcu...
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//Chapter 14 //Example 14_10 //Page 371 clear;clc; v=400; ph_v=230; r=0.2; i=30; pfr=-0.866; pfy=0.866; pfb=1; ar=0; ay=-120; ab=120; //referring to the phasor diagram given in the text book air=-30; aiy=-90; aib=120; vr=ph_v*(cos(0)-%i*sin(0)); vy=ph_v*(cos(-120*%pi/180)-%i*sin(-120*%pi/180)); vb=ph_v*(cos(120*%pi/...
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//ques-16.15 //Calculating activation energy for the reaction clc k2=5.16*10^4; k1=3.76*10^3; T2=1125; T1=1085;//temperature (in K) Ea=(log10(k2/k1)*2.303*8.314*T1*T2)/(T2-T1);//activation energy printf("Activation energy for the reaction is %d J/mol.",Ea);
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clear; clc; V=230; E=150; R=8; th1=asind(E/(sqrt(2)*V)); I_o=(1/(2*%pi*R))*(2*sqrt(2)*230*cosd(th1)-E*(%pi-2*th1*%pi/180)); printf("avg charging curent=%.4f A",I_o); P=E*I_o; printf("\npower supplied to the battery=%.2f W",P); I_or=sqrt((1/(2*%pi*R^2))*((V^2+E^2)*(%pi-2*th1*%pi/180)+V^2*sind(2*th1)-4*...
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//Chapter 3 //Example 3_2 //Page 50 clear; clc; cl = 43; max_dem = 20; ugpa = 61.5e6; //Demand factor and load factor printf("Demand factor = %.3f \n\n", max_dem/cl); avg_dem = ugpa/8760; printf("Load facor = %.3f \n\n", avg_dem/max_dem/1000);
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clc; clear all; disp("heat loss per hour") r1=120;//mm r2=r1+50;//mm r3=r2+40;//mm kA=0.092;//W/(m*C) kB=0.062;//W/(m*C) t1=390;// degree C t3=40;// degree C L=210;//m Q1=2*3.1416*L*(t1-t3)/((log(r2/r1))/kA+(log (r3/r2))/kB); Q=(3600/1000)*2*3.1416*L*(t1-t3)/((log(r2/r1))/kA+(log (r3/r2))/kB); disp("kJ/hr...
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//Displacement //After 0.5 sec theta=5.236*0.5*(180/%pi) //displacement x=0.75*sind(150) //m //Velocity vx=0.75*5.236*cosd(150) ax=0.375*5.236^2 //m/sec^2 printf("\nx=%.3f m\nvx=%.3f m/sec\nax=%.2f ",x,vx,ax)
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plot1 = fscanfMat('plot_tf.txt'); plot2 = fscanfMat('plot_gs.txt'); plot3 = fscanfMat('plot_plain.txt'); fs = 4; scf(1); plot(plot1(:, 1), plot1(:, 2), '-'); plot(plot2(:, 1), plot2(:, 2), '--'); plot(plot3(:, 1), plot3(:, 2), '-.');
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// Example 7.3 clc; clear; close; // Given data format('v',7); f_H= 1*10^3;//cut-off frequency in Hz C= 0.0047*10^-6;// in F R= 1/(2*%pi*f_H*C);// in Ω R= (R*10^-3);// in kΩ R1= 30;// in kΩ (assume) Rf= 0.586*R1;// in kΩ C= C*10^6;// in µF disp("The value of R''= R= "+string(R)+" kΩ (standard value 33 kΩ)"...
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clear; clc; //Example - 3.9 //Page number - 93 printf("Example - 3.9 and Page number - 93\n\n"); //Given m_ice = 1000;//[g] - Mass of ice m_water = 1000;//[g] - Mass of water T_ice = 273.15;//[K] - Temperature of ice T_water = 373.15;//[K] - Temperature of water L = 79.71;//[cal/g] - Latent heat of meltin...
