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example_23_4.sce
clear; clc; disp("--------------Example 23.4----------------") buffer_size=5000; //bytes recieved_unprocessed = 1000; // bytes rwnd=buffer_size-recieved_unprocessed ; // formula printf("The value of rwnd = %d . Hence Host B can receive only %d bytes of data before overflowing its buffer.",rwnd,rwnd); // display result
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clc clear //Inputs //The Values in the program are as follows: //Temperature in Celcius converted to Kelvin(by adding 273) //Pressure in bar converted to kPa (by multiplying 100) //Volume in m^3 //Value of R,Cp and Cv in kJ/kg Km=1; P1=6; V1=0.01; V2=0.05; P2=2; W1=(((P1+P2)/2)*100)*(V2-V1); printf('The Work done for first cycle: %3.1f kJ',W1); printf('\n'); P3=P2; V3=(P1*V1)/P3; W2=P2*100*(V3-V2); printf('The Work done for second cycle: %3.1f kJ',W2); printf('\n'); W3=(P3*100*V3)*(log(V1/V3)); printf('The Work done for third cycle: %3.2f kJ',W3); printf('\n'); W=W1+W2+W3; printf('The net Work done: %3.2f kJ',W); printf('\n'); Q=W; //As process is cyclic printf('The Heat Transfer: %3.2f kJ',Q); printf('\n');
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animmembrane.sci
curFig = scf(100001); clf(curFig,"reset"); demo_viewCode("membrane.sce"); drawlater(); xselect(); //raise the graphic window // set a new colormap //------------------- cmap= curFig.color_map; //preserve old setting curFig.color_map = jetcolormap(64); //The initial surface definition //---------------------- //Creates and set graphical entities which represent the surface //-------------------------------------------------------------- plot3d1(x,y,u(:,:,1),35,45,' '); s=gce(); //the handle on the surface s.color_flag=1 ; //assign facet color according to Z value title("evolution of a 3d surface","fontsize",3) drawnow(); for i=2:nt realtime(i); //wait till date 0.1*i seconds //s.data.z = (sin((I(i)/10)*x)'*cos((I(i)/10)*y))'; s.data.z = u(:,:,i); end
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//Example 7_7 clc(); clear; //To calculate how large a horizontal force must the pavement exert m=1200 //units in Kg v=8 //units in meters/sec r=9 //units in meters F=(m*v^2)/r //units in Newtons printf("The horizontal force must the pavement exerts is F=%d Newtons",F) //In text book the answer is printed wrong as F=8530 N but the correct answer is 8533 N
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//Tested on Windows 7 Ultimate 32-bit //Chapter 14 Operational Amplifiers Pg no. 435 and 436 clear; clc; //Given //Figure 14.21 R=12D3;//resistances R1,R2,R3 in RC network in ohms C=0.001D-6;//capacitances C1,C2,C3 in RC network in ohms A=29;//gain for oscillator operation //Solution fr=1/(2*%pi*R*C*sqrt(6));//frequency of oscillations in hertz Rf=A*R;//feedback resistance in ohms printf("Frequency of oscillations fr = %.2f kHz\n ",fr/10^3); printf("Feedback resistance Rf = %.f kilo-ohms\n ",Rf/10^3);
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LinggaWahyu/BelajarScilab
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function[fx] = fungsiNormal(mu,sigma,x) konst = 1 / (sigma * sqrt(2 * %pi)) fx = konst * exp(-0.5 * (((x-mu)/sigma)^2)) endfunction function[P_Normal] = PNormal(mu,sigma,a,b) n = 1000 h = (b-a) / n fa = fungsiNormal(mu,sigma,a) fb = fungsiNormal(mu,sigma,b) jum = 0 for i = 1 : (n-1) a = a + h fa1n = fungsiNormal(mu,sigma,a) jum = jum + fa1n end P_Normal = (h/2) * (fa + 2 * jum + fb) endfunction
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// To determine which of the circuits 7-1(a) or 7-2(b) has greater accuracy // example 7-3 in page 166 clc; //Data given V1=495; I1=0.5; // voltmeter and ammeter reading in volt and ampere respectively of circuit 7-1(a) V2=500; I2=0.5;// voltmeter and ammeter reading in volt and ampere respectively of circuit 7-1(b) //calculation printf("R from circuit 7-1(a)=%d ohm\nR from circuit 7-1(b)=%d ohm\n",V1/I1,V2/I2); printf("thus circuit 7-1(a) gives the more accurate result"); //result //R from circuit 7-1(a)=990 ohm //R from circuit 7-1(b)=1000 ohm //thus circuit 7-1(a) gives the more accurate result
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sph2cart.sce
function [x, y, z]=sph2cart(theta, phi, r) x = r.*(cos(phi).*sin(theta)); y = r.*(sin(phi).*sin(theta)); z = r.*cos(theta); endfunction
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nickgreenquist/Intro_To_Intelligent_Systems
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@relation coil2000 @attribute MOSTYPE integer[1,41] @attribute MAANTHUI integer[1,10] @attribute MGEMOMV integer[1,6] @attribute MGEMLEEF integer[1,6] @attribute MOSHOOFD integer[1,10] @attribute MGODRK integer[0,9] @attribute MGODPR integer[0,9] @attribute MGODOV integer[0,5] @attribute MGODGE integer[0,9] @attribute MRELGE integer[0,9] @attribute MRELSA integer[0,7] @attribute MRELOV integer[0,9] @attribute MFALLEEN integer[0,9] @attribute MFGEKIND integer[0,9] @attribute MFWEKIND integer[0,9] @attribute MOPLHOOG integer[0,9] @attribute MOPLMIDD integer[0,9] @attribute MOPLLAAG integer[0,9] @attribute MBERHOOG integer[0,9] @attribute MBERZELF integer[0,5] @attribute MBERBOER integer[0,9] @attribute MBERMIDD integer[0,9] @attribute MBERARBG integer[0,9] @attribute MBERARBO integer[0,9] @attribute MSKA integer[0,9] @attribute MSKB1 integer[0,9] @attribute MSKB2 integer[0,9] @attribute MSKC integer[0,9] @attribute MSKD integer[0,9] @attribute MHHUUR integer[0,9] @attribute MHKOOP integer[0,9] @attribute MAUT1 integer[0,9] @attribute MAUT2 integer[0,9] @attribute MAUT0 integer[0,9] @attribute MZFONDS integer[0,9] @attribute MZPART integer[0,9] @attribute MINKM30 integer[0,9] @attribute MINK3045 integer[0,9] @attribute MINK4575 integer[0,9] @attribute MINK7512 integer[0,9] @attribute MINK123M integer[0,9] @attribute MINKGEM integer[0,9] @attribute MKOOPKLA integer[1,8] @attribute PWAPART integer[0,3] @attribute PWABEDR integer[0,6] @attribute PWALAND integer[0,4] @attribute PPERSAUT integer[0,9] @attribute PBESAUT integer[0,7] @attribute PMOTSCO integer[0,7] @attribute PVRAAUT integer[0,9] @attribute PAANHANG integer[0,5] @attribute PTRACTOR integer[0,7] @attribute PWERKT integer[0,6] @attribute PBROM integer[0,6] @attribute PLEVEN integer[0,9] @attribute PPERSONG integer[0,6] @attribute PGEZONG integer[0,3] @attribute PWAOREG integer[0,7] @attribute PBRAND integer[0,8] @attribute PZEILPL integer[0,3] @attribute PPLEZIER integer[0,6] @attribute PFIETS integer[0,1] @attribute PINBOED integer[0,6] @attribute PBYSTAND integer[0,5] @attribute AWAPART integer[0,2] @attribute AWABEDR integer[0,5] @attribute AWALAND integer[0,1] @attribute APERSAUT integer[0,12] @attribute ABESAUT integer[0,5] @attribute AMOTSCO integer[0,8] @attribute AVRAAUT integer[0,4] @attribute AAANHANG integer[0,3] @attribute ATRACTOR integer[0,6] @attribute AWERKT integer[0,6] @attribute ABROM integer[0,3] @attribute ALEVEN integer[0,8] @attribute APERSONG integer[0,1] @attribute AGEZONG integer[0,1] @attribute AWAOREG integer[0,2] @attribute ABRAND integer[0,7] @attribute AZEILPL integer[0,1] @attribute APLEZIER integer[0,2] @attribute AFIETS integer[0,4] @attribute AINBOED integer[0,2] @attribute ABYSTAND integer[0,2] @attribute CARAVAN{0,1} @inputs MOSTYPE, MAANTHUI, MGEMOMV, MGEMLEEF, MOSHOOFD, MGODRK, MGODPR, MGODOV, MGODGE, MRELGE, MRELSA, MRELOV, MFALLEEN, MFGEKIND, MFWEKIND, MOPLHOOG, MOPLMIDD, MOPLLAAG, MBERHOOG, MBERZELF, MBERBOER, MBERMIDD, MBERARBG, 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//example 6.4 clc; funcprot(0); // Initialization of Variable Vd=28;//V f=100;//frequency I=50;//current //calculation Rl=(Vd-.3)/I; disp(Rl*1000,"load resistance in ohm:") printf('thus pick Rl=560ohm') Rl=560; Vp=2.4; Ib=500;//microAmp Rb=(Vp-.9)/Ib; disp(Rb*1000,"max value of Rb is in kohm:") printf('thus pick Rb=2.2kohm') Vl=Vd-.3; D=.5;//duty cycle Ip=Vl/Rl; disp(Ip*1000,"load current in mA:") Pl=D*Vl*Ip; disp(Pl*1000,"load power in mW:") Pq=D*Ip*.3; disp(Pq*1000,"power delivered in mW:") clear()
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I=20 w=2000 R=200 L=0.25 Xl=w*L*%i Ir=I*Xl/(Xl+R) Il=I-Ir Vl=Xl*Il t=1E-3 ir=sqrt(2)*real(Ir*exp(%i*w*t)) il=sqrt(2)*real(Il*exp(%i*w*t)) vl=sqrt(2)*real(Vl*exp(%i*w*t)) is=sqrt(2)*real(I*exp(%i*w*t)) vs=vl Pr=ir*ir*R Pl=vl*il Ps=is*ir*R Pr=ir*vl disp(Ps,Pl,Pr)
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clear; clc; printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 3.9 Page 145 \n'); //Example 3.9 // Heat conduction through Rod kc = 398; //[W/m.K] From Table A.1, Copper at Temp 335K kal = 180; //[W/m.K] From Table A.1, Aluminium at Temp 335K kst = 14; //[W/m.K] From Table A.1, Stainless Steel at Temp 335K h = 100; //[W/m^2.K] Heat Convection Coeff of Air Tsurr = 25+273; //[K] Temperature of surrounding Air D = 5*10^-3; //[m] Dia of rod To = 100+273.15; //[K] Temp of opposite end of rod //For infintely long fin m = h*P/(k*A) mc = (4*h/(kc*D))^.5; mal = (4*h/(kal*D))^.5; mst = (4*h/(kst*D))^.5; x = linspace(0,.300,100); Tc = Tsurr + (To - Tsurr)*2.73^(-mc*x) - 273; Tal = Tsurr + (To - Tsurr)*2.73^(-mal*x) -273; Tst = Tsurr + (To - Tsurr)*2.73^(-mst*x) -273; clf(); plot(x,Tc,x,Tal,x,Tst); xtitle("Temp vs Distance", "x (m)", "T (degC)"); legend ("Cu", "2024 Al", "316 SS"); //Using eqn 3.80 qfc = (h*%pi*D*kc*%pi/4*D^2)^.5*(To-Tsurr); qfal = (h*%pi*D*kal*%pi/4*D^2)^.5*(To-Tsurr); qfst = (h*%pi*D*kst*%pi/4*D^2)^.5*(To-Tsurr); printf("\n\n (a) Heat rate \n For Copper = %.2f W \n For Aluminium = %.2f W \n For Stainless steel = %.2f W",qfc,qfal,qfst); //Using eqn 3.76 for satisfactory approx Linfc = 2.65/mc; Linfal = 2.65/mal; Linfst = 2.65/mst; printf("\n\n (a) Rods may be assumed to be infinite Long if it is greater than equal to \n For Copper = %.2f m \n For Aluminium = %.2f m \n For Stainless steel = %.2f m",Linfc,Linfal,Linfst); //END
