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// Example 5.1 // Computation of minimum value of (a) Locked rotor torque (b) Breakdown torque // (c) Pull up torque // Page No. 173 clc; clear; close; // Given data f=60; // Frequency in Hz p=6; // Number of poles hp=10; // Horsepower n=1150; // Rated speed of machine ns=120*f/p; // (a) Locked rotor torque Trated=hp*5252/n; // Rated torque Tlockedrotor=2.25*Trated; // (b) Breakdown torque Tbreakdown=1.90*Trated; // (c) Pull up torque Tpullup=1.65*Trated; // Display result on command window printf("\n Locked rotor torque = %0.1f lb-ft ",Tlockedrotor); printf("\n Breakdown torque = %0.1f lb-ft ",Tbreakdown); printf("\n Pull up torque = %0.1f lb-ft",Tpullup);
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clc clear //DATA GIVEN m=1000; //mass of steam generated in kg/hr p=16; //pressure of steam in bar x=0.9; //dryness fraction Tsup=380+273; //temp. of superheated steam in K Tfw=30; //temp. of feed water in deg. celsius Cps=2.2; //specific heat of steam in kJ/kg //At 16 bar, from steam tables Ts=201.4+273; //in K hf=858.6; //kJ/kg hfg=1933.2; //kJ/kg Hs=m*[(hf+x*hfg)-1*4.187*(Tfw-0)]; //heat supplied to feed water per hr to produce wet steam Ha=m*[(1-x)*hfg+Cps*(Tsup-Ts)]; //heat absorbed by superheater per hour printf('(i) The Heat supplied to feed water per hour to produce wet steam is: %4.2f*10^3 kJ. \n',(Hs/1000)); printf('(ii) The Heat absorbed by superheater per hour is: %3.2f*10^3 kJ. \n',(Ha/1000));
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s=%s; sys1=syslin('c',5/((1-s)*(s))) nyquist(sys1)
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clc; //page no 122 //prob no. 3.9 // refer fig 3.14 // from spectrum we can see that each of the two sidebands is 20dB below the ref level of 10dBm. Therefore each sideband has a power of -10dBm i.e. 100uW. power_of_each_sideband = 100; Total_power = 2* power_of_each_sideband; disp('uW',Total_power,'The total power is'); div=4; freq_per_div=1; sideband_separation = div * freq_per_div; f_mod= sideband_separation/2; disp('kHz',f_mod,'The modulating freq is '); // Even if this siganl has no carrier, it still has a carrier freq which is midway between the two sidebands. Therefore carrier_freq = 10; disp('MHz',carrier_freq,'The carrier freq');
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//heat produced in a reaction clear; clc; printf("\t Example 6.3\n"); mSO2=74.2;//mass in g SO2=64.07;//molar mass in g nSO2=mSO2/SO2;//moles of SO2 deltaH=-99.1;//heat produced for 1 mol, in kJ/mol Hprod=deltaH*nSO2;//heat produced in this case, in kJ/mol printf("\t the heat produced in a reaction is : %4.0f kJ\n",Hprod); //End
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function [x,y]=Henon_orbit(NbrIti,A,X0,B,Y0,NbrItTrans) if ~isdef('NbrIti','local')... then NbrIti=1000, end; Commandline='henon -l'+string(NbrIti), if isdef('A','local')... then Commandline=Commandline+' -A'+string(A), end; if isdef('B','local')... then Commandline=Commandline+' -B'+string(B), end; if isdef('X0','local')... then Commandline=Commandline+' -X'+string(X0), end; if isdef('Y0','local')... then Commandline=Commandline+' -Y'+string(Y0), end; if isdef('NbrItTrans','local')... then Commandline=Commandline+' -x'+string(NbrItTrans), end; mdelete('tmp.ahm') , Commandline=Commandline+' > temp.ahm'; test=host(Commandline); if test~=0... then disp('l''utilitaire Henon non trouvé'); else orbit=read('temp.ahm',-1,1,'(a)'), orbit=strsubst(orbit,'1.#INF','10e308'); orbit=evstr(orbit), x=orbit(:), y=orbit(:,2), x=x.', y=y.', end; endfunction
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// Scilab code Ex5.8: Pg:218 (2008) clc;clear; Lambda = 5000; // Wavelength of spectral line, Angstorm n = 1; // First order principal maxima n = 3; // Third order principal maxima aplusb = 18000; // Grating element where a is the width of slit and b is the width of opaque region in a grating, cm n = 1; // First order diffraction tl_ratio_1 = 1/sqrt((aplusb/n)^2-Lambda^2); // Angular dispersion produced by a grating around a mean wavelength lambda, radian per angstorm n = 3; // Second order diffraction tl_ratio_3 = 1/sqrt((aplusb/n)^2-Lambda^2); // Angular dispersion produced by a grating around a mean wavelength lambda, radian per angstorm printf("\nThe dispersive powers of first and third order spectra of diffraction grating are %4.2e rad/angstrom and %3.1e rad/angstrom", tl_ratio_1, tl_ratio_3); // Result // The dispersive powers of first and third order spectra of diffraction grating are 5.78e-005 rad/angstrom and 3.0e-004 rad/angstrom
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//Chapter 10 //range of Differentiation & Sketch the output Waveform //page no. 365 //Example10_5 //Figure 10.19 //Given clc; clear; Ri=100; //in Ohm Ci=10^-8; //in farad Rf=5000; //in Ohm Cf=10^-10; //in farad fhf=1/(2*%pi*Rf*Cf); fh_in=1/(2*%pi*Ri*Ci); printf("\n Fhigh(f dbk)=%.0f Hz",fhf); printf("\n Fhigh(in)=%.0f Hz",fh_in); //graph is drawn taking function sin(t) t=[0:0.01:15]; Vi=sin(t); plot(2*Vi); plot(diff(-1.885*100*Vi)); xtitle("Partial Differentiator of sin(t)","t","V"); xgrid;
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errcatch(-1,"stop");mode(2);//Ex:1.19 ; ; v=3;//in volts i=1.5;//in amperes p=v*i; printf("Power supplied = %f watts",p); exit();
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//<z,zd,u,ar>=exar2() //<z,zd,u,ar>=exar2() // // Exemple de processus ARMAX ( K.J. Astrom) // On simule une version bidimensionnelle // de l'exemple exar1(); //! a=[1,-2.851,2.717,-0.865].*.eye(2,2) b=[0,1,1,1].*.[1;1]; d=[1,0.7,0.2].*.eye(2,2); sig=eye(2,2); ar=armac(a,b,d,2,1,sig); write(%io(2),"Simulation of the ARMAX process :"); armap(ar); u=-prbs_a(300,1,int([2.5,5,10,17.5,20,22,27,35]*100/12)); zd=narsimul(a,b,d,sig,u); z=narsimul(a,b,d,0.0*sig,u); write(%io(2),"Least square identification ARX :"); [la,lb,sig,resid]=armax(3,3,zd,u,1,1); //end
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clear //Given u0=-200.0 //cm f0=30.0 //cm fe=3 //Calculation v0=1/((1/f0)+1/u0) a=v0+fe //Result printf("\n Separation between the objective and eyepiece is %0.1f cm",a)
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xdel(winsid()); clear; File = mgetl("frequency_double_measurement.txt"); Vector = evstr(File); len= size(Vector); frequency = len(2)/20; time_stamp = 10000/frequency; File = mgetl("Experiment_2_top.txt"); Vector = evstr(File); Vector=Vector(131:length(Vector)); time_axis = 0:time_stamp/10000:length(Vector)*time_stamp/10000/2; f = linspace(0,frequency,length(time_axis)); n = length(time_axis); top=Vector(1:2:length(Vector)); bottom=Vector(2:2:length(Vector)); plot(time_axis(1:length(top)),top, 'b'); a=gca(); a.font_size=3; xgrid(1, 1, 7); mtlb_axis([0,5,-80,80]); title("Wykres sygnału górnego akcelerometru", "fontsize",5); xlabel("Czas [s]", "fontsize",5); ylabel("Przyśpieszenie [m/s^2]", "fontsize",5); figure; plot(time_axis(1:length(bottom)),bottom, 'r'); title("Wykres sygnału dolnego akcelerometru", "fontsize",5); a=gca(); a.font_size=3; xgrid(1, 1, 7); mtlb_axis([0,5,-80,90]); xlabel("Czas [s]", "fontsize",5); ylabel("Przyśpieszenie [m/s^2]", "fontsize",5); figure; X=fft(top)./(length(top)/2); plot(f(1:n/2),abs(X(1:n/2)), 'b'); a=gca(); a.font_size=3; xgrid(1, 1, 7); title("Charakterystyka częstotliwościowa sygnału", "fontsize",5); xlabel("Częstotliwość [Hz]", "fontsize",5); ylabel("Moduł widma", "fontsize",5); figure; hz = iir(8,'lp','butt',6/frequency,[]); [hzm,fr]=frmag(hz,256); fr2 = fr.*frequency; plot(f(1:n/2),abs(X(1:n/2)),fr2,hzm); a=gca(); a.font_size=3; xgrid(1, 1, 7); title("Charakterystyka częstotliwościowa sygnału oraz filtra", "fontsize",5); xlabel("Częstotliwość [Hz]", "fontsize",5); ylabel("Moduł widma", "fontsize",5); figure; y = flts(top,hz); Y = fft(y)./(length(top)/2); plot(f(1:n/2),abs(Y(1:n/2))); a=gca(); a.font_size=3; xgrid(1, 1, 7); xlabel("Częstotliwość [Hz]", "fontsize",5); ylabel("Moduł widma", "fontsize",5); title("Charakterystyka częstotliwościowa przefiltrowanego sygnału", "fontsize",5); figure; plot(time_axis(1:length(y)), y); a=gca(); a.font_size=3; xgrid(1, 1, 7); mtlb_axis([0,5,-80,80]); xlabel("Czas [s]", "fontsize",5); ylabel("Przyśpieszenie [m/s^2]", "fontsize",5); title("Wykres przefiltrowanego sygnału górnego akcelerometru", "fontsize",5); figure; plot(time_axis, sin(time_axis));
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//Given that m=.4 //in kg Vi = .5 //in m/s k = 750 //in N/m //Sample Problem 7-8 printf("**Sample Problem 7-8**\n") //Using work energy theorem //Wnet = Kf - Ki //Kf = 0 //.5*k*x^2 = Ki x = sqrt(m*Vi^2/k) printf("The compression in the spring is %em", x)
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\test \test \test A TAB character: assdf Two TABS: asdf 3 TABS: sdfa<f 12345678123456781234567812345678 <an empty line follows> <last line> 
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//data Lp=10^(-5) //H Cp=10^(-11) //F Rp=10^5 //ohm RL=10^5 //ohm //formula and result printf("\nresult:-") Wo=1/sqrt(Lp*Cp) printf("\nWo=1/sqrt(Lp*Cp)=%.0e rad/s",Wo) Q=Rp/(Wo*Lp) printf("\nQ=Rp/(Wo*Lp)=%.0f",Q) Qe=RL/(Wo*Lp) printf("\nQe=RL/(Wo*Lp)=%.0f",Qe) QL=Q*Qe/(Q+Qe) printf("\nQL=Q*Qe/(Q+Qe)=%.0f",QL)
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isEpipoleInImage.sci
