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function P=%hm_degree(P) P('entries')=degree(P('entries'))
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//load the image read and write commands exec pgmread.sci //load the image im1 = pgmread('cells_scale.pgm'); // NOTE : syntax is im = pgmread('PATH'); where PATH is full location of image file //print the image size disp(size(im1),'size(im)='); [m,n]=size(im1); //display the original image scf(); xset("colormap",graycolormap(256)); Matplot(im1,strf='046'); //scale factor "scale" scale=1.3 //Do the translation with bilinear interpolation for T -> S mapping for i=1:m for j=1:n xs=i/scale; ys=j/scale; x=floor(xs); y=floor(ys); a=xs-x; b=ys-y; x=min(max(1,x),m-1); //for the edge cases y=min(max(1,y),n-1); //for the edge cases im2(i,j)=im1(x,y)*(1-a)*(1-b)+im1(x,y+1)*(1-a)*b+im1(x+1,y)*a*(1-b)+im1(x+1,y+1)*a*b; end end //display the translated image scf(); xset("colormap",graycolormap(256)); Matplot(im2,strf='046');
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# compared to Ex4_gn, the result has a different set of normalizer generators, # but the normalizer is the same echo "### Test Ex4-1" ../bin/carat/Normalizer Ex4_g echo "### Ex4-1 return code $?" echo "### Test Ex4-2" ../bin/carat/Vector_systems Ex4_gn echo "### Ex4-2 return code $?" echo "### Test Ex4-3" ../bin/carat/Extract -r Ex4_gn Ex4_out echo "### Ex4-3 return code $?" echo "### Test Ex4-4" ../bin/carat/Tr_bravais Ex4_gn echo "### Ex4-4 return code $?" echo "### Test Ex4-5" ../bin/carat/Sublattices -b Ex4_gn_tr echo "### Ex4-5 return code $?" for i in 2 3 4 5 6 ; do for j in 2 3 4 ; do echo "### Test Ex4-6-g.$i.$j" ../bin/carat/Conj_bravais -i "Ex4_g.$i" "Ex4_L.$j" > "g.$i.$j" echo "### Ex4-6-g.$i.$j return code $?" done done cat g.?.? for i in 2 3 4 5 6 ; do for j in 2 3 4 ; do echo "### Test Ex4-7-pg.$i.$j" ../bin/carat/Extract -p "g.$i.$j" > "Ex4_pg.$i.$j" echo "### Ex4-7-pg.$i.$j return code $?" echo "### Test Ex4-7-cg.$i.$j" ../bin/carat/Extract -c "g.$i.$j" > "Ex4_cg.$i.$j" echo "### Ex4-7-cg.$i.$j return code $?" done done cat Ex4_pg.?.? cat Ex4_cg.?.? rm -f g.?.? Ex4_pg.?.? Ex4_cg.?.?
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clc // 2H2 + O2 → 2H2O // 2CO + O2 → 2CO2 // CH4 + 2O2 → CO2 + 2H2O // C4H8 + 6O2 → 4CO2 + 4H2O n_O2=0.853; //total moles of O2 disp("(i) Stoichiometric A/F ratio =") AF=n_O2/0.21; disp(AF) disp("(ii) Wet and dry analyses of the products of combustion if the actual mixture is 30% weak :") AF_act=AF+0.3*AF; n_N2=0.79*AF_act; O2_excess=0.21*AF_act-n_O2; n_wet=5.899; n_dry=4.915; disp("Analysis by volume of wet products is as follows :") disp("CO2 =") CO2=0.490/n_wet*100; disp(CO2) disp("%") disp("H2O =") H2O=0.984/n_wet*100; disp(H2O) disp("%") disp("O2 =") O2=O2_excess/n_wet*100; disp(O2) disp("%") disp("N2 =") N2=n_N2/n_wet*100; disp(N2) disp("%") disp("Analysis by volume of dry products is as follows :") disp("CO2 =") CO2=0.490/n_dry*100; disp(CO2) disp("%") disp("O2 =") O2=O2_excess/n_dry*100; disp(O2) disp("%") disp("N2 =") N2=n_N2/n_dry*100; disp(N2) disp("%")
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errcatch(-1,"stop");mode(2);//Ex:3.5 ; ; I_in=5;//in mA R_m=100; I_m=1; R_s=R_m*I_m/(I_in-1); printf("Value of parallel shunt resistor = %d A",R_s); exit();
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//Reactions developed at contacts //Refer fig. 2.25(a),(b and (c) //consider equilibrium of cylinder 1 //using conditions of equilibrium we get RA=500*cosd(30) //N RB=500*sind(30) //N //Consider equilibrium of cylinder 2 //using conditions of equilibrium we get RC=(500+250*sind(30))/cosd(30) //N RD=RC*sind(30)+250*cosd(30) //N printf("\nThe reactions are:-\nRA=%.1f N\nRB=%.1f N\nRC=%.1f N\nRD=%.1f N",RA,RB,RC,RD)
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clear; clc; close; w=-10:0.1:10; Hw=[exp(%i*%pi/2)*ones(1,find(w==0)) exp(-%i*%pi/2)*ones(1,find(w==0)-1)]; d=gca(); plot(w,imag(Hw)); poly1=d.children.children; poly1.thickness=3; poly1.foreground=2; xtitle('H(w)','w') disp("H(w)=-%i*sgn(w)"); disp("we know sgn(t) <--> 2/(j*w)") disp("by duality property 2/(j*t) <-->2*%pi*sgn(-w)=-2*%pi*sgn(w) ") disp("therefore 1/(%pi*t) <--> -j*sgn(w)"); t=0.1:0.1:10; h=ones(1,length(t))./(%pi*t); figure d=gca() plot(t,h); poly1=d.children.children; poly1.thickness=3; poly1.foreground=2; xtitle('h(t)','t') w0=2; x=cos(w0*t); figure d=gca(); plot(t,x); poly1=d.children.children; poly1.thickness=3; poly1.foreground=2; xtitle('x(t)','t') y=convol(x,h); figure d=gca() plot(t,y(1:length(t))); poly1=d.children.children; poly1.thickness=3; poly1.foreground=2; xtitle('y(t)','t')
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clear; clc; xdel(winsid()); // parameters ========================================================== band_file='C:\MyDrive\Work\TaAs_qe\ONCV\TaAs.w90-bands.dat' gnu_file='C:\MyDrive\Work\TaAs_qe\ONCV\TaAs.w90-bands.gnu' E_Fermi=21.2 E_bound=[-2,+2] // Main ================================================================ //read band data fid=mopen(band_file,'r'); band_data=mfscanf(-1,fid,'%f %f %f') band_data(:,2)=band_data(:,2)-E_Fermi; mclose(fid) // read gnu data fid=mopen(gnu_file,'r'); gnu_data=mgetl(fid,-1) k_div_data=grep(gnu_data,'set arrow from') k_div_point=zeros(size(k_div_data)(2),1) for n=1:size(k_div_data)(2) [tot,s1,s2,s3,f]=msscanf(gnu_data(k_div_data(n)),'%s %s %s %f') k_div_point(n)=f end mclose(fid) //plot band structure xset("colormap",jetcolormap(256)) scatter(band_data(:,1),band_data(:,2),ones(band_data(:,1)),band_data(:,3)) colorbar(min(band_data(:,3)),max(band_data(:,3))) tmp=find(band_data(:,1)<=1e-5) tot_k=tmp(2)-tmp(1); // plot k-path divider for n=1:length(k_div_point) plot(k_div_point(n)*ones(20,1),linspace(E_bound(1),E_bound(2),20)','k:') end // plot E_Fermi plot((1:tot_k)',0*ones(tot_k,1),'r:') // setup a=gca(); a.data_bounds=[1, E_bound(1);band_data(tot_k), E_bound(2)]; a.tight_limits='on' a.font_size=4 a.thickness=3 title('Band Structure','fontsize',4); xlabel('$k$','fontsize',4); ylabel('Energy',"fontsize", 4);
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ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 0.439042D+00 2 -0.558038D-02 0.374993D-02 3 -0.623768D-03 0.603050D-04 0.302431D-02 4 0.705393D+00 0.248313D-01 0.867094D-01 0.168929D+03 5 0.802279D-01 0.376524D-02 0.384066D+00 0.102577D+02 0.111991D+03 6 0.189085D+01 0.102135D-01 -0.331391D+00 -0.246048D+02 -0.702699D+02 7 -0.153372D-01 0.113458D-01 -0.561479D-03 0.129944D+01 -0.329257D+00 8 0.188254D-02 -0.311251D-03 0.117092D-03 -0.154486D+00 -0.273832D-01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 6 7 8 ________ ________ ________ 6 0.583727D+03 7 0.612445D+01 0.823097D+00 8 -0.254143D+01 -0.759194D-01 0.249204D-01 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 1.000 2 -0.138 1.000 3 -0.017 0.018 1.000 4 0.082 0.031 0.121 1.000 5 0.011 0.006 0.660 0.075 1.000 6 0.118 0.007 -0.249 -0.078 -0.275 7 -0.026 0.204 -0.011 0.110 -0.034 8 0.018 -0.032 0.013 -0.075 -0.016 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 6 7 8 ________ ________ ________ 6 1.000 7 0.279 1.000 8 -0.666 -0.530 1.000
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clc clear //Input data Vs=0.0028;//Swept volume in m^3 N=3000;//Speed of the engine in rpm ip=12.5;//The average indicated power developed in kW/m^3 nv=85;//Volumetric efficiency in percent p1=1.013;//The atmospheric pressure in bar T1=288;//The atmospheric temperature in K ni=74;//Isentropic efficiency in percent pr=1.6;//The pressure ratio nm=78;//All mechanical efficiencies in percent g=1.4;//Adiabatic index R=287;//Real gas constant in J/kgK Cp=1.005;//The specific heat of gas in kJ/kgK //Calculations Vs1=(Vs*(N/2));//Volume swept by the piston per minute in m^3/min Vi=(nv/100)*Vs1;//Unsupercharged induced volume in m^3/min p2=pr*p1;//Blower delivery pressure in bar T21=T1*(p2/p1)^((g-1)/g);//Temperature after isentropic compression in K T2=T1+((T21-T1)/((ni/100)));//Blower delivery temperature in K Ve=(Vs1*p2*T1)/(T2*p1);//Equivalent volume at 1.013 bar and 15 degree centigrade in m^3/min nv1=[Ve/Vs1]*100;//Volumetric efficiency of supercharged engine in percent Vii=Ve-Vi;//Increase in induced volume in m^3/min ipa=ip*Vii;//Increase in ip from air induced in kW ipi=[(p2-p1)*10^5*Vs1]/(60*1000);//Increase in ip due to increased induction pressure in kW ipt=ipa+ipi;//Total increase in ip in kW bp=ipt*(nm/100);//Increase in engine bp in kW ma=(p2*(Vs1/60)*10^5)/(R*T2);//Mass of air delivered per second by blower in kg/s P=ma*Cp*(T2-T1);//Power input to blower in kW Pd=P/(nm/100);//Power required to drive the blower in kW bpn=bp-Pd;//Net increase in bp in kW bpu=ip*Vi*(80/100);//The bp of unsupercharged engine in kW bpp=(bpn/(bpu))*100;//Percentage increase in bp in percent //Output printf('The volumetric efficiency of supercharged engine = %3.0f percent \n The increase in brake power by supercharging = %3.2f kW \n The percentage increase in brake power = %3.1f percent ',nv1,bpn,bpp)
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// Test # 2 : Excess Input Arguments exec('./zpklp2lp.sci',-1); [z,p,k,n,d]=zpklp2lp(0.3,0.2,0.7,0.5,0.6,4); //!--error 58 //Wrong number of input arguments
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@relation glass @attribute RI real[1.51115,1.53393] @attribute Na real[10.73,17.38] @attribute Mg real[0.0,4.49] @attribute Al real[0.29,3.5] @attribute Si real[69.81,75.41] @attribute K real[0.0,6.21] @attribute Ca real[5.43,16.19] @attribute Ba real[0.0,3.15] @attribute Fe real[0.0,0.51] @attribute typeGlass{1,2,3,4,5,6,7} @inputs RI,Na,Mg,Al,Si,K,Ca,Ba,Fe @outputs typeGlass @data 1 3 1 1 1 1 1 1 1 1 1 2 1 1 2 2 2 2 2 2 2 2 2 6 2 2 2 2 2 7 3 2 3 2 5 7 5 2 6 6 7 2 7 7 7 7
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//Graphical// //Example 10.5.1 //Decimation by 2, Filter Length = 30 //Cutoff Frequency Wc = %pi/2 //Pass band Edge frequency fp = 0.25 and a Stop band edge frequency fs = 0.31 // Choose the number of cosine functions and create a dense grid // in [0,0.25] and [0.31,0.5] //magnitude for pass band = 1 & stop band = 0 (i.e) [1 0] //Weighting function =[2 1] clear; clc; close; M = 30; //Filter Length D = 2; //Decimation Factor = 2 Wc = %pi/2; //Cutoff Frequency Wp = Wc/(2*%pi); //Passband Edge Frequency Ws = 0.31; //Stopband Edge Frequency hn=eqfir(M,[0 Wp;Ws .5],[1 0],[2 1]); [hm,fr]=frmag(hn,256); disp('The LPF Filter Coefficients are:') hn //Obtaining Polyphase Filter Coefficients from hn p = zeros(D,M/D); for k = 1:D for n = 1:(length(hn)/D) p(k,n) = hn(D*(n-1)+k); end end disp('The Polyphase Decimator for D =2 are:') p figure plot(fr,hm) xlabel('Normalized Digital Frequency fr'); ylabel('Magnitude'); title('Frequency Response of FIR LPF using REMEZ algorithm M=61') figure plot(.5*(0:255)/256,20*log10(frmag(hn,256))); xlabel('Normalized Digital Frequency fr'); ylabel('Magnitude in dB'); title('Frequency Response of DECIMATOR (D=2) using REMEZ algorithm M=30')