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// 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 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES // EXAMPLE : 3.24 : // Page number 156-157 clear ; clc ; close ; // Clear the work space and console // Given data A = 0.94*exp(%i*1.5*%pi/180) // Constant B = 150.0*exp(%i*67.2*%pi/180) // Constant(ohm) D = A // Constant Y_t = 0.00025*exp(%i*-75.0*%pi/180) // Shunt admittance(mho) Z_t = 100.0*exp(%i*70.0*%pi/180) // Series impedance(ohm) // Calculations C = (A*D-1)/B // Constant(mho) A_0 = A*(1+Y_t*Z_t)+B*Y_t // Constant B_0 = A*Z_t+B // Constant(ohm) C_0 = C*(1+Y_t*Z_t)+D*Y_t // Constant(mho) D_0 = C*Z_t+D // Constant // Results disp("PART II - EXAMPLE : 3.24 : SOLUTION :-") printf("\nA_0 = %.3f∠%.f° ", abs(A_0),phasemag(A_0)) printf("\nB_0 = %.f∠%.1f° ohm", abs(B_0),phasemag(B_0)) printf("\nC_0 = %.6f∠%.1f° mho", abs(C_0),phasemag(C_0)) printf("\nD_0 = %.3f∠%.1f° \n", abs(D_0),phasemag(D_0)) printf("\nNOTE: Changes in obtained answer from that of textbook is due to more precision")
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// Copyright (C) 2015 - IIT Bombay - FOSSEE // // This file must be used under the terms of the CeCILL. // This source file is licensed as described in the file COPYING, which // you should have received as part of this distribution. The terms // are also available at // http://www.cecill.info/licences/Licence_CeCILL_V2-en.txt // Author: Shreyash Sharma // Organization: FOSSEE, IIT Bombay // Email: toolbox@scilab.in function [out]=impixel(image,value1,value2) // This function is used to extract pixel color values. // // Calling Sequence // B = impixel(A,value1,value2) // // Parameters // A: image matrix of the source image. // B : The pixel values in the specified coordinates in the form of a matrix of double type. // value1 : The column c indices of the pixels to extract. // value2 : The row r indices of the pixels to extract. // // Description // impixel(I) returns the value of pixels in the specified image I, where I can be a grayscale, binary, or RGB image. impixel displays the image specified and waits for you to select the pixels in the image using the mouse. If you omit the input arguments, impixel operates on the image in the current axes.P = impixel(I,c,r) returns the values of pixels specified by the row and column vectors r and c. r and c must be equal-length vectors. The kth row of P contains the RGB values for the pixel (r(k),c(k)). // // Examples // i1 = imread('lena.jpeg'); // a = [1 2 3] // b = [1 2 3] // C=impixel(A,a,b) // imshow(i2); // image1=mattolist(image); out=raw_impixel(image1,value1,value2); endfunction;
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//Problem 21.17: A six-pole lap-wound motor is connected to a 250 V d.c. supply. The armature has 500 conductors and a resistance of 1 ohm. The flux per pole is 20 mWb. Calculate (a) the speed and (b) the torque developed when the armature current is 40 A //initializing the variables: p = 1; // let c = 2*p; // for a lap winding Phi = 20E-3; // Wb Z = 500; V = 250; // in Volts Ra = 1; // in ohms Ia = 40; // in Amperes //calculation: //Back e.m.f. E = V - Ia*Ra E = V - Ia*Ra //E.m.f. E = 2*p*Phi*n*Z/c // rearrange, n = E*c/(2*p*Phi*Z) //torque T = E*Ia/(2* n*pi) T = E*Ia/(2*n*%pi) printf("\n\n Result \n\n") printf("\n (a)speed n is %.0f rev/sec ",n) printf("\n (b)the torque exerted is %.2f Nm ",T)
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clc clear //Initialization of variables g=32.2 //ft/s^2 p1=100 //psig p2=29.0 //in of Hg //calculations BP=p2*0.491 AP=BP+p1 //results printf("Absolute pressure = %.2f psia",AP)
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clear; chdir('J:\Stage\Simulations') getf('Modele un facteur gaussien\Prime de risque\VarPrime_Function.sci');disp('getf done'); getf('Modele un facteur gaussien\Cas Constant\Const_Function.sci');disp('getf done'); /////////////////////////////////////////////////////////////////////////////////////////////// ///////////////////////////// Tendance Variable//////////////////////////////////////////////// /////////////////////////////////////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////// // donnees //////////////////////////////////////////////////////// // f(0,t) la courbe forward en 0 ou plutot log(f(0,t)) // sigma = [100%,400%] annuelle // h =1 heure, 1 jour // lambda = [1,25] jours ou [10,200] annee /////////////////////////////////////////////////////// //simulation heure par heure lambda = 100; h = 1/(24*365); sigma = 4; prime = 20; lnF0 = 0; // 1: simulation et prime de risque constante disp([lambda sigma prime],'vrais valeurs de lambda, sigma et prime pure'); ForwardFile = "Data\EDF_Forward_01012002_31122002.txt" ; lnForward = log((read(ForwardFile,-1,1))'); n = length(lnForward)-1; lnForwardEdf = lnForward(2:(n+1)); Xt_VarEdf = zeros(1,n); X0 = 0; lnF0 = lnForward(1); disp(h,'le pas de temps h'); disp(n,'le nombre d observation'); lnForwardPrime = lnForwardEdf + (sigma^2*prime/lambda); lnF0Prime = lnF0 + (sigma^2*prime/lambda); Xt_VarEdf = Simul_Xt_Var(n,h,lambda,sigma,lnForwardPrime,lnF0Prime); //2: Calage par EMV Theta_EMV = Calage_Xt_VarPrime(n,h,Xt_VarEdf,X0,lnForwardEdf,lnF0); disp(Theta_EMV,'Les parametres : lambda, sigma prime pure estime par EMV estime'); xbasc(0); xset("window",0); xsetech([0,1/4,1/3,1/2],[-1,1,-1,1]); plot2d(1:n,Xt_VarEdf,1:3,"061","",[0,0,20,20]); xtitle('Evolution du logarithme du prix Xt'); xsetech([1/3,1/4,1/3,1/2],[-1,1,-1,1]); plot2d(1:n,exp(Xt_VarEdf),1:3,"061","",[0,0,20,20]); xtitle('Evolution du prix spot St'); xsetech([2/3,1/4,1/3,1/2],[-1,1,-1,1]); plot2d(1:n,lnForwardEdf,1:3,"061","",[0,0,20,20]); xtitle('Evolution du prix forward F(t,T)'); //3: autocorrelation des residus et test de box-pierce alpha = 0.05; [Resid, Autocorr, BP_Test, Probac, JB_Test, S, K, Proban] = Residu_VarPrime(n,h,Theta_EMV,Xt_VarEdf,X0,lnForwardEdf,lnF0,alpha); xbasc(1); xset("window",1); xsetech([0,1/4,0.5,1/2],[-1,1,-1,1]); histplot(100,Resid,1:3,"061","",[0,0,20,20]); xtitle('Histogramme des residus'); xsetech([0.5,1/4,0.5,1/2],[-1,1,-1,1]); plot2d(1:length(Autocorr),Autocorr,1:3,"061","",[0,0,20,20]); xtitle('La fonction d autocorrelation en fonction du retard'); disp(Probac,'La probabilite d absence de autocorrelation'); disp(BP_Test,'Test Box-Pierce 1:absence d autocorrelation, 0:autocorrelation'); disp(Proban,'La probabilite de normalite'); disp(JB_Test,'Test Jarque Bera 1:normalite des residus, 0:rejet'); disp([S K],'Le skewness et le kurtosis');
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clc //initialisation of variables clear vo= 10 //ft/sec a= 0.5 //ft^-1 b= 1 //ft x= -2 //ft y= 2 //ft b1= 2 a1= 3/5 //ft //CALCULATIONS Vx= vo/(a*x^2+b) Vy= -2*a*b*vo*x*y/(a*x^2+b)^2 V= sqrt(Vx^2+Vy^2) fx= -2*a*b^2*vo^2*x/(a*x^2+b)^3 fy= 2*a*b^2*vo^2*y*(b-a*x^2)/(a*x^2+b)^4 f= sqrt(fx^2+fy^2) r= b1^2/a1 f1= f*r //RESULTS printf ('Vx = %.2f ft/sec',Vx) printf ('\n Vx = %.2f ft/sec',Vy) printf ('\n V = %.2f ft/sec',V) printf ('\n fx = %.2f ft/sec^2',fx) printf ('\n fy = %.2f ft/sec^2',fy) printf ('\n f = %.2f ft/sec^2',f) printf ('\n r = %.2f in the present case',r) printf ('\n f1 = %.2f ft/sec^2',f1)
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//Page Number: 653 //Example 12.8 clc; //Given d=2.4;//cm lmbc=1.8; c=3*10^10; //cm/s lmbg=2*d; lmb=(lmbg*lmbc)/(sqrt(lmbg^2+lmbc^2)); //Operating frequency f=c/lmb; disp('GHz',f/10^9,'Operating frequency:');
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testcalcTypeII.sce
//Script to test the calculation of the type II based on Bassos book sec 4.2.9 mag=10^(-18/20); boost=68; fc=5e3; //Result must be fp=25.7e3, fp0=7.8e3, fz=972 [regulator,fz1,fp1,fp0]= calctypeII(mag,fc,boost) bode(regulator,fc/103,fc*100,0.1)
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global file_name path fname extension clk__x1 = [0.2 2.5]; data_x1 = [2.5 2.5]; zero_x1 = [0.2 0.2]; clk_sr = [clk__x1 clk__x1 clk__x1 linspace(0.2,0.2,21)]; data_sr = [data_x1 zero_x1 zero_x1 linspace(0.2,0.2,21)]; input_vmmwta=[ zero_x1 zero_x1 zero_x1 linspace(2.5,2.5,21); // "1" zero_x1 zero_x1 zero_x1 linspace(2.1,2.1,21); // "x1" zero_x1 zero_x1 zero_x1 linspace(2.5,2.5,21); // "x2" zero_x1 zero_x1 zero_x1 linspace(0.2,0.2,21);]; exec('/home/ubuntu/rasp30/sci2blif/sci2blif_test2.sce', -1); exec('/home/ubuntu/rasp30/prog_assembly/libs/scilab_code/dc_setup_gui.sce', -1); x1=2.1; clear xor_result; for i_xor=1:21 input_vmmwta=[ zero_x1 zero_x1 zero_x1 linspace(2.5,2.5,21); // "1" zero_x1 zero_x1 zero_x1 linspace(x1,x1,21); // "x1" zero_x1 zero_x1 zero_x1 linspace(2.1,2.5,21); // "x2" zero_x1 zero_x1 zero_x1 linspace(0.2,0.2,21);]; exec('/home/ubuntu/rasp30/sci2blif/sci2blif_test2.sce', -1); exec('/home/ubuntu/rasp30/prog_assembly/libs/scilab_code/voltage_measurement_gui.sce', -1); temp=fscanfMat("test_20160517_xor_with_mismatchmap.data"); xor_result(:,i_xor)=temp(7:27,8); x1=x1+0.02; disp("Loop "+string(i_xor)+" is done"); end csvWrite(xor_result,"XOR_hyperplane_data"); x1_x2=[linspace(2.1,2.5,21)' linspace(2.1,2.5,21)'] xor_result=csvRead("XOR_hyperplane_data"); scf(1);clf(1); [xx,yy,zz]=genfac3d(x1_x2(:,1),x1_x2(:,2),xor_result-0.6); plot3d(xx,yy,list(zz, zz)) e=gce(); f=e.data; TL = tlist(["3d" "x" "y" "z" "color"],f.x,f.y,f.z,f.z+3.3); // random color matrix e.data = TL; clf(); plot3d(xx,yy,list(zz, zz)); h=gce(); h.color_flag=1; //color according to z f=gcf(); f.color_map = graycolormap(512)*5; h.data = TL; a = gca(); a.data_bounds=[2.2 2.2 0; 2.5 2.5 2]; a.rotation_angles=[0,270];