// Copyright (C) 2015 - 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: Shreyash Sharma,Suraj Prakash // Organization: FOSSEE, IIT Bombay // Email: toolbox@scilab.in function [isepi, varargout ] = isEpipoleInImage(fundamental_matrix, imagesize) // Find whether image contains epipole. // // Calling Sequence // isepi = isEpipoleInImage(F, imagesize) // [isepi, epipole] = isEpipoleInImage(F, imagesize) // // Parameters // F : A 3 * 3 fundamental matrix computed from stereo images. It should be double or single // imagesize : The size of the image // isepi : Logical value true / false denoting whether the image contains epipole // epipole : Location of the epipole. It is 1 * 2 vector. // // Description // The function determines whether the image with fundamental matrix F contains the epipole or not. It also gives the position of the epipole. // // Examples // i = imread('left11.jpg',0); // i1 = imread('right11.jpg',0); // new1 = detectCheckerboardCorner(i1,[7,10]); // new1 = detectCheckerboardCorner(i,[7,10]); // new2 = detectCheckerboardCorner(i1,[7,10]); // f1 = estimateFundamentalMat(new1,new2); // [isep isep2] = isEpipoleInImage(f1,[360 640]); // [ lhs, rhs ] = argn(0) if lhs > 2 then error(msprintf("Too many output arguments")); end /// If there is more than one output parameter [rows cols] = size(fundamental_matrix) if rows ~= 3 | cols ~=3 then error(msprintf("Invalid size of fundamental matrix\n")); end // [rows1 col2] = size(imagesize) //if rows1 ~=1 | cols ~= 2 then // error(msprintf("Invalid image size matrix\n")); //end if lhs == 2 then [isepi, temp ] = raw_isEpipoleInImage(fundamental_matrix, imagesize); varargout(1) = temp; /// if there is only one output parameter else isepi = raw_isEpipoleInImage(fundamental_matrix, imagesize); end endfunction
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clear // // // //Variable declaration lamda=51*10**-6 //wavelength(cm) D=200 //separation between screen and slit(cm) beta1=1 //fringe width(cm) n=10 //Calculation d=lamda*D/beta1 //slit separation(cm) //Result printf("\n slit separation is %0.3f m",d*100)
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histogrammeFct.sci
// Fonction pour faire l'histogramme function histo=histogrammeFct(image) SizeX = size(image, 1); //On récupère la longueur de l'image à modifier. SizeY = size(image, 2); //On récupère la largeur de l'image à modifier. histo = zeros(1, 256); //On crée une matrice nulle qui va contenir l'image modifiée (ici une matrice ligne). for Y = 1:SizeY, //On parcourt la matrice. for X = 1:SizeX, histo(image(X, Y)+1) = histo(image(X, Y)+1)+1; //On incrémente à chaque nouveau pixel de même intensité. end end endfunction
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clear close clc s = poly(0,'s'); w = 1:2:9; // Natural frequencies for i = w g = i^2/(s^2 +2*(0.5)*i*s+ i^2); // xi = 0.5 G = syslin('c', g); t = 0:0.0001:8; gs = csim('step',t,G); plot2d(t,gs, style = i) disp(i ,'The natural frequency is ') //Percentage Overshoot g_steady = gs($); g_max = max(gs); per_OS = (g_max-g_steady)/g_steady *100; disp(per_OS, 'The percentage peak overshoot is '); //Peak time // Infinite for overdamped ssytem disp('The peak time is infinite') //Settling time times = find(abs(gs - .98*g_max)<0.001) disp(t(times($)),"The settling time is "); //Rise time ten_percent_indx_g = find((abs(gs - 0.1*g_steady)<0.001))($) ninety_percent_indx_g = find(abs(gs - 0.9*g_steady)<0.001)(1) rise_time_g = t(ninety_percent_indx_g) - t(ten_percent_indx_g); //Delay time for i = 1:length(t) if (abs(gs(i)- 0.5*g_steady) < 0.001) t_delay = t(i); break; end end //delay_t = t(find(gs == 0.5*g_max)) disp(t_delay,'The delay time is ') end h = legend(['omega_n = 1','omega_n = 3','omega_n = 5','omega_n = 7','omega_n = 9']) xlabel('Time','fontsize',4) ylabel('Amplitude','fontsize',4) title('Step response of the system', 'fontsize',4)
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//Example 5.41 //Piecewise Cubic Hermite Interpolation Method //Page no. 182 clc;close;clear; x=[0,1] y=[1,3] y1=[0,6] x0=poly(0,'x') printf('\tx\ty=f(x)\n-----------------------\n') for i=1:2 printf('x%i\t%i\t %i\n',i-1,x(i),y(i)) end p=1;p1=1;i=1; for k=1:2 for j=1:2 if k~=j then p=p*(x0-x(j)) p1=p1*(x(k)-x(j)) end end L(k)=p/p1 p=1;p1=1; end p=0; L1=[-1,1] for i=1:2 disp(L(i),"L(x) = ") p=p+(1-2*L1(i)*(x0-x(i)))*L(i)^2*y(i)+(x0-x(i))*((L(i))^2)*y1(i) end disp(p,"P2(x) = ") printf('\n\n\n\n\n') x=[1,2] y=[3,21] y1=[6,36] x0=poly(0,'x') printf('\tx\ty=f(x)\n-----------------------\n') for i=1:2 printf('x%i\t%i\t %i\n',i-1,x(i),y(i)) end p=1;p1=1;i=1; for k=1:2 for j=1:2 if k~=j then p=p*(x0-x(j)) p1=p1*(x(k)-x(j)) end end L(k)=p/p1 p=1;p1=1; end p=0; L1=[-1,1] for i=1:2 disp(L(i),"L(x) = ") p=p+(1-2*L1(i)*(x0-x(i)))*L(i)^2*y(i)+(x0-x(i))*((L(i))^2)*y1(i) end disp(p,"P3(x) = ")
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errcatch(-1,"stop");mode(2);// Exa 4.30 ; ; // Given data I_C = 10;// in mA I_B = 0.1;// in mA bita = I_C/I_B; disp(bita,"The current gain is"); exit();
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//Exa 2.14 clc; clear; close; //Given data : N=700;//No. of modes d=30;//in um a=d/2;//in um NA=0.62;//Numerical Aperture //Formula : v=2*sqrt(N) and v=2*%pi*a*NA/lambda lambda=2*%pi*a*NA/(2*sqrt(N));//in um disp(lambda,"Wavelength of light propagating in fibre in micro meter : ");
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//PRACTICAL-2_Q2 clc; clear all; close; x1=[1,3,7,-2,5]; h=[2,-1,0,3]; y=xcorr(x1,h); disp(y,"IS THE REQUIRED CORRELATION!"); l=length(y); t=0:l-1; plot2d3(t,y); xlabel("n"); ylabel("AMPLITUDE"); title("CORRELATION-1"); x1=[1,3,7,-2,5]; h1=[3,0,-1,2]; y=xcorr(x1,h1); disp(y,"IS THE REQUIRED CORRELATION!"); l=length(y); t=0:l-1; plot2d3(t,y); xlabel("n"); ylabel("AMPLITUDE"); title("CORRELATION-2"); //STABILITY Maximum_Limit=10; sum1=0; for n=0:Maximum_Limit-1 sum1=sum1+(n+6) end if (sum1 > Maximum_Limit) disp('WE HAVE AN UNSTABLE SYSTEM'); disp('THE SUM OF RESPONSES RUN OFF THROUGH '); disp(sum1); else disp('WE HAVE A STABLE SYSTEM'); disp('THE SUM OF RESPONSE ARE LIMITED TO '); disp(sum1); end
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//***** Shift register 1input 16outputs (non vecterized version) ******* if (blk_name.entries(bl) =='sr_1i_16o_nv') then addvmm = %t; mputl("# Shift register 1input 16outputs",fd_w); for ss=1:scs_m.objs(bl).model.ipar(1) sr_1i_16o_nv_str= ".subckt in2in_x1 in[0]=fbout_"+string(internal_number)+"_"+string(ss)+" in[1]=net"+string(blk(blk_objs(bl),5))+"_"+string(ss)+" in[2]=net"+string(blk(blk_objs(bl),5))+"_internal_"+string(ss)+" out[0]=fbout_"+string(internal_number)+"_"+string(ss); mputl(sr_1i_16o_nv_str,fd_w); mputl(" ",fd_w); sr_1i_16o_nv_str= ".subckt sftreg2 in[0]=net"+string(blk(blk_objs(bl),2))+"_1"+" in[1]=net"+string(blk(blk_objs(bl),3))+"_1"+" in[2]=net"+string(blk(blk_objs(bl),4))+"_1"+" out[0]=net"+string(blk(blk_objs(bl),5))+"_internal_"+string(ss)+" out[1]=net"+string(blk(blk_objs(bl),2+numofip))+"_"+string(ss)+" out[2]=net"+string(blk(blk_objs(bl),3+numofip))+"_"+string(ss)+" out[3]=net"+string(blk(blk_objs(bl),4+numofip))+"_"+string(ss)+" out[4]=net"+string(blk(blk_objs(bl),5+numofip))+"_"+string(ss)+" out[5]=net"+string(blk(blk_objs(bl),6+numofip))+"_"+string(ss)+" out[6]=net"+string(blk(blk_objs(bl),7+numofip))+"_"+string(ss)+" out[7]=net"+string(blk(blk_objs(bl),8+numofip))+"_"+string(ss)+" out[8]=net"+string(blk(blk_objs(bl),9+numofip))+"_"+string(ss)+" out[9]=net"+string(blk(blk_objs(bl),10+numofip))+"_"+string(ss)+" out[10]=net"+string(blk(blk_objs(bl),11+numofip))+"_"+string(ss)+" out[11]=net"+string(blk(blk_objs(bl),12+numofip))+"_"+string(ss)+" out[12]=net"+string(blk(blk_objs(bl),13+numofip))+"_"+string(ss)+" out[13]=net"+string(blk(blk_objs(bl),14+numofip))+"_"+string(ss)+" out[14]=net"+string(blk(blk_objs(bl),15+numofip))+"_"+string(ss)+" out[15]=net"+string(blk(blk_objs(bl),16+numofip))+"_"+string(ss)+" out[16]=net"+string(blk(blk_objs(bl),17+numofip))+"_"+string(ss)+" out[17]=net"+string(blk(blk_objs(bl),18+numofip))+"_"+string(ss)+" out[18]=net"+string(blk(blk_objs(bl),19+numofip))+"_"+string(ss)+" out[19]=net"+string(blk(blk_objs(bl),20+numofip))+"_"+string(ss)+" #sftreg2_fg =0"; mputl(sr_1i_16o_nv_str,fd_w); mputl(" ",fd_w); end internal_number=internal_number+1; end
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clear; function y = ramp(n, m, d) N = length(n); y = zeros(1, N); for i = 1 : N if n(i) >= -d y(i) = m*(n(i)+d); else end end endfunction clf; dn = 1; n = -6 : dn : 6; y1 = ramp(n, 1, 1); plot2d3(n, y1);
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//Example No. 5.30 clc; clear; close; format('v',9); //Given Data : V=230;//V N=870;//rpm Ia=100;//A Ra=0.05;//ohm T=400;//N-m E=V-Ia*Ra;//V Vgen=V+Ia*Ra;//V N2=N*Vgen/E;//rpm disp(N2,"Motor speed in rpm : ");
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//example2.6 clc disp("Consider a short shunt generator as shown in the fig 2.32") disp("R_a=0.04 ohm, R_sh=90 ohm, R_se=0.02 ohm") disp("V_t=225 V , I_L=75 A") disp("I_a = I_L + I_sh") disp("Now, E=(V_t)+[(I_a)*(R_a)]+[(I_L)*(R_se)]") disp("and drop across armature terminals is,") disp("E-[(I_a)*(R_a)]=(V_t)+[(I_t)*(R_se)]") e=225+(75*0.02) disp(e,"Therefore, E-[(I_a)*(R_a)]=") disp("Therefore, I_sh=[E-(I_a)(R_a)]/(R_sh)=[(V_t)+(I_L)(R_se)]/(R_sh)") i=226.5/90 format(7) disp(i,"Therefore, I_sh(in A)=") i=75+2.5167 disp(i,"Therefore, I_a=I_L+I_sh=") disp("Therefore, E=V_t+[(I_a)*(I_sh)]+[(I_L)*(R_se)]") e=225+(77.5167*0.04)+(75*0.02) format(6) disp(e,"E(in V)=")
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clear; clc; //Example 6.3 Vdd=10; R1=70.9;//(Kohm) R2=29.1;//(Kohm) Rd=5;//(Kohm) Vtn=1.5; Kn=0.5;//(mA/V^2) //lambda=y y=0.01;//V^-1 Rsi=4;//(Kohm) Vgsq=Vdd*R2/(R1+R2); printf('\ngate to source voltage=%.2f V\n',Vgsq) Idq=Kn*(Vgsq-Vtn)^2; printf('\ndrain current=%.3f mA\n',Idq) Vdsq=Vdd-Idq*Rd; printf('\ndrain to source voltage=%.2f V\n',Vdsq) g_m=2*Kn*(Vgsq-Vtn); printf('\ntransconductance=%.3f mA/V\n',g_m) ro=(y*Idq)^-1; printf('\noutput resistance=%.2f KOhm\n',ro) Ri=R1*R2/(R1+R2); printf('\namplifier input resistance=%.2f Kohm\n',Ri) Av=-g_m*(ro*Rd/(ro+Rd))*Ri/(Ri+Rsi); printf('\nsmall signal voltage gain=%.2f\n',Av) printf('\namplifier input resistance=%.2f Kohm\n',Ri) Ro=Rd*ro/(Rd+ro); printf('\namplifier output resistance=%.2f Kohm\n',Ro)
