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//(Design against Fluctuating Load) Example 5.7 //Ultimate tensile strength of 40C8 Sut (N/mm2) Sut = 600 //Minimum value of N is Nmin Nmin = (10^3) //Maximum value of N is Nmax Nmax = (10^6) //Reversed bending stress acting on the bar Sf (N/mm2) Sf = 250 //Diameter of the bar d (mm) d = 50 //Reliability (%) reliability = 90 //Operation //Nomenclature: //1 - Ground //2 - Machined or cold drawn //3 - Hot-rolled //4 - Forged op = 4
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clc clear cd C:\Users\ybozorgi.NEOTEKENERGYDAL\Desktop\Yaser\codes\TruthData M = read_csv("WeeklyReport 12-20-16.csv"); exec('C:\Users\ybozorgi.NEOTEKENERGYDAL\Desktop\Yaser\codes\TruthData\TruthDataAnalysis\TruthData.sci', -1) M = TruthData(M); //parsing and organizing the data s=size(M); StartDAYY=M(1,3); StartTIMEE=M(1,4) StopDAYY=M(s(1),6); StopTIMEE=M(s(1),7) //start=strcat([StartDAYY ' ' StartTIMEE]); start=strcat([StartDAYY+' '+StartTIMEE]); //stop=strcat([StopDAYY ' ' StopTIMEE]); stop=strcat([StopDAYY+' '+StopTIMEE]); exec('C:\Users\ybozorgi.NEOTEKENERGYDAL\Desktop\Yaser\codes\TruthData\TruthDataAnalysis\ParseWFADate.sci', -1) on= ParseWFADate(start); initialTime=on; off = ParseWFADate(stop); elapsedDAY=abs(etime(on,off))/60/60/24; //based on truth data for i=1:1:s(1) StartDAY(i)=M(i,3); StartTIME(i)=M(i,4) StopDAY(i)=M(i,6); StopTIME(i)=M(i,7) //start=strcat([StartDAY(i) ' ' StartTIME(i)]); start=strcat([StartDAY+' '+StartTIME]); //stop=strcat([StopDAY(i) ' ' StopTIME(i)]); stop=strcat([StopDAY+' '+StopTIME]); on= ParseWFADate(start); off = ParseWFADate(stop); dur(i)=etime(off,initialTime)/60/60/24; //startime is based on traces (it is one number) //on is beased on truth data x(i)=dur(i) oil(i)=strtod(M(i,11)); water(i)=strtod(M(i,12)); gas(i)=strtod(M(i,13)) end PP=tabul(M(:,2)); Wells=PP(1); //tabulate all the wells FrequencyOfWells=PP(2); //how many times each well is repeated L=size(PP(2)); //number of wells for i=1:1:L(1) J=grep(M(:,2),Wells(i)); S(i,1:1:FrequencyOfWells(i))=J; end j=0 for i=1:1:L(1) for k=1:1:max(PP(2)) if (S(i,k) ~=0) then j=j+1; N(1,j) = S(i,k); //organizing the ID of wells in order end end end x=0 for i=1:1:L(1) x=FrequencyOfWells(i)+x X(i)=x; end disp ("Number of Wells is: ") disp( L(1)) plot(1:1:j,(oil(N)./(oil(N)+water(N)))','-o') title('Oil Fraction in Liquid'); a=get("current_axes")//get the handle of the newly created axes a.axes_visible="on"; // makes the axes visible a.font_size=4; //set the tics label font size plot([X X]', [0 max(oil(N)./(oil(N)+water(N)))]','k-.') scf() plot(1:1:j,(oil(N)./(1000*.178*gas(N)+oil(N)+water(N)))','-o') title('Oil Fraction in Total') a=get("current_axes")//get the handle of the newly created axes a.axes_visible="on"; // makes the axes visible a.font_size=4; //set the tics label font size plot([X X]', [0 max(oil(N)./(1000*.178*gas(N)+oil(N)+water(N)))]','k-.') scf() plot(1:1:j,(oil(N)+water(N))','-o') title('Total Liquid Production') a=get("current_axes")//get the handle of the newly created axes a.axes_visible="on"; // makes the axes visible a.font_size=4; //set the tics label font size plot([X X]', [0 max(oil(N)+water(N))]','k-.') scf() plot(1:1:j,oil(N)','-o') title('Oil Production') a=get("current_axes")//get the handle of the newly created axes a.axes_visible="on"; // makes the axes visible a.font_size=4; //set the tics label font size plot([X X]', [0 max(oil(N))]','k-.') scf() plot(1:1:j,water(N)','-o') title('Water Production') a=get("current_axes")//get the handle of the newly created axes a.axes_visible="on"; // makes the axes visible a.font_size=4; //set the tics label font size plot([X X]', [0 max(water(N))]','k-.') //plot([10 10]', [-100 100000]','k-.') scf() plot(1:1:j,gas(N)','-o') title('gas Production') a=get("current_axes")//get the handle of the newly created axes a.axes_visible="on"; // makes the axes visible a.font_size=4; //set the tics label font size plot([X X]', [0 max(gas(N))]','k-.') scf() plot(1:1:j,(gas(N)./oil(N))','-o') title('gas/oil Production') a=get("current_axes")//get the handle of the newly created axes a.axes_visible="on"; // makes the axes visible a.font_size=4; //set the tics label font size plot([X X]', [0 200]','k-.')
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Vo = 8.4; Vh = 14; n1 = Vo/22.4; n2 = Vh/22.4; R = 8.31; x1 = n1/(n1+n2); x2 = n2/(n1+n2); S = -R*(n1*log(x1)+n2*log(x2)); disp("J/K",S,"Entropy change for the process is")
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//Caption:Determine point of attachment & length of stub. //Exa 3.6 clc; clear; close; Z_l=100;//in ohms Z_o=600;//in ohms f=100*10^6;//in Hz wl=(3*10^8)/f; //Position of stub is : m=((Z_l*Z_o)/(Z_l-Z_o))^0.5; pos={wl/(2*%pi)}*atan((Z_l/Z_o)^0.5);//in meters l={wl/(2*%pi)}*{atan(m)};//in meters disp(pos,"Position of stub (in meters) ="); disp(abs(l),"Length of stub (in meters) =");
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//Chapter 9 Ex 8 clc; clear; close; n=1; //assuming expr=((243^(n/5))*(3^(2*n+1)))/((9^n)*(3^(n-1))); mprintf("The value of expression is %d",expr);
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Probabilistic Output. Iris-setosa,Iris-versicolor,Iris-virginica, 1.0, 0.0, 0.0, 1.0, 0.0, 0.0, 1.0, 0.0, 0.0, 1.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.8, 0.2, 0.0, 1.0, 0.0, 0.0, 1.0, 0.0, 0.0, 1.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.2, 0.8, 0.0, 0.0, 1.0, 0.0, 0.8, 0.2, 0.0, 0.0, 1.0, 0.0, 0.2, 0.8,
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//Example 1.9(b) clear; clc; rd=2*10^6;//Input Resistance ro=75;//Output Resistance a=200*10^3;//Open loop Gain printf("Note (as mentioned in the book): Because of much larger value, we simply ignore the exact calculations and use only the approximations."); //R12=infinity R2=0; //b2=R12/(R12+R22) (Feedback Factor)will be equal to 1 as R12 tends to infinity and R22 is 0 b=1;//Feedback Factor T=a*b;//reteurn ratio or loop gain //Aapprox=(1+(R22/R12))*(1/(1+(1/T2)))(Approximate Gain) but R22/R12=0 Trec=1/T; Aden=(1+Trec); Anum=1; Aapprox=Anum/Aden;//Approximate Gain Riapprox=rd*(1+T);//Approximate Input Resistance Roapprox=ro/(1+T);//Approximate Output Resistance //Ideal Value of input resistance Ri2 is infinity and ideal value of output resistance Ro2 is 0. printf("\nApproximate Value of A is =%.f V/V",Aapprox); printf("\nIdeal Value of A is =%.2f V/V",1); printf("\nApproximate Value of Ri is =%.3f Gohms",Riapprox/10^9); printf("\nIdeal Value of Ri is infinity"); printf("\nApproximate Value of Ro is =%.3f uohms",Roapprox*10^6); printf("\nApproximate Value of Ro is =%.f ohms",0);
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clear; clc; l = 50;// feet b = 4;// feet P = 3;// tons per foot run t = 1/2;// inches b1 = 4;// inches d1 = 4;// inches h1 = 1/2;// inches d = 7/8;// inches f_s = 6;// tons/in^2 f_b = 12;// tons/in^2 P_s = 2*0.25*%pi*d^2 *f_s;// tons P_b = t*d*f_b;// tons R = P_b;// tons F = l*P*2/d1;// tons p_min = R*(l-0.5*b1)/F ;// inches printf('The minimum pitch required is p = %.2f inches, say %d inches',p_min,p_min);
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class hola: x=10 y=11 z=x-y def main: for x in y...15: if x>3 or x<4: print m elif x<0: print y else: print m x=9 print "dios" def hola(x): main()
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//Section-9,Example-3,Page no.-E.6 //To determine the concentration of Cd2+ ions in the given electrochemical cell. E=-0.02 E_0=0.04 n=2 C_Fe=0.1 x=C_Fe/(10^(n*(E_0-E))/(0.0591)) disp(x,'Concentration of Cd2+ ions in the given electrochemical cell(M)')
