pierreqi/scilab_code · Datasets at Hugging Face

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riccati.sce

clear,clc global A B D S xi Xt Xbox Tbox i h hit A = [-1 -1;-0 -2]; B = [0;1]; D = [1 0;0.4 1]; R = 1; Q = [0.79552 -0.0077911;-0.0077911 1.2702]; E = D*inv(R)*D'; P = ricc(A,E,Q,'cont') S = inv(R)*B'*P

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Ex5_34.sce

//=========================================================================== //chapter 5 example 34 clc; clear all; //variable declaration e0 =8.854*10^-12; d =0.05; er = 1; a = 0.25; V1 = 12000; //voltage in V V2 = 32000; //voltage in V //calculations //x-x0 = (1/2)*((V^2)/k)*(dc/dx) //C =(2*e0*er*A)/d //dC =(2*e0*er*a*x)/d // y = dC/dx = (2*e0*er*a)/d y = (2*e0*er*a)/d; X1 = 0.25/4; // A =x1+x01 = (1/2)*((V1^2)/k)*(dc/dx) X2 = 0.25/2; //B = x2+x01 = (1/2)*((V2^2)/k)*(dc/dx) //C = B/A =(V2/V1)^2 C = (V2/V1)^2; x01 = (X2-(C*X1))/(1-C); k = ((1/2)*((V1^2))*(y))/(X1-x01); X3 = (3/4)*0.25; V = sqrt(((X3-x01)*2*k)/y); //voltage in V //result mprintf("voltage required to pull the plate three quarte way in = %3.3f KV",(V*10^-3));

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connector.tst

# Cross connecter thermo test units SI $thermo = VirtualMaterials.Peng-Robinson / -> $thermo thermo + PROPANE ISOBUTANE n-BUTANE n-PENTANE WATER # lets have some streams for this test coldInlet = Stream.Stream_Material() hotInlet = Stream.Stream_Material() cd hotInlet.In T = 200 P = 150 Fraction = .01 .02 .01 0 1 MoleFlow = 500 cd / cd /coldInlet.In Fraction Fraction = .75 15 .08 .02 0 VapFrac = 0 P = 300 T = MoleFlow = 1000 cd / coldOutlet = Stream.Stream_Material() exch = Heater.HeatExchanger() exch cd exch DeltaPC = 10 DeltaPH = 50 DeltaTHO = 5 K cd / # hot side will use steam property package $thermo1 = VirtualMaterials.Steam95 exch.HotSide -> $thermo1 exch.HotSide.thermo1 + water # create hot outlet and assign the hot inlet thermo hotOutlet = Stream.Stream_Material() hotOutlet -> $thermo1 # create CrossConnector xc = CrossConnector.CrossConnector() hotInlet.Out -> xc.In xc.In xc.Out #connect things coldInlet.Out -> exch.InC exch.OutC -> coldOutlet.In xc.Out -> exch.InH exch.OutH.T exch.OutH -> hotOutlet.In # results coldInlet coldInlet.Out coldOutlet.Out hotInlet.Out hotOutlet.Out exch.ColdSide.InQ # one more stream and connector hotOut2 = Stream.Stream_Material() xc2 = CrossConnector.CrossConnector() xc2.Out -> hotOut2.In hotOut2.In copy / paste / cd /RootClone coldInlet coldInlet.Out coldOutlet.Out hotInlet.Out hotOutlet.Out exch.ColdSide.InQ

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// Chapter 9 example 9 //------------------------------------------------------------------------------ clc; clear; // Given Data fc1 = 495; // freq in Mhz fc2 = 505; // freq in Mhz // Calculations fo = (fc1 + fc2)/2; // Center of spectrum in Mhz BW = fc2 - fc1; // Bandwidth in Mhz PW = 1/BW; // compressed pulse width in us // Output mprintf('Center of spectrum = %d Mhz\n Matched Bandwidth = %d Mhz\n Compressed Pulse width = %3.1fus',fo,BW,PW); //------------------------------------------------------------------------------

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Huiping\Huiping_010.jpg Huiping\Huiping_017.jpg Don\Don_014.jpg Don\Don_008.jpg Shirley\Shirley_012.jpg Shirley\Shirley_009.jpg Kiran\Kiran_005.jpg Kiran\Kiran_013.jpg Allison\Allison_011.jpg Allison\Allison_002.jpg Amit\Amit_011.jpg Amit\Amit_001.jpg Gang\Gang_002.jpg Gang\Gang_015.jpg Ethan\Ethan_002.jpg Ethan\Ethan_004.jpg Rob\Rob_009.jpg Rob\Rob_004.jpg Nara\Nara_007.jpg Nara\Nara_008.jpg Weihong\Weihong_007.jpg Weihong\Weihong_011.jpg Dave\Dave_008.jpg Dave\Dave_006.jpg

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slr_convert_arglist.tst

// Author : Pierre Vuillemin (2017) // <-- NO CHECK REF --> mode(-1); attr_list = list('attr1',1,'attr2','value2') s = slr_convert_arglist(attr_list) assert_checktrue(typeof(s) == 'st') assert_checkequal(s.attr1, attr_list(2)) assert_checkequal(s.attr2, attr_list(4)) // mlist_type = 'myType' m = slr_convert_arglist(attr_list, mlist_type) assert_checktrue(typeof(m) == mlist_type) assert_checkequal(m.attr1, attr_list(2)) assert_checkequal(m.attr2, attr_list(4))

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clc //initialisation of variables P=81.5 P2=0.0019 //lbf/in^2 T1=360//F T2=420//F //CALCULATIONS p=P2/P//lbf/in^2 //RESULTS printf('The presure of water =% f lbf/in^2',p)

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Example13_7.sce

close(); clear; clc; //from solved example 13.6 V = 120; //V Pc = 75; //W Rc = V^2/Pc; I = 1.5; //A pf = 0.417; Im = sqrt(I^2 - (I*pf)^2); //A Ic = V/Rc; //A Xm = V/Im; //ohm mprintf("Rc = %d ohm\nXm = %d ohm",Rc,Xm);

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//chapter3 //example3.1 //page41 Ib1=10 // mA Eb1=100 // V Ib2=20 // mA // Ib is proportional to Eb^(3/2) // so we can say Ib1/Ib2 = Eb1^1.5/Eb2^1.5 //thus we can write log_Eb2=(2/3)*log(Eb1^1.5*Ib2/Ib1) Eb2=exp(log_Eb2) printf("required plate voltage = %.3f V",Eb2)

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//finding SOP// //example 37// clc //clears the command window// clear //clears// disp('given f=AC') disp('f=AC(B+B'')(D+D'')') disp('f=ACBD+ACBD''+ACB''D+ACB''D''');//required sum of minterms// disp('f=1111+1110+1011+1010') disp('required SOP form:') disp('f=summation(10,11,14,15)')

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// ---------------------------------------------------------------------------- // This file was automatically generated by SWIG (http://www.swig.org). // Version 4.0.0 // // Do not make changes to this file unless you know what you are doing--modify // the SWIG interface file instead. // ----------------------------------------------------------------------------- */ libinc_path = get_absolute_file_path('loader.sce'); [bOK, ilib] = c_link('libinc'); if bOK then ulink(ilib); end list_functions = [.. 'inc_Init'; .. 'SWIG_this'; .. 'SWIG_ptr'; .. 'inc'; .. ]; addinter(fullfile(libinc_path, 'libinc' + getdynlibext()), 'libinc', list_functions); clear libinc_path; clear bOK; clear ilib; clear list_functions;

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int start = 0; int end = 100; bool next = true; void main() { sum = start; while(next){ sum = sum + i; if(sum > end){ next = false; } } print("sum: ",sum,"\n"); }

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clear;lines(0); c=part(['a','abc','abcd'],[1,1,2])

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5_1.sce

clc //initialization of varaibles V1=10 //cu ft P1=15 //psia V2=5 //cu ft H=34.7 //Btu //calculations W=P1*(V2-V1)*144 dE=-H-W/778 //results printf("Internal energy change = %.1f Btu",dE)

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//Example No. 5.18 clc; clear; close; format('v',6); //Given Data : V=400;//V f=50;//Hz I=50;//A Ra=0.1;//ohm K=0.3;//V/rpm Ia=5;//A alfa=30;//degree Vavg=3*sqrt(3)*V*sqrt(2)/sqrt(3)/2/%pi*(1+cosd(alfa));//V Eb=Vavg-Ia*Ra;//V N=Eb/K;//rpm disp(N,"No load speed in rpm : "); Speed=1600;//rpm Eb=Speed*K;//V Vin=Eb+I*Ra;//V alfa=acosd(Vin/3/sqrt(3)/V/sqrt(2)*sqrt(3)*2*%pi-1);//degree disp(alfa,"Fringe angle in degree : ");

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//scilab 5.4.1 //Windows 7 operating system //chapter 19 VLSI Technology and Circuits clc clear ID=50*10^-6//ID=drain current in amperes k=25*10^-6//k=ue/D in A/V^2 VDS=0.25//VDS=drain-to-source voltage VGS=5//VGS=gate-to-source voltage VTH=1.5//VTH=threshold voltage w=ID/(k*(VGS-VTH)*VDS)//w=W/L format("v",5) disp(w,"W/L=") P=VDS*ID//P=power dissipated by the transistor disp("micro Watt",P*10^6,"The dissipated power is=") VDD=5//VDD=drain supply voltage of given NMOS transistor R=(VDD-VDS)/ID//R=load resistor to be connected in series with the drain disp("kilo ohm",R/1000,"The load resistance is=")

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Ex4_10.sce

// Ex4_10 clc; // Given E=2.5;// in MeV // Solution: k=0.693/(5.27*3.16*10^7);// decay constant A=k*0.1*6.022*10^23;// atoms/s A1=3.6*10^3*A;// atoms /hr E1=A1*E*1.6*10^-13*10^-3;//Energy in KJ/hr printf("The total energy dissipate per hour is = %f KJ",E1)