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// Exa 1.21.7 clc; clear; close; // Given data lembda = 1.54;// in Å lembda= lembda*10^-8;// in cm At = 63.54;// atomic weight density = 9.024;// in gm/cc n = 1; N_A = 6.023*10^23; m = At/N_A;// mass a =(density*m)^(1/3);// in cm h = 1; k = 0; l = 0; d = a/(sqrt( ((h)^2) + ((k)^2) + ((l)^2) ));// in cm n = 1; //Formula 2*d*sin(theta) = n*lembda; theta = asind( (lembda)/(2*d) );// in degree disp(theta,"The glancing angle in degree is");
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//page 166 clear; close; clc; A=[1 0 1;1 0 0;2 1 0];//independent vectors stored in columns of A disp(A,'A='); [m,n]=size(A); for k=1:n V(:,k)=A(:,k); for j=1:k-1 R(j,k)=V(:,j)'*A(:,k); V(:,k)=V(:,k)-R(j,k)*V(:,j); end R(k,k)=norm(V(:,k)); V(:,k)=V(:,k)/R(k,k); end disp(V,'Q=')
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// Calculating the armature voltage drop clc; disp('Example 9.12, Page No. = 9.49') // Given Data P = 300;// Power rating (in kW) V = 500;// Voltage rating (in volts) a = 6;// Number of parallel paths (Since lap winding) p = 0.021;// resistivity (in ohm mm square) Ns = 150;// Number of slots Lmt = 2.5;// Length of mean turn (in meter) az = 25;// Area of each conductror (in mm square) // Calculation of the armature voltage drop Z = Ns*8;// Number of armature conductors. Since 8 conductors per slot ra = Z*p*Lmt/(2*a*a*az);// Resistance of armature (in ohm) Ia = P*10^(3)/V;// Armature current disp(Ia*ra,'Armature voltage drop (Volts) ='); //in book answer is 21 (Volt). The answers vary due to round off error
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// To calculate em power developed,mech power fed, torque provided by primemover clc; phi=32*10^-3; //flux/pole n=1600; //speed in rpm Z=728; //no of conductors p=4; A=4; E_a=phi*n*Z*(p/A)/60; I_a=100; P_e=E_a*I_a; disp(P_e,'electromagnetic power(W)'); P_m=P_e; disp(P_m,'mechanical power(W) fed'); w_m=2*%pi*n/60; T=P_m/w_m; disp(T,'primemover torque(Nm)');
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// 08.05.16 Point style // 08.05.17 MixL // 08.05.19 Changed // 08.05.21 Eps=0.1 // 08.08.16 square // 09.11.04 window option // 09.11.07 gcf used // 09.12.25 // 10.01.01 // 10.01.09 ' ' // 10.01.10 drw one by one // 10.02.08 revised for the case "a" // 10.02.13 square debugged // 10.02.21 square debugged (2) // 10.02.23 case of mixlength=3 // 10.04.10 Origin bug // 11.05.29 Flattenlist used function Windisp(varargin) global XMIN XMAX YMIN YMAX GENTEN Nargs=length(varargin); Tp=varargin(Nargs); Ch=''; if type(Tp)~=10 scf(); else Ch=part(Tp,1); Nargs=Nargs-1; select Ch; case 'N' then scf(); case 'n' then scf(); case 'C' then Tmp=gcf();clf('clear'); case 'c' then Tmp=gcf();clf('clear'); case 'A' then Tmp=gcf(); case 'a' then Tmp=gcf(); else Tmp1=part(Tp,length(Tp)); Tmp2=part(Tp,1:length(Tp)-1); Tmp3=evstr(Tmp2); scf(Tmp3); if Tmp1=='C'|Tmp1=='c' clf('clear'); end; end; end; Eps=0.1; Tmp=Doscaling([XMIN,YMIN]); Xm=Tmp(1); Ym=Tmp(2); Tmp=Doscaling([XMAX,YMAX]); XM=Tmp(1); YM=Tmp(2); C=[(Xm+XM)/2,(Ym+YM)/2]; H=max(XM-Xm,YM-Ym)/2; R1=C-H; R2=C+H; R1=Unscaling(R1); R2=Unscaling(R2); if (Ch~='A')&(Ch~='a') Tmp1=R1(1); Tmp2=R2(1); Tmp3=R1(2); Tmp4=R2(2); square(Tmp1,Tmp3,Tmp2,Tmp4);// P=Framedata([XMIN,XMAX],[YMIN,YMAX]); Tmp1=P(:,1)'; Tmp2=P(:,2)'; plot2d(Tmp1,Tmp2); PtO=GENTEN; // 10.04.10 if (PtO(2)>=YMIN) & (PtO(2)<=YMAX) // P=Listplot([XMIN,PtO(2)],[XMAX,PtO(2)]); Tmp1=P(:,1)'; Tmp2=P(:,2)'; plot2d(Tmp1,Tmp2,style=[3]); end; if (PtO(1)>=XMIN) & (PtO(1)<=XMAX) P=Listplot([PtO(1),YMIN],[PtO(1),YMAX]); Tmp1=P(:,1)'; Tmp2=P(:,2)'; plot2d(Tmp1,Tmp2,style=[3]); end; end; for I=1:Nargs Tmp=varargin(I); Pdata=Flattenlist(Tmp); // for II=1:length(Pdata) Tmp=Op(II,Pdata); P=MakeCurves(Tmp,0); P=Unscaling(P); Ndm=Dataindex(P); for J=1:size(Ndm,1) Q=P(Ndm(J,1):Ndm(J,2),:) if size(Q,1)==1 for K=1:2:length(Q) plot2d(Q(K),Q(K+1),style=[-3]); end else Tmp1=Q(:,1)'; Tmp2=Q(:,2)'; plot2d(Tmp1,Tmp2); end end end // for II=2:length(Pdata) // Tmp=Op(II,Pdata); // Windisp(Tmp,'a'); // end; end for I=2:Nargs Pdata=varargin(I); Windisp(Pdata,'a'); end; endfunction;
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clear; lambda=45 d=50 // *** データの作成 a** X = linspace(-180,180,360); //Y = log(X); // *** 解くべき関数の定義 *** function y=func(x) y=asind((80-x*sind(45))/x)-asind((40-x*sind(45))/x)-8; endfunction Y=func(X) // *** 非線形方程式ソルバ *** yp = 0; // yp = f(x) x0 = 1000; // ソルバ―の初期値 // 非線形方程式を解く xp = fsolve(x0, func) // 誤差 //err = abs(xp - exp(yp)) // *** グラフのプロット *** plot(X, Y, '-b'); plot(X, X*0, '--k'); plot(xp, yp, 'or'); //plot()
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mPTS_test_5ch.sce
#################################################### 1. SDL HEADER ################################################### scenario = "BrainEXPain_sound"; active_buttons = 1; button_codes = 1; response_matching = simple_matching; #default_formatted_text = true; default_background_color = 70,70,70; default_text_color = 0, 255, 255; default_font = "Tahoma"; default_font_size = 30; channels = 6; # 1-8 channels (2 = stereo) bits_per_sample = 16; # the amount of data in each digital sound sample given in number of bits sampling_rate = 48000; # sample rate in Hertz ################################################## 2. SDL Part ######################################## begin; ## instructie begin picture { text {caption = "TEST - press space"; font_size = 30;}; x = 0; y = 0; } instructie_pic; ## fixatie picture { text{caption = "*"; font_size = 60; description = "Fix"; }; x = 0; y = 0; } fix_pic; picture { text{caption = "*"; font_size = 60; font_color = 255, 0, 0; description = "Fix"; }; x = 0; y = 0; } stim; picture { text{caption = "thumb tip"; font_size = 60; font_color = 255, 0, 0; description = "Fix"; }; x = 0; y = 0; } stim1; picture { text{caption = "index tip"; font_size = 60; font_color = 255, 0, 0; description = "Fix"; }; x = 0; y = 0; } stim2; picture { text{caption = "middle tip"; font_size = 60; font_color = 255, 0, 0; description = "Fix"; }; x = 0; y = 0; } stim3; picture { text{caption = "ring tip"; font_size = 60; font_color = 255, 0, 0; description = "Fix"; }; x = 0; y = 0; } stim4; picture { text{caption = "pinky tip"; font_size = 60; font_color = 255, 0, 0; description = "Fix"; }; x = 0; y = 0; } stim5; picture { text{caption = "ring base"; font_size = 60; font_color = 255, 0, 0; description = "Fix"; }; x = 0; y = 0; } stim6; #picture { text{caption = "index base"; font_size = 60; font_color = 255, 0, 0; description = "Fix"; }; x = 0; y = 0; #} stim4; #picture { text{caption = "middle base"; font_size = 60; font_color = 255, 0, 0; description = "Fix"; }; x = 0; y = 0; #} stim5; #wavefile { filename = "25HzSine_6ch_15s_ch1.wav";}soundstim; #wavefile { filename = "audio5CH.wav";}soundstim; #wavefile { filename = "25HzSine_1ch_1s.wav";}soundstim; #wavefile { filename = "30_5_unpredictable.wav";}soundstim; #FOR TESTING #wavefile { filename = "25HzSine_1ch_2-9s.wav";}soundstim; #FOR TESTING #wavefile { filename = "5_8_regularinterruptions.wav";}soundstim; #FOR TESTING #wavefile { filename = "5_8_nointerruptions.wav";}soundstim; #FOR TESTING #wavefile { filename = "14_8_regularinterruptions.wav";}soundstim; #FOR TESTING #wavefile { filename = "14_8_nointerruptions.wav";}soundstim; #FOR TESTING #wavefile { filename = "predictable.wav";}soundstim; #FOR TESTING #wavefile { filename = "unpredictable.wav";}soundstim; #FOR TESTING #wavefile { filename = "7_9_unpredictable.wav";}soundstim; #FOR TESTING #wavefile { filename = "15_9_unpredictable.wav";}soundstim; #FOR TESTING #wavefile { filename = "chirp_32-1Hz.wav";}soundstim; wavefile { filename = "30_5_unpredictable.wav";}soundstim; array { sound { wavefile soundstim; description = "module1"; default_code = "module1"; }module1_audio;#d1 speaker 3 sound { wavefile soundstim; description = "module2"; default_code = "module2"; }module2_audio;#d2 speaker 6 sound { wavefile soundstim; description = "module3"; default_code = "module3"; }module3_audio;#d3 speaker 5 sound { wavefile soundstim; description = "module4"; default_code = "module6"; }module4_audio;#d4 speaker 2 sound { wavefile soundstim; description = "module5"; default_code = "module5"; }module5_audio;#d5 speaker 1 sound { wavefile soundstim; description = "module6"; default_code = "module4"; }module6_audio;# }sounds; ### trials ### ## instructie begin trial {trial_duration = forever; trial_type = specific_response; terminator_button = 1; picture instructie_pic; code = "Instructie_hand"; } instructie1_trial; ## grijze rechthoek, eerste baseline trial { trial_duration = stimuli_length; stimulus_event { picture fix_pic; code = "First_fix"; } first_fix_stimulus; } fix_trial; ## stimulation trial; wordt aangepast op basis van randomisatie trial { stimulus_event { sound module1_audio; code = "sound"; deltat = 0; }stim_snd; stimulus_event { picture stim; code = "stim"; }stim_pic; }stimulation_trial; ############################################################# 3. PCL ################################################### begin_pcl; int total_nr_trials = 5; ## veranderen op basis van randomisatie!! array <int> finger [total_nr_trials] = { 1, 2, 3, 4, 5}; ## start of actual trials instructie1_trial.present(); fix_trial.set_duration(3000); fix_trial.present(); int i = 1; loop until i > total_nr_trials begin if finger[i] == 1 then stim_snd.set_stimulus(module1_audio); stim_snd.set_event_code("thumb tip"); stim_pic.set_stimulus(stim1); stim_pic.set_event_code("thumb tip"); elseif finger[i] == 2 then stim_snd.set_stimulus(module2_audio); stim_snd.set_event_code("index tip"); stim_pic.set_stimulus(stim2); stim_pic.set_event_code("index tip"); elseif finger[i] == 3 then stim_snd.set_stimulus(module3_audio); stim_snd.set_event_code("middle tip"); stim_pic.set_stimulus(stim3); stim_pic.set_event_code("middle tip"); elseif finger[i] == 4 then stim_snd.set_stimulus(module4_audio); stim_snd.set_event_code("ring tip"); stim_pic.set_stimulus(stim4); stim_pic.set_event_code("ring tip"); elseif finger[i] == 5 then stim_snd.set_stimulus(module5_audio); stim_snd.set_event_code("pinky base"); stim_pic.set_stimulus(stim5); stim_pic.set_event_code("pinky base"); elseif finger[i] == 6 then stim_snd.set_stimulus(module6_audio); stim_snd.set_event_code("ring base"); stim_pic.set_stimulus(stim6); stim_pic.set_event_code("ring base"); end; stimulation_trial.present(); fix_trial.set_duration(3000); fix_trial.present(); i = i + 1; end;