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clc //initialisation of variables p2=150//lbf/in^2 v=20 //f h1=79.385 //lbf/in^2 g=144//in^2/ft^3 T=778//F p=35.736//lbf/in^2 s1=0.16719 //btu/lbm-R T2=122.9 //F h2=90.330 //lbf/in^2 v2=0.28270 //Btu/lbm //CALCULATIONS u1=h1-(p*g*1.0988/T)//Btu/lbmu u2=h2-(p2*g*v2/T//Btu/lbm W=u1-u2 //Btu/lbm //RESULT...
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clear; clc; a=5*10^-2,b=4*10^-2,c=10*10^-2,C=5.8*10^7,Uo=4*%pi*10^-7; f101=3.335*10^9; d=sqrt(1/(%pi*f101*Uo*C)); Q=(a*a+c*c)*a*b*c/(d*(2*b*(a^3 + c^3)+a*c*(a*a+c*c))); disp(Q,'Quality factor of TE101 = ');
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//Ex 1.2 clc;clear;close; format('v',5); I=1.5;//A R1=2;//ohm R2=3;//ohm R3=8;//ohm V1=I*R1;//V V2=I*R2;//V V3=I*R3;//V disp(V1,"Voltage across R1(V)"); disp(V2,"Voltage across R2(V)"); disp(V3,"Voltage across R3(V)"); V=V1+V2+V3;//V(Supply voltage) disp(V,"Supply Voltage(V)");
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clc; XL=3160; Rw=25; Q=XL/Rw; disp(' ',Q,"Q=");//The answers vary due to round off error
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// Scilab Code Ex16.4 Behaviour of paramagnetic salt when placed in uniform magnetic field: Page-514 (2010) k = 1.38e-023; // Boltzmann constant, joule per mole per kelvin T = 300; // Thermal energy of specimen, joule mu_B = 9.27e-024; // Bohr's magneton, ampere per metre square mu_0 = 4*%pi*1e-07; // M...
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रोना V;PROG;MASC;SG;2;LGSPEC3 तैरना V;HAB;MASC;SG;3;LGSPEC3 मिटना V;PRF;FEM;PL;3;LGSPEC4 लौटना V;PRF;FEM;SBJV;SG;2 उठना V;PRF;FEM;PL;1;LGSPEC3 बरसना V;PROG;MASC;PL;1;LGSPEC4 लड़ना V;PROG;MASC;SG;3;LGSPEC4 भागना V;PRF;FEM;PL;2;PST मुड़ना V;HAB;FEM;SG;2;PRS बेचना V;HAB;FEM;SBJV;SG;1 घूमना V;PRF;MASC;SBJV;PL;2 प्राप्त करन...
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clc clear //Input data T2=298;//Maximum temperature at which CO2 machine works in K T1=268;//Minimum temperature at which CO2 machine works in K sf1=-0.042;//Liquid entropy at 268 K in kJ/kg K hfg1=245.3;//Latent heat of gas at 268 K in kJ/kg sf2=0.251;//Liquid entropy in kJ/kg K hfg2=121.4;//Latent heat of gas...
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//To find speed of shaft clc //Given: N1=150 //rpm d1=750, d2=450, d3=900, d4=150 //mm //Solution: //Calculating the speed of the dynamo shaft when there is no slip N41=N1*(d1*d3)/(d2*d4) //rpm //Calculating the speed of the dynamo shaft whne there is a slip of 2% at each drive s1=2, s2=2 //% N42=N1*(d1*d3)/(...
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clc; //Assuming SI units for all quantities //given signal is x(t)=1 //energy of signal x(t) t0=0;t1=5; x=1; y=integrate('x^2','t',t0,t1); disp(+'joule',y,'energy of signal x(t)='); //to find correlation coefficient we have to calculate the energies of different given signals e5=integrate('((%e)^(-t))^2','t',t0,t1); d...