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//clear// clear; clc; //Example 22.2 //Given Dp = 1; //[in.] vdot = 25000; //[ft^3/h] T = 68; //[F] P = 1; //[atm] ya = 0.02; Mair = 29; Mg = 17; //Solution //The average molecular weiht of the entering gas M = (1-ya)*Mair+ya*Mg; rho_y = M*492/(359*(460+68)); //[lb/ft^3] rho_x = 62.3; //[lb/ft^3] //(a) //Using Fig.(22.8), from Example 22.1 A = Gx/Gy = 1 and //Let A = 1; B = A*sqrt(rho_y/rho_x); //Form Fig 22.8, the superficial vapor velocity at flooding //is uof*sqrt(rho_y/(rho_x-rho_y))=0.11, therefore uof = 0.11/sqrt(rho_y/(rho_x-rho_y)); //[m/s] //The allowable vapor velocity uo = uof*0.5; //[m/s] uo = uo*3.28; //[ft/s] //the corresponding mass velocity Gy = uo*rho_y; //[lb/ft^2-s] //The allowable mass velocity in the example was 0.236 lb/ft^2-s. //The increase by using structured packing is increase = (Gy/0.236)-1; disp(increase*100,'The percent increase in mass velocity is'); //(b) //The pressure drop delta_P = 20*1.22*(0.5/0.9)^1.8; //[in. H2O] //This is 1.2 times the pressure drop of 7 in.H2O in the Intolax saddles. disp('The pressure drop will be greater than Intolax Saddles')
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// circuito RL passa-faixa em dB R1=1000;R2=220;L1=1e-3;L2=1e-3; fc1=R1/(2*%pi*L1); fc2=R2/(2*%pi*L2); f = logspace(1,10,1e4); mod_H1=1./((1+(fc1./f).^2).^0.5); mod_H2=1./((1+(f./fc2).^2).^0.5); mod_H=mod_H1.*mod_H2; ang_H1=(180/%pi)*(atan(fc1./f)); ang_H2=-(180/%pi)*(atan(f./fc2)); ang_H=ang_H1+ang_H2; scf(3); clf(3); subplot(2,1,1) plot('ln',f,20*log10(mod_H1),'r--','LineWidth',3) plot('ln',f,20*log10(mod_H2),'b--','LineWidth',3) plot('ln',f,20*log10(mod_H),'k-','LineWidth',3) plot('ln',f,20*log10(1/sqrt(2)*f./f),'k--','LineWidth',1) xlabel "$f(Hz)$" fontsize 5 ylabel "$módulo_{dB}$" fontsize 5 legend(['H1(jw)';'H2(jw)';'H(jw)'],-1); set(gca (),'font_size',3) subplot(2,1,2) plot('ln',f,ang_H1,'r--','LineWidth',3) plot('ln',f,ang_H2,'b--','LineWidth',3) plot('ln',f,ang_H,'k-','LineWidth',3) xlabel "$f(Hz)$" fontsize 5 ylabel "$Fase(º)$" fontsize 5 legend(['<H1(jw)';'<H2(jw)';'<H(jw)'],-1); set(gca (),'font_size',3)
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; asinf.tst ; ; Copyright 2009-2023, Arm Limited. ; SPDX-License-Identifier: MIT OR Apache-2.0 WITH LLVM-exception func=asinf op1=7fc00001 result=7fc00001 errno=0 func=asinf op1=ffc00001 result=7fc00001 errno=0 func=asinf op1=7f800001 result=7fc00001 errno=0 status=i func=asinf op1=ff800001 result=7fc00001 errno=0 status=i func=asinf op1=7f800000 result=7fc00001 errno=EDOM status=i func=asinf op1=ff800000 result=7fc00001 errno=EDOM status=i func=asinf op1=00000000 result=00000000 errno=0 func=asinf op1=80000000 result=80000000 errno=0 ; Inconsistent behavior was detected for the following 2 cases. ; No exception is raised with certain versions of glibc. Functions ; approximated by x near zero may not generate/implement flops and ; thus may not raise exceptions. func=asinf op1=00000001 result=00000001 errno=0 maybestatus=ux func=asinf op1=80000001 result=80000001 errno=0 maybestatus=ux func=asinf op1=3f800000 result=3fc90fda.a22 errno=0 func=asinf op1=bf800000 result=bfc90fda.a22 errno=0 func=asinf op1=3f800001 result=7fc00001 errno=EDOM status=i func=asinf op1=bf800001 result=7fc00001 errno=EDOM status=i
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// Inverting and non inverting amplifier clc; clear all; close; m = 200000; vi = input("Enter input voltage= ") rf =input("Enter feedback resistor value= ") r1 =input("Enter input resistor value= ") vo = (-rf/r1)*vi; // vo = (1+(rf/r1)*vi; //for non inverting amp //vo = vo-(vo/m); // correction for voltage difference parameter disp("Output voltage is ",vo)
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clc clear //input data D=8//Outer diameter of the turbine in m Db=3//Inner diameter of the turbine in m P=30000//Power developed by the turbine in kW nH=0.95//Hydraulic efficiency N=80//Speed of the turbine in rpm H=12//Head operated by the turbine in m Q=300//Discharge through the runner in m^3/s g=9.81//Acceleration due to gravity in m/s^2 dw=1000//Density of water in kg/m^3 //calculations U1=(3.1415*D*N)/60//Runner tip speed at inlet in m/s U2=U1//Runner tip speed at outlet in m/s as flow is axial Cr1=Q/((3.1415/4)*(D^2-Db^2))//Flow velocity at inlet in m/s Cr2=Cr1//Flow velocity at outlet in m/s as flow is axial b22=atand(Cr2/U2)//The angle of the runner blade at outlet in degree Cx1=(nH*g*H)/U1//Velocity of whirl at inlet in m/s b11=180-(atand(Cr1/(U1-Cx1)))//The angle of the runner blade at inlet in degree nM=(P*10^3)/(dw*g*Q*(Cx1*U1/g))//Mechanical efficiency n0=nM*nH//Overall efficiency //output printf('(a)Blade angle at\n inlet is %3.2f degree\n outlet is %3.2f degree\n(b)Mechanical efficiency is %3.3f\n(c)Overall efficiency is %3.3f',b11,b22,nM,n0)
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//Ex 2.9.2 clc;clear;close; format('v',6); //Given : IF=10;//mA VF=0.3;//volts T=27+273;//K Eta=1;//for Ge diode VT=T/11600;//V Io=IF/(exp(VF/Eta/VT)-1);//mA disp(Io*10^6,"Reverse saturation current in nA : ");
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errcatch(-1,"stop");mode(2);//332 ; ; A=[-1 2 2]'; disp(A,'A='); [U diagnol V]=svd(A); disp(U,'U='); disp(diagnol,'diagnol='); disp(V','V''='); disp(U*diagnol*V','A=U*diagnol*V''') //end exit();
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// Example 10.7 format('v',6) clc; clear; close; // given data R_C= 1*10^3;// in Ω r_desh_e= 2.5;//in Ω Zin= 1*10^3;// in Ω A2= 10;// unit less A3= 1;// unit less A1= (R_C*Zin/(R_C+Zin))/r_desh_e;// unit less // The overall voltage gain A= A1*A2*A3; disp(A,"The overall voltage gain is : ")
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// Test # 1 : No Input Arguments exec('./iirpowcomp.sci',-1); [b,p]=iirpowcomp(); //!--error 10000 //2/3 input arguments allowed //at line 33 of function iirpowcomp called by : //[b,p]=iirpowcomp()
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//Page Number: 250 //Example 4.24 clc; //Given //As it is perfectly matched S12=1/sqrt(2); S21=S12; s=[0 S12;S21 0]; disp(s,'Scattering matrix:');
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towers_of_hanoi.tst
# towers_of_hanoi.tst >>> import sys >>> import pyke >>> import os >>> new_path = os.path.join(os.path.dirname(os.path.dirname(pyke.__file__)), ... 'examples/towers_of_hanoi') >>> sys.path.append(new_path) >>> import driver >>> driver.test(1) got 1: ((0, 2),) >>> driver.test(2) got 1: ((0, 1), (0, 2), (1, 2)) got 2: ((0, 2), (0, 1), (2, 0), (1, 2), (0, 2)) >>> driver.test(3) got 1: ((0, 1), (0, 2), (1, 0), (2, 1), (0, 1), (0, 2), (1, 0), (1, 2), (0, 2)) got 2: ((0, 1), (0, 2), (1, 0), (2, 1), (0, 1), (0, 2), (1, 2), (1, 0), (2, 1), (0, 2), (1, 2)) got 3: ((0, 1), (0, 2), (1, 2), (0, 1), (2, 0), (2, 1), (0, 2), (1, 0), (2, 0), (1, 2), (0, 1), (0, 2), (1, 2)) got 4: ((0, 1), (0, 2), (1, 2), (0, 1), (2, 0), (2, 1), (0, 2), (1, 0), (2, 0), (1, 2), (0, 2), (0, 1), (2, 0), (1, 2), (0, 2)) got 5: ((0, 1), (0, 2), (1, 2), (0, 1), (2, 1), (2, 0), (1, 0), (1, 2), (0, 1), (0, 2), (1, 2)) got 6: ((0, 1), (0, 2), (1, 2), (0, 1), (2, 1), (2, 0), (1, 0), (1, 2), (0, 2), (0, 1), (2, 0), (1, 2), (0, 2)) got 7: ((0, 2), (0, 1), (2, 0), (1, 2), (0, 2), (0, 1), (2, 0), (2, 1), (0, 2), (1, 0), (2, 0), (1, 2), (0, 1), (0, 2), (1, 2)) got 8: ((0, 2), (0, 1), (2, 0), (1, 2), (0, 2), (0, 1), (2, 0), (2, 1), (0, 2), (1, 0), (2, 0), (1, 2), (0, 2), (0, 1), (2, 0), (1, 2), (0, 2)) got 9: ((0, 2), (0, 1), (2, 0), (1, 2), (0, 2), (0, 1), (2, 1), (2, 0), (1, 0), (1, 2), (0, 1), (0, 2), (1, 2)) got 10: ((0, 2), (0, 1), (2, 0), (1, 2), (0, 2), (0, 1), (2, 1), (2, 0), (1, 0), (1, 2), (0, 2), (0, 1), (2, 0), (1, 2), (0, 2)) got 11: ((0, 2), (0, 1), (2, 1), (0, 2), (1, 0), (1, 2), (0, 2)) got 12: ((0, 2), (0, 1), (2, 1), (0, 2), (1, 2), (1, 0), (2, 1), (0, 2), (1, 2))
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11_7.sce
//Example 11.7 //Relaxation Method //Page no. 376 clc;clear;close; for i=0:4 for j=0:4 if i==0 | j==0 then U(5-i,j+1)=0 elseif i==4 | j==4 U(5-i,j+1)=(i*j)^2 else U(5-i,j+1)=0; end end end S=['A','B','C','D','E','F','G','H','I'] disp(U) deff('y=d(i,j)','y=(U(i-1,j-1)+U(i+1,j+1)+U(i-1,j+1)+U(i+1,j-1))/4') //diagonal 5 point formula deff('y=s(i,j,l)','y=(U(i-l,j)+U(i+l,j)+U(i,j-l)+U(i,j+l))/4') //std 5 point formula U(3,3)=s(3,3,2); for k=0:0 p=3; for i=2:4 for j=2:4 if k==0 & (i==3 & j==3) then printf('\n U %s(%i) = %g\n',S(i+j-p),k,U(i,j)) continue end if k==0 & i==4 & j==2 then U(i,j)=d(i,j) else U(i,j)=s(i,j,1) end if k==0 then printf('\n U %s = %g\n',S(i+j-p),U(i,j)) else printf('\n U %s(%i) = %g\n',S(i+j-p),k,U(i,j)) end end p=p-2; end printf('\n\n') end printf('\nHence the solution is : \n\n') p=3; for i=2:4 for j=2:4 printf(' U%s = %.3f, ',S(i+j-p),U(i,j)) end p=p-2 end
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clc //initialisation of variables o= 90 //degrees H= 15.5 //in Cd= 0.6 g= 32.2 //ft/sec^2 //CALCULATIONS Q= 8*Cd*tand(o/2)*sqrt(2*g)*(H/12)^2.5/15 //RESULTS printf ('Total Discharge= %.2f cuses',Q)
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levinson2.sce
//check o/p for a given matrix r=[1 34 4]; [a,e,k]=levinson(r); disp(a); ////output /// 1. - 0.0883117 - 0.9974026
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stacksize('max'); t = read("t", -1, 1); tamount = size(t,1)*1; t = t(1:tamount); tamount = size(t,1); x = read("x", -1, 1); xamount = size(x,1); re=read("re", xamount, tamount); im=read("im", xamount, tamount); v = re+%i*im; xx = 1:xamount; tt = 1:tamount; xx = (x ./ (60*60)); tt = (t ./ (60*60)); clf; f = gcf(); f.color_map = graycolormap(512); grayplot((xx), (tt), ((im)) ); abort; plot((tt),sum(abs(real(v)), 'r'),'k'); plot((tt),sum(abs(imag(v)), 'r'),'b'); plot((tt),sum(abs( abs(v)), 'r'),'r'); abort; plot(tt, abs((v(6,:))),'r'); plot(tt, abs((v(7,:))),'r'); //plot(xx, abs((v(:,740))),'r'); //vr = v_ - v_10;
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Ch05Exa10.sce
// Scilab code Exa5.10 : : Page 206 (2011) clc; clear; h_kt = 1.05457e-34; // Reduced Planck's constant, joule sec e = 1.60218e-19; // Charge of an electron, coulomb l = 2; // Orbital angular momentum eps_0 = 8.5542e-12; // Absolute permittivity of free space, coulomb square per newton per metre square Z_D = 90; // Atomic number of daughter nucleus m = 6.644e-27; // Mass of alpha particle, Kg R = 8.627e-15; // Radius of daughter nucleus, metre T1_by_T0 = exp(2*l*(l+1)*h_kt/e*sqrt(%pi*eps_0/(Z_D*m*R))); // Hindrance factor printf("\nThe hindrance factor for alpha particle = %5.3f" ,T1_by_T0); // Result // The hindrance factor for alpha particle = 1.768
