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//Exa Misc 8.2 clc; clear; close; //given data format('v',6); fo=10;//in KHz R1=25;//in kohm R2=60;//in kohm Rc=40;//in kohm R=7.1;//in kohm hie=1.8;//in kohm C=1/(2*%pi*fo*10^3*R*10^3*sqrt(6+4*Rc/R));//in F disp(C*10^9,"Value of Capacitor(in nF) :"); hfe=23+29*R/Rc+4*Rc/R;//unitless disp("Value of hfe is ≥ "+string(hfe));
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function [nsize,nodes]=max_clique(g,ind) // Copyright INRIA [lhs,rhs]=argn(0) if rhs>2|rhs<1 then error(39) elseif rhs==1 then ind=0 end; // check g check_graph(g) // check ind if (ind<>0&ind<>1) then error('The second argument must be 0 or 1') end m=prod(size(g('tail'))) n=g('node_number') head=g('head'); tail=g('tail'); if (ind==1) then tta=[tail head];hhe=[head tail]; xm=sparse([tta' hhe'],ones(tta)',[n,n]); [ij,v,mn]=spget(xm); [lp,la,ls]=adj_lists(1,n,ij(:,1)',ij(:,2)'); m2=2*m;np1=n+1;nwk=10*m; [wcl]=m6clique1(n,m,m2,np1,nwk,lp,ls); ii=find(wcl == 0); if (ii<>[]) then wcl(ii)=[]; end; nodes=wcl;nsize=size(wcl,2); else xm=sparse([tail' head'],ones(tail'),[n,n]); xadj=full(xm+xm'); mn=m*n; [clmax,clnod,bestn]=m6clique(n,m,mn,xadj); nsize=clmax; ii=find(bestn==0); if (ii<>[]) then bestn(ii)=[]; end; nodes=clnod(bestn); end;
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clear; clc; c=3*(10^8); f=3000*(10^8); lo=c/f; l=lo*(10^4); m=1;n=0;a=7.62; lc=2*a; printf("-Critical wavelength = %f cm\n",lc); lg=sqrt((l*l*lc*lc)/((lc*lc)-(l*l))); printf("-Guide wavelength = %f cm",round(lg*10)/10);
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//Chemical Engineering Thermodynamics //Chapter 14 //Thermodynamics of Chemical Reactions //Example 14.6 clear; clc; //Given //SO2 + (1/2)O2 - SO3 //Basis: 1 Kgmole of SO2 n_SO2 = 1;// SO2 fed in Kgmole //From table 14.1 (page no 301) //alpha values for the following components are given as a_SO2 = 7.116; a_O2 = 6.148; a_SO3 = 6.077; //beta values for the following components are given as b_SO2 = 9.512*10^-3; b_O2 = 3.102*10^-3; b_SO3 = 25.537*10^-3; //Standard enthalpy of the following components at 25 deg cel in Kcal/Kgmole are given as H_SO2 =-70960; H_O2 = 0; H_SO3 = -94450; //Standard free energy of the following components at 25 deg cel in Kcal/Kgmole K are given as F_SO2 = -71680; F_O2 = 0; F_SO3 = -88590; n_O2 = n_SO2;//O2 fed in Kgmole; since 50 mole percent mixture of SO2 &amp; O2 is fed n_SO3 = n_SO2;//SO3 formed in Kgmole n_O2_e = n_O2-(n_O2/2);//Kgmoles of O2 in exit gas n_O2_r = n_O2/2;//Kgmoles of O2 reacted R = 1.98;//gas constant in Kcal/Kgmole K //To show the variation of the standard heats of reaction with temperature and the equilibrium constant with temperature graphically in the given temperature range //(i)Variation of the standard heats of reaction with temperature del_H = (n_SO3*H_SO3)-(n_O2_r*H_O2)-(n_SO2*H_SO2);// in Kcal/Kgmole del_F = (n_SO3*F_SO3)-(n_O2_r*F_O2)-(n_SO2*F_SO2);// in Kcal/Kgmole //From equation 14.10 (page no 301) del_a = (n_SO3*a_SO3)-(n_O2_r*a_O2)-(n_SO2*a_SO2); del_b = (n_SO3*b_SO3)-(n_O2_r*b_O2)-(n_SO2*b_SO2); //In equation 14.11 (page no 302), substituting del_H at T = 298;//in deg cel I = del_H - del_a*T - (del_b*(T^2)/2);// integrating constant mprintf('(i)The standard heat of reaction at any tempperature can be calculated by the relation:'); mprintf('\n del_Ht = %fT + %fT^2 %f',del_a,del_b/2,I); //(ii)Variation of the equilibrium constant with temperature //K1 = lnKa (say) K1 = -del_F/(R*T); //From equation 14.42 (page no 316); M1 = M/R (say) M1 = K1-(del_a/R)*log(T)-(del_b/(2*R))*T+(I/(T*R)); //Let us assume the temperature in the range 800K to 1500K as Ta = [700 800 825 850 900 1000 1100 1300 1500]; for i = 1:9 Ka(i) = %e^((del_a/R)*log(Ta(i))+(del_b*Ta(i)/(2*R))-(I/(Ta(i)*R))+M1); end clf; plot(Ta,Ka); xtitle(" ","Temperature in K","equilibrium constant K"); mprintf('\n\n(ii)From the graph it can be seen that as temperature increases Ka decreases exponentially,so the reaction is exothermic.'); //end
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errcatch(-1,"stop");mode(2);//Exa4.12 ; ; //given data Vz=5;//in volts to=25;//in degree centigrade t=100;//in degree centigrade Vdrop=4.8;//in Volts delVz=Vdrop-Vz;//in Volts delt=t-to;//in degree centigrade TempCoeff=delVz*100/(Vz*delt); disp(TempCoeff,"Temperature coefficient f zener diode in % : "); exit();
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clc clear //Initialization of variables n1=8 n2=9 n3=1 n4=12.5 U11=3852 U12=115 U21=3009 U22=101 U31=24773 U32=640 U41=2539 U42=83 H=-2203389 //calculations dU1=n1*(U11-U12)+n2*(U21-U22) dU2=n3*(U31-U32)+n4*(U41-U42) Q=H+dU1-dU2 //results printf("Heat of reaction = %d Btu",Q)
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// This file was created by Evyatar. load EQ2.asm, output-file EQ2.out, compare-to EQ2.cmp, output-list RAM[0]%D2.6.2 RAM[256]%D2.6.2; set RAM[0] 256, // initializes the stack pointer repeat 200 { // enough cycles to complete the execution ticktock; } output; // the stack pointer and the stack base
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clear clc w1=1;//amount of glucose C6H12O6 in gm w2=1;//amount of sucrose C12H12022 in gm n=(w1/180)+(w2/342);//amount of solute R=8.314;//in J/Kmol T=25;//in C V=1000;//volume of water in gm P=(n*R*(T+273))/(V*10^-6);//osmotic pressure of solution printf('P=%.3f *10^4 N/m^2',P/10^4) w=(w1+w2);//weight of solute M=(w*R*(T+273))/(P*(V*10^-3));//molar mass of solute printf('\nM=%.4f kg/mol',M) Mn=((w1*10^-3)+(w2*10^-3))/(n);//average molar mass in Kg/mol printf('\nMn=%.4f kg/mol',Mn) //page 67
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// ELECTRICAL MACHINES // R.K.Srivastava // First Impression 2011 // CENGAGE LEARNING INDIA PVT. LTD // CHAPTER : 3 : TRANSFORMERS // EXAMPLE : 3.10 clear ; clc ; close ; // Clear the work space and console // GIVEN DATA V1 = 440; // HV Side Voltage Rating of the Transformer in Volts V2 = 220; // LV Side Voltage Rating of the Transformer in Volts pf_o = 0.2; // No-load Power factor lagging pf_l = 0.8; // Load Power factor lagging I_o = 5; // No-load current in Amphere I_2 = 120; // Load current in Amphere // CALCULATIONS a = V1/V2; // Turns ratio of the two winding Transformers theta_o = acosd(pf_o); // No load power factor of the two winding Transformers in Degrees Io = I_o * exp(-(%i*theta_o*%pi/180)); // No load current of the two winding Transformers (minus because lagging) in Amphere theta = acosd(pf_l); // load power factor of the two winding Transformers in Degrees I2 = I_2 * exp(-(%i*theta*%pi/180)); // secondary load current of the two winding Transformers (minus because lagging) in Amphere I21 = I2/a; // Secondary referred to the primaryin Amphere I1 = Io + I21; // Primary current in Amphere I1_mag = abs(I1); // Primary current magnitude inj Amphere theta_1 = atand( imag(I1),real(I1)); // Primary current angle in Degree // DISPLAY RESULTS disp("EXAMPLE : 3.10 : SOLUTION :-") ; printf("\n (a) Primary current , I1 = %.2f < %.1f A \n ",I1_mag,theta_1);
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//Exa 2.6 clc; clear; close; //given data ni=1.4*10^18;//in atoms/m^3 Nd=1.4*10^24;//in atoms/m^3 n=Nd;//in atoms/m^3 p=ni^2/n;//in atoms/m^3 ratio=n/p;//unitless disp(p,"Concentration of holes in per m^3 : "); disp(ratio,"Ratio of electron to hole concentration : ");
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// Example 22_15 clc;funcprot(0); //Given data T_1=400;// °C p_1=40;// bar p_2=2;// bar p_3=0.5;// bar p_4=0.05;// bar n_t1=75/100;// The isentropic efficiency of the first stage of the turbine n_t2=80/100;// The isentropic efficiency of the second stage of the turbine n_t3=85/100;// The isentropic efficiency of the third stage of the turbine m_s=10;// The steam flow in kg/sec // Calculation // From h-s chart: h_1=3210;// kJ/kg h_2a=2562;// kJ/kg h_2=h_1-((n_t1)*(h_1-h_2a));// kJ/kg h_3a=2508;// kJ/kg h_3=h_2-((n_t2)*(h_2-h_3a));// kJ/kg h_4a=2232;// kJ/kg h_4=h_3-((n_t3)*(h_3-h_4a));// kJ/kg // From steam tables h_f8=502;// kJ/kg(2 bar) h_f10=h_f8;// kJ/kg h_f6=339;// kJ/kg(0.5 bar) h_f7=h_f6;// kJ/kg h_f9=h_f6;// kJ/kg h_f5=136;// kJ/kg(0.05 bar) //Assume m_1=y(1);m_2=y(2) function[X]=mass(y) X(1)=(y(1)*(h_2-h_f10))-((1-y(1))*(h_f8-h_f7)); X(2)=(y(2)*(h_3-h_f9))-((1-y(1)-y(2))*(h_f6-h_f5)); endfunction y=[0.01 0.01]; z=fsolve(y,mass); m_1=z(1);// kJ/kg m_2=z(2);// kJ/kg W=(h_1-h_2)+((1-m_1)*(h_2-h_3))+((1-m_1-m_2)*(h_3-h_4));// kJ/kg P=W*m_s;// Power developed by the turbine in kW Q_s=h_1-h_f10;// Heat supplied per kg of steam in kJ/kg n_th=(W/Q_s)*100;// Thermal efficiency of the cycle in % printf('\n(a)Steam bled for regenerative heaters per kg of steam to turbine,m_1=%0.4f kJ/kg & m_2=%0.4f kJ/kg \n(b)Power developed by the turbine=%0.0f kW \n(c)Thermal efficiency of the cycle=%0.2f percentage',m_1,m_2,P,n_th); // The answer provided in the textbook is wrong
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//EXAMPLE 2-108 PG NO-146 I1=(100/1.414)^2; I2=(10/1.414)^2; R.M.S=(I1+I2)^0.5; disp('R.M.S VALUE is = '+string(R.M.S)+' A');
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function [pt]=smooth(ptd,pas) // [lhs,rhs]=argn(0) [m,n]=size(ptd) d=splin(ptd(1,:),ptd(2,:)) if rhs=1 then l=abs(ptd(1,n)-ptd(1,1));pas=l/100;end pt=[ptd(1,1)+pas:pas:ptd(1,n)] pt=[ptd(:,1) [pt;interp(pt,ptd(1,:),ptd(2,:),d)] ptd(:,n)]
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Ex5_14.sce
//Chapter-5,Example 5_14,Page 5-31 clc() //Given Values: m=1.676*10^-27 //mass of neutron h=6.625*10^-34 //Planck's constant lam=1*10^-10 //wavelength of neutron //Calculations: C=3*10^8 //Velocity of light Ep1=h*C/lam //Energy of photon in joules E1=Ep1/(1.6*10^-19) //Energy of photon in electron-Volts printf('Energy of photon is =%.2f eV \n \n',E1) //we know, lam=h/sqrt(2*m*E) //de Broglie wavelength En1=h^2/(2*m*lam^2) //Energy of neutron in joules E2=En1/(1.6*10^-19) //Energy of neutron in electron-Volts printf(' Energy of neutron is =%.3f eV \n \n',E2) R=E1/E2 //Ratio of energies of proton to neutron printf(' Ratio of energies of proton to neutron is =%.0f \n \n',R)
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Ex14_5.sce
clc// // // //Variable declaration V=20*15*5; //volume(m^3) T=3.5; //time(s) A=950; //surface area(m^2) //Calculation sigma_as=0.165*V/T; //total absorption in the hall(OWU) ac=sigma_as/A; //average absorption coefficient //Result printf("\n total absorption in the hall is %0.3f OWU",sigma_as) printf("\n average absorption coefficient is %0.3f sabine/m^2",ac)
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9_06.sce