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n=10 S11=(n^2-1)/(n^2+1) S21=2*n/(n^2+1) S22=(1-n^2)/(1+n^2) S12=2*n/(n^2+1) A=[S11,S12;S21,S22] disp(A)
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H0NH3 = -11020; // cal/moleN2 H0H2 = 0; HN2 = 0; CpNH3 = 8.92;// cal/moleH2.K CpH2 = 6.992;// cal/moleN2.K CpN2 =6.984;// cal/moleNH3.K T = 423; //K TR = 298; //K
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clear;lines(0); ny=2;nu=3;nx=4;P=ssrand(ny,nu,nx);[A,B,C,D]=abcd(P); Kc=-ppol(A,B,[-1,-1,-1,-1]); //Controller gain Kf=-ppol(A',C',[-2,-2,-2,-2]);Kf=Kf'; //Observer gain cl=P/.(-obscont(P,Kc,Kf));spec(cl('A')) //closed loop system [J,r]=obscont(P,Kc,Kf); Q=ssrand(nu,ny,3);Q('A')=Q('A')-(maxi(real(spec(Q('A'))))+0.5)*eye(Q('A')) //Q is a stable parameter K=lft(J,r,Q); spec(h_cl(P,K)) // closed-loop A matrix (should be stable);
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clear; clc; printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 14.6 Page 909 \n')// Example 14.6 // Maximum Thickness of a bacteria laden biofilm, that may be siccessfully treated Dab = 2*10^-12 ;//[m^2/s] Diffusion coefficient Ca0 = 4*10^-3 ;//[kmol/m^3] Fixed Concentration of medication Na = -.2*10^-3 ;//[kmol/m^3.s] Minimum consumption rate of antibiotic k1 = .1 ;//[s^-1] Reaction Coefficient //For firsst order kinetic reaction Equation 14.74 m = (k1/Dab)^.5; L = m^-1*acosh(-k1*Ca0/Na); printf('\n Maximum Thickness of a bacteria laden biofilm, that may be siccessfully treated is %.1f pico-m',L*10^6); //END
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//Chapter-1, Example 1.57, Page 67 //============================================================================= clc; clear; //INPUT DATA I1=25;//current source in A I2=6;//current source in A I3=5;//current source in A RAB=5;//Resistance in ohms RAC=10;//Resistance in ohms RBC=2;//Resistance in ohms //let currents across AC and BC and AB are Ix,Iy and Iz respectively //applying kirchoff's current law at node A //-I1+Ix+I3+Iz=0------eqn(1) //applying kirchoff's current law at node B //-Iz-I3+Iy+I2=0------eqn(2) //CALCULATIONS [a]=[((1/RAC)+(1/RAB)),(-1/RAB);(-1/RAB),((1/RAB)+(1/RBC))]; [b]=[20;-1]; [c]=inv(a)*(b) VA=c(1);//voltage at node A VB=c(2);//voltage at node B //OUTPUT mprintf("Thus voltages at node A and B are %2.1f V and %2.1f V",-VA,VB); //=================================END OF PROGRAM==============================
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//Example 17_9 page no:847 clc; R0=800; D=20; N=10^(D/20); R1=R0*(N-1)/(N+1); R2=R0*(N+1)/(N-1); disp(R1,"the resistance R1 is (in ohm)"); disp(R2,"the resistance R2 is (in ohm)");
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clc //initialisation of variables H= 8 //in b= 6 //in d= 4 //in H1= 5 //in //CALCULATIONS Ia1= ((b*H^3)/12)+b*H*d^2 Ia2= %pi*((d/2)^2)+%pi*((d/2)^2)*(H1)^2 I= Ia1-Ia2 //RESULTS printf ('I= %.2f in^4',I)
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//chapter 25 ex 8 clc; clear; close; a=sqrt(1734/6); //given surface area=1734 and by formula surface area=6*a^2 volume=a^3; printf("The volume of cube is %d cubic cm",volume);
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clear // //The pulley C is in equilibrium under the action of tensile forces in CA and CB and vertical downward load 200 N. The tensile forces in segment CA and CB are the same since the pulley is frictionless. Now consider the equilibrium of pulley C //sum of vertical Fy & sum of horizontal forces Fx is zero //Assume direction of Fx is right //Assume direction of Fy is up //variable declaration L=200.0 //suspended load at C,N AB=3.0 BI=1.0 ACB=5.0 //Length of cord,m DE=3.0 BE=4.0 theta=asin(4.0/5.0) //assume T is tension in string making angle theta1 & theta2,solving horizontal we find theta1=theta2,lets called them theta ,as triangleCFD=triangle=CFA.so, CD=AC HI=BI*DE/BE AH=DE-HI x=AH/2 printf("\n x= %0.3f m",x) T=L/(2*sin(theta)) printf("\n T= %0.0f N",T)
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deff('y=u(x)','y=(1-exp(100*x))/(1-exp(100))'); function x=shufle(m, d) a = m(1,:); b = m(2,:); c = m(3,:); s = length(b); et(1) = 0; ks(1) = 0; for i=1:s ks(i+1) = (-c(i)) / (a(i)*ks(i) + b(i)); et(i+1) = (d(i) - a(i) * et(i)) / (a(i) * ks(i) + b(i)); end x(s+1) = 0; for j=0:s-1 x(s-j) = ks(s-j+1)*x(s-j+1) + et(s-j+1); end x(s+1) = []; endfunction // аппроксимация для (5) function [m, d] = makeLinearSystem1 (s,h) a = [0]; b = [1]; c = [0]; dd = [0]; k1 = -(1/h^2 + 100/(2*h)); k2 = 2/h^2; k3 = (100/(2*h) - 1/h^2); for i=2:s-1 a(i) = k1; b(i) = k2; c(i) = k3; dd(i) = 0; end a(s) = 0; b(s) = 1; c(s) = 0; dd(s) = 1; m = [a';b';c']; d = dd'; endfunction // аппроксимация для (6) function [m, d] = makeLinearSystem2 (s,h) a = [0]; b = [1]; c = [0]; dd = [0]; k1 = -(1/h^2 + 100/(h)); k2 = 2/h^2 + 100/h; k3 = -1/h^2; for i=2:s-1 a(i) = k1; b(i) = k2; c(i) = k3; dd(i) = 0; end a(s) = 0; b(s) = 1; c(s) = 0; dd(s) = 1; m = [a';b';c']; d = dd'; endfunction function [e1,e2,Uex,y1,y2] = makeApp (X) h = 1/X; x = 0:h:1; Uex = u(x); s = length(x); [m1,d1] = makeLinearSystem1(s, h); [m2,d2] = makeLinearSystem2(s, h); tmp = shufle(m1, d1); y1 = tmp'; tmp = shufle(m2, d2); y2 = tmp'; e1 = y1 - Uex; e2 = y2 - Uex; endfunction N = 1000; [e1,e2,Uex,y1,y2] = makeApp(N); ee1(1) = max(abs(e1)); ee2(1) = max(abs(e2)); for i = 2:2 N = 2*N; X = N; [e1,e2,Uex,y1,y2] = makeApp(X); ee1(i) = max(abs(e1)); ee2(i) = max(abs(e2)); x = 0:1/X:1; plot(x,Uex,'-b'); plot(x, y1,'-g'); plot(x,y2,'-r'); printf("ee1(%d)=%f,ee1(%d)=%f, p = %f\n",i-1,ee1(i-1),i,ee1(i), abs(log2(ee1(i-1)/ee1(i)))); printf("ee2(%d)=%f,ee2(%d)=%f, p = %f\n",i-1,ee2(i-1),i,ee2(i), abs(log2(ee2(i-1)/ee2(i)))); end
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//This is script to define a fuction function y = functwovar(t) x = t(1); z = t(2); y = [x^2+z^2-1;x^2*exp(-2*x) + z] endfunction
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//===================================================================================== //Chapter 12 example 18 clc;clear all; //variable declaration R3 = 2000; //resistance of arm in Ω R4 = 2950; //resistance of arm in Ω R2 = 5; //resistance of arm in Ω r2 = 0.4; //resistance in Ω C2 = 0.5*10^-6; //capacitance in F f = 450; //frequency in Hz //calculations r1 = (R3*(r2+R2))/R4 //resistance in Ω C1 = ((R4/R3)*C2) //capacitance in F tand = 2*(%pi)*f*C1*r1 //dissipation power ,C1 in uF //result mprintf("resistace = %3.2f Ω",r1); mprintf("\ncapacitance = %3.2e uF",(C1*10^6)); mprintf("\ndissipation factor = %3.2e ",(tand));
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clear // // // //Variable declaration N=3*10**25 //number of atoms epsilon0=8.85*10**-12 r=0.2*10**-9 //radius(m) E=1 //field //Calculation p=4*%pi*epsilon0*r**3 //dipole moment per unit electric field(F-m**2) P=N*p //polarisation(C-m) epsilonr=1+(4*%pi*r**3*N/E) //dielectric constant alphae=epsilon0*(epsilonr-1)/N //polarisability(Fm**2) //Result