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clc //initialisation of variables h1=176.48 //under -25 degrees temp in kj/kg h2=215.17 //kj/kg h4=74.59 //kj/kg //CALCULATIONS re=h1-h4 w=h2-h1 cop=re/w //RESULTS printf('the refrigeration effect is %2fkj/kg',re) printf('\ncoefficient of performance is %2f',cop)
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clc(); clear; //To determine the length of a box h=6.626*10^-34; //plancks constant n=3; //for second excited state m=1.67*10^-27; //mass of proton E=0.5; //energy ...
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// Calculate surface energy of copper crystal of type {111} clc E = 56.4 // bond energy in KJ N_a = 6.023e23 // Avogadro’s number n = 12 // number of bonds m = 3 // number of broken bonds N = 1.77e19 // number of atoms in copper crystal of type {111} per m^2 printf("\n Example 6.5") b_e = 1/2*E*1e3*n/N_a // bo...
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clc; Te=200; s=0.04; c=4; //given multiplying factor of leakage reactance //3V*V=a*WS a=Te*s*(((1+(1/s))*(1+(1/s)))+((c+c)*(c+c))); Test=a*(1/((1+1)*(1+1)+(c+c)*(c+c))); Tem=a*(1/2)*(1/(1+sqrt((1)*(1)+(c+c)*(c+c)))); mprintf('The starting torque is %f Nm \n The maximun Torque is %f Nm',Test,Tem);
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//Chapter-5, Illustration 6, Page 254 //Title: Air Compressors //============================================================================= clc clear //INPUT DATA D=0.2;//Bore in m L=0.3;//Stroke in m r=0.05;//Ratio of clearance volume to stroke volume P1=1;//Pressure at point 1 in bar T1=293;//Temperatu...
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// Ex 21 Page 363 clc;clear;close; // Given fc=25*10**6;//Hz fm=400;//Hz Vm=4;//V del=10*10**3;//Hz wc=2*%pi*fc;//rad/s wm=2*%pi*fm;//rad/s m=del/fm;//modulation index mf=m;mp=m;//modulation index printf("General equation of FM wave is:") printf("\n v=%d*sin(%.2e*t+%d*sin(%d*t)",Vm,wc,mf,wm) printf("\n\n General equat...
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//Load scripts from folder funcprot(0) getd("../scripts"); //Global variables imgPos = "../images/"; //The position of the source images renderPos = "render/"; //The folder where the render images will be saved //Load image imgin = readpbm(imgPos+"Earth.pbm"); //Assign a "seuillage" filter to the image imgout=seuil...
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// ELECTRICAL MACHINES // R.K.Srivastava // First Impression 2011 // CENGAGE LEARNING INDIA PVT. LTD // CHAPTER : 3 : TRANSFORMERS // EXAMPLE : 3.4 clear ; clc ; close ; // Clear the work space and console // GIVEN DATA S = 1.5; // Transformer Rating in KVA E1 = 220; ...
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// Example 1_5 clc;funcprot(0); // Given data D=0.07;// The diameter in m R=D/2;// The radius in m h=0.15;// The height in m L=(3/4)*h;// m rho=1000;// The density in kg/m^3 M=18;// kg/kg mole // Calculation m=(%pi*R^2*L)*rho;// The mass of water in the glass in kg n=m/M;// The number of moles in kg moles ...