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clc Cd=0.62; g=9.81; // m/s^2 d=0.1; // m d0=0.06; // m d1=0.12; // m rho=1000; // kg/m^3 rho_m=13600; // kg/m^3 rho_f=0.86*10^3; //kg/m^3 A0=%pi/4*d0^2; A1=%pi/4*d1^2; p_diff=(rho_m-rho_f)*g*d; h=p_diff/rho_f/g; Q=Cd*A0*((2*g*h)/(1-(A0/A1)^2))^(1/2); m=rho_f*Q; disp("Mass flow rate = ") disp(m) disp("kg/s")
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12_2.sce
clc //initialisation of variables a=27.6//sq ft h=1.37//ft d=1.53*(27.9)^0.38*(1.36)^0.24//ft //CALCULATIONS R=d/4//ft A=(%pi*d^2)/4//sq ft //RESULTS printf('The diameter hydraulics radius and area of the hydraulically equivalent circular conduit=% f sq ft',A)
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clear //Given f=4.80 //cm a=1.20 v=-24.0 //cm //Calculation D=f/(a-1) u=1/((1/v)-1/f) //Result printf("\n (i) The least distance of distinct vision is %0.3f cm",D) printf("\n (ii) Distance from the lens is %0.3f cm",-u)
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clear;lines(0); xbasc(); s=poly(0,'s'); h=syslin('c',(s^2+2*0.9*10*s+100)/(s^2+2*0.3*10.1*s+102.01)); comm='(s^2+2*0.9*10*s+100)/(s^2+2*0.3*10.1*s+102.01)'; nyquist(h,0.01,100,comm); h1=h*syslin('c',(s^2+2*0.1*15.1*s+228.01)/(s^2+2*0.9*15*s+225)) xbasc(); nyquist([h1;h],0.01,100,['h1';'h']) xbasc();nyquist([h1;h])
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ex6_6.sce
// Exa 6.6 clc; clear; close; format('v',7) // Given data A = 2500;// open loop gain // Desensitivity of transfer gain trnsfr_gain_densitivity = 40;// in dB trnsfr_gain_densitivity = 10^(trnsfr_gain_densitivity/20); Af = A/trnsfr_gain_densitivity;// unit less disp(Af,"The gain with feed back is"); I = A/Af;// assumed disp("The input for same output will become "+string(I)+" times the input without feedback.")
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// Theory and Problems of Thermodynamics // Chapter 10 // Chemical Thermodynamics // Example 9 clear ;clc; //Given data T = 1000 // reaction temperature in K Tf = 298 // standard heat temperature in K del_H_f = -241.82 // heat of formation of H2O in kJ a1 = 28.85*1e-3 // constant 'a' for water in heat capacities b1 = 12.06*1e-6 // constant 'b' for water in heat capacities a2 = 30.25*1e-3 // constant 'a' for oxygen in heat capacities b2 = 4.21*1e-6 // constant 'b' for oxygen in heat capacities a3 = 27.27*1e-3 // constant 'a' for nitrogen in heat capacities b3 = 4.93*1e-6 // constant 'b' for nitrogen in heat capacities // Reaction //H2(g)+ 0.5*1.5*O2(g)+ 0.5*1.5*3.76*N2 => H2O(g)+ 0.25*O2(g)+ 2.82*N2(g) del_a = a1 + 0.25*a2 + 2.82*a3 del_b = b1 + 0.25*b2 + 2.82*b3 // Reactants(298 K) => Products(298 K) => Products(P) // complete combustion => del_H_T = 0 // del_H_T = del_H_f + del_H_P_T // 0 = del_H_F + del_H_P_T (A) // del_H_P_T = del_a*(T-298) + del_b/2*(T-298)^2 (B) // solving Equation A and B deff('y=temp(T)', 'y = del_H_f + del_a*(T-Tf) + del_b*((T^2)-(Tf^2))/2') T = fsolve(10,temp) // Output Results mprintf('Maximum flame temperature attained in welding torch = %4.0f K' , T);
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// ARX model parameter estimation // Computes Covariance matrix // Computes Noisevariance of a process // // Authors: // Inderpreet Arora // Ashutosh Kumar function [sys] = arx(varargin) [lhs,rhs] = argn(); data = varargin(1) if ( rhs < 2 ) then errmsg = msprintf(gettext("%s: Unexpected number of input arguments : %d provided while should be 4"), "arx", rhs); error(errmsg) end if ((~size(data,2)==2) & (~size(data,1)==2)) then errmsg = msprintf(gettext("%s: input and output data matrix should be of size (number of data)*2"), "arx"); error(errmsg); end if (~isreal(data)) then errmsg = msprintf(gettext("%s: input and output data matrix should be a real matrix"), "arx"); error(errmsg); end n = varargin(2) //pause na = n(1);nb = n(2);nk = n(3) if (size(find(n<0),"*") | size(find(((n-floor(n))<%eps)== %f))) then errmsg = msprintf(gettext("%s: values of order and delay matrix [na nb nk] should be nonnegative integer number "), "arx"); error(errmsg); end az = max(na,nb+nk-1); zer = zeros(az,1); zd = data; // Zeros appended zd1(:,1) = [zer; zd(:,1)]; zd1(:,2) = [zer; zd(:,2)]; [r,c] = size(zd1); t = az+1:r; yt = zd1(:,1); ut = zd1(:,2); yt1 = yt'; ut1 = ut'; // row vector len1 = length(yt1); yt2 = zeros(1,len1-az); ut2 = zeros(1,len1-az); // arx(Data,[na nb nk]) for i=1:na yt2 = [yt2; -yt1(t-i)]; end; for i=nk:nb+nk-1 ut2 = [ut2; ut1(t-i)]; end; [r1_a,c1_a] = size(yt2); [r2_a,c2_a] = size(ut2); phit = [yt2(2:r1_a,:); ut2(2:r2_a,:)]; m1 = phit*phit'; [qm,rm] = qr(m1); m2 = phit*zd(:,1); thetaN = inv(rm)*qm'*m2; [r1,c1] = size(thetaN); a = thetaN(1:na); b = thetaN(na+1:r1); bpol = poly([repmat(0,nk,1);thetaN(1+na:na+nb)]',"q","coeff"); apol = poly([1; thetaN(1:na)]',"q","coeff"); sys = idpoly(coeff(apol),coeff(bpol),1,1,1,1) endfunction;
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// Example 4.8, page no-212 clear clc T=0.5 sg1=1.02 sg2=0.98 wt=1000*10^-6 v=T/((sg1-sg2)*wt) v=ceil(v) printf("V=%d cm^3",v)
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clc //given that del_x = 1 // let uncertainty in position is unity m_e = 9.1e-31 // mass of electron in kg m_p = 1.67e-27 // mass of proton in kg h = 6.63e-34 // Plank constant printf("Example 2.7") h_bar = h / (2*%pi) // constant del_v_ratio = m_p/m_e // calculation in uncertainties in the velocity of electron and proton printf("\n Ratio of uncertainties in the velocity of electron to proton is %d.\n\n\n",del_v_ratio)
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//i/p args contain imaginary elements b = [(1/3)*%i 1/4 1/5 1]; a = b(:,$:-1:1); flag = islinphase(b,a); disp(flag); //output // 0
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clc; clear all; Pavg=375;//given average power PEP=Pavg;//peak envelope power disp(PEP,'peak envelope voltage is=');
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c=0.1;//concentration of the solution// Kb=1.8*10^-5; printf('The value of a should be calculated first using Kb=(c*a^2)/(1-a)\nThis gives rise to a quadratic equation which can be solved to obtain the value of a.'); printf('\nUsually it is permissible to use approximation methods if K<10^-5\nOne can neglect a in comparison to 1 and solve for a.\nA better way is to use the method of succesive approximations.\nThis will be illustrated using the above equation'); printf('\nFirst find the approximate value of a by neglecting the value of a in comparison with 1.\nLet the approximate value be a1'); a1=1.342*10^-2; a2=1.332*10^-2; printf('\nWe repeat this procedure till 2 consecutive values of a do not differ significantly.'); a3=1.332*10^-2; OH=a3*c;//concentration of OH- in the solution// printf('\nSince the values of a2 and a3 are the same the correct value of a=1.332*10^-2\nThe approximate value is greater than the correct value by about 1percent.'); printf('\nThe concentration of OH- =%f=1.332*10^-3M',OH);
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clear //Given A_1 = 0.125 //sq.in , The area of the crossection of AB A_2 = 0.219 //sq.in , The area of the crossection of BC l_1 = 3*(5**0.5) //in , The length of AB l_2 = 6*(2**0.5) //in , The length of BC p = 3 //k , Force acting on the system E = 10.6*(10**3) //ksi - youngs modulus of the material p_1 = (5**0.5)*p/3 //P, The component of p on AB p_2 = -2*(2**0.5)*p/3 //P, The component of p on AB e = p_1*l_1*p_1/(p*E*A_1) + p_2*l_2*p_2/(p*E*A_2) //in, By virtual deflection method printf("\n The deflection is %0.3f in",e)
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function [out]=integralKernel(InputArraypoints) [t1 t2 t3 t4 t5]= opencv_integralKernel(InputArraypoints); out=struct("size1",t1,"size2",t2,"center1",t3,"center2",t4,"angle",t5); endfunction;
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//Example 12.13 //Find the required probabilities. disp('P(X>0.6)=1-F(0.6)=') disp(%e^(-1.2),'1-(1-e^(-1.2))=e^(-1.2)=') disp(1-%e^(-0.5),'P(X<=0.25)=(1-e^(2*(-1.2))=1-e^(-0.5)=') disp('P(0.4<X<=0.8)=F(0.8)-F(0.4)=') disp((1-%e^(-1.6))-(1-%e^(-0.8)),'(1-e^(-1.6))-(1-e^(-0.8))=')
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exec('Load.sci'); m=fscanfMat('../result.dat'); q=m(1:$,2:22)'; Visu(q);
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function s=complexstring(a) if imag(a)>=0 then s=sprintf('%g+%gi',real(a),imag(a)) else s=sprintf('%g%gi',real(a),imag(a)) end funcprot(0) endfunction
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<cmd> ../build/42sh</cmd> <ref> bash</ref> <stdin> a=1 b=2 c=3 echo "we are not$((($a + $b) + $c))not the $a $(($a + $b))champions" </stdin>
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//ques-25.4 //Calculating kinetic energy of an ideal gas and temperature required clc T1=273;//temperature (in K) n1=1; n2=3;//number of moles KE=(3/2)*n1*8.314*T1; T2=KE/((3/2)*8.314*n2); printf("The kinetic energy of the ideal gas is %.3f kJ/mol and the temperature required for 3 moles of gas is %d K.",KE/1000,T2);