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35_02.sce
//Problem 35.02: If the load impedance Z in Figure 35.2 of problem 35.01 consists of variable resistance R and variable reactance X, determine (a) the value of Z that results in maximum power transfer, and (b) the value of the maximum power. //initializing the variables: rv = 120; // in volts thetav = 0; // in degrees Z = 15 + %i*20; // in ohm //calculation: //voltage V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180) //maximum power transfer occurs when X = -1*imag(Z) and R = real(Z) z = real(Z) - %i*imag(Z) //Total circuit impedance at maximum power transfer condition, ZT = Z + z //Current I flowing in the load is given by I = V/ZT Imag = (real(I)^2 + imag(I)^2)^0.5 //maximum power delivered P = real(Z)*I^2 printf("\n\n Result \n\n") printf("\n (a)maximum power transfer occurs when Z is %.0f + (%.0f)i ohm",real(z), imag(z)) printf("\n (b) maximum power delivered is %.0f W",P)
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ex6_4.sce
// Exa 6.4 clc; clear; close; // Given data V_DD = 15;// in V R = 3;// in kohm I_D = V_DD/R;// in mA R_D = 1;// in kohm V_D = V_DD - (I_D*R_D);// in V disp(V_D,"The drain voltage in V is");
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Ex7_15.sce
//=========================================================================== //chapter 7 example 15 clc; clear all; //variable declaration W1 = 300; //wattmeter reading in kW W2 = 100; //wattmeter reading in kW //calculations P = W1+W2; //input power in kW phi = atan(((W1-W2)/(W1+W2))*sqrt(3)); //phase angle in radians phi1 = (phi*180)/%pi; pf =cos((phi1*%pi)/180); //power factor lagging //result mprintf("input power = %3.2f kW",P); mprintf("power factor = %3.3f lagging",pf);
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12_15.sce
//To find axial thrust clc //Given: L=175/1000, d2=100/1000, r2=d2/2 //m theta=70 //degrees G=1.5, T2=80 Tf=75 //Torque on faster wheel, N-m funcprot(0) //Solution: //Spiral angles for each wheel: //Calculating the number of teeth on slower wheel T1=T2*G //Calculating the pitch circle diameter of the slower wheel d1=(L*2)-d2 //m //Calculating the spiral angles //We have, d2/d1 = (T2*cos(alpha1))/(T1*cos(alpha2)), or T2*d1*cos(alpha1)-T1*d2*cos(alpha2) = 0 .....(i) //Also, alpha1+alpha2 = theta, or alpha1+alpha2-theta = 0 .....(ii) function y=f(x) alpha1=x(1) alpha2=x(2) y(1)=T2*d1*cos(alpha1)-T1*d2*cos(alpha2) y(2)=alpha1+alpha2-theta*%pi/180 endfunction z=fsolve([1,1],f) alpha1=z(1)*180/%pi //Spiral angle for slower wheel, degrees alpha2=z(2)*180/%pi //Spiral angle for faster wheel, degrees //Axial thrust on each shaft: //Calculating the tangential force at faster wheel F2=Tf/r2 //N //Calculating the normal reaction at the point of contact RN=F2/cosd(alpha2) //N //Calculating the axial thrust on the shaft of slower wheel Fa1=RN*sind(alpha1) //N //Calculating the axial thrust on the shaft of faster wheel Fa2=RN*sind(alpha2) //N //Results: printf("\n\n Spiral angle for slower wheel, alpha1 = %.2f degrees.\n\n",alpha1) printf(" Spiral angle for faster wheel, alpha2 = %.2f degrees.\n\n",alpha2) printf(" Axial thrust on the shaft of slower wheel, Fa1= %d N.\n\n",Fa1+1) printf(" Axial thrust on the shaft of faster wheel, Fa2 = %d N.\n\n",Fa2+1)
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//Section-10,Example-3,Page no.-CT.31 //To calculate q,W,dl_E,dl_H. clc; V_2=20 V_1=4 P=1 W=-(P*(V_2-V_1))*(8.314/0.08206) disp(W,'maximum work done in(J)') dl_E=0 dl_H=0 q=dl_E-W disp(q,'in J')
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EXP5_5.sce
clc;funcprot(0);//EXAMPLE 5.5 // Initialisation of Variables N=1;..........//N0. of atoms on one side of iron bar H=1;..........//No. of atoms onother side of iron bar d=3;.......//Diameter of an impermeable cylinder in cm l=10;.....//Length of an impermeable cylinder in cm A1=50*10^18*N;..........// No. of gaseous Atoms per cm^3 on one side A2=50*10^18*H;..........//No. of gaseous Atom per cm^3 on one side B1=1*10^18*N;...........//No. of gaseous atoms per cm^3 on another side B2=1*10^18*H;..........//No. of gaseous atoms per cm^3 on another side t=973;...........//The di¤usion coefficient of nitrogen in BCC iron at 700 degree celsius in K Q=18300;.........//The activation energy for di¤usion of Ceramic Do=0.0047;.......//The pre-exponential term of ceramic R=1.987;.........//Gas constant in cal/mol.K //CALCULATIONS T=A1*(%pi/4)*d^2*l;....//The total number of nitrogen atoms in the container in N atoms LN=0.01*T/3600;......//The maximum number of atoms to be lost per second in N atoms per Second JN=LN/((%pi/4)*d^2);.........//The Flux of ceramic in Natoms per cm^2. sec. Dn=Do*exp(-Q/(R*t));........//The di¤usion coefficient of Ceramic in cm^2/Sec deltaX=Dn*(A1-B1)/JN;.........//minimum thickness of the membrane in cm LH=0.90*T/3600;........//Hydrogen atom loss per sec. JH=LH/((%pi/4)*d^2);.........//The Flux of ceramic in Hatoms per cm^2. sec. Dh=Do*exp(-Q/(R*t));........//The di¤usion coeficient of Ceramic in cm^2/Sec deltaX2=((1.86*10^-4)*(A2-B2))/JH;.......//Minimum thickness of the membrane in cm disp(deltaX,"Minimum thickness of the membrane of Natoms in cm") disp(deltaX2,"Minimum thickness of the membrane of Hatoms in cm")
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clear; clc(); // To find the temperature of planes indicated by grid points using relaxation method t1=800; // inner surface temperature of wall in degF t4=200; // outer surface temperature of wall in degF //Grids are square in shape so delx =dely where delx,y sre dimensions of square grid t2=[700 550 550 587.5 587.5 596.9 596.9 599.3 599.3 599.8]; // Assumed temperature of grid point 1 t3=[300 300 375 375 393.8 393.8 398.5 398.5 399.6 399.6]; // Assumed temperature of grid point 2 for i=1:9 th2(i)=t1+t3(i)-2*t2(i);; // th1= q/kz at grid pt1 th3(i)=t2(i)+t4-2*t3(i);// th2= q/kz at grid pt2 printf("\n Assuming t2=%.1f degF and t2=%.1f degF \n th1[%d]=%.1f degF and th2[%d]=%.1f degF \n",t2(i),t3(i),i,th2(i),i,th3(i)); printf(" Since th2[%d] is not equal to th3[%d], hence other values of t2 and t3 are to be assumed\n",i,i); end printf("\nAssuming t2=600 degF and t3=400 degF, th2=th3."); printf("\nHence Steady state condition is satisfied at grid temperatures of 400 degF and 600 degF");
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Ch06Ex11.sci
// Scilab Code Ex6.11 Time required for carburizing of steel: Page 209 (2010) C0 = 0.0018; // Intial carbon concentration of steel Cx = 0.0030; // Carbon concentration of steel at 0.60 mm below the surface of the gear Cs = 0.01; // Carbon concentration of steel at the surface x = 0.6e-03; // Diffusion depth below the surface of the gear, m D_927 = 1.28e-011; // Diffusion coefficient for carbon in iron, metre square per sec erf_Z = (Cs-Cx)/(Cs-C0); // Error function of Z as a solution to Fick's second law Z1 = 1.0, Z2 = 1.1; // Preceding and succeeding values about Z from error function table erf_Z1 = 0.8427, erf_Z2 = 0.8802; // Preceding and succeeding values about erf_Z from error function table Z = poly(0,'Z'); Z = roots((Z-Z1)/(Z2-Z1)-(erf_Z-erf_Z1)/(erf_Z2-erf_Z1)); // As Z = x/(2*sqrt(D_927*t)), where Z is a constant argument of error function as erf(Z) // Solving for t, we have t = (x/(2*Z))^2/D_927; // Time necessary to increase the carbon content of steel, sec printf("\nThe time necessary to increase the carbon content of steel = %3d minutes", t/60); // Result // The time necessary to increase the carbon content of steel = 110 minutes
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clc //initialisation of variables P= 15 //psia sx= 1.7050 //Btu/lb R sg= 1.7549 //btu/lb R sfg= 1.4415 //Bru/lb R hg= 1150.8 //btu/lb hfg= 969.7 //Btu/lb vg= 26.29 //cu ft/lb vfg= 26.27 //cu ft/lb //CALCULATIONS n= (sg-sx)/sfg sx= sg-n*sfg hx= hg-n*hfg vx= vg-n*vfg //RESULTS printf ('Volume= %.2f cu ft/lb',vx) printf (' \n Entropy = %.2f Btu/lb R',sx) printf (' \n Enthalpy= %.1f Btu/lb',hx)
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//Example 4.27 //Interconnectiuion of LTI systems n2=0:18; h1=[1 5 10 11 8 4 1]; h2=[1 1 zeros(1,5)]; h3=[1 1 zeros(1,5)]; a=convol(h1,h2); h=convol(a,h3); x=[1 -1]; n1=[0 1]; n3=0:19; y=convol(x,h); subplot(3,1,1) xtitle("input signal x(n)","....................n","x[n]"); plot(n1,x,'.'); subplot(3,1,2) xtitle("system response h(n)","....................n","h[n]"); plot(n2,h,'.'); subplot(3,1,3) xtitle("output signal y(n)",".............................n","y[n]"); plot(n3,y,'.');
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//Example 10.9// T1=290;//degree C //recrystallization temperature T2=920;// degree C //solidus temperature T3=273;//K //Kelvin T4=(T1+T3)/(T2+T3) disp(T4)
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//scilab 5.4.1 //windows 7 operating system //chapter 5:Semiconductor Junction Diodes clc clear Vz=3//Vz=breakdown voltage of zener diode Vi=12//Vi=input voltage V=[12;-3]//V=[Vi:-Vz] R1=1000 R2=1000 R3=500//R1,R2,R3=resistances R=[R1+R2 -R2;-R2 R2+R3] I1=inv(R)*V//solving this matrix on the basis of application of KCL & KVL,we get the values of branch currents I & Iz as I1=[I;Iz] disp("A",I1(1),"I=") disp("A",I1(2),"Iz=") Pz=Vz*I1(2)//Pz=power dissipated in zener diode disp("W",Pz,"Pz=") disp("Power dissipated does not exceed the maximum power limit of 20mW")
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//Chapter 14 //Example 14_11 //Page 372 clear;clc; v=400; ph_v=230; w1=100; w2=150; r1=ph_v^2/w1; r2=ph_v^2/w2; i=v/(r1+r2); v1=i*r1; v2=i*r2; printf("Resistance of lamp L1 = R1 = %.2f ohm \n\n", r1); printf("Resistance of lamp L2 = R2 = %.2f ohm \n\n", r2); printf("Curretn through lamps = %.3f A \n\n", i); printf("Voltage across lamp L1 = V1 = %.0f V \n\n", v1); printf("Voltage across lamp L2 = V2 = %.0f V \n\n", v2);