clear clc FAo=1000;//mol/min //Drawing trial operating lines with a slope of 1/72 and for each evaluating integral dXA/-rA //From graph Area=1.72; V=FAo*Area; printf("\n The volume of adiabatic plug flow reactor is %f",V) printf("litres")
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example3.sce
// to find load voltage and load current using ideal diode // Example 3-3, page 65 clear;clc; close; // Given data // diode is forward biased, equivalent to a closed switch. // Calculations V=10;// load voltage in volts R=1000;// load resistance in ohms I=V/R;// all the source voltage appears across the load resistor disp("Amperes",I,"Load Current=") disp("Volts",V,"Load Voltage=")
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Ex8_4.sce
//Caption:Design a square wave generator using 7555 CMOS //Ex8.4 clc; clear; close; V=5//Supply voltage(in volts) f1=1//Frequency(in khz) f2=3//Frequency(in khz) C=0.01//Capacitance(in micro farad) Ra=47//Choosed resistor(in kilo ohm) t1=1/(2*f1) t2=1/(2*f2) R=t1/(0.693*C) Rb=R-Ra disp(C,Rb,Ra,'Components required to design the circuit are Ra,Rb(in kilo ohm) and Capacitance(in micro farad)=')
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Ex20_6.sce
//chapter20 //example20.6 //page441 Vin_min=22 // V Vout=15 // V Il_max=0.1 // A // for maximum series resistance, we consider the case when input voltage is minimum and load current is maximum because then zener current drops to minimum.Thus, Rs_max=(Vin_min-Vout)/Il_max printf("required series resistance = %.3f ohm \n",Rs_max)
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Morisita_RP-003.tst
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example25.sce
//Example 2.5 clc; Max_Limit=10; h=ones(1,Max_Limit); N2=0:length(h)-1; a=0.5;//constant a>0 for t=1:Max_Limit x(t)=exp(-a*(t-1)); end N1=0:length(x)-1; y=convol(x,h)-1; N=0:length(x)+length(h)-2; figure a=gca(); plot2d(N2,h) xtitle('Impulse Response','t','h(t)'); a.thickness=2; figure a=gca(); plot2d(N1,x) xtitle('Input Response','t','x(t)'); a.thickness=2; figure a=gca (); plot2d(N(1:Max_Limit),y(1:Max_Limit)) xtitle('Output Response','t','y(t)'); a.thickness=2;
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17_5.sce
clc //initialisation of variables M1= 18.02 //gms m1= 0.965 //gms m2= 0.035 //gms M2= 58.5 //gms R= 8.314 //J/mol K M= 18.02 //kg T= 20 //C vf= 0.001002 //m^3 x21= 0.021856 //m^3 //CALCULATIONS n1= m1/M1 n2= m2/M2 x1= n1/(n1+n2) x2= n2/(n2+n1) P= R*(273.15+T)*x2/(M*vf) P1= R*(273.15+T)*x21/(M*vf) //RESULTS printf (' Osmotic pressure= %.1f kpa',P) printf (' \n Osmotic pressure= %.1f kpa',P1)
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2_19.sce
clc,clear printf('Example 2.19\n\n') V=110 P=4 R_a = 0.1,R=0.01 //A resistance of 0.01 ohms R_se=R+R //case(i) I_1=50, I_a1=I_1 N_1=700 E_b1 = V -I_a1*(R_a + R_se) //T (prop.) phi*I_a from torque equation (1) //phi_1 (prop.) I_a1 (2) //case(ii) when I_a2 gets divided to half //phi_2 (prop.) I_a2/2 (3) //combining (1)(2)(3) and T1=T2 I_a2 = sqrt(2*I_a1^2) R_se_eqvt=(R*R)/(R+R) //Equavalent of parallel combination E_b2 = V - I_a2*R_a - I_a2* R_se_eqvt //Using speed equation N (prop.) E_b / phi and using (2) and (3) N_2 = N_1 *( E_b2/E_b1) *(I_a1/(I_a2/2)) printf('Speed after reconnection = %.3f r.p.m\n\n',N_2)
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CH18Exa15.sce
// Scilab code Exa18.15 : : Page-768 (2011) clc; clear; I_1 = 3/2; // Isospin for delta(1232) I_2 = 1/2; // Isospin for delta 0 delta_ratio = sqrt((2/3)^2)/sqrt((1/3)^2); // Branching ratio printf("\nThe branching ratio for a resonance with I = 1/2 is %d", delta_ratio); // Result // The branching ratio for a resonance with I = 1/2 is 2
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Ch3_3_54.sce
clc disp("Example 3.54") printf("\n") disp(" Draw a DC load line for the base bias circuit neglecting Vbe") printf("Given\n") //given betadc=100 Rc=5*10^3 Rb=1.5*10^6 Vc=30 //to find Ib Ib=Vc/Rb //from ciruit //Ic value Icq=betadc*Ib //Vce value Vceq=Vc-(Icq*Rc) //to draw DC load line Ic1=Vc/Rc Vce1=Vc Vce=[Vc Vceq 0] Ic=[0 Icq Ic1] printf("Q(%f volt,%f ampere)\n",Vceq,Icq) plot2d(Vce, Ic) xlabel("Vce in volt") ylabel("Ic in ampere") xtitle("DC load line for base bias circuit")
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Ex_9_2.sce
// Example 9.2;//photocurrent clc; clear; close; R=0.85;//responsivity in ampere per watt Po=1;//output power in milli watt Ip= R*Po;//photocurrent in milli ampere disp(Ip,"photocurrent in milli ampere is ")
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7_10.sce
clc; b=0.228; a=1-b; c=[1+(2*0.455)-b-2*a]/2 n2=a+b+c+1.709; p1=8.28; T2=555; n1=1+0.455+1.709; T1=2968; p2=p1*(n2/n1)*(T1/T2); p=1; K=a/b*[n2*p/(c*p2)]^0.5; disp(log(K),"log(K) is:"); disp("2968","from tables it is proved that temperatur is:")
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// Variable declaration val = 0 data1 = [27,23,64,44,30,75,26,124,54,30,14] data2 = [15,13,22,29,31,64,30,64,56,20,21] y = [1,1,1,1,1,1,1,1,1,1.1,1] d = [] alpha = 0.05 // level of significance for i = 1:11 d(i) = data1(i)-data2(i) end n = length(d) Mean = sum(d)/(length(d)) // Calculation for i = 1:length(d) val = val + (Mean-d(i))^2 end val = val/(n-1) deg = n-1 t = 2.228 // t-value for 10 degree of freedom y1 = (Mean - t*(sqrt(val/n))) // Lower limit y2 = (Mean + t*(sqrt(val/n))) // upper limit // Result printf ( "95%% confidence interval: ( %.2f , %.2f )",y1,y2) disp(length(d),length(y)) plot(d,y,"ro") title("Dot Diagram")
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r_lineal.sce
clear /////////////////////////////////////////////////////// // r_lineal.sce // // Este programa obtiene los coefientes de una ecuacion // lineal por medio de sus fórmulas de regresión, dado // un conjunto de pares coordenados. // // José Elías Garza Vázquez // 22 / 01 / 20 version 1.0 ////////////////////////////////////////////////////// ////////////////////////////////////////////////////// // regresionLineal // // Funcion que calcula los coeficientes de una ecuacion // de primer grado por medio de sus formulas de regresión // aplicadas a un conjunto de pares coordenados contenidos // por una matriz. // // Parametros: // pares es la matriz que contiene a los pares // coordenados // Regresa: // coeficientes es un vector que contiene los 2 // coeficientes a0 y a1 de la ecuación. // ///////////////////////////////////////////////////// function coeficientes = regresionLineal(pares) // Obtengo el numero de datos n = size(pares)(1) // Obtengo la sumatoria de todas las x xSum = sum(pares(:, 1)) // Obtengo la sumatoria de todas las y ySum = sum(pares(:, 2)) // Elaboro ciclo para obtener la sumatoria de todos los productos x*y xySum = 0 for i = 1 : n xySum = xySum + pares(i, 1)*pares(i, 2) end // Obtengo la sumatoria de todas las x elevadas al cuadrado x2Sum = sum(pares(:, 1)^2) // Obtengo el promedio de x meanX = mean(pares(:, 1)) // Obtengo el promedio de y meanY = mean(pares(:, 2)) // Uso la ecuacion para calcular el primer coeficiente con los datos anteriores a1 = (n*xySum - xSum*ySum)/(n*x2Sum - (xSum)^2) // Uso la ecuacion para calcular el segundo coeficiente con los datos anteriores y el primer coeficiente a0 = meanY - a1*meanX // Genero el vector con los dos coeficientes de la ecuacion coeficientes = [a0, a1] endfunction /////// Programa Principal // pido los valores n = input("Inserte el número de pares coordenados que desea utilizar: ") // Creo una matriz con el número de espacios que el usuario pidio data = zeros(n, 2) // Elaboro ciclo para pedir los pares for i = 1 : n for j = 1 : 2 if(j == 1) data(i, j) = input("inserta x: ") else data(i, j) = input("inserta y: ") end end end //despliego los coeficientes resultantes de la regresion A = regresionLineal(data) disp("a0 = " + string(A(1)) + ", a1 = " + string(A(2)))
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clc clear //INPUT DATA n=2//no.of atoms in BCC structure d=7.86*10^6//density of iron of FCC structure in kg/m^3 AW=55.85//atomic weight of Fe N=6.023*10^23//Avogadro's Number per Kg mol //CALCULATION a=(((n*AW)/(d*N))^(1/3))/10^-10//The lattice constant in m r=((a*sqrt(3))/4)//The atomic radius of Fe which has BCC structure in armstrong*10^-10 //OUTPUT printf('The atomic radius of Fe which has BCC structure is %3.3f armstrong',r)
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clc //Variable Initialisation Ea=230//Input Voltage of motor in volts Ia=50//Armature Current in Ampere N1=800//Rated Speed of Motor in rpm Ra=0.4//Armature resistance in ohm d1=0.3//Duty ratio for Motoring Operation d2=0.6//Duty ratio for Motoring Operation d3=0.7//Duty ratio for Braking Operation d4=0.4//Duty ratio for Braking Operation //Solution E01=d1*Ea Eb1=Ea-(Ia*Ra) Eb2=E01-(Ia*Ra) N2=(Eb2/Eb1)*N1 E02=d2*Ea Eb3=E02-(Ia*Ra) N3=(Eb3/Eb1)*N1 E03=d3*Ea Eb4=E03+(Ia*Ra) N4=(Eb4/Eb1)*N1 E04=d4*Ea Eb5=E04+(Ia*Ra) N5=(Eb5/Eb1)*N1 printf('\n\n Motor speed for Motoring Operation 1 =%0.1f rpm\n\n',N2) printf('\n\n Motor speed for Motoring Operation 2=%0.1f rpm\n\n',N3) printf('\n\n Motor speed for Braking Operation 1=%0.1f rpm\n\n',N4) printf('\n\n Motor speed for Braking Operation 2=%0.1f rpm\n\n',N5)
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//chapter 12 //example 12.3 //page 371 clear all; clc ; //given Vcc=20;//supply voltage Rl=500;//load resistance Av=75;//closed loop voltage gain f1=30;// i/p signal frquency Hz hfe=80; hie=1.4; Ve=5;Vce=3; VR2=5.7; Vbe=5; VR3=Vcc-Vce-Ve; R3=Rl/10; Ic=VR3/R3*10^3; printf("\ncollector current = %d microA,\ntoo small for operations of transistor,\nSo Ic=1 mA",Ic) Ic=1; R3=VR3/Ic; printf('\nR3=%d kohm which is standard value',R3) R45=(Ve/Ic); R4=1000*((R3*Rl)/(R3+Rl))/Av; printf("\nR4=%d ohm ,use standard value 150 ohm",R4);R4=150; R5=R45-R4/1000; printf("\nR5=%.2f kohm ,use standard value 5.6 kohm",R5); R5=5.6;//kohm //potential divider current and resistors I2=Ic/10; Ve=0.7; R2=(Vbe+Ve)/I2; printf("\nR2=%d kohm,use standard value 56 kohm",R2);R2=56; R1=(Vcc-VR2)/I2; printf("\nR1=%d kohm,use standard value 150 kohm",R1);R1=150; Zb=hie+(1+hfe)*R4/1000; //input impedance Zi=1/((1/Zb)+(1/R1)+(1/R2)); printf("\ninput impedance=%.2f kohm",(Zi)); //capacitance XC1=Zi/10; XC1=1; C1=10^3/(2*%pi *f1*XC1); printf("\ncapacitor C1=%.1f microF,use standard value 5.6 microF",(C1)) XC2=R4; C2=10^6/(2 * %pi *f1* XC2); printf("\ncapacitor C2=%.1f microF,use standard value 39 microF",C2) C3=10^3/(2*%pi *f1*Rl/10); printf("\ncapacitor C3=%.3f microF,use standard value 0.1 microF",C3)