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clear ; clc; // Example 6.6 printf('Example 6.6\n\n'); printf('Page No. 149\n\n'); // given F = 1;// Fuel feed in kg C = 0.86;// Mass of Carbon in kg H2 = 0.05;// Mass of Hydrogen in kg S = 0.01;// Mass of Sulphur in kg O2 = 0.08;// Mass of Oxygen in kg w_C = 12; // mol. weight of C w_H2 = 2; //mol. weight of H2 w_O2 = 32; // mol. weight of O2 w_S = 32; //mol. weight of S //Basis- Per kg of fuel mol_C = C / w_C;// kmol of C mol_H2 = H2 /w_H2;//kmol of H2 mol_O2 = O2 /w_O2;//kmol of O2 mol_S = S /w_S;//kmol of S //By kmol of product CO2 = mol_C*1;// CO2 formed by the reaction C + O2 -> CO2 H2O = mol_H2*1;// H2O formed by the reaction H2 + (1/2)O2 -> H2O SO2 = mol_S*1;// SO2 formed by the reaction S + O2 -> SO2 Pdt = CO2 + H2O + SO2;// Total kmol of combustion products in kmol //Calculation of excess air C_req = mol_C*1;//O2 required by entering C given by reaction C+O2->CO2 in kmol H_req = mol_H2*0.5;//O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol S_req = mol_S*1;//O2 required by entering S given by reaction S+O2->SO2 in kmol O2_req = (C_req + H_req + S_req) - mol_O2// Total number of kmol of O2 required per kg of fuel in kmol N2 = (O2_req*79)/21;// in kmol (considering air consists of 79% N2 and 21% O2 by moles) Wet_pdts = Pdt + N2;// Wet combustion products in kmol //Considering air as an ideal gas,calculating volume of air by ideal gas equation-P*V = n*R*T R = 8310;//Universal gas constant in J/kmol-K T = (273+0);// in K P = 1.013*10^5;// in N/m^2 n_wet = Wet_pdts;// in kmol V_wet = (n_wet*R*T)/P;// In m^3 n_dry = n_wet - H2O;//in kmol V_dry = (n_dry*R*T)/P;// In m^3 printf('Volume of wet flue gas is %3.2f m^3 \n',V_wet) printf('Volume of dry flue gas is %3.2f m^3',V_dry)
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; @Harness: simulator ; @Format: atmel ; @Arch: avr ; @Purpose: "Test the CP (compare two registers) instruction" ; @Result: "flags.h=0, flags.s=0, flags.v=1, flags.n=1, flags.z=0, flags.c=1, r16 = 0" start: ldi r16, 0b00000000 ldi r17, 0b10000000 cp r16, r17 end: break
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Ex13_14.sce
// chapter 13 // example 13.14 // Determine total average power loss and percentage increase in deivce rating // page-840-841 clear; clc; // given T_J=150; // in degree C (maximum junction temperature) Q_JC=0.015; // in degree C/W Q_CS=0.08; // in degree C/W Ta_1=60; // in degree (heat sink temperature1) Ta_2=50 // in degree (heat sink temperature2) // calculate Pav1=(T_J-Ta_1)/(Q_JC+Q_CS); Pav2=(T_J-Ta_2)/(Q_JC+Q_CS); percent_increase=((sqrt(Pav2)-sqrt(Pav1))/sqrt(Pav1))*100; printf("\nThe total average power loss is \t\t\t Pav1=%.2f W",Pav1); printf("\nThe required percentage increase in deivce rating is \t %.2f %%",percent_increase); // Note :The answer vary slightly due to precise calculation
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//example 1 //chapter 46 theta=acosd(1/sqrt(2)) disp(180-theta,"Polarization angle theta=")
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clear clc //Example 17.3 disp('Example 17.3') z=%z; Gz=(-0.3225*z^-2+0.5712*z^-3)/(1-0.9744*z^-1+0.2231*z^-2); G=tf2ss(Gz) n=10; u=ones(1,n); y=dsimul(G,u); disp(y','y=') mprintf('\n\nAlternatively the simulation can also be done\n... using syslin(d,Gz) and flts(u,Gz)\n\n') Gz2=syslin('d',Gz); y2=flts(u,Gz2) disp(y2','y2=')
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clc; //Drawing of shear and bending moment diagram printf("Given problem is for drawing diagram, this diagram is drawn by step by step manner.\n "); F_B=500;//N, force applied at B F_C=500;//N, force applied at C. F_DE=2400;//N/m, distributed load applied at D to E AB=0.4;//m, perpendicular distance between A and B BC=0.4;//m, perpendicular distance between C and B CD=0.4;//m, perpendicular distance between C and D DE=0.3;//m, perpendicular distance between E and D F_E=F_DE*DE;//N, force exerted at DE/2 from E //By free body of entire beam //By sum(m_D)=0 A=(CD*F_C+(BC+CD)*F_B-F_E*DE/2)/(AB+BC+CD);//N, Reaction at A //By sum(Fy)=0 Dy=F_C+F_B+F_E-A;//N,Y component of Reaction at D //By sum(Fx)=0 Dx=0;//N,Y component of Reaction at D //For section 1 //Applying sum(Fy)=0 V1=A;//N, shear force from A to B //For section 2 //Applying sum(Fy)=0 V2=A-F_B;//N, shear force from B to C //For section 3 //Applying sum(Fy)=0 V3=A-F_B-F_C;//N, shear force from C to D //For section 4 //Applying sum(Fy)=0 V4=A-F_B-F_C+Dy;//N, shear force At D //For section 5 //Applying sum(Fy)=0 V5=0;//N, shear force at A //Area under bending curve is change in bending moment of that 2 points MA=0;//N.m MB=MA+V1*AB;//N.m MC=MB+V2*BC;//N.m MD=MC+V3*CD;//N.m ME=MD+1/2*V4*AB;//N.m X=[0,0.4,0.4,0.8,0.8,1.2,1.2,1.5]; V=[V1,V1,V2,V2,V3,V3,V4,V5];//Shear matrix, plot(X,V);//Shear diagram X=[0,AB,AB+BC,AB+BC+CD,AB+BC+CD+DE]; M=[MA,MB,MC,MD,ME];//Bending moment matrix plot(X,M,'r');//Bending moment diagram
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clc,clear printf('Example 2.30\n\n') V=220 I_L=52 N_1=750, N_2=600 R_a=0.2, R_sh = 110 //armature and shunt field resistance I_sh=V/ R_sh I_a1= I_L - I_sh I_a2=I_a1//T (prop.) I_a and T is constant E_b1 = V - I_a1*R_a //N (prop.) E_b/phi (prop.) E_b E_b2 = E_b1*(N_2/N_1) R_x = (V- E_b2)/I_a2 -R_a //Because E_b2 = V - I_a2*(R_a+R_x) printf('Resistance to be connected in series = %.2f ohms\n',R_x) //After R_x gets connected in series with armature and 110 ohms in series with field winding N_1=600 I_sh2=V /(R_sh+110) I_a1=50,I_sh1=2,I_sh2=1 //T (prop.) I_a*I_sh and T doesn't vary I_a2 = I_a1*(I_sh1/I_sh2) E_b1 = V - I_a1*(R_a+R_x) E_b2 = V - I_a2*(R_a+R_x) N_2 = N_1*(E_b2/E_b1)*(I_sh1/I_sh2) //Because N (prop.) E_b/I_sh printf('New speed= %.3f rpm',N_2)
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@start @mob1 name=モブ1 @mob2 name=モブ2 @m1 こんにちは。 @m1 私の名前は、@mob1 name= で指定したものになります。 @m1 name=一時的な名前 name= を設定すると、一時的にその名前になります。 @m1 face=Actor1 index=2 顔グラも設定できます。 @m2 モブその2です @m2 color=2 モブその2です。色付き。 @hide
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//Example 26.3 d_i=(-30+1.5)*10^-2;//Image distance (m) d_o=%inf;//Object distance (m) P=1/d_o+1/d_i;//Power (D) printf('Power of spectacle lens required = %0.2f D',P) //Openstax - College Physics //Download for free at http://cnx.org/content/col11406/latest
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//Initilization of variables F1=100 //lb F2=80 //lb F3=120 //lb F4=150 //lb theta1=((60*%pi)/180) //radians theta2=((45*%pi)/180) //radians theta3=((90*%pi)/180) //radians theta4=((75*%pi)/180) //radians //Distance from point O x1=-5 //ft y1=20 //ft x2=10 //ft y2=10 //ft x3=25 //ft y3=25 //ft x4=35 //ft y4=15 //ft //Calculations Fx=F1*cos(theta1)+F2*cos(theta2)+F4*cos(theta4) //lb Fy=-F1*sin(theta1)+F2*sin(theta2)-F3-F4*sin(theta4) //lb R=sqrt(Fx^2+Fy^2) //lb theta=atan(Fy/Fx) //radians theta_x=(theta*180)/%pi //degrees M_O=-(F1*cos(theta1)*y1)+(-x1)*(F1*sin(theta1))-(x2)*(F2*cos(theta2))+(y2)*(F2*sin(theta2))-(x3*F3)-(y4*F4*cos(theta4))-(x4*F4*sin(theta4)) //lb-ft //Applying varignons theorem x=M_O/Fy //ft y=-M_O/Fx //ft //Results clc printf('The resultant of the force system is:%f lb\n',R) //lb printf('The resultant acts at %f degrees with respect to X-Axis \n',theta_x) //degrees printf('The moment about point O is:%f lb-ft \n',M_O) //lb-ft printf('The x intercept of resultant is:%f ft\n',x) //ft printf('The y intercept of resultant is:%f ft\n',y) //ft //Answer for angle should be negative which has not been mentioned in the tectbook but a schematic shows the angle in fourth quadrant to clarify the doubt
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yN2=0.75 yH2=1-yN2 P=800 //atm T= -70+273.2 //K TcH2=33 //K TcN2=126.2 //K PcH2=12.8 //atm PcN2=33.5 //atm