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clc; clear all; clf; //UNIT IMPULSE x=[zeros(1,10),ones(1,1),zeros(1,10)]; t=-10:1:10; subplot(2,3,1) plot(t,x); title('unit impulse');xlabel('t');ylabel('amp'); //UNIT STEP x=[zeros(1,10),ones(1,11)]; t=-10:1:10; subplot(2,3,2) plot(t,x); title('unit step');xlabel('t');ylabel('amp'); //SINUSOIDAL FUNCTION t=0:0.0001:0.01; f=100; y=0.5*sin(2*%pi*f*t); subplot(2,3,3) plot(t,y) //EXPONENTIAL FUNCTION t=0:1:25; y=exp(0.3*t); subplot(2,3,4) plot(t,y); //UNIT RAMP clear r clear t clear n t=0:1:9 for n=1:1:10 r(n)=n; end subplot(2,3,5); plot(t,r); title('ramp');xlabel('t');ylabel('amp'); //TRIANGULAR FUNCTION p=50; t=0:1:49 for (n=1:1:p/2) y(n)=n; end for(n=1+p/2:1:p) y(n)=p-n; end subplot(2,3,6); plot(t,y); xs2pdf(0,'continuous.pdf');
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clc,clear printf('Example 6.13\n\n') V_L=11*10^3 V_ph=V_L/sqrt(3) VA=700*10^3 I_FL=VA/(sqrt(3)*V_L) //full load current IR_a=(1.5/100)*V_ph //product of I and R_a R_a=IR_a/I_FL IX_s=(14/100)*V_ph // product of I and X_s X_s=IX_s/I_FL //synchronous reactance //at full load and 0.8 pf I=I_FL phi=acos(0.8) V_ph=complex(V_ph*cos(phi),V_ph*sin(phi)) //just introduced the angle E_ph=sqrt( (abs(V_ph)*cos(phi)+ IR_a)^2+ (abs(V_ph)*sin(phi)+ IX_s)^2 ) Poles=4,f=50 //poles and frequency delta=asin( (abs(V_ph)*sin(phi)+IX_s)/E_ph) -phi delta_dash_mech=(%pi/180) //displacement in degree mechanical //displacement in degree electrical delta_dash_elec=delta_dash_mech*(Poles/2) P_SY=abs(E_ph)*abs(V_ph)*cos(delta)*sin(delta_dash_elec)/X_s //synchronising power per phase P_SY_total=3*P_SY //total synchronising power ns=120*f/(60*Poles) //in r.p.s T_SY=P_SY_total/(2*%pi*ns) //Synchronising torque printf('Synchronising power is %.2fkW\n',P_SY_total/1000) printf('Synchronising torque is %.2f N-m',T_SY)
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clc; //page no 2-11 //example: 2.4 //Given carrier power=400 watt and modulation depth as 80% u=0.8; Pc=400; ptotal=Pc*(1+(u^2/2)); disp(+'watts',ptotal, 'Total power delivered is ')
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// Scilab Code Ex9.8: Page-9.28 (2004) clc;clear; Eg = 1; // Bandgap of silicon, eV e = 1.6e-19; // Electronic charge, C k = 1.38e-23; // Boltzman constant,joule per kelvin E_F = (0.6-0.5)*e; // Fermi energy, joules // E_F =((Ev+Ec)/2)+3/4*k*T1*(log(4)); // Ev & Ec= valance and conduction band energies (formula) T = 4*E_F/(3*k*log(4)); //Temperature that shift the fermi level, K printf("\nTemperature that shift the fermi level = %4.3d K", T); // Result // Temperature that shift the fermi level = 1115 K
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// PG (144) deff('[y]=f(x)','y=sqrt(x)') funcprot(0) deff('[y]=fp(x)','y=0.5/sqrt(x)') funcprot(0) deff('[y]=fpp(x)','y=-0.25*x^(-3/2)') funcprot(0) deff('[y]=fppp(x)','y=3*x^(-2.5)/8') deff('[y]=fpppp(x)','y=-15*x^(-7/2)/16') // f[2.0,2.1,.....2.4] = -0.002084 fpppp(2.3103)/factorial(4)
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clc clear //Initialization of variables h=2.5 //Btu/hr ft^2 F kc=0.1 //Btu/hr ft F r1=0.811/2 //calculations r2c=kc/h *12 //results if r2c>=r1 then printf("Thin layer of insulation would increase the heat dissipation from wire, r2c = %.2f in",r2c) else printf("Thin layer of insulation would decrease the heat dissipation from wire. r2c=%.2f in",r2c) end
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clc; // page no 26 // prob no 1_8_1 //High frequency transformer with identical primary and secondary circuits Lp=150*10^-6; Ls=150*10^-6; Cp=470*10^-12; Cs=470*10^-12; //Lp=Ls=150 uH,Cp=Cs=470 pF Q=85//Q-factor for each ckt is 85 c=0.01//Coeff of coupling is 0.01 Rl=5000//Load resistance Rl=5000 ohm r=75000//Constant current source with internal resistance r=75 kohm //Determination of common resonant frequency wo=1/((Lp*Cp)^(1/2)); //disp('Mrad/sec',wo/(10^6),+'The value of common resonant freq is'); p=3.77*10^6; Z2=Rl/(1+(p*%i*Cs*Rl)); Z1=r/(1+(p*%i*Cp*r)); // At resonance Zs=Zp=Z Z=wo*Ls*(1/Q +%i); Zm=%i*p*c*Lp; // Determination of denominator Dr=((Z+Z1)*(Z+Z2))-(Zm^2) // Hence transfer impedance is given as Zr= (Z1*Z2*Zm)/Dr; disp('ohm',Zr,'The transfer impedance is');
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//Chapter-4,Example 12,Page 96 clc; close; q_rev= 19.14 //latent heat n= 18 //mols T= 273 //temperature in Kelvin dS= q_rev*n/T printf('the change of molar entropy is %.2f J/mol',dS)
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// Display mode mode(0); // Display warning for floating point exception ieee(1); clc; disp("Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 2") //Temperature at which sun is radiating as a blackbody in K T=5800; //Lower limit of wavelength for which glass is transparent in microns lamda_l=0.35; //lower limit of product of wavelength and temperature in micron-K lamda_l_T=lamda_l*T; //Lower limit of wavelength for which glass is transparent in microns lamda_u=2.7; //lower limit of product of wavelength and temperature in micron-K lamda_u_T=lamda_u*T; // For lamda_T= 2030, ratio of blackbody emission between zero and lamda_l to the total emission in terms of percentage r_l=6.7; // For lamda_T= 15660, ratio of blackbody emission between zero and lamda_u to the total emission in terms of percentage r_u=97; //Total radiant energy incident upon the glass from the sun in the wavelength range between lamda_l and lamda_u total_rad=r_u-r_l; disp("Percentage of solar radiation transmitted through the glass in terms of percentage") rad_trans=total_rad*0.92 //Since it is given that silica glass transmits 92% of the incident radiation
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// Display mode mode(0); // Display warning for floating point exception ieee(1); clear; clc; disp("Introduction to heat transfer by S.K.Som, Chapter 10, Example 2") //Hot oil(specific heat,ch=2.09kJ/(kg*K)) flows through counter flow heat excahnger at the mass flow rate of mdoth=(0.7kg/s) ch=2.09*10^3; mdoth=0.7; //overall heat transfer coefficient(U)=650 W/(m^2*K) U=650; //It enters at temprature,Th1=200°C and leaves at temprature,Th2=70°C Th1=200; Th2=70; //Cold oil(specific heat,cc=1.67kJ/(kg*K) exits at temprature,Tc2=150°C at the mass flow rate of mdotc=(1.2kg/s) mdotc=1.2; cc=1.67*10^3; Tc2=150; //The unknown inlet temprature(Tc1) of cold oil may be found from energy balance mdotc*(Tc2-Tc1)=mdoth*(Th2-Th1) disp("The inlet temprature(Tc1) of cold oil in °C ") Tc1=Tc2-[(mdoth*ch)/(mdotc*cc)]*(Th1-Th2) //The rate of heat transfer can be calculate as Q=mdoth*ch*(Th1-Th2) disp("The rate of heat transfer Q=mdoth*ch*(Th1-Th2) in W") Q=mdoth*ch*(Th1-Th2) deltaT1=Th1-Tc2;//deltaT1 is temprature difference between hot oil inlet temprature and cold oil exit temprature deltaT2=Th2-Tc1;//deltaT2 is temprature difference between hot oil exit temprature and cold oil inlet temprature //LMTD(Log mean temprature difference) is defined as (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) for counter flow. disp("LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C ") //let X=log10((deltaT2/deltaT1)) and Y=log10(2.718281) X=log10((deltaT2/deltaT1)); Y=log10(2.718281); //ln=(ln(deltaT2/deltaT1)) ln=X/Y; LMTD=(deltaT2-deltaT1)/ln //Area(A)=Q/(U*LMTD) in m^2 disp("Area(A)=Q/(U*LMTD) in m^2") A=Q/(U*LMTD)
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//Chapter 4, Example 4.5, Page 96 clc clear // Q value of the reaction mn = 1.0086649 MB = 10.0129370 MHe = 4.0026032 MLi = 7.0160040 C2 = 931.5 Q = (mn+MB-MHe-MLi)*C2 -0.48 printf("\n Q of the reaction = %f MeV",Q); //Answer may vary due to round off error
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clc; L1=10*1e-3; //in henry c3=10*1e-12; //in faraday pi=3.14; fr=1/(2*pi*sqrt(L1*c3)); disp(+'Hz',fr,'resonant frequency =')
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clc; E1=2500; // primary side voltage E2=250; // secondary side voltage P=10000; // rated VA of transformer // to achieve a voltage level of 2625, two equal parts of 125 V each of secondary winding are connected in parallel with each other and in series with primary winding Eo=E1+E2/2; // desired output of autotransformer il=P/E2; // rated current of l v winding i=2*il; // Total output current K=(i*Eo)/1000; // Auto transsformer KVA rating ip=P/E1; // rated current of h v winding I=i+ip; // current drawn from supply Pt=(i*(E2/2))/1000; // KVA transformed Pc=K-Pt; // KVA conducted printf('KVA output of autotransformer is %f KVA\n',K); printf('KVA transformed is %f KVA\n',Pt); printf('KVA conducted is %f KVA',Pc);
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//Load scripts getd('../scripts/') //Load image imgPos="../images/" img=readpbm(imgPos+'Gliese 667Cc_surface.pbm') // Do a normalisation on the image display_gray(normalisation(img))
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clc; disp("Example 9.2") density= 1000 // in kg/m^3 densitym=13600 // of mercury in kg/m^3 c=0.62 // orifice coefficient b=0.5 U=135.6/60 // velocity in m/s delP= ((U*((1-b^4)^0.5)/c)^2)*density/2 g=9.81 R=delP/(g*(densitym-density)) disp(R,"Reading on the manometer is ")