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function [res, tab] = particionUniforme(f, a, b, rect) h = (b - a)/rect tab = [] x = a res = 0 for i=1:1:rect tab(1, i) = x; tab(2, i) = f(x); x = x + h; res = res + tab(2, i); end res = res * h; endfunction
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//clc() Ps = 100;//kPa A1 = 13.8587;//(1 = n-heptane) A2 = 13.8216;//(2 = n-hexane) B1 = 2911.32; B2 = 2697.55; C1 = 56.51; C2 = 48.78; //lnPs = A - B / ( T - C) T1 = B1 / (-log(Ps)+A1) + C1; T2 = B2 / (-log(Ps)+A2) + C2; x2 = 0.25;
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clc; clear; old=5000; bank=200; new=old-bank; mprintf("new fault =%dMVA",new);
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A=100;//cross sectional area of a stringer r=381;//radius of the fuselage,given in mm t=0.8;//thickness,given in mm Sy=100*10^3;//given in N d=150;//distence of action of load from center,given in mm
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clc //initialisation of variables clear w= 35.46 //gms T= 298.2 //K Qc= 4.03 //CALCULATIONS S= 4.576*(1.5*log10(w)+2.5*log10(T)+log10(Qc)-0.5055) //RESULTS printf ('Standard entropy = %.1f cal deg^-1 g atom^-1',S)
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[Read-the-docs-example-6] # Read the docs - example 6 user_indice: {'indice_name': 'my_indice', 'calc_operation': 'min','logical_operation': 'gt', 'thresh': 273.15, 'date_event': True} in_files: ['tas_day_MPI-ESM-LR_historical_r1i1p1_19500101-19591231.nc', 'tas_day_MPI-ESM-LR_historical_r1i1p1_19600101-19691231.nc', 't...
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function [t,y] = ode_euler_PC(f,tspan,y0,N) if (~exists("N", "local")) | (N <= 0) N = 100 end if ~exists("tspan","local") y0 = 0 end h = (tspan(2) - tspan(1))/N t = tspan(1) + [0:N]'*h // check y0, transpose if needed if size(y0,1) ~= 1 if size(y0,2) > 1 y0 = y0' end ...
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s=[0:1:9]; a = 1^s; a=a($:-1:1); b = 2^s; b=b($:-1:1); c = 3^s; c=c($:-1:1); d = 4^s; d=d($:-1:1); e = 5^s; e=e($:-1:1); f = 6^s; f=f($:-1:1); g = 7^s; g=g($:-1:1); h = 8^s; h=h($:-1:1); i = 9^s; i=i($:-1:1); j = 10^s; j=j($:-1:1); a b c d e f g h i j //A3 = [a3; b3; c3]; //AA = A3(1:2,$-1:$); //B3 = ([1:1:3]^3)' //BB ...
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clc //Chapter 1 //Ex1.12, page no 43 //Given mprintf('The given function is:\n f(t)=exp(j*wo*t) (for -inf<t<inf)\n')// Displaying the given function disp('exp(j*wo*t)= cos(wo*t)+j*sin(wo*t)') disp('F[exp(j*wo*t)]=F[cos(wo*t)]+F[j*sin(wo*t)]') disp('f[exp(j*wo*t)]=pi*[d(w-wo)+d(w+wo)-d(w+wo)+d(w-wo)]') disp('the...
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(unwatch all) (clear) (setgen 1) (dribble-on "Actual//dftmpfun.out") (batch "dftmpfun.bat") (dribble-off) (clear) (open "Results//dftmpfun.rsl" dftmpfun "w") (load "compline.clp") (printout dftmpfun "dftmpfun.bat differences are as follows:" crlf) (compare-files "Expected//dftmpfun.out" "Actual//dftmpfun.out" dftmpfun)...
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//Exa 2.20 clc; clear; close; //Given data : N=742;//No. of guided modes(unitless) n1=1.5;//unitlessnm alfa=2;//characteristic index profile NA=0.3;//unitless d=70;//in um a=d/2;//in um alfa=2;//Graded index profile for parabolic //Formula : N=(alfa/(alfa+2))/(v^2/2) v=sqrt(N*((alfa+2)/alfa)*2);//Unitless ...
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// chapter 7 example 13 //----------------------------------------------------------------------------- clc; clear; // given data f = 60*10^6; // frequency in Hz c = 3*10^8 // velocity of EM wave in m/s // Calculations lamda = c/f; // wavelength in m l_dipole= lamda/2 ...