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; @Harness: simulator ; @Format: atmel ; @Arch: avr ; @Purpose: "Test the BST (bit store to register T) instruction" ; @Result: "flags.t = 0, r17 = 11" start: set ldi r17, 0b1011 bst r17, 2 end: break
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begin; array { 1; 2; 3; 4; } firstArray; array { 5; 6; 7; 8; } secArray; array { 9; 10; 11; 12; } thirdArray; array { 13; 14; 15; 16; } fourthArray; begin_pcl;
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// Scilab Code Ex1.10 Heat equivalent of mass: Pg:23 (2008) c = 3e+08; // Speed of light, m/s J = 4.18; // Joule's equivalent of heat, joule per calorie m = 4.18e-03; // Mass of the substance, kg U = m*c^2; // Energy equivalent of mass, J Q = U/J; // Heat equivalent of mass, calorie printf("\nThe heat equivalent of mass = %1.0e cal", Q); // Result // The heat equivalent of mass = 9e+013 cal
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function Y=f0(X) Y=[1/2 0;0 1/2]*X; endfunction function Y=f1(X) Y=[1/2 0;0 1/2]*X+[1/2;0]; endfunction function Y=f2(X) Y=[1/2 0;0 1/2]*X+[1/4;sqrt(3)/4]; endfunction N=input("Entrez n, le nombre de points a dessiner : "); T=zeros(2,N); T(:,1)=[0,0]'; //initialisation de l'ensemble de N points for i=2:N t=floor(3*rand(1)+1); //selection aleatoire de la fonction select t case 1 then T(:,i)=f0(T(:,i-1)); case 2 then T(:,i)=f1(T(:,i-1)); case 3 then T(:,i)=f2(T(:,i-1)); end end clf; isoview(0,1,0,1); plot(T(1,:),T(2,:),".",'markersize',1); //affichage
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//pathname=get_absolute_file_path('9.15.sce') //filename=pathname+filesep()+'9.15-data.sci' //exec(filename) //Ambient temperature(in K): T1=17+273 //Temperature at state 3(in K): T3=1400 T5=420 //Ambient pressure(in bar): p1=1 //As pressure ratio is 10, pressure at state 2(in bar): p2=10 p3=10 p4=1 //Pressure in HSRG(in kPa): ph=6000 //Condensor pressure(in kPa): pc=15 //Combined cycle output(in MW): O=37.3 //Adiabatic index of compression: r=1.4 //Value of Cp(in kJ/kg.K): Cp=1.0032 //From steam tables: ha=3177.2 //kJ/kg sa=6.5408 //kJ/kg.K sb=sa x=0.7976 hb=2118.72 //kJ/kg hc=225.94 //kJ/kg vc=0.001014 //m^3/kg //Temperature at state 2(in K): T2=T1*(p2/p1)^((r-1)/r) //Temperature at state 4(in K): T4=T3*(p4/p3)^((r-1)/r) //Compressor work per kg(in kJ/kg): Wc=Cp*(T2-T1) //Turbine work per kg(in kJ/kg): Wt=Cp*(T3-T4) //Heat added in combustion chamber(in kJ/kg): qa=Cp*(T3-T2) //Net gas turbine output(in kJ/kg air): WnetGT=Wt-Wc //Heat recovered in HSRG for steam generation(in kJ/kg): qHSRG=Cp*(T4-T5) //Enthalpy at exit of feed pump(in kJ/kg): hd=vc*(ph-pc)*10^2 //Heat added per kg of steam(in kJ/kg): had=ha-hd //Mass of steam generated per kg of air: m=qHSRG/had //Net steam turbine cycle output(in kJ/kg): WnetST=ha-hb-(hd-hc) //Steam cycle output per kg(in kJ/kg air): sco=WnetST*m //Total combined output(in kJ/kg air): tco=WnetGT+sco //Combined cycle efficiency: ncc=tco/qa //Gas turbine efficiency: ngt=WnetGT/qa printf("\nRESULT") printf("\nOverall efficiency = %f percent",ncc*100) printf("\nSteam per kg of air =%f kg steam/kg air",m)
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ar_t=[0,6,7,11,14,20,25] sr_t=[2,3,1,1,1,1,2] sr_b=[0,0,0,0,0,0,0] sr_e=[2,0,0,0,0,0,0] SI=[0,0,0,0,0,0,0] cw=[0,0,0,0,0,0,0] for i=1:6 if ar_t(i+1)>=sr_e(i) then //when the server is available cw(i+1)=0 //customer wait sr_b(i+1)=ar_t(i+1) SI(i+1)=sr_b(i+1)-sr_e(i) // server idle time sr_e(i+1)=sr_b(i+1)+sr_t(i+1) else //when the server is occupied cw(i+1)=sr_e(i)-ar_t(i+1) SI(i+1)=0 sr_b(i+1)=sr_e(i) sr_e(i+1)=sr_b(i+1)+sr_t(i+1) end disp(cw(i),sr_b(i),sr_e(i),ar_t(i),sr_t(i),SI(i)) disp(mean(cw),mean(SI)) end
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//Chapter 3: Thermodynamic and Chemical Equilibrium //Problem: 13 clc; //Declaration of Constants l_v = 101 //in cal /g, Latent headt of vap. mwt = 78 // molecular weight of benzene //Declaration of Variable m = 2 Tb = 80.2 // C, boiling point of benzene // Solution Tb = Tb + 273 // K d_h = l_v * mwt d_s = d_h / Tb d_g = d_h - Tb * d_s mprintf("d_s = %.2f cal / K\n",d_s) mprintf(" d_g = d_a = %d", d_g)
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clc //initialisation of variables clear r=1 r1=1 //CALCULATIONS e1= r+r1 e2= r-r1 //RESULTS printf ('vorticity = %.f*k',e1) printf ('\n vorticity = %.f',e2)
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sys=ssrand(1,1,4); scf(1); h=rlocusplot(sys); h.background=-3;h.font_size=2;
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// EXAMPLE (PG 512) A = [1 2 1;2 2 3;-1 -3 0] // Coefficient matrix b = [0 3 2]' // Right hand matrix [l,u] = lu(A) // l is lower triangular matrix & u is upper triangular matrix l*u if(A==l*u) disp('A = LU is verified') end det(A) det(u) if(det(A)==det(u)) disp('Determinant of A is equal to that of its upper triangular matrix') // Product rule of determinants is verified
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//Example 21.2 R1=1;//Resistance of resistor 1 (ohm) R2=6;//Resistance of resistor 2 (ohm) R3=13;//Resistance of resistor 3 (ohm) R_p=1/(1/R1+1/R2+1/R3);//Equivalent resistance for parallel combination (ohm) printf('a.Total resistance = %0.3f ohm',R_p) V=12;//Voltage (V) I=V/R_p;//Current (A) printf('\nb.Current = %0.2f A',I) I1=V/R1;//Current through resistor 1 (A) printf('\nc.Current through resistor 1 = %0.1f A',I1) I2=V/R2;//Current through resistor 2 (A) printf('\n Current through resistor 2 = %0.2f A',I2) I3=V/R3;//Current through resistor 3 (A) printf('\n Current through resistor 3 = %0.2f A',I3) printf('\nDiscussion:\n Total current = %0.2f A',I1+I2+I3) if (I1+I2+I3)==I printf('\n It is equal to the current output of the source') else printf('\n It is not equal to the current output of the source') end P1=V^2/R1;//Power dissipated in resistor 1 (W) printf('\nd.Power dissipated in resistor 1 = %0.1f W',P1) P2=V^2/R2;//Power dissipated in resistor 2 (W) printf('\n Power dissipated in resistor 2 = %0.1f W',P2) P3=V^2/R3;//Power dissipated in resistor 3 (W) printf('\n Power dissipated in resistor 3 = %0.1f W',P3) P=I*V;//Power output of source (W) printf('\ne.Power output of source = %0.1f W',P) if abs((P1+P2+P3)-P)<0.1 printf('\nDiscussion:\nIt is equal to the total power dissipated by the resistors, (P1+P2+P3)') else printf('\nDiscussion:\nIt is not equal to the total power dissipated by the resistors, (P1+P2+P3)') end //Disregarding a variation of less than 0.1W for sub-question(e) //Answer varies due to round off error //Openstax - College Physics //Download for free at http://cnx.org/content/col11406/latest
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function [s]=%lss_f_s(s1,d2) //operation s=[s1;d2] //! // origine s. steer inria 1987 // Copyright INRIA [a1,b1,c1,d1,x1,dom1]=s1(2:7) [n1,m1]=size(c1);[p2,m2]=size(d2); s=tlist(['lss','A','B','C','D','X0','dt'],a1,b1,[c1;0*ones(p2,m1)],[d1;d2],x1,dom1)
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function y =t0(x) y = -481 +408 +589.64 * x + (-2349.163 * x^2)/2 endfunction function y=dv0(x) y=589.64 -2349.163*x endfunction x = 0 contador=0 while(1) contador = contador +1 xOld = x; x= abs(x - t0(x)/dv0(x)) Er = abs((x-xOld)/x) if (Er < 10^-3) then break end end
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Ex10_7.sce
clear // // // //Variable declaration d=50*10**-6 //diameter(m) NA=0.2 //numerical aperture(m) lamda=1*10**-6 //wavelength(m) //Calculation N=4.9*(d*NA/lamda)**2 //total number of guided modes //Result printf("\n total number of guided modes is %0.3f",N)