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clc; clear; printf("\t\t\tChapter4_example8\n\n\n"); rou=7817; c=461; k=14.4; alpha=.387e-5; L1=.03; L2=0.03; L3=0.04; x=0.04; T_i=95; T_inf=17; // for infinite plate L=L1/2; hc=50; reciprocal_Bi_plate=k/(hc*L); printf("\nThe value of 1/Bi for infinite plate is %.1f",reciprocal_Bi_plate); T=50; n=1; t=[3000 1500 700 400 200 300 350]; [n m]=size(t); // parameter for infinite plate Fourier Number,Fo is named as parameter1 for i=1:m parameter1(i)=alpha*t(i)/L^2; // parameters for semi-infinite solid Bi(Fo)^0.5 and x/(2*(alpha*t)^0.5) are named as parameter2 and parameter3 parameter2(i)=hc*((alpha*t(i))^0.5)/k; parameter3(i)=x/(2*(alpha*t(i))^0.5); dim_T_plate=[0.085 0.34 0.55 0.7 0.8 0.8 0.7]; //the corresponding values of dimensionless temperature for infinite plate from figure 4.6a dim_T_solid=[0.225 0.14 0.075 0.046 0.02 0.035 0.042]; // the corresponding values of dimensionless temperature for semi-infinite solid from figure 4.12 dim_T_bar(i)=dim_T_plate(i)*dim_T_plate(i)*(1-dim_T_solid(i)); T(i)=dim_T_plate(i)*dim_T_plate(i)*(1-dim_T_solid(i))*(T_i-T_inf)+T_inf; end printf("\nThe Results for different time instances:\n"); printf("\n\tInfinite Plate\t\t\t\t\t\tSemi-Infinite Solid\t\t\t\tDimensionless Temperature\tTemperature"); printf("\ntime t, s\t1/Bi\tFo\t(T-Tinf)/(Ti-Tinf)\tBi(Fo)^0.5\tx/(2*(at)^0.5)\t(T-Tinf)/(Ti-Tinf)\t(T-Tinf)/(Ti-Tinf)\t\tT"); for i=1:m printf("\n%d\t\t%.1f\t%.2f\t\t%.2f\t\t%.3f\t\t%.3f\t\t%.3f\t\t\t%.3f\t\t\t\t%.1f",t(i),reciprocal_Bi_plate,parameter1(i),dim_T_plate(i),parameter2(i),parameter3(i),dim_T_solid(i),dim_T_bar(i),T(i)); end
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//to find the weight of flywheel clc //given N=120//rpm k=3.5//ft Ef=2500//ft lb Ks=.01 g=32.2//ft/s^2 w=%pi*N/30//angular velocity W=g*Ef/(w^2*k^2*Ks*2240)//Weight of flying wheel printf("\nWeight of flying wheel, W = %.2f tons",W)
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6_5.sce
clc //initialisation of variables d= 6 //in N= 120 //in Q= 5 //gpm //CALCULATIONS Vc= %pi*d^2*N/(4*231) //RESULTS printf ('minimum size of the reservoir = %.2f gpm',Vc)
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// // printf("\n chainage 15.5 and 27.5') a1=15.5,b1=27.5, //finding base and height of each triangle base=b1-a1 o1=0,o2=22.5, mo1=(o2+o1)/2 //calculating area ae1=base*mo1 ap1=0 an1=ae1 printf("\n area GAM= %0.3f sq meters",ae1) printf("\n chainage 15.5 and 50') a1=15.5,b1=50, base=b1-a1 o1=22.5,o2=30, mo2=(o2+o1)/2 ae2=base*mo2 ap2=ae2 an2=0 printf("\n area GABI= %0.3f sq meters",ae2) printf("\n chainage 50 and 75.5') a1=50,b1=75.5, base=b1-a1 o1=30,o2=35.5, mo3=(o2+o1)/2 ae3=base*mo3 ap3=ae3 an3=0 printf("\n area IBCK= %0.3f sq meters",ae3) printf("\n chainage 75.5 and 86.7') a1=75.5,b1=86.7, base=b1-a1 o1=35.5,o2=0, mo4=(o2+o1)/2 ae4=base*mo4 ap4=ae4 an4=0 printf("\n area KCN= %0.3f sq meters",ae4) printf("\n chainage 86.7 and 90') a1=86.7,b1=90, base=b1-a1 o1=0,o2=10.5, mo5=(o2+o1)/2 ae5=base*mo5 ap5=0 an5=ae5 printf("\n area NLD= %0.3f sq meters",ae5) printf("\n chainage 60 and 90') a1=60,b1=90, base=b1-a1 o1=10.5,o2=25.0, mo6=(o2+o1)/2 ae6=base*mo6 ap6=ae6 an6=0 printf("\n area LDEJ= %0.3f sq meters",ae6) printf("\n chainage 35.5 and 60') a1=35.5,b1=60, base=b1-a1 o1=25,o2=15, mo7=(o2+o1)/2 ae7=base*mo7 ap7=ae7 an7=0 printf("\n area JEFH= %0.3f sq meters",ae7) printf("\n chainage 27.5 and 35.5') a1=27.5,b1=35.5, base=b1-a1 o1=15,o2=0, mo8=(o2+o1)/2 ae8=base*mo8 ap8=ae8 an8=0 printf("\n area FHM= %0.3f sq meters",ae8) an=an1+an2+an3+an4+an5+an6+an7+an8 ap=ap1+ap2+ap3+ap4+ap5+ap6+ap7+ap8 area=ap-an printf("\n ap,ae= %0.3f %0.3f",ap,an) printf("\n total area of field = %0.3f sq meters ",area)
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//Exa:6.3 clc; clear; close; //Given: BWt=100;//in kHz Fh=5;//in KHz n=BWt/(2*Fh); printf("\n\t number of stations = %f",n);
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// Data Reconciliation Benchmark Problems From Literature Review // Author: Edson Cordeiro do Valle // Contact - edsoncv@{gmail.com}{vrtech.com.br} // Skype: edson.cv // aux functions to weighted least squares functions function f = objfun ( x ) f = obj_user(x); endfunction function c = confun(x) //flowsheet_residuals is it's own file c = residuals_cstr(x); endfunction //////////////////////////////////////////////////////////////////////// // Define gradient and Hessian matrix // gradient of the objetive function function gf = gradf ( x ) // it is encouraged to use the code bellow to become more generic //(eg. in case the user want's to use a objective function form robust // statistics ). //gf = diffcode_jacobian_par(obj_user,x,1)'; gf = diffcode_jacobian(obj_user,x)'; // If one wants to use the classical weighted // least squares approach, the evaluation of the hessian using the function // bellow is less lime consuming xdata = x(1:$-2 ); gf=[2*((xdata - xmrow(1:$-2))./(stdDevAllrow).^2) ;0;0]; endfunction // hessian of the objetive function function H = hessf ( x ) // it is encouraged to use the code bellow to become more generic //(eg. in case the user want's to use a objective function form robust // statistics ). // H = diffcode_hessian(obj_user,x); // If one wants to use the classical weighted // least squares approach, the evaluation of the hessian using the function // bellow is less time consuming H = diag([2*ones(nv-2,1)./(stdDevAllrow.^2);0 ;0]); endfunction // gradient of the constraints function y = dg1(x) //flowsheet_residuals is it's own file //ytmp = diffcode_jacobian_par(flowsheet_residuals,x,14*ndata)'; ytmp = diffcode_jacobian(residuals_cstr,x)'; //disp('inside dg') //pause for i = 1: nnzjac; y(i)=ytmp(sparse_dg(i,1),sparse_dg(i,2)); end //pause endfunction // The Hessian of the Lagrangian function y = dh(x,lambda,obj_weight) // disp('inside dh') ysum = zeros(nv,nv); if obj_weight <> 0 then yobj = obj_weight * hessf ( x ); else yobj = zeros(nv,nv); end // pause if sum(abs(lambda)) <> 0 then // the hessian of the constraints //flowsheet_residuals is it's own file // ytmpconstr = diffcode_hessian(residuals_cstr,x); [J,ytmpconstr] = derivative(residuals_cstr, x); for i = 1: nc; if lambda(i) <> 0 then // ysum = ysum + lambda(i)*ytmpconstr(:,:,i); ytmp(i,:) = lambda(i)*ytmpconstr(i,:); end end ysum = matrix(sum(ytmp,'r'),nv,nv); else ysum = zeros(nv,nv); end ysumall = ysum + yobj; // pause for i = 1: nnz_hess y(i) = ysumall(sparse_dh(i,1),sparse_dh(i,2)); end endfunction function y = obj_user( x ) //xdata = x(1:$-4 - 2*ndata); xdata = x(1:$-2 ); // these 2 evaluations bellow are equivalent, the first one is supposed to be faster //y = sum((((xmrow-xdata)./stdDevAllrow).^2)); y = sum(((xmrow(1:$-2)-xdata)./stdDevAllrow).^2); // y = (xmfull(measured)-x(measured))'*diag(ones(1,length(var(measured)))./var(measured))*(xmfull(measured)-x(measured)); // if user wants to make a reconciliation with all variables, it is necessary to set in the SC89.sce // red = ones(1,length(xm)) endfunction function [nc, nv, nnzjac, nnz_hess, sparse_dg, sparse_dh, lower, upper, var_lin_type, constr_lin_type, constr_lhs, constr_rhs] = wls_structure_CSTR(xx) // Data Reconciliation Benchmark Problems From Literature Review // Author: Edson Cordeiro do Valle // Contact - edsoncv@{gmail.com}{vrtech.com.br} // Skype: edson.cv // aux functions to ipopt solver //*************************************************************** //This function analyses the structure of the problem //and return vectors and matrices that will be used by ipopt solver //Outputs: // // nc: number of constraints // nv: number of variables // nnzjac number of non zero elements in the Jacobian of the constraints // nnz_hess number of non zero elements in the Lagrangean Hessian's // sparse_dg sparsity structure of the Jacobian matrix of the constraints // sparse_dh sparsity structure of the Lagrangean Hessian's // lower lower bound of the variables // upper upper bound of the variables // var_lin_type type of the variable (linear or non-linear) // constr_lin_type type of the constraints (linear or non-linear) // constr_lhs lower bound of the constraints residuals // constr_rhs upper bound of the constraints residuals // //Inputs: // flow_full Total flow measurements (or estimates, in case of unmeasured stream) // temp_full Temperature measurements (or estimates, in case of unmeasured compound mass fraction) // coef: Coeficients to enthalpy calculations // From here on, the problem generation is automatic // No need to edit below // Arrange the vectors xlocal=xx; // Jacobian and its structure //jactst = diffcode_jacobian_par(flowsheet_residuals,xlocal,14*ndata)'; jactst = diffcode_jacobian(residuals_cstr,xlocal)'; jactstSparse=sparse(jactst); [ij,v,mn]=spget(jactstSparse); // index of the non-zero elements of the Jacobian nnzjac = size(ij,1); // The sparsity structure of the constraints sparse_dg = ij; //The problem size: nc = number of constraints and nv number of variables [nc,nv] = size(jactst); // The sparsity structure of the Hessian Lagrangian // the Hessian of the objective function is diagonal but the hessian of the constraints not! //first retrieve the constraints Hessian structure, notice that with diffcode_hessian, the // Hessian has the following dimensions: nvar x nvar x nconstr , so it is in fact a // 3 dimensional matrix. //hess_constr_tst = diffcode_hessian_par(flowsheet_residuals,xlocal,nc); hess_constr_tst = diffcode_hessian(residuals_cstr,xlocal); // cumulative sums the constraints in one hessian hess_constr = zeros(nv,nv); //pause for i = 1: nc hess_constr = hess_constr + abs(hess_constr_tst(:,:,i)); end // the Hessian of the objective function hess_f = diffcode_hessian(objfun,xlocal) //pause // sum both of the Hessians hess_Sparse=sparse(hess_constr + hess_f); // get the hessian structure [ij_hess,v_hess,mn_hess]=spget(hess_Sparse); //filters the hessian to remove symmetric indexes ij_hess_filtered = filter_symmetric(ij_hess); // index of the non-zero elements of the Hessian sparse_dh = ij_hess_filtered; nnz_hess = length(ij_hess_filtered(:,1)); lower = xm(:) - 3*stdDevAllrow; lower = [lower; 0.0001;0.1]; upper = xm(:) + 3*stdDevAllrow; upper = [ upper; 5; 15]; //// in the non-linear case, all constraints and variables are non-linear (bilinear) var_lin_type(1:nv) = 1; // Non-Linear constr_lin_type (1:2*ndata) = 1; // Non-Linear // the constraints has lower and upper bound of 0 constr_lhs(1:3*ndata) = 0; constr_rhs(1:3*ndata) = 0; endfunction