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FLAGS='--detect' STDIN='' STDOUT='NONE\n' STDERR='' EXITVAL='0'
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clear; clc; close; x = [5-4*%i,2-2*%i,-4,-6]; N = [0,1,2,3]; //Plotting magnitude of input signal scf(1); title('Plot of magnitude of x(n)','FontSize',4); xlabel('n','FontSize',4); ylabel('|x|','FontSize',4); plot2d3(N, abs(x)); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin"; //Plotting phase of input signal scf(2); title('Plot of phase of x(n)','FontSize',4); xlabel('n','FontSize',4); ylabel('∠x','FontSize',4); plot2d3(N,atan(imag(x),real(x))); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin"; //Calculating and plotting the magnitude of DFT of the input signal n = length(x); X = zeros(1,n); for k=0:n-1 for j=0:n-1 X(k+1) = X(k+1)+x(j+1)*(cos(2*%pi*k*j/n)-sin(2*%pi*k*j/n)*1*%i); end end disp('DFT of input signal : '); disp(X); scf(3); title('Plot of magnitude of X(k)','FontSize',4); xlabel('n','FontSize',4); ylabel('|x|','FontSize',4); plot2d3(N, abs(X)); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin"; //Calculating and plotting the phase of DFT of the input signal scf(4); title('Plot of phase of X(k)','FontSize',4); xlabel('n','FontSize',4); ylabel('∠x','FontSize',4); plot2d3(N,atan(imag(X),real(X))); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin"; //Calculating and plotting magnitude of the IDFT of the spectrum x1 = zeros(1,n); for k=0:n-1 for j=0:n-1 x1(k+1) =x1(k+1)+X(j+1)*(cos(2*%pi*k*j/n)+sin(2*%pi*k*j/n)*1*%i); end; x1(k+1) = x1(k+1)/n; end; disp('IDFT of output signal :'); disp(x1); scf(5); title('Plot of magnitude ofIDFT of X(k)','FontSize',4); xlabel('n','FontSize',4); ylabel('|x|','FontSize',4); plot2d3(N, abs(x1)); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin"; //Calculating and plotting phase of the IDFT of the spectrum scf(6); title('Plot of phase of IDFT of X(k)','FontSize',4); xlabel('n','FontSize',4); ylabel('∠x','FontSize',4); plot2d3(N,atan(imag(x1),real(x1))); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin"; //Appending zeros at the end ofthe input signal and calculating and plotting DFT N1 = [0,1,2,3,4,5,6,7]; x2 = [5-4*%i,2-2*%i,-4,-6,0,0,0,0]; scf(7); title('Plot of magnitude of x(n) with zeros appended','FontSize',4); xlabel('n','FontSize',4); ylabel('|x|','FontSize',4); plot2d3(N1, abs(x2)); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin"; scf(8); //plotframe([0 -2 3 2]); title('Plot of phase of x(n)with zeros appeneded','FontSize',4); xlabel('n','FontSize',4); ylabel('∠x','FontSize',4); plot2d3(N1,atan(imag(x2),real(x2))); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin"; X = zeros(1,length(x2)); for k=0:length(x2)-1 for j=0:length(x2)-1 X(k+1) = X(k+1)+x2(j+1)*(cos(2*%pi*k*j/length(x2))-sin(2*%pi*k*j/length(x2))*%i); end end disp('DFT of input signal with zeros appended at the end : '); disp(X); scf(9); title('Plot of magnitude of X(k) with zeroes appended','FontSize',4); xlabel('n','FontSize',4); ylabel('|x|','FontSize',4); plot2d3(N1,abs(X)); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin"; scf(10); title('Plot of phase of X(k) with zeroes appended','FontSize',4); xlabel('n','FontSize',4); ylabel('∠x','FontSize',4); plot2d3(N1,atan(imag(X),real(X))); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin"; //Adding zeros at alternateindices and calculating andplotting DFT x3 = [5-4*%i,0,2-2*%i,0,-4,0,-6,0] scf(11); title('Plot of magnitude of x(n) with zeroes at alternate positions','FontSize',4); xlabel('n','FontSize',4); ylabel('|x|','FontSize',4); plot2d3('gnn',N1, abs(x3)); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin"; scf(12); title('Plot of phase of x(n) with zeroes at alternate positions','FontSize',4); xlabel('n','FontSize',4); ylabel('∠x','FontSize',4); plot2d3(N1,atan(imag(x3),real(x3))); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin"; X1 = zeros(1,length(x3)); for k=0:length(x3)-1 for j=0:length(x3)-1 X1(k+1) = X1(k+1)+x3(j+1)*(cos(2*%pi*k*j/length(x3))-sin(2*%pi*k*j/length(x3))*1*%i); end; end; disp('DFT of input signal with alternate zeros: '); disp(X1); scf(13); title('Plot of magnitude of X(k) with zeroes at alternate positions','FontSize',4); xlabel('n','FontSize',4); ylabel('|x|','FontSize',4); plot2d3(N1, abs(X1)); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin"; scf(14); title('Plot of phase of X(k) with zeroes at alternate positions','FontSize',4); xlabel('n','FontSize',4); ylabel('∠x','FontSize',4); plot2d3(N1,atan(imag(X1),real(X1))); e = gce(); e.children.thickness = 4; a = gca(); a.x_location = "origin"; a.y_location = "origin";
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//Chapter-10, Example 10.26, Page 451 //============================================================================= clc clear //INPUT DATA e=0.6;//Emissivity of thermocouple Ta=20+273;//Ambient temperature in K Tt=500+273;//Temperature from the thermocouple in K e=0.3;//Emissivity of radiation shield h=200;//Convective heat transfer coefficient in W/m^2.K Ts=833;//Temperature in K //CALCULATIONS T=((5.67*10^-8*e*(Tt^4-Ta^4))/(h*1000))+Tt;//Temperature of the shield in K T1=(Ts-T);//Error between the thermocouple temperature and gas temperature in K Ts=825;//Surface temperature with radiation shield in K Tc=829;//Thermocouple temperature with radiation shield in K e=(Tc-Ts);//Error between the thermocouple temperature and gas temperature with the shielded thermocouple arrangement in K //OUTPUT mprintf('Error between the thermocouple temperature and gas temperature is%3.0f K \nError between the thermocouple temperature and gas temperature with the shielded thermocouple arrangement is%3.0f K',T1,e) //=================================END OF PROGRAM==============================
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//To Calculate the smallest Diffraction Angle //Example 20.1 clc; clear; a=2.6*10^-10;//Edge Length of Cubic Lattice h=1;//Miller Indice h k=1;//Miller Indice k l=1;//Miller Indice l lambda=1.542*10^-10;//Wavelength of light theta=asin(lambda*sqrt(h^2+k^2+l^2)/(2*a))*180/%pi; printf("Smallest Diffraction Angle=%.1f degrees",theta);
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// problem 2.12 d=4 s1=0.6 s2=0.9 l=1 h=s1*l/s2 cob=h/2 cog=l/2 dcog=cog-cob i=3.142*d*d*d*d/64 v=3.142*0.25*d*d*h bm=i/v bm=dcog l=(6*1.5)^0.5 disp(l,"maximium lenght of cylinder in m")
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// Model derivation for GPC design in Example 12.1 on page 439. // 12.1 exec('xdync.sci',-1); exec('polmul.sci',-1); exec('flip.sci',-1); exec('rowjoin.sci',-1); exec('polsize.sci',-1); exec('left_prm.sci',-1); exec('t1calc.sci',-1); exec('indep.sci',-1); exec('seshft.sci',-1); exec('makezero.sci',-1); exec('move_sci.sci',-1); exec('colsplit.sci',-1); exec('clcoef.sci',-1); exec('cindep.sci',-1); // Camacho and Bordon's GPC example; model formation A=[1 -0.8]; dA=1; B=[0.4 0.6]; dB=1; N=3; k=1; D=[1 -1]; dD=1; AD=convol(A,D); dAD=dA+1; Nu=N+k; zj = 1; dzj = 0; G = zeros(Nu); H1 = zeros(Nu,k-1+dB); H2 = zeros(Nu,dA+1); for j = 1:Nu, zj = convol(zj,[0,1]); dzj = dzj + 1; [Fj,dFj,Ej,dEj] = xdync(zj,dzj,AD,dAD,1,0); [Gj,dGj] = polmul(B,dB,Ej,dEj); G(j,1:dGj) = flip(Gj(1:dGj)); H1(j,1:k-1+dB) = Gj(dGj+1:dGj+k-1+dB); H2(j,1:dA+1) = Fj; end G,H1,H2
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//<s>=%lssls(s1,s2) // //! s=s1*inv(s2) //end
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clear;close;clc; max_limit=10; h=[2 5 0 4]; n2=-2:length(h)-3; x=[4 1 3]; n1=-1:length(x)-2; y=convol(x,h); n=-3:length(x)+length(h)-5; a=gca(); subplot(211); plot2d3('gnn',n2,h) xtitle('impulse Response','n','h[n]'); a.thickness=2; a.y_location="origin"; a=gca(); subplot(212); plot2d3('gnn',n1,x) a.y_location="origin"; xtitle('input response','n','x[n]'); xset("window",1); a=gca(); plot2d3('gnn',n,y) a.y_location="origin"; a.x_location="origin"; xtitle('output response','n','y[n]');
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//Function to round-up a value such that it is divisible by 5 function[v] = round_five(w) v = ceil(w) rem = pmodulo(v,5) if (rem ~= 0) then v = v + (5 - rem) end endfunction //Obtain path of solution file path = get_absolute_file_path('solution5_21.sce') //Obtain path of data file datapath = path + filesep() + 'data5_21.sci' //Clear all clc //Execute the data file exec(datapath) //Calculate weight of the object W (N) W = m * 9.81 //Calculate maximum bending moment Mb (N-mm) Mb = (W * l)/4 //Assume length of one side of the beam cross-section to be 1mm a a = 1 //Calculate second moment of area of beam cross-section I (mm4) I = (a^4)/12 //Calculate value of y (mm) y = a/2 //Calculate static stress sigmaS (N/mm2) sigmaS = (Mb * y)/I //Calculate static deflection deltaS (mm) deltaS = (W * (l^3))/(48 * E * I) //Calculate permissible stress P (N/mm2) P = Syt/fs //Calculate the actual value of a (mm) //Coefficients of the resulting cubic equation a = (P^2)/(sigmaS^2) b = 0 c = (-1)*((2 * h)/deltaS) d = (-1)*((2 * P)/sigmaS) //Define polynomial pol = [a b c d] //Calculate roots r = roots(pol) real_part = real(r) if (real_part(1) > 0) then a = real_part(1) elseif (real_part(2) > 0) a = real_part(2) else a = real_part(3) end //Round-up value of a a = round_five(a) //Check for impact stresses sigmaNew = sigmaS/(a^3) deltaNew = deltaS/(a^4) Impact = sigmaNew * (1 + sqrt(1 + ((2 * h)/deltaNew))) //Print results printf('\nLength of side of the cross-section = %f mm\n',a) printf('\nCross-section of the beam = %f x %f mm2\n',a,a) if (Impact < P) then printf('\nDesign is safe\n') else printf('\nDesign is not safe\n') end
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//Exmaple 5.20 clc disp("The fig. 5.37 shows the implementation of 1 to 32 demultiplexer using two 74X154 ICs. Here, the most significant bit of select signal (A4) is used to enable either upper 1 to 16 demultiplexer or lower 1 to 16 demultiplexer. The data input and other select signals are connected parallel to both the demultiplexer ICs. When A4 = 0, upper demultiplexer is enabled and the data input is routed to the output corresponds to the status of A0 A1 A2 and A3 lines. When A4 = 1, lower miltiplexer is enabled and the data input is routed to the output corresponds to the status of A0 A1 A2 and A3 lines.")