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//Example 6.6 G=6.67*10^-11;//Universal gravitational constant (N.m^2/kg^2) M=5.98*10^24;//Mass of the Earth (kg) r=3.84*10^8;//Radius of Moon's orbit (m) g=G*M/r^2;//Acceleration due to gravity (m/s^2) printf('a.Acceleration due to Earth''s gravity at the distance of the moon = %0.2e m/s^2',g) delta_theta=2*%pi;//One complete rotation of Moon's orbit (rad) delta_t=27.3*(1*24*60*60);//Period to make one complete rotation of Moon's orbit = 27.3 days,converted to seconds omega=delta_theta/delta_t;//Angular velocity (rad/s) a_c=r*omega^2;//Centripetal acceleration (m/s^2) printf('\nb.Centripetal acceleration = %0.2e m/s^2',a_c) printf('\nDiscussion: Centripetal acceleration found in (b.) differs from acceleration due to Earth''s gravity found in (a.) \nby %0.2f%%',(a_c-g)/g*100) //Discussion : In agreement with answer in textbook; less than 1% //Openstax - College Physics //Download for free at http://cnx.org/content/col11406/latest
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//Ex8_3 PG-8.6 clc Amid=100;//mid-band gain f1=1e3;//loer cut-off frequency f=20;//frquency at which the gain of the amplifier should be found A=Amid/sqrt(1+(f1/f)^2) printf("\n Therefore the gain of the amplifier at f = 20Hz is %.0f \n",A)
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10_1_1.sce
//CAPTION:Conventional_Magnetron //chapter_no.-10, page_no.-448 //Example_no.10-1-1 clc; //(a) Calculate_the_cyclotron_angular_frequency em=1.759*(10^11);//em=e/m=charge_is_to_mass_ratio B0=.336;//Magnetic_flux_density wc=(em)*B0; disp(wc,'The_cyclotron_angular_frequency(in rad)is ='); //(b) Calculate_the_cutoff_voltage_for_a_fixed_B0 a=5*(10^-2);//radius_of_cathode_cylinder b=10*(10^-2);//radius_of_vane_edge_to_centre Voc=(em*(B0^2)*(b^2)*((1-((a/b)^2))^2))/8; Voc=Voc/(10^5); disp(Voc,'the_cutoff_voltage_for_a_fixed_B0(in KV)is ='); //(c) Calculate_the_cutoff_magnetic_flux_density_for_a_fixed_V0 V0=26*(10^3);//Anode_voltage Boc=(((8*V0)/em)^(1/2))/(b*(1-((a/b)^2))); Boc=Boc*1000; disp(Boc,'the_cutoff_magnetic_flux_density_for_a_fixed_V0(in mWb/m^2)is =');
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errcatch(-1,"stop");mode(2);//Ex:13.3 ; ; wl=8; f=30*10^6; v=f*wl; printf("The veocity of propagation =%d m/s",v); exit();
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//example 5.16 //calculate transmissibility of acquifer //draw daown in main well clc; //given H=25; //static water level rw=0.15; //radius of well Q=5400; //discharge(litre/min) t=24; //time of discharge r1=30; //distance of first well s1=1.11; //drawdown h1=H-s1; r2=90; //distance of second well s2=0.53; //drawdown h2=H-s2; k=(Q*2.303*log10(r2/r1))/(%pi*(h2^2-h1^2)*60000); T=k*H; T=round(T*10000)/10000; mprintf("transmissibility of acquifer=%f cumec/sec.",T); hw=(h2^2-(Q*2.303*log10(r2/rw))/(%pi*k*60000))^0.5; sw=H-hw; sw=round(sw*100)/100; printf("\ndraw daown in main well=%f m.",sw);
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////Ex 11.1 clc; clear; close; format('v',5); Vin=12.5;//V Ri=10;//kohm IS=10^-13;//A T=27;//degree C VT=26;//mV Vref=Ri*IS*1000;//V Vout=-VT*10^-3*log(Vin/Vref);//V disp(Vout,"Output Voltahe, Vout(V)");
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Ex7_33.sce
//Example 7.33 // DC servo system redesign with modified with dominant second // order pole locations. xdel(winsid())//close all graphics Windows clear; clc; //------------------------------------------------------------------ // State space representation //Transfer function model for DC Servo s=poly(0,'s'); num=10; den=s*(s+2)*(s+8); Gs=syslin('c',num/den); // State space representation F=[-10 1 0;-16 0 1;0 0 0] G=[0 0 10]'; H=[1 0 0]; J=0; n=sqrt(length(F)); //Desired poles for the DC Servo system. Pc=[-1.41+1.41*%i -1.41-1.41*%i -8] // State feedback gain K=ppol(F,G,Pc) disp(K,'K=',"State feedback gain") //Estimator - error roots are at Pe=[-4.24+4.24*%i -4.24-4.24*%i -8] exec .\acker_dk.sci; Lt=ppol(F',H',Pe); L=clean(Lt'); disp(L,'L=',"Observer gain") //Error in book, Gain values are different in book. //------------------------------------------------------------------ //Compensator Design DK=-K*inv(s*eye(n,n)-F+G*K+L*H)*L; DK=syslin('c',DK) exec('./zpk_dk.sci', -1); [pl,zr,Kp]=zpk_dk(DK*10); disp(zr,"zeros",pl,"Poles",Kp*10,"Gain(includung system gain)") Dcs=poly(zr,'s','roots')/poly(pl,'s','roots') disp(Dcs,'Dcs=',"Compensator transfer function") //------------------------------------------------------------------
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// Ex2_2 Page:31 (2014) clc; clear; h = 6.626e-034; // Planck's constant, Js c = 3e+08; // Speed of light, m/s e = 1.6e-019; // Energy equivalent of 1 eV, J m = 9.1e-031; // Mass of an electron, kg w0 = 2.28; // Work function of sodium, eV lambda = 400e-09; // Wavelength of light, m nu = c/lambda; // Frequency of light, Hz KE_max = h*nu/e - w0; // Maximum kinetic energy of photoelectrons, eV v = sqrt(2*KE_max*e/m); // Speed of photoelectrons, m/s printf("\nThe maximum kinetic energy of photoelectrons = %5.3f eV", KE_max); printf("\nThe speed of photoelectrons = %4.2e m/s", v); // Result // The maximum kinetic energy of photoelectrons = 0.826 eV // The speed of photoelectrons = 5.39e+05 m/s
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//Example 1.3 (b) //MAXIMA SCILAB TOOLBOX REQUIRED FOR THIS PROGRAM //Calculate Following Summations clear; clc ; close ; syms n; X= symsum (%e^(2*n),n ,0, 0); //Display the result in command window disp (X,"The Value of summation comes out to be:");
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//Variable declaration R=20 //resistance across which voltage is to be calculated(ohms) //Calculations a = [35 -20 ; -20 50] //solving two linear mesh equations b = [50;-100] x = a\b i=x(1)-x(2) //current through 20 ohms resistor(ohms) V=20*i //voltage across 20 ohms(V) //Results printf ("i is %.2f",i) printf ("voltage across 20 ohms is %.1f V",V)
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//<u>=prbs_a(n,nc,ids) //<u>=prbs_a(n,nc,[ids]) // Tirage de PRBS // tirage de u=<u0,u1,...,u_(n-1)>; // a valeurs dans {-1,1} et changeant nc fois de signe // au plus. // Si on veut fixer les dates de changement de signe on peut // fournir une variable de plus ids qui est un vecteur // donnant les indices de changement de signe de u (ordre quelconque) //! [lhs,rhs]=argn(0) if rhs <=2, rand('uniform'); yy= int(mini(maxi(n*rand(1,nc),1*ones(1,nc)),n*ones(1,nc))); ids=sort(yy);ids=[n,ids,1]; else [n1,n2]=size(ids); ids=[n,mini(n*ones(ids),maxi(sort(ids),1*ones(ids))),1]; end u=0*ones(1,n); [n1,n2]=size(ids); val=1; for i=1:n2-1, if ids(i)<>ids(i+1); u(ids(i+1):ids(i))=val*ones(ids(i+1):ids(i));val=-1*val; end end //end
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clear // //Initilization of Variables L=9 //m //Length of rigid bar L_b=3000 //Length of bar A_b=1000 //mm**2 //Area of bar E_b=1*10**5 //N/mm**2 //Modulus of Elasticity of brasss bar L_s=5000 //mm //Length of steel bar A_s=445 //mm**2 //Area of steel bar E_s=2*10**5 //N/mm**2 //Modulus of elasticity of steel bar P=3000 //N //Load //Calculations //From static equilibrium Equation of the rod after appliying Load is //P_b+P_s=P ......................(1) //P_b=1.8727*P_s ..................(2) //NOw substituting equation 2 in equation 1 we get P_s=P*2.8727**-1 P_b=P-P_s d=P_s*L*P**-1 //Result printf("\n Distance at which Load applied even after which bar remains horizontal is %0.2f m",d)