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newtonGregoryForward.sci

/* Author : Bagas Adi Firdaus Deskripsi : Program Metode Newton-Gregory Forward */ clear; clc; printf('\nProgram Metode Newton-Gregory Forward\n'); x=0.45; X = [0.1 0.3 0.5 0.7 0.9 1.1]; Y = [0.003 0.067 0.148 0.248 0.370 0.518]; printf('Diketahui Data Berikut:\n'); //Menampilkan data for i=1:6 printf('n = %d\t x = %.6f\t f(x) = %.6f\n',i, X(i),Y(i)); end x=input('Masukkan nilai x yang akan dicari f(x)nya = '); //Fungsi pencari faktorial function faktorial=faktorial(p) fak=1; for k=2 : p fak=fak*k; end faktorial=fak; endfunction //Menyimpan Y[k] pada kolom 1 matriks TS (tabel selisih) for k=1:6 TS(k,1)=Y(k); end //Membentuk tabel selisih for k=2:6 for i=1:7-k TS(i,k)=TS(i+1,k-1)-TS(i,k-1); end end //Jarak antar titik h=X(2)-X(1); //Menghitung p(x) //Orde yang digunakan adalah orde terbesar (orde 5) s=(x - X(1))/h; jumlah=TS(1,1); for i=2:6 suku=TS(1,i); for k=0:i-2 suku=suku*(s-k); end suku=suku/faktorial(i-1); jumlah=jumlah+suku; end hasil=jumlah; printf('Orde yang digunakan adalah orde terbesar yaitu orde 5\nJadi nilai f(%.2f) adalah : %.6f', x, hasil);

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clc //Initialization of variables n1=2 n2=10 n3=3 P=720 //mm of Hg //calculations n=n1+n2+n3 x1=n1/n P1=x1*P x2=n2/n P2=x2*P x3=n3/n P3=x3*P //results printf("\n Partial pressure of N2 = %d mm",P1) printf("\n Partial pressure of O2 = %d mm",P2) printf("\n Partial pressure of CO2 = %d mm",P3)

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%rnp.sci

function [r]=%rnp(l1,l2) //%rnp(l1,l2) <=> l1<>l2 r=degree(l1(3))==0 if r then r=l1(2)./coeff(l1(3))==l2,end r=~r

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exp_11_5.sce

clear; clc; Xtf=.2+.2+(.3*.6/0.9); pi=0.9; po=pi; del1=asin(Xtf*pi/(1.2*1)); Pm=1.2*1/Xtf; //fault condition Xtf1=(.4*.3+.3*.3+.3*.4)/.3; Pm1=1.2*1/Xtf1; //post fault condition Xtf2=.2+.2+.3; Pm2=1.2*1/Xtf2; delm=(%pi-(asin(pi/Pm2))); delc=acos((pi*(delm-del1)+Pm2*cos(delm)-Pm1*cos(del1))/(Pm2-Pm1)); mprintf("rotor angle is %.3f radian \n",del1); mprintf("Critical clearing angle is %.3f radian",delc);

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ex1_30.sce

//Chapter-1, Example 1.30, Page 43 //============================================================================= clc; clear; //INPUT DATA P=70;//total power dissipated in circuit in watts V1=6;//since applied voltage E is 6V,as per the characteristics of parallel circuit P.D across R1 is V2=6;//V1=V2,in volts R1=12;//resistance1 in parallel combination in ohms R2=6;//resistance2 in parallel combination in ohms R3=6.25//resistance3 in series with parallel combination in ohms I1=V1/R1;// current through the resistance R1 in Amps I2=V2/R2;//current through the resistance R2 in Amps r=0.25;//internal resistance in ohm //CALCULATIONS I=I1+I2;//total current through parallel combination E=(I*r)+(I*R3)+V2;//emf of battery in Volts //OUTPUT mprintf("Thus the value of emf of battery in Volts is %2.2f volts ",E); //=================================END OF PROGRAM==============================

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Chapter6_Example3.sce

clc clear //Input data C=[1.5145,1.5170,1.5230]//Refractive index of the crown glass for C,D and F line respectively F=[1.6444,1.6520,1.6637]//Refractive index of the flint glass for C,D and F line respectively //Calculations w1=(C(3)-C(1))/(C(2)-1)//Dispersive power of the first lens w2=(F(3)-F(1))/(F(2)-1)//Dispersive power of the second lens //Output printf('The dispersive power for crown glass is %3.4f \n The dispersive power for the flint glass is %3.5f',w1,w2)

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clear // //Given //Variable declaration sigma1=120 //Major tensile stress in N/sq.mm sigma2=-90 //Minor compressive stress in N/sq.mm sigma_gp=150 //Greatest principal stress in N/sq.mm //Calculation //case(a):Magnitude of the shearing stresses on the two planes tau=(sqrt(((sigma_gp-((sigma1+sigma2)/2))**2)-(((sigma1-sigma2)/2)**2))) //case(b):Maximum shear stress at the point sigmat_max=int((sqrt((sigma1-sigma2)**2+(4*tau**2)))/2) //Result printf("\n Shear stress on the two planes = %0.3f N/mm^2",tau) printf("\n Maximum shear stress at the point = %0.3f N/mm^2",sigmat_max)

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src = imread("../images/image_0197.jpg"); [rows cols] = size(src) mask = zeros(rows,cols); mask(1:rows,1:(cols/2)) = 255; output = textureFlattening(src,mask,12,180, 5 ); imshow(output);

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2 0 0 0 0 0 0 0 0 0 1e-08 1e-08 0.5 Central 1 0 0 0 9.48027 0 0 0 0 1e-14 -1e-14 0.1 Satélite -20 -20 -20 20 20 20 0 0 0 0 0 0 299792448 4 0

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clc //to calculate fringe width mu=1.5 //refractive index (unitless) alpha=%pi/180 //refracting angle in radian Y1=20*10^-2 //distance between the source and the biprism in m Y2=80*10^-2 //distance in m D=Y1+Y2 // distance in m lambda=6900*10^-10 //wavelength in m twod=2*(mu-1)*alpha*Y1 omega=D*lambda/twod disp("the fringe width is omega="+string(omega)+"m")

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commentsTestCase.tst

load 0 Reg0 # Comment load 255 Reg0 # Comment load 0 Reg1# Comment load 255 Reg1 move Reg0 # This is a comment at the end of the line move Reg1# Comment # This is me commenting out a section of code # load 0 Reg0# Comment add Reg0# Comment add Reg1# Comment sub Reg0# Comment sub Reg1# Comment sr# Comment sl# Comment and Reg0# Comment and Reg1# Comment or Reg0# Comment or Reg1# Comment #Insert block comment # # # # # # # # # # # # # # inv# Comment j 0# Comment j 22# Comment jaz 0# Comment jaz 22# Comment jal 0# Comment jal 22# Comment jr# Comment wri Reg0# Comment wri Reg1# Comment wri P1# Comment wri P2# Comment wri Tx# Comment # Comment str Reg0# Comment str Reg1# Comment

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Ex22_17.sce

// Example 22_17 clc;funcprot(0); //Given data P=500;// Plant capacity in kW T_1=300;// °C p_4=30;// bar p_5=7;// bar p_6=0.04;// bar dT=5;// The rise in cooling water temperature in °C C_pw=4.2;// kJ/kg.°C // Calculation // From h-s chart: h_4=3000;// kJ/kg h_5=2700;// kJ/kg h_6=1970;// kJ/kg // From steam tables h_f1=121.4;// kJ/kg(at 0.04 bar) h_f2=697;// kJ/kg(at 7 bar) function[X]=mass(y) X(1)=((y(1)*h_5)+((1-y(1))*h_f1))-(1*h_f2); endfunction y=[0.1]; z=fsolve(y,mass); m=z(1);// kg W=(1*(h_4-h_5))+((1-m)*(h_5-h_6));// kJ/kg Q_s=h_4-h_f2;// Heat supplied in kJ/kg n_s=(W/Q_s)*100;// Efficiency in % m_s=(P/W)*3600;//Steam generated per second in kg/hr m_w=((h_6-h_f1)*(m_s/3600)*(1-m))/(C_pw*dT);// kg/sec // If there ie no feed water,then W_1=h_4-h_6;// kJ/kg Q_s=h_4-h_f1;// kJ/kg n=(W_1/Q_s)*100;// Efficiency in % m_s1=(P/W_1)*3600;//Steam generated per second in kg/hr m_w1=((m_s/3600)*(h_6-h_f1))/(C_pw*dT);// The amount of cooling water in kg/sec printf('\n(a)The rankine efficiency=%0.1f percentage \n Steam generation rate of boiler=%0.1f kg/hr \n The amount of cooling water=%0.2f kg/sec \n(b)The rankine efficiency=%0.1f percentage \n Steam generation rate of boiler=%0.1f kg/hr \n The amount of cooling water=%0.2f kg/sec',n_s,m_s,m_w,n,m_s1,m_w1); // The answer vary due to round off error

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9_16.sci

// calculate the open circuit voltage clc; P=200*10^3; R=70*10^-3; v=0.25; t=1*10^-3; r=60*10^-3; E=200*10^9; Sr=[3*P*R^2*v/(8*t^2)]*{(1/v+1)-(3/v+1)*(r/R)^2}; St=[3*P*R^2*v/(8*t^2)]*{(1/v+1)-(1/v+3)*(r/R)^2}; Sta2=(Sr-v*St)/E; Sta3=(Sr-v*St)/E; r0=10*10^-3; Sr1=[3*P*R^2*v/(8*t^2)]*{(1/v+1)-(3/v+1)*(r0/R)^2}; St1=[3*P*R^2*v/(8*t^2)]*{(1/v+1)-(1/v+3)*(r0/R)^2}; Sta1=(Sr1-v*St1)/E; Sta4=(Sr1-v*St1)/E; Gf=1.8; ei=12; eo=(Sta1+Sta4-Sta2-Sta3)*Gf*ei/4; disp(eo,'output voltage (V)')

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clc h2=1597; //kJ/kg h3=1790; //kJ/kg h4=513; //kJ/kg h1=h4; t3=58; //0C x1=0.13; tc=27; //0C capacity=10.5; //tonnes disp("(i) Condition of the vapour at the outlet of the compressor =") t=t3-tc; disp(t) disp("°C") disp("(ii) Condition of vapour at entrance to evaporator =") disp(x1) disp("COP =") COP=(h2-h1)/(h3-h2); disp(COP) disp("(iv) Power required =") P=capacity*14000/COP/3600; disp(P) disp("kW")