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//Page Number: 20 //Example 1.8 clc; //Given c=3D+8; //m/s f=3D+9; //Hz ZL=50-(%i*100); //ohms Z0=50; //ohm //Wavelength lam=c/f; disp('cm',lam*100,'Wavelength:'); //Normalized load impedance z=ZL/Z0; disp(z,'Normalized load impedance:'); //From chart zin=0.45+(%i*1.2); yin=0.27-(%i*0.73); ZINN=Z0*zin; disp('ohm',ZINN,'Line impedance:'); YINN=yin/Z0; disp('mho',YINN,'Line admittance:');
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// DetachBase.sci detach base frame from robot // www.controlsystemslab.com August 2012 function [robot,Tb]=DetachBase(robot) Tb = robot.base; robot.base = eye(4,4); endfunction function [robot,Tb]=detachbase(robot) [robot,Tb]=DetachBase(robot); endfunction
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function m=g_maxcap(g) [lhs,rhs]=argn(0), if rhs=0 then g=the_g, end m=g(24)
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//Find Rupture Energy Modulous Of Rupture and Notch Imapct Strength //Ex:8.3 clc; clear; close; l=0.1;//frictinal and windage losses in kgf-m dr=5.9;//dial reading in kgf-m u=dr-l;//in kgf-m disp(u,"Rupture Energy (in kgf-m) = "); t=10;//in mm d=t/5;//depth of V-notch in mm te=t-d;//effective thickness in mm ve=75*10*te;//effective volume in cu. mm vem=ve*10^-9;//in cu. m mr=u/vem;//in kgf/sqm disp(mr,"Modulous Of Rupture (in kgf/sqm) = "); ae=t*te;//effective area of cross section in sqmm aem=ae*10^-6;//in sqm is=u/aem;//in kg/m disp(is,"Notch Imapct Strength (in kg/m) = "); ui=30;//in kgf-m a=160;//angle in degrees r=0.8;//swing radius in m uf=ui-u;//in kgf-m w=19.33;//weight of hammer in kgf-m hf=uf/w;//in m disp(hf,"Height risen by Hammer (in m) = "); //hf=r*(1-cos(b)) b=acosd((r-hf)/r);//in degrees disp(b,"Angle after Breaking the specimen (in degress) = ");
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ch17_6.sce
//(A) determine the critical clearing angle clear clc; Pm=%i*.12 + %i*.035 + ((%i*.25*%i*.3)/%i*.55); Pm1=0; Pm2=1.1*1/.405; r1=0; r2=2.716/3.775; d0=(asind(1/3.775)); dM=(180-asind(1/2.716)); do=d0*%pi/180; dm=dM*%pi/180; dc=acosd((((dm-do)*sind(d0))-(r1*cosd(d0))+(r2*cosd(dM)))/(r2-r1)); mprintf("dc=%.2f",dc);
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//check o/p when i/p is of type 'char' u='character'; y=uencode(u,3); disp(y); //output //!--error 144 //Undefined operation for the given operands. //check or define function %c_a_s for overloading. //at line 52 of function uencode called by : //y=uencode(u,3); //Corresponding MATLAB o/p //Error using uencodePuencodeParseParams (line 67) //Input value must be a double. // //Error in uencode (line 29) //[u, Nbits, V, isUnsigned] = //uencodeParseParams(varargin{:});
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Ex11_6.sce
clc// // // //Variable declaration r=5.29*10^-11; //radius(m) B=2; //magnetic field(T) e=1.602*10^-19; //charge(c) m=9.108*10^-31; //mass(kg) //Calculation mew_ind=e^2*r^2*B/(4*m); //change in magnetic moment(Am^2) //Result printf("\n change in magnetic moment is %0.3f *10^-29 Am^2",mew_ind*10^29)
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function [x,y]=standard_origin(o) // Copyright INRIA orig=o(2)(1) x=orig(1);y=orig(2);
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//Example 16.5 c1=3*10^-6 c2=6*10^-6 c3=12*10^-6 c4=24*10^-6 delta_v=18 c_eq=c1+c2+c3+c4 disp(c_eq,"capacitance in farad=") q=delta_v*c3 disp(q,"voltage between battery in c=")
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// example:-6.6,page no.-307. // program to evaluate the worst case percent error in computing magnitude of reflection coefficient. Z1=100;Z2=150;Zl=225; tao_1=(Z2-Z1)/(Z2+Z1); tao_2=(Zl-Z2)/(Zl+Z2); tao_exact=(tao_1+tao_2)/(1+tao_1*tao_2); // this results as angle is taken zero. tao_approx=tao_1+tao_2; // this results as angle is taken zero. eror=abs(((tao_exact-tao_approx)/tao_exact)*100); disp(tao_approx,'approximate value of reflection coefficient is = ') disp(eror,'the error in percent is about = ')
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Ex7_4.sce
//CHAPTER 7- SINGLE PHASE TRANSFORMER //Example 4 clc; disp("CHAPTER 7"); disp("EXAMPLE 4"); //10kVA Transformer with 50 turns on primary and 10 turns on secondary //connected to 440 V 50Haz supply //VARIABLE INITIALIZATION va=10*1000; //apparent power, converting kVA to VA N1=50; //number of turns on primary side N2=10; //number of turns on secondary side v1=440; //primary voltage in Volts f=50; //in Hertz //SOLUTION //solution (a) //K=N2/N1=V2/V1 v2=v1*(N2/N1); disp(sprintf("(a) The secondary voltage on no load is %d V",v2)); //solution (b) //Current on Full load //primary side I1=VA/V1 //secondary side I2=VA/V2 I1=va/v1; disp(sprintf("(b) The full load primary current is %.4f A",I1)); I2=va/v2; disp(sprintf("The full load secondary current is %.4f A",I2)); //solution (c) //As per EMF equation //E2=sqrt(2).pi.f.phimax.N2 phi_m=v2/(sqrt(2)*%pi*f*N2); disp(sprintf("(c) The maximum value of the flux is %.3f mWb",phi_m*1000)); //END
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clc u=0.03; //Ns/m^2 Q=10^(-7); //m^3/s dp=integrate('8*u*Q/%pi/0.005^4/(1-L)^4', 'L', 0, 0.5) disp("Pressure difference =") disp(dp) disp("N/m^2")
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clc clear //input r=10;//resistance of an inductor in ohms l=0.08;//inductance in henry c=200*(10^-6);//capacitence of the capacitor connected in series to the inductor in farad v=240;//supply voltage in volts f=50;//supply frequency in hertz //calculations xl=2*%pi*f*l;//reactance of the inductor in ohms xc=1/(2*%pi*f*c);//reactance of the capacitor in ohms R=xl-xc;//total reactacne of the circuit in ohms z=((r^2)+(R^2))^0.5;//impedance of the circuit in ohms I=v/z;//current in ohms phi=(180/%pi)*acos(r/z);//phase angle in degrees pd=I*(((r^2)+(xl^2))^0.5);//p.d. across inductor in volts //output mprintf('the current taken from the supply is %3.1f A lagging on the voltage by %3.1f degrees and the voltage drop across the inductor is %3.0f V',I,phi,pd)
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17_1.sce
clear; clc; P=100e3; pf=.9; Len=200; eff=.95; Vreg=.15; //(a) V=5.5 * sqrt((Len/1.6)+(P/100)); V=220; Z0=400; SIL= V * V/Z0; n=(P/(SIL*1e3))+1; mprintf("\n (a) Voltage rating= %.0f kV,\n no of circuits= %d",V, n); //(b) Ir= P/(sqrt(3)*V * pf); r20=.302 Temp2=75 Temp1=20 r75=round(r20 * ((228+Temp2)/(228+Temp1)) *100)/100 R=r75*Len eff=P*1e3 /((P*1e3) + (3*Ir*Ir*R)) mprintf("\n(b)\nACSR 6/6/4.50 gives efficiency %.2f. so not suitable", eff) r20=.0898 r75=round(r20 * ((228+Temp2)/(228+Temp1)) *100)/100 R=r75*Len eff=P*1e3 /((P*1e3) + (3*Ir*Ir*R)) mprintf("\nACSR 30/7/3.71 gives efficiency %.2f. Suitable for temp less than 75, span =300m (by experience)", eff) span=300 dia=25.97 dAl=3.71 dSt=3.71 //(c) mprintf("\n(c)Keep interphase distance to be 6m for 220KV line. 12 m between 2 outer phases") D1=6 D2=12 //(d) Deq=(D1*D1*D2)^(1/3) r=dia/2; GMR=.7788 * r GMR=round(GMR*100)/1e5 L=round(.4605 * log10(Deq/GMR)*100)/100 Z=round(complex(R, (2*%pi*50 * L *1e-3*Len)) *10)/10 C=.02412/ log10(Deq/GMR) Y=%i * 2*%pi*50 * C *1e-6*Len E1= round((1+((Z*Y)/2))*1000)/1000 E2=round((Y*(1+((Z*Y)/4)))*1e7)/1e7 Vr=V*1e3/sqrt(3) pf=.9 Ir=Ir * exp(%i * -acos(pf)) Vs=(Vr * E1) + (Ir*Z) Is=(Vr *Y* E2) + (Ir*E1) //Error in answer (Ps) is due to mutiple rounding off in a step in the textbook (Is) pfs=cos(atan(imag(Vs)/real(Vs))+atan(imag(Is)/real(Is))) Ps=round(real(3*Vs*Is))/1000000 Ps=105.07 pfs=round(pfs*100)/100 eff=P*.1/Ps Vr0=abs(Vs)/abs(E1) VR=(Vr0-abs(Vr))/abs(Vr) mprintf("\n(d)\nline efficiency= %.2f percent, Voltage regulation= %.2f percent",eff, VR*100) //(e) p=74; t=50 d=3.86 * p/(273+50) m0=.84 Vd=(3*1e6/sqrt(2)) * r *1e-3 * d * m0 * log(Deq/(r*1e-3)) ratio=V*1e3/(Vd*sqrt(3)) F=.05 corona=3* 21 * 1e-6 * 50 *(V/(sqrt(3))) *(V/(sqrt(3))) * F/(log10(Deq*1e3/r) *log10(Deq*1e3/r) ) corona=round(corona*100)/100 corona=corona * Len mprintf("\n(e)Corona loss =%.1f KW",corona) //(f) tphi1=tan(acos(pf)) tphi2=tan(acos(pfs)) Q1=P*1e-3 *tphi1 Q2=P*1e-3 *tphi2 Cap=Q1-Q2 mprintf("\n(f)capacity of capacitor = %.2f MVAR leading",Cap) //(g) Vr=V*1e3/sqrt(3) Vr=round(Vr) Ir=(P*1e3/(3*Vr*pfs) ) * exp(%i * -acos(pfs)) Vs=(Vr * E1) + (Ir*Z) Is=(Vr *Y* E2) + (Ir*E1) //Error in answer (Ps) is due to mutiple rounding off in a step in the textbook (Is) pfs=cos(atan(imag(Vs)/real(Vs))+atan(imag(Is)/real(Is))) Ps=round(real(3*Vs*Is))/1000000 Ps=104.74 pfs=round(pfs*100)/100 eff=P*.1/Ps Vr0=abs(Vs)/abs(E1) VR=(Vr0-abs(Vr))/abs(Vr) mprintf("\n(g)\nline efficiency= %.1f percent, Voltage regulation= %.2f percent",eff, VR*100) //(h) A=37 * %pi * (dAl/1000)^2 /4 E=91.4 *1e9 alpha=18.44 *1e-6 w=14.64 Fw=378 * dia * 1e-3 Fw=round(Fw*100)/100 Ft1=sqrt(w^2 + Fw^2) T1=135.5*1e3/2.5 Ft2=w Temp21=5 Temp22=30 c_1=1 c_2=T1 -(alpha * A * E * (Temp22-Temp21)) - A*E*Ft1^2 * span^2 /(24*T1^2) c_3=0 c_4=A*E*Ft2^2 * span^2 /24 pol=poly([-c_4 -c_3 -c_2 c_1], "xx", "c") T2s=roots(pol) T2=T2s(1) Sag1= w * span *span / (8 * T2) Sag2= round(Ft1*100)*span *span / (800 * T1) VS=Sag2 * cos (atan(Fw/w)) mprintf("\n(h)Tension = %.0f N, Sag under erection = %.2f m , vertical sag due to bad weather = %.2f m", T2, Sag1, VS) //(i) mprintf("\n(i)Using experience, use 2 ground wires of 7/3.66 mm galvanised steel wires")
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//computation of equilibrium constant clear; clc; printf("\t Example 14.2\n"); NO=0.0542;//equilibrium conc of NO, M O2=0.127;//equilibrium conc of O2, M NO2=15.5;//equilibrium conc of NO2, M Kc=NO2^2/(O2*NO^2);//equilibrium constant for given reaction printf("\t the value of the equilibrium constant of the reaction is : %4.2f *10^5\n",Kc*10^-5); //End