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clear all; clc; disp("Scilab Code Ex 6.21 : ") //Given: M = 2; //kNm Ew = 12; //GPa Est = 200; //GPa bw = 150/1000; //m t = 20/1000; //m rib = 9/1000;//m //Section Properties: n = (Ew/Est); bst = n*bw; y1 = t/2; A1 = t*bw; y2 = bw/2 + t; A2 = rib*bw; y_bar = (y1*A1 +y2*A2)/(A1+A2); I1 = (1/12)*(bw)*(t^3) + A1*(y_bar - y1)^2; I2 = (1/12)*(rib)*(bw^3) + A2*(y2-y_bar)^2; Ina = I1+I2; //Normal Stress: sigma_B = (M*(bw+t-y_bar))/(Ina*1000); sigma_C = (M*(y_bar))/(Ina*1000); //Normal Stress in the wood: sigmaB = n*sigma_B; //Display: printf("\n\nThe normal stress at point B = %1.1f MPa',sigma_B); printf("\nThe normal stress at point C = %1.2f MPa',sigma_C); printf("\nThe normal stress at point B in the wood = %1.2f MPa',sigmaB); //------------------------------------------------------------------------END---------------------------------------------------------------------------------------
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[A2,b2,x2]=make_matrix(250) [A5,b5,x5]=make_matrix(500); [A10,b10,x10]=make_matrix(1000); [t2,d2,l2,u2,rho2]=make_t(A2,2,0.9); [t5,d5,l5,u5,rho5]=make_t(A5,2,0.9); [t10,d10,l10,u10,rho10]=make_t(A10,2,0.9); [x2sol, nit2, rh2, time2]=gauss_seidel(A2,b2,x2,100,1e-5); [x5sol, nit5, rh5, time5]=gauss_seidel(A5,b5,x5,100,1e-5); [x10sol, nit10, rh10, time10]=gauss_seidel(A10,b10,x10,100,1e-5); log_plot(rh2,"Gauss-Seidel 250x250"); log_plot(rh5,"Gauss-Seidel 500x500"); log_plot(rh10,"Gauss-Seidel 1000x1000"); tic();[x] = gauss(A10,b10);toc() tic();[x] = gauss(A5,b5);toc() tic();[x] = gauss(A2,b2);toc()
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// Example 5.4: VCEQ, ICQ clc, clear VBE=0.7; // in volts betaf=45; // From Fig. 5.14 VEE=9; // in volts RB=100e3; // in ohms RC=1.2e3; // in ohms // Applying KVL in the clockwise direction base emitter loop IB=(VEE-VBE)/RB; // in amperes IC=betaf*IB; // in amperes // Writing KVL for the collector loop VCE=VEE-IC*RC; // in volts IC=IC*1e3; // in mili-amperes disp(VCE,"VCEQ (V) ="); disp(IC,"ICQ (mA) =");
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//A Textbook of Chemical Engineering Thermodynamics //Chapter 6 //Thermodynamic Properties of Pure Fluids //Example 30 clear; clc; //Given: M = 24.32; //molecular wt of solid magnesium T = 300; //temperature in K P = 10; //pressure in bar Po = 1; //reference state pressure (bar) d = 1.745*10^3; //density of Mg at 300 K in kg/m^3 //To determine the ativity of solid magnesiun //Using eq. 6.149 (Page no. 237) ln_a = (M/(d*10^3*R*T))*(P-Po)*10^5; a = %e^ln_a; mprintf('Acivity of solid magnesium at 300 K and 10 bar is %f',a); //end
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# test ability to recall old cases # oldcase_v1 is a gas plant created with an old version of VMGSim # and is a file version 1 case recall oldcase_v1.s42 #print a couple of product streams units Field /S6.Out /T1.LiquidDraw_9_reboilerL # change feed temperature - HX1 has temp cross S1.In.T = 55 /S6.Out /T1.LiquidDraw_9_reboilerL # Fix temp cross /Hx1.DeltaTHO = None /Hx1.DeltaTHI = 5 /S6.Out /T1.LiquidDraw_9_reboilerL clear # oldcase_v1 is the gasplant.tst case stored at file version 8 recall oldcase_v8.s42 /overhead.Out /bottoms.Out /Gas-Gas.OutC # change feed temp /Feed.In.T = 60 /overhead.Out /bottoms.Out /Gas-Gas.OutC copy / paste / /RootClone.overhead.Out
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//Exa 4.13 clc; clear; close; format('v',5) //Given data : r=5/2;//mm rdash=2.176*r*10^-3;//m daa_dash=sqrt(6^2+16^2);//m dbb_dash=6;//m dcc_dash=daa_dash;//m dab=8;//m dab_dash=sqrt(6^2+8^2);//m dbc=8;//m dbc_dash=sqrt(6^2+8^2);//m dca=16;//m dca_dash=6;//m Dsa=sqrt(rdash*daa_dash);//m Dsb=sqrt(rdash*dbb_dash);//m Dsc=sqrt(rdash*dcc_dash);//m Ds=(Dsa*Dsb*Dsc)^(1/3);//m disp(Ds,"GMD(m) : "); Dab=(dab*dab_dash)^(1/2);//m Dbc=(dbc*dbc_dash)^(1/2);//m Dca=(dca*dca_dash)^(1/2);//m Dm=(Dab*Dbc*Dca)^(1/3);//m disp(Dm,"Deq or Dm(m) : "); L=0.2*log(Dm/Ds);//mH/km L=L*10^-3*100;//H(for 100 km line) disp(L,"Inductance of 100 km line(H)"); ///Alternate method is given below d1=dab;//m d2=dca_dash;//m L=0.2*log(2^(1/6))*sqrt(d1/rdash)*((d1^2+d2^2)/(4*d1^2+d2^2))^(1/6);//mH L=L*10^-3*100;//H(for 100 km line) disp(L,"Using Alternate method, Inductance of 100 km line(H)");
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//Example 9.2 // current clc; clear; close; //given data : n=10; // dry cells of emf E=1.5; // emf in volts R=4.9; // resistance in ohm r=1; // internal resistance in ohm I=(n*E)/((n*R)+(r)); disp(I,"current flowing,I(A) = ")
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function [s]=%pflss(d1,s2) //[s]=%pflss(d1,s2) <=> s=[d1,s2] //! [a2,b2,c2,d2,x2,dom2]=s2(2:7) [n2,m2]=size(c2);[p1,m1]=size(d1) s=tlist(['lss','A','B','C','D','X0','dt'],a2,b2,[0*ones(p1,m2);c2],[d1;d2],x2,dom2)
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//Ex:87 clc; clear; close; t_e1=100;// in K t_e2=60;// in K t_e3=20;// in K G1=10^6; G2=10^4; t_e=t_e1+(t_e2/G1)+(t_e3/G1*G2); printf("The equivalent noise temperature=%d K",t_e);
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clc //Intitalisation of variables clear p= 795 //mm v= 0.501 //lit w= 1.35 //gms m= 92 //gms R= 0.082 //lit-atm mole^-1 K^-1 T= 45 //C //CALCULATIONS a= ((p/760)*v/((w/m)*R*(273+T)))-1 //RESULTS printf ('Fraction of N2O4 dissociated into NO2 = %.3f',a)
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l=10*10^(-6); rs=5; c=0.01*10^(-6); fr0=sqrt(1-c*rs^2/l)/(2*%pi*sqrt(l*c)); x_l=2*%pi*fr0*l; disp("Part a"); r=sqrt(l/c); rmin=r-rs; disp("the minimum resistance (in Ω) to be added is"); disp(rmin); disp("Part b"); f=100*10^3; fr=sqrt(1-c*rs^2/l)/(2*%pi*sqrt(l*c)); q=fr/f; fr1=sqrt(q^2/(1+q^2))/(2*%pi*sqrt(l*c)); x_l1=2*%pi*fr1*l; q1=fr1/f; rs1=x_l1/q1; rmin1=r1-rs; disp("the resiatance (in Ω) to be added in seriesis"); disp(rmin1); disp("Part c"); rp=(rs1^2+x_l^2)/rs1; z=(rs^2+x_l^2)/rs; r2=1/(1/rp-1/z); disp("the shunting resistance (in Ω) to be connected is"); disp(r2);
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clc clear printf("example 3.28 page number 115\n\n") //to find the top and bottom product,condenser duty,heat input to rebpoiler hF = 171 //in kJ/kg hD = 67 //in kJ/kg hL = hD; hW = 200 //in kJ/kg H = 540 //in kJ/kg disp('part 1') F = 1000 //in kg/h xF = 0.40 xW = 0.02; xD = 0.97; D = F*(xF-xW)/(xD-xW); W = F-D; printf("bottom product = %f kg/hr",W) printf("\ntop product = %f kg/hr\n\n",D) disp('part 2') L = 3.5*D; V = L+D; Qc = V*H-L*hL-D*hD; printf("condenser duty = %f KJ/hr\n\n",Qc) disp('part 3') Qr = Qc - 24200; printf("rate of heat input to reboiler = %f kJ/hr",Qr)
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// Scilab Code Ex9.2.16: Page-420(2008) clc; clear; m = 3; n = 3; p = 2; // 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 mul_fact = double(lcm(int32([m,n,p]))); // Find l.c.m. of m,n and p m1 = m_inv*mul_fact; // Clear the first fraction m2 = n_inv*mul_fact; // Clear the second fraction m3 = p_inv*mul_fact; // Clear the third fraction printf("\nThe miller indices for planes with set of intercepts (%da, %db, %dc) are (%d %d %d) ", m, n, p, m1, m2, m3); m = 1; n = 2; p = %inf; // 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 mul_fact = double(lcm(int32([m,n]))); // Find l.c.m. of m,n and p m1 = m_inv*mul_fact; // Clear the first fraction m2 = n_inv*mul_fact; // Clear the second fraction m3 = p_inv*mul_fact; // Clear the third fraction printf("\nThe miller indices for planes with set of intercepts (%da, %db, %dc) are (%d %d %d) ", m, n, p, m1, m2, m3); // Result // The miller indices for planes with set of intercepts (3a, 3b, 2c) are (2 2 3) // The miller indices for planes with set of intercepts (1a, 2b, Infc) are (2 1 0)