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//To determine the max regulation and the pf at which it occurs clc; clear; Vr=2.5; Vx=5; printf('The expression for voltage requlation is y= %g cos(phi) + %g sin(phi) \n',Vr,Vx ) printf('Differenciating w.r.t phi and equating it to zero, we get the power factor angle \n') printf('We get tan(phi)=> Vr/Vx => 5/2.5 => 2 \n \n') phi=atand(Vx/Vr); // power factor angle y= Vr*cosd(phi)+Vx*sind(phi); // Max Volatge regulation printf('The maximum regulation is %g percent \n and the power factor at which it occurs is %g degrees \n',y,phi)

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// design of bersia power station // ex 7.6 pgno.178 clc p=24.7e6// power watt h=26.5 //m N=187.5 // rev/min Q=104 //m3/s w=(2*N*%pi)/60 g=9.8 mprintf('\n w= %f rad/s',w) wt=(w*sqrt(p/10^3))/(g*h)^(5/4) mprintf('\n wt =%f',wt) Ns=(N*sqrt(p/10^3))/(h^(5/4)) // speed mprintf('\n Ns =%f',Ns) n0=p/(9800*Q*h) // overall efficiency mprintf('\n Overall efficiency n0= %f percentage',n0*100) mprintf('\n Based on specific speed values obtained kaplan turbine is selected with an overall efficiency of %f percentage',n0*100)

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style.fontSize=14; style.displayedLabel="CurrentstarvedInverter" pal5=xcosPalAddBlock(pal5,"CurrentstarvedInverter",[],style);

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EX25_7.sce

//EXAMPLE 25.7 //COUPLED COILS clc; funcprot(0); //Variable Initialisation x=poly(0,"x"); L11=1+(1/x); L22=0.5+(1/x); L12=1/x; L21=1/x; ic1=20;.........//First coil is excited by constant current in Amperes ic2=-10;.........//Second coil is excited by constant current in Amperes x1=0.5;........//Displacement in Centi Meter x2=1;...........//Displacement in Centi Meter Wfd=((1/2)*L11*ic1^2)+(L12*ic1*ic2)+((1/2)*L22*ic2^2);........//Function of mechanical work done F1=(-1)*derivat(Wfd);..........//Function for energy stored disp(F1,"Energy stored :"); deff('y=f(x)','y=50/(x^2)'); dWmech=intg(0.5,1,f);.................//Mechanical work if x ranges from 0.5 to 1.0cm in Joules disp(dWmech,"(a).Mechanical work if x ranges from 0.5 to 1.0cm in Joules:"); y1=(L11*ic1)+(L12*ic2);........//Leakage flux for coil 1; disp(y1,"(b).Function for leakage flux for coil 1"); y1x1=20+10/(x1);....//at x1 y1x2=20+10/(x2);....//at x2 dwelec1=ic1*(y1x2-y1x1);........//Energy supplid by coil 1 in Joules disp(dwelec1,"Energy supplid by coil 1 in Joules:"); y2=(L12*ic1)+(L22*ic2);.......//Leakage flux for coil 2; disp(y2,"Leakage flux for coil 2"); y2x1=-5+10/(x1);....//at x1 y2x2=-5+10/(x2);....//at x2 dwelec2=ic2*(y2x2-y2x1);........//Energy supplid by coil 1 in Joules disp(dwelec2,"Energy supplid by coil 2 in Joules:");

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/NLP_Project/test/tweet/bow/bow.5_15.tst

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mandar15/NLP_Project

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2020-05-20T13:36:05.842840

2013-07-31T06:53:59

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bow.5_15.tst

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clear; clc; //Example - 6.8 //Page number - 225 printf("Example - 6.8 and Page number - 225\n\n"); //Given T_1 = 298.15;//[K] - Standard temperature T_2 = 973.15;//[K] - Reaction temperature //At 298.15 K delta_H_CH4_for_298 = -17.889*10^(3);//[cal/mol] - Enthalpy of formation of CH4 at 298.15 K delta_H_C_for_298 = 0.00;//[cal/mol] - Enthalpy of formation of C (s, graphite) at 298.15 K delta_H_H2_for_298 = 0.00;//[cal/mol] - Enthalpy of formation of H2 at 298.15 K delta_G_CH4_for_298 = -12.140*10^(3);//[cal/mol] - Gibbs free energy change for formation of H2 at 298.15 K delta_G_C_for_298 = 0.00;//[cal/mol] - Gibbs free energy change for formation of C (s, graphite) at 298.15 K delta_G_H2_for_298 = 0.00;//[cal/mol] - Gibbs free energy change for formation of H2 at 298.15 K ///Standaerd heat capacity data in cal/mol-K are given below, T is in K //Cp_0_CH4 = 4.75 + 1.2*10^(-2)*T + 0.303*10^(-5)*T^(2) - 2.63*10^(-9)*T^(3) //Cp_0_C = 3.519 + 1.532*10^(-3)*T - 1.723*10^(5)*T^(-2) //Cp_0_H2 = 6.952 - 0.04576*10^(-2)*T + 0.09563*10^(-5)*T^(2) - 0.2079*10^(-9)*T^(3) //Therefore standard heat capacity of reaction is given by, //Cp_0_rkn = 2*Cp_0_H2 + Cp_0_C - Cp_0_CH4 //On simplification,we get the relation //Cp_0_rkn = 12.673 - 0.0113832*T - 1.1174*10^(-6)*T^(2) + 2.2142*10^(-9)*T^(3) - 1.723*10^(5)*T^(-2) delta_H_rkn_298 = -delta_H_CH4_for_298;//[cal] - Enthalpy of reaction at 298.15 K delta_G_rkn_298 = -delta_G_CH4_for_298;//[cal] - Gibbs free energy of the reaction at 298.15 K delta_H_rkn_973 = delta_H_rkn_298 + integrate('12.673-0.0113832*T-1.1174*10^(-6)*T^(2)+2.2142*10^(-9)*T^(3)-1.723*10^(5)*T^(-2)','T',T_1,T_2);//[cal] printf(" Standard enthalpy change of reaction at 973.15 K is %f cal\n\n",delta_H_rkn_973); //Now determining the standard entropy change of reaction at 298.15 K delta_S_rkn_298 = (delta_H_rkn_298 - delta_G_rkn_298)/298.15;//[cal/K] delta_S_rkn_973 = delta_S_rkn_298 + integrate('(12.673-0.0113832*T-1.1174*10^(-6)*T^(2)+2.2142*10^(-9)*T^(3)-1.723*10^(5)*T^(-2))/T','T',T_1,T_2);//[cal/K] //Therefore,the standard Gibbs free energy change of the reaction is given by, delta_G_rkn_973 = delta_H_rkn_973 - 973.15*delta_S_rkn_973;//[cal] printf(" Standard Gibbs free energy change of reaction at 973 K is %f cal\n",delta_G_rkn_973);

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//Example5.29 // To find Slew rate of an op-amp clc; clear; close; Iq = 15 ; // uA // bias current Cm = 30 ; // pF // internal frequency compensated capacitor Slewrate = (Iq/Cm); disp('the Slew rate of an op-amp is = '+string(Slewrate)+' V/u sec');

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disp('The recurrence relation t[n]=4(t[n-1]-t[n-2])') x=poly(0,'x'); disp(g=x^2-4*x+4,'characterstic polynomial equation for the above recurrence relation') j=[]; j=roots(g); disp(j,'roots of the characterstic equation j1,j2') disp('the general solution is t[n]=n*2^n) disp('initial condition at n=0 and n=1 respectively are:') t0=1; t1=1; //putting the values of t0 and t1 we get the equations to solve D=[1 0;2 2] K=[1 1]' c=linsolve(D,K) D=[1 0;2 2] K=[1 1]' c=[]; c=D\K; c1=c(1) c2=c(2) disp('thus the solution is t{n}=2*n-n*2^(n-1)')

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-- VectorCAST 18.sp3 (10/18/18) -- Test Case Script -- -- Environment : MANAGER_DATABASE -- Unit(s) Under Test: database manager -- -- Script Features TEST.SCRIPT_FEATURE:C_DIRECT_ARRAY_INDEXING TEST.SCRIPT_FEATURE:CPP_CLASS_OBJECT_REVISION TEST.SCRIPT_FEATURE:MULTIPLE_UUT_SUPPORT TEST.SCRIPT_FEATURE:MIXED_CASE_NAMES TEST.SCRIPT_FEATURE:STATIC_HEADER_FUNCS_IN_UUTS -- -- Unit: manager -- Subprogram: Clear_Table -- Test Case: Clear_Table.001 TEST.UNIT:manager TEST.SUBPROGRAM:Clear_Table TEST.NEW TEST.NAME:Clear_Table.001 TEST.EXPECTED:manager.Clear_Table.return:0 TEST.END -- Subprogram: Get_Check_Total -- Test Case: Get_Check_Total.001 TEST.UNIT:manager TEST.SUBPROGRAM:Get_Check_Total TEST.NEW TEST.NAME:Get_Check_Total.001 TEST.COMPOUND_ONLY TEST.VALUE:manager.Get_Check_Total.Table:1 TEST.EXPECTED:manager.Get_Check_Total.return:24 TEST.END -- Subprogram: Place_Order -- Test Case: chicken TEST.UNIT:manager TEST.SUBPROGRAM:Place_Order TEST.NEW TEST.NAME:chicken TEST.COMPOUND_ONLY TEST.VALUE:manager.Place_Order.Table:1 TEST.VALUE:manager.Place_Order.Seat:1 TEST.VALUE:manager.Place_Order.Order.Entree:CHICKEN TEST.EXPECTED_USER_CODE:<<testcase>> TEST.END_EXPECTED_USER_CODE: TEST.END -- Test Case: steak TEST.UNIT:manager TEST.SUBPROGRAM:Place_Order TEST.NEW TEST.NAME:steak TEST.COMPOUND_ONLY TEST.VALUE:manager.Place_Order.Table:1 TEST.VALUE:manager.Place_Order.Seat:2 TEST.VALUE:manager.Place_Order.Order.Entree:STEAK TEST.END -- COMPOUND TESTS TEST.SUBPROGRAM:<<COMPOUND>> TEST.NEW TEST.NAME:Test_Order TEST.REQUIREMENT_KEY:FR17 TEST.REQUIREMENT_KEY:FR22 TEST.SLOT: "1", "manager", "Place_Order", "1", "chicken" TEST.SLOT: "2", "manager", "Place_Order", "1", "steak" TEST.SLOT: "3", "manager", "Get_Check_Total", "1", "Get_Check_Total.001" TEST.END --