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//Example 5_2 page no:193 clc R=1*10^3//resistance in ohm L=50*10^-3//inductance in henry V=10 f=10*10^3//frequency in Hz Xl=2*%pi*f*L Z=R+(%i*Xl) Z=sqrt(R^2+Xl^2) disp(Z,"impedence is (in ohm)") I=V/Z I=I*1000//converting to milli ampere disp(I,"current is (in mA)") angle=atand(Xl/R) disp(angle,"the phase angle is (in degree)") Vr=I*10^-3*R//current in milli ampere disp(Vr,"Voltage across resistance is (in volts)") Vl=I*10^-3*Xl//current in milli ampere disp(Vl,"Voltage across inductive reactance is (in volts)") //the values varies slightly with text book hence values are rounded off in text book
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clear clc //For system when P_NH3=P_HCl r=1;//no.of equations C=3;//no. of constituents Z1=1;//no. of restricting equations C1=C-r-Z1;//no. of components printf('C1=%.1d',C1) //For system when P_NH3 not equal P_HCl Z2=0;//no. of restricting equations C1=C-r-Z2 printf('\nC1=%.1d',C1) //page 103
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//Chapter 16: Practical Design Considerations of Large Aperture Antennas //Example 16-2.2 clc; //Variable Initialization del_phi = 36.0 //rms phase error (degrees) n_irr = 100.0 //Number of irregularities //Calculations max_side = tan(del_phi*%pi/180)**2 max_side = -10*log10(max_side) //Maximum side-lobe level (dB) ran_side = (1/n_irr)*tan(del_phi*%pi/180)**2 ran_side = -10*log10(ran_side) //Random side-lobe level (dB) //Result mprintf("The maximum side lobe level from main lobe is %.1f dB", max_side) mprintf("\nThe random side lobe level from main lobe is %.1f dB", ran_side)
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clc clear //Input data l=1*10^-6//Wavelength of light used in m n1=1.45//Refractive index of the core n2=1.448//Refractive index of the cladding d=6*10^-6//Diamter of the core in m //Calculations NA=sqrt(n1^2-n2^2)//Numerical aperture N=4.9*(d*NA/l)^2//Number of modes propagating through the fibre //Output printf('The number of modes that can be allowed through the fibre is %i. \n It is a single-mode fibre',N)
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// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal // Chapter 1-Properties of Fluid // Problem 1.7 //Given Data Set in the Problem Area=0.8*0.8 theta=%pi/6 W=300 du=0.3 dy=1.5/1000 //Calculations W_alongPlane=W*cos(%pi/2-theta) Shear_Force=W_alongPlane ss=Shear_Force/Area visc=ss/(du/dy) //Shear Stress+Viscosity * Velocity Gradient mprintf("The Dynamic Viscosity of the Oil is %f poise",visc*10)
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//clc() //Fe(s) + 2HCl(aq) = FeCl2(aq) + H2(g) MFe = 55.847; m = 1;//kg Nfe = m * 10^3/MFe; Nh2 = Nfe;//(since 1 mole of Fe produces 1 mole of H2) T = 300;//K R = 8.314; //the change in volume is equal to the volume occupied by hydrogen produced PV = Nh2 * R * T; W = PV; disp("kJ",W,"Work done = ")
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clear; clc; close; dt=.1; t0=1;//positive number t=-10:dt:10; for i=1:length(t) if modulo(t(i),t0)==0 then x(i)=1; else x(i)=0; end end a=gca(); plot2d3(t,x); plot(t,x,'r.') poly1=a.children.children; poly1.thickness=3; poly1.foreground=2; xtitle('x(t)','t') wmax=10; w=-wmax:0.1:wmax; Xw=x'*exp(-%i*(w'*t))*dt; figure a=gca(); plot2d3(w,round(abs(Xw))); poly1=a.children.children; poly1.thickness=2; poly1.foreground=2; xtitle('X(w)','w') //or the fourier series is doesnt work //ck=1/t0; //k=-10:0.1:10; //x=ck*(exp(%i*2*%pi*t.*k/t0)); //wmax=10; //w=-wmax:0.1:wmax; //Xw=x*exp(-%i*(w'*t))*dt; //clf(); //plot2d3(k,x);
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// // 12.01.07 kirikomi(Cut) // 12.08.26 debugged // 15.06.11 debugged // 15.10.24 debugged function Arrowline(varargin) global YaSize YaAngle YaPosition YaThick YaStyle Kirikomi; Nargs=length(varargin) P=varargin(1); Q=varargin(2); Futosa=YaThick; Ookisa=YaSize; Hiraki=YaAngle; Yapos=YaPosition; Cutstr="Cut=0"; Str=YaStyle; Flg=0; for I=3:Nargs Tmp=varargin(I); if type(Tmp)==10 Equal=mtlb_findstr(Tmp,'='); // 12.01.07 from if length(Equal)>0 Tmp2=strsplit(Tmp,Equal-1); if (Tmp2(1)=="Cut") | (Tmp2(1)=="cut") Tmp="Cut"+Tmp2(2); Cutstr=Tmp; end else Str=Tmp; // 12.01.07 upto (debugged on 12.08.26) end; end; if (type(Tmp)==1) & (length(Tmp)==1) if Flg==0 Ookisa=Ookisa*Tmp; end if Flg==1 if Tmp<5 Hiraki=Tmp*Hiraki; else Hiraki=Tmp; end end if Flg==2 Yapos=Tmp; end if Flg==3 Futosa=Tmp; end Flg=Flg+1; end end R=P+Yapos*(Q-P); Tmp=Q-Unscaling(0.2*Ookisa/2*(Q-P)/norm(Q-P)); // 15.10.24 Drwline(Listplot([P,Tmp]),Futosa); Arrowhead(R,Q-P,Ookisa,Hiraki,Futosa,Cutstr,Str); // 12.01.07 endfunction
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** File Info Version: 1.0 Num Logs = 0 Num Trans = 0 Num Writers = 0 Init Tranlog = 0 Total Entries = 14 Tranlog Offset = 0 Transaction Id = 11 Index Free List = 12 Total Size of Data = 428 Data Transformation Id = 9 Index Transformation Id = 57 ** Freelist Info First freelist entry = 12 Iterating over freelist...(OK) Final freelist entry = 13 Total freelist entries = 2
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clear; clc; // Stoichiometry // Chapter 2 // Basic Chemical Calculations // Example 2.25 // Page 40 printf("Example 2.25, Page 40 \n \n"); // solution p = 4 //[bar] T = 773.15 //[K] R = .083145 V = R*T/p // [l/mol] molar volume printf("Molar volume = "+string(V)+" l/mol.\n \n \n") // using appendix III // calculating Tc and Pc of different gases according to their mass fractions Tc1 = .352*32.20 // H2 Tc2 = .148*190.56 // methane Tc3 = .128*282.34 //ethylene Tc4 = .339*132.91 // CO Tc5 = .015*304.10 // CO2 Tc6 = .018*126.09 // N2 Tc = Tc1+Tc2+Tc3+Tc4+Tc5+Tc6 // Tc of gas // similarly finding Pc Pc1=.352*12.97 Pc2=.148*45.99 Pc3=.128*50.41 Pc4=.339*34.99 Pc5=.015*73.75 Pc6=.018*33.94 Pc=Pc1+Pc2+Pc3+Pc4+Pc5+Pc6 // Pc of gas a = (27*R^2*Tc^2)/(64*Pc) // [bar/mol^2] b = (R*Tc)/(8*Pc) // l/mol // substituting these values in vanderwall eq and solving by Newton Rhapson method we get V = 15.74 // [l/mol] printf("by Vanderwall eq molar volume = "+string(V)+" l/mol")
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//Example 9.5.2;average voltage clc; clear; close; format('v',7) a1=30;//in degree a2=75;//in degree b1=60;//in degree ia=10;//in amperes vsrms=230;//in volts b3=180-a1;// a3=180-b1;// b2=180-a2;// alfa=0;// vldc=((vsrms*sqrt(2))/%pi)*(cosd(a1)-cosd(b1)+cosd(a2)-cosd(b2)+cosd(a3)-cosd(b3));// disp(vldc,"average voltage in volts is") Is_rms=ia*((1/180)*(b1-a1+b2-a2+b3-a3))^(1/2);// disp(Is_rms," Is_rms(A) = ") I1_rms=((sqrt(2)*ia)/(%pi))*(cosd(a1)-cosd(b1)+cosd(a2)-cosd(b2)+cosd(a3)-cosd(b3));// disp(I1_rms," I1_rms(A) = ") fi=alfa; FPF=cosd(fi); disp(FPF,"FPF = ") DF=I1_rms/Is_rms; disp(DF," DF = ") PF=DF*FPF; disp(PF," PF(lag)= ") HF=sqrt((1/DF^2)-1); disp(HF*100," HF(%) = ")
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clc // Given that lambda = 6e-7 // wavelength of light in meter f = 0.6 // focal length of convex lens in meter n = 1 // no. of half period zone // Sample Problem 2 on page no. 2.38 printf("\n # PROBLEM 2 # \n") Rn = sqrt(n * lambda * f)// calculation for radius of half period zone printf("Standard formula used \n f = Rn^2/(n*lambda)\n") printf("\n Radius of half period zone = %f mm ",Rn*1000)
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clear //Initialization ni=26 //Decimal number //Calculation bini = 0 i = 1 while (ni > 0) rem = ni-int(ni/2)*2 ni = int(ni/2) bini = bini + rem*i i = i * 10 end w= bini //Declaration printf("\n Binary Equivalent = %d",w)
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function tapis(N,a,b,c,d) if N<1 then plot([a(1),b(1)],[a(2),b(2)]); plot([b(1),c(1)],[b(2),c(2)]); plot([c(1),d(1)],[c(2),d(2)]); plot([d(1),a(1)],[d(2),a(2)]); else //calcul des nouveaux sommets e=[a(1),a(2)+2*(d(2)-a(2))/3]; f=[a(1),a(2)+(d(2)-a(2))/3]; g=[a(1)+(b(1)-a(1))/3,a(2)]; h=[a(1)+2*(b(1)-a(1))/3,a(2)]; i=[b(1),a(2)+(c(2)-b(2))/3]; j=[b(1),a(2)+2*(c(2)-b(2))/3]; k=[a(1)+2*(b(1)-a(1))/3,c(2)]; l=[a(1)+(b(1)-a(1))/3,c(2)]; m=[a(1)+(b(1)-a(1))/3,a(2)+2*(d(2)-a(2))/3]; n=[a(1)+(b(1)-a(1))/3,a(2)+(d(2)-a(2))/3]; o=[a(1)+2*(b(1)-a(1))/3,a(2)+(c(2)-b(2))/3]; p=[a(1)+2*(b(1)-a(1))/3,a(2)+2*(c(2)-b(2))/3]; //appels de tapis pour chacun des carres retenus tapisSierpinski(N-1,a,g,n,f); tapisSierpinski(N-1,f,n,m,e); tapisSierpinski(N-1,e,m,l,d); tapisSierpinski(N-1,l,k,p,m); tapisSierpinski(N-1,p,j,c,k); tapisSierpinski(N-1,j,p,o,i); tapisSierpinski(N-1,b,h,o,i); tapisSierpinski(N-1,n,o,h,g); end endfunction A=[0,0]; B=[1,0]; C=[1,1]; D=[0,1]; //points du premier carre N=input("Entrez n, le nombre de niveaux a dessiner : "); tapis(N,A,B,C,D) clf isoview(0,1,0,1); //affichage
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clear; clc; //Example - 3.19 //Page number - 113 printf("Example - 3.19 and Page number - 113\n\n"); //Given T_1 = 600;//[C] - Temperature at entry P_1 = 15;//[MPa] - Pressure at entry T_2 = 400;//[K] - Temperature at exit P_2 = 100;//[kPa] - Pressure at exit A_in = 0.045;//[metre square] - flow in area A_out = 0.31;//[metre square] - flow out area m = 30;//[kg/s] - mass flow rate. //At 15 MPa and 600 C,it has been reported in the book that the properties of steam are, Vol_1 = 0.02491;//[m^(3)/kg] - Specific volume H_1 = 3582.3;//[kJ/kg] - Enthalpy // m = den*vel*A = (Vel*A)/Vol, substituting the values vel_1 = (m*Vol_1)/A_in;//[m/s] - Velocity at point 1. printf(" The inlet velocity is %f m/s\n",vel_1); //At 100 MPa (saturated vapour),it has been reported in the book that the properties of steam are, T_sat = 99.63 C, and Vol_vap_2 = 1.6940;//[m^(3)/kg] - specific volume of saturated vapour. H_vap_2 = 2675.5;//[kJ/kg] - Enthalpy os saturated vapour. vel_2 = (m*Vol_vap_2)/A_out;//[m/s] - Velocity at point 2. printf(" The exit velocity is %f m/s\n",vel_2); //From first law we get, q - w =delta_H + delta_V^(2)/2 //q = 0, therefore, -w = delta_H + delta_V^(2)/2 delta_H = H_vap_2 - H_1;//[kJ/kg] - change in enthalpy. delta_V_square = (vel_2^(2) - vel_1^(2))/2;//[J/kg] delta_V_square = delta_V_square*10^(-3);//[kJ/kg] w = -(delta_H + delta_V_square);//[J/kg] W_net = w*m;//[kW] W_net = W_net*10^(-3);//[MW] - power produced. printf(" The power that can be produced by the turbine is %f MW",W_net);