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// Theory and Problems of Thermodynamics // Chapter 4 // Energy Analysis of Process // Example 10 clear ;clc; //Given data V = 2 // volume in m^3 gam = 1.67 // gamma ideal gas coefficient P0 = 0.1 // Initial Pressure in MPa T0 = 300 // Initial Temperature in K Pf = 3 // Final Pressure in MPa T1 = 500 // Final Temperature in K R = 8.314 // Gas constant // Calculation // the first law of thermodynamics for a transient flow and after rewriting Tf = Pf/(((Pf-P0)/(gam*T1))+(P0/T0)) // Final temperature Pf = Pf * 1e6 // units conversion from MPa to Pa P0 = P0 * 1e6 // units conversion from MPa to Pa q = (V/R)*(Pf/Tf - P0/T0) // Output Results mprintf('Temperature of helium in the tank = %6.2f K',Tf) mprintf('\nQuantity of helium entered into tank = %6.2f mol',q)
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k=1; cp=zeros(1,991); c=zeros(1,991); noe=zeros(1,991); for n=10:1000 noe(k)=n; a=round(rand(1,n)*100); [cmp_h,ab] = quick(a,1,n,0); [cmp_r,ar] = sorttest_quick(a,1,n,0); cp(k) = cmp_r; c(k) = cmp_h; k=k+1; end plot(noe,cp,"m"); plot(noe,c); xlabel("ARRAY SIZE"); ylabel("NUMBER OF COMPARISON"); title("FOR REVERSE DATA"); xgrid(2); legend("NORMAL QUICK","HYBRID QUICK",2);
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disp('chapter 5 ex5.5') disp('given') disp('709 op-amp is used to design an noninverting amplifier') disp('voltage gain Av=50') Av=50 disp('voltage gain Av=50=34dB') disp('compensation components are listed for Av=20dB and for Av=40dB') disp('for over compensation use components for Av=20dB') disp('C1=500pF R1=1.5kohms C2=20pF')
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try type(cdo_factor); delmenu('CDO'); catch // exec builder.sce; exec loader.sce; end; addmenu('CDO',['Product';'Model';'Computation';'Last computation';'Numdef';'Losses']); CDO=['exec(''product.sci'',-1)';'exec(''model.sci'',-1)';'exec(''computation.sci'',-1)';'exec(''last_computation.sci'',-1)';'exec(''animate_numdef'',-1)';'exec(''animate_losses.sci'',-1)']; unsetmenu('CDO',5); unsetmenu('CDO',6); exec('product.sci',-1); exec('model.sci',-1); exec('computation.sci',-1);
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//problem 16 pagenumber 2.99 //given f1=500;//hz f2=2.2e3;//hz a=5; c=0.1e-6;//farad rf1=10e3;//ohm //determine r1 r2 R1=1/(2*3.14*f1*c); R2=1/(2*3.14*f2*c); Ri=2e3;//ohm assuming Rf=(a-1)*Ri;format(6); disp('R = '+string(R1/10^3)+' Kohm'); disp('R2 = '+string(R2/10^3)+' Kohm');//error in book disp('R1 = '+string(Ri/10^3)+' Kohm'); disp('Rf = '+string(Rf/10^3)+' Kohm');
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clc; clear; funcprot(0); function V0 = UpOutPut_BS_MC_Richardson (S0, r, sigma, T, K, B, M, m) delta_t = T/(2*m); // Fine mash // generate a Matrix of mxM normaly distributed r.v. with mean = 0 and ... // sd = sqrt(delta_t) delta_W1 = grand(m, M, "nor", 0, sqrt(delta_t)); delta_W2 = grand(m, M, "nor", 0, sqrt(delta_t)); // Initialize matrices for no barrier hit information nobarrier_hit_fine = ones(1,M); nobarrier_hit_coarse = ones(1,M); // Set initial stock values for first time t=0 Euler_fine = S0; Euler_coarse = S0; // Since in the BS model dSt = r*St*dt + sigma*St*delta_W for i=1:m // Calculation matrices of values for fine grid and for coarse grid // Since grid of fine calculation is twice as fine, twice as many // calculation steps are needed Euler_fine = Euler_fine + r*Euler_fine*delta_t +... sigma.*Euler_fine.*delta_W1(i,:); nobarrier_hit_fine = nobarrier_hit_fine.*(Euler_fine<B); Euler_fine = Euler_fine + r*Euler_fine*delta_t +... sigma.*Euler_fine.*delta_W2(i,:); nobarrier_hit_fine = nobarrier_hit_fine.*(Euler_fine<B); // For the coarse grid, stepsize is twice as large: 2*delta_t // and also both brownian motions are needed for this stepsize Euler_coarse = Euler_coarse + r*Euler_coarse*2*delta_t +... sigma.*Euler_coarse.*(delta_W1(i,:)+delta_W2(i,:)); nobarrier_hit_coarse = nobarrier_hit_coarse.*(Euler_coarse<B); end // Define put option function g function y = g(x) y = max((K - x), 0); endfunction // V_fine = nobarrier_hit_fine.*(g(Euler_fine)*exp(-r*T)); V_coarse = nobarrier_hit_coarse.*(g(Euler_coarse)*exp(-r*T)); // Calculating final result for option price at time t=0 by averaging over... // all simulated prices with the MC simulation V0 = mean(2*V_fine-V_coarse); endfunction // Function from Exercise 12 function V_0 = UpOutPut_BinMod(S_0, r, sigma, T, K, B, M) delta_t = T/M; //calculation of delta_t Beta = (exp(-r*delta_t)+exp((r+sigma^2)*delta_t))/2; //calculation of beta for CRR u=Beta+sqrt((Beta^2)-1); //so u>d is true d=u^-1; //because ud=1 q=(exp(r*delta_t)-d)/(u-d); //calculation of succes probability (u) S=zeros(M+1,M+1); //creation stock price matrix S(1,1)=S_0; //Setting stock price at t=0 as initial price in the stock matrix for i=2:M+1 //Initializing algo for computation of stock price for j=1:i S(j,i)=S(1,1)*u^(j-1)*d^(i-j); //with j upwards and i-j downwards movements end end V=-ones(M+1,M+1); //creating option value matrix V(:,M+1)=max((K-S(:,M+1)), 0); //calculation of option values for last column for i=M:-1:1 //Initializing algo for computation of option price //option value is zero if at current point stock price is higher than the barrier V(1:i,i)=exp(-r*delta_t)*(q*V(2:i+1,i+1)+(1-q)*V(1:i,i+1)).*(S(1:i,i) < B); end V_0 = V(1,1); //setting of first element of the option value matrix as option price at time t=0 endfunction // Set values for the input variables S0 = 100; S_0 = S0; r = 0.05; sigma = 0.2; T = 1; K = 100; B = 110; M = 10000; m = 250; // Run functions and display result V0 = UpOutPut_BS_MC_Richardson (S0, r, sigma, T, K, B, M, m) M = 1000; V_0 = UpOutPut_BinMod(S_0, r, sigma, T, K, B, M) disp("Price of UpOutPut BS_MC_Richardson: "+string(V0)) disp("Price of UpOutPut BinMod: "+string(V_0))
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//Example 1-3, Page No - 15 clear clc wavelength_feet=75 wavelength_meter= 75/3.28 c=300000000 frequency=c/wavelength_meter printf('The signal frequncy is %.3f Megahertz',frequency/1000000)
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//Book Name:Fundamentals of Electrical Engineering //Author:Rajendra Prasad //Publisher: PHI Learning Private Limited //Edition:Third ,2014 //Ex8_4.sce clc; clear; p=4; s=21; Cs_per_slot=4; Cs=Cs_per_slot*s; printf("\n Number of coil sides=%d \n",Cs) C=Cs/2; printf("\n Number of coils=%d \n",C) Yc1=(C+1)/(p/2); Yc2=(C-1)/(p/2); C=41; //Simplex wave winding is not possible with 42 coils.Therefore active coils are 42 Yc=(C+1)/(p/2); printf("\n Commutator pitch=%d \n",Yc) Y=((2*C)+2)/(p/2); printf("\n Winding pitch=%d \n",Y) Yb=Y/2; printf("\n Back pitch=%d \n",Yb) Yf=Yb; printf("\n Full pitch=%d \n",Yf) //This value of Yb also satisfies the condition to avoid split winding
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clear clc t = [0:0.001:4] //tempo l1 = 1 //tamanho do braco 1 l2 = .5 theta1 = %pi*t/8 //theta1 em funcao do tempo theta2 = %pi*(t.^2)/16 // theta2 em funcao do tempo dtheta1 = %pi/8 dtheta2 = -%pi*t/8 ddtheta1 = 0 ddtheta2 = -%pi/8 //Movimento x = l1*cos(theta1) + l2*cos(theta1+theta2) y = l1*sin(theta1) + l2*sin(theta1+theta2) //Velocidade vx = -l2*(dtheta1 + dtheta2).*sin(theta2-theta1) - l1.*sin(theta1).*(dtheta1) vy = 12*(dthet1+dtheta2).*cos(theta2+theta1) + l1.*cos(theta1).*(dtheta1) //Aceleracao ax = -12*(ddtheta2 + ddtheta1).*sin(theta2-theta1) - 12*((dtheta2+dtheta1).^2).*cos(theta2-theta1) - l1*sin(theta1).*ddtheta1 - l1*cos(theta1).*(dtheta^2) ay = -12((dtheta2 + dtheta1.^2).*sin(theta2+theta1)
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//Example6.22 // To determine the output voltage clc; clear; close; Vin = 2 ; R2 = 20*10^3 ; R1 = 2*10^3 ; // the output voltage of follower Vo1 is Vo1 = Vin ; disp('the output voltage of follower Vo1 is = '+string(Vo1)+ ' V'); // the output voltage of an inverting amplifier Vo = -(R2/R1)*Vo1 ; disp('The output voltage of an inverting amplifier is = '+string(Vo)+ ' V ');
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clear all; clc; disp("From figure 8.14c we have Pl=620hp at N=18400rpm. Pick a point on the curve of N/(T01^0.5)=18400/(530^0.5)=800") disp("In figure 8.14a,say p02/p01=5") disp("So we have") disp("m(T01^0.5)/p01 ETAc p02(psia) m(lbm/s) p03(psia) p03/p04") disp(" 5.7 0.85 73.5 3.64 71.5 4.86") disp("where p02=5*14.7") p02=5*14.7 printf(" Thus p02= %0.2f",p02) disp("m=5.7*14.7/(530^0.5)") m=5.7*14.7/(530^0.5) printf(" m= %0.2f",m) disp("73.5-2") p03=73.5-2 printf(" m= %0.2f psia",p03) //Let i=p03/p04 i=71.5/14.7 printf("\n p03/p04= %0.2f",i) disp("Then from figure 8.14b,with p03/p04 and m3=m2 we have") disp("m(T01^0.5)/p01 T03(R) N/(T03^0.5) ETAt") disp("2.56 2528 366 0.87") disp("where T03=(2.56*71.5/3.64)^2") T03=(2.56*71.5/3.64)^2 printf(" T03= %0.2f",T03) disp("N/T03=18400/(2528^0.5)") //let k=N/T03 k=18400/(sqrt(2528)) printf(" Thus T03= %0.2f",k) disp("So from equations (8.1),(7.4) and(8.11) we have:") disp("delta_T034=ETAt*T03*[1-(p04/p03)^((k-1)/k)]") delta_T034=0.87*2528*[1-(4.86)^(-0.248)] printf(" delta_T034 = %0.0f R",delta_T034) disp("delta_T012=(T01/ETAc)*[(p02/p01)^((k-1)/k)-1]") delta_T012=(530/0.85)*[(5^0.2857)-1] printf(" delta_T012= %0.0fR",delta_T012) P0=3.64*(0.28*713-0.24*(364/0.95)) printf(" \n and P0=3.64*(0.28*713-0.24*(364/0.95))= %0.0f Btu/s=554hp, which is less than Pl",P0) disp("So we pick another point on the same curve , say p02/p01=5.2, and repeat the calculations ") disp("m(T01^0.5)/p01 ETAc p02(psia) m(lbm/s) p03(psia) p03/p04 m(T01^0.5)/p01 T03(R) N/(T03^0.5) ETAt") disp("5.6 0.88 76.4 3.57 74.4 5.06 2.55 2824 346 0.85") delta_T034=0.85*2824*[1-(5.06^(-0.248))] printf("\n The new delta_T034= %0.2fR",delta_T034)//the book has rounded off the value to 794R,the value calculated in this code is more accurate delta_T012=(530/0.88)*[(5.2^0.2857)-1] printf("\n delta_T012= %0.0fR",delta_T012) P0=3.57*(0.28*794-0.24*362/0.95) printf("\n Net output power P0= %0.0f Btu/s=660hp, which is much greater than Pl \n\n",P0) disp("Pick another point say p02/p01=5.15") disp("m(T01^0.5)/p01 ETAc p02(psia) m(lbm/s) p03(psia) p03/p04 m(T01^0.5)/p01 T03(R) N/(T03^0.5) ETAt") disp("5.65 0.87 75.7 3.61 73.7 5.01 2.55 2710 353 0.86") delta_T034=0.86*2710*[1-(5.01^(-0.248))] printf("\n\n From new values delta_T034= %0.0f R",delta_T034) delta_T012=(530/0.87)*[(5.15^(0.2857))-1] printf("\n and delta_T012= %0.0f R",delta_T012) P0=3.61*(0.28*768-0.24*(364/0.95)) printf("also P0= %0.1f Btu/s =628hp",P0) disp("P0 is close to Pl") disp("So the running point is around p02/p01=5.15, m(T01^0.5)/p01=5.65") disp("and N/(T01^0.5)=800 on the compressor characteristics") disp("It is not too close to the surge line and hence is safe.")