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clc clear v=(4*10^5)//Velocity of the positively charged particle in m/s E=300//Intensity of electric field in N/C e=(1.6*10^-19)//Charge of the positively charged particle in C m=(1.67*10^-27)//Mass of the positively charged particle in kg q=35//Angle made by the particle in degrees //Calculations t=((v*sind...
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CAMERA Point(50, 50, 200) LookAt(50, 25, 0) FOV(60) Up(UP) LIGHT Type(OMNI) Point(50, 95, 50) Color(155, 155, 155) LIGHT Type(OMNI) Point(5, 95, 100) Color(255, 220, 200) LIGHT Type(OMNI) Point(95, 5, 100) Color(50, 50, 100) MATERIAL Color(255, 255, 255) Type(ORIGINAL) Metal(0) Specular(0) Lambert(0.85) Ambient(0.05)...
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//Transport Processes and Seperation Process Principles //Chapter 9 //Example 9.7-2 //Drying of Process Materials //given data X=(1/1000)*[195 150 100 65 50 40] R=0.01*[151 121 90 71 37 27] Rinv=[] for i=1:6 Rinv(i)=1/R(1,i); end v=inttrap(X,Rinv) X1=0.38; XC=0.195; X2=0.040; Rc=1.51; A=18.5...
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clc //initialisation of variables r=4 //CALCULATIONS w=log(r) //results printf(' change in entropy = % 1f R/J cal for each',w)
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//chapter11 //example11.11 //page210 V_Rc=1 gain_beta=45 Rc=1 // kilo ohm Ic=V_Rc/Rc //since gain_beta=Ic/Ib Ib=Ic/gain_beta printf("base current = %.3f mA",Ib)
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//Example5.6 // find the input offset current of an op-amp circuit clc; clear; close; Vo = 12*10^-3; // V // output voltage R1 = 2*10^3 ; // ohm // input resistence R2 = 150*10^3; // ohm // feedback resistence // the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is // Vo = R2*...
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//Exa 1.39 clc; clear; close; format('v',7); //Given Data : V=0.01653;//m^3 m=5.6;//Kg M=28;//Kg/Kgmol p=200;//bar Z=0.605; Rdash=8314.3;//J/Kgk R=Rdash/M;//J/Kgk //p*V=m*Z*R*T T=p*10^5*V/m/Z/R;//K disp(T,"Temperature in K : ");
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clear; clc; // Illustration 7.1 // Page: 222 printf('Illustration 7.1 - Page: 222\n\n'); // Solution // ****Data****// Temp1 = 273+26.1;// [K] P1 = 100;// [mm Hg] Temp2 = 273+60.6;// [K] P2 = 400;// [mm Hg] P = 200;// [mm Hg] //*****// deff('[y] = f12(T)','y = ((1/Temp1)-(1/T))/((1/Temp1)-(1/Temp...
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s = poly (0 , 's') ; G1 = 2/( s +2) ; G1 = syslin ('c', G1 ) ; G2 = 1/( s ^2 + 10* s + 1) ; G2 = syslin ('c', G2 ) ; t = 0:0.003:45; y1 = csim ('step', t , G1 ) ; y2 = csim ('step', t , G2 ) ; plot2d (t , [y1',y2']) ; legend([ 'first order'; 'second order'] ) ; xlabel ('Time (in secs )') ; ylabel...
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[[16,16,3],[0,8,4],[16,16,5]] [35,12,37] [99,20,101] [195,28,197] [323,36,325] [483,44,485] [675,52,677] [899,60,901] [1155,68,1157] [1443,76,1445] [1763,84,1765] [2115,92,2117] [2499,100,2501] [2915,108,2917] [3363,116,3365] [3843,124,3845] [[-16,16,-3],[0,8,-4],[-16,16,-5]] [-3,4,-5] [-35,12,-37] [-99,20,-101] [-195...