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clc; clear; printf("\n Example 6.6\n"); G=15; //Mass flow rate of organic liquid printf("\n Given:\n Mass flow rate of organic liquid = %d kg/s",G) L_ow=2;//Length of the weir printf("\n Length of the weir = %.1f m",L_ow); rho_l=650; printf("\n Density of liquid = %d kg/m^3",rho_l); Q=G/rho_l; //Use is made of the Francis formula (equation 6.43), h_ow=(2/3)*(Q/L_ow)^(2/3); printf("\n\n Calculations:\n Height of liquid flowing over the weir = %.2f mm",h_ow*1e3);
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// Get the dcm encoded by the quaternion // // The computed dcm is the matrix such that // ____ // q_AB * v_B * q_AB = dcm_AB * v_B // // INTPUT // - q_AB: input quaternion // // OUTPUT // - dcm_AB: dcm B --> A // // USAGE // [dcm_AB] = quat2mat(q_AB); // // HISTORY // 09/02/2015: T. Pareaud - Creation function [dcm_AB] = quat2dcm(q_AB) q0 = q_AB(1,1); q1 = q_AB(2,1); q2 = q_AB(3,1); q3 = q_AB(4,1); dcm_AB = [ (q0^2 + q1^2 - q2^2 - q3^2) 2*(q1*q2 - q0*q3) 2*(q0*q2 + q1*q3) ; 2*(q1*q2 + q0*q3) (q0^2 - q1^2 + q2^2 - q3^2) 2*(q2*q3 - q0*q1) ; 2*(q1*q3 - q0*q2) 2*(q0*q2 + q1*q3) (q0^2 - q1^2 - q2^2 + q3^2) ; ]; endfunction
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//i/p arg x is a vector x=[1 2 3 4 5 7 89 8]; fc=100; fs=500; y = modulate(x,fc,fs,'pm'); disp(y); //output // column 1 to 3 // // 0.9993771 0.2411607 - 0.8666131 // // column 4 to 6 // // - 0.7182491 0.4712022 0.9696279 // // column 7 to 8 // // - 0.3090170 - 0.9407611 //
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function y = polynomial(x, a) n = size(x, 'r') m = size(a, 'r') y = zeros(n, 1) for i=1:n y(i) = a(1) for j=1:m-1 y(i) = y(i) + a(j + 1)*(x(i)^j) end y(i) = y(i) + (2 * rand() - 1) * 10^-6 end endfunction
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function scs_m=do_icon_edit(scs_m) // do_block - edit a block icon // Copyright INRIA while %t [btn,xc,yc,win,Cmenu]=getclick() if Cmenu<>[] then Cmenu=resume(Cmenu) end K=getblock(scs_m,[xc;yc]) if K<>[] then break,end end gr_i=scs_m(K)(2)(9) if type(gr_i)<>15 then gr_i=list(gr_i,[],list('sd',[0 0 1 1])) end oldwin=xget('window') win=winsid() if win==[] then win=0 else win=max(win)+1 end xset('window',win) xselect() coli=gr_i(2) sd=gr_i(3) sd=gr_menu(sd);xdel(win) txt=sd2sci(sd,['sz(1)','sz(2)'],['orig(1)','orig(2)']) txt(1)=[] gr_i=['thick=xget(''thickness'')'; 'pat=xget(''pattern'')'; 'fnt=xget(''font'')'; txt 'xset(''thickness'',thick)' 'xset(''pattern'',pat)' 'xset(''font'',fnt(1),fnt(2))'] xset('window',oldwin) mac=null();deff('[]=mac()',gr_i,'n') if check_mac(mac) then o=scs_m(K) drawblock(o) o(2)(9)=list(gr_i,coli,sd) drawblock(o) scs_m(K)=o break end
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clc clear //INPUT DATA //C8H18+12.5(O2+3.773N2)=8 CO2 +9 H2O +47.16 N2 ;//FUEL COMPOSITION n=60.66;//number of moles of air //CALCULATIONS n1=8+9+47.16;//number of moles of air and product xs= 15.14/1;//air fuel ratio xs1=1/xs;//fuel air ratio Mr=(1/n)*(114.15+59.66*28.96);//Molecular weights of reactants Mp=(1/n1)*(8*44+9*18+47.16*28);//Molecular weights of products //OUTPUT printf('(i)number of moles of air and product %3.2f \n (ii)(A/F)s %3.2f \n (F/A)s %3.2f \n (iii)Molecular weights of reactants %3.2f \n Molecular weights of products %3.2f',n1,xs,xs1,Mr,Mp)
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/* PROGRAM 3 : //HLL int i = 1 ; int sum = 0 ; while (i < 100) { sum = sum + i ; i = i + 1 ; } //endHLL i : RAM16K[16] sum : RAM16K[17] */ load HackComputer.hdl, //loading hdl file output-file loop.out, //declaring output file output-list RAM64[16]%D1.10.1 RAM64[17]%D1.10.1 ; //output list format ROM32K load loop.hack ; set reset 1, //reset is set to 1 tick, tock , output ; set reset 0, //reset is now set to 0 repeat 1420 { //min clock cycles required=1400 (divided into 100 iterations of 14 clock cycles each) n>1400 will do tick, tock, } output;
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// Example 5.2 l=4; // Layers of Solenoid w=350; // turns Winding s=0.5; // Length of Solenoid n=(l*w)/s; // No.Of turns I=6; // Current in the Solenoid mo=4*%pi*10^-7; // Permeability of free Space B=mo*n*I; // Formula for Megnetic Field at the centre disp('(a) Megnitude of field near the Centre of Solenoid = '+string(B)+' Tesla'); B1=B/2; // Formula for Megnetic Field at the end disp('(b) Megnitude of field at the end of Solenoid = '+string(B1)+' Tesla'); disp('(c) Megnetic Field outside the solenoid is Negligible'); // p 188 5.2
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//Données graphe aléatoire NPE=2; //Nombre de paramètres environnementaux (déchets) //Choix d'un jeu de paramètres de lois gamma ks=[25 35]; //Shape parameters mu=[6 7]; //Moyennes Beta=ks./mu; //Rate parameters NQ=7; //Nombre de quantiles extraits + 1 Q=zeros(NPE,NQ-1); for i=1:NPE for l=1:(NQ-1) Q(i,l)=cdfgam("X",ks(i),Beta(i),l/NQ,1-l/NQ); end end //Grille des indices des paramètres GI=list(); l=ones(1,NPE); i=1; GI(i)=l; h=1; while sum(l)<(NQ-1)*NPE //Construction de la grille des indices l(h)=l(h)+1; if l(h)>NQ-1 l(h)=1; end while l(h)==1 h=h+1; l(h)=l(h)+1; if l(h)>NQ-1 l(h)=1; end end h=1; i=i+1; GI(i)=l; end //Grille des paramètres G=list(); for i=1:size(GI) G(i)=zeros(1,NPE); for h=1:NPE G(i)(h)=Q(h,GI(i)(h)); end end //Voisins (de niveau 1, en indices "uniques") //V1=list(); //Delta=PH(NPE); //for i=1:size(G) // V1(i)=[]; // for j=1:size(Delta) // if min(GI(i)+Delta(j))>=1 & max(GI(i)+Delta(j))<=NQ-1 then // V1(i)=[V1(i) GIinv(NPE,NQ-1,GI(i)+Delta(j))]; // end // end //end //Voisins (de niveau 1, en indices "simples") VP=list(); NMV=1; //Niveau maximal de voisins for i=1:size(G) VP(i)=list(); for k=1:NMV VP(i)(k)=[]; end end Delta=PH(NPE); for i=1:size(G) for j=1:size(Delta) //Voisins de niveau 1 if min(GI(i)+Delta(j))>=1 & max(GI(i)+Delta(j))<=NQ-1 then VP(i)(1)=[VP(i)(1) GIinv(NPE,NQ-1,GI(i)+Delta(j))]; end end for j=1:size(G) //Voisins de niveaux 2 et plus if norm(GI(i)-GI(j),'inf')<=NMV & norm(GI(i)-GI(j),'inf')>=2 VP(i)(norm(GI(i)-GI(j),'inf'))=[VP(i)(norm(GI(i)-GI(j),'inf')) j]; end end end //Matrice des distances (déterministe) c=-1*ones(17,17); for k=2:7 //1=dépôt c(1,k)=2; end for k=2:6 c(k,k+1)=2; end c(2,7)=2; c(2,10)=2; c(7,10)=2; c(7,11)=2; c(6,11)=2; c(3,8)=2; c(4,8)=2; c(4,9)=2; c(5,9)=2; c(8,9)=5; c(10,11)=5; c(8,12)=2; c(3,12)=2; c(12,13)=2; c(3,13)=2; c(2,13)=2; c(13,14)=2; c(2,14)=2; c(10,14)=2; c(9,15)=2; c(5,15)=2; c(15,16)=2; c(5,16)=2; c(6,16)=2; c(16,17)=2; c(6,17)=2; c(11,17)=2; NS=size(c,1); for i=2:NS //Symétrisation for j=1:(i-1) c(i,j)=c(j,i); end end a=bool2s(c>0); //Matrice d'adjacence //Altitudes alt(1)=4; alt(2)=2.5; alt(3)=2; alt(4)=2; alt(5)=3.5; alt(6)=3; alt(7)=3.5; alt(8)=1.5; alt(9)=2.5; alt(10)=2.5; alt(11)=3.5; alt(12)=1; alt(13)=1.5; alt(14)=2; alt(15)=3; alt(16)=3.5; alt(17)=4; //Quantités de déchets P=size(G); //Taille de la grille=nombre de fourmis dans une colonie NS=17; q=zeros(NS,NS,NPE); //Coordonnées de déchets en composantes principales (ici: choix aléatoire) q(1,2,1)=1.3447899; q(1,2,2)=0.4034345; q(1,3,1)=0.7823148; q(1,3,2)=1.6600633; q(1,4,1)=1.175744; q(1,4,2)=0.9658359; q(1,5,1)=0.4465730; q(1,5,2)=1.6801771; q(1,6,1)=0.2411992; q(1,6,2)=0.5710728; q(1,7,1)=1.7215029; q(1,7,2)=1.6988203; q(2,3,1)=1.0514122; q(2,3,2)=1.986242; q(2,7,1)=1.2977126; q(2,7,2)=1.9846382; q(2,10,1)=0.1000840; q(2,10,2)=1.4971013; q(2,13,1)=0.8208118; q(2,13,2)=1.2169053; q(2,14,1)=1.7088422; q(2,14,2)=0.1285293; q(3,4,1)=1.6558166; q(3,4,2)=1.8524688; q(3,8,1)=1.1334423; q(3,8,2)=1.1423278; q(3,12,1)=1.6320221; q(3,12,2)=0.1137856; q(3,13,1)=1.1191873; q(3,13,2)=0.2498681; q(4,5,1)=1.4558445; q(4,5,2)=0.5355533; q(4,8,1)=1.093067; q(4,8,2)=1.9770815; q(4,9,1)=1.4791313; q(4,9,2)=0.0074346; q(5,6,1)=1.1801146; q(5,6,2)=0.6192935; q(5,9,1)=0.5104411; q(5,9,2)=1.2503759; q(5,15,1)=0.2314835; q(5,15,2)=1.2234008; q(5,16,1)=1.3567913; q(5,16,2)=0.6640191; q(6,7,1)=0.0517420; q(6,7,2)=1.0348936; q(6,11,1)=0.7833746; q(6,11,2)=0.4827077; q(6,16,1)=1.012887; q(6,16,2)=0.8472204; q(6,17,1)=0.5787455; q(6,17,2)=0.1775864; q(7,10,1)=1.2425764; q(7,10,2)=0.6909969; q(7,11,1)=1.4129735; q(7,11,2)=1.0422945; q(8,9,1)=0.5740802; q(8,9,2)=1.300559; q(8,12,1)=0.1762670; q(8,12,2)=0.8997527; q(9,15,1)=1.4454506; q(9,15,2)=1.7953593; q(10,11,1)=0.4855644; q(10,11,2)=0.8675442; q(10,14,1)=1.9354106; q(10,14,2)=1.0137069; q(11,17,1)=1.0465953; q(11,17,2)=1.1193895; q(12,13,1)=1.1234614; q(12,13,2)=0.9363520; q(13,14,1)=1.5589093; q(13,14,2)=1.5802144; q(15,16,1)=1.9617084; q(15,16,2)=1.6374132; q(16,17,1)=1.9617084; q(16,17,2)=1.6374132; for i=2:NS //Symétrisation for j=1:(i-1) q(i,j,:)=q(j,i,:); end end q2=zeros(NS,NS,P); //Déchets par paramètres for k=1:P l=GI(k); for i=1:NS for j=1:NS q2(i,j,k)=0; for h=1:NPE q2(i,j,k)=q2(i,j,k)+q(i,j,h)*Q(h,l(h)); end end end end q2A=q2; N=3; //Nombre de tournées C=170; //Capacité d'un camion