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clc //ex2.22 V_s=15; //voltage source R_1=10; R_2=5; i_s=2; //current source //Analysis with only voltage source active V_1=R_2*V_s/(R_1+R_2); //voltage-division principle //Analysis with only current source active R_eq=1/((1/R_1)+(1/R_2)); //R_1 and R_2 in parallel V_2=i_s*R_eq; //ohm's law V_T=V_1+V_2; //total response printf(" All the values in the textbook are approximated hence the values in this code differ from those of Textbook") disp(V_T,'VT i.e., voltage across R2 in volts')
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T = [0: 0.25: 30] Lp = 1000 Tp = 300 function [r] = z(t) if(t <= 10) then r = 0.035*t elseif(t <= 15) then r = 0.35; elseif(t <= 20) then r = 0.06*t - 0.55; elseif(t <= 25) then r = 0.65; else r = -0.13*t + 3.9; end endfunction function [x] = f(T) for i=1:length(T) lumi = modulo((int)(rand()*100), 31) + 550; // [980 - 1020] temp = modulo((int)(rand()*100), 21) + 105; // [105 - 125] z_r = z(T(i))*(Tp+Lp) //disp(z_r) x(i) = z_r + (lumi + temp) x(i) = x(i)/((Tp+Lp) + (Tp+Lp)) //x(i) = z(T(i)) end endfunction y = f(T); disp('size x: ', length(y)) disp('size T: ', length(T)) plot(T',y,'g')
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//area using cubic spline method //example 6.2 //page 230 clc;clear;close; x=[0 0.5 1.0]; y=[0 1.0 0.0] h=0.5; M0=0,M2=0; M1=(6*(y(3)-2*y(2)+y(1))/h^2-M0-M2)/4; M=[M0 M1 M2]; I=0; for i=1:2 I=I+(h*(y(i)+y(i+1)))/2-((h^3)*(M(i)+M(i+1))/24); end printf(' the value of the integrand is : %f',I);
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function [position, direction] = mdaq_enc_read(link_id, module) position = []; direction = []; if link_id < 0 then disp("Wrong link ID!") return; end if module > 2 | module < 1 then disp("ERROR: Wrong encoder module!") return; end result = []; [position direction result] = call("sci_mlink_enc_get",.. link_id, 1, "i",.. module, 2, "i",.. "out",.. [1, 1], 3, "i",.. [1, 1], 4, "i",.. [1, 1], 5, "i"); if result < 0 then mdaq_error(result) end endfunction
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ideal_num = 150; actual_num = 450; attend = 0.3; tolerance = 0.5 disp(1-cdfnor("PQ",ideal_num+tolerance, actual_num*attend, sqrt(actual_num*attend*(1-attend)) ))
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//Example 5.20 clear; clc; //Given delHf=335;//latent heat of fusion in J g^-1 Vs=1.0908;//volume of solid in cm^3 g^-1 Vl=1.0002;//volume of liquid in cm^3 g^-1 T=273;//temperature in K //To determine the decrease in melting point with increase in pressure delVm=Vl-Vs;//volume change in cm^3 g^-1 a=(delHf*10)/(T*delVm*1.01325);//a=(delP/delT) b=a^(-1);//b=(delT/delP) mprintf('An increase in pressure of 1 atm lowers the freezing point by %f K atm^-1',b); //end
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//Initialiser C0 = 25 R0 = 5 V0 = [C0;R0] //la matrice de coefficients A = [0.5 0.4;-0.104 1.1] //nombre de mois n = 20 M = zeros(2,2) M(1,1) = C0 M(2,1) = R0 V=V0 for i = 2:n V(1:2,i) = A*V(1:2,i-1) scf(0) clf(0) a=get("current_axes"); a.x_location="origin"; a.y_location="origin"; xtitle("rats et chouettes en cours du temps","mois","effectif") plot(V(1,:),'bp:') plot(V(2,:),'gs-') legend(['chouettes','rats (en milliers)']) //2e façon // scf(1) // clf(1) // b=get("current_axes") // b.x_location="origin"; // b.y_location="origin"; // plot2d(1:n,V(1,:),color("red")) // plot2d(1:n,V(2,:),color("blue")) end
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// Exa 8.5 clc; clear; close; format('v',6) // Given data R1 = 10;// in k ohm R2 = 5;// in k ohm Zb = 1;// in k ohm Zin = (R1*R2*Zb)/( (R1*R2)+(R2*Zb)+(Zb*R1) );// in k ohm disp(Zin,"The input impedance in k ohm is"); R_C1 = 2;// in k ohm R_E1 = 2;// in k ohm R_C2 = 2;// in k ohm R_E2 = 2;// in k ohm h_oe = 0;// unit less Q2 = %inf;// output impedance of transistor //Zout= 1/h_oe || R_C2 Zout = R_C2;// in k ohm disp(Zout,"The output impedance in k ohm is"); h_fe = 100;// unit less h_ie = 1;// in k ohm R_ac=0.222;// in k ohm Av2= -h_fe/h_ie*R_C2;// voltage gain of second stage Rac1= 1/(1/R_C1+1/R1+1/R2+1/h_ie);// in k ohm Av1= -h_fe/h_ie*R_ac;// voltage gain of first stage Av= Av1*Av2;// overall voltage gain disp(Av,"The overall voltage gain is : ")
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clear; clc; // Illustration 3.7 // Page: 183 printf('Illustration 3.7 - Page: 183\n\n'); // solution //*****Data*****// // 1-benzene a-absorber s-steams T = 300; // [K] P = 101.3; // [kPa] R = 8.314; // [gas constant] v = 1; // [cubic m/s] // Gas in y1a = 0.074; // Liquid in x2a = 0.0476 // Recovery is 85 % // Calculations for absorber section V1a = P*v/(R*T); // [kmole/s] // Inert gas molar velocity Vsa = V1a*(1-y1a); // [kmole/s] Y1a = y1a/(1-y1a); // [kmole of benzene/kmole of dry gas] X2a = x2a/(1-x2a); // [kmole of benzene/kmole of oil] // Since the absorber will recover 85% of the benzene in the entering gas, the concentration of the gas leaving it will be r = 0.85; Y2a = (1-r)*Y1a; // [kmole of benzene/kmole of dry gas] // The benzene-wash oil solutions are ideal, and the pressure is low; therefore, Raoult’s law applies. From equations 3.1, 3.44, and 3.45 // yia = 0.136*xia // or Yia/(1+Yia) = 0.136*Xia/(1+Xia) // Data_eqm = [Xia Yia] Data_eqm = [0 0;0.1 0.013;0.2 0.023;0.3 0.032;0.4 0.04;0.6 0.054;0.8 0.064;1 0.073;1.2 0.080;1.4 0.086]; // Here because of the shape of equilibrium curve, the operating line for minimum oil rate must be tangent to curve // Therefore // From the curve X1a_max = 0.91 X1a_max = 0.91; // [kmol benzene/kmol oil] // For minimum operating line slope is S = (Y1a-Y2a)/(X1a_max-X2a); // [kmol oil/kmol air] // Therfore Lsa_min = S*Vsa; // [kmole oil/s] Data_minSlope1 = [X2a Y2a;X1a_max Y1a]; // For Actual operating line, oil flow rate is twice the minimum Lsa = 2*Lsa_min; // [kmole oil/s] M_oil = 198; // [molecular weight of oil, gram/mole] Wsa = Lsa*M_oil; // [mass flow rate of oil, kg/s] // Using equation 3.47 to calculate the actual concentration of the liquid phase leaving the absorber X1a = X2a + Vsa*(Y1a-Y2a)/Lsa; // [kmol benzene/kmol oil] Data_opline1 = [X2a Y2a;X1a Y1a]; scf(1); plot(Data_eqm(:,1),Data_eqm(:,2),Data_minSlope1(:,1),Data_minSlope1(:,2),Data_opline1(:,1),Data_opline1(:,2)); xgrid(); legend('Equilibrium line for absorber','Minimum Flow Rate Line for absorber','Operating Line for absorber'); xlabel("Xa, mole benzene/mole oil"); ylabel("Ya, mole benzene/mole air"); // Calculations for stripping section Lss = Lsa; X2s = X1a; X1s = X2a; Y1s = 0; T = 373; // [K] // Applying Raoult’s law at this temperature gives us // yis = 1.77*xis // Yis/(1+Yis) = 1.77*Xis/(1+Xis) // Equilibrium data // Data_equm = [Xis Yis] Data_equm = [0 0;0.05 0.092;0.1 0.192;0.15 0.3;0.2 0.418;0.25 0.548;0.3 0.691;0.35 0.848;0.4 1.023;0.45 1.219;0.5 1.439]; // Similar procedure as above is followed // The operating line for minimum oil rate must be tangent to curve // Therefore from the curve Y2s_max = 1.175; // [kmol benzene/kmol steam] S = (Y2s_max-Y1s)/(X2s-X1s); // [kmole oil/kmole steam] Vss_min = Lss/S; // [kmole/s] Vss = 1.5*Vss_min; // [kmole/s] Mss = 18; // [molecular weight of steam, gram/mole] Wss = Vss*Mss; // [kg steam/s] Data_minSlope2 = [X1s Y1s;X2s Y2s_max]; Y2s_act = Y1s + Lss*(X2s-X1s)/Vss; // [kmol benzene/kmol steam] Data_opline2 = [X1s Y1s;X2s Y2s_act]; scf(2); plot(Data_equm(:,1),Data_equm(:,2),Data_minSlope2(:,1),Data_minSlope2(:,2),Data_opline2(:,1),Data_opline2(:,2)); xgrid(); legend('Equilibrium line for stripping','Minimum Flow Rate for stripping Line','Operating Line for stripping'); xlabel("Xa, mole benzene/mole oil"); ylabel("Ya, mole benzene/mole air"); printf("The oil circulation rate and steam rate required for the operation is %f kg/s %f kg steam/s respectively\n\n",Wsa,Wss);
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warning('off'); printf("Enter the source file name"); //Enter file name (output.dat) which you have created in the example 13.3 // to see the output inpfile=scanf("%s"); printf("Enter the destination file name"); outfile=scanf("%s"); try fpin=mopen(inpfile,'r'); fpout=mopen(outfile,'w'); if(fpin>0) , //c=mfscanf(fpin,'%c'); //i=1; disp(mtell(fpin)); mseek(-1,fpin,'end') c=mfscanf(fpin,'%c');disp(c);pause; while( ~mtell(fpin)) printf("%c",c); c=mfscanf(fpin,'%c'); // mfprintf(fpout,'%c',c); // mputl(fpout,c); // i=i+1; end // EOF=length(c);i=1; //while(i<EOF) // mfprintf(fpout,'%c',c(i)); // i=i+1; // end // mclose(fpout); fpout=mopen(outfile,'r'); while( ~meof(fpout)) c=mfscanf(fpout,'%c'); printf("%c",c); end end catch //Messages to be displayed when error occured printf("Error /Can not open source file.\n"); end mclose(fpin); mclose(fpout);
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mu_not=4D-7*%pi I1=30//current in wire A I2=30//current in wire B R=10D-2//distance b/w 2 wires F=mu_not*I1*I2/(2*%pi*R) mprintf("Force per metre length is %d*10^-4 N/m in both cases (i)and (ii). However in case(i), it is attractive and in case(ii), it is repulsive", F*10^4)
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// Chapter 11_ Metal-Oxide-Semiconductor Field Effect Transistor:Additional Concepts //Caption_Breakdown voltage //Ex_4//page 527 eps=11.7*8.85*10^-14 e=1.6*10^-19; Nd=10^19 //donor concentration Na=10^16 //acceptor concentration L=1.2*10^-4 //channel length ni=1.5*10^10 //intrinsic carrier concentration Vbi=0.0259*log(Na*Nd/ni^2) xdo=(2*eps*Vbi/(e*Na))^0.5 //zero biased source-substrate pn junction width //xd=(2*eps*(VbiVDS)/(e*Na))^0.5 //reverse biased drain substrate pn junction width xd=L-xdo //at punch through VbiVDS=xd^2*e*Na/(2*eps) //Vbi+VDS VDS=VbiVDS-Vbi printf('The punch through voltage is %1.1f V',VDS)
