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//Example 12.3 //Program to Calculate the change in overall Gain of the Feedback //Amplifier with given Gain reduction clear; clc ; close ; //Given Circuit Data A=1000; //60dB, Voltage Gain B=0.005; //Negative Feedback dAbyA=-0.12; //dA/A = 12 % //Calculation dAfbyAf=1/(1+A*B)*dAbyA; //dAf/Af=1/(1+A*B)*dA/A //Displaying The Results in Command Window printf("\n\t The change in overall Gain of the Feedback Amplifier is, dAf/Af = %f which is equivalent to %f percent.",dAfbyAf,dAfbyAf*-100);
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//fiber optic communications by joseph c. palais //example 8.7 //OS=Windows XP sp3 //Scilab version 5.4.1 //given clc clear all Ps=3//source power in dBm NA_glass=0.24//numerical aperture for glass NA_plastic=0.48//numerical aperure for plastic loss_glass=12.4//loss for glass fiber in dB loss_plastic=6.4//loss for plastic fiber in dB reflectn_loss=0.2// reflection losses in dB atten_glass=5//attenuation in glass dB/Km atten_plastic=200//attenuation in plastic dB/Km L1=10*10^-3//fiber length in Km L2=100*10^-3//fiber length in Km //to find glass_coup_loss=Ps-(reflectn_loss + loss_glass)//Glass fiber coupling Loss in dBW mprintf('Coupling glass loss=%fdBW',glass_coup_loss) plastic_coup_loss=Ps-(reflectn_loss + loss_plastic)//plastic coupling fiber loss in dBW mprintf('\nCoupling loss in plastic=%fdBW',plastic_coup_loss) glass_cp= glass_coup_loss-atten_glass*L1//Coupled power in glass in dBW for 10m mprintf('\nCoupled power in glass=%fdBW after 10m',glass_cp) plastic_cp=plastic_coup_loss-atten_plastic*L1//Coupled power in plastic in dBW for 10m mprintf('\nCoupled power in plastic=%fdBW after 10m',plastic_cp) glass_cp= glass_coup_loss-atten_glass*L2// Coupled power in glass in dBW for 100m mprintf('\nCoupled power in glass=%fdBW after 100m',glass_cp) plastic_cp=plastic_coup_loss-atten_plastic*L2 // Coupled power in plastic in dBW for 100m mprintf('\nCoupled power in plastic=%fdBW after 100m',plastic_cp)
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//Ex8.5.1.;calculate: steam flow rate,cooling water flow,plant efficiency,Heat rate //Enthalpy at point 1 at (31 kg/cm^2)=669.6 kcal/kg //H1=H2=H3,enthalpy remain constant during throttling H1=669.7;//unit= kcal/kg H2=669.7;//unit= kcal/kg H3=669.7;//unit= kcal/kg //At point 3, P3=9.55;//unit= kg/cm^2 //specific volume vs3=0.22;//unit=m^3/kg //Entropy S3=1.580 T3=190;//unit=degree C,(degree of superheat=13 degree C) //S4_s at 0.34 kg/cm^2=S3 //x4_s=0.838 //and H4_s=hs+xL H4_s=72+(0.838*556) printf(" H4_s=%f kcal/kg",H4_s) //Isentropic turbine work=H3-H4_s ITW=H3-H4_s; printf("\n Isentropic turbine work=%f kcal/kg",ITW); //Actual turbine work ATW=0.80*ITW; printf("\n Actual turbine work=%f kcal/kg",ATW); H4=669.7-ATW; printf("\n H4=%f kcal/kg",H4) h5_6=72;//unit= kcal/kg; (Ignoring pump work) //sensible heat h7=h5=25 kcal/kg h5=25;//unit=kcal/kg h7=25;//unit=kcal/kg //Turbine steam flow TSF=(250*0.860*10^6)/(ATW*0.9); printf("\n Turbine steam flow=%f kg/hr",TSF); //let m4=TSF; //Turbine volume flow TVF=(TSF/60)*vs3; printf("\n Turbine volume flow=%f m^3/min",TVF); //cooling water flow m7:m7(h5_6-h7)=m4(H4-h5_6) m7=((H4-h5_6)/(h5_6-h7))*m4; printf("\n cooling water flow m7=%f kg/hr",m7); Heat_added=H1-h5_6; printf("\n Heat_added=%f kcal/kg",Heat_added); //plant efficiency=(Actual Turbine work*nmg)/Heat added //nmg=combined mechanical and electrical efficiency of turbine-generator nmg=0.90; Plant_efficiency=(ATW*nmg)/Heat_added; plant_efficiency=Plant_efficiency*100; printf("\n Plant Efficiency nplant=%f persent",plant_efficiency); //Plant heat rate=(860*Heat_added)/net_work //net_work=105.36*0.90 Plant_heat_rate=(860/Plant_efficiency); printf("\n Plant heat rate=%f kcal/kWH",Plant_heat_rate); //The value of "turbine steam flow" is wrong due to calculating mistak in textbook,due to which the further value related with it is given wrong //The values are corrected in this program
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clear;lines(0); A=diag([1,2,3]);B=[1;1;1];C=[2,2,2]; sys=syslin('c',A,B,C); sys("A") sys("C") [A1,B1,C1,D1]=abcd(sys); A1 systf=ss2tf(sys); [a,b,c,d]=abcd(systf) spec(a) c*b-C*B c*a*b-C*A*B
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// 10.01.21 function Out=Texvctr(N) Out='\value{'+Texctr(N)+'}'; endfunction;
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clc Db=10 disp("Db = "+string(Db)+" cm^2/s") //initializiation the value of one of parametere of the transistor. Bt=0.95 disp("Bt = "+string(Bt)) //initializiation the value of base transport factor of the transistor. tb=10^-7 disp("tb = "+string(tb)+" s") //initializiation the value of one of parametere of the transistor. Lp=(sqrt(Db*tb)) disp("Lp=(sqrt(Db*tb)))="+string(Lp)+" cm")//calculation WB=(Lp*(acosh(1/Bt))) disp("WB=(Lp*(acosh(1/Bt)))="+string(WB)+" cm")//calculation
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//CHAPTER 2- STEADY-STATE ANALYSIS OF SINGLE-PHASE A.C. CIRCUIT //Example 11 disp("CHAPTER 2"); disp("EXAMPLE 11"); //VARIABLE INITIALIZATION vdc=100; //DC voltage in Volts vac=100; //AC voltage in Volts f=50; //in Hertz I1=10; //in Amperes I2=5; //in Amperes //SOLUTION r=vdc/I1; z=vac/I2; xl=sqrt((z^2)-(r^2)); L=xl/(2*%pi*f); pf=r/z; disp(sprintf("The inductance of the coil is %f H",L)); disp(sprintf("The power factor of the coil is %f (lagging)",pf)); //END
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clc //initialisation of variables s= 1.03 w= 64.3 //lbf/ft^3 dg= 14 //ft A= 40 //ft^2 b= 5 //ft d= 8 //ft b1= 2.5 //ft y= 10 //ft x= 3 //ft z= 4 //ft //CALCULATIONS F= w*dg*A*16.05/36000 r= ((b*d^3/12)/(A*dg)) F3= F*b1/b F1= (F*(x-r)-(F/2)*x)/(y-z) F2= F-(F1+F3) //RESULTS printf (' force= %.3f tonf',F3) printf (' \n force= %.3f tonf',F2)
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function [acepted, rejected] = pca_recognizer(trainSamples, testSamples) brain = pca_train(trainSamples); acepted = list(); rejected = list(); for i=1: size(testSamples) sampl = testSamples(i); c = pca_test(imread(sampl.path), brain); if c == sampl.class then acepted($+1) = sampl; else rejected($+1) = sampl; end end endfunction function out = pca_train(samples) brain = tlist(["brain", "class", "samples"], [], list()); for i=1:Nc brain.class(i) = iClass(i); brain.samples(i) = list(); end for i=1:size(samples) sampl = samples(i); index = find(brain.class==sampl.class); class = brain.samples(index); class($+1) = imread(sampl.path); brain.samples(index) = class; end compiledBrain = tlist(["brain", "class", "sampl"], [], list()); for i=1:Nc nClass = brain.class(i); nSamples = brain.samples(i); A = zeros(Np*Cm,Ni); k=1; for j=1:Ni img = nSamples(j); A(:,k) = img(:); k = k+1; end compiledBrain.class(i) = nClass; compiledBrain.sampl(i) = pca_compress(A,d); end out = compiledBrain; endfunction function [class] = pca_test(imagem, brain) nc = size(brain.class, 1); for i=1:nc c = brain.sampl(i); y = c\imagem(:); diffs(i) = pca_diffImagem(imagem, c*y); end [j,k] = min(diffs); class = brain.class(k); endfunction function out=pca_compress(A, d) [U S V] = sva(A, d); out = U*S; endfunction function out = pca_diffImagem(A,B) out = norm(A(:) - B(:)); endfunction