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function CheckMatrix(Input, Name) //////////////////////////////////////////////////////////////////////////// // IPD - Image Processing Design Toolbox // // Copyright (c) by Dr. Eng. (J) Harald Galda, 2009 - 2011 // // This program is free software; you can redistribute it and/or modify // it under the terms of the GNU General Public License as published by // the Free Software Foundation; either version 3 of the License, or // (at your option) any later version. // // This program is distributed in the hope that it will be useful, // but WITHOUT ANY WARRANTY; without even the implied warranty of // MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the // GNU General Public License for more details. //////////////////////////////////////////////////////////////////////////// global TYPE_INT; global TYPE_DOUBLE; global TYPE_BOOLEAN; if length(size(Input)) ~= 2 then error('Parameter ''' + Name + ''' must be a 2D matrix.'); end; DataType = type(Input); if (DataType ~= TYPE_DOUBLE) & (DataType ~= TYPE_INT) & (DataType ~= TYPE_BOOLEAN) error('Parameter ' + Name + ' must be numeric or boolean.'); end; endfunction
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//Harriot P.,2003,Chemical Reactor Design (I-Edition) Marcel Dekker,Inc.,USA,pp 436. //Chapter-1 Ex1.4 Pg No. 23 //Title: Activation energy from packed bed data //========================================================================================================= clear clc clf // COMMON INPUT L= [0 1 2 3 4 5 6 9];//Bed length in feet(ft) T=[330 338 348 361 380 415 447 458 ] //Temperature Corresponding the bed length given (°C) R=1.98587E-3;//Gas constant (kcal/mol K) //CALCLATION (Ex1.4.a) //Basis is 1mol of feed A(Furfural) X moles reacted to form Furfuran and CO x=(T-330)./130;//Conversion based on fractional temperature rise n=length (T);//6 moles of steam per mole of Furfural is used to decrease temperature rise in the bed P_mol=x+7;//Total No. of moles in product stream for i=1:(n-1) T_avg(i)= (T(i)+T(i+1))/2 P_molavg(i)= (P_mol(i)+P_mol(i+1))/2 delta_L(i)=L(i+1)-L(i) k_1(i)=((P_molavg(i))/delta_L(i))*log((1-x(i))/(1-x(i+1))) u1(i)=(1/(T_avg(i)+273.15)); end v1=(log(k_1)); i=length(u1); X1=[u1 ones(i,1) ]; result1= X1\v1; k_1_dash=exp(result1(2,1)); E1=(-R)*(result1(1,1)); //OUTPUT (Ex1.4.a) //Console Output mprintf('\n OUTPUT Ex1.4.a'); mprintf('\n========================================================================================\n') mprintf('L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_1') mprintf('\n(ft) \t \t (°C) \t\t \t\t (°C) \t ') mprintf('\n========================================================================================') for i=1:n-1 mprintf('\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1)) mprintf('\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_1(i)) end mprintf('\n\nThe activation energy from the slope =%f kcal/mol',E1 ); //===================================================================================================== //Title: II Order Reaction //========================================================================================================= //CALCULATION (Ex 1.4.b) for i=1:(n-1) T_avg(i)= (T(i)+T(i+1))/2 P_molavg(i)= (P_mol(i)+P_mol(i+1))/2 delta_L(i)=L(i+1)-L(i) k_2(i)=((P_molavg(i))/delta_L(i))*((x(i+1)-x(i))/((1-x(i+1))*(1-x(i)))) u2(i)=(1/(T_avg(i)+273.15)); end v2=(log(k_2)); plot(u1.*1000,v1,'o',u2.*1000,v2,'*'); xlabel("1000/T (K^-1)"); ylabel("ln k_1 or ln k_2"); xtitle("ln k vs 1000/T "); legend('ln k_1','ln k_2'); j=length(u2); X2=[u2 ones(j,1) ]; result2= X2\v2; k_2_dash=exp(result2(2,1)); E2=(-R)*(result2(1,1)); //OUTPUT (Ex 1.4.b) mprintf('\n OUTPUT Ex1.4.b'); mprintf('\n========================================================================================\n') mprintf('L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_2') mprintf('\n(ft) \t \t (°C) \t\t \t\t (°C) \t ') mprintf('\n========================================================================================') for i=1:n-1 mprintf('\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1)) mprintf('\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_2(i)) end mprintf('\n\nThe activation energy from the slope =%f kcal/mol',E2 ); //FILE OUTPUT fid= mopen('.\Chapter1-Ex4-Output.txt','w'); mfprintf(fid,'\n OUTPUT Ex1.4.a'); mfprintf(fid,'\n========================================================================================\n') mfprintf(fid,'L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_1') mfprintf(fid,'\n(ft) \t \t (°C) \t\t \t\t (°C) \t ') mfprintf(fid,'\n========================================================================================') for i=1:n-1 mfprintf(fid,'\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1)) mfprintf(fid,'\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_1(i)) end mfprintf(fid,'\n\nThe activation energy from the slope =%f kcal/mol',E1 ); mfprintf(fid,'\n\n========================================================================================\n') mfprintf(fid,'\n OUTPUT Ex1.4.b'); mfprintf(fid,'\n========================================================================================\n') mfprintf(fid,'L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_2') mfprintf(fid,'\n(ft) \t \t (°C) \t\t \t\t (°C) \t ') mfprintf(fid,'\n========================================================================================') for i=1:n-1 mfprintf(fid,'\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1)) mfprintf(fid,'\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_2(i)) end mfprintf(fid,'\n\nThe activation energy from the slope =%f kcal/mol',E2 ); mclose(all); //============================================================END OF PROGRAM=========================================== //Disclaimer (Ex1.4.a):The last value of tavg and k_1 corresponding to L=9 in Table 1.6 (Pg No. 25)of the textbook is a misprint. // The value should be 452.5 and 4.955476 respectively instead of 455 and 18.2 as printed in the textbook. //Hence there is a change in the activation energy obtained from the code // The answer obtained is 21.3935 kcal/mol instead of 27 kcal/mol as reported in the textbook. //Figure 1.8 is a plot between ln k_1 vs 1000/T instead of k_1 vs 1000/T as stated in the solution of Ex1.4.a //========================================================================================================= //Disclaimer (Ex1.4.b): There is a discrepancy between the computed value of activation energy and value reported in textbook // Error could have been on similar lines as reported for example Ex.1.4.a // Further, intermeidate values for Ex.1.4.b is not available/ reported in textbook and hence could not be compared. //Figure 1.8 is a plot between ln k_2 vs 1000/T instead of k_2 vs 1000/T as stated in the solution of Ex1.4.b
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// sum 25-2 clc; clear; N=800; P=6000; n=200; Cs=1.4; sigb=150; FOS=2; Zp=18; Zg=Zp*N/n; Y=%pi*(0.154-(0.912/Zp)); p=[1 0 -9.5846 -38.135]; function r= myroots (p) a= coeff (p ,0); b= coeff (p ,1); c= coeff (p ,2); d= coeff (p, 3); r(1)=( -b+ sqrt (b^2 -4*a*c ))/(2* a); r(2)=( -b- sqrt (b^2 -4*a*c ))/(2* a); endfunction m=roots(p); m=4.5; dp=m*Zp; dg=m*Zg; // printing data in scilab o/p window printf("dp is %0.0f mm ",dp); printf("\n dg is %0.0f mm ",dg);
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clear // //given i=35 v=220 ra=0.15 n1=1600 //when motor is running at 1200rpm the back emf eb1 is given by eb1=v-(35*0.15) eb1=214.75 //flux phy1 is proportional to armature current ia.Thus, at ia1=35 and ia2=15 n is proportional to eb/phy //2=(eb2*phy1)/(phy2*eb1) //therefore eb2=184.07 //case a //resistance to be connected in series is rse ohm ia2=15 rse=((v-eb2)/ia2)-ra printf("\n rse= %0.1f ohm",rse) //case b eb2=0.5*1.15*214.75 ia2=50 rse=((v-eb2)/ia2)-ra phy1=35 eb2=220-50*0.15 n2=(n1*eb2*phy1)/(1.15*phy1*eb1) printf("\n n2= %0.1f rpm",n2)
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syms s G= 100/(s^2*(s+2)*(s+5)) syms s Kp=limit(s*y/s,s,0) //Kp= position error coefficient Kv=limit(s*G*H,s,0) //Kv= velocity error coefficient Ka=limit(s^2*G*H,s,0) //Ka= accelaration error coefficient disp(Ka ,"Ka = ") disp(Kv ,"Kv = ") disp(Kp ,"Kp = ") Ess=1/(1+Kp) + (1/Kv) + (4/Ka) disp(Ess, "Ess = ")
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//pagenumber 553 example 1 clear slope1=130; trivol=15;//volt d=0.5;//watts ig=sqrt(d/slope1); vg=slope1*ig; r=(trivol-vg)/ig; disp("source resistance = "+string((r))+"ohm");
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// page no 310 // example 10.1 // BCD TO BINARY // BCD into its binary equivalent. // given BCD no is 72 clc; a=72; x=modulo(a,10); // seperating the units digit printf('Unpacked BCD1 ') disp(dec2bin(x,8)); a=a/10; // seperating the tens place digit a=floor(a); printf('\n \n Unpacked BCD2'); disp(dec2bin(a,8)); printf('\n \n Multiply BCD2 by 10 and add BCD1');
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// To calculate frequency, instantaneous voltage and time of a voltage wave clc; clear; // The volatage eqaution is v= 0.02 sin (4000t + 30(degress)). Vm=0.02; deff('a=vol(b)','a=Vm*sind(((4000*b)*(180/%pi))+30)'); // Function for voltage equation t=320*(10^-6); w=4000; // angular frequency // General expression for voltage is given by V=Vm sin ()(2*pi*f*t)+theta) // Comparing both the eqautions we get 2*pi*f=4000 f=w/(2*%pi); v=vol(t); // 360degress is equal to 1/f s. //Refer the diagram with this code to understand better. // 30degress is t30=30/(f*360); disp('Hz',f,'The frequency of the voltage wave =') disp('V',v,'The instantaneous voltage at t= 320 micro seconds =') disp('s',t30,'The time represented by 30 degrees phase difference =')
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// Exa 8.14 clc; clear; close; // Given data Rf = 250;// in kohm Vo= '-5*Va+3*Vb';// given expression // But output voltage of difference amplifier is // Vo= -Rf/R1*Va+(R2/(R1+R2))*(1+Rf/R1)*Vb (i) // By comparing (i) with given expression R1 = Rf/5;// in kohm disp(R1,"The value of R1 in kΩ is : "); // (R2/(R1+R2))*(1+Rf/R1)= 3 R2= 3*R1^2/(R1+Rf-3*R1);// in kΩ disp(R2,"The value of R2 in kΩ is : ") // Note: There is calculation error to find the value of R2 in the book.