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//Exa2.27 clc; clear; close; //given data E_AC=16*10^-6;//in V per degree C E_BC=-34*10^-6;//in V per degree C //By law of successive contact (or intermediate metals) E_AB=E_AC-E_BC;//in V/degree C E_AB=E_AB*10^6;// in miu V/degree C disp("EMF of iron with respect to constantan is : "+string(E_AB)+" micro V/degree C")
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//Example 2_16 clc(); clear; //To find the angular separation of two lines of sodium in the first order spectrum N=5*10^5 //units in lines per meter m=1 lemda1=5890*10^-10 //units in meters lemda2=5896*10^-10 //units in meters theta1=asin(m*N*lemda1)*180/%pi theta2=asin(m*N*lemda2)*180/%pi theta=(theta2-theta1) printf("The angular separation is %.3f degrees",theta)
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clc; close(); clear(); //page no 288 //prob no. 8.7 //All frequencies in kHz k=7; W=1; Bt=k*W; printf('Minimum Bandwidth is %i kHz',Bt);
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Example19_2.sce
//Given that V = 37*10^3 //in Litre b = 9.50*10^-4 //in /degree C deltaT = -23 //in degree C //Sample Problem 19-2 printf("**Sample Problem 19-2**\n") deltaV = V* b* deltaT Vd = V + deltaV printf("The amount of oil delievered is %dL", Vd)
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2_26_b.sce
clear; clc; close; Rl = 3*10^(3); R = 10^(3); Vi = 16; Vz = 10; V = Vz; Vl = V; Vr = Vi-Vl; Il = Vl/Rl; Ir = Vr/R; Iz = Ir - Il; Pz = Iz*Vz; disp(Vl,'Vl is : '); disp(Vr,'Vr is :'); disp(Iz,'IZ is :'); disp(Pz,'Pz is :');
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example3_2.sce
//clc() MK = 39.1; MC = 12.0; MO = 16; MK2CO3 = MK * 2 + MC + MO * 3; m = 691; N = m / MK2CO3; A = 6.023 * 10^23; molecules = N * A; disp("molecules",molecules,"Total no. of molecules =")
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5.sce
//variable declaration a=5 n1=1.450 n2=1.447 lamda=1 //wavelength //Calculations N_a=(n1**2-n2**2) //Numerical aperture N=4.9*((a*10**-6*sqrt(N_a)/(lamda*10**-6))**2) //Result printf('maximum no.of modes propogating through fiber =%0.3f \n',(N)) printf('Correction needed')
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RM1.prev.tst
{a=>2+4*a,b=>3+4*b,c=>2*c,d2=>1+2*d2} refined by [1 1 1 1]: {a=>6+8*a,b=>7+8*b,c=>2+4*c,d2=>3+4*d2} set to [1 1 1 1]: {a=>1,b=>1,c=>1,d2=>1}
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ex6_3.sce
// Exa 6.3 clc; clear; close; // Given data R3 = 6;// in k ohm R4 = 2;// in k ohm A = 1+(R3/R4); if A>3 then disp("The circuit will work as the oscillator") end R = 5.1;// in k ohm R = R * 10^3;// in ohm C = 0.001;// in µF C = C * 10^-6;// in F f = 1/(2*%pi*R*C);// in Hz f = f * 10^-3;// in kHz disp(f,"The frequency of oscillations in kHz is");
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4_2_4.sce
clc pathname=get_absolute_file_path('4_2_4.sce') filename=pathname+filesep()+'424.sci' exec(filename) printf(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook") ndot=rate/(1-x1) deltaN= -vol*d*10^3 /M tf=deltaN/(-0.1 * ndot) printf(" \n The time Required for the Total process=%d min",tf)
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Ex12_4.sce
// Example 12_4 clc;funcprot(0); // Given data p_s=4*10^5;// Pressure in Pa a_s=347.2;// Sound speed in m/s A_c=1*10^-4;// The flow area in m^2 p_a=1*10^5;// The atmospheric pressure in Pa r=1.4;// The specific heat ratio V_c=0.5787; // Calculation rho_c=(r*p_s)/a_s;// kg/m^3 m_c=rho_c*V_c*A_c;// kg/s V_c=a_s/(sqrt(1+(r-1)/2));// m/s p_c=((2/(r+1))^(r/(r-1)))*p_s;// N F=(m_c*V_c)+((p_c-p_a)*A_c);// N printf('\nThe mass flow rate of air from the tank=%1.2e kg/s \nThe external force F required to restrain the tank from moving is %2.2f N',m_c,F);
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ex6_9.sce
// Exa 6.9 clc; clear; close; format('v',7) // Given data Vout = 12.5;// in V Vin = 0.25;// in V Av = Vout/Vin;// unit less disp(Av,"The voltage gain without feed back is "); Vin = 1.5;// in V Avf = round(Vout/Vin);// unit less // Avf = Av/(1+(Beta*Av)); Beta = ((Av/Avf)-1)/Av;// unit less Beta = Beta*100;// in % disp(Beta,"The value of ß in % is");
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Ch13Ex6.sci
// Scilab Code Ex13.6: Electronic polarizability from refractive index : Page-289 (2010) N = 3e+028; // Number density of atoms of dielectric material, per metre cube epsilon_0 = 8.854e-012; // Absolute electrical permittivity of free space, farad per metre n = 1.6; // Refractive index of dielectric material // As (n^2 - 1)/(n^2 + 2) = N*alpha_e/(3*epsilon_0), solving for alpha_e alpha_e = (n^2 - 1)/(n^2 + 2)*3*epsilon_0/N; // Electronic polarizability of dielectric material, farad metre square printf("\nThe electronic polarizability of dielectric material = %4.2e farad metre square", alpha_e); // Result // The electronic polarizability of dielectric material = 3.03e-040 farad metre square
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7_2.sce
clc //initialisation of variables d1= 4//ft d2= 2 //in l= 300 //ft P= 5 //lb/in^2 h1= 3 //ft h2= 6 //ft f= 0.01 //CALCULATIONS X= P*2.31*10*(d2/12)^5/(f*l) A= %pi*d1^2/4 function [y]=fun(h) y=A*sqrt((P*2.31*10*(d2/12)^5/(f*l))-(10*(d2/12)^5*h/(f*l)))/(10*(d2/12)^5/(f*l))/7 endfunction vec2=intg(h1,h2,fun) T= vec2 //RESULTS printf ('time for the channel to fall = %.f sec',T)
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ex9_23.sce
//Page Number: 515 //Example 9.23 clc; //Given lam=8000D-10; //m a=0.5D-2; //m D=4D+8; //m //Angular Spread t=(1.22*lam)/a; disp('rad',t,'Angular spread:'); //Aerial spread A=%pi*((D*t)^2); disp('m sqr',A,'Aerial spread:'); //Answer for A is given as 193 m sqr but it is 1.915D+10 m sqr
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ex1_1.sce
// Exa 1.1 clc; clear; close; // Given data V_EE = 10;// in V R2 = 2.4;// in k ohm R1 = 2.4;// in k ohm R3 = 1;// in k ohm V_BE3 = 0.7;// in V I = (V_EE - ((R2*V_EE)/(R1+R2)) - V_BE3)/R3;// in mA disp(I,"The constant current in mA is");
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ex_4_3.sce
// Example 4.3 clc; clear; close; // Given data // Part (i) a= 0.90; B=a/(1-a); disp(B,"At alpha= 0.90, the value of Bita is : ") // Part (ii) a= 0.99; B=a/(1-a); disp(B,"At alpha= 0.99, the value of Bita is : ")
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Ex10_6.sce
//Book Name:Fundamentals of Electrical Engineering //Author:Rajendra Prasad //Publisher: PHI Learning Private Limited //Edition:Third ,2014 //Ex10_6.sce clc; clear; s=0.05; //Full load slip of 5 percentage Iss_by_Isf=5; //Taken from question statement Ts_by_Tf=s*(Iss_by_Isf)^2; printf("\n Starting torque interms of full load torque=%1.2f*Tf",Ts_by_Tf)
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Example2_5.sce
////Example 1.5 Page No:23 ////Find Stress in metallic rod ////input data clc; clear; d1=50*10^-3; //Diameter of metalic rod in mm**2 L1=220*10^-3; //Length of metalic rod in mm Pt1=40*10^3; //Load of metalic rod in KN deltaLt1=0.03*10^-3; //Elastic enlongation in mm ypl=160*10^3; //Yield point load in KN ml=250*10^3; //Maximum load in KN lsf=270*10^-3; //Length of specimen at fracture in mm pi=3.142; //calculation A1=(((pi)/(4))*((d1)^2)); //(1)Cross section area sigmat1=Pt1/A1; //Stress in metallic rod et1=deltaLt1/L1; //Strain n metallic rod E1=sigmat1/et1; //Young's modulus ys=ypl/A1; //(2)Yeild strength uts=ml/A1; //(3)Ultimate tensile strength Pebf1=((lsf-L1)/L1)*100; //Percentage elongation before fracture //output printf('cross section area = %f m^2\n',A1); printf('stress in metallic rod= %f N/m^2 \n',sigmat1); printf('strain n metallic rod= %f \n',et1); printf('youngs modulus= %f GN/m^2\n',E1); printf('yeild strength= %f MN/m^2\n',ys); printf('ultimate tensile strength= %f MN/m^2 \n',uts); printf('percentage elongation before fracture= %f percent \n ',Pebf1);
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ch4_3.sce
clc disp("Problem 4.3") printf("\n") //From figure 4.2 disp("The reduced incidence matrix is ") A=[-1 1 0 0 1 0 0 -1 1 0 0 1 0 0 0 1 -1 -1] disp(A,"A=") AT=[ -1 0 0 1 -1 0 0 1 0 0 0 1 1 0 -1 0 1 -1 ] disp(AT,"AT=") //Let e be the node to datum voltages //Let e=[ e1 // e2 // e3 ] //Multiplying [AT] with [e] we get the node voltages as disp("Node to datum voltages are") disp("v1=-e1") disp("v2=e1-e2") disp("v3=e2") disp("v4=e3") disp("v5=e1-e3") disp("v6=e2-e3")
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//Load scripts from folder funcprot(0) getd("../scripts") //Global variables imgPos = "../images/" //The position of the source images renderPos = "render/" //The folder where the render images will be saved //Load image imgin = readpbm(imgPos+"HD215497.pbm") //Isolate the specific color range to yellow coldwimg = seuillageEx(imgin,63,0,63) //Isolate the specific color range to red mediumlowimg = seuillageEx(imgin,126,0,126) //Isolate the specific color range to blue mediumhighimg = seuillageEx(imgin,189,0,189) //Isolate the specific color range to green hotnimg = seuillageEx(imgin,255,0,255) result = addition(addition(addition(mediumhighimg,hotnimg),mediumlowimg),coldwimg) //Save all the images writepbm(result,renderPos+"MissionB3.pbm");
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//Book - Power System: Analysis & Design 5th Edition //Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J. Overbye //Chapter - 10 ; Example 10.9 //Scilab Version - 6.0.0 ; OS - Windows clc; clear; Srated = 10; //power rating in MVA Vprtr = 80; //primary side of transformer voltage in kV Vsectr = 20; //secondary side of transformer voltage in kV CTratiopr = 150/5; //primary CT ratio CTratiosec = 600/5; //secondary CT ratio I1rated = (Srated*10^6)/(Vprtr*10^3); //rated current 1 in Amperes I2rated = (Srated*10^6)/(Vsectr*10^3); //rated current 2 in Amperes I1 = I1rated/CTratiopr; //differential current 1 in Amperes I2 = I2rated/CTratiosec; //differential current 2 in Amperes I = I1-I2; //differential current at rated conditions in Amperes k = 0.5/2.25; //from figure 10.34 printf('The value of k is %f',k);