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//Chemical Engineering Thermodynamics //Chapter 10 //Compressor //Example 10.5 clear; clc; //Given P1 = 1;//Initial pressure in Kgf/sq cm P4 = 200;//Final pressure in Kgf/sq cm n = 4;//no of stages //To find out the presure between stages r = (P4/P1)^(1/n);//Compression ratio P2 = r*P1; mprintf('The pressure after 1st stage is %f Kgf/sq cm',P2); P3 = r*P2; mprintf('\n The pressure after 2nd stage is %f Kgf/sq cm',P3); P4 = r*P3; mprintf('\n The pressure after 3rd stage is %f Kgf/sq cm',P4); //end

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function [s]=%lsssr(s1,s2) //s=%lsssr(s1,s2) <=> s=s1-s2 // s1 : state-space // s2 : transfer matrix //! [s1,s2]=sysconv(s1,s2) s=s1-s2

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x = 0:10; ReLU = x figure(1) plot(x,ReLU) a = 0.1 x = -10:0 ReLU = a*x plot(x,ReLU)

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//Exa 4.8 clc; clear; close; //Given data : r=(20/2)/10;//cm d1=4*100;//cm d2=5*100;//cm d3=6*100;//cm rdash=0.7788*r;//cm L=0.2*log((d1*d2*d3)^(1/3)/rdash);//mH disp(L,"Inductance per phase(mH)");

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//define problem parameters ni=1.5e10*1e6; //intrinsic carrier concentration in Si [m^(-3)] Na=1e15*1e6; //acceptor doping concentration [m^(-3)] Nd=5e15*1e6; //donor concentration [m^(-3)] A=1e-4*1e-4; //cross sectional area [m^2] eps_r=11.9; //cross sectional area [m^2] //define physical constants (SI units) q=1.60218e-19; //electron charge k=1.38066e-23; //Boltzmann's constant eps0=8.85e-12; //permittivity of free space eps=eps_r*eps0; T=300; //temperatuure //compute diffusion barrier voltage Vdiff=k*T/q*log(Na*Nd/ni^2) //junction capacitance at zero applied voltage C0=A*sqrt(q*eps/(1/Na+1/Nd)/2/Vdiff) //extents of the space charge region dn=sqrt(2*eps*Vdiff/q*Na/Nd/(Na+Nd)); dp=sqrt(2*eps*Vdiff/q*Nd/Na/(Na+Nd)); //define range for applied voltage VA=-5:0.1:Vdiff; //compute junction capacitance C=C0*(1-VA/Vdiff).^(-1/2); plot(VA,C/1e-12); title('Junction capacitance of abrupt Si pn-contact'); xlabel('Applied junction voltage V_A, Volts'); ylabel('Junction capacitance C, pF');

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clc; A=[0 1;-1/8 3/4]; B=[0;1]; C=[-1/8 3/4]; D=[1]; Hz=ss2tf(syslin('d',A,B,C,D)); disp(Hz,"H(z)="); z = %z; syms n z1; //To f i n d out I n v e r s e z t r a n s f o rm z must //be l i n e a r z = z1 X =z ^2 /((z -(1/2) )*(z -(1/4) )) X1 = denom (X); zp = roots (X1); X1 = z1 ^2 /(( z1 -(1/4) )*(z1 -(1/2) )) F1 = X1 *( z1 ^(n -1) )*(z1 -zp (1) ); F2 = X1 *( z1 ^(n -1) )*(z1 -zp (2) ); h1 = limit (F1 ,z1 ,zp (1) ); disp (h1 , ' h1 [ n]= ' ) h2 = limit (F2 ,z1 ,zp (2) ); disp (h2 , ' h2 [ n]= ' ) h = h1+h2; disp ('for n>=0',h, ' h [ n]= ' )

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function [z]=zeta(k,t) z=(1-exp(-k*t))/k endfunction function [m, v, dm, dv]=Z_esp_var(x_0, x_inf, k, sigma, t) m=(x_0-x_inf)*zeta(k,t)+x_inf*t v=sigma**2*(t-zeta(k,t))/k**2 -sigma**2*zeta(k,t)**2/(2*k) dm = (x_0-x_inf)*exp(-k*t)+x_inf dv = sigma**2*(1-exp(-k*t))/k**2 - sigma**2*zeta(k,t)*exp(-k*t)/(2*k) endfunction function [theta1, theta2]=thetas(x_0, x_inf, k, sigma, t, eta) [m, v, dm, dv]=Z_esp_var(x_0, x_inf, k, sigma, t) theta1 = exp((1+eta)*m + (1+eta)**2 *v/2) theta2 = -eta*(dm+(1+eta)*dv/2)*theta1 endfunction //discretisation: calcul de phi(t,T) function [phi_t_T] = discrete_phi_t_T(t, T, N_pas, x_0, x_inf, k, sigma, eta, mu_i) phi_t_T = 0 delta_t = int((T-t)/N_pas) t_i = t for i=0:(T-t) [theta1, theta2]=thetas(x_0, x_inf, k, sigma, t_i - t, eta) phi_t_T = phi_t_T + delta_t* exp(-mu_i*(t-t_i))*(mu_i*theta1+theta2) t_i = t_i + delta_t end [theta1, theta2]=thetas(x_0, x_inf, k, sigma, T - t, eta) phi_t_T = phi_t_T + exp(-mu_i*(T-t))*theta1 endfunction //Input: //Taux d'interet risque neutre par calibration k_r = 0.2080239; r_inf = 0.0402051; r_0 = 0.005; sigma_r = 0.0018370; //Processus x_t k_x = 0.3; x_inf = -0.01; x_0 = 0.02; sigma_x = 0.008; //Log return of the asser mu_a = 0.04; sigma_a = 0.06; //Correlation rho_xr = 0.; rho_as = 0.95; rho_ar = 0.25; //Parametres taux de rachat alpha = -0.05; beta_ = -0.01; gamma_ = 0.01; delta = 0.03; mu_min = -0.05; mu_max = 0.3; //Rachat structurel mu_i = 0.05; eta = 2; //Taux minimum garanti TMG = 0.015; //Maturité T = 10 ; //Nb de pas de temps N = 15; function [f]=f(x,r) f = x+r endfunction function [g]=g(x) g = mu_i - eta*x endfunction M=10000;//taille échantillon Monte Carlo A_0 = 101479200;//ACTIF: obligations, actions, immobilier à t=0 E_0 = 57238200;//PASSIF: dettes vis-à-vis des actionnaires //L0 = ;//PASSIF: dettes vis-à-vis des assurés //INUTILE P = 36000000; //Deux gaussiennes indépendantes Gr = rand(M,1,"normal"); Gx = rand(M,1,"normal"); Ga = rand(M,1,"normal"); //Modélisation des dynamiques r_1 et x_1 r_1 = r_0*exp(-k_r) + r_inf*(1-exp(-k_r)) + sigma_r*sqrt((1-exp(-2*k_r))/(2*k_r))*Gr; x_1 = x_0*exp(-k_x) + x_inf*(1-exp(-k_x)) + sigma_x*sqrt((1-exp(-2*k_x))/(2*k_x))*((rho_as/sqrt(1-rho_ar*rho_ar))*Ga + sqrt((1-rho_ar**2- rho_as**2)/(1-rho_ar*rho_ar))*Gx); R_1 = mu_a + rho_ar*sigma_a*Gr + sqrt(1-rho_ar**2)*sigma_a*Ga; A_1 = A_0*(1 + R_1); PM_1 = P*exp(f(x_0,r_0) - g(x_0)) t = 1 BE_1 = PM_1*discrete_phi_t_T(t, T, N, x_0, x_inf, k_x, sigma_x, eta, mu_i) test = discrete_phi_t_T(t, T, N, x_0, x_inf, k_x, sigma_x, eta, mu_i) BE = BE_1*ones(M,1) //Calcul de E_1 F_1 = P*g(x_0) E_1 = A_1 - BE - F_1; //Calcul de L L = exp(-r_0).*E_1; //Statistique d'ordre de L (trier L) L = gsort(L,'g','i'); mprintf("VaR = %f \n",L(ceil(M*0.005))); //Calcul de SCR_0 SCR_0 = E_0 - L(ceil(M*0.005)); mprintf("SCR_0 = %f \n",SCR_0); Nb_simul = linspace(1,M,M)' plot(Nb_simul, E_0-L) //hold on plot(M*0.005, SCR_0, 'r*') xlabel('N',"fontsize",6) ylabel('$ E_0 - e^{-r_0}E_1 $',"fontsize",6)

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//Example 6_1_u2 clc(); clear; //To calculate the di-electric constant eo=8.85*10^-12 //units in F/meter alphae=36*10^-40 //units in meter^3 n=5*10^28 //units in meter^-3 er=((30*eo)+(2*n*alphae))/((30*eo)-(n*alphae)) printf("The di-electric constant is er=%.2f",er)

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// Example 1.13 page no-34 clear clc l=1.27 //cm D=19.4 //cm s=0.475 //cm Va=400 //volts Se=l*D*10^-2/(2*s*Va) Se=ceil(Se*10^5) printf("\nS_E=%.2f mm/v",Se/100) v=30 //volt e=1.6*10^-19 //C m=9.1*10^-31 //kg x=sqrt(m/e) B=(x*0.65*30*sqrt(2*Va))/(l*D) printf("\nB=%.2f*10^-5 wb/m^2",B*10^5)//answer not matches with given answer

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// Scilab Code Ex8.4: Page-430 (2011) clc;clear; n = 1e+006;....// Frequency of Ultrasonic waves, Hz C = 2.5e-014;....// Capcitance of capacitor, F // Frequency of elecric oscillations is given by n = 1/(2*%pi)*sqrt(1/(L*C)), solving for L L = 1/(4*%pi^2*n^2*C); // The inductance of an inductor to produce ultrasonic waves, henry printf("\nThe inductance of an inductor to produce ultrasonic waves = %d henry", L); // Result // The inductance of an inductor to produce ultrasonic waves = 1 henry