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clc; N=14.01; //mass of N H=1.008; //mass of H m=N+(3*H); //calculating mass moles=1/m //cal moles v=moles*22.4; //cal vol disp(v,"Volume = "); //displaying result disp((1*1.32*373)/(1.2*273),"V2 = "); //displaying result
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clc //Initialization of variables k=516 //N/m m=1.67*10^-27 //kg //calculations v=sqrt(k/m) /(2*%pi) E=6.624*10^-34 *v //results printf("Separation between adjacent levels frequency, %.2e Hz",v) printf("\n Energy = %.2e",E)
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//Cálculo de impedâncias de entrada e saída do bloco filtro PB j = %i; f = 112.5; w = 2*%pi*f; Ro = 10; R1 = 18e3; R2 = 22e4; R3 = 39e3; C1 = 1.2e-9; C2 = 39e-3; Xc1 = 1/(j*w*C1); Xc2 = 1/(j*w*C2); Zeq1 = 1/((1/(R3+Xc1)) + (1/(R2))); Zeq2 = Zeq1 + Ro; Zeq3 = 1/((1/(Xc2))+(1/(Zeq2))); Zin = R1 + Zeq1 + Ro; Zeq4 = Zeq1 + Ro; Zout = 1/((1/(Xc2))+(1/(Zeq4)));
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//Example 22.2 clc syms T tau z n disp("f(t)=exp(-t/tau)") ft=exp(-n*T/tau)*z^(-n); Z=symsum(ft,n,0,%inf)
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function [x,iter] = falsapo(a,b,e,n) iter = 0; x = a; erro = 1; while erro > e & iter < n xant = x; x = a-(b-a)*h(a)/(h(b)-h(a)); if h(a)*h(x) < 0 b = x; else a = x; end iter = iter + 1; erro = abs((x-xant)/x); end endfunction
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errcatch(-1,"stop");mode(2);//Initilization of variables lAB=0.4 //m lBC=0.3 //m //Calculations alpha=atand(lAB/lBC) //degree //Results printf('The angle wich the force should make with the horizontal to keep the edge AB of the body vertical %f degree \n',alpha) //here alpha=theta exit();
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//Program for example 6 chapter 3 clear t='y' clc disp("Example 6: Simulate a calculator to evaluate mathematical expressions") disp("*****************************************************************") disp("Answer : ") disp("Continue...??? ") printf("Enter the expressions to be evaluated one by one\n") halt("") //clc() printf("Calculator simulation using bc command in Unix\n\n") while t=='y' xt=input("Expression :: ","string") if xt=="^c" then break end printf("\n%d \n\n",evstr(xt)) end //clc() printf("______ \n") printf(" | | | /\\ |\\ | | / | |\n") sleep(300) printf(" | |----| /--\\ | \\ | |< | |\n") sleep(300) printf(" | | | / \\ | \\| | \\ [___]\n") sleep(300) disp("Thank You") disp("*****************************************************************")
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//ANALOG AND DIGITAL COMMUNICATION //BY Dr.SANJAY SHARMA //CHAPTER 10 //DIGITAL MULTIPLEXERS clear all; clc; printf("EXAMPLE 10.1(PAGENO 469)"); //given X_1 = 4*10^3//first analog signal in Hz X_2 = 4.5*10^3//second analog signal in Hz //calculation //the highest frequency cmponent of the composite signal consisting among two signal is X_2 f_sMIN = 2*X_2; printf("\n\nThe minimum value of permissible sampling rate = %2f Hz",f_sMIN);
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//Chapter 12 : Solutions to the Exercises //Scilab 6.0.1 //Windows 10 clear; clc; //Solution for 4.14 A=[2 2 -1 -1;-1 0 0 0;-1 -1 1 0;0 1 -1 1] disp(A^-1)
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disp('To show that the given matrix P is orthogonal.') p=[.6 .8;.8 -.6] disp(p,'P=') disp('P is composed of two vectors.') p1=[.6;.8] p2=[.8;-.6] disp(p2,'p2=',p1,'p1=') disp('To show that the columns are orthonormal') disp('p1.p2=') s=p1'*p2 r=p1' disp(p2,'*',r,'=') disp(s,'=') if(s==0) disp('The columns of P are othonormal') end disp('||p1||=') disp(sqrt(p(1,1)^2+p(2,1)^2)) disp('||p2||=') disp(sqrt(p(1,2)^2+p(2,2)^2)) disp('Hence, ||p1||=||p2||=1. Thus P is an orthogonal matrix')
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clc clear printf("example 6.3 page number 215\n\n") //to find the flux and pressure difference D_AB=6.75*10^-5 //in m2/s Z=0.03 //in m R=8314 p_A1=5.5*10^4 //in Pa p_A2=1.5*10^4 //in Pa T=298 //in K N_A=D_AB*(p_A1-p_A2)/(R*T*Z); printf("flux = %f kmol/sq m s",N_A) //for partial pressure Z=0.02; //in m p_A2=p_A1-((N_A*R*T*Z)/D_AB); printf("\n\npressure = %f Pa",p_A2)
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function [new_descendants] = mutation(descendants, prob_indiv, prob_gene) [drows, dcols] = size(descendants); descend_matrix = descendants; for i = 1:drows prob_random = rand(); if (prob_random < prob_indiv) for j = 1:dcols prob_random = rand(); if (prob_random < prob_gene) if (descendants(i,j) == 1) then descend_matrix(i,j) = 0; else descend_matrix(i,j) = 1; end end end end end new_descendants = return(descend_matrix); endfunction
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//Example 16.9 I=700;//Intensity of sunlight (W/m^2) A=0.500;//Area of solar collector (m^2) t=4;//Time (h) t=t*60*60;//Time (s) E=I*A*t;//Energy (J) printf('a.Energy falling on solar collector = %0.2e J',E) R_area=200;//Ratio of old area to new area I_new=R_area*I;//New intensity after derivation (W/m^2) printf('\nb.Intensity of concentrated sunlight = %0.2e W/m^2',I_new) //Openstax - College Physics //Download for free at http://cnx.org/content/col11406/latest
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clc clear //INPUT DATA w=5.0;//weight of lead ball in lb cp=0.032;//specific heat of lead in Btu/lbdeg.F h=50;//height at which ball thrown in feets v=20;//vertical speed in ft/sec g=32;//accelararion due to gravity in ft/sec^2 //CALCULATIONS //half the kinetic energy is converted into heat after instant impact with ground u=(v^2)+2*g*h ke=(w/2*(u));//kinetic energy of the ball at ground T=ke/(2*32*778*w*cp);//rise of temperature in deg.F //OUTPUT mprintf('the rise in temperature is %3.2f deg.F',T)
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clear [x_1k, fs] = wavread("./week1/sample01.wav"); t = (0:length(x_1k) - 1) / fs; partition = 2; x_1k_part = x_1k(1:length(x_1k) / partition); t_part = (0:length(x_1k_part) - 1)/fs; plot(t, x_1k) xgrid(); legend("x_1k", 1); title("音声波形") xlabel("時間(s)") ylabel("振幅") figure; plot(t_part, x_1k_part) xgrid(); legend("x_1k_part", 1); title("音声波形") xlabel("時間(s)") ylabel("振幅")
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function GetPicInfo(currentfolder,picname) { rs=Map; rs.tpe='side'; rs.col=color(1,1,1); filecontent=''; if FileIsPresent(currentfolder+"\"+picname+".txt") then filecontent=readtextfile(currentfolder+"\"+picname+".txt"); if filecontent.length==0 then if FileIsPresent(currentfolder+"\_settings.txt") then filecontent=readtextfile(currentfolder+"\_settings.txt"); if filecontent.length>0 then { while filecontent.length>0 do { line=filecontent.split("~n"); id=line.split('='); if id=='TYPE' then rs.tpe=line; if id=='COLOR' then rs.col=color(ToScalar(line.split(',')),ToScalar(line.split(',')),ToScalar(line.split(','))); } } return(rs); }