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clear; clc; //Example 8.9 Iso=3*10^-14; Isq=10^-13; b=75; Vt=0.026; Rl=8; P=5; Vp=sqrt(2*Rl*P); printf('\npeak voltage Vp=%.2f V\n',Vp) Vcc=Vp/0.8; printf('\nsupply voltage=%.2f V\n',Vcc) Ien=Vp/Rl; printf('\nemitter current=%.3f A\n',Ien) Ibn=Ien/(1+b); Ibn=Ibn*1000;//mA printf('\nbase current=%.2f mA...
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// Theory and Problems of Thermodynamics // Chapter 10 // Chemical Thermodynamics // Example 1 clear ;clc; //Given data // Combustion Reaction // C2H2 + 2.5*O2 + 3.76*2.5*N2 => 2*CO2 + H2O + 9.4*N2 afr1 = (2.5+3.76*2.5)/1 // air fuel ratio = (O2+N2)/C2H2 M_C2H2 = 26; M_air = 28.67; a...
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clc //initialisation of variables BeamB= 200 //mm BeamH= 200 //mm BeamT= 20 //mm BeamT2= 20 //mm sigmaxtop= 150 //N/mm^2 sigmaxbottom= -150 //N/mm^2 //CALCULATIONS y= ((BeamB*BeamT*BeamT/2)+(BeamH*BeamT2*((BeamH/2)+BeamT)))/(BeamB*BeamT+BeamH*BeamT2) Iz= ((BeamB*(y^3))/3)-(((BeamB-BeamT2)*(y-BeamT)^3)/3)+(Bea...
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// Example 1.58 clear; clc; close; format('v',7); // Given data IscByIfl=3*180/100;//ratio TstByTfl=0.35;//ratio X=80/100;//tapping //Calculations //Formula : TstByTfl=1/3*(IscByIfl^2)*Sfl Sfl=TstByTfl/IscByIfl^2*3;//slip at full load IstByIsc=X^2;//ratio IstByIfl=IstByIsc*IscByIfl;//ratio disp("Star...
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//Book - Power System: Analysis & Design 5th Edition //Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J. Overbye //Chapter - 12 ; Example 12.3 //Scilab Version - 6.0.0 ; OS - Windows clc; clear; M=500; //MVA rating of the generator f=60; //requency in Hertz R=0.05; ...
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// Scilab Code Ex9.4 Page:280 (2006) clc;clear; e = 1.6e-019; // Energy equivalent of 1 eV, J/eV E_g = 3.4e-04; // Energy gap of aluminium, eV v_F = 2.02e+08; // Fermi velocity of aluminium, cm/sec h_bar = 1.05e-034; // Planck's constant L = h_bar*v_F/(2*E_g*e); // Coherence Length of aluminium, cm ...
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//(14.9) As a result of heating, a system consisting initially of 1 kmol of CO2,.5 kmol of O2, and 5 kmol of N2 forms an equilibrium mixture of CO2, CO, O2, N2, and NO at 3000 K, 1 atm. Determine the composition of the equilibrium mixture. //solution //The overall reaction can be written as //CO2 + .5O2 + ...
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clc // Fundamental of Electric Circuit // Charles K. Alexander and Matthew N.O Sadiku // Mc Graw Hill of New York // 5th Edition // Part 1 : DC Circuits // Chapter 1 : Basic Concepts // Example 1 - 7 clear; clc; close; // // Given data V1 = 20.0000; V2 = 1...
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function [varargout] = mean3(varargin) % MEAN : Calculate the mean value. % For vectors: returns a value. % For matrices: returns the mean value of each column. for ii = 1:nargin %ii x=varargin{ii}; [ m , n ] = size ( x ); if m == 1 m = n ; end varar...