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8_18.sce
//Fuels and Combustion// //Example 8.18// C=624;//weight of carbon in 1kg of coal sample in grams// O=69;//weight of oxygen in 1kg of coal sample in grams// S=8;//weight of Sulphur in 1kg of coal sample in grams// N=12;//weight of nytrogen in 1kg of coal sample in grams// H=41;//weight of hydrogen in 1kg of coal sample in grams// CO2=129;//weight of CO2 in 1kg of coal sample in grams// CO=2;//weight of CO in 1kg of coal sample in grams// MO=C*32/12+H*16/2+S*32/32-O;//minimum weight of oxygen needed in grams// MA=MO*0.1/23;//minimum weight of air needed in kilograms// printf('minimum amount of air needed=MA=%fkg',MA); WC=CO2*(12/44)+CO*(12/28);//weight of C in fuel gas/kg// printf('\nWeight of C in fuel gas/kg=WC=%fg',WC); WF=C/WC;//Weight of fuel gas/kg of coal in g// printf('\nweight of fuel gas/kg of coal=WF=%fg',WF); O2=2*16/28;//O2 needed to convert CO to CO2 in Kg// RWO2=(61-O2)/1000;//remaining weight of O2/kg of fuel gas in Kg// printf('\nRemaining weight of O2/kg of fuel gas=RWO2=%fkg',RWO2); WO2=WF*RWO2;//weight of O2 obtained by burning 1kg coal in kg// printf('\nWeight of O2 obtained by burning coal=WO2=%fkg',WO2); AR=WO2*100/23;//air required in kilograms// printf('\nAir required=AR=%fkg',AR); WAS=MA+AR;//weight of air actually supplied/kg coal burnt in kg// printf('\nWeight of air actually supplied/kg coal burnt=WAS=%fkg',WAS);
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//Example 15.11 //Midpoint Method //Page no. 518 clc;clear;close; deff('y=f(x,y)','y=y+x') y=1; h=0.2; printf('i\txi\tyi\tslope1\tslope2\ty(i+1)\n-----------------------------------------------\n') for i=1:3 x=(i-1)*h s1=f(x,y); s2=f(x+h/2,y+s1*h/2); printf(' %i\t%g\t%g\t%g\t%g',i-1,x,y,s1,s2) y=y+s2*h; printf('\t%g\n',y) end
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ex2_11.sce
clc printf("\n") //let //S=displacement of car from rest with uniform acceleration a, the engine torque T assumed to remain ocnstant //v=final speed ofcar //G=gear ratio //r=effective radius //n=efficiency of transmission //M=mass of the car //Ia and Ib=moments of inertia of road whels and engine //formulas => F=29.5nG ; Me= 1648+$.54nG^2 ; a=32.2 F/Me //given G1=22.5 G2=12.5 G3=7.3 G4=5.4 n=.82//for 1st ,2nd and 3rd gear n4=.9//for 4th gear F1=29.5*n*G1 F2=29.5*n*G2 F3=29.5*n*G3 F4=29.5*n4*G4 //on reduction and putting values we get Me1=1648+4.54*n*G1^2 Me2=1648+4.54*n*G2^2 Me3=1648+4.54*n*G3^2 Me4=1648+4.54*n4*G4^2 a1=32.2*F1/Me1 a2=32.2*F2/Me2 a3=32.2*F3/Me3 a4=32.2*F4/Me4 printf("Maximum acceleration of car on top gear is %.2f ft/s^2 \n",a4) printf("Maximum acceleration of car on third gear is %.2f ft/s^2 \n",a3) printf("Maximum acceleration of car on second gear is %.2f ft/s^2 \n",a2) printf("Maximum acceleration of car on first gear is %.2f ft/s^2 \n",a1)
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FLAGS='--detect --list' STDIN='' STDOUT='' STDERR='!USAGE!' EXITVAL='1'
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Example11_1.sce
// A Texbook on POWER SYSTEM ENGINEERING // A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar // DHANPAT RAI & Co. // SECOND EDITION // PART II : TRANSMISSION AND DISTRIBUTION // CHAPTER 4: OVERHEAD LINE INSULATORS // EXAMPLE : 4.1 : // Page number 183 clear ; clc ; close ; // Clear the work space and console // Given data V_1 = 9.0 // Potential across top unit(kV) V_2 = 11.0 // Potential across middle unit(kV) n = 3.0 // Number of disc insulators // Calculations // Case(a) K = (V_2-V_1)/V_1 // Ratio of capacitance b/w pin & earth to self capacitance // Case(b) V_3 = V_2+(V_1+V_2)*K // Potential across bottom unit(kV) V = V_1+V_2+V_3 // Voltage between line and earth(kV) V_l = 3**0.5*V // Line voltage(kV) // Case(c) eff = V/(n*V_3)*100 // String efficiency(%) // Results disp("PART II - EXAMPLE : 4.1 : SOLUTION :-") printf("\nCase(a): Ratio of capacitance b/w pin & earth to self-capacitance of each unit, K = %.2f ", K) printf("\nCase(b): Line voltage = %.2f kV", V_l) printf("\nCase(c): String efficiency = %.f percent", eff)
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test_mnprobln.sci
exec("mnprobln.sci",2); function ok=test_mnprobln() mu=[1;-1]; Sigma=[5 6;6 9]; n1=51; y1=linspace(-10,10,n1); n2=51; y2=linspace(-10,10,n2); y=zeros(2,n1*n2); for i1=1:n1 b=(i1-1)*n2; y(1,b+1:b+n2)=y1(i1); y(2,b+1:b+n2)=y2; end dy=((y1(n1)-y1(1))/(n1-1))*((y2(n2)-y2(1))/(n2-1)); p=exp(mnprobln(y,mu,Sigma))*dy; m0=sum(p); m1=zeros(2,1); m1(1)=sum(p .* y(1,:))/m0; m1(2)=sum(p .* y(2,:))/m0; m2=zeros(2,2); for i=1:2 for j=1:2 m2(i,j)=sum(p .* (y(i,:)-mu(i)).*(y(j,:)-mu(j)))/m0; end end assert_checkalmostequal(m0,1,0.025); assert_checkalmostequal(m1,mu,0.025); assert_checkalmostequal(m2,Sigma,0.025); ok=%T; endfunction
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Ex12_20.sce
//Initilization of variables theta=45 //degrees l=0.5 //m w=10 //rad/s //Calculations //PART a //Here the theta derivative with respect to time is angular speed w Vp1=l*(secd(theta)^2)*w //m/s //Part b //Radial Component r=l*secd(theta)*tand(theta)*w //m/s //Transverse Component t=l*secd(theta)*w //m/s //Total Vp2=sqrt(r^2+t^2) //m/s //Result clc printf('The velocity is:%fm/s\n',Vp1) printf('The velocity in part b is %fm/s',Vp2)
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//ex_23 product of discrete signal and unit step function clear; clc; close; t=-3:3; x=[3 2 1 0 1 2 3]; //u[1-n] for i=1:length(t); if t(i)<=1 then u1(i)=1; else u1(i)=0; end end y=x.*u1'; figure a=gca(); plot2d3(t,y) plot(t,y,'r.') xtitle('y[n]','n') a.y_location='origin' //u[n+2]-u[n] for i=1:length(t); if t(i)<1 & t(i)>=-2 then u2(i)=1; else u2(i)=0; end end z=x.*u2'; figure a=gca(); plot2d3(t,z) plot(t,z,'r.') xtitle('z[n]','n') a.y_location='origin' //$[n-1] for i=1:length(t); if t(i)==1 then del(i)=1; else del(i)=0; end end p=x.*del'; figure a=gca(); plot2d3(t,p) plot(t,p,'r.') xtitle('y[n]','n') a.y_location='origin'
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// Example 3.2: calculate the input impedence, output impedence, voltage gain and current gain clc, clear; Hie=3.2; // the h-parameters of the transistor in kilo-ohm Hfe=100; // the h-parameters of the transistor R1=40; // resistance in kilo-ohm R2=4.7; // resistance in kilo-ohm; Rc=4; // resistance in kilo-ohm; Re=1.2; // resistance in kilo-ohm; Rb=(R1*R2)/(R1+R2); zi=(Rb*Hie)/(Rb+Hie); zo=Rc; Av=-(Hfe*Rc)/Hie; Ai=-(Rb*Hfe)/(Rb+Hie); Vcc=16; // voltage in volts Vbe=0.6; // voltage in volts Beta=100; // for transistor Vb=(R2*Vcc)/(R1+R2); Ib=(Vb-Vbe)/(Rb+(1+Beta)*Re); Ic=Beta*Ib; Ie=Ic; re=26/Ie; Zin=(Rb*(Beta*re*10^-3))/(Rb+(Beta*re*10^-3)); Zout=Rc; Av1=-(Rc*10^3)/re; Ai1=-(Beta*(Rb*10^3))/((Rb*10^3)+(Beta*re)); disp("part 1 -from h-parameter model") disp(Rb,"base resistance (kilo-ohm) = ") disp(zi,"input impedence (kilo-ohm) = ") disp(zo,"ouput impedence (kilo-ohm) = ") disp(Ai,"current gain (unitless) = ") disp(Av,"voltage gain (unitless) = ") disp("part 2 -from re model") disp(Vb,"base voltage (V) = ") disp(Ib,"base curret (mA) = ") disp(Ic,"collector curret (mA) = ") disp(Ie,"emitter curret (mA) = ") disp(re,"resistance = ") disp(Zin,"input impedence (kilo-ohm) = ") disp(Zout,"ouput impedence (kilo-ohm) = ") disp(Ai1,"current gain (unitless) = ") disp(Av1,"voltage gain (unitless) = ")
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Ex2_9.sce
// chapter 2 // example 2.9 // Fig. 2.17 // Calculate the value of commutating capacitance // page-54 clear; clc; // given Edc=120; // in V (source voltage) I=20; // in A (current) t_off=60; // in us (turn-off time of both SCR) // calculate R1=Edc/I; // calculation of R1 R2=R1; // calculation of R2 C=1.44*t_off/R1; // calculation of commutating capacitance printf("\n\nThe commutating capacitance for succcessful commutation is \tC= %.1f uF",C);
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clear; clc; ld=100; //load in KW v=500;............// voltage in volts res=1.75*(.000001);...........//restivity in milli ohm per cm sq nos_cores=2;.......//number of feeder core l=0.8;.............//length of tx line in km area=1;...........//area in cm sq cost_of_energy=0.12;..//cost in Rs per unit dep=0.1..............//depreciation percentage is 10% flc=(ld*1000)/500;..//full load current ra=(res*l*1000*ld);..//resistance* area in ohm-m pow_loss=2*(flc*flc*ra*.001);...//(power loss*area) for the two cores in the cable ann_en_loss=pow_loss*365*24;..//annual energy loss in KWH cost=ann_en_loss*cost_of_energy;..//(cost*area) of annual energy loss in Rs ann_dep=6*l*1000*dep;........//(ann_dep*area) in Rs c=(cost/ann_dep); area=sqrt(c);.......//area in cm sq d=(area*4/(%pi)); dia=sqrt(d);..........//diameter in cm printf("\n the most economical size is: %.2f cm\n ",dia);