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// Scilab Code Ex3.22:: Page-3.45 (2009) clc; clear; lambda1 = 5890e-008; // Wavelength of D1 line of Na, cm lambda2 = 5896e-008; // Wavelength of D2 line of Na, cm N = 3000/0.5; // No. of lines per cm of grating, lines/cm a_plus_b = 1/N; // Grating element, cm n = 1; // Order of diffraction for principal maxima // As (a+b)*sin(theta1) = n*lambda, solving for theta1 theta1 = asind(n*lambda1/(a_plus_b)); // Angle of diffraction for the principal maxima of D1 line, degrees theta2 = asind(n*lambda2/(a_plus_b)); // Angle of diffraction for the principal maxima of D2 line, degrees printf("\nThe angle of diffraction for the principal maxima of D1 line = %5.2f degrees", theta1); printf("\nThe angle of diffraction for the principal maxima of D2 line = %5.2f degrees", theta2); // Result // The angle of diffraction for the principal maxima of D1 line = 20.70 degrees // The angle of diffraction for the principal maxima of D2 line = 20.72 degrees
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disp('chapter 12 ex12.2') disp('given') disp("output =12V") Vo=12 disp("max load current=50mA") Il=.05 disp("Vsmin=Vo+3 V") Vsmin=Vo+3 disp( 'volts',Vsmin) disp("allowing Vrs=2V(p to p)") Vrs=2 disp("Vs=Vsmin+Vrs/2") Vs=Vsmin+Vrs/2 disp('volts',Vs) disp("let Vz=Vs/2") Vz=Vs/2 disp('volts',Vz) disp("Iz=20mA") Iz=.02 disp("R1=(Vs-Vz)/Iz") R1=(Vs-Vz)/Iz disp('ohms',R1) disp("R1=390 ohm std value") R1=390 disp("let I2>>Ibmax I2=50uA") I2=50*10^(-6) disp("R2=(Vo-Vz)/I2") Vz=8.2 R2=(Vo-Vz)/I2 disp('ohms',R2) disp("R2=68kohm std value") R2=68000 disp("I2=(Vo-Vz)/R2") I2=(Vo-Vz)/R2 disp('amperes',I2) disp("R3=Vz/Iz") R3=Vz/I2 disp('ohms',R3) disp("use 150 k ohm std value") R3=150000 disp("select C1=50uF") C1=50*10^(-6) disp("Q1 specification") disp("Vcemax=Vsmax=Vs+Vrs/2") Vcemax=Vs+Vrs/2 disp('volts',Vcemax) Ie=Il disp("P=Vce*Il=(Vs-Vo)*Il") P=(Vs-Vo)*Il disp('watts',P) disp("A 2N718 is a suitable device")
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@relation car @attribute Buying{vhigh,high,med,low} @attribute Maint{vhigh,high,med,low} @attribute Doors{2,3,4,5more} @attribute Persons{2,4,more} @attribute Lug_boot{small,med,big} @attribute Safety{low,med,high} @attribute Acceptability{unacc,acc,vgood,good} @inputs Buying,Maint,Doors,Persons,Lug_boot,Safety @outputs Acceptability @data unacc ? unacc ? unacc ? unacc ? unacc unacc unacc unacc unacc unacc unacc unacc unacc ? unacc ? unacc unacc unacc unacc unacc acc unacc acc unacc acc unacc acc unacc acc acc acc unacc unacc unacc unacc unacc unacc unacc unacc acc acc unacc unacc unacc unacc acc acc unacc unacc unacc unacc unacc unacc acc acc acc acc unacc ? acc ? unacc ? unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc ? unacc ? unacc ? acc ? unacc unacc unacc unacc unacc unacc unacc unacc unacc acc unacc acc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc acc acc acc acc acc acc unacc unacc unacc unacc unacc unacc unacc unacc unacc acc acc acc unacc unacc acc acc unacc unacc acc acc acc acc acc acc acc vgood unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc acc acc acc acc acc acc unacc unacc acc acc unacc unacc unacc unacc acc acc unacc ? acc vgood unacc unacc acc ? unacc unacc unacc unacc unacc unacc acc acc unacc unacc unacc unacc unacc unacc acc acc acc acc acc acc acc acc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc acc vgood unacc unacc unacc unacc acc acc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc unacc acc acc vgood acc unacc unacc unacc unacc unacc unacc good acc unacc unacc unacc ? unacc unacc unacc unacc good acc unacc ? unacc ? acc ? unacc unacc unacc unacc unacc unacc unacc unacc acc acc unacc ? unacc ? acc ? unacc ? unacc unacc unacc unacc unacc unacc acc acc unacc unacc acc acc vgood vgood unacc unacc acc acc acc acc unacc unacc good good good acc unacc unacc unacc unacc good acc vgood vgood unacc unacc unacc unacc vgood vgood unacc ? vgood ? vgood ? unacc unacc unacc unacc unacc unacc good good vgood vgood unacc ?
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clc; //page no 273 // prob no 6.3 // from the expresion given on the page no 272; (So/No)=(a+6n) dB where a=10log[3/[ln(1+u)]^2] //check the ollowing code for L=64 and L=256 L=input("enter the value of L = "); B=input("enter the bandwidth of signal in hertz = "); n=log2(L); Bt=n*B; u=100;//given a=10*log10(3/[log(1+u)]^2); SNR=(a+(6*n)); disp(SNR,"SNR ratio is = "); // Here the SNR ratio for the two cases are found out. The difference between the two SNRs is 12dB which is the ratio of 16. Thus the SNR for L=256 is 16 times the SNR for L=64. The former requires just about 33% more bandwidth compared to the later.
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function graph=Histogramme(image, nom_histogramme) //Création de la base du graphique (nom et couleur de fond) graph=figure("figure_name", nom_histogramme, "backgroundcolor",[1,1,1]) //Récupération taille de l'image [hauteur, largeur]=size(image) //Initialisation du tableau contenant les valeurs de l'histogramme nombre_pixels=zeros(256, 2) nombre_pixels(1,2)=0 for x=2:255 nombre_pixels(x,2)=x end //Incrémentation des pixels du tableau comportant les valeurs de l'histogramme for x=1:hauteur for y=1:largeur nombre_pixels(image(x,y)+1,1)=(nombre_pixels(image(x,y)+1,1))+1 end end //Créer la légende xtitle("Histogramme", "Niveau de gris", "Nombre de pixels") //Afficher le graphique plot2d(nombre_pixels) endfunction
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// Example No.1.5. // Page No.30. clc;clear; Y = 7.9*10^(10);//Young's modulus of quartz -[N/m^2]. d = 2650;//Density of quartz -[kg/m^3]. v = sqroot(Y/d);//Velocity of ultrasonic wave. printf("\nThe velocity of the ultrasonic waves is %.2f m/s",v);
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clear clc integrate('x*sin(x)^8*cos(x)^4','x',0,%pi)
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// @Harness: verifier // @Purpose: "Test for unresolved types" // @Result: "UnresolvedType @ 7:16" architecture unr_type_11 { operand-type A[5]: int [0,31]; subroutine foo(a: A): void { write : duck(a, 0); } }
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clc //to calculate increase in radius q=4.8*10^-8 //charge in coulomb r=10*10^-2 //radius in m epsilon0=8.85*10^-12 //C^2/N-m^2 P=10^5 //N/m^2 dr=(q^2)/(96*((%pi)^2)*(r^3)*epsilon0*P) disp("increase in radius is dr="+string(dr)+"m")
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//Example 7.25 clear; clc; //Given R=0.082;//gas constant in atm dm^3 K^-1 mol^-1 delTf=0.3;//freezing temperature depression in K Kf=1.86;//freezing point depression constant of waater m=0.1;//molality of acid solution in mol kg^-1 //To determine the degree of dissociation a=(delTf/(Kf*m))-1;//degree of dissociation mprintf('The degree of dissociation = %f',a); //end
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//Diameter of chimney(in m): D=1; //Height of chimney(in m): L=25; //Speed of wind(in kmph): s=50; //Density of air(in kg/m^3): d=1.23; //Viscosity of air(in kg/(m-s)): u=1.79*10^-5; //Pressure(in kPa): p=101;
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clc //ex7.3 N=343.392; //convert the integer and decimal parts into binary form separately B_1='101010111'; //for 343 from ex7.1 B_2='0.011001'; //for 0.392 from ex7.2 //combining these two B='101010111.011001'; //for N, given number disp(B,'binary form of 343.392')