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//Exa 7.14 clc; clear; close; //Given data : format('v',7); //Applying KVL on +ve side V1=200-(600*0.015)-(100)*0.03;//in volt disp(V1,"Voltage at +ve side(in V): "); //Applying KVL on -ve side V2=200-(-100*0.03)-500*0.0015;//in volt disp(V2,"Voltage at -ve side(in V): "); //Note : answer of 2nd part is wr...
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//example 7 //heat transfer rate in aftercooler clear clc V1=0 //we assume initial velocity to be zero because its given that it enters with a low velocity V2=25 //final velocity with which carbon dioxide exits in m/s h2=401.52 //final specific enthalpy of heat when carbon dioxide exits in kJ/kg h1=198 //initial...
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//Example 1.61 //MAXIMA SCILAB TOOLBOX REQUIRED FOR THIS PROGRAM //Fourier transform of (3)^n u(n) clear; clc ; close ; syms n; x =(3) ^n; X= symsum (x,n ,0, %inf ) //Display the result in command window disp (X,"The Fourier Transform does not exit as x(n) is not absolutely summable and approaches infinity i...
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function [Tn]=chepol(n,var) //Recursive implementation of Chebychev polynomial // n :Polynomial order // var :Polynomial variable (character string) // Tn :Polynomial in var // //! //Author F.D. // Copyright INRIA T1=poly(0,var); T0=1+0*T1; if n==0 then, Tn=T0, return, end, if n==1 then, Tn=T1, ...
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// Exa 6.7 clc; clear; close; // given data h_f = 720.7;// in kJ h_fg = 2046.6;// in kJ v_g = 0.2405;// in m^3 x = 0.9; P = 8*10^2;// in kN/m^2 U_sat = h_f+x*h_fg-x*v_g*P;// in kJ disp(U_sat,"When the steam is wet, the internal energy in kJ is"); En = 2767.3;// Enthalpy of dry saturated stream U_sat = En-(...
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//Chapter 8 //Example 8_16 //Page 185 clear;clc; pc=53; v=106; loss=98; pl=110.9; cv=113; sq=sqrt(loss/pc); vc=(sq*v/sqrt(3)-pl/sqrt(3))/(sq-1); w=(cv/sqrt(3)-vc)^2/(v/sqrt(3)-vc)^2*pc; printf("Critical disruptive voltage = %.2f kV \n\n", vc); printf("Power loss at %.0f kV = %.0f kW \n\n", cv, w);
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//MissionB1 function missionB1(imgB1) disp("Résultat de la mission B1 :") //Version 1 //x=histplot(255,imgB1) //imgB1=imgB1*10 //Version 2: //On cherche les pixels qui se raprochant le plus de 255 greaterShadeofgray = max(imgB1); disp("Valeur de gris la plus haute: " + string(greaterShadeofgray)) ...
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Df=100//in mm bf=1250//in mm bw=250//in mm d=550//in mm Mu=400//in kN-m fy=415//in MPa fck=15//in MPa Asf=0.446*fck*(bf-bw)*Df/0.87/fy//in sq mm Muf=0.446*fck*(bf-bw)*Df*(d-Df/2)/10^6//in kN-m Muw=Mu-Muf//in kN-m //using Cu=Tu, 0.36 fck bw Xu = 0.87 fy Ast, Xu = a Asw a=0.87*fy/0.36/fck/bw //Muw=0.87 fy Asw...
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u=[2,3,-4]; v=[1,-5,8]; u+v 5*u -v 2*u-3*v u.*v; k=sum(u.*v); disp(k,'dot product of the two vectors') l=norm(u); disp(l,'norm or length of the vector u')
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//Finding the Recovery Energy in an Inductor with feedback Diode //Example 2.8 (Page No- 62) clc clear //given data Lm = 250*10^-6; //H N1 = 10; N2 = 100; Vs = 220; //V t1 = 50*10^-6;// s a= N2/N1; //part (a) vd = Vs*(1+a); printf('Reverse voltage of diode: %d V \n',vd); //part(b) I0 = (Vs/Lm)*t1; p...