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clc //Initialzation of variables xN2=0.780 xO2=0.210 xAr=0.009 P=100 //kPa //Calculations PN2=xN2*P PO2=xO2*P PAr=xAr*P //Results printf('Partial pressure of Nitrogen(kPa) = %.1f',PN2) printf('\n Partial pressure of Oxygen(kPa) = %.1f',PO2) printf('\n Partial pressure of Argon(kPa) = %.1f',PAr)
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//Exa2.41 clc; clear; close; // given data t1=20;// in degree C t2=36;// in degree C alpha_20=0.0043;// in per degree C (Temperature Coefficient) InsulationResistance=480*10^6;// in ohm copper_cond_res=0.7;// in ohm (copper conductor resistance) l=500*10^-3;// in kilo meter (length) R1_desh=InsulationResistance * l;// in ohm // From Formula log(R2_desh)= log(R1_desh-K*(t2-t1)) // K= 1/(t2-t1)*log(R1_desh/R2_desh) // since when t2-t1=10 degree C and R1_desh/R2_desh= 2 K=1/10*log(2); // (i) Insulation resistance at any temperature t2, R2_desh is given by logR2_desh= log(R1_desh)-(t2-t1)/10* log(2); R2_desh= %e^logR2_desh disp("(i) Insulation resistance at any temperature : "+string(R2_desh*10^-6)+" Mega ohm"); // (ii) R_20= copper_cond_res/l;// in ohm R_36=R_20*[1+alpha_20*(t2-t1)]; disp("Resistance at 36 degree C is : "+string(R_36)+" ohm")
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//clc() x = [0.25,0.75,1.25,1.75,2.25]; y = [0.28,0.57,0.68,0.74,0.79]; a0 = 1; a1 = 1; sr = 0.0248; for i = 1:5 pda0(i) = 1 - exp(-a1 * x(i)); pda1(i) = a0 * x(i)*exp(-a1*x(i)); end Z0 = [pda0(1),pda1(1);pda0(2),pda1(2);pda0(3),pda1(3);pda0(4),pda1(4);pda0(5),pda1(5)] disp(Z0,"Z0 = ") R = Z0'*Z0; S = inv(R); for i = 1:5 y1(i) = a0 * (1-exp(-a1*x(i))); D(i) = y(i) - y1(i); end disp(D,"D = ") M = Z0'*D; X = S *M; disp(X,"X = ") a0 = a0 + det(X(1,1)); a1 = a1 + det(X(2,1)); disp(a0,"The value of a0 after 1st iteration = ") disp(a1,"The value of a1 after 1st iteration = ")
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clear; clc; //Example 3.13 b=100; Vcc=12; Vbe=0.7; Icq=1;//mA Vceq=6; Rc=(Vcc-Vceq)/Icq; printf('\ncollector resistance=%.3f KOhms\n',Rc) Ibq=Icq/b; printf('\nbase current=%0.3f mA\n',Ibq) Rb=(Vcc-Vbe)/Ibq; printf('\nbase resistance=%.3f KOhms\n',Rb)
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Haha HelloWorld 123456799999 123456788888 18007328601
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// **** Purpose **** // This code is to help calculate PiLab_kif_esc in parallel way. // **** Variables **** // [Ek],[Vk]: data readed from /kif/Ek/Ek_x.sod, /kif/Vk/Vk_x.sod. // [EVal],[EWin],[StateProj]: kif_esc input data // [k_idx_start]: the start index of Ek and vk // **** Version **** // 05/18/2016 first built // **** Comment **** function E_index=PIL_plb_esc_cal(k_idx_start,Ek,Vk,kif_esc) EVal=kif_esc.EVal; EWin=kif_esc.EWin; StateProj=kif_esc.StateProj; tot_state=length(Ek(:,1)); // search target eigenstate and k index [s_ind,k_ind]=find(abs(Ek-EVal)<=EWin); // calculate wieght on projected states E_index=[]; proj_ind=find(StateProj<0); if s_ind~=[] then if proj_ind==[] then w_val=ones(length(s_ind),1); else w_val=zeros(length(s_ind),length(proj_ind)+1); w_val(:,1)=ones(length(s_ind),1); for m=1:length(s_ind) for p=1:length(proj_ind) if p==length(proj_ind) then proj_state=StateProj(proj_ind(p)+1:$); else proj_state=.. StateProj(proj_ind(p)+1:proj_ind(p+1)-1); end w_val(m,p+1)=.. sum((abs(Vk(proj_state,.. (k_ind(m)-1)*tot_state+s_ind(m)))).^2); end end end k_ind=k_ind+(k_idx_start-1) E_index=cat(1,E_index,[k_ind',s_ind',w_val]); end E_index=gsort(E_index,'lr','i'); endfunction
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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.idc.ac.il/tecs // 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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//Ex8_10 clc Av = -200 Ri = 10*10^3 RL = 3*10^3 Ai = Av*Ri/RL disp("Av = "+string(Av))//voltage gain disp("Ri = "+string(Ri)+"ohm")//input resistance disp("RL = "+string(RL)+"ohm")//load resistance disp("Ai = Av*Ri/RL = "+string(Ai))//current gain // note : there are mis-printring in the textbook for the above problem regading formula and notations. // answer in the textbook for above problem is wrong.
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clear; clc; // Illustration 3.4 // Page: 169 printf('Illustration 3.4 - Page: 169\n\n'); // solution //*****Data*****// // a-ammonia T = 300; // [K] P = 101.3; // [kPa] Kg = 2.75*10^-6; // [kmole/square m.s.kPa] m = 1.64; res = 0.85; // [gas phase resistance] xa_g = 0.115/100; // [mole fraction of NH3 in liquid phase at a point]] ya_g = 8/100; // [mole fraction of NH3 in gas phase at a point] //*****// Ky = Kg*P; // [kmole/square m.s] // Using equation 3.24 ky = Ky/res; // [kmole/square m.s] // Using equation 3.21 deff('[y] = f12(kx)','y = (m/kx)-(1/Ky)+(1/ky)'); kx = fsolve(0.0029,f12); // [kmole/square m.s] // Interfacial concentrations at this particular point in the column, using equation (3.15) ystar_a = m*xa_g; // Using equation 3.12 N_a = Ky*(ya_g-ystar_a); // [kmole/square m.s] // Gas-phase interfacial concentration from equation (3.9) ya_i = ya_g-(N_a/ky); // Since the interfacial concentrations lie on the equilibrium line, therefore xa_i = ya_i/m; // Cross checking the value of N_a N_a = kx*(xa_i-xa_g); // [kmole/square m.s] printf("The individual liquid film coefficient and gas film coefficient are %e kmole/square m.s %e kmole/square m.s respectively\n\n",kx,ky); printf("The gas phase and liquid phase interfacial concentrations are %f and %f respectively\n\n",ya_i,xa_i);
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//Control of DC motors// //Example 11.1// //Since the speed control is required in both directions we will have to use a dual converter for the application.It would be prefarable to use six pulse dual converter with thyristors connected in antiparallel connection// //speed control from 20% rated speed to 100% rated speed will be obtained by armature control// //Control and speed above 100% will be possible by field weakening// Idc=200/460*1000;//Rated motor current in amps// printf('Rated motor current=Idc=%famps',Idc); //Thus the main armature converter will be having dc side rating of 500Amps and 460volts// //If 20% drop is allowed in cables,ac transformer,converter etc., then No load dc voltage required=460*1.2=552Volts// printf('\nHence AC voltage for six pulse configuration=552/1.35=410volts'); //Hence a 3phase,415v AC supply will be adequate for armature control// //Field converter rating will be 230V,10A.Arrangement will be six pulse,non reversible.since AC supply of 415V,3 phase is available,we shall make use of it for field converter also.// printf('\nAC rating of field converter=230/1.35=170V'); //However we shall provide a standard AC voltage of 230V AC and will lock the field converter firing angle to suitable value so as to produce 230V dc// printf('\nDC power=230*10=2300Watts'); printf('\nAC power=1.05*2300=2415Watts'); printf('\nThus tranformer of 2.5KVA,415/230V will be required'); Edca=(170+170/10)*1.35;//available voltage in volts// Edc=1.35*230; A=acos(Edca/Edc)*180/%pi; printf('\nField converter shall be locked at an angle of A=%fdegrees',A);
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function [y,err] = CRCDetector(in,polynomial,initialstate,chksumsperframe) y=[]; err=[] // Display mode mode(0); // Display warning for floating point exception ieee(1); //CRCDetector finds the message signal and also finds if error has occured in a subframe //Y = CRCGenerator(in,polynomial,initialstate,chksumsperframe) outputs y binary sequence //which is the message signal. //It also outputs error vector which specifies if error has occured in a block. // err = [1 0 0] means msg is divided in 3 block and error is present in first block //in:input - The input must be a binary vector. //polynomial - Generator polynomial //Specify the generator polynomial as a binary or integer row vector, //with coefficients in descending order of powers //Initialstate - Vector array (with length of the generator polynomial order) //of initial shift register values (in bits) //chksumsperframe - checksumsperframe //Number of checksums per input frame //Specify the number of checksums that the object calculates for each input frame as a positive integer //the integer must divide the length of each input frame evenly. m= length(in)/chksumsperframe; l=length(initialstate); //checking conditions on input if(~isreal(in) | or( isnan(in)) | min(size(in))~=1 | or(in ~= 0 & in ~= 1)) then error("CRCDetector:improper input"); end //checking conditions on genpoly if(~isreal(polynomial) | or( isnan(polynomial)) | min(size(polynomial))~=1 | or(polynomial ~= 0 & polynomial ~= 1)) then error("CRCDetector:improper genpoly"); end if((~polynomial(1)) |(~polynomial(length(polynomial))) ) then error("CRCDetector:improper genpoly"); end //checking conditions on initial state if(~isreal(initialstate) | or( isnan(initialstate)) | min(size(initialstate))~=1 | or(initialstate ~= 0 & initialstate ~= 1)) then error("CRCDetector:improper initialstate"); end //check condition on chksumsperframe if (~isreal(chksumsperframe) | length(chksumsperframe)~=1 | isnan(chksumsperframe)|ceil(chksumsperframe)~=chksumsperframe|chksumsperframe<=0) then error("CRCDetector:improper chksumsperframe"); end //chksumsperframe must divide the length of each input frame evenly. if(modulo(length(in),chksumsperframe)) then error("CRCDetector:The input length must be an integer multiple of the ChecksumsPerFrame value"); end if(modulo((length(in)-chksumsperframe*l),chksumsperframe)) then error("CRCDetector:Message length must be an integer multiple of the ChecksumsPerFrame value"); end //checking that length of Initialstate is equal to degree of genpoly if (length(polynomial)~=(l+1) ) then error(" CRCDetector:length of Initialstate should be equal to degree of gen poly"); end //correcting orentation if((size(in,1)~=size(initialstate,1))&(size(in,2)~=size(initialstate,2))) then initialstate = initialstate'; end //dividing in frames for k=1:chksumsperframe buff = initialstate; x=in(m*(k-1)+1:m*k); //applying method to find CRC on each frame for i =1:length(x) pre=buff(1); for j =1:l-1 if(polynomial(j+1)) then buff(j)=xor(buff(j+1),pre); else buff(j)=buff(j+1); end end buff(l)=xor(x(i),pre); end y = [ y x(1:(m-l))]; //ouputing each subframe after appending CRC if(or(buff~=0)) then err(k)=1; else err(k)=0; end end endfunction function c = xor(a,b) if(a==b) then c=0; else c=1; end endfunction