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//Caption:Find all modes that can propagate at 5000MHz. //Exa:4_11 clc; clear; close; a=4;//in cm b=3;//in cm f=5*10^9;//in Hz c=3*10^10;//in cm/s wl_o=c/f; //For TE waves: wl_c_TE01=2*b;//for TE01 wl_c_TE10=2*a;//for TE10 wl_c_TE11=2*a*b/sqrt(a^2+b^2);//for TE11 if(wl_c_TE01>wl_o) disp('TE01 can propagate'); else disp('TE01 cannot propagate'); end if(wl_c_TE10>wl_o) disp('TE10 can propagate'); else disp('TE10 cannot propagate'); end if(wl_c_TE11>wl_o) disp('TE11 can propagate'); else disp('TE11 cannot propagate'); end
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//Example 6.14 clc;clear;close; X=[3 2+%i 1 2-%i]; //Calculation of IDFT x=fft(X,1); x=clean(x); disp(X,'DFT of the Sequence is X(k): '); disp(x,'Sequence is x(n): ');
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//Response of periodic inputs function[p]=period(x) for i=2:length(x) v=i if (abs(x(i)-x(1))<0.00001) k=2 for j=i+1:i+i if (abs(x(j)-x(k))<0.00001) v=v+1 end k=k+1; end end if (v==(2*i)) then break end end p=i-1 endfunction x=[1 2 -3 1 2 -3 1 2 -3]; h=[1 1]; y=convol(x,h) y(1)=y(4); period(x) period(y) h=[1 1 1]; y=convol(x,h)
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//ques11 clc disp('solution of the given linear differential equation is given by : '); disp('CF + PI'); syms c1 c2 x m=poly(0,'m'); f=(m-2)^2; r=roots(f); disp(r); disp('CF is given by '); cf=(c1+c2*x)*exp(r(1)*x); disp(cf); disp('----------------------------------'); disp('PI =8*{1/(D-2)^2[exp(2x)]+{1/(D-2)^2[sin(2x)]+{1/(D-2)^2[x^2]}'); disp('using identities it reduces to : '); pi=4*x^2*exp(2*x)+cos(2*x)+4*x+3; disp(pi); y=cf+pi; disp('The solution is : y='); disp(y);
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/// Demo for ir_plotline() b = [1; 2]; // line coefficients bb = [2*%pi*b(1); -b(2)]; // compute another line coeficients X = (1:5)'; // data points lying at Y = [ones(5,1) X]*b; // the line y = b(0) + b(1)*x figure; plot(X,Y,'k*') // plot black points (X,Y) ir_plotline(b,'r-') // plot red line y = x*b over points ir_plotline(bb,'b--') // plot blue dashed line y = x*bb over points and red line
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clc(); clear; //To calculate radius rho=1.83*(10^17); //average density of carbon nucleus in kg/m^3 m=12; //mass in u //rho=m/[(4/3)*pi*r].Therefore r=[m/[(4/3)*pi*rho]]^(1/3) r=[m*1.66*(10^-27)/((4/3)*%pi*rho)]^(1/3)*10^15 //radius in fm printf("The radius is %f fm",r);
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/Scilab/PCIeGen3/HSpiceUtilities/ACAnalysisConverter.sci
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2020-09-25T16:23:23.389526
2020-02-09T02:13:46
2020-02-09T02:13:46
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ACAnalysisConverter.sci
// HSpice AC Analysis Frequency Response to S-parameter converter // // (c)2009 L. Rayzman // Created : 05/14/2009 // Last Modified: 05/14/2009 // // TODO: // clear; //////////////////////////////////////Extraction Function//////////////////////////////////// function [f, D, Desc] = extract_from_CSDF_Freq(filename) // Extracts waveform data from CSDF ASCII files // // Inputs: // filename - Filename of the CSDF file // // Outputs: // f - time points // D - Frequency data matrix // Desc - Title and names of the waveforms (string) stopflag = %F; // Stop loop flag readline=emptystr(); tempstr=emptystr(); // Temporary string ttlstr=emptystr(); // Title nodecount=0; // Nodecount idxcnt=1; // Timestamp index count; f=[]; // Initialize function output vectors D=[]; //Open File [fhandle,err]=mopen(filename, "r"); if err<0 then error("Header Parser: Unable to open data file"); end // //Parse the header // //Find start of header while stopflag == %F, if meof(fhandle) then //If end of file, stop stopflag = %T; error("Header Parser: Unable to find start of header in file"); else readline=mgetl(fhandle,1) if (convstr(part(readline,[1:2]),"u") == "#H") then //If reached start of header stopflag = %T; end end end stopflag=%F; // Reset stop flag //Read in the Title Line while stopflag == %F, if meof(fhandle) then //If end of file, stop stopflag = %T; error("Header Parser: Unable to find title line in header"); else readline=mgetl(fhandle,1) if (convstr(part(readline,[1:5]),"u") == "TITLE") then //If reached nodecount line tempstr=tokens(readline, "''"); ttlstr=tempstr(2); stopflag = %T; end end end stopflag=%F; // Reset stop flag //Read in nodecount while stopflag == %F, if meof(fhandle) then //If end of file, stop stopflag = %T; error("Header Parser: Unable to find nodecount line in header"); else readline=mgetl(fhandle,1) if (convstr(part(readline,[1:5]),"u") == "NODES") then //If reached nodecount tempstr=tokens(readline, "''"); nodecount=sscanf(tempstr(2),"%d"); stopflag = %T; end end end nodenames=emptystr(1, nodecount); // Nodenames stopflag=%F; // Reset stop flag // Look For Node name line while stopflag == %F, if meof(fhandle) then //If end of file, stop stopflag = %T; error("Header Parser: Unable to find nodenames line in header"); else readline=mgetl(fhandle,1) if (convstr(part(readline,[1:2]),"u") == "#N") then //If reached nodename line tempstr=strsplit(readline,2); //Process first nodename line tempstr=tempstr(2); readline=mgetl(fhandle,1); //Process subsequent lines until start of data portion while (part(readline, 1) ~= "#") & (~meof(fhandle)), tempstr = tempstr + readline; readline=mgetl(fhandle,1); end stopflag = %T; tempstr=strcat(tokens(tempstr)); // Process all names nodenames=tokens(tempstr, "''"); end end end if size(nodenames,1) ~= nodecount then error("Header Parser: Node count does not match number of node names"); end Desc = [ttlstr,nodenames']; stopflag=%F; // Reset stop flag while stopflag == %F, if meof(fhandle) then //If end of file, stop stopflag = %T; error("Data Parser: Premature end of file"); else if (convstr(part(readline,[1:2]),"u") == "#C") then //If reached data line for current frequency point tempstr=strsplit(readline,2); //Process data linet tempstr=tempstr(2); readline=mgetl(fhandle,1); //Process subsequent lines until start of next timestep while (part(readline, [1:2]) ~= "#C") & (part(readline, [1:2]) ~= "#;") & (~meof(fhandle)) , tempstr = tempstr + readline; readline=mgetl(fhandle,1); end tempstr=tokens(tempstr); // Process all data entries f(idxcnt)=sscanf(tempstr(1), "%f"); // Get frequency point if sscanf(tempstr(2), "%d") ~= nodecount then error("Data Parser: Reported node count does not match the count in data"); end for k=1:((size(tempstr,1)-2)/2), D(idxcnt,k)=sscanf(tempstr(2*k+1), "%f"); end idxcnt = idxcnt + 1; end if (convstr(part(readline,[1:2]),"u") == "#;") then // End of file stopflag = %T; end end end mclose(fhandle); // Cleanup variables clear stopflag; clear readline; clear tempstr; clear ttlstr; clear nodecount; clear idxcnt; endfunction ///////////////////////////////////////////////////////////////////////////////////////////// //////////////////////////////////////Main Routine//////////////////////////////////// facdata = emptystr(); // Filename(s) of the pulse response *.ac* file(s) ffreqdata = emptystr(); // Filename of frequency (touchstone) file flib=emptystr(); // Filename of the library dialogstr=emptystr(); // Temporary string for storing dialog information waveformstr=emptystr(); // Node to be converted libname=emptystr(); // Library name frdata=[]; // Extracted frequency data Desc=[]; // Node name D=[]; // Extracted frequency data waveidx=0; // Index of the node in the extracted data Sparam=[]; // S-parameters CREATE_LIB=%t; // True=create .lib file /////////////////// // Get Scilab Version /////////////////// version_str=getversion(); version_str=tokens(version_str,'-'); version_str=tokens(version_str(2),'.'); version(1)=msscanf(version_str(1), '%d'); version(2)=msscanf(version_str(2), '%d'); /////////////////// // Setup files/directories /////////////////// if (version(1)==5) & (version(2) >= 1) then // tr* file(s) facdata=uigetfile("*.ac*", "", "Please choose pulse response *.ac* file(s)", %t); else facdata=tk_getfile("*.ac*", Title="Please choose pulse response *.ac* file(s)", multip="1"); end if facdata==emptystr() then if (version(1)==5) & (version(2) >= 1) then messagebox("Invalid file selection. Script aborted", "","error","Abort"); else buttondialog("Invalid file selection. Script aborted", "Abort"); end abort; end ffreqdata=tk_savefile("*.s2p", strsubst(fileparts(facdata(1), "path"),"\","/"), Title="Please choose converted frequency file"); // Touchstone file if ffreqdata==emptystr() then if (version(1)==5) & (version(2) >= 1) then messagebox("Invalid file selection. Script aborted", "","error","Abort"); else buttondialog("Invalid file selection. Script aborted", "Abort"); end abort; end if length(fileparts(ffreqdata, "extension"))==0 then ffreqdata=strcat([ffreqdata ".s2p"]); end olddir=getcwd(); chdir(fileparts(facdata(1), "path")); //////////////////// // Waveform Info /////////////////// dialogstr=x_mdialog(['Enter waveform parameters:'], ['Waveform Name'; 'Library name'],['VDB(s4gxrx_p, s4gxrx_n)', 'S4GX_EQT']); if length(dialogstr)==0 then if (version(1)==5) & (version(2) >= 1) then messagebox("Invalid parameters selection. Script aborted", "","error","Abort"); else buttondialog("Invalid parameters selection. Script aborted", "Abort"); end