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//Example 10.7 //Iterative Method //Page no. 326 clc;clear;close; x=[0;0;0]; printf('n\txn\t\tyn\t\tzn\n---------------------------------------------------------\n') for i=1:7 printf(' %i\t%.10f\t%.10f\t%.10f\n',i-1,x(1),x(2),x(3)) x(1)=0.7+0.1*x(2)^2-0.05*x(3)^2 x(2)=0.5-0.3*x(1)^2+0.1*x(1)*x(3) x(3)=1.2-0.4*x(2)^2-0.1*x(1)*x(2) end printf('\n\nThe solution is x = %.10f, y = %.10f and z = %.10f',x(1),x(2),x(3))
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//ques-5.10 //Calculating pH of ammonium hydroxide clc c=0.002;//normality of NH4OH d=2.3;//Percentage dissociation c1=c*(d/100);//hydroxide content c2=10^-14/c1;//hydrogen content p=-log10(c2);//pH printf("pH of ammonium hydroxide is %.4f.",p);
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//Exa 7.1 clc; clear; close; //given data lamda=2*10^-6;// in m C1=0.374*10^-15; T=2000+273;// in K' C2=1.4388*10^-2; //(a) // Formula Eb_lamda= (C1*lamda^-5)/[exp(C2/(lamda*T))-1] Eb_lamda= (C1*lamda^-5)/[exp(C2/(lamda*T))-1]; disp(Eb_lamda,"Monochromatic emissive power at 2 micro wavelength in W/m^2 is :"); //(b) // Formula lamda_max * T =2898 // in micro m K lamda_max= 2898/T;// in micro m disp(lamda_max,"Wave-length at which the emission is maximum in micro m"); //(c) Elamdab_max=1.285*10^-5*T^5;// in W/m^2-m disp(Elamdab_max,"Maximum emissive power in W/m^2-m : "); //(d) sigma=5.67*10^-8; E=sigma*T^4; disp(E,"Total emissive power in W/m^2 :"); //Note: Answer of part (a) in the book is wrong
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//Example 6.8, page 141 clc //part a w=(2*%pi*50)/3//angular f, rad/sec k=6000/w kw=6000//n-m, initial brakin torque Tf=300//n-m, fictional torque j=540//kg-m2 tr=(j/k)*log((kw+Tf)/Tf) //disp(tr) s=%e^((-k*tr)/j) //disp(s) temp=((j/k)*(kw+Tf)*(1-s))-((Tf*tr)) Nr=(1/(2*%pi*k))*temp //disp(Nr) printf("Time taken for rheostatic braking is %f s",Nr) //part b beta=3600/j motor_rest_time=w/6.67 //disp(motor_rest_time) rev=(1000/60)*.5*(motor_rest_time) printf("Number of revolutions made is %f",rev)
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clc //initialisation of variables R=8400//j/mole T1=273//k g=1.66 r=2 //CALCULATIONS T2=T1*r^(g-1) w=(R*(T1-T2))/(22400*(g-1)) wi=R*T1*log(1/r)/22400 //results printf(' \n amount of work done= % 1f J',w) printf(' \n isothermal work done= % 1f J',wi)
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<?xml version="1.0" encoding="utf-8"?> <test> <description>Channel Flow P=3 with convective link outflow BC</description> <executable>IncNavierStokesSolver</executable> <parameters>ChanFlow_m3_ConOBC.xml</parameters> <files> <file description="Session File">ChanFlow_m3_ConOBC.xml</file> </files> <metrics> <metric type="L2" id="1"> <value variable="u" tolerance="5e-6">1.03944e-05</value> <value variable="v" tolerance="2e-6">4.98402e-06</value> <value variable="p" tolerance="1e-6">2.85015e-05</value> </metric> <metric type="Linf" id="2"> <value variable="u" tolerance="4e-5">6.46273e-05</value> <value variable="v" tolerance="1e-5">1.86564e-05</value> <value variable="p" tolerance="1e-5">4.56732e-05</value> </metric> </metrics> </test>
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//Ex 4.3 clc;clear;close; format('v',6); fc=1;//MHz fm=5;//kHz m=60/100;//Modulation Pc=6;//kW RL=50;//W Pavg=Pc*(1+m^2/2);//kW(Average power delivered to load) disp("Part(a)"); disp(Pavg,"Average power of modulated signal(kW)"); PdB=10*log10(Pavg*1000);//dB disp(PdB,"Average power of modulated signal(dB)"); PdBm=10*log10(Pavg*10^6);//dBm disp(PdBm,"Average power of modulated signal(dBm)"); disp("Part(b)"); VS_RMS=sqrt(2*RL*Pavg*1000)/1000;//kV disp(VS_RMS,"RMS voltage of modulated signal(kV)"); Vp=sqrt(2)*VS_RMS;//V disp(Vp,"Peak value of modulated signal(kV)"); //Answer is wrong in the book.
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clc //initialisation of variables R= 8.31 //J/mol K T= 25 //C F= 96500 //coloums c= 0.02 //molar c1= 0.1 //molar c2= 1 //molar c3= 1 //molar E1= 1.486 //v //CALCULATIONS E= E1-R*(273+T)*2.3*log10(c*c1^2/(c2*c3))/(2*F) //RESULTS printf (' potential of the cell = %.3f v',E)
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(set-strategy depth) (unwatch all) ; smfmmix.clp test (clear) (open "sfmfmix.rsl" sfmfmix "w") (load "compline.clp") (dribble-on "sfmfmix.out") (load "sfmfmix.clp") (rules) (dribble-off) (printout sfmfmix "sfmfmix.clp differences are as follows:" crlf) (compare-files sfmfmix.exp sfmfmix.out sfmfmix) ; close result file (close sfmfmix)
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// Example 10-2 // Gain matrix using ppol and Ackermanns formula clear; clc; xdel(winsid()); //close all windows // please edit the path // cd "<path to dependencies>"; // exec("ackermann.sci"); A = [0 1 0; 0 0 1;-1 -5 -6]; B = [0; 0; 1]; P = [-2 + %i*4 , -2 - %i*4, -10]; K = ackermann(A,B,P);disp(K,'using ackermanns formula K = '); K = ppol(A,B,P); disp(K,'using ppol function K = ') // ackermann's formula is computationally tedious // and hence avoided
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errcatch(-1,"stop");mode(2);//Example 7.7 :stress and strain ; ; format('v',6) //given data : sigma=450;//in MPa epsilon=0.63; sigma_t=sigma*(1+epsilon); disp(sigma_t,"true stress,sigma_t(MPa) = ") epsilon_t=log(1+epsilon); disp(epsilon_t,"true strain,epsilon_t = ") exit();
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//while loop txt=''; i=0; while %T i=i+1; txt=txt+string(i)+' '; if i>=10 then break end end i,txt //for loop txt=''; for i=1:10 txt=txt+string(i)+' '; if i>=5 then break end end i,txt
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//Example 1.5 clear; clc; Rf=100*10^3;//Assuming Feedback Resistance Rf Vee=-15; //Imposing Vo=-(Rf/R1)Vi-(Rf/R2)(-15)=-10Vi+5 R1=Rf/10; R2=(Rf*15)/5; printf("Designed Function Generator :"); printf("\n R1=%.2f kohms",(R1/1000)); printf("\n R2=%.2f kohms",(R2/1000)); printf("\n Rf=%.2f kohms",(Rf/1000));
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clear //Given m=0.3 //Kg a=30 //degree B=0.15 //T g=9.8 //m/s**2 //Calculation // I=(m*g*tan(a*3.14/180.0))/B //Result printf("\n value of current is %0.2f A",I)
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12_3.sce
clc //Intitalisation of variables clear pa= 114.9 //mm pb= 238.3 //mm xa= 0.525 xb= 0.475 //CALCULATIONS xa1= xa*pa/((xa*pa)+(xb*pb)) xb1= 1-xa1 //RESULTS printf ('Mole fraction CCl4 = %.3f ',xa1) printf ('\n Mole fraction of SiCl4 = %.3f ',xb1)
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ex_4_2.sce
errcatch(-1,"stop");mode(2);//Example 4.2;//frational power ; ; format('v',4) p01=0.11;//from the graph p11=0.347;//from the graph disp(p01*100,"power for LP01 mode is (%) ") disp(p11*100,"power for LP11 mode is (%)" ) exit();
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sgolay.sci