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//CHAPTER 1- D.C. CIRCUIT ANALYSIS AND NETWORK THEOREMS //Example 9 disp("CHAPTER 1"); disp("EXAMPLE 9"); //VARIABLE INITIALIZATION r1=1; //in Ohms r2=2; //in Ohms r3=3; //in Ohms r4=1; //in Ohms //SOLUTION //delta values rab=r1; //between points a and b rac=r2; //between points a and c rbc=r3; //between points b and c //coverting delta abc into star //star values r=rab+rbc+rac; ra=(rab*rac)/r; rb=(rab*rbc)/r; rc=(rbc*rac)/r; req1=r1+r4; req2=rb+r2; req3=(req1*req2)/(req1+req2); req4=ra+req3; disp(sprintf("The equivalent input resistance is %f Ω",req4)); //END
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function [io,s]=syssize(sys) //Old stuff // io=syssize(sys) // [io,ns]=syssize(sys) // // sys : syslin list // io : io=[nout,nin] // nout: nb. ouputs // nin : nb. inputs // s : nb states. // Copyright INRIA select type(sys) case 1 then io=size(sys) s=[] case 16 then sys1=sys(1) select sys1(1) case 'lss' then io=size(sys('D')), [s,s]=size(sys('A')) case 'r' then io=size(sys('den')) [lhs,rhs]=argn(0); if lhs==2 then sys=tf2ss(sys);[s,s]=size(sys('A')),end else error(97,1) end else error(97,1), end
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//Example No. 4.6 clc; clear; close; format('v',6); //Given Data : cycle1=50;//hp t1=20;//sec cycle2=100;//hp t2=20;//sec cycle3=150;//hp t3=10;//sec cycle4=120;//hp t4=20;//sec cycle5=0;//hp t5=15;//sec hp_rms=sqrt((cycle1^2*t1+cycle2^2*t2+cycle3^2*t3+cycle4^2*t4+cycle5^2*t5)/(t1+t2+t3+t4+t5));//hp disp(hp_rms,"hp(rms) for the motor : "); disp("We should choose 100hp motor.")
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//Chapter-7,Example7_1,pg 7-13 N=100 B=0.15 A=10*8*10^-6 I=5*10^-3 Td=N*B*A*I//deflecting torque K=0.2*10^-6//spring const. theta=Td/K//deflecting angle printf("deflection theta=%.2f deg",theta)
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//Example 13.3 //Program to calculate the number of received photons per bit for: //(a)ASK heterodyne synchronous detection //(b)ASK heterodyne asynchronous detection //(c)PSK homodyne detection clear; clc ; close ; //Given data eeta=1; //*100 percent - EFFICIENCY FOR IDEAL RECEIVER BER=10^(-9); //BIT ERROR RATE //Number of received photons per bit for: printf("\n\n\t Number of received photons per bit for:"); //(a)ASK heterodyne synchronous detection Np=(erfinv(1-2*BER))^2*4/eeta; //erfc(x)=1-erf(x) //Displaying the Result in Command Window printf("\n\n\t (a)ASK heterodyne synchronous detection = %1.0f.",Np/2); //(b)ASK heterodyne asynchronous detection Np=-log(2*BER)*4/eeta; //Displaying the Result in Command Window printf("\n\n\t (b)ASK heterodyne asynchronous detection = %1.0f.",Np/2); //(c)PSK homodyne detection Np=(erfinv(1-2*BER))^2/2; //erfc(x)=1-erf(x) //Displaying the Result in Command Window printf("\n\n\t (c)PSK homodyne detection = %1.0f.",Np);
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Pin=60D+3//power input p=1D+3//stator losses Pg=Pin-p//air gap power s=3/100//slip Pm=Pg*(1-s) Prcu=s*Pg mprintf("Mechanical power developed by the motor=%f kW with rotor copper losses of %f kW",Pm/1000,Prcu/1000)
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//Example 7.1(c) clear; clc; fL=0.1; fH=1*10^6; enw=20*10^(-9); fce=200; En=enw*sqrt((fce*log(fH/fL))+fH-fL); printf("Estimated RMS input voltage=%.1f uV",En*10^6);
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clc //initialisation of variables clear M2= 92 //gms M1= 78 //gms pb= 118.2 //mm pt= 36.7 //mm //CALCULATIONS n1= M2/(M1+M2) n2= 1-n1 p1= n1*pb p2= n2*pt w= p1*M1/(p2*M2) //RESULTS printf ('partial pressure of benzene = %.f mm',p1) printf ('\n partial pressure of toulene = %.1f mm',p2) printf ('\n weight proportions = %.2f ',w)
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function header = http_header(keys, values) function [keys, values] = http_header(header) function header = http_header() // This function does the following: // 1. Given a set of key-value pairs, generates HTTP Header text // 2. Given HTTP Header text, returns key-value pairs as Scilab vectors // 3. Given no input, returns a default header endfunction
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deff('x=f(y,t)','x=y*((sin(t))^3)') // Euler Modified y=1; y1=1; printf('x. y') for i=0:0.1:3 printf('\n %4.4f. %4.4f',i,y1) n = 10*i + 1 abc(n) = y1 t=y1; y=y+(0.1*f(i,y)); y1=y1+((0.1/2)*(f(i+0.05,y+0.01*f(i+0.1,y)))); y=y1; end // Runge Kutta y=1 printf('\n\nx. y') for i=0:0.1:3 printf('\n %4.4f. %4.4f',i,y) xyz(n) = y k1=0.1*f(i,y); k2=0.1*f(i+(0.1/2),y+(k1/2)); k3=0.1*f(i+(0.1/2),y+(k2/2)); k4=0.1*f(i+0.1,y+k3); k=(1/6)*(k1+(2*k2)+(2*k3)+k4); y=y+k; end xdata = linspace(0, 3 , length(abc)); ydata = abc plot ( xdata , ydata , "+-" ) ydata2 = xyz plot ( xdata , ydata2 , "o-" )
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function [x,y,typ] = mdaq_mem_read(job,arg1,arg2) mem_write_desc = ["This block reads data from MicroDAQ memory."; "Block with mdaqMemWrite function can be used to"; "change Standalone and Ext model parameters. "; "Mode parameter sets block read behaviour."; "If Trigger input is enabled, rising"; "edge on trigger input will reset data "; "index to defined start index."; "Block can read up to 250000 values. Block "; "memory read size can be calculated by:"; "Number of vectors * Vector Size "; ""; "Init value:"; "Initializes memory with provided value"; ""; "Mode:"; "0 - single read, ignore init value"; "1 - circular read, use init value"; "2 - signle read, use init value"; "3 - circular read, ignore init value"; ""; "Trigger input:"; "0 - disabled"; "1 - enabled"; ""; "Set block parameters:"]; x=[];y=[];typ=[]; select job case 'set' then x=arg1 model=arg1.model; graphics=arg1.graphics; exprs=graphics.exprs; while %t do try [ok,start_idx, vec_num, vec_size,init_value,read_mode,trigger_input,exprs]=.. scicos_getvalue(mem_write_desc,.. ['Start index:'; 'Number of vectors:'; 'Vector size:'; 'Init value:'; 'Mode:'; 'Trigger input:'],.. list('vec',1,'vec',1,'vec',1,'vec',-1,'vec',1,'vec',1),exprs) catch [ok,start_idx, vec_num, vec_size,init_value,read_mode,trigger_input,exprs]=.. scicos_getvalue(mem_write_desc,.. ['Start index:'; 'Number of vectors:'; 'Vector size:'; 'Init value:'; 'Mode:'; 'Trigger input:'],.. list('vec',1,'vec',1,'vec',1,'vec',-1,'vec',1,'vec',1),exprs) end; if ~ok then break end //~1MB = 1 000 000B = 250 000 floats MEM_MAX_DATA_SIZE = 250000; max_data_size = MEM_MAX_DATA_SIZE-start_idx+1; data_size = vec_size*vec_num; if start_idx < 1 | start_idx > MEM_MAX_DATA_SIZE then ok = %f; message("Incorrect memory start index - use index from 1 to "+string(MEM_MAX_DATA_SIZE)); end if vec_size > 10000 | vec_size < 1 then ok = %f; message("Wrong vector size - use 10000 max!"); end if data_size < 1 | data_size > max_data_size then ok = %f; message("Incorrect data size (min 1 / max "+string(max_data_size)+")"); end if read_mode > 3 | read_mode < 0 then ok = %f; message("Use 0-3 to setup read mode."); end if ok then init_data_size = size(init_value, '*'); if init_data_size > 1 then if init_data_size <> data_size then message('Initial