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clc //Initialization of variables T1=400+460 //R P1=100 //lb/in^2 P2=20 //lb/in^2 T2=140+460 //R Cp=50 //calculations Pratio=P1/P2 Tratio=T1/T2 C=log(Tratio) /log(Pratio) n=1/(1-C) v1=Cp*T1/(144*P1) v2=Cp*T2/(144*P2) w=144*P1*v1^n function[p]=fun(v) p=w/v^n endfunction Work=intg(v1,v2,fun) //results printf("Work done = %.d ft-lb/lbm",Work) //The answers in the textbook varies a bit due to rounding off errors
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function a=%sp_sum(a,flag) // Copyright INRIA [m,n]=size(a); if flag==2|flag=='c' then a=sparse(a*ones(n,1)) end if flag==1|flag=='r' then a=sparse(ones(1,m)*a); end
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//Example 20_3 clc(); clear; //To find the current through the inductor f=60 //Units in Hz l=15*10^-3 //Units in H xl=2*%pi*f*l //Units in Ohms v=40 //Units in V i=v/xl //Units in A printf("The current in the inductor when frequency=60 Hz is I=%.2f A",i) f=6*10^5 //Units in Hz l=15*10^-3 //Units in H xl=2*%pi*f*l //Units in Ohms v=40 //Units in V i=v/xl //Units in A printf("\nThe current in the inductor when frequency=6*10^2 Hz is I=%.6f A",i)
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7412d9c017b1ddb206797b5ae4ec7734dd7146d4
c3280ada260999123d75347caeaad1c7fc9f8266
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[]
no_license
dikisp/Kuliah-MetNum
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7775f5e0251457702fb3e24c88483df0ff37fee7
refs/heads/master
2020-03-23T06:41:49.307866
2018-07-17T03:13:25
2018-07-17T03:13:25
null
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sce
kali.sce
function total = hitung(atas) jum = 0; i=1; while i < atas if i modulo 5 = 0 then jum = jum + i; end i=i+1; end total = jum; endfunction
cd9910c3be89fba2e76def203cd9526f33b88908
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[]
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fernandotugu/estudos
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e73d25e7e76aa77de59130fdd95a394ab7baeb7f
refs/heads/master
2023-01-06T22:59:07.256808
2020-01-15T19:54:03
2020-01-15T19:54:03
199,859,219
0
0
null
2023-01-05T03:11:47
2019-07-31T13:13:53
Vue
UTF-8
Scilab
false
false
33
tst
nginx.tst
rewrite ^(.+)$ /index.html last;
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/1853/CH1/EX1.7/Ex1_7.sce
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[]
no_license
FOSSEE/Scilab-TBC-Uploads
948e5d1126d46bdd2f89a44c54ba62b0f0a1f5e1
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37,975,407
3
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null
null
null
null
UTF-8
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sce
Ex1_7.sce
//calculate the current I1.I2 I=12 R1=6 R2=8 I1=I*R2/(R1+R2) I2=I*R1/(R1+R2) disp('I1='+string(I1)+'amps' , 'I2 ='+string(I2)+'amps')
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FOSSEE/Scilab-TBC-Uploads
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2020-04-09T02:43:26.499817
2018-02-03T05:31:52
2018-02-03T05:31:52
37,975,407
3
12
null
null
null
null
UTF-8
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sce
15_2.sce
clc //initialisation of variables Ksp= 2*10^-12 M= 8.84*10^-5 //molar n= 2 //CALCULATIONS r= (Ksp/(n^2*M^3))^(1/3) //RESULTS printf ('mean ionic activity coefficient = %.1f ',r)
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FOSSEE/Scilab-TBC-Uploads
948e5d1126d46bdd2f89a44c54ba62b0f0a1f5e1
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2020-04-09T02:43:26.499817
2018-02-03T05:31:52
2018-02-03T05:31:52
37,975,407
3
12
null
null
null
null
UTF-8
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sce
eg_4_10.sce
clc; k=[1 1 0 1; 0 1 1 1]; karmap3(k);
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449d555969bfd7befe906877abab098c6e63a0e8
/213/CH12/EX12.4/12_4.sce
dcb5d38f3b5b2f5a16f72b7f4f1dc827d06cd68e
[]
no_license
FOSSEE/Scilab-TBC-Uploads
948e5d1126d46bdd2f89a44c54ba62b0f0a1f5e1
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refs/heads/master
2020-04-09T02:43:26.499817
2018-02-03T05:31:52
2018-02-03T05:31:52
37,975,407
3
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null
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null
null
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sce
12_4.sce
//To find angle and maximum velocity clc //Given: phi=20 //degrees t=20, G=2 m=5 //mm v=1.2 //m/s addendum=1*m //mm //Solution: //Angle turned through by pinion when one pair of teeth is in mesh: //Calculating the pitch circle radius of pinion r=m*t/2 //mm //Calculating the pitch circle radius of wheel R=m*G*t/2 //mm //Calculating the radius of addendum circle of pinion rA=r+addendum //mm //Calculating the radius of addendum circle of wheel RA=R+addendum //mm //Calculating the length of path of approach KP=sqrt(RA^2-R^2*(cosd(phi))^2)-R*sind(phi) //mm //Calculating the length of path of recess PL=sqrt(rA^2-r^2*(cosd(phi))^2)-r*sind(phi) //mm //Calculating the length of path of contact KL=KP+PL //mm //Calculating the length of arc of contact Lac=KL/cosd(phi) //mm //Calculating the angle turned by the pinion angle=Lac*360/(2*%pi*r) //Angle turned by the pinion, degrees //Maximum velocity of sliding: //Calculating the angular speed of pinion omega1=v*1000/r //rad/s //Calculating the angular speed of wheel omega2=v*1000/R //rad/s //Calculating the maximum velocity of sliding vS=(omega1+omega2)*KP //mm/s //Results: printf("\n\n Angle turned through by pinion when one pair of teeth is in mesh = %.2f degrees.\n\n",angle) printf(" Maximum velocity of sliding, vS = %.1f mm/s.\n\n",vS)
c7d5e1b11d16111f5ccc9eae332593e58df606a0
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/test/unite.tst
a6bd7f597d4302d7a2c5ee45149ae5348e8d390c
[ "BSD-3-Clause", "BSD-2-Clause" ]
permissive
dougsong/reposurgeon
394001c0da4c3503bc8bae14935808ffd6f45657
ee63ba2b0786fa1b79dd232bf3d4c2fe9c22104b
refs/heads/master
2023-03-09T15:22:45.041046
2023-02-25T08:33:06
2023-02-25T08:33:06
280,299,498
1
0
NOASSERTION
2023-02-25T08:33:08
2020-07-17T01:45:32
Go
UTF-8
Scilab
false
false
87
tst
unite.tst
## Test of the unite feature read <bzr.fi read <testrepo.fi unite bzr testrepo write -
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/1439/CH2/EX2.4/2_4.sce
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[]
no_license
FOSSEE/Scilab-TBC-Uploads
948e5d1126d46bdd2f89a44c54ba62b0f0a1f5e1
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2020-04-09T02:43:26.499817
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2018-02-03T05:31:52
37,975,407
3
12
null
null
null
null
UTF-8
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clc //initialisation of variables R= 0.08205 //l-atm deg^-1 mole^-1 T= 25 //K n= 1 //mole V= 0.5 //lit b= 0.04267 //lit mole^-1 a= 3.592 //lit^2 atm mol^-2 //CALCULATIONS P= R*(273.15+T)/V P1= (R*(273.15+T)/(V-b))-(a/V^2) //RESULTS printf ('pressure calculated using ideal gas law= %.1f atm',P) printf ('\n pressure calculated using vander wals equation= %.1f atm',P1)
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THE OPTIMIZATION ALGORITHM HAS CHANGED TO THE EM ALGORITHM. ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 0.392222D+00 2 -0.181207D-02 0.320877D-02 3 0.778638D-01 -0.123696D-02 0.346914D+00 4 -0.217820D-02 0.540345D-03 -0.326635D-02 0.269617D-02 5 0.287357D-02 0.153158D-03 0.386547D-03 0.271808D-03 0.280372D-02 6 0.128765D-02 0.837698D-04 0.191652D-02 -0.285775D-04 0.178009D-03 7 0.363841D-02 0.125532D-03 0.385156D-02 0.499000D-04 0.960668D-03 8 0.125776D-02 0.105493D-03 0.879214D-03 0.911894D-05 0.295441D-03 9 -0.198090D+00 0.359036D-01 -0.260684D+00 0.156248D-01 0.929418D-01 10 0.124136D+00 0.124765D-01 0.243819D+00 0.192241D-01 0.138199D+00 11 -0.336344D+00 0.209016D-01 -0.212080D+00 0.267959D-02 -0.819912D-02 12 -0.390098D+00 0.232064D-02 -0.135127D+01 0.486141D-01 -0.161560D-02 13 0.221704D+00 0.713195D-02 0.151251D+00 0.853271D-02 0.572121D-01 14 0.386655D-01 0.123158D-01 -0.307228D+00 0.115187D-01 0.379904D-01 15 -0.100610D+01 -0.753929D-01 -0.439232D+00 -0.175553D-01 -0.147556D+00 16 -0.408993D-01 -0.997008D-02 -0.226009D-01 -0.270024D-02 -0.341144D-02 17 -0.662610D-02 -0.729953D-03 -0.177068D-02 -0.217760D-03 -0.571579D-03 18 -0.934302D+00 -0.292553D-01 0.908277D-01 -0.524748D-01 0.910759D-02 19 -0.166638D+00 0.436321D-02 0.227984D+00 -0.406460D-02 0.650274D-02 20 -0.473567D+00 -0.365187D-01 -0.351682D+01 -0.485252D-01 -0.167575D-01 21 0.133839D+00 -0.739353D-02 -0.180650D+00 -0.206366D-02 -0.678085D-02 22 0.159621D-02 -0.302356D-03 -0.403007D-02 0.627117D-03 -0.475239D-03 23 0.322072D-01 -0.213202D-02 -0.257863D-01 -0.122237D-01 -0.279451D-02 24 -0.392981D-03 0.689969D-03 0.630446D-02 -0.133319D-03 -0.788199D-04 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 0.109667D-02 7 0.146359D-02 0.397705D-02 8 0.276692D-03 0.436147D-03 0.284663D-02 9 0.188016D-01 0.746891D-01 0.253571D-01 0.588583D+02 10 0.228684D-01 0.732937D-01 0.230293D-01 0.209222D+01 0.195424D+02 11 0.201738D-01 0.844365D-02 -0.128581D-01 0.740504D+01 -0.142705D+01 12 -0.414191D-03 -0.391511D-03 -0.581719D-01 0.356450D+01 -0.125351D+01 13 0.888706D-01 0.182589D+00 0.353754D-01 0.275293D+01 