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function [result] = f(x) result = 3*sin(x-1) + 5*x - 7.8 endfunction function [raiz,it] = bissecao(a,b,e,N) it = 0 erro = 100 raiz = a while(erro > e & it < N) raiz_anterior = raiz raiz = (a+b)/2 if(f(raiz) * f(a) < 0) then b = raiz else a = raiz end erro = abs((raiz-raiz_anterior)/raiz) it = it + 1 end endfunction

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//fminunc function y=fun(x) y=100*(x(2)-x(1)^2)^2 + (1-x(1))^2 endfunction function y=grad(x) //y=[3*x(1)^2,3*x(2)^2]; y= [-400*x(1)*x(2) + 400*x(1)^3 + 2*x(1)-2, 200*(x(2)-x(1)^2)]; endfunction function y=hess(x) //y=[6*x(1),0;0,6*x(2)] y= [1200*x(1)^2- 400*x(2) + 2, -400*x(1);-400*x(1), 200 ]; endfunction //FAILURE CASES //Fails sometimes if starting point is a stationary point i.e. f'=0. (works for x1^2+x2^2 and (0,0) but fails for (x1-1)^2+(x2-1)^2 and (1,1)) //Fails when it converges to point of inflecion. So if function has a point of inflection nearby when compared to the local minimum the function may fail to find the optimal value. pt=[1,1.1]; options=list("MaxIter", [1500], "CpuTime", [500], "Gradient", "OFF", "Hessian", "OFF"); [x,f,e,s,g,h]=fminunc(fun,pt,options) //fminbnd function y=fun1(x) y=(x-1)^3 endfunction //FAILURE CASES //Fails if point of inflection is x=0 and is included in the bounded interval options1=list("MaxIter",[15000000],"CpuTime", [10000]) [x1,f1,e1,s1]=fminbnd(fun1,-10,12,options1)

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s=%s sys=syslin('c',((k)*(s+2)*(s+3))/((s+1)*(s))) clf evans(sys)

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concept('Good'). concept('Patricide'). role(hasChild). equiv(and([some(arole(hasChild), and([aconcept('Patricide'), some(arole(hasChild), not(aconcept('Patricide')))]))]), aconcept('Good')). cassertion(not(aconcept('Patricide')), i12). cassertion(aconcept('Patricide'), i2). rassertion(arole(hasChild), i11, i12). rassertion(arole(hasChild), i10, i11). rassertion(arole(hasChild), i9, i10). rassertion(arole(hasChild), i8, i9). rassertion(arole(hasChild), i7, i8). rassertion(arole(hasChild), i6, i7). rassertion(arole(hasChild), i5, i6). rassertion(arole(hasChild), i4, i5). rassertion(arole(hasChild), i3, i4). rassertion(arole(hasChild), i2, i3). rassertion(arole(hasChild), i1, i11). rassertion(arole(hasChild), i1, i10). rassertion(arole(hasChild), i1, i9). rassertion(arole(hasChild), i1, i8). rassertion(arole(hasChild), i1, i7). rassertion(arole(hasChild), i1, i6). rassertion(arole(hasChild), i1, i5). rassertion(arole(hasChild), i1, i4). rassertion(arole(hasChild), i1, i3). rassertion(arole(hasChild), i1, i2). query(instances(aconcept('Good')), [i1]). query(instance(i1, aconcept('Good'))). %default answer: true query(instance(i2, aconcept('Good')), false). query(roleFillers(i1, arole(hasChild)), [i2, i3, i4, i5, i6, i7, i8, i9, i10, i11]). query(relatedIndividuals(arole(hasChild)), [i1-i2, i1-i3, i1-i4, i1-i5, i1-i6, i1-i7, i1-i8, i1-i9, i1-i10, i1-i11, i2-i3, i3-i4, i4-i5, i5-i6, i6-i7, i7-i8, i8-i9, i9-i10, i10-i11, i11-i12]).

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PL/SQL Developer Test script 3.0 53 begin -- Call the procedure insertions.set_mascota_adoptar(nombre_m => 'Sparky', raza_m => 'Husky Siberiano', tamano1 => 'Grande', imagen_m => 'husk.jpg', chip_ident => 'HS4875', color_m => 'Gris', estado_m => 'Adoptar', pais1 => 'Costa Rica', provincia1 => 'Cartago', canton1 => 'La Unión', distrito1 => 'Tres Rios', detalle_direc => '500 metros norte del parque', recompensa1 => '10000', descripcion => 'Muy bueno para cuidar propiedades', fecha => '13/11/2014', username => 'emmanuelrs'); insertions.set_mascota_adoptar(nombre_m => 'Colin', raza_m => 'Rottweiler', tamano1 => 'Grande', imagen_m => 'rot.jpg', chip_ident => 'rt0101', color_m => 'Negro', estado_m => 'Adoptar', pais1 => 'Costa Rica', provincia1 => 'San José', canton1 => 'Moravia', distrito1 => 'San Vicente', detalle_direc => 'contiguo a la panaderia San Marcos', recompensa1 => '5000', descripcion => 'Muy jugueton, no es bravo', fecha => '1/11/2014', username => 'Dani'); insertions.set_mascota_adoptar(nombre_m => 'Dino', raza_m => 'Schnauzer', tamano1 => 'Mediana', imagen_m => 'schna.jpg', chip_ident => 'SH4875', color_m => 'Gris', estado_m => 'Adoptar', pais1 => 'Costa Rica', provincia1 => 'Heredia', canton1 => 'Belén', distrito1 => 'Asunción', detalle_direc => '200 metros este del parque', recompensa1 => '0', descripcion => 'muy jugueton', fecha => '28/09/2014', username => 'Dani'); end; 16 nombre_m 0 -5 raza_m 0 -5 tamano1 0 -5 imagen_m 0 -5 chip_ident 0 -5 color_m 0 -5 estado_m 0 -5 pais1 0 -5 provincia1 0 -5 canton1 0 -5 distrito1 0 -5 detalle_direc 0 -5 recompensa1 0 -5 descripcion 0 -5 fecha 0 -5 username 0 -5 0

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clc; // page no 407 // prob no 12_1 //A radio channel with BW=10KHz and SNR=15 dB B=10*10^3; snr=15; //converting dB in power ratio SNR=10^(snr/10); //a)Determination of theoretical max data rate C1=B*log2(1+SNR); disp('kb/s',C1/1000,'a)The theoretical max data rate is'); //b)Determination of data rate with 4 states i.e M=4 M=4; C2=2*B*log2(M); disp('kb/s',C2/1000,'b)The data rate for 4 states is');

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//chapter-3 page 47 example 3.1 //============================================================================= clc; clear; Z0=100;//Characteristic Impedance in ohms S=5;//Voltage Standing Wave Ratio(VSWR) //CALCULATION Zm=Z0*S;//Termainating impedance at a max of the voltage standing wave Zl=Zm;//Loading Impedance //OUTPUT mprintf('Terminating impedance at a maximum of the voltage standing wave is Zl= %3.0f ohms',Zl); //====================END OF PROGRAM========================================

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exec("swigtest.start", -1); p = getnull(); checkequal(SWIG_this(p), 0, "SWIG_this(p)"); checkequal(funk(p), %T, "funk(p)"); exec("swigtest.quit", -1);

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// problem 7.4 b=3.5 i=1/1000 d=1.5 C=60 y=60 x=1.5/tand(y) w=b+x*2 A=(w+b)*0.5*d P=b+2*((x*x+d*d)^0.5) m=A/P Q=A*C*((m*i)^0.5) disp(Q*1000,"discharge carried by the canal in litres/sec")

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//control systems by Nagoor Kani A //Edition 3 //Year of publication 2015 //Scilab version 6.0.0 //operating systems windows 10 // Example 3.7 clc; clear; s=%s p=poly([1 0.4 0 ],'s','coeff') q=poly([0 0.6 1],'s','coeff') g=p./q disp(g,'the given transfer function is') c=g/(1+g) disp(c,'the closed loop transfer function is') u=c/s disp(u,'the in put is unit step signal') //standard form od second order system is w^2/s^2+2*zeta*w*s+w^2 //compaing h with the standard form w=1//natural frequency of oscillation disp(w,'natural frequency of oscillation in rad/sec') zeta=1/(2*w) disp(zeta,'the damping ratio is') mp=exp((-zeta*%pi)/sqrt(1-(zeta)^2))*100//percentage peak overshoot disp(mp,'percentage peak overshoot in percentage') tp=%pi/(w*sqrt(1-(zeta)^2)) disp(tp,'peak time in seconds')

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Example9_2.sce

clear; clc; // Illustration 9.2 // Page: 354 printf('Illustration 9.2 - Page: 354\n\n'); // solution //****Data****// // a:water b:ethylaniline Pt = 760; // [mm Hg] ma1 = 50;// [g] mb1 = 50;// [g] //*******// // Data = [Temp Pa(mm Hg) Pb(mm Hg)] Data = [38.5 51.1 1;64.4 199.7 5;80.6 363.9 10;96.0 657.6 20;99.15 737.2 22.8;113.2 1225 40]; Ma = 18.02;// [kg/kmol] Mb = 121.1;// [kg/kmol] for i = 1:6 p = Data(i,2)+Data(i,3); if p = = Pt pa = Data(5,2);// [mm Hg] pb = Data(i,3);// [mm Hg] T = Data(i,1);// [OC] end end ya_star = pa/Pt; yb_star = pb/Pt; ya1 = ma1/Ma;// [g mol water] yb1 = mb1/Mb;// [g mol ethylalinine] Y = ya1*(yb_star/ya_star);// [g mol ethylalinine] printf("The original mixture contained %f g mol water and %f g mol ethylalinine\n",ya1,yb1); printf("The mixture will continue to boil at %f OC, where the equilibrium vapour of the indicated composition,until all the water evaporated together with %f g mol ethylalinine\n",T,Y); printf("The temparature will then rise to 204 OC, and the equilibrium vapour will be of pure ethylalinine");