values don''t mach vector data size (vector number * vector size)!') ok = %f; end init_value = init_value'; end end if size(init_value, '*') > 1 then if read_mode <> 1 then ok = %f; message("To use Init Value as a vec type change mode paramter to 1."); end end trigger_input_size = 1; if trigger_input <> 1 then trigger_input_size = []; trigger_input = 0 end if ok then [model,graphics,ok] = check_io(model,graphics, trigger_input_size, vec_size, 1, []); graphics.exprs = exprs; model.rpar = init_value; model.ipar = [(start_idx-1);vec_size;read_mode;data_size;0;init_data_size;trigger_input]; model.dstate = []; x.graphics = graphics; x.model = model; break end end case 'define' then start_idx = 1; vec_size = 1; init_value = 0; vec_num = 1; read_mode = 1; init_data_size = 1; trigger_input = 0; model=scicos_model() model.sim=list('mdaq_mem_read_sim',5) model.in =[] model.out=vec_size model.out2=1 model.outtyp=1 model.evtin=1 model.rpar=[]; model.ipar=[(start_idx-1);vec_size;read_mode;vec_num;0;init_data_size;0] model.dstate=[]; model.blocktype='d' model.dep_ut=[%t %f] exprs=[sci2exp(start_idx);sci2exp(vec_num);sci2exp(vec_size);sci2exp(init_value);sci2exp(read_mode);sci2exp(trigger_input)] gr_i=['xstringb(orig(1),orig(2),['''' ; ],sz(1),sz(2),''fill'');'] x=standard_define([4 3],model,exprs,gr_i) x.graphics.in_implicit=[]; x.graphics.exprs=exprs; end endfunction
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//chapter 12 //example 12.1 //page 364 clear all; clc ; //given Av=75;//voltage gain R10=12;//coupling resistor in kohm R7=R10; R9=1000*R10/Av; printf("\nFeedback component R9 is %d ohm,select 150ohm standard value to make Av \nslightly greater than specified.",R9); R9=150; f1=100; XC6=1000*R10/100; C6=10^6/(2*%pi *f1*XC6); printf("\ncapacitor C6 is %.1f microF,use 15 microF standard value",C6); XC2=R9; C2=10^6/(2*%pi *f1*XC2); printf("\ncapacitor C2 is %.1f microF,use 15 microF standard value",C2);
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function convCAF(celcius) fahrenheit=1.8*celcius+32 printf("La conversion de grados Celsius a Fahrenheit es : %f", fahrenheit) endfunction function convFAC(fahrenheit) celcius=(fahrenheit-32)/1.8 printf("La conversion de grados Fahrenheit a Celsius es : %f",celcius) endfunction
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//Chapter 9 Example 3// clc clear // radius of each conductor=r,distance beween the seperation of conductors=d// r=2;// in cm// d=300;// in cms// // inductance of conductor=l// r1=0.7788*r; l1=2*10^-7*(log(d/r1)); printf("\n Inductance of one conductor = %.9f H/m\n",l1); // to convert it into mH/km multiply by 10^6// // loop inductance =l// l=2*l1*10^6;// 10^6 conversion factor// printf("\n Loop Inductance = %.3f mH/km\n",l);
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clc clear n=1.2; m=5; R=0.287; T2=107+273; T1=27+273; IP=[n/(n-1)]*[m/60]*[R*(T2-T1)]; printf('Air Power: %3.2f kW',IP); printf('\n'); BP=14; Em=IP*100/BP; printf('Mechanical Efficiency: %3.0f Percent',Em); printf('\n');
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//Ex 2.5 page 69 clc; clear; close; // VG=10*IG -- eqn(1) PGM=5;// W PGav=.5;// W VS=12;// V Tp=20;// us // PGM = VG*IG where VG=10*IG IG=sqrt(PGM/10);// A VG=10*IG;// V // During the application of pulse VS = VG+(IG*RS) RS=(VS-VG)/IG ;// ohm f=PGav/(PGM*Tp*10**-6)/1000;// kHz delta=f*1000*Tp*10**-6;// Duty Cycle printf('Value of resistance to be connected in series = %.2f ohm',RS) printf('\n Triggering frequency = %.2f kHz',f) printf('\n Duty Cycle = %.1f ',delta) // Note : ans in the textbook is not accurate.
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sauravdekhtawala/FOSSEE-Signal-Processing-Toolbox
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// Copyright (C) 2018 - IIT Bombay - FOSSEE // // This file must be used under the terms of the CeCILL. // This source file is licensed as described in the file COPYING, which // you should have received as part of this distribution. The terms // are also available at // http://www.cecill.info/licences/Licence_CeCILL_V2-en.txt // Author:[insert name] // Organization: FOSSEE, IIT Bombay // Email: toolbox@scilab.in //Calling sequence: // y = upsample (x, n) // y = upsample (x, n, offset) // Upsample the signal, inserting n-1 zeros between every element. // // If x is a matrix, upsample every column. // // If offset is specified, control the position of the inserted sample in // the block of n zeros. // //Testcases: //1.upsample([1,3,5],2) //EXPECTED OUTPUT:[1,0,3,0,5,0] //2.upsample([1;3;5],2) //EXPECTED OUTPUT:[1;0;3;0;5;0] //3.upsample([1,2;5,6;9,10],2) //EXPECTED OUTPUT:[1,2;0,0;5,6;0,0;9,10;0,0] // // function y = upsample (x,n,phase ) [nargout,nargin]=argn() if (nargin<2 | nargin>3), error("wrong no. of input arguments") end if nargin==2 phase=0; else if phase > n-1 warning("This is incompatible with Matlab (phase = 0:n-1). See octave-forge signal package release notes for details." ) end end [nr,nc] = size(x); if (nc==1 | nr==1) then if ( nc==1) y = zeros(n*nr*nc,1); y(phase+1:n:$) = x; end if (nr==1) y = zeros(n*nr*nc,1); y(phase+1:n:$) = x'; y = y.'; end else y = zeros(n*nr,nc); y(phase + 1:n:$,:) = x; end endfunction
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FOSSEE/Scilab-TBC-Uploads
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clc //initialisation m=2*10^-3//kg R=8.31//j/mol/k p=2*10^5 v=8.2*10^-4 a=0.136//pa m^6 M=28*10^-3//kg/ //CALCULATIONS t=(p*v*M)/(R*m) T=(M/(m*R))*(p+(m*m*a/(M*M*v*v)))*(v-(m*b/M)) //results printf(' \n temperature for a perfect gas= % 1f k',t) printf(' \n temperature for vanderwaals gas= % 1f k',T)
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int f(float x, int y), g(int z) /* ошибка */ { }
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19_3.sce
clear// //Variables V1 = 230 //Primary voltage (in volts) N2byN1 = 1.0/2.0 //Turns ratio RL = 200 //Resistance (in ohm) //Calculation V2 = V1 * N2byN1 //Secondary voltage (in volts) Vm = 2**0.5 * V2 //Maximum value of secondary voltage (in volts) Im = Vm / RL //Maximum value of load current (in Ampere) Pm = Im**2 * RL //Maximum value of load power (in watt) Vdc = 0.318 * Vm //Average value of load power (in watt) Idc = Vdc / RL //Average value of load current (in Ampere) Pdc = Idc**2 * RL //Average value of load power (in watt) //Result printf("\n Maximum value of load power is %0.1f W.",Pm) printf("\n Average value of load power is %0.1f W.",Pdc)
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Chapter2_Example23.sce
clc clear //INPUT DATA //CH4+2O2=CO2+2H2O ;//STOICHIOMETRIC REACTION //CASE I //H2O in the products is liquid //CASE II //H2O in the products is gas Hr=-74.87;//enthalpy of reactants Hp1=-964.2;//enthalpy of products Hp2=-876.18;//enthalpy of products R=8.314*10^-3;//gas constant t=298;//initial temperature in K //CLCULATIONS dH1=Hp1-Hr;//Enthalpy increase in MJ/kmol dH2=Hp2-Hr dU1=dH2-((1-3)*R*t);//internal energy in MJ/kmol dU=Hp2;//internal energy in MJ/kmol //OUTPUT printf('Enthalpy increase is %3.2f MJ/kmol of CH4 \n internal energy increase is %3.2f MJ/kmol of CH4 \n',dH2,dU1) printf('H2O in the products and internal energy increase are same \n') printf('np=nr')
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Ex88.sce
//Ex:88 clc; clear; close; F1=2; F2=10; F3=15; F4=20; G1=100; G2=10; G3=10; F=F1+(F2-1)/G1+(F3-1)/(G1*G2)+(F4-1)/(G1*G2*G3); f=10*log(F)/log(10);//noise figure in db printf("The noise figure=%f db",f);