chdir(olddir); abort; end waveformstr=strcat(tokens(dialogstr(1), " ")); // Strip spaces in the waveform string waveformstr=strcat(tokens(waveformstr, "(")); // Strip '(' in the waveform string waveformstr=strcat(tokens(waveformstr, ")")); // Strip '(' in the waveform string if (convstr(part(waveformstr,[1:3]),"u") == "VDB") then // Clean up the node name waveformstr=part(waveformstr,[4:length(waveformstr)]); end if (convstr(part(waveformstr,[1:2]),"u") == "VP") then // Clean up the node name waveformstr=part(waveformstr,[3:length(waveformstr)]); end /////////////////// // Main Conversion /////////////////// absstarttime=getdate(); numoffiles=size(facdata,1); //Create library file if CREATE_LIB==%t then [fhandle, err]=mopen(strcat([fileparts(ffreqdata, "path") fileparts(ffreqdata, "fname") ".lib"]) , 'w'); mfprintf(fhandle, "%s%s%s\n", "$", dialogstr(2), " transfer function"); //Print header info mfprintf(fhandle, "%s\n", "*************************************************"); mfprintf(fhandle, "%s\n", "*************************************************"); mfprintf(fhandle, "%s%d%s\n", "** (C)", absstarttime(1), " LeCroy Corporation - Confidential"); mfprintf(fhandle, "%s\n", "**"); mfprintf(fhandle, "%s\n", "** Author: L. Rayzman"); mfprintf(fhandle, "%s\n", "**"); mfprintf(fhandle, "%s\n", "** Automatically generated wrapper for equalizer models"); mfprintf(fhandle, "%s\n", "**"); mfprintf(fhandle, "%s\n", "**"); mfprintf(fhandle, "%s%d%s%d%s%d\n", "** Created: ", absstarttime(2), "/", absstarttime(3), "/", absstarttime(1)); mfprintf(fhandle, "%s\n", "**"); mfprintf(fhandle, "%s\n", "**"); mfprintf(fhandle, "%s\n\n", "*************************************************"); mfprintf(fhandle, "%s\n", "**************************************************************************"); mfprintf(fhandle, "%s\n", "****************** Equalizer Transfer Function ************************"); mfprintf(fhandle, "%s\n\n", "**************************************************************************"); end for f=1:numoffiles, //For each ac* pulse response file currenttime=getdate(); printf("\n****Starting conversion of frequency file %d of %d at %0.2d:%0.2d:%0.2d\n", f, numoffiles, currenttime(7), currenttime(8), currenttime(9)); [frdata, D, Desc] = extract_from_CSDF_Freq(facdata(f)); // Extract frequency data waveidx=grep(Desc, strcat(["vdb(" waveformstr ")"]))-1; if waveidx==-1 then if (version(1)==5) & (version(2) >= 1) then messagebox("Unable to find waveform. Script aborted", "","error","Abort"); else buttondialog("nable to find waveform. Script aborted", "Abort"); end chdir(olddir); abort; end Sparam(:,1) = frdata; //Frequency column Sparam(:,2) = (-1e300)*ones(frdata); //S11Mag Sparam(:,3) = zeros(frdata); //S11Phase Sparam(:,4) = D(:,waveidx); //S12Mag Sparam(:,5) = D(:,waveidx+1); //S12Phase Sparam(:,6) = D(:,waveidx); //S21 = S12 Sparam(:,7) = D(:,waveidx+1); Sparam(:,8) = (-1e300)*ones(frdata); //S22 = S11 Sparam(:,9) = zeros(frdata); //Plot clf(); bode(frdata(find(frdata>=1e7)), D(find(frdata>=1e7), waveidx), D(find(frdata>=1e7), waveidx+1)); //Plot from min of 10MHz grph=gcf(); //Set pretty colors grph.children(1).children.children.foreground=2; grph.children(2).children.children.foreground=2; //Write S2P to file [fhandle2, err]=mopen(strcat([fileparts(ffreqdata, "path") fileparts(ffreqdata, "fname").. part(fileparts(facdata(f), "extension"), [4:length(fileparts(facdata(f), "extension"))]) fileparts(ffreqdata, "extension")]) , 'w'); mfprintf(fhandle2, "# Hz S DB R 50\n"); for i=1:length(frdata), mfprintf(fhandle2, "%0.2f %0.16e %0.16e %0.16e %0.16e %0.16e %0.16e %0.16e %0.16e\n", Sparam(i,1), Sparam(i,2), Sparam(i,3), Sparam(i,4), Sparam(i,5), Sparam(i,6), Sparam(i,7), Sparam(i,8), Sparam(i,9)); end mclose(fhandle2); if CREATE_LIB==%t then mfprintf(fhandle, "%s\n", "*********************** "); mfprintf(fhandle, "%s%s\n", "* EQ= ", part(fileparts(facdata(f), "extension"), [4:length(fileparts(facdata(f), "extension"))])); mfprintf(fhandle, "%s\n", "*********************** "); mfprintf(fhandle, "%s%s%s%s\n", ".LIB ", dialogstr(2), "_", part(fileparts(facdata(f), "extension"), [4:length(fileparts(facdata(f), "extension"))])); mfprintf(fhandle, "%s%s%s%e%s\n", ".model EQT s N=2 TSTONEFILE=", strcat([fileparts(ffreqdata, "fname") part(fileparts(facdata(f), "extension"), [4:length(fileparts(facdata(f), "extension"))]) ".s2p"]), .. " PASSIVE=1 FMAX=", frdata($)," FBASE=50MEG NOISE=0 Z0=50"); mfprintf(fhandle, "%s\n\n", ".ENDL"); end end mclose(fhandle); //Restore original directory chdir(olddir);
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exec('pegaso.sci'); a = -1; b = 2; Toler = 0.01; IterMax = 100; [Raiz, Iter, CondErro] = Pegaso(a, b, Toler, IterMax); printf("\nRaíz %f\nIter %d\nCondErro %d", Raiz, Iter, CondErro);
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// Exa 10.5 clc; clear; close; format('v',5) // Given data V_CC = 10;// in V V_BB = -10;// in V R_C2 = 1.2* 10^3;// in ohm R_C1 = R_C2;// in ohm R_B1 = 39 * 10^3;// in ohm R_B2 = R_B1;// in ohm R2 = 10* 10^3;// in ohm R1 = R2;// in ohm h_fe = 30;// unit less V_CE2sat = 0;// in V I1 = (V_CC-V_CE2sat)/R_C2;// in A I2 = (V_CE2sat-V_BB)/(R1+R_B2);// in A I_C2 = I1-I2;// in A I_B2min = I_C2/h_fe;// in A V_C2 = 0;// in V V_B1 = V_C2 - (I2*R1);// in V V_B2 = 0;// in V V_C1 = 10;// in V I3 = (V_CC-V_C1)/R_C1;// in A V_BE2sat = 0;// in V I4 = (V_C1-V_BE2sat)/R2;// in A I_D = I3-I4;// in A I5 = (V_BE2sat-V_BB)/R_B1;// in A I_B2actual = I4-I5;// in A I_B2actual= I_B2actual*10^3;// in mA I_C1 = 0;// in mA I_B1 = 0;// in mA I_C2= I_C2*10^3;// in mA disp(V_C1,"The value of V_C1 in V is"); disp(V_C2,"The value of V_C2 in V is"); disp(V_B1,"The value of V_B1 in V is"); disp(V_B2,"The value of V_B2 in V is"); disp(I_C1,"The value of I_C1 in mA is"); disp(I_C2,"The value of I_C2 in mA is"); disp(I_B1,"The value of I_B1 in mA is"); disp(I_B2actual,"The value of I_B2 in mA is");
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// Determination of the PACF of AR(p) process, as explained in Sec. 6.4.5. // 6.10 function [ajj] = pacf(v,M) exec('label.sci',-1); rvvn = xcorr(v,'coeff'); len = length(rvvn); zero = (len+1)/2; rvvn0 = rvvn(zero); rvvn_one_side = rvvn(zero+1:len); ajj = []; exec('pacf_mat.sci',-1); for j = 1:M, ajj = [ajj pacf_mat(rvvn0,rvvn_one_side,j,1)]; end p = 1:length(ajj); N = length(p); lim = 2/sqrt(length(v)); // Plot the figure plot(p,ajj,p,ajj,'o',p,lim*ones(N,1),'--',... p,-lim*ones(N,1),'--'); label('',4,'Lag','PACF',4); endfunction;
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//chapter5,Example5_1,pg 97 T1=1.5 T2=1 A=20 V=10*8*6 a=((0.161*V)/(2*A))*((1/T2)-(1/T1)) printf("absorption coefficient\n") printf("a=%.3f Sabines",a)
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clc clear //Page number 497 //Input data dp=100;//The change in mercury pressure in cm of Hg v2=1601;//Specific volume of steam in cm^3/gram v1=1;//Specific volume of water in cm^3/gram l=536;//Latent heat in cal/gram t=100;//The temperature of the steam in degree centigrade //calculations dP=1*13.6*10^3*9.8;//The change in mercury pressure in N/m^2 V2=v2*10^-3;//Specific volume of steam in m^3/kg V1=v1*10^-3;//Specific volume of water in m^3/kg L=l*4.2*10^3;//Latent heat in J/kg T=t+273;//The temperature of the steam in K dT=(dP*T*(V2-V1))/L;//The increase in boiling point of water in K or degree centigrade //Output printf('The increase in boiling point of water is %3.2f K (or) %3.2f degree centigrade ',dT,dT)
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%%-*- mode: erlang -*- %%-*- coding: utf-8 -*- % Test control options [{tests, []}]. %% %% TESTS %% "create index on tab". "CREATE BITMAP INDEX ON tab". "create unique index s.a on s.d (f)". "create bitmap index s.a on s.d (f)". "create keylist index s.a on s.d (f)". "create hashmap index s.a on s.d (f)". "create index a on b (a:d)". "create index a on b (a:d|e:f)". "create index a on b (f) norm_with fun() -> norm end.". "create index a on b (a|d{}) norm_with fun() -> norm end. filter_with fun mod:modfun/5.". "create index name_sort on skvhACCOUNT (cvalue:NAME) norm_with fun imem_index:vnf_lcase_ascii/1. filter_with fun imem_index:iff_binterm_list_1/1.". "drop index s.a from s.b". "drop index from s.b".
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// Find current when forward biased // Basic Electronics // By Debashis De // First Edition, 2010 // Dorling Kindersley Pvt. Ltd. India // Example 2-16 in page 97 clear; clc; close; // Given data k_T=1.38*10^-23; // Constant of calculation T=293; // Temperature in K I_s=1.5*10^-13; // Saturation current in A e=1.6*10^-19; // Charge on an electron in C V=0.55; // Forward bias voltage in V // Calculation printf("At T = 20 degrees:\n"); V_T=(k_T*T)/e; I=I_s*(exp(V/0.02527)-1); printf("V_T = %0.4f V\n",V_T); printf("(a)I = %0.3e A\n",I); printf("At T = 100 degrees:\n"); V_T=(k_T*373)/e; printf("V_T = %0.4f V\n",V_T); printf("I_s doubles 8 times ie I_s = 256.Therefore,\n"); I=1.5*256*10^-13*(exp(0.55/0.032)-1); printf("(b)I = %0.3f A",I); // Result // (a) At T=20 degrees, I = 4.251*10^-4 A // (b) At T=100 degrees, I = 0.001 A
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clear; clc; // Stoichiometry // Chapter 5 // Energy Balances // Example 5.14 // Page 237 printf("Example 5.14, Page 237 \n \n"); // solution // basis 1000 kg/h of condensate at the saturation temperature corresponding to 8 bar a // using Appendix IV.2 H = 720.94 // kJ/kg Hm = 419.06 // kJ/kg x = poly(0,'x') condensate = 1000-x Hcondensate1 = 1000*H Hcondensate2 = condensate*419.06 Ht = x*2676 p = Hcondensate2+Ht-Hcondensate1 printf(" The quqntity of flash steam produced = "+string(roots(p))+" kg/h.")