function F = sgolay (p, n, m, ts) //This function computes the filter coefficients for all Savitzsky-Golay smoothing filters. //Calling Sequence //F = sgolay (p, n) //F = sgolay (p, n, m) //F = sgolay (p, n, m, ts) //Parameters //p: polynomial //n: odd integer value, larger than polynomial p //m: positive integer less than 2^31 or logical //ts: real or complex value //Description //This is an Octave function. //This function computes the filter coefficients for all Savitzsky-Golay smoothing filters of order p for length n (odd). //m can be used in order to get directly the mth derivative; ts is a scaling factor. //Examples //y = sgolay(1,3,0) //y = // 0.83333 0.33333 -0.16667 // 0.33333 0.33333 0.33333 // -0.16667 0.33333 0.83333 funcprot(0); rhs = argn(2) if(rhs<2 | rhs>4) error("Wrong number of input arguments.") end select(rhs) case 2 then F = callOctave("sgolay",p,n) case 3 then F = callOctave("sgolay",p,n,m) case 4 then F = callOctave("sgolay",p,n,m,ts) end endfunction
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//Example 4.9.b:resolution clc; clear; close; st=15;//steps r=5;//ohm rsw=5.5;//ohm tr=(st*r)+rsw;//ohm vr=1.61;//V i=vr/tr;//A e2=1.61;//V e1=2.40;//V rh=(e1-e2)/i;//ohm trn=11;//turns slwr=rsw/trn;//ohm vd=slwr*i;//V dv=100;//divisions vedv=(1/dv)*vd;// rs=vedv/5;// disp(rs*10^6,"resolution of the instrument is,(micro-V)=")
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ex6_5.sce
// Example 6.5 format('v',5) clc; clear; close; // given data R1= 10*10^3;// in Ω R2= 2.2*10^3;// in Ω R_C= 3.6*10^3;// in Ω V_CC= 10;// in V I_C= 1.1*10^-3;// in A // The base voltage V_B= R2*V_CC/(R1+R2);// in V // The collector voltage V_C= V_CC-I_C*R_C;// in V disp(V_B,"The base voltage in V is : ") disp(V_C,"The collector voltage in V is : ")
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Four Axis SCARA Robot.sce
// Code for 4 Axis SCARA Robot // Initialise clc; clear all; disp('For 4 axis SCARA Robot: ') for i=1:4 disp('For Joint No.') disp(i) theta(i)= input('Enter joint angle : '); d(i)=input('Enter joint distance : '); alpha(i)=input('Enter link twist angle: '); a(i)=input('Enter link length : '); end // Matrix Initialisation T=[cosd(theta(1)-theta(2)-theta(4)) sind(theta(1)-theta(2)-theta(4)) 0 a(1)*cosd(theta(1))+a(2)*cosd(theta(1)-theta(2)); sind(theta(1)-theta(2)-theta(4)) -cosd(theta(1)-theta(2)-theta(4)) 0 a(1)*sind(theta(1))+a(2)*sind(theta(1)-theta(2)); 0 0 -1 d(1)-d(3)-d(4); 0 0 0 1 ] disp('T=') disp(T)
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taylorSeno.sci
function [a] = taylorSeno(x0 , t , x) a = sin(x0); for i = 2 : t if modulo(i,4) == 0 a = a + (-cos(x0)*(x - x0)^(i -1))/ factorial(i -1) elseif modulo(i,4) == 1 a = a + (sin(x0)*(x - x0)^(i -1))/ factorial(i -1) elseif modulo(i,4) == 2 a = a + (cos(x0)*(x - x0)^(i -1))/ factorial(i -1) elseif modulo(i,4) == 3 a = a + (-sin(x0)*(x - x0)^(i -1))/ factorial(i -1) end end endfunction
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clear; clc; disp("--------------Example 8.2---------------") printf("Consider a circuit-switched network that connects computers in two remote offices of a private company. The offices are\nconnected using a T-1 line leased from a communication service provider. There are two 4 X 8 (4 inputs and 8 outputs)\nswitches in this network. For each switch, four output ports are folded into the input ports to allow communication between\ncomputers in the same office. Four other output ports allow communication between the two offices."); // display example explanation // display the figure clf(); xname("--------------Example 8.2----------------"); xset("font size",3); xstring(0,.9,"Circuit-switched network"); xstring(.12,.67,"4x8 switch"); xstring(.62,.67,"4x8 switch"); xstring(.3,.63,"T-1 line with 1.544 Mbps"); xrects([0 .1 .6;.89 .8 .8;.8 .1 .1;.6 .3 .3]); xpoly([.23 .28 .23],[.73 .68 .63],"lines",1); xpoly([.57 .52 .57],[.73 .68 .63],"lines",1); for i=0:2 xpoly([.2 .23],[.72-(i/25) .72-(i/25)]); xpoly([.57 .6],[.72-(i/25) .72-(i/25)]); end for i=0:3 xpoly([.11+(i/45) .11+(i/45)],[.5 .45-(i/25)]); xpoly([.11+(i/45) .09-(i/45)],[.45-(i/25) .45-(i/25)]); xpoly([.09-(i/45) .09-(i/45)],[.45-(i/25) .79-(i/12)]); xpoly([.69-(i/45) .69-(i/45)],[.5 .45-(i/25)]); xpoly([.69-(i/45) .71+(i/45)],[.45-(i/25) .45-(i/25)]); xpoly([.71+(i/45) .71+(i/45)],[.45-(i/25) .79-(i/12)]); xpoly([-.05 .1],[.79-(i/12) .79-(i/12)]); xpoly([.7 .85],[.79-(i/12) .79-(i/12)]); xstring(-.13,.79-(i/12),"Computer"); xstring(.86,.79-(i/12),"Computer"); xfarc(.085-(i/45),.795-(i/12),.01,.01,0,64*360); xfarc(.705+(i/45),.795-(i/12),.01,.01,0,64*360); end xset("thickness",2.5); xpoly([.28 .52],[.68 .68]);
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//EXAMPLE 8-22 PG NO-537-538 L=60*10^-3; C=150*10^-9; Fo=1/[2*%pi*(L*C)^0.5]; disp('ii) FREQUENCY (Fo) is = '+string (Fo) +' Hz ') R=670; B.W=R/L; disp('ii) BAND WIDTH (B.W) is = '+string (B.W) +' rad/sec ') FL=Fo-(1777.22/2); disp('ii) Lower Cut of Frequency (FL) is = '+string (FL) +' Hz ') Fu=Fo+(1777.22/2); disp('ii) Upper Cut of Frequency (Fu) is = '+string (Fu) +' Hz ')
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o//Book name: Fundamentals of electrical drives by Mohamad A. El- Sharkawi //chapter 3 //example 3.1 //edition 1 //publisher and place:Nelson Engineering clc; clear; Vrms=110; //source voltage of the circuit in volts alpha=90; //triggering angle in degree Vm=Vrms*(2)^(1/2); //maximum voltage in volts Vave=(Vm/(2*%pi))*(1+cosd(alpha)); R=(0.2*(Vave)^(2))+5; //load resistance in ohm Iave=Vave/R; //average current of the load disp(Iave,'The average current when the triggering angle 90 degree in ampere is:')
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27_11.sce
clc; clear; function yp=predprey(t,y) yp=[1.2*y(1)-0.6*y(1)*y(2);-0.8*y(2)+0.3*y(1)*y(2)]; endfunction t=0:10; y0=[2 1]; sol=ode(y0,0,t,predprey); count=0; for i=1:2:22 disp(count,"istep=") disp(count,"time=") disp(sol(i),"y1=") disp(sol(i+1),"y2=") disp("---------------------------------------------------") count=count+1; end
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clc T=300 //K k=8.617*10^-5 //eV/K q=1.6*10**-19 //C NA=10^18 //cm^-3 ND=10^15//cm^-3 ni=9.65*10^9 Vbi=(k*T)*log(NA*ND/ni^2) disp(Vbi,"the built in potential in V=")
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//given poles are -1, -2+%i , -2-%i and zero is s=-3 num=poly([-3],'s','roots'); den=poly([5 9 5 1 ],'s','coeff'); G=k*num/den; disp(G,"G(s)=") //to find k //G(0)=10 given k=(10*(0+1)*(0+0+5))/3 disp(k,"value of k is") disp(G,"transfer function is")
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example15.sce
//Chapter-11 example 15 //============================================================================= clc; clear; //input data Te = 0.2*10^-3;//echo time in sec PRF = 1000;//pulse repetitive Frequency in Hz Vo = 3*10^8;//Velocity of EM wave in m/s //Calculations R = (Vo*Te)/2;//Range of the target in m Runamb = (Vo/(2*PRF));//Maximum unambiguous Range in m //Output mprintf('Target range is %g Km\n Maximum Unambiguous Range is %g Km',R/1000,Runamb/1000); //=================end of program==============================================
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//Exa 2.6 clc; clear; close; //Given data deltaVBE=200;//in mVolt deltaIB=100;//in uA ri=deltaVBE*10^-3/(deltaIB*10^-6);//in Ohm disp(ri/1000,"Input resistane of transistor in kohm :");
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function [stk,txt,top]=sci_semilogy() // Copyright INRIA txt=[] RHS=[] for k=1:rhs RHS=[stk(top)(1),RHS] top=top-1 end stk=list('mtlb_semilogy'+'('+makeargs(RHS)+')','0','?','?','?')