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// Display mode mode(0); // Display warning for floating point exception ieee(1); clear; clc; disp("Turbomachinery Design and Theory,Rama S. R. Gorla and Aijaz A. Khan, Chapter 1, Example 4") disp ("Theoritical Question") //liquid discharge rate Q; head H;specific weight of the liquid is w. disp("Expression for Pumping power is P = kwQH")
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function r = biseccion(f, a, b, e) c = (a + b) / 2 if b - c <= e then r = c else if f(b)*f(c) <= 0 then r = biseccion(f, c, b, e) else r = biseccion(f, a, c, e) end end endfunction function y = p(x) y = (x - 3) * (x + 3) endfunction assert_checkalmostequal(biseccion(p, -4, -2, 0.1), -3, 0.1) assert_checkalmostequal(biseccion(p, 2, 4, 0.1), 3, 0.1)
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Vdot=20 //CC/min x=0.015 MH2O=18.02 //g DH2O=1 //g/CC x1=0.2
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6_6.sce
clc; clear; //Example 6.6 mf_dot=5000 //[kg/h] ic=0.01 //Initial concentration [kg/h] fc=0.02 //Final concentration [kg/h] T=373 //Boiling pt of saturation in [K] Ts=383 //Saturation temperature of steam in [K] mdash_dot=ic*mf_dot/fc //[kg/h] mv_dot=mf_dot-mdash_dot //Water evaporated in [kg/h] Hf=125.79 //[kJ/kg] Hdash=419.04 //[kJ/kg] Hv=2676.1 //[kJ/kg] lambda_s=2230.2 //[kJ/kg] ms_dot=(mdash_dot*Hdash+mv_dot*Hv-mf_dot*Hf)/lambda_s //Steam flow rate in [kg/h] eco=mv_dot/ms_dot //Steam economy Q=ms_dot*lambda_s //Rate of heat transfer in [kJ/h] Q=Q*1000/3600 //[J/s] dT=Ts-T //[K] A=69 //Heating area of evaporator in [sq m] U=Q/(A*dT) //Overall heat transfer coeff in [W/sq m.K] printf("\nSteam economy is %f\n",eco); printf("\n\nOverall heat transfer coefficient is %d W/sq m.K",round(U));
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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 // Original Source : https://octave.sourceforge.io/signal/ // Modifieded by:Sonu Sharma, RGIT Mumbai // Organization: FOSSEE, IIT Bombay // Email: toolbox@scilab.in function [n, Wp] = ellipord(Wp, Ws, Rp, Rs) //Minimum filter order of a digital elliptic or Cauer filter with the desired response characteristics. //Calling Sequence //[n] = ellipord(Wp, Ws, Rp, Rs) //[n, Wp] = ellipord(Wp, Ws, Rp, Rs) //Parameters //Wp: scalar or vector of length 2 (passband edge(s)), all elements must be in the range [0,1] //Ws: scalar or vector of length 2 (stopband edge(s)), all elements must be in the range [0,1] //Rp: passband ripple in dB. //Rs: stopband attenuation in dB. //n: Minimum order of filter satisfying given specs. //Description //This function computes the minimum filter order of an elliptic filter with the desired response characteristics. //Stopband frequency ws and passband frequency wp specify the the filter frequency band edges. //Frequencies are normalized to the Nyquist frequency in the range [0,1]. //Rp is measured in decibels and is the allowable passband ripple and Rs is also measured in decibels and is the minimum attenuation in the stop band. //If ws>wp then the filter is a low pass filter. If wp>ws, then the filter is a high pass filter. //If wp and ws are vectors of length 2, then the passband interval is defined by wp and the stopband interval is defined by ws. //If wp is contained within the lower and upper limits of ws, the filter is a band-pass filter. If ws is contained within the lower and upper limits of wp, the filter is a band-stop or band-reject filter. //Examples //Wp = [60 200]/500; //Ws = [50 250]/500; //Rp = 3; //Rs = 40; //[n,Wp] = ellipord(Wp,Ws,Rp,Rs) //Output : // Wp = // // 0.12 0.4 // n = // // 5. funcprot(0); [nargout nargin] = argn(); if (nargin ~= 4) error("ellipord: invalid number of inputs"); else validate_filter_bands ("ellipord", Wp, Ws); end // sampling frequency of 2 Hz T = 2; Wpw = tan(%pi.*Wp./T); // prewarp Wsw = tan(%pi.*Ws./T); // prewarp // pass/stop band to low pass filter transform: if (length(Wpw)==2 & length(Wsw)==2) wp=1; w02 = Wpw(1) * Wpw(2); // Central frequency of stop/pass band (square) w3 = w02/Wsw(2); w4 = w02/Wsw(1); if (w3 > Wsw(1)) ws = (Wsw(2)-w3)/(Wpw(2)-Wpw(1)); elseif (w4 < Wsw(2)) ws = (w4-Wsw(1))/(Wpw(2)-Wpw(1)); else ws = (Wsw(2)-Wsw(1))/(Wpw(2)-Wpw(1)); end elseif (Wpw > Wsw) wp = Wsw; ws = Wpw; else wp = Wpw; ws = Wsw; end k=wp/ws; k1=sqrt(1-k^2); q0=(1/2)*((1-sqrt(k1))/(1+sqrt(k1))); q= q0 + 2*q0^5 + 15*q0^9 + 150*q0^13; //(....) D=(10^(0.1*Rs)-1)/(10^(0.1*Rp)-1); n=ceil(log10(16*D)/log10(1/q)); endfunction
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//Exa 10.8 clc; clear; close; //Given data : wc=1;//kg/m L=280;//m D=20;//mm r=10;//mm Pw=40;//kg/m^2(Wind pressure) rho_i=910;//kg/m^3(density of ice) U_stress=10000;//kg/cm^2 SF=2;//factor of safety wi=rho_i*%pi*r*10^-3*(D+r)*10^-3;//kg w_w=Pw*(D+2*r)*10^-3;//kg wr=sqrt((wc+wi)^2+w_w^2);//kg(Resultant force per m length of conductor) T=U_stress/SF;//kg Smax=wr*L^2/8/T;//msag in air disp(Smax,"Maximum Sag(meter)");
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Ex4_4.sce
clear // //Initilization of Variables //Flange (Top) b1=80 //mm //Width t1=40 //mm //Thickness //Flange (Bottom) b2=160 //mm //width t2=40 //mm //Thickness //web d=120 //mm //Depth t3=20 //mm //Thickness D=200 //mm //Overall Depth sigma1=30 //N/mm**2 //Tensile stress sigma2=90 //N/mm**2 //Compressive stress L=6000 //mm //Span //Calculations //Distance of centroid from bottom fibre y_bar=(b1*t1*(D-t1*2**-1)+d*t3*(d*2**-1+t2)+b2*t2*t2*2**-1)*(b1*t1+d*t3+b2*t2)**-1 //mm //Moment of Inertia I=1*12**-1*b1*t1**3+b1*t1*(D-t1*2**-1-(y_bar))**2+1*12**-1*t3*d**3+t3*d*(d*2**-1+t2-(y_bar))**2+1*12**-1*b2*t2**3+b2*t2*(t2*2**-1-(y_bar))**2 //Extreme fibre distance of top and bottom fibres are y_t and y_c respectively y_t=y_bar //mm y_c=D-y_bar //mm //Moment carrying capacity considering Tensile strength M1=sigma1*I*y_t**-1*10**-6 //KN-m //Moment carrying capacity considering compressive strength M2=sigma2*I*y_c**-1*10**-6 //KN-m //Max Bending moment in simply supported beam 6 m due to u.d.l //M_max=w*L*10**-3*8**-1 //After simplifying further we get //M_max=4.5*w //Now Equating it to Moment carrying capacity, we get load carrying capacity w=M1*4.5**-1 //KN/m //Result printf("\n Max Uniformly Distributed Load is %0.3f KN/m",w)
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// Scilab Code Ex14.3 Calculating electric polarizability of a molecule from its susceptibility: Page-464 (2010) NA = 6.023e+023; // Avogadro's number epsilon_0 = 8.85e-012; // Electrical permittivity of free space, coulomb square per newton per metre chi = 0.985e-03; // Electrical susceptibility of carbon-dioxide molecule rho = 1.977; // Density of carbon-dioxide, kg per metre cube M = 44e-03; // Molecular weight of CO2, kg N = NA*rho/M; // Number of molecules per unit volume, per metre cube alpha = epsilon_0*chi/N; // Total electric polarizability of carbon-dioxide, farad-metre square printf("\nThe total electric polarizability of carbon-dioxide = %4.2e farad-metre square", alpha); // Result // The total electric polarizability of carbon-dioxide = 3.22e-040 farad-metre square
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clc; //ex2.12 Vb=0.7; //volt If=[0.001 0.005]; //Ampere Rb=5; //ohm Vf1=Vb+If(1,1)*Rb; //Volt//VF=VB+If*Rb; Vf2=Vb+If(1,2)*Rb; //Volt//VF=VB+If*Rb; disp('mV',Vf1,"Vf1="); disp('mV',Vf2,"Vf2=");
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//Example No. 14.6.2 clc; clear; close; format('v',7); f=150;//MHz(frequency) c=3*10^8;//m/s(speed of light) GT=1.64;//dB(Transmitter gain) PT=20;//W(Transmitted power) d=50;//km(distance) lambda=c/(f*10^6);//m(Wavelength) E=sqrt(30*GT*PT)/(d*1000);//V/m(emf induced) le=lambda/%pi;//m(Effective length) Voc=E*le;//V/m(Open circuit voltage) disp(Voc*10^6,"Open circuit voltage in micro Volt : ");
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sce
Ex1_2_17.sce
//Section-1,Example-4,Page no.-AC.205 //To find the air required for the perfect combustion of 1 m^3 of the given gas. clc; T=0.22 L_O2=0.02 Net_O2=0.2 Plus_CO2=0.05 T_CO2=T+L_O2+Plus_CO2 T_N2=1.6 T_O2=Net_O2*(40/100) T_W=T_CO2+T_N2+T_O2 M_Q=Net_O2*(100/21) P_CO2=(T_CO2/T_W)*100 disp(P_CO2,'Percentage composition of CO_2') P_N2=(T_N2/T_W)*100 disp(P_N2,'Percentage composition of N_2') P_O2=(T_O2/T_W)*100 disp(P_O2,'Percentage composition of O_2')