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Example4_8.sce

//Gerschgorin's first theorem clc; clear; close(); //find the eigen values lying [0,4] with an error of 0.25 //taking p at mid point of the interval C=[2,-1,0;-1,2,-1;0,-1,1]; p=2; f(1)=1; f(2)=C(1,1)-p; count = 0; if f(1)*f(2)>0 then count = 1; end for r=3:4 br=C(r-2,r-1); f(r)=-br^2*f(r-2)+(C(r-1,r-1)-p)*f(r-1); if f(r)*f(r-1)>0 then count = count+1; // elseif f(r-1)==0 && f(r-1)* ?????? check for sign when f(r)=zero end end disp(f,'Sturm sequences') disp(count,'Number of eigen values strickly greater than 2 : ') p=1; f(1)=1; f(2)=C(1,1)-p; count1 = 0; if f(1)*f(2)>0 then count1 = 1; end for r=3:4 br=C(r-2,r-1); f(r)=-br^2*f(r-2)+(C(r-1,r-1)-p)*f(r-1); if f(r)*f(r-1)>0 then count1 = count1+1; end end disp(f,'Sturm sequences') disp(count1,'Number of eigen values strickly greater than 1 : ') p=1.5; f(1)=1; f(2)=C(1,1)-p; count2 = 0; if f(1)*f(2)>0 then count2 = 1; end for r=3:4 br=C(r-2,r-1); f(r)=-br^2*f(r-2)+(C(r-1,r-1)-p)*f(r-1); if f(r)*f(r-1)>0 then count2 = count2+1; end end disp(f,'Sturm sequences') disp(count2,'Number of eigen values strickly greater than 1.5 : ') disp(p+0.25,'Eigen value lying between [1.5,2] ie with an error of 0.25 is : ')

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/experiments/adaboost/adaboost/results/Ignore-MV.AdaBoost.NC-C.vehicle/result2s0.tst

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result2s0.tst

@relation vehicle @attribute COMPACTNESS integer[73,119] @attribute CIRCULARITY integer[33,59] @attribute DISTANCECIRCULARITY integer[40,112] @attribute RADIUSRATIO integer[104,333] @attribute PRAXISASPECTRATIO integer[47,138] @attribute MAXLENGTHASPECTRATIO integer[2,55] @attribute SCATTERRATIO integer[112,265] @attribute ELONGATEDNESS integer[26,61] @attribute PRAXISRECTANGULAR integer[17,29] @attribute LENGTHRECTANGULAR integer[118,188] @attribute MAJORVARIANCE integer[130,320] @attribute MINORVARIANCE integer[184,1018] @attribute GYRATIONRADIUS integer[109,268] @attribute MAJORSKEWNESS integer[59,135] @attribute MINORSKEWNESS integer[0,22] @attribute MINORKURTOSIS integer[0,41] @attribute MAJORKURTOSIS integer[176,206] @attribute HOLLOWSRATIO integer[181,211] @attribute class{van,saab,bus,opel} @inputs COMPACTNESS,CIRCULARITY,DISTANCECIRCULARITY,RADIUSRATIO,PRAXISASPECTRATIO,MAXLENGTHASPECTRATIO,SCATTERRATIO,ELONGATEDNESS,PRAXISRECTANGULAR,LENGTHRECTANGULAR,MAJORVARIANCE,MINORVARIANCE,GYRATIONRADIUS,MAJORSKEWNESS,MINORSKEWNESS,MINORKURTOSIS,MAJORKURTOSIS,HOLLOWSRATIO @outputs class @data van van bus saab saab van opel saab saab saab bus saab opel saab van van van van bus saab opel saab van van van van bus saab van van bus saab opel saab van van saab saab opel saab bus saab van van opel saab bus saab bus saab opel van saab saab van van van van van van saab saab saab saab saab saab bus saab bus saab opel saab opel saab opel van van van bus saab bus van bus saab van van bus saab saab saab van van bus saab bus saab saab saab saab saab van van opel saab opel saab saab saab van van saab saab opel saab opel van opel saab van van opel saab saab saab saab saab bus saab bus saab saab saab opel saab saab saab saab saab bus saab saab saab opel saab saab saab bus saab saab saab van van opel saab saab saab bus saab van saab opel saab bus saab opel saab saab saab van van

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//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 14.3 //calculation of the elastic potential energy stored in the stretched steel wire //given data l=2//length(in m) of the steel wire A=4*10^-6//cross sectional area(in m^2) of the steel wire dl=2*10^-3//increase in the length(in m) Y=2*10^11//Young modulus(in N/m^2) //calculation St=dl/l//strain in the wire Ss=Y*St//stress in the wire V=A*l//volume of the steel wire U=Ss*St*V/2 printf('the elastic potential energy stored in the stretched steel wire is %3.1f J',U)

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ch12_18.sce

clear; clc; w_m=0; printf("lower limit of speed control=%.0f rpm",w_m); I_a=25; r_a=.2; V_s=220; K_m=0.08; a=(K_m*w_m+I_a*r_a)/V_s; printf("\nlower limit of duty cycle=%.3f",a); a=1; printf("\nupper limit of duty cycle=%.0f",a); w_m=(a*V_s-I_a*r_a)/K_m; printf("\nupper limit of speed control=%.1f rpm",w_m);

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Example12_19.sce

//Example 12.19 //Program to estimate the maximum system length for satisfactory //performance clear; clc ; close ; //Given data SNR_dB=17; //dB - REQUIRED SNR L=100*10^3; //metre - INTERVAL SPACING K=4; //FOR AMPLIFIER h= 6.626*10^(-34); //J/K - PLANK's CONSTANT c=2.998*10^8; //m/s - VELOCITY OF LIGHT IN VACCUM B=1.2*10^(9); //bit/s - TRANSMISSION RATE Pi_dBm=0; //dBm - INPUT POWER Lambda=1.55*10^(-6); //metre - OPERATING WAVELENGTH alpha_fc=0.22; //dB/km - FIBER CABLE ATTENUATION alpha_j=0.03; //dB/km - SPLICE LOSS //Calculation of SNR and Pi SNR=10^(SNR_dB/10); Pi=10^(Pi_dBm/10)*10^(-3); //Maximum system length Lto=(Pi*Lambda*10^(-(alpha_fc+alpha_j)*L/10/10^3)/(K*h*c*B))/SNR*L; //Displaying the Result in Command Window printf("\n\n\t Maximum system length for satisfactory performance is %1.0f X 10^4 km.",Lto/10^7);

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cha2_8.sce

Pri=1330;Sec=230;Zl=0.12+%i*0.25;Phase=3;V=230; Z=0.8+%i*5;Power=27; Zz=0.003+%i*0.015;Pf=0.9 A=(Pri/Sec)^2*(Zl) Req=4.01; Xeqh=8.36; a=(sqrt(Phase)*Pri)/V Reql=0.8; Xeql=5; Rr=0.003; Xx=0.015; R=(Reql+Req)*(1/10^2)+Rr X=(Xeql+Xeqh)*(1/10^2)+Xx Vl=V/sqrt(Phase) Il=(Power*10^3)/(Phase*133) Angle=-acos(%pi*Pf/180)

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11_4_1.sce

//CAPTION:Characteristic_Impedance_of_a_Shielded_Strip_Line //chapter_no.-11, page_no.-508 //Example_no.11-4-1 clc; //(a)Calculate_the_K_factor er=2.56//relative_dielectric_constant w=25;//strip_width t=14;//strip_thickness d=70;//shield_depth K=1/(1-(t/d)); disp(K,'the_K_factor is ='); //(b)Calculate_the_fringe_capacitance Cf=((8.854*er)*((2*K*log(K+1))-((K-1)*log((K^2)-1))))/%pi; disp(Cf,'the_fringe_capacitance(in pF/m)is ='); //(c) Calculate_the_characteristic_impedance_of_the_line Z0=94.15/((((w/d)*K)+(Cf/(8.854*er)))*(sqrt(er))); disp(Z0,'the_characteristic_impedance_of_the_line(in ohms)is =');

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/interpolationBaseCanonique.sci

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interpolationBaseCanonique.sci

function a = lagBaseCanonique(f,n) h = 1/n; X = [0:h:1]'; A = []; for i=0:n A=[A X.^i] end y = f(X); a = A\y x = [0:0.01:1]; plot2d(x,f(x)); p=0; for i=0:n p=p + a(i+1)*x.^i end plot2d(x,p,5); endfunction function y=f(x) y = sin(10*x.*cos(x)); endfunction

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/1793/CH11/EX11.6/11Q6.sce

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11Q6.sce

clc T=8.5 eo=0.8 Cc=0.28 To=2650 T1=970 C1=0.02 t2=5 t1=1.5 H=8.5*12 epr=Cc*log10((To+T1)/To) ep=eo-epr C2=C1/(1+ep) Sc=epr*H/(1+eo) Ss=C2*H*log10(t2/t1) TS=Sc+Ss printf('Total consolidation settlement of the clay = %f in',TS)

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SCB_1.sce

// sum 17-1 clc; clear; Ta=22; u=7/10^9; nj=20; r=25; l=2*r; Ao=30000; Uo=15.3/10^3; c=0.025; //specific weight of the material is rho rho=8.46*(10^-6); Cp=179.8; Tf=Ta+(16*%pi^3*u*nj^2*l*r^3/(Uo*Ao*c)); // avg mean film temperature is Tav Tav=(Tf-Ta)/2; x= l*c*rho*%pi*r*nj*Cp*10^3; y=Ao*Tav*Uo; delT=y/x; // printing data in scilab o/p window printf("Tav is %0.2f degC ",Tav); printf("\n delT is %0.1f degC ",delT);

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/3204/CH21/EX21.1/Ex21_1.sce

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Ex21_1.sce

// Initilization of variables N=1800 // r.p.m // Speed of the shaft t=5 // seconds // time taken to attain the rated speed // case (a) T=90 // seconds // time taken by the unit to come to rest // case (b) // Calculations omega=(2*%pi*N)/(60) // (a) // we take alpha_1,theta_1 & n_1 for case (a) alpha_1=omega/t // rad/s^2 // theta_1=(omega^2)/(2*alpha_1) // radian // Let n_1 be the number of revolutions turned, n_1=theta_1*(1/(2*%pi)) // (b) // similarly we take alpha_1,theta_1 & n_1 for case (b) alpha_2=(omega/T) // rad/s^2 // However here alpha_2 is -ve theta_2=(omega^2)/(2*alpha_2) // radians // Let n_2 be the number of revolutions turned, n_2=theta_2*(1/(2*%pi)) // Results clc printf('(a) The no of revolutions the unit turns to attain the rated speed is %f \n',n_1) printf('(b) The no of revolutions the unit turns to come to rest is %f \n',n_2)