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//Caption: Linear regression and Time series //Simple regression //Example10.9 //Page389 clear; clc; x = [1994,1995,1996,1997,1998,1999,2000,2001,2002];//Year X = x-1998; y = [10,12,15,27,33,38,44,49,60];//sales [b,a,sig]=reglin(X,y);//Linear Regression disp(b,'Regression coefficient b=') disp(a,'Regression coefficient a=') D = 2006; Y = a+b*(D-1998); disp(Y*100000,'Sales for the year 2006 in lakhs of tons Y=') //Result //Regression coefficient b= // // 6.3333333 // // Regression coefficient a= // // 32. // // Sales for the year 2006 in lakhs of tons Y= // // 8266666.7
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//convolution by polynomial method x=[4 1 3]; h=[2 5 0 4]; z=%z; n=length(x)-1:-1:0; X=x*z^n'; n1=length(h)-1:-1:0; H=h*z^n1'; y=X*H //effect of zero insertion on convolution h=[2 0 5 0 0 0 4]; x=[4 0 1 0 3]; y=convol(x,h) //effect of zero padding on convolution h=[2 5 0 4 0 0]; x=[4 1 3 0]; y=convol(x,h)
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//A Textbook of Chemical Engineering Thermodynamics //Chapter 5 //Some Applications of the Laws of Thermodynamics //Example 6 clear; clc; //Given: //The given numerical is theoretical and does not involve any numerical computation //end
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organfilename = xgetfile('*.smb', GL_SYS_DIR+'smb',title='open smb file'); [path,organfname,extension]=fileparts(organfilename); disp(organfname) Flag_windows=1; //open the smb file [fid, err] = mopen(organfilename, 'rb',0); if err //disp("File "+filename+" does not exist!"); return else [P,nbfaces,Faces] = Draw_read_organ(fid,Flag_windows); L=20; O=[0,0,0]; //test different rotation matrice w3=3.14159/2; Vz=[cos(w3) -sin(w3) 0;sin(w3) cos(w3) 0; 0 0 1]; ap=3.14159/2; Vy=[cos(ap) 0 sin(ap);0 1 0;-sin(ap) 0 cos(ap)]; //y-axis rotation matrix for axis angle Vx=[1 0 0; 0 cos(ap) sin(ap); 0 -sin(ap) cos(ap)]; V=Vx; V=eye(3,3); //V=[0.866 -0.5 0; 0 0 -1; 0.5 -0.866 0] C=[0;0.6;0]; [X,Y,Z,CP]=Draw_SMB(P,nbfaces,Faces,L,O,V,C); g=gcf(); //g=scf(); //g.figure_name=smb; g.visible='off'; drawlater(); for i=1:length(X) if size(X(i),2)>1 then //a single facet has problem of 'monotonous' plot3d(X(i),Y(i),list(Z(i),CP(i))) h=get("hdl"); h.hiddencolor=-CP(i); end end g.visible='on'; a=get("current_axes"); a.rotation_angles = [90,-90] ; drawnow(); end //close the file mclose(fid);
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// Example 11.3 : Time required for v_B to reach 4.5V // Consider sense-amplifier circuit uC_n=50*10^-6; //uC_n=u_n*C_ox (A/V^2) uC_p=20*10^-6; //uC_p=u_p*C_ox (A/V^2) W_n=12*10^-6; // (m) L_n=4*10^-6; // (m) W_p=30*10^-6; // (m) L_p=4*10^-6; // (m) v_B=4.5; // (V) C_B=1*10^-12; // (F) V_GS=2.5; // (V) V_t=1; // (V) deltaV=0.1; // (V) g_mn=uC_n*(W_n/L_n)*(V_GS-V_t); // (A/V) g_mp=uC_p*(W_p/L_p)*(V_GS-V_t); // (A/V) G_m=g_mn+g_mp; // (A/V) T=C_B/G_m; // (s) deltat=T*(log(v_B/V_GS)-log(deltaV)); disp(deltat,"The time for v_B to reach 4.5V (s)")
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<cmd> ./main_test/lexer_tst "echo just a check to test with spaces and "$'\t'"tabs"</cmd> <ref> cat ./lexer/test6.ref</ref> <ret> 0</ret>
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//Gaussian Quadrature Rule clc; clear; close(); format('v',10); funcprot(0); disp('Integral 0 to 2 exp(x)dx'); deff('[y]=f(t)','y=exp(t+1)'); b = 1; a = -1; x = poly(0,'x'); p = x^4 - 6*x^2/7+3/35; x1 = roots(p); A = [1 1 1 1;x1';(x1.^2)';(x1.^3)']; B = [(b-a);(b^2-a^2)/2;(b^3-a^3)/3;(b^4-a^4)/4]; C = inv(A)*B; I = C(1)*f(x1(1))+C(2)*f(x1(2))+C(3)*f(x1(3))+C(4)*f(x1(4)); disp(I,'Calculated integration : '); exact = integrate('exp(x)','x',0,2); disp(exact,'The exact value of intergation is :'); err = exact - I ; disp(err,'Error : ' );
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//Example 7.8. refer fig.7.16. clc format(6) IDSS=10*10^-3 VP=-3.5 Rth=120*10^3 //R1+R2=120 k-ohm ID=5*10^-3 VDS=5 RS=0.5*10^3 disp("Assume that the JFET is biased in the saturation region. Then the dc drain current is given by") disp(" ID = IDSS*(1-(VGS/VP))^2") VGS=VP*(1-(sqrt(ID/IDSS))) disp(VGS,"Therefore, VGS(V) =") // textbook answer is wrong disp("The voltage at the source terminal is") VS=(ID*RS)-5 disp(VS," VS(V) = (ID*RS) - 5 =") disp("The gate voltage is") VG=VGS+VS disp(VG," VG(V) = VGS + VS =") disp("The gate voltage is") disp(" VG = ((R2 / (R1 + R2))*10) - 5") R2=(Rth*(VG+5))/10 R2_1=R2*10^-3 disp(R2_1,"Therefore, R2(k-ohm) =") // textbook answer is wrong R1=Rth-R2 R1_1=R1*10^-3 disp(R1_1,"and R1(k-ohm) =") // textbook answer is wrong disp("The drain-to-source voltage is") disp("VDS = 5 - ID*RD - ID*RS - (-5)") RD=(10-VDS-(ID*RS))/ID RD1=RD*10^-3 disp(RD1," RD(k-ohm) = ") format(5) x=VGS-VP disp(x,"VGS - VP = ") // textbook has taken a different value hence the wrong answer in textbook disp("Here, since VDS > (VGS-VP), the JFET is biased in the saturation region, which satisfies the initial assumption")
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frame_left_estr.enable ="off" // Escolha dos nós BCtoDel = SelectNodesInMat(Restricoes.data) // Restrições de apoio a deletar Restricoes = DeleteItemInArrayStructure(BCtoDel,Restricoes) if isempty(Restricoes) then BotoesGeomet(4).enable='off' end frame_left_estr.enable ="on" if ~isempty(BCtoDel) then BotoesAnalise(2:3).enable = "off"; nao_calculado = 1; BotoesMatrizes.enable = "off" for i=1:3; frequencias(i).string = ""; end; end
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// Scilab Code Ex8.4: Page-398 (2008) clc; clear; NA = 0.5; // Numerical aperture of the optical fibre n1 = 1.54; // Refractive index of the core material n2 = sqrt(n1^2-NA^2); // Refractive index of the cladding in an optical fibre printf("\nThe refractive index of the cladding in the optical fibre = %4.2f", n2); // Result // The refractive index of the cladding in the optical fibre = 1.46
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function dout = rw(reg_info,din) m = size(reg_info,2); addr = reg_info(1:m-2); mask = reg_info(m-1); w1 = zeros(size(addr)); if mask == 255 if argn(2) == 1 dout = spi_read(addr(1),1); else spi_write(addr(1),din); dout = []; end else for offset = 0:7 if bitand(mask,2^offset) break; end end //r = 0; //for n = 1:size(addr,2) // r = r + spi_read(addr(n),1) * 2^(8*(n-1)); //end r = spi_read(addr(1),size(addr,2)) r = sum(r .* 2^(8*(0:size(addr,2)-1))); if argn(2) == 1 dout = bitand(r,mask)/2^offset; else maskb = 2^(8*size(addr,2))-1 - mask; w = bitand(din*2^offset,mask) + bitand(r,maskb); for n = 1:size(addr,2) w1(n) = modulo(floor(w/2^(8*(n-1))), 2^8) //spi_write(addr(n), modulo(floor(w/2^(8*(n-1))), 2^8)); end spi_write(addr(1),w1); dout = []; end end endfunction