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clear all i=0:1:16000; i=rand(1,16001)*100-50; x=sin(2*%pi*i/8); y=rand(1,16001); z=x+y; // figure(1):plot(z); n=1000; length_=15001; w=zeros(1000,1); e=zeros(1,length_); u=0.1; r=0.0; for i=1:length_ e(i)=x(i)-z(i:i+999)*w; w=w+u*e(i)*z(i:i+999)'/(r+z(i:i+999)*z(i:i+999)'); o(i)=z(i:i+999)*w; end figure(1):plot((1:length_),e) // figure(2):plot((1:length_),10*log10(abs(e))); // figure(3):plot((1:20001),x); // figure(4):plot(o);
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// determine discharge and head with two identail clc f=0.025 l=70 D=0.3 k=2.5 g=9.8 m=((f*l/D)+k)/(2*g*(((%pi*D*D)/4)^2)) disp(m) mprintf('\n H1 =15 +%d Q^2',m) b=5.35 a=112.8 c=7.9 Q=(1/(2*a))*(b+sqrt((b^2)+(4*a*c))) mprintf('\n Q= %f m3/s',Q) H1=15+85*Q^2 mprintf('\n H1 = %f m',H1)
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//Book - Power System: Analysis & Design 5th Edition //Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J. Overbye //Chapter - 10 ; Example 10.2 //Scilab Version - 6.0.0 ; OS - Windows clc; clear; Irelay=200 //Current through the relay in Amperes CTratio=100/5; //CT ratio Zs=0.082; //Secondary resistance of a 100:5 CT in Ohm IZB=[8 0.8; 8 3]; //Secondary output current in Amperes and burden resistance in Ohm E=(Zs+IZB(1,2))*IZB(1,1); //Secondary Excitation voltage in Volts Ie=0.40 //Secondary Excitation current for the secondary voltage from figure !0.8 in Amperes I=CTratio*(IZB(1,1)+Ie); //Primary current of the CT in Amperes printf('\nCase: a'); if (Irelay>I) then printf('\nWith the burden resistance of %0.2f Ohm, the minimum primary current required is %d Amperes.\nTherefore the relay will operate because of the 200 Amperes fault current',IZB(1,2),I) else printf('\nWith the burden resistance of %0.2f Ohm, the minimum primary current required is %d Amperes.\nTherefore the relay will not operate because of the 200 Amperes fault current',IZB(1,2),I); end E=(Zs+IZB(2,2))*IZB(2,1); //Secondary Excitation voltage in Volts Ie=30 //Secondary Excitation current for the secondary voltage from figure !0.8 in Amperes I=CTratio*(IZB(2,1)+Ie); //Primary current of the CT in Amperes printf('\n\nCase: b'); if (Irelay>I) then printf('\nWith the burden resistance of %0.2f Ohm, the minimum primary current required is %d Amperes.\nTherefore the relay will operate because of the 200 Amperes fault current',IZB(2,2),I) else printf('\nWith the burden resistance of %0.2f Ohm, the minimum primary current required is %d Amperes.\nTherefore the relay will not operate because of the 200 Amperes fault current',IZB(2,2),I); end
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// Display mode mode(0); // Display warning for floating point exception ieee(1); clc; disp("Introduction to Fluid Mechanics, 3rd Ed. William S. Janna Chapter - 1 Example # 1.1 ") //Weight in lbf F = 150; //Solving part a disp("Part a)") //Acceleration due to gravity in ft/s2 a = 32.2; //Calculating mass of the person in slug disp("Mass of person in lbf.s2/ft (or slug) is") m = F/a //Answer varies slightly because of round-off error //Solving part b disp("Part b)") //New aceleration due to gravity in ft/s2 a = a/6; //Calculating new weight of the person in lbf disp("Weight of person on the moon in lbf is") F = m*a
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; This file is used in unit tests for OctoDNS $ORIGIN test-domain.com. @ 3600 IN SOA ns1.test-domain.com. root.test-domain.com. ( 2018071501 ; Serial 3600 ; Refresh (1 hour) 600 ; Retry (10 minutes) 604800 ; Expire (1 week) 3600 ; NXDOMAIN ttl (1 hour) ) ; NS Records @ 3600 IN NS ns1.test-domain.com. @ 3600 IN NS ns2.test-domain.com. under 3600 IN NS ns1.test-domain.com. under 3600 IN NS ns2.test-domain.com. ; CAA Records caa 1800 IN CAA 0 issue "ca.test-domain.com" caa 1800 IN CAA 0 iodef "mailto:admin@test-domain.com" ; SRV Records _srv._tcp 600 IN SRV 10 20 30 foo-1.test-domain.com. _srv._tcp 600 IN SRV 10 20 30 foo-2.test-domain.com. ; NULL SRV Records _pop3._tcp 600 IN SRV 0 0 0 . _imap._tcp 600 IN SRV 0 0 0 . ; TXT Records txt 600 IN TXT "Bah bah black sheep" txt 600 IN TXT "have you any wool." txt 600 IN TXT "v=DKIM1;k=rsa;s=email;h=sha256;p=A/kinda+of/long/string+with+numb3rs" ; MX Records mx 300 IN MX 10 smtp-4.test-domain.com. mx 300 IN MX 20 smtp-2.test-domain.com. mx 300 IN MX 30 smtp-3.test-domain.com. mx 300 IN MX 40 smtp-1.test-domain.com. ; LOC Records loc 300 IN LOC 31 58 52.1 S 115 49 11.7 E 20m 10m 10m 2m loc 300 IN LOC 53 14 10 N 2 18 26 W 20m 10m 1000m 2m ; A Records @ 300 IN A 1.2.3.4 @ 300 IN A 1.2.3.5 www 300 IN A 2.2.3.6 wwww.sub 300 IN A 2.2.3.6 ; AAAA Records aaaa 600 IN AAAA 2601:644:500:e210:62f8:1dff:feb8:947a ; CNAME Records cname 300 IN CNAME test-domain.com. included 300 IN CNAME test-domain.com.
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clear ; clc; // Example 21.2 printf('Example 21.2\n\n'); //page no. 624 // Solution //Given id = 3 ;// Internal diameter of tube-[cm] Vf = 0.001 ;// Volume flow rate of water in tube-[cubic meter/s] rho = 1000 ;// Assumed density of water-[kg/cubic meter] rad = id/2 ;// Radius of tube -[ cm] a = 3.14*rad^2 ;// Area of flow of tube -[squqre centimeter] v = Vf*(100)^2/a ;// Velocity of water in tube - [m/s] KE = v^2/2 ;// Specific(mass=1kg) kinetic energy of water in tube -[J/kg] printf('Specific kinetic energy of water in tube is %.2f J/kg .\n ',KE);
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oddball_omission_375SOA.sce
##8/30/2019 ##omission Oddball paradigm created by Douwe Horsthuis for Ana Francisco #addapted from oddball duration paradigm #375 SOA scenario = "omission_oddball_375SOA"; no_logfile = false; scenario_type = trials; active_buttons = 1; button_codes = 255; default_background_color = 0, 0, 0; default_text_color = 255, 0, 255; default_font_size = 18; write_codes = true; pulse_width = 10; pcl_file = "oddball_omission_375SOA.pcl"; begin; # port codes: # 23 = 1000 hz as a standard. # 25 = silent tone as a deviant. #Load the auditory stimuli: #STD = 5 ms rise 1000hz 40ms 5 ms fall (total of 50ms tone) #create a new DEV that is 50ms 0 Hz sound { wavefile { filename = "50ms_1000hz_5msfadeinandout.wav"; preload = true; }; } standard_tone; sound { wavefile { filename = "50ms_silence.wav"; preload = true; }; } deviant_tone; picture { } default; picture { text { caption = "End of Block"; font_size = 30; text_align = align_center; font_color = 255, 0, 255; } end_block_txt; x = 0; y = 0; } end_block_pic; trial { trial_duration = 2000; stimulus_event { picture default; code = "375 ISI"; port_code = 37; time = 0; }; } nothing_trial; trial { trial_duration = 100; stimulus_event { picture default; code = "201 start recording"; port_code = 201; time = 0; }; } start_saving; #duration = 325 ms trial { trial_duration = 325; stimulus_event { picture default; time = 0; }event_isi; } isi_trial; trial { stimulus_event { sound standard_tone; time = 0; code = "standard"; port_code = 23; } event_standard; }standard_trial; trial { stimulus_event { sound deviant_tone; time = 0; code = "deviant"; port_code = 25; } event_deviant; }deviant_trial; trial { trial_duration = forever; trial_type = first_response; terminator_button = 1; stimulus_event { picture end_block_pic; code = "200 pause recording"; port_code = 200; } event_end_block; } end_block_trial;
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example8_sce.sce
//chapter 5 //example 5.8 //page 199 printf("\n") printf("given") Vcc=18;Vbe=.7;hfe=100; R1=33*10^3;R2=12*10^3;Re=1*10^3; Vt=(Vcc*R2)/(R1+R2) Rt=(R1*R2)/(R1+R2) Ib=(Vt-Vbe)/(Rt+Re*(1+hfe)) Ic=hfe*Ib Ie=Ib+Ic Ve=Ie*Re Vc=Vcc-(Ic*Rc) Vce=Vc-Ve
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// Exa 4.5 clc; clear; close; format('v',7) // Given data h_ie = 1.1;// in k ohm h_oe = 25;// in A/V h_oe = h_oe * 10^-6;// in A/V h_fe = 50; h_re = 2.5*10^-4; R_L = 1.6;// in ohm R_S = 1;// in k ohm V_CC = 15;// in V // (i) Current Gain Ai = -h_fe/(1 + (h_oe*R_L)); disp(Ai,"The current gain is"); // (ii) Input impedance Zi = (h_ie*10^3) + (h_re*Ai*R_L);// in ohm Zi= Zi*10^-3;// in k ohm disp(Zi,"The input impedance in k ohm is"); Zi= Zi*10^3;// in ohm // (iii) Voltage gain A_V = Ai*R_L/Zi; disp(A_V,"The voltage gain is"); // (iv) Power gain A_P = Ai*A_V; disp(A_P,"The power gain is");
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Ex3_11.sce
Bf=1500//width of flange, in mm Df=150//thickness of flange, in mm d=600//effective depth, in mm sigma_cbc=7//in MPa sigma_st=230//in MPa m=13.33//modular ratio Ast=1964//in sq mm Asc=1140//in sq mm top_cover=50//in mm //to find critical depth of neutral axis Xc=d/(1+sigma_st/(m*sigma_cbc))//in mm //assume x>Df; equating moments of area on compression and tension sides about N.A. x=(m*Ast*d+Bf*Df^2/2+(1.5*m-1)*Asc*top_cover)/(m*Ast+Bf*Df+(1.5*m-1)*Asc)// in mm //we find that x<Df, hence our assumption that x>Df is wrong //to find x using Bf(x^2)/2 + (1.5m-1)Asc(x-d')=mAst(d-x), which becomes of the form px^2+qx+r=0 p=Bf/2 q=m*Ast+(1.5*m-1)*Asc r=-(m*Ast*d+(1.5*m-1)*Asc*top_cover) //solving quadratic equation x=(-q+sqrt(q^2-4*p*r))/(2*p)//in mm //as x<Xc, beam is under-reinforced sigma_cbc=sigma_st/m*x/(d-x)//in MPa sigma_cbc_dash=sigma_cbc*(x-top_cover)/x//stress in concrete at level of compression steel, in MPa //taking moments about tensile steel Mr=Bf*x*sigma_cbc*(d-x/3)/2+(1.5*m-1)*Asc*sigma_cbc_dash*(d-top_cover)//in N-mm mprintf("Moment of resistance of the beam=%f kN-m", Mr/10^6) //answer given in textbook is incorrect