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/1208/CH1/EX1.8/Exa8.sce

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Exa8.sce

//Exa8 clc; clear; close; //given data : R1=5000;//in Rs R2=10000;//in Rs R3=15000;//in Rs R4=10000;//in Rs R5=8000;//in Rs r=10;//in % per annum i=r/100; n=5;//in years //formula Vn=R1*(1+i)^(n-1)+R2*(1+i)^(n-2)+.............+Rn-1*(1+i)+Rn V5=R1*(1+i)^(n-1)+R2*(1+i)^(n-2)+R3*(1+i)^(n-3)+R4*(1+i)^(n-4)+R5; disp(V5,"The future value of this series of payments(in Rs) will be : ")

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//Example 1_7 clc; clear; close; format('v',5); //given data : //11*I2+8*I3=4 from loop GDAG //8*I2+11*I3=6 from loop HDAH A=[11 8;8 11];//coefiicient matrix B=[4;6];//coefiicient matrix X=A^-1*B;//Matrix multiplication I2=X(1);//A I3=X(2);//A I8=I2+I3;//A disp(I8,"Current in 8 ohm resistor(A)");

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serveur1_repart.sce

clf; clear; clc; load('C:\Users\tangu\OneDrive\Documents\GitHub\Modelisation\TD4\NetworkData.sod') // Extraction des temps de service index_bool = ( data(:, 3) == 1 ) tabS1 = data(index_bool, :) t_s1 = tabS1(1:$,4); tab = tabul(t_s1,'i') tab(:,2) = tab(:,2)/length(t_s1) F = cumsum(tab(:,2)) plot2d2(tab(:,1),F) legend("Fonction de répartitions des temps de service") // Définition des paramètres d'affichages a=gca(); a.x_location = "origin"; a.grid=[5,5];

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//Variable declaration p = 0.5 // Probability of getting head in one flip //Results printf ( "Probability: %.2f",p*p)

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clear //------------------------------------------------------------------------------ Descender_PMF = 0.79; Descender_deltaV=2143; TUG_deltaV=864; g=9.81; mw_ascender= 9939; for k=320:0.5:420 Descender_ISP(k-319)=k; end ISP_Tug=440; //------------------------------------------------------------------------------ PMFstart=80; for i=1:100 TUG_PMF(i)=(i/10+PMFstart)/100; for j=1:1:length(Descender_ISP) E_Descender=exp(Descender_deltaV/(Descender_ISP(j)*g)); mw_descender(i,j)=mw_ascender*(1-E_Descender)/(E_Descender*(1-Descender_PMF)-1); E_Tug=exp(TUG_deltaV/(ISP_Tug*g)); mw_stage(i,j)=(mw_descender(i,j)+mw_ascender)*(1-E_Tug)/(E_Tug*(1-TUG_PMF(i))-1); mNRHO(i,j)=mw_descender(i,j)+mw_ascender+mw_stage(i,j); end end //------------------------------------------------------------------------------ // //------------------------------------------------------------------------------ f = gcf(); Sgrayplot(TUG_PMF*100,Descender_ISP,mNRHO); //grayplot(TUG_PMF*100,Descender_ISP,mNRHO); //f.color_map = rainbowcolormap(64); f.color_map=jetcolormap(64); colorbar(20000,35000); xlabel("TUG PMF [%]"); ylabel("Descender ISP [s]"); //------------------------------------------------------------------------------

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errcatch(-1,"stop");mode(2);//Example 3.10 : largest diameter ; ; //given data : format('v',6) a=3.61; // edge length in angstrum r=(a*sqrt(2))/4; d=2*r; disp(d,"largest diameter,d(angstrom) = ") exit();

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clc //initialisation n=400 a1=300 b1=100 a2=200 b2=200 r=2 //CALCULATIONS p1=factorial(n)/(factorial(a1)*factorial(b1)*r^n) p2=factorial(n)/(factorial(a2)*factorial(b2)*r^n) w=p1/p2 //results printf(' \n ratio of probabilities= % 1e ',w)

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function out=a(n) if (n==0) out=2*lambda/%pi; else out=-2/(n*%pi)*sin(n*(%pi-lambda)); end endfunction N=100; t=linspace(0.01,1,N); lambda=%pi/2; theta=linspace(0,2*%pi,N); p=20; somme=zeros(N,length(x)); for i=1:N somme(i,:)=a(0)/2; for n=1:p somme(i,:)=somme(i,:)+a(n)*cos(n*theta)*exp(-n^2*t(i)); end end clf; set(gcf(),"color_map",jetcolormap(128)) surf(theta,t,somme); set(gce(),"color_flag",3) xlabel theta; ylabel t; zlabel u; title("Courbe de la chaleur en fonction de theta et de t",'fontsize',4);

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Chapter10_Example24.sce

//Chapter-10, Example 10.24, Page 448 //============================================================================= clc clear //INPUT DATA Q=79;//Reduction in net radiation from the surfaces e1=0.05;//Emissivity of the screen e2=0.8;//Emissivity of the surface //CALCULATIONS n=(((Q*((2/e2)-1))-((2/e2)+1))/((2/e1)-1));//Number of screens to be placed between the two surfaces to achieve the reduction in heat exchange //OUTPUT mprintf('Number of screens to be placed between the two surfaces to achieve the reduction in heat exchange is%3.0f',n) //=================================END OF PROGRAM==============================

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//Ex 6.17 clc; clear; close; format('v',4); //Data given Ls=3;//H Cs=0.05;//pF Rs=2;//kohm Cm=10;//pF fS=1/2/%pi/sqrt(Ls*Cs*10^-12)/1000;//kHz disp(fS,"Series resonant frequency(kHz)"); CT=Cm*Cs/(Cm+Cs);//pF////Equivalent capacitance fP=1/2/%pi/sqrt(Ls*CT*10^-12)/1000;//kHz disp(fP,"Parallel resonant frequency(kHz)");

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//Network Theorem 1 //page no-2.19 //example 2.10 disp("Applying KVL to mesh 1"); disp("7*I1-I2=10");....//equation 1 disp("Applying KVL to mesh 2"); disp("-I1+6*I2-3*I3=0");....//equation 2 disp("Applying KVL to mesh 3"); disp("-3*I2+13*I3=-20");....//equation 3 disp("Solving the three equations"); A=[7 -1 0;-1 6 -3;0 -3 13];//solving the equations in matrix form B=[10 0 -20]' X=inv(A)*B; disp(X); disp("I1=1.34 A"); disp("I1=-0.62 A"); disp("I3=-1.68 A"); disp("I2ohm=1.34 A");

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//Page Number: 11.15 //Example 11.7 clc; //(a)Channel Matrix //Given Py1byx1=0.9; Py2byx1=0.1; Py1byx2=0.2; Py2byx2=0.8; PYbyX=[Py1byx1 Py2byx1;Py1byx2 Py2byx2]; disp(PYbyX,'Channel Matrix,P(Y/X):'); //(b)Py1 and Py2 //Given Px1=0.5; Px2=Px1; //As P(Y)=P(X)*P(Y/X) PX=[Px1 Px2]; PY=PX*PYbyX; disp(PY,'P(y1) P(y2):'); //(c)Joint Probabilities P(x1,y2) and P(x2,y1) //Diagonalizing PX PXd=diag(PX); PXY=PXd*PYbyX; disp(PXY(2,1),PXY(1,2),'P(x1,y2) P(x2,y1)');

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clc;clear; //Example 11.5 //given data m=0.1; T1=0+460; T3=80+460;//converting into R from F //from Table A–17E // at T1 h1=109.90; Pr1=.7913; //pressure ratio at compressor is 4 Pr2=4*Pr1; //at Pr2 h2=163.5; T2=683; //at T3 h3=129.06; Pr3=1.3860; //pressure ratio at compressor is 4 Pr4=Pr3/4; //at Pr4 h4=86.7; T4=363; //calculations qL=h1-h4; Wout=h3-h4; Win=h2-h1; COPR=qL/(Win-Wout); Qrefrig=m*qL; disp((T4-460),'the minimum temperatures in the cycle in F'); disp((T2-460),'the maximum temperatures in the cycle in F'); disp(COPR,'the coefficient of performance'); disp(Qrefrig,'the rate of refrigeration for a mass flow rate of 0.1 lbm/s. in Btu/s')

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Chapter7_Exampl4.sce

clc clear //INPUT DATA p=760;//pressure in mm of Hg t=30;//dry bulb temperature in Degree c p2=0.04246*10^5;//pressure in N/m^2 cp=1.005;//specific pressure hfg=2500;//Specific enthalpy in kJ/kgw.v. cpv=1.88;//specific pressure //CALCULATIONS ps=(p2/133.5);//pressure in mm of Hg ws=(0.62*(ps/(p-ps)));//Specific humidity in kg w.v./kg d.a h=(cp*t)+ws*(hfg+(cpv*t));//Enthalpy of air per kg of dry air in kJ/kg d.a //OUTPUT printf('(a)Accorrding to steam tables The vapour pressure is %3.2f mm Hg \n (b)Specific humidity %3.4f kg w.v./kg d.a \n (c)Enthalpy of air per kg of dry air is %3.2f kJ/kg d.a ',ps,ws,h)

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//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 20.2 //calculation of the dispersive power of the material of the lens //given data fr=90//focal length(in cm) for the red light fv=86.4//focal length(in cm) for the violet light //calculation //(1/f) = (mu-1) * ((1/R1) - (1/R2)) //muv - 1 =K/fv.....and.....mur - 1 = K/fr //let m = muv - mur and K = 1 m=((1/fv)-(1/fr)) //muy - 1 = ((muv + mur)/2) - 1 = (K/2)*((1/fv) - (1/fr)) //let n = muy -1 and K = 1 n=(1/2)*((1/fv)+(1/fr)) //w = (muv-mur)/(mu-1).........definition of the dispersive power w=m/n printf('the dispersive power of the material of the lens is %3.3f',w)