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//Problem 40.21: The total loop inductance of an isolated twin power line is 2.185 μH/m. The diameter of each conductor is 12 mm. Determine the distance between their centres. //initializing the variables: L = 2.185E-6; // in H/m u0 = 4*%pi*1E-7; ur = 1; a = 0.012/2; // in m //calculation: //distance D D = a*%e^((L*%pi)/(u0*ur) - 0.25) printf("\n\n Result \n\n") printf("\ndistance D is %.2f m",D)
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parseRecurrence("2*n*a(n) +(-23*n+36)*a(n-1) +6*(-2*n+3)*a(n-2)=0") -> "2*n*a_0+(-23*n+36)*a_1+6*(-2*n+3)*a_2=0
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// Exa 1.22 clc; clear; close; // given : A=1+3*%i // complex no. A c=conj(A) // conjugate of complex no. A magnitude=sqrt((real(A))^2+(imag(A))^2) // magnitude of complex number A phi=atand(imag(A)/real(A)) // phase of complex number A in degrees disp(magnitude,"magnitude of complex number A:") disp(phi,"phase of complex number A in degrees:") disp(c,"conjugate of complex no. A:")
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vr=30; vl=40; v=sqrt((vr^2)+(vl^2)); theta=atan(vl/vr); disp("the polar form of the total voltage has a magnitude (in V) of") ; disp(v); disp("with a phase angle (in degrees) of"); disp(theta*180/%pi);
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Ex18_5.sce
//Introduction to Fiber Optics by A. Ghatak and K. Thyagarajan, Cambridge, New Delhi, 1999 //Example 18.5 //OS=Windows XP sp3 //Scilab version 5.5.2 clc; clear; //given V=2.64e-4;//Verdet constant for silica in deg/A N=30;//Number of turns of fiber I=1;//Current through the fiber in A Theta=V*N*I;//Corresponding rotation of plane of polarization in deg mprintf("\n Theta= %.2e deg",Theta);
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Ex6_15.sce
clear //Given l1=33.7 l2=51.9 //Calculation S1=l1/(100-l1) s11=l2/(100-l2) s=((s11*12)/S1)-12 R=s*S1 //Result printf("\n Value of R is %0.2f ohm \nValue of S is %0.1f ohm",R,s)
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Ex4_31.sce
clc clear //At 9.2 bar pressure x1=0.96; //Dryness Fraction Sf1=2.1038; //in kJ/kg K Sg1=6.6151; //in kJ/kg K //At 3.55 bar pressure Sf2=1.7327; //in kJ/kg K Sg2=6.9358; //in kJ/kg K Vg2=0.5173; //in m^3/kg //Now at 0.36 bar pressure Vg3=4.408; //in m^3/kg S1=Sf1+(x1*(Sg1-Sf1)); //As process is adiabatic S2=S1; //From steam table, Sg=6.9358 > S2 x2=(S2-Sf2)/(Sg2-Sf2); V2=x2*Vg2; //As volume remains constant V3=V2; x3=V3/Vg3; printf('The dryness fraction of steam: %2.3f',x3); printf('\n');
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Ptstart.sci
function P=Ptstart(Fig) P=Fig(1,:) endfunction
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Ex4_8.sce
clc //initialisation of variables p=1.227//kg v=90.27//m/s^2 p1=1.01*10^5//N/m^2 //CALCULATIONS Ps=p1+(1/2)*(p*v)//N/m^2 //RESULTS printf('The stagnation pressure is=% f N/m^2',Ps)
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Ex15_3.sce
// Iintilization of variables r=250 // m // radius of the curved road a_t=0.6 // m/s^2 // tangential acceleration a=0.75 // m/s^2 // total acceleration attained by the car // Calculations a_n=sqrt(a^2-a_t^2) // m/s^2 v=sqrt(a_n*r) // m/s // Using v=u+a*t u=0 t=v/a_t // seconds // Now using v^2-u^2=2*a*s s=v^2/(2*a_t) // m // Results clc printf('The distance traveled by the car is %f m \n',s) printf('The time for which the car travels is %f seconds \n',t)
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function Y=Xmin() global XMIN Y=XMIN; endfunction
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ch10_2.sci
// to calculate parameters of the ckt model, line current, power factor, shaft torque and efficiency clc; V_0=215; I_0=3.9; P_0=185; R_1=1.6; V_sc=85; I_sc=9.8; P_sc=390; X=(V_0/I_0)*2; //magnetisation reactance phi_sc=acosd(P_sc/(V_sc*I_sc)); I_e=V_sc/complex(0,X); I_SC=I_sc*complex(cosd(phi_sc*(-1)),sind(phi_sc*(-1))); I_m=I_SC-I_e; Z=V_sc/I_m; R_2=real(Z)-R_1; //real(Z)=R=R1+R2 disp(R_2,'R_2(ohm)'); disp(imag(Z),'X_1+X_2(ohm)'); n=1500; nn=1440; s=(n-nn)/n; a=1.55/s; b=1.55/(2-s); Z_ftot=(complex(0,X/2))*(complex(a+.8,imag(Z)/2))/((complex(0,X/2))+(complex(a+.8,imag(Z)/2))); Z_btot=(complex(0,X/2))*(complex(b+.8,imag(Z)/2))/((complex(0,X/2))+(complex(b+.8,imag(Z)/2))); Z_tot=Z_ftot+Z_btot; I_m=V_0/Z_tot; I_L=abs(I_m);disp(I_L,'line current(A)'); pf=cosd(atand(real(I_m)/imag(I_m))); disp(pf,'pf'); P_in=V_0*I_L*pf; I_mf=I_m*complex(0,X/2)/complex(39.55,59.12); I_mb=I_m*complex(0,X/2)/complex(1.59,59.12); T=(1/157.1)*(abs(I_mf)^2*38.75-abs(I_mb)^2*.79); P_m=157.1*(1-s)*T; P_L=185; P_out=P_m-P_L; eff=P_out/P_in; disp(eff*100,'efficiency(%)'); T_shaft=P_out/157.1; disp(T_shaft,'shaft torque(Nm)');
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Ch3_3_25.sce
clc disp("Example 3.25") printf("\n") disp("Find the DC collector voltage & voltage gain of circuit for Vi=50mV") printf("Given\n") //base current for Vbe=0.7 Ib=30*10^-6 Vbe=0.7 beta=80 //collector current Ic=beta*Ib //given from ckt Vcc=20 Rc=5.8*10^3 //writing KVL for Common Emitter circuit Vc=Vcc-(Ic*Rc) //for input characteristics delVi=delVb=50mV Vi=50*10^-3 delIb=5*10^-6 Ic1=beta*delIb //output voltage Vo=Ic1*Rc //voltage gain Av=Vo/Vi printf("Dc collector voltage \n%f volt\n",Vc) printf("voltage gain \n%f\n",Av)
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rout-mcast13.tst
description multicast routing with pim over bier addrouter r1 int eth1 eth 0000.0000.1111 $1a$ $1b$ ! vrf def v1 rd 1:1 label-mode per-prefix exit router lsrp4 1 vrf v1 router 4.4.4.1 bier 256 10 1 exit router lsrp6 1 vrf v1 router 6.6.6.1 bier 256 10 1 exit int lo1 vrf for v1 ipv4 addr 2.2.2.1 255.255.255.255 ipv6 addr 4321::1 ffff:ffff:ffff:ffff:ffff:ffff:ffff:ffff router lsrp4 1 ena router lsrp6 1 ena ipv4 pim ena ipv6 pim ena ipv4 pim join lo1 ipv6 pim join lo1 ipv4 pim bier 1 ipv6 pim bier 1 exit int lo2 vrf for v1 ipv4 addr 2.2.2.11 255.255.255.255 ipv6 addr 4321::11 ffff:ffff:ffff:ffff:ffff:ffff:ffff:ffff exit int eth1 vrf for v1 ipv4 addr 1.1.1.1 255.255.255.0 ipv6 addr 1234::1 ffff:: mpls enable mpls ldp4 mpls ldp6 router lsrp4 1 ena router lsrp6 1 ena ipv4 pim ena ipv6 pim ena ipv4 pim join lo1 ipv6 pim join lo1 ipv4 pim bier 1 ipv6 pim bier 1 exit router bgp4 1 vrf v1 address uni multi local-as 1 router-id 4.4.4.1 neigh 2.2.2.4 remote-as 1 neigh 2.2.2.4 update lo1 red conn exit router bgp6 1 vrf v1 address uni multi local-as 1 router-id 6.6.6.1 neigh 4321::4 remote-as 1 neigh 4321::4 update lo1 red conn exit ! addrouter r2 int eth1 eth 0000.0000.2222 $1b$ $1a$ int eth2 eth 0000.0000.2222 $2a$ $2b$ ! vrf def v1 rd 1:1 label-mode per-prefix exit router lsrp4 1 vrf v1 router 4.4.4.2 bier 256 10 2 red conn exit router lsrp6 1 vrf v1 router 