0.544117D+01 14 0.592130D-01 0.111035D+00 0.269634D+00 0.729509D+01 0.478625D+01 15 -0.663522D-01 -0.128638D+00 -0.896748D-01 -0.160089D+02 -0.116847D+02 16 -0.951995D-03 -0.222579D-02 -0.352780D-02 0.820918D+00 -0.400724D+00 17 0.639016D-04 -0.173257D-03 0.252127D-03 -0.881177D-01 -0.509979D-01 18 -0.945106D-01 -0.192073D+00 -0.450816D-01 -0.989001D+01 -0.374632D+00 19 -0.118236D-02 0.157322D-01 -0.128780D-02 -0.243145D+01 0.595639D+00 20 -0.999096D-01 -0.206362D+00 -0.178186D+00 0.317339D+01 0.494525D-01 21 -0.595117D-03 -0.188201D-01 0.642220D-03 0.222404D+01 -0.827548D+00 22 -0.151768D-04 0.997691D-05 0.186035D-03 0.393527D-01 -0.335286D-01 23 0.702514D-03 0.222424D-03 0.138398D-02 -0.311141D+00 -0.108359D+00 24 0.170063D-03 0.431924D-03 -0.192785D-03 -0.307207D-01 -0.193037D-01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 0.289542D+02 12 0.894261D+01 0.892475D+02 13 -0.784660D+00 0.114011D+01 0.181735D+02 14 0.991617D+00 -0.380945D+01 0.776706D+01 0.583777D+02 15 -0.390862D+01 -0.101895D+00 -0.817002D+01 -0.109291D+02 0.347934D+03 16 0.213536D+00 0.630556D+00 -0.148859D+00 -0.420948D-01 0.292971D+01 17 0.484524D-01 -0.531543D-02 -0.146292D-01 0.454718D-01 -0.144965D+01 18 -0.159045D+01 -0.389107D+01 -0.119121D+02 -0.113948D+02 0.996371D+02 19 -0.102651D+00 -0.204770D+01 0.676177D-01 -0.418956D+00 0.783826D+01 20 -0.975864D+00 -0.344465D+01 -0.107825D+02 -0.256995D+02 0.763812D+02 21 0.585499D+00 0.162271D+01 -0.345379D+00 0.559554D+00 -0.738923D+01 22 -0.460569D-01 0.810672D-02 0.309571D-02 0.491965D-01 -0.441284D+00 23 0.526456D-01 -0.384553D+00 0.420477D-01 0.279734D+00 0.188321D+00 24 -0.153614D-01 -0.896729D-01 0.147729D-01 -0.585337D-01 -0.214903D+00 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 0.557279D+00 17 -0.239118D-01 0.162288D-01 18 0.497796D+00 -0.508437D+00 0.261248D+03 19 0.106180D-01 -0.485961D-01 0.849230D+01 0.517602D+01 20 0.807282D+00 -0.385086D+00 0.103296D+03 0.423897D+00 0.407990D+03 21 0.186529D+00 0.378221D-01 -0.448466D+01 -0.469781D+01 -0.292818D+00 22 -0.556613D-02 0.454205D-02 -0.114402D+01 -0.566988D-01 -0.411484D+00 23 0.255326D-01 0.497836D-02 -0.957477D+00 -0.186697D+00 0.318220D+01 24 0.198788D-02 0.178198D-02 -0.310382D+00 0.180802D-01 -0.179880D+01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 0.548469D+01 22 0.233928D-01 0.107491D-01 23 0.360451D+00 -0.880703D-02 0.522233D+00 24 -0.489617D-01 0.212866D-02 -0.313304D-01 0.181023D-01 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 1.000 2 -0.051 1.000 3 0.211 -0.037 1.000 4 -0.067 0.184 -0.107 1.000 5 0.087 0.051 0.012 0.099 1.000 6 0.062 0.045 0.098 -0.017 0.102 7 0.092 0.035 0.104 0.015 0.288 8 0.038 0.035 0.028 0.003 0.105 9 -0.041 0.083 -0.058 0.039 0.229 10 0.045 0.050 0.094 0.084 0.590 11 -0.100 0.069 -0.067 0.010 -0.029 12 -0.066 0.004 -0.243 0.099 -0.003 13 0.083 0.030 0.060 0.039 0.253 14 0.008 0.028 -0.068 0.029 0.094 15 -0.086 -0.071 -0.040 -0.018 -0.149 16 -0.087 -0.236 -0.051 -0.070 -0.086 17 -0.083 -0.101 -0.024 -0.033 -0.085 18 -0.092 -0.032 0.010 -0.063 0.011 19 -0.117 0.034 0.170 -0.034 0.054 20 -0.037 -0.032 -0.296 -0.046 -0.016 21 0.091 -0.056 -0.131 -0.017 -0.055 22 0.025 -0.051 -0.066 0.116 -0.087 23 0.071 -0.052 -0.061 -0.326 -0.073 24 -0.005 0.091 0.080 -0.019 -0.011 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 1.000 7 0.701 1.000 8 0.157 0.130 1.000 9 0.074 0.154 0.062 1.000 10 0.156 0.263 0.098 0.062 1.000 11 0.113 0.025 -0.045 0.179 -0.060 12 -0.001 -0.001 -0.115 0.049 -0.030 13 0.630 0.679 0.156 0.084 0.289 14 0.234 0.230 0.661 0.124 0.142 15 -0.107 -0.109 -0.090 -0.112 -0.142 16 -0.039 -0.047 -0.089 0.143 -0.121 17 0.015 -0.022 0.037 -0.090 -0.091 18 -0.177 -0.188 -0.052 -0.080 -0.005 19 -0.016 0.110 -0.011 -0.139 0.059 20 -0.149 -0.162 -0.165 0.020 0.001 21 -0.008 -0.127 0.005 0.124 -0.080 22 -0.004 0.002 0.034 0.049 -0.073 23 0.029 0.005 0.036 -0.056 -0.034 24 0.038 0.051 -0.027 -0.030 -0.032 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 1.000 12 0.176 1.000 13 -0.034 0.028 1.000 14 0.024 -0.053 0.238 1.000 15 -0.039 -0.001 -0.103 -0.077 1.000 16 0.053 0.089 -0.047 -0.007 0.210 17 0.071 -0.004 -0.027 0.047 -0.610 18 -0.018 -0.025 -0.173 -0.092 0.330 19 -0.008 -0.095 0.007 -0.024 0.185 20 -0.009 -0.018 -0.125 -0.167 0.203 21 0.046 0.073 -0.035 0.031 -0.169 22 -0.083 0.008 0.007 0.062 -0.228 23 0.014 -0.056 0.014 0.051 0.014 24 -0.021 -0.071 0.026 -0.057 -0.086 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 1.000 17 -0.251 1.000 18 0.041 -0.247 1.000 19 0.006 -0.168 0.231 1.000 20 0.054 -0.150 0.316 0.009 1.000 21 0.107 0.127 -0.118 -0.882 -0.006 22 -0.072 0.344 -0.683 -0.240 -0.196 23 0.047 0.054 -0.082 -0.114 0.218 24 0.020 0.104 -0.143 0.059 -0.662 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 1.000 22 0.096 1.000 23 0.213 -0.118 1.000 24 -0.155 0.153 -0.322 1.000
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clc //initialisation of variables p= 800 //atm P= 10000 //psi x= 14.696 //psi/atm //CALCULATIONS P1= p*x //RESULTS if (P1>P) then disp("Salesman is honest")
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FLAGS='--list --print NONE' STDIN='' STDOUT='' STDERR='!USAGE!' EXITVAL='1'
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clear; clc; v=400; Xl=.30; Xc=3.75*10^(-6); l=300; L=Xl/(2*%pi*50); C=Xc/(2*%pi*50); Zc=sqrt(L/C); mprintf(" surge impedence is %.3f ohm\n",Zc); phase=2*%pi*50*sqrt(L*C); pconstant=%i*phase; mprintf("propagation constant is %.4f\n",imag(pconstant)); A=cos(phase*l); D=A; B=%i*Zc*sin(phase*l); C=(%i*sin(phase*l))/Zc; lamda=(3*10^(8))/50; SIL=v^(2)/Zc; mprintf("A,B,C and D parameters are respectively\n") disp(A,B,C,D); mprintf("Lamda and SIL is %.5fm and %.5fMW",lamda,SIL);
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x = 0:5:35; y = [0 20 43 59 89 124 154 200]; h = 5; x = 25/h + 1 velocityAt25seconds = (y(x + 1) - y(x - 1)) /(2* h) ; accelerationAt25seconds = (y(x + 1) - (2 * y(x)) + y(x - 1))/(h^2) x = 1; velocityAt0seconds = (y(x + 1) - y(x)) /(h) ;
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//Mass flow// pathname=get_absolute_file_path('12.02.sce') filename=pathname+filesep()+'12.02-data.sci' exec(filename) //Checking for chocking: c=pb/p0; if(c<=0.528) //choked else //Not choked //Therefore pressure at exit = back pressure pe=pb; //Mach number at exit: Me=(((p0/pe)^((k-1)/k)-1)*(2/(k-1)))^0.5 //Temperature at exit(in K): Te=T0/(1+(k-1)/2*Me^2) //Velocity at exit(in m/sec): Ve=Me*sqrt(k*R*Te) //Density at exit(in kg/m^3): de=pe*10^3/R/Te //Mass flow rate of air(kg/sec): m=de*Ve*Ae end; printf("\n\nRESULTS\n\n") printf("\n\nMach number at exit: %.3f\n\n",Me) printf("\n\nMass flow rate of air: %.3f kg/sec\n\n",m)
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THE OPTIMIZATION ALGORITHM HAS CHANGED TO THE EM ALGORITHM. ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 0.303401D+00 2 -0.247993D-02 0.232315D-02 3 0.340604D-01 -0.685035D-03 0.213163D+00 4 -0.130137D-03 0.821904D-04 -0.174102D-02 0.182530D-02 5 -0.550776D-03 0.133076D-03 0.118883D-02 0.431554D-04 0.345639D-02 6 0.407618D-03 -0.788687D-04 0.399060D-03 -0.991487D-04 -0.323277D-03 7 0.565044D-03 0.460162D-04 0.110999D-03 0.115608D-03 0.691443D-03 8 -0.405103D-03 0.123556D-03 -0.373069D-03 -0.153680D-04 0.808340D-04 9 -0.312135D+00 0.871458D-02 -0.782795D-01 -0.138469D-01 0.365374D-01 10 -0.159170D+00 -0.743475D-02 0.945541D-01 -0.720885D-02 0.111161D+00 11 -0.150916D-01 -0.449102D-02 0.366519D-01 0.816983D-02 0.407277D-01 12 0.309817D+00 0.152856D-01 -0.410695D+00 0.455652D-01 -0.192799D-01 13 -0.254838D-01 -0.129019D-03 0.874900D-01 -0.758872D-02 0.272646D-02 14 -0.187140D+00 0.237234D-01 -0.459302D+00 0.362370D-02 0.154618D-01 15 -0.230190D+01 -0.253264D-01 -0.472893D+00 0.155612D-01 -0.118580D+00 16 -0.626858D-01 -0.264366D-02 0.170622D-02 -0.426516D-03 -0.878664D-03 17 0.984018D-02 -0.538205D-03 0.255795D-02 0.740191D-04 -0.367325D-03 18 -0.861715D-01 0.154356D-01 -0.181478D-01 -0.126820D-01 -0.228255D-01 19 0.263177D-02 0.462749D-02 0.255441D-01 -0.963622D-03 0.448316D-02 20 0.153318D+00 -0.203372D-01 -0.497674D+00 -0.225761D-01 -0.118430D-01 21 0.243690D-01 -0.695945D-02 -0.140297D-01 0.390953D-02 -0.626761D-02 22 -0.149193D-02 -0.471717D-04 0.108122D-02 0.319289D-03 -0.698137D-04 23 0.162088D-01 0.132355D-03 0.145657D-02 -0.476181D-02 -0.363003D-03 24 -0.519034D-03 0.302249D-03 -0.154409D-02 0.124626D-03 0.703638D-04 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 0.919271D-03 7 0.509448D-03 0.248753D-02 8 0.500240D-04 0.223001D-03 0.237134D-02 9 0.101411D-01 -0.174562D-01 0.514973D-02 0.451935D+02 10 0.123242D-02 0.238520D-01 0.121560D-01 0.449744D+01 0.184801D+02 11 0.197239D-01 0.520424D-01 0.165216D-02 0.270916D+01 -0.188317D+01 12 -0.643725D-01 -0.687643D-02 0.562203D-01 -0.221679D+01 0.126919D+01 13 0.525642D-01 0.807760D-01 0.172337D-01 0.179647D+01 0.132721D+01 14 -0.250154D-02 0.266932D-01 0.176204D+00 0.129235D+01 0.346716D+01 15 -0.165669D-01 -0.615116D-01 -0.404714D-01 -0.915028D+01 -0.120854D+02 16 -0.273164D-03 -0.127542D-02 0.299219D-03 0.830265D+00 -0.140196D+00 17 0.138506D-05 0.939403D-04 -0.721396D-04 -0.163549D+00 -0.527152D-02 18 -0.433919D-01 -0.512046D-01 -0.221434D-01 -0.163713D+01 -0.559053D+00 19 -0.823600D-02 0.824046D-02 -0.106662D-01 -0.163366D+01 -0.617223D+00 20 0.263851D-02 -0.101426D-01 -0.118692D+00 -0.192849D+01 -0.155353D+01 21 0.925505D-02 -0.720501D-02 0.942590D-02 0.188186D+01 0.452801D+00 22 -0.885653D-04 -0.442422D-03 -0.934072D-04 -0.429194D-02 -0.223692D-01 23 -0.603244D-03 -0.886918D-03 0.862255D-03 -0.226944D+00 0.142308D-01 24 0.562268D-05 -0.165600D-03 -0.375589D-03 0.349344D-01 -0.502152D-02 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 0.278842D+02 12 -0.200025D+01 0.125916D+03 13 -0.933144D+00 -0.768438D-01 0.110912D+02 14 -0.783777D+00 0.953788D+01 0.819402D+00 0.442142D+02 15 0.679090D+00 0.981243D+01 -0.125605D+01 -0.238871D+01 0.243570D+03 16 -0.158930D+00 0.187194D+00 -0.161004D-01 0.412438D-01 0.172552D+01 17 0.112572D-01 0.291726D-01 -0.810618D-02 -0.321791D-01 -0.103591D+01 18 -0.248469D+01 0.688975D+01 -0.477486D+01 -0.322117D+01 0.200639D+02 19 0.671277D+00 0.621586D+00 -0.940015D+00 -0.147441D+01 -0.141183D+00 20 0.119984D+01 -0.245831D+02 -0.332896D+01 -0.167547D+02 0.333359D+01 21 -0.239428D+00 -0.682059D+00 0.898222D+00 0.128798D+01 -0.100333D+01 22 -0.428155D-01 0.336431D-01 -0.721415D-03 -0.810789D-02 -0.246233D-01 23 -0.466483D-01 0.444405D+00 -0.434297D-01 0.130919D+00 -0.450552D+00 24 -0.926477D-02 -0.246692D-01 0.280066D-02 -0.546840D-01 0.555859D-01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 0.475753D+00 17 -0.231875D-01 0.139935D-01 18 -0.472227D+00 -0.641895D-01 0.955696D+02 19 -0.388902D-01 0.178028D-01 0.195850D+01 0.314615D+01 20 -0.354941D+00 0.587348D-01 -0.297252D+01 0.805177D+00 0.133999D+03 21 0.174030D+00 -0.186240D-01 -0.100638D+01 -0.295978D+01 -0.933749D+00 22 0.481236D-03 0.157608D-02 -0.394783D+00 -0.116115D-01 0.130625D-01 23 -0.617817D-02 0.493967D-02 -0.171630D+00 0.577171D-01 0.158434D+01 24 0.400447D-02 -0.859340D-03 0.257354D-01 -0.868711D-02 -0.615275D+00 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 0.348881D+01 22 -0.902024D-02 0.487393D-02 23 -0.515198D-02 -0.242284D-02 0.238883D+00 24 0.311077D-02 0.483380D-03 -0.199390D-01 0.740460D-02 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 1.000 2 -0.093 1.000 3 0.134 -0.031 1.000 4 -0.006 0.040 -0.088 1.000 5 -0.017 0.047 0.044 0.017 1.000 6 0.024 -0.054 0.029 -0.077 -0.181 7 0.021 0.019 0.005 0.054 0.236 8 -0.015 0.053 -0.017 -0.007 0.028 9 -0.084 0.027 -0.025 -0.048 0.092 10 -0.067 -0.036 0.048 -0.039 0.440 11 -0.005 -0.018 0.015 0.036 0.131 12 0.050 0.028 -0.079 0.095 -0.029 13 -0.014 -0.001 0.057 -0.053 0.014 14 -0.051 0.074 -0.150 0.013 0.040 15 -0.268 -0.034 -0.066 0.023 -0.129 16 -0.165 -0.080 0.005 -0.014 -0.022 17 0.151 -0.094 0.047 0.015 -0.053 18 -0.016 0.033 -0.004 -0.030 -0.040 19 0.003 0.054 0.031 -0.013 0.043 20 0.024 -0.036 -0.093 -0.046 -0.017 21 0.024 -0.077 -0.016 0.049 -0.057 22 -0.039 -0.014 0.034 0.107 -0.017 23 0.060 0.006 0.006 -0.228 -0.013 24 -0.011 0.073 -0.039 0.034 0.014 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 1.000 7 0.337 1.000 8 0.034 0.092 1.000 9 0.050 -0.052 0.016 1.000 10 0.009 0.111 0.058 0.156 1.000 11 0.123 0.198 0.006 0.076 -0.083 12 -0.189 -0.012 0.103 -0.029 0.026 13 0.521 0.486 0.106 0.080 0.093 14 -0.012 0.080 0.544 0.029 0.121 15 -0.035 -0.079 -0.053 -0.087 -0.180 16 -0.013 -0.037 0.009 0.179 -0.047 17 0.000 0.016 -0.013 -0.206 -0.010 18 -0.146 -0.105 -0.047 -0.025 -0.013 19 -0.153 0.093 -0.123 -0.137 -0.081 20 0.008 -0.018 -0.211 -0.025 -0.031 21 0.163 -0.077 0.104 0.150 0.056 22 -0.042 -0.127 -0.027 -0.009 -0.075 23 -0.041 -0.036 0.036 -0.069 0.007 24 0.002 -0.039 -0.090 0.060 -0.014 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 1.000 12 -0.034 1.000 13 -0.053 -0.002 1.000 14 -0.022 0.128 0.037 1.000 15 0.008 0.056 -0.024 -0.023 1.000 16 -0.044 0.024 -0.007 0.009 0.160 17 0.018 0.022 -0.021 -0.041 -0.561 18 -0.048 0.063 -0.147 -0.050 0.132 19 0.072 0.031 -0.159 -0.125 -0.005 20 0.020 -0.189 -0.086 -0.218 0.018 21 -0.024 -0.033 0.144 0.104 -0.034 22 -0.116 0.043 -0.003 -0.017 -0.023 23 -0.018 0.081 -0.027 0.040 -0.059 24 -0.020 -0.026 0.010 -0.096 0.041 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 1.000 17 -0.284 1.000 18 -0.070 -0.056 1.000 19 -0.032 0.085 0.113 1.000 20 -0.044 0.043 -0.026 0.039 1.000 21 0.135 -0.084 -0.055 -0.893 -0.043 22 0.010 0.191 -0.578 -0.094 0.016 23 -0.018 0.085 -0.036 0.067 0.280 24 0.067 -0.084 0.031 -0.057 -0.618 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 1.000 22 -0.069 1.000 23 -0.006 -0.071 1.000 24 0.019 0.080 -0.474 1.000
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clear; clf; dt = 1/10000; //Generating a sine wave and then passing it through a quadratic function t = -0.01:dt:0.01; x = sin(200.*(%pi).*t); a = 3; b = 5; y = a.*x + b.*x.*x; subplot(2,1,1); plot(t,x); xgrid(1); xlabel("T", "fontsize", 3); ylabel("X", "fontsize", 3); title("Sine wave: x", "fontsize", 3); subplot(2,1,2); plot(t,y); xgrid(1); xlabel("T", "fontsize", 3); ylabel("X", "fontsize", 3); title("y = ax + bx^2", "fontsize", 3);
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function f = soal_06(z) epsilon0 = 0.885e-12 F = 0.3 Q = 9.4e-6 q = 2.4e-5 R = 0.1 c = 1 - z/sqrt(z^2 + R^2) f = Q*q*c/(2*epsilon0) - F endfunction function F = eval_F(z) epsilon0 = 0.885e-12 Q = 9.4e-6 q = 2.4e-5 R = 0.1 c = 1 - z/sqrt(z^2 + R^2) F = Q*q*c/(2*epsilon0) endfunction function do_plot() z1 = 0.1 z2 = 2.0 Npoints = 100 z = linspace(z1, z2, Npoints) f = zeros(1,Npoints) for i = 1:Npoints f(i) = soal_06(z(i)) printf("%18.10f %18.10f\n", z(i), f(i)) end clf() plot(z, f) xgrid() xs2pdf( gcf(), "soal_06.pdf" ) endfunction // do_plot() exec("bisection.sce", -1) root = bisection( soal_06, 1.4, 1.6 ) printf("At root = %18.10f\n", eval_F(root)) exec("regula_falsi.sce", -1) root = regula_falsi( soal_06, 1.4, 1.6 ) printf("At root = %18.10f\n", eval_F(root)) exec("ridder.sce", -1) root = ridder( soal_06, 1.4, 1.6 ) printf("At root = %18.10f\n", eval_F(root)) if getscilabmode() ~= "STD" quit() end
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//Ex 3.4 clc; clear; close; format('v',5); SR=0.5;//V/micro second Vm=10;//V f=100;//kHz fm=(SR/10^-6)/(2*%pi*Vm);//Hz disp(fm/1000,"Maximum frequency, fm is(kHz)");
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clc; close(); clear(); //page no 379 //prob no. 11.8 B=400*64; //Mb/s mprintf('Memory bus bandwidth = %i Mb/s or %i MB/s',B,B/8);
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// Copyright (C) 2018 - IIT Bombay - FOSSEE // // This file must be used under the terms of the CeCILL. // This source file is licensed as described in the file COPYING, which // you should have received as part of this distribution. The terms // are also available at // http://www.cecill.info/licences/Licence_CeCILL_V2-en.txt // Author:Sonu Sharma, RGIT Mumbai // Organization: FOSSEE, IIT Bombay // Email: toolbox@scilab.in // This is a supporting function function y = prepad(x, n, varargin) if argn(2) > 3 | argn(2) < 2 then error("prepad : wrong number of input argument ") elseif argn(2) == 2 c = 0 ; else c = varargin(1); end y = x; for i = 1:(n-length(x)) y = [c y]; end endfunction
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clear// //Variables VCC = -18.0 //Source voltage (in volts) RC = 4.3 //Collector resistance (in kilo-ohm) RE = 1.0 //Emitter resistance (in kilo-ohm) beta = 200.0 //Common emitter current gain VBE = -0.7 //Emitter-to-Base Voltage (in volts) R1 = 39.0 //Resistance (in kilo-ohm) R2 = 8.2 //Resistance (in kilo-ohm) RL = 3.0 //Load resistance (in kilo-ohm) //Calculation Vth = VCC * R2 / (R1 + R2) //Thevenin's voltage (in volts) Rth = R1 * R2 / (R1 + R2) //Thevenin's equivalent voltage (in volts) IC = (Vth - VBE)/(RE + Rth/beta) //Collector current (in milli-Ampere) IE = -IC //Emitter current (in milli-Amper) r1e = 30.0/IE * 10**-3 //a.ac resistance (in kilo-ohm) Ris = Rth * beta*r1e/(Rth + beta*r1e) //input resistance of the stage (in ohm) rL = RC * RL / (RC + RL) //a.c. load resistance (in kilo-ohm) Av = rL / r1e //Voltage gain //Result printf("\n Voltage gain is %0.1f .",Av)
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//example 12.9 clc; funcprot(0); // Initialization of Variable k1=0.88;//p2/poy; k2=0.628;//poy/pox pox=100;//pressure R=1545; T2=494; k=1.4; M2=0.24; A=2.4; V2=M2*sqroot(k*R/28.97*T2*32.2); mdot=95.9*A*V2/T2/R*28.97; disp(mdot,"mass flow rate in lb/s"); p2=k1*k2*pox; disp(p2,"pressure in lbf/in^2") clear()
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// Exa 1.22 clc; clear; close; // Given data miu_p= 0.05;// in m^2/v-sec rho=5*10^28/10^8;// in /m^3 q= 1.6*10^-19;// in C sigma= q*rho*miu_p;// in (Ωm)^-1 disp(sigma,"The conductivity of silicon material in (Ωm)^-1 is : ")