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// This file is part of www.nand2tetris.org // and the book "The Elements of Computing Systems" // by Nisan and Schocken, MIT Press. // File name: projects/03/a/Register.tst load Register32.hdl, output-file Register32.out, output-list time%S1.4.1 in1%D1.6.1 in2%D1.6.1 load%B2.1.2 out1%D1.6.1 out2%D1.6.1; set in1 0, set in2 0, set load 0, tick, output; tock, output; set in1 0, set in2 0, set load 1, tick, output; tock, output; set in1 -32123, set in2 -32123, set load 0, tick, output; tock, output; set in1 11111, set in2 11111, set load 0, tick, output; tock, output; set in1 -32123, set in2 -32123, set load 1, tick, output; tock, output; set in1 -32123, set in2 -32123, set load 1, tick, output; tock, output; set in1 -32123, set in2 -32123, set load 0, tick, output; tock, output; set in1 12345, set in2 12345, set load 1, tick, output; tock, output; set in1 0, set in2 0, set load 0, tick, output; tock, output; set in1 0, set in2 0, set load 1, tick, output; tock, output; set in1 %B0000000000000001, set in2 %B0000000000000001, set load 0, tick, output; tock, output; set load 1, tick, output; tock, output; set in1 %B0000000000000010, set in2 %B0000000000000010, set load 0, tick, output; tock, output;

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//chapter 11 //example 11.5 //page 456 printf("\n") printf("given") Yfs=3000*10^-6;rd=50*10^3;Rs=3.3*10^3;Rd=4.7*10^3;Rl=50*10^3;rs=600; Zs=1/Yfs Zi=((1/Yfs)*Rs)/((1/Yfs)+Rs) Zd=rd Zo=(Rd*rd)/(Rd+rd) Av=Yfs*((Rd*Rl)/(Rd+Rl)) disp("overall volateg gain") Av=(Yfs*((Rd*Rl)/(Rd+Rl))*Zi)/(rs+Zi)

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errcatch(-1,"stop");mode(2);// to calculate no of parrallel path ; S=12; //no of commutator segments P=4; Y_cs=S/P; //slots Y_b=2*Y_cs+1; y_f=Y_b-2; disp(y_f,'no of parralel path'); exit();

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clc //Example 16.5 disp('Given') disp('R=5 ohm L=100mH w=100 rad/s') Rs=5; Ls=100*10^-3 ;w=100; //Let Xs be the capacitive and inductive reactance Xs=w*Ls Q=Xs/Rs //As Q is greater than 5 we can approximate as Rp=Q^2*Rs Lp=Ls printf("The parallel equivalent is \n"); printf("Rp= %d ohm \t Lp=%d mH",Rp,Lp*10^3);

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//developed in windows XP operating system //platform Scilab 5.4.1 clc;clear; //example 3.8 //calculation of the load voltage and the load current //given data Vs=10//source voltage(in V) Rl=10//value of resistance(in ohm) Vd=0.7//voltage drop(in V) across diode Rb=0.23//value of diode resistance(in ohm) //calculation Rt=Rl+Rb//total resistance Vt=Vs-Vd//total voltage across Rt //from the equation of ohm's law.....I=V/R Il=Vt/Rt Vl=Il*Rl printf('the load voltage is %3.2f V',Vl) printf('\nthe load current is %3.3f A',Il)

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// Grob's Basic Electronics 11e // Chapter No. 31 // Example No. 31_2 clc; clear; // Claculate the following AC quantities Av, Vout, Pl, Pcc and percent efficiency. Also calculate the endpoints of ac loadline // Given data Icq = 7.91*10^-3; // Collector Currect(Q-point)=7.91 mAmps Rl = 1.5*10^3; // Load Resistor=1.5 kOhms Rc = 1*10^3; // Collector Resistor=1 kOhms Vin = 25*10^-3; // Input Voltage=25 mVolts(p-p) R1 = 18*10^3; // Resistor 1=18 kOhms R2 = 2.7*10^3; // Resistor 2=2.7 kOhms Vcc = 20; // Supply Voltage(Collector)=20 Volts Vceq = 10.19; // Voltage Colector-Emitter(Q-point)=10.19 Volts rc = (25*10^-3)/Icq; rl = (Rc*Rl)/(Rc+Rl) Av = rl/rc; disp (Av,'The Voltage Gain Av is') disp ('Appox 190') Vout = Av*Vin; disp (Vout,'The Output Voltage in Volts') disp ('Appox 4.75 Volts') Pl = (Vout*Vout)/(8*Rl); disp (Pl,'The Load Power in Watts') disp ('i.e Appox 1.88 mWatts') Ivd = Vcc/(R1+R2); // Ic = Icq Icc = Ivd+Icq; Pcc = Vcc*Icc; disp (Pcc,'The Dc Input Power in Watts') disp ('i.e Appox 177.4 mWatts') efficiency = ((Pl/Pcc)*100); disp (efficiency,'The Efficiency in % is') disp ('Appox 1%') // Endpoints of AC load line icsat = Icq+(Vceq/rl); disp (icsat,'The Y-axis Value of AC Load-line is ic(sat) in Amps') disp ('i.e 24.89 mAmps') vceoff = Vceq+Icq*rl; disp (vceoff,'The X-axis value of AC Load-line is vce(off) in Volts') // For AC load line Vce1=[vceoff Vceq 0] Ic1=[0 Icq icsat] //To plot AC load line printf("Q(%f,%f)\n",Vceq,Icq) plot2d(Vce1, Ic1) plot(Vceq,Icq,".r") plot(0,Icq,".r") plot(Vceq,0,".r") plot(0,icsat,".b") plot(vceoff,0,".b") xlabel("Vce in volt") ylabel("Ic in Ampere") xtitle("AC Load-line for Common-Emitter Class A Amplifier Circuit")

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ex_15_9.sce

//find thickness of vessel,dia of screw,xsec of beam at A,dia of pins at C and D,dia at E and f...G and H.....xsec at E and F clc //soltuion //given //ref fig 15.24 p=0.2//N/mm^2 d=600//mm ftc=17.5//N/mm^2 fts=52.5//N/mm^2 fcs=52.5//N/mm^2 ts=42//n/mm^2 //let t be thickness of vessel //t=(p*d)/(2*ftc)//mm printf("the thickness of vessel si,%f mm\n",(p*d)/(2*ftc)) printf("the thickness can not be less then 6mm,therfore we take 6 as thickness\n") t=6//mm //let dc be core dia W=p*(%pi*d^2)/4//N //let dc be core dia //W=(%pi/4)*dc^2*fts=41.24*dc^2 dc=(W/41.24)^(0.5)//mm printf("we shall use standard size of screw M48 with core dia 41.5mm and outer dia 48mm\n") //let t1 be thickness and b1 be width //b1=2*t1 Rc=W/2//N Rd=W/2//N l=750//mm M=W*l/4//N-mm //Z=(1/6)*t1*b1^2 //Z=0.66*t1^3 //fts=M/Z t1=(M/(52.5*0.66))^(1/3) b1=2*t1//mm printf("thickness and width of beamA is,%f mm\n,%f mm\n",t1,b1) //let d1 be dia of pi at C and D //Rc=2*(%pi/4)*d1^2*ts d1=sqrt(Rc/66)//mm printf("the dia of pin at C and D is,%f mm\n",d1) printf("since load at E and F IS SAME AS THAT OF C AND D,therefr dia of pins at E and F is 21 mm\n ") //let d2 be dia at G and H Rg=W/2//N //Rg=(%pi/4)*d2^2*fts d2=(Rg/41)^(0.5)//mm printf("the dia at G and H is,%f mm\n",d2) //let t2 be support thickness and b2 be width of support x=375-(300+t) M2=Rc*x//N-mm //b2=2t2 //Z=(1/6)*t2*b2^2=0.66t2^3 //ftc=M/Z t2=[M2/(0.66*17.5)]^(1/3)//mm b2=2*t2 printf("the thickness and wdth of support at E and F is,%fmm\n,%f mm\n",t2,b2)

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Ex2_11.sce

errcatch(-1,"stop");mode(2);//caption:Find unknown resistance,percent error and error in ohm //Ex2.11 P=100//resistance of arm of wheatstone bridge(in ohm) ep=0.5//error in P(in %) Q=50//resistance of arm of wheatstone bridge(in ohm) eq=0.5//error in Q(in %) S=75.5//resistance of arm of wheatstone bridge(in ohm) es=0.5//error in S(in %) X=(P*S)/Q disp(X,'unknown resistance(in ohm)=') xo1=ep+es-eq disp(xo1,'percent error when Q is taken positive(in %)=') ex1=(xo1*X)/100 disp(ex1,'error in ohm(in ohm)=') xo2=ep+es+eq disp(xo2,'percent error when Q is taken negative(in %)=') ex2=(xo2*X)/100 disp(ex2,'error in ohm(in ohm)=') exit();

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typecheck.sci

function out = typecheck(A,B,C,D,F1) if(A==1) if(C==1 | F1==1) out = "oe" else out = "bj" end elseif(D==1 & F1==1) if(C==1)' out = "arx" else out = "armax" end else out = "idpoly" end endfunction

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/3415/CH11/EX11.5/EX11_5.sce

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EX11_5.sce

//fiber optic communications by joseph c. palais //example 11.5 //OS=Windows XP sp3 //Scilab version 5.4.1 clc clear all //given RL=100//load in ohm T=300//temperature in kelvin lambda=0.82*10^-6//wavelength in um e=1.6e-19//charge of electron in colums k=1.38e-23//boltzman constant h=6.63e-34//plancks constant deltaf=1e6//link bandwidth in Hz Error_rate=10^-4//desired error rate eta=1//quantum efficiency c=3*10^8//speed of light in m/s snr=17.5//Signal to noise ratio from plot correspnding to error rate of 10^-4 in dB SNR=10^(snr/10)//signal to noise ratio in normal scale tau=10^-6//bit interval in Sec //to find f=c/lambda//optic frequency in Hz P=(h*f/(eta*e) )*sqrt((4*k*T*deltaf)/RL)*sqrt(SNR)//Optic power incident in Watts mprintf("Optic power incident=%fnW",P*10^9)//multiplication by 10^9 is to convert the unit from W to nW i=eta*e*P/(h*f)//current in Amperes mprintf("\nCurrent=%fnA",i*10^9)//multiplication by 10^9 is to convert the unit from A to nA np=P/(h*f)*tau// No. of photons per bit mprintf("\nNo. of Photons per bit=%fx10^5 photons/bit",np/10^5)//multiplication by 10^5 is to convert the unit x10^5