6.6.6.2 bier 256 10 2 red conn exit int lo1 vrf for v1 ipv4 addr 2.2.2.2 255.255.255.255 ipv6 addr 4321::2 ffff:ffff:ffff:ffff:ffff:ffff:ffff:ffff exit int eth1 vrf for v1 ipv4 addr 1.1.1.2 255.255.255.0 ipv6 addr 1234::2 ffff:: mpls enable mpls ldp4 mpls ldp6 router lsrp4 1 ena router lsrp6 1 ena exit int eth2 vrf for v1 ipv4 addr 1.1.2.2 255.255.255.0 ipv6 addr 1235::2 ffff:: mpls enable mpls ldp4 mpls ldp6 router lsrp4 1 ena router lsrp6 1 ena exit ! addrouter r3 int eth1 eth 0000.0000.3333 $2b$ $2a$ int eth2 eth 0000.0000.3333 $3a$ $3b$ ! vrf def v1 rd 1:1 label-mode per-prefix exit router lsrp4 1 vrf v1 router 4.4.4.3 bier 256 10 3 red conn exit router lsrp6 1 vrf v1 router 6.6.6.3 bier 256 10 3 red conn exit int lo1 vrf for v1 ipv4 addr 2.2.2.3 255.255.255.255 ipv6 addr 4321::3 ffff:ffff:ffff:ffff:ffff:ffff:ffff:ffff exit int eth1 vrf for v1 ipv4 addr 1.1.2.3 255.255.255.0 ipv6 addr 1235::3 ffff:: mpls enable mpls ldp4 mpls ldp6 router lsrp4 1 ena router lsrp6 1 ena exit int eth2 vrf for v1 ipv4 addr 1.1.3.3 255.255.255.0 ipv6 addr 1236::3 ffff:: mpls enable mpls ldp4 mpls ldp6 router lsrp4 1 ena router lsrp6 1 ena exit ! addrouter r4 int eth1 eth 0000.0000.4444 $3b$ $3a$ ! vrf def v1 rd 1:1 label-mode per-prefix exit router lsrp4 1 vrf v1 router 4.4.4.4 bier 256 10 4 exit router lsrp6 1 vrf v1 router 6.6.6.4 bier 256 10 4 exit int lo1 vrf for v1 ipv4 addr 2.2.2.4 255.255.255.255 ipv6 addr 4321::4 ffff:ffff:ffff:ffff:ffff:ffff:ffff:ffff router lsrp4 1 ena router lsrp6 1 ena ipv4 pim ena ipv6 pim ena ipv4 pim join lo1 ipv6 pim join lo1 ipv4 pim bier 4 ipv6 pim bier 4 exit int lo2 vrf for v1 ipv4 addr 2.2.2.14 255.255.255.255 ipv6 addr 4321::14 ffff:ffff:ffff:ffff:ffff:ffff:ffff:ffff exit int eth1 vrf for v1 ipv4 addr 1.1.3.4 255.255.255.0 ipv6 addr 1236::4 ffff:: mpls enable mpls ldp4 mpls ldp6 router lsrp4 1 ena router lsrp6 1 ena ipv4 pim ena ipv6 pim ena ipv4 pim join lo1 ipv6 pim join lo1 ipv4 pim bier 4 ipv6 pim bier 4 exit router bgp4 1 vrf v1 address uni multi local-as 1 router-id 4.4.4.4 neigh 2.2.2.1 remote-as 1 neigh 2.2.2.1 update lo1 red conn exit router bgp6 1 vrf v1 address uni multi local-as 1 router-id 6.6.6.4 neigh 4321::1 remote-as 1 neigh 4321::1 update lo1 red conn exit ipv4 multi v1 join 232.2.2.2 2.2.2.1 ipv6 multi v1 join ff06::1 4321::1 ! r1 tping 100 20 2.2.2.2 /vrf v1 /int lo1 r1 tping 100 20 4321::2 /vrf v1 /int lo1 r1 tping 100 20 2.2.2.3 /vrf v1 /int lo1 r1 tping 100 20 4321::3 /vrf v1 /int lo1 r1 tping 100 20 2.2.2.4 /vrf v1 /int lo1 r1 tping 100 20 4321::4 /vrf v1 /int lo1 r2 tping 100 20 2.2.2.1 /vrf v1 /int lo1 r2 tping 100 20 4321::1 /vrf v1 /int lo1 r2 tping 100 20 2.2.2.3 /vrf v1 /int lo1 r2 tping 100 20 4321::3 /vrf v1 /int lo1 r2 tping 100 20 2.2.2.4 /vrf v1 /int lo1 r2 tping 100 20 4321::4 /vrf v1 /int lo1 r3 tping 100 20 2.2.2.1 /vrf v1 /int lo1 r3 tping 100 20 4321::1 /vrf v1 /int lo1 r3 tping 100 20 2.2.2.2 /vrf v1 /int lo1 r3 tping 100 20 4321::2 /vrf v1 /int lo1 r3 tping 100 20 2.2.2.4 /vrf v1 /int lo1 r3 tping 100 20 4321::4 /vrf v1 /int lo1 r4 tping 100 20 2.2.2.1 /vrf v1 /int lo1 r4 tping 100 20 4321::1 /vrf v1 /int lo1 r4 tping 100 20 2.2.2.2 /vrf v1 /int lo1 r4 tping 100 20 4321::2 /vrf v1 /int lo1 r4 tping 100 20 2.2.2.3 /vrf v1 /int lo1 r4 tping 100 20 4321::3 /vrf v1 /int lo1 r1 tping 100 20 2.2.2.14 /vrf v1 /int lo2 r1 tping 100 20 4321::14 /vrf v1 /int lo2 r4 tping 100 20 2.2.2.11 /vrf v1 /int lo2 r4 tping 100 20 4321::11 /vrf v1 /int lo2 r1 tping 100 10 232.2.2.2 /vrf v1 /int lo1 r1 tping 100 10 ff06::1 /vrf v1 /int lo1
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14_1.sce
clc //initialisation of variables clear H1=26 //B.Th.U /lb H2= -27.5 //BThU/lb //CALCULATIONS H= H1-H2 t=-83//f s=3.43//ft^3/lb v=224*sqrt(H) //RESULTS printf ('Final temperature= %.f F',t) printf (' \n Final specific volume = %.2f ft^3/Lb',s) printf (' \n Hd = %.1f B.Th.U/lb',H) printf (' \n velocity of the air = %.f ft/sec',v+7)
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Ex6_28.sce
clear // // // //Variable declaration delta_t=10^-12; //life time(s) hby2pi=1.054*10^-34; e=1.6*10^-19; //charge of electron(c) //Calculations deltaE=hby2pi/(2*e*delta_t); //uncertainity in energy(eV) //Result printf("\n uncertainity in energy is %0.1f *10^-4 eV",deltaE*10^4)
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Example11_13.sce
//Example 11.13 clc; clear; close; format('v',7); //Given data : w=1.5;//m d=0.75;//m Cd=0.64;//Coeff of discharge QT=45;//cumec h=8;//meter A=w*d;//m^2 g=9.81;//gravity acceleration Q=Cd*A*sqrt(2*g*h);//m^3/sec n=QT/Q;//no. of spillways disp(round(n),"No. of spillways : ");
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Ex_8_3_b.sce
// Example 8.3.b //duration of time slot clc; clear; close; wd=8;//bit wide ts=32;//time slots nb=ts*wd;//no. of bits in a frame nf=8*10^3;//no. of frames tr=nf*nb;//transmission rate bdr1=1/tr;//bit duration bdr=bdr1*wd;// disp(bdr*10^6,"duration of time slot in micro seconds")
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// problem 9.7 Hs=5 Ls=10 D=0.15 d=0.1 N=30/60 s=0.15 g=9.81 W=2*3.142*N w=9810 ha=10.3 z=(D/d)^2 H=(Ls*z*W*W*s/g) Ph=Hs+H Phabs=ha-Ph f=0.01 Hfs=(4*f*Ls/(d*2*g))*((z*W*s)^2) H1=Hs+Hfs H1abs=ha-H1 H2=Hs-H H2abs=ha-H2 Hd=15 Ld=25 H11=(Ld*z*W*W*s/g) H12=H11+Hd H12abs=ha+H12 Hfd=(4*f*Ld/(d*2*g))*((z*W*s)^2) H22=Hd+Hfd H22abs=ha+H22 H3=Hd-H11 H3abs=ha+H3 a=3.142*D*D/4 Q=a*s*2*N power=(w*Q*(Hs+Hd+(0.6666*Hfs)+Hfd*0.6666))/1000 disp(H2abs,H1abs,"pressure head at middle and end of suction stroke") disp(H3abs,H22abs,H12abs,"pressure head at beginning,middle,end of suction stroke") disp(power,"power in Kw required to drive the pump")
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/2273/CH6/EX6.13/ex6_13.sce
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ex6_13.sce
//Find characteristics impedance and propogation constant and ABCD constants clear; clc; //soltion //FUNCTIONS function [z]=rxr(A,B)//Function for the multiplication of rectangular z(1)=A(1)*B(1) z(2)=A(2)+B(2) endfunction function [z]=rdr(A,B)//Function for the division in rectangular form z(1)=A(1)/B(1) z(2)=A(2)-B(2) endfunction function [v]=p2r(q)//Function for polar to rectangular v(1)=abs(q) v(2)=atand(imag(q)/real(q)) endfunction //given Z=complex(14.1,51.48); Y=complex(0,1.194*10^-3); l=200;//length of the line z=Z/l; y=Y/l; Zc=p2r(sqrt(z/y)); printf("Characteristics Impedance= %d∠%.2f° ohm\n",ceil(Zc(1)),Zc(2)); P=sqrt(z*y);//propogation constant printf("Propagation constants= %f + i%f\n",real(P),imag(P)); al=real(P)*l; bl=imag(P)*l; yl=P*l; A=p2r(cosh(yl)); printf("A = D = %.4f∠%.2f°\n",A(1),A(2)); B=rxr(Zc,p2r(sinh(yl))); printf("B= %.2f∠%.2f° ohm\n",B(1),B(2)); C=rdr(p2r(sinh(yl)),Zc); printf("C= %.6f∠%.2f° mho\n",C(1),C(2));