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clc; //Example 10.3 //Page No 400 disp("Given: For figure, Analog sample voltage of +1.07 V. ;"); //solution a=1.07/1 disp(round(a)); disp("The quantization error is the difference between the original sample voltage and the quantized level, or"); q=1.07-1; disp(q,"Qe = "); disp("From Table 10-2, the PCM code for 1 is 101");
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a=1 theta=0:%pi/64:2*%pi; r=a*(1-cos(theta)) polarplot(theta,r) r=a*sin(theta) polarplot(theta,r)
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// here is an example use of the while statement // which is used for finding the root of a polynomial // which is known to lie within a certain interval. // a is the lower value of the range // b is the upper value of the range a= 0; fa = -%inf; b =3 ; fb = %inf ; while abs(b-a) > %eps*b x = ( a + b ) / 2; fx = x^3 - 2*x - 5 ; if sign(fx) == sign(fa) a=x; fa = fx ; else b = x ; fb = fx; end end disp ( ' The root is :' ); disp (x) ;
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//Ex 6.4 clc;clear;close; format('v',5); Range=0:10;//range //(i)2-bit DAC n=2;//no. of bits step=max(Range)/2^n;//V reading='10';//input in binary Vo=step*bin2dec(reading);//V disp(Vo,"(i) Output Voltage(V)"); //(ii)4-bit DAC n=4;//no. of bits step=max(Range)/2^n;//V reading='0110';//input in binary Vo=step*bin2dec(reading);//V disp(Vo,"(ii) Output Voltage(V)"); //(i)8-bit DAC n=8;//no. of bits step=max(Range)/2^n;//V reading='10111100';//input in binary Vo=step*bin2dec(reading);//V disp(Vo,"(iii) Output Voltage(V)");
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EX20_39.sce
clc;funcprot(0);//EXAMPLE 20.39 // Initialisation of Variables N=10000;.................//Compressor rpm v=660;............//Volume of air delivered in m^3/min p1=1;.................//Inlet pressure in bar t1=293;.............//Inlet temperature in K rp=4;.............//Pressure ratio etaisen=0.82;........//Isentropic efficiency Cf2=62;...............//Flow velocity in m/s rr=2;.............//Ratio of outer radius of impeller to inner radius of impeller ka=0.9;..............//Blade area co efficient fis=0.9;...........//Slip factor cp=1.005;..............//Specific heat at constant pressure in kJ/kgK ga=1.4;............//Ratio of specific heats R=287;..............//Gas constant for air in kJ/kgK //Calculations t21=t1*(rp^((ga-1)/ga));Cf1=Cf2; t2=t1+((t21-t1)/etaisen);..........//Final temperature of air m=(p1*10^5*v/60)/(R*t1);...............//Mass flow rate in m^3/s P=m*cp*(t2-t1);.........//Theoretical power in kW disp(t2,"Final temperature of air in Kevin:") disp(P,"Theoretical power in kW:") Cbl2=sqrt(1000*cp*(t2-t1)/fis); d2=60*Cbl2/(%pi*N);..........//Impeller diameter at outlet in m d1=d2/rr;...............//Impeller diameter at inlet in m disp(d2*100,"Impeller diameter at outlet in cm") disp(d1*100,"Impeller diameter at inlet in cm") b1=(v/60)/(2*%pi*(d1/2)*Cf1*ka);.........//Breadth of impeller at inlet in m disp(b1*100,"Breadth of impeller at inlet in cm:") Cbl1=Cbl2/rr; beta1=(atan(Cf1/Cbl1))*180/%pi; al2=(atan(Cf2/(fis*Cbl2)))*180/%pi; disp(beta1,"Impeller blade angle at inlet in degrees:") disp(al2,"Diffuser blade angle at inlet in degrees")
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Ex1_52.sce
errcatch(-1,"stop");mode(2);//Example 1_52 ; ; //To find diameter of 25th ring dm=0.62 //units in cm ds=0.3 //units in cm d25=2*(dm^2-ds^2)+ds^2 //units in cm^2 d25=sqrt(d25) //units in cm printf("Diameter of 25th ring is %.3f cm",d25) exit();
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Ex5_18.sce
//Variable Declaration x = [0.1 0.4 0.1 ; 0.2 0.2 0] //Calculation p1 = x(2,1)/sum(x(2,1:3)) // for x1=0 & x2=1 p2 = x(2,2)/sum(x(2,1:3)) // for x1=1 & x2=1 p3 = x(2,3)/sum(x(2,1:3)) // for x1=2 & x2=1 // Result printf ( "P(0|1): %.2f , P(1|1): %.2f , P(2|1) : %.2f",p1,p2,p3) printf ( "As P(0|1) is not equal to P(0) (0.5!=0.3) i.e. X1 & X2 are dependent" )
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Ex10_11.sce
clear // // // //Variable declaration epsilon0=8.854*10^-12; //relative permeability(F/m) alphae=10^-40; //dielectric polarizability(Fm^2) N=3*10^28; //number of atoms //Calculations epsilonr=1+(N*alphae/epsilon0); //dielectric constant //Result printf("\n dielectric constant is %e ",epsilonr)
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Ch12Exa6.sci
Thalf1= 2.5*(10^5); //half-life of U-234, y AtomicRatio= 1.8*(10^4); //atomic ratio of u-238 and U-234 in the sample Thalf2= AtomicRatio*Thalf1; //using Eqn12.9 disp(Thalf2,"The half-life of Uranium-238, in years, is: ") //Result // The half-life of Uranium-238, in years, is: // 4.500D+09
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5_16.sce
clear all; clc; disp("Ex 5_16") disp("Free body diagram is as shown in fig 5-30b") //Initilization of variables F=[0,0,-200] //matrix notation //Calculations //Applying summation of forces along all axes and equating them to zero //Applying moment about point X,Y,Z and equating it to zero //We get six equations and we solve for each component //Solving by matrix method to obtain solution A=[1,0,0,0,1;0,1,0,1,0;0,0,1,0,0;0,0,0,2,0;0,0,0,1,-2] B=[0;0;200;200;0] C=inv(A) D=C*B Ax=D(1) Ay=D(2) Az=D(3) Td=D(4) Te=D(5) //Result printf('\n The values are \n') printf('\n Ax=%0.0f \n Ay=%0.0f \n Az=%0.0f \n Td=%0.0f \n Te=%0.0f \n All values are in N',Ax,Ay,Az,Td,Te)
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10_7.sce
//chapter 10 //example 10.7 //page 303 clear all; clc ; //given R1=1 ; R2=3.9;//in Mohm Rl=56; Rs=2.2; Rd=6.8;//in kohm gm=4.5;//mA/V rd=100;//kohm vi=50;//mV rs=200;//ohm Rp=1/gm;//in kohm Zi=1000*(Rs*Rp)/(Rs+Rp); Zo=Rd*rd/(Rd+rd); Av=gm*(Rd*Rl)/(Rd+Rl); vo=Av*vi*Zi/(rs+Zi); printf('\nInput Impedance(Zi)=%d ohm',ceil(Zi)) printf('\nOutput Impedance(Zo)=%.1f kohm',(Zo)) printf('\nVoltage Gain:\nAv=%.1f ',Av) printf('\nOutput Voltage(vo)=%d mV',ceil(vo))
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// This file is part of the materials accompanying the book // "The Elements of Computing Systems" by Nisan and Schocken, // MIT Press. Book site: www.nand2tetris.org // File name: projects/00/And.tst load And.hdl, output-file And.out, compare-to And.cmp, output-list a%B3.1.3 b%B3.1.3 out%B3.1.3; set a 0, set b 0, eval, output; set a 0, set b 1, eval, output; set a 1, set b 0, eval, output; set a 1, set b 1, eval, output;
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Ex2_23.sce
//Force P required to hold the system //Refer to fig. 2.28 (a),(b)&(c) //Applying Lami's Theorem at A we get C=4000*sind(180-60)/sind(60+90-15) //N //Applying equilibrium conditions B P=(-2000*cosd(45)+C*cosd(60))/cosd(15) //N printf("P=%.1f N is required to hold the system in given position.",P)
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4_15.sce
//pathname=get_absolute_file_path('4.15.sce') //filename=pathname+filesep()+'4.15-data.sci' //exec(filename) //Maximum temperature(in K): T1=500+273 //Minimum temperature(in K): T2=200+273 //Temperature of the body(in K): T3=450+273 //Efficiency: n=1-T2/T1 //Ratio of W to Q1: r1=n //COP of pump: COP=T3/(T3-T2) //Ratio of Q3 to W: r2=COP*2/3 //Ratio of Q3 to Q1: r3=r1*r2 printf("\nRESULT\n") printf("\nRatio of heat rejected to body at 450ºC to the heat supplied by the reservoir = %f",r3)
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// File name: projects/07/MemoryAccess/BasicTest/BasicTestVME.tst load BasicTest.vm, output-file BasicTest.out, compare-to BasicTest.cmp, output-list RAM[256]%D1.6.1 RAM[300]%D1.6.1 RAM[401]%D1.6.1 RAM[402]%D1.6.1 RAM[3006]%D1.6.1 RAM[3012]%D1.6.1 RAM[3015]%D1.6.1 RAM[11]%D1.6.1; set sp 256, set local 300, set argument 400, set this 3000, set that 3010, repeat 25 { vmstep; } output;
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Ex1_22.sce
clear // //applying kcl to circuit at node b i3+i4=6-4=2 vb=8 vba=2 //voltage drop across nodes b and a va=6 //potential of node a w.r.t note c i2=3 //applying kcl to node a isa=1 vs=va+2*isa printf("\n vse= %0.1f V",vs)
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//Example 12.1 //Program to Calculate the Gain of a Negative Feedback Amplifier with //Given Specifications clear; clc ; close ; //Given Circuit Data A=100; //Internal Gain B=1/10; //Feedback Factor //Calculation Af=A/(1+A*B); //Displaying The Results in Command Window printf("\n\t The Value of the Gain of Feedback Amplifier is, Af = %f .",Af);
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example3_5.sce
//example 3.5 //calculate Field Capacity clc; //Given d=2; //root zone depth Wc=0.05; //existing water content gammad=15; //dry density of soil gammaw=9.81; //unit weigth of water Vw=500 //water applied to the soil Wl=0.1; //water loss A=1000; //area of plot Vu=Vw*0.9; //volume of water used in soil Wu=Vu*gammaw; //weigth of water used in soil Ws=A*d*gammad; //total dry weigth of soil Wa=Wu*100/Ws; //percent water added Fc=Wc*100+Wa; Fc=round(Fc*100)/100; mprintf("The Field Capacity of soil is=%f percent.",Fc);