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//finding required data// //example 36// clc //clears the command window// clear //clears// disp('f=A''B''C+A''BC''+AB''C+ABC''');//since f is 1 at positions 1,2,5,6;this is the required SOP expression// disp('f=(A+B+C)(A+B''+C'')(A''+B+C)(A''+B''+C'')');//since f is 0 at 0,3,4,7;this is the required POS express...
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clc,clear printf('Example 3.16\n\n') I=4 //no load current in amperes V=500 motor_input_no_load=I*V //no load motor input R_a=0.5,R_sh=250//resistance of armature and shunt field resistnace I_sh=V/R_sh I_a=I-I_sh arm_cu_loss_noload=R_a*I_a^2 //No-load armature copper losses constant_loss=motor_input_no_...
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//Example 7_8 clc(); clear; //To find out the tension in the string when the ball is at point A //As (T+W)=((m*v^2)/r) printf("Tension in the string is T=m*((v^2/r)-g)\n") printf("If v^2/r==g then the tension in the string is zero\n") printf("If v<sqrt(r*g) then the required centripetal force is less than the ba...
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// fsolve // The “fsolve” function solves systems of n nonlinear equations and // n unknowns. // lets solve an equation: // -2*x**3 + 5*x**2 + x = 2 // todo: plot eq x=<-1,3> // Where are roots approx? function y = problem(x) y = -2*x**3 + 5*x**2 + x - 2 // note the form: -2*x**3 + 5*x**2 + ...
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clc; pathname=get_absolute_file_path('8_5_soln.sce') filename=pathname+filesep()+'8_5_data.sci' exec(filename) // Solution: // capacity coefficient in English Units, Cv=Q/sqrt(del_p/SG_oil); //gpm/sqrt(psi) // capacity coefficient in Metric Units, Cv1=Q1/sqrt(del_p1/SG_oil); //Lpm/sqrt(kPA) // Results: printf("\n Re...
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//chapter 2 //half wave dipole printf("\n"); dl=1/15;//assume lamda=1; Rloss=1.5; Rrad=80*(%pi)^2*(1/15)^2; n=Rrad/(Rrad+Rloss); printf("the radiation efficiency is %g",n);
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//check o/p when the i/p vector contains only zeros k = [0 0 0 0 0]; r0 = 0.1; a = rc2ac(k,r0); disp(a); //output // 0.1 // 0. // 0. // 0. // 0. // 0. //
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clc; g=9.8; //gravitational constant in m/sec square m=5; //mass in kg F=100; //force in Newton disp(m*g,"Weight in Newton = "); a=F/m; //calculating acc. disp(a,"Accelaration in m/sec square = "); //displaying result
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clc A=[4 2 1 3;6 3 4 7;2 1 0 1] disp('rank of A is') p=rank(A) if p==4 then disp('equations have only a trivial solution:x=y=z=0') else disp('equations have infinite no. of solutions.') end
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//Example 3.9 clc; clear; close; format('v',9); //Given data : g=9.81;//gravity GH=4;//meter IJ=4;//meter IC=2;//meter GC=3;//meter AG=(10-4)/2;//meter BH=(10-4)/2;//meter EI=AG*IC/GC;//meter JF=AG*IC/GC;//meter EF=EI+IJ+JF;//meter A=(8+4)/2*2;//in m^2 a=4;//meter b=8;//meter d=2;//meter xbar=(2*a+b...
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//Ex:4.6 clc; clear; close; x=2;// index profile dl=0.0126;// index difference a=(85/2)*10^-6;// core radius R=2*10^-3;// curve of radius n1=1.45;// core refractive index k=6.28; y=850*10^-9;// wavelength in m A=(x+2)/(2*x*dl); B=(2*a/R); C=(3*y/(2*k*R*n1))^(2/3); D=B+C; E=A*D; F=1-E; Lm=-10*log(1-A*(B...