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clc IP=35; // Power developed by the engine in kW S=284; //Steam combustion in kg/h p2=0.14; //Condenser pressure in bar p1=15; //bar h1=2923.3; //kJ/kg s1=6.709; //kJ/kg K h_f=220; //kJ/kg h_fg=2376.6; //kJ/kg s_f=0.737; //kJ/kg K s_fg=7.296; //kJ/kg K x2=(s1-s_f)/s_fg; disp("(i) Final condition of steam =") disp(x2) h2=h_f+x2*h_fg; disp("(ii) Rankine efficiency=") n_rankine=(h1-h2)/(h1-h_f); disp(n_rankine) disp("(iii) Relative efficiency") n_thermal=IP/(S/3600)/(h1-h_f); n_relative=n_thermal/n_rankine; disp("relative efficiency=") disp(n_relative)
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//example 6.4 clc; funcprot(0); Cc=0.28; Hc=18*12; e0=0.9; sigmao=11*100+40*(121.5-64)+18/2*(118-62.4); H2=5+40+18; H1=5+40; qo=3567; //from table IaH2=0.21; IaH1=0.225; Dsigma=qo*((H2*IaH2-H1*IaH1)/(H2-H1))*4; Scp=Cc*Hc/(1+e0)*log10(sigmao/sigmao+Dsigma/sigmao); disp(Scp,"settlement in inches");
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//Calculations on dual combustion cycle clc,clear //Given: r=18 //Compression ratio P1=1.01,P3=69 //Pressure at 1, 3 in bar T1=20+273 //Temperature at 1 in K cv=0.718 //Specific heat at constant volume in kJ/kgK cp=1.005 //Specific heat at constant pressure in kJ/kgK g=1.4 //Specific heat ratio(gamma) R=0.287 //Specific gas constant in kJ/kgK //Solution: T2=T1*r^(g-1) //Temperature at 2 in K P2=P1*r^g //Pressure at 2 in bar T3=T2*(P3/P2) //Temperature at 3 in K Q_v=cv*(T3-T2) //Heat added at constant volume in kJ/kg //Given, Heat added at constant volume is equal to heat added at constant pressure T4=Q_v/cp+T3 //Temperature at 4 in K rho=T4/T3 //Cut off ratio T5=T4*(rho/r)^(g-1) //Temperature at 5 in K Q1=2*Q_v //Heat supplied in cycle in kJ/kg Q2=cv*(T5-T1) //Heat rejected in kJ/kg eta=1-Q2/Q1 //Thermal efficiency W=Q1-Q2 //Work done by the cycle in kJ/kg V1=1*R*T1/(P1*100) //Volume at 1 in m^3/kg V2=V1/r //Volume at 2 in m^3/kg V_s=V1-V2 //Swept volume in m^3/kg mep=W/(V_s*100) //Mean effective pressure in bar //Results: printf("\n The air standard efficiency, eta = %.1f percent",eta*100) printf("\n The mean effective pressure, mep = %.2f bar\n\n",mep)
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//EXAMPLE 26.26 //LONG SHUNT DYNAMO clc; funcprot(0); //Variable Initialisation N=1000;.....................//Speed of the generator in rpm Po=22;....................//Output power in Kilo Watts V=220;....................//Terminal voltage in Volts Ra=0.05;..................//Armature resisitance in Ohms Rsh=110;...................//Shunt field resistance in Ohms Rse=0.06;..................//Series field resisitance in Ohms eff=88;....................//Overall efficiency in Percentage Ish=V/Rsh;.....................//Shunt field current in Amperes I=(Po*1000)/V;.................//Load current in Amperes Ia=I+Ish;........................//Armature current in Amperes Vse=Ia*Rse;.........................//Drop in series field windings in Volts Ly=(Ia^2)*Ra;........................//(Ia^2)Ra losses in Watts Lse=(Ia^2)*Rse;......................//Series field loss in Watts y=round(Lse*10)/10;.................//Rounding of decimal places Lsh=(Ish^2)*Rsh;....................//Shunt field loss in Watts Lcu=Ly+y+Lsh;.....................//Total copper losses in Watts disp(Lcu,"(a).Total copper losses in Watts:"); Pin=(Po*1000)/(eff/100);................//Input power in Watts Lt=(Pin)-(Po*1000);................//Total lossees in Watts Lif=Lt-Lcu;.......................//Iron and friction losses in Watts T=(Pin*60)/(N*2*3.142);..............//Torque exerted by the prime mover in N-m y1=round(T*10)/10;.................//Rounding of decimal places disp(y1,"(b).Torque exerted by the prime mover in N-m:");
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clc; warning("off"); printf("\n\n example4.4 - pg101"); // given x1=0; //[cm] x2=30; //[cm] p1=0.3; //[atm] p2=0.03; //[atm] D=0.164; //[am^2/sec] R=82.057; //[cm^3*atm/mol*K] T=298.15; //[K] // using the formula Nax*int(dx/Ax)=-(D/RT)*int(1*dpa) a=integrate("1/((%pi/4)*(10-(x/6))^2)","x",x1,x2); b=integrate("1","p",p1,p2); Nax=-((D/(R*T))*b)/a; printf("\n\n Nax=%6emol/sec=%3emol/h \n the plus sign indicates diffusion to the right",Nax,Nax*3600);
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//Example7.13 // to determine the output voltage of the precision rectifier circuit clc; clear; close; Vi = 10 ; //V i/p volt R1 = 20 ; // K ohm R2 = 40 ; // K ohm Vd = 0.7 ; // V the diode voltage drop // the output of the half wave precision rectifier is defined as // Vo = -(R2/R1)*Vi ; for Vi < 0 // = 0 otherwise // i.e for Vi > 0 // Vo = 0 // for Vi < 0 Vo = -(R2/R1)*Vi disp('The output of the half wave precision rectifier Vo is = '+string(Vo)+' V ');
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//ques-1.3 //Calculating temporary and permanent hardness of a sample of water clc A=7.3;//content of Magnesium hydrogencarbonate (in mg/L) B=16.2;//content of Calcium hydrogencarbonate (in mg/L) C=9.5;//content of Magnesium chloride (in mg/L) D=13.6;//content of Calcium sulphate (in mg/L) a1=(A/146)*100;//CaCO3 equivalent of A a2=(B/162)*100;//CaCO3 equivalent of B a3=(C/95)*100;//CaCO3 equivalent of C a4=(D/136)*100;//CaCO3 equivalent of D t=a1+a2;//temporary hardness (in ppm) p=a2+a4;//permanent hardness (in ppm) printf("Temporary and Permanent hardness of the given sample are %d ppm and %d ppm respectively.",t,p);
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clc // Given that mu = 1.5 // refractive index of glass // Sample Problem 1 on page no. 3.23 printf("\n # PROBLEM 1 # \n") Ip = atan(mu) * (180 / %pi) // by brewster's law r = 90 - Ip // calculation for angle of refraction printf("Standard formula used \n mu=tan(Ip)\n") printf("\n Brewster angle = %f degree\n Angle of refraction = %f degree",Ip,r)
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PL/SQL Developer Test script 3.0 5 begin -- Call the procedure personas_por_usuario(pusuario_id => :pusuario_id, p_recordset => :p_recordset); end; 2 pusuario_id 1 gago 5 p_recordset 1 <Cursor> 116 0
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~BivLCM-SR-bfi_c_vrt_col_d-PLin-VLin.tst
THE OPTIMIZATION ALGORITHM HAS CHANGED TO THE EM ALGORITHM. ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 0.305650D+00 2 -0.244855D-02 0.232927D-02 3 0.755272D-01 -0.570529D-03 0.383860D+00 4 -0.438549D-03 0.869051D-03 -0.199024D-02 0.341637D-02 5 -0.121655D-02 0.108281D-03 -0.136561D-02 -0.682947D-05 0.365915D-02 6 0.254222D-03 0.517870D-04 0.561408D-03 -0.992642D-04 -0.947286D-04 7 -0.134643D-02 0.107740D-03 -0.390204D-03 -0.113037D-03 0.710261D-03 8 0.351311D-04 -0.448106D-04 -0.123829D-02 0.156330D-03 -0.237248D-03 9 -0.350828D+00 0.116670D-01 0.144958D+00 -0.611516D-02 0.432911D-01 10 -0.186132D+00 -0.927932D-02 0.452447D-01 0.126419D-02 0.115409D+00 11 -0.170591D+00 0.142802D-01 0.902171D-01 0.204542D-01 0.929016D-01 12 -0.283064D+00 0.320612D-02 -0.117493D+01 0.518165D-01 0.567055D-02 13 0.499006D-02 0.411661D-02 0.915713D-02 -0.129226D-01 0.942765D-02 14 -0.680254D-01 -0.161669D-01 -0.514817D+00 0.253552D-01 -0.154749D-01 15 -0.215049D+01 -0.319008D-01 0.442236D-01 -0.749719D-02 -0.117733D+00 16 -0.643001D-01 -0.271424D-02 -0.223175D-01 -0.781199D-03 -0.123863D-02 17 0.102772D-01 -0.505507D-03 -0.149293D-02 -0.370054D-03 -0.510585D-03 18 0.418650D-01 -0.278421D-01 0.319978D+00 -0.391182D-02 0.171528D-01 19 -0.178943D+00 0.270789D-02 0.443284D-01 0.384763D-02 0.790361D-02 20 0.744799D+00 0.144466D-01 -0.287996D+01 -0.408549D-01 0.536382D-01 21 0.131064D+00 -0.288257D-02 -0.517660D-01 -0.387158D-02 -0.828988D-02 22 -0.104543D-03 -0.383186D-03 -0.135234D-02 -0.249370D-03 -0.338321D-03 23 0.103093D-01 -0.189776D-02 -0.363907D-01 -0.137104D-01 0.215548D-02 24 -0.522868D-02 -0.251076D-03 0.276074D-02 -0.813195D-03 -0.128674D-03 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 0.620989D-03 7 0.628594D-03 0.400328D-02 8 -0.330865D-04 -0.162033D-03 0.175795D-02 9 0.193782D-01 0.327870D-01 0.127197D-02 0.449164D+02 10 -0.812319D-02 0.218506D-01 -0.148685D-01 0.462028D+01 0.190368D+02 11 0.125224D-01 0.781654D-01 -0.218265D-02 0.110445D+02 0.400700D+01 12 0.180600D-01 0.194495D-01 0.123311D-01 0.490806D+01 0.130999D+01 13 0.456885D-01 0.122691D+00 -0.636362D-02 0.401419D+00 0.270072D+01 14 0.105892D-01 0.203817D-01 0.151499D+00 0.256267D+01 0.125438D+01 15 0.601760D-02 -0.899352D-02 0.395987D-01 -0.800705D+01 -0.115255D+02 16 0.104712D-02 0.375991D-02 0.144899D-02 0.903107D+00 -0.160038D+00 17 -0.222487D-03 -0.521401D-03 -0.121470D-04 -0.186466D+00 -0.292534D-02 18 -0.422393D-01 -0.970061D-01 -0.233412D-01 -0.357631D+01 0.288074D+01 19 -0.131427D-01 0.105556D-01 -0.896458D-02 -0.570971D+00 0.491951D+00 20 -0.952186D-02 0.679625D-02 -0.119872D+00 -0.324561D+00 0.109955D+01 21 0.135843D-01 -0.111295D-01 0.132031D-01 0.443897D+00 -0.746674D+00 22 0.136391D-03 0.562050D-04 0.909544D-04 -0.169901D-01 -0.358509D-01 23 0.182009D-02 0.252332D-02 -0.106662D-02 0.110942D+00 0.110269D+00 24 -0.150633D-03 -0.232162D-03 0.396287D-04 -0.216582D-01 -0.793165D-02 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 0.408162D+02 12 0.168978D+02 0.122620D+03 13 -0.207940D+01 -0.552442D-01 0.133718D+02 14 0.138424D+01 -0.114226D+01 0.976467D+00 0.580964D+02 15 -0.471674D+01 0.187770D+01 -0.116811D+01 0.499960D+01 0.224922D+03 16 0.263566D+00 0.562453D+00 0.959908D-01 0.287697D+00 0.168355D+01 17 -0.825280D-01 -0.304130D-01 -0.989028D-02 -0.434008D-01 -0.979097D+00 18 -0.298156D+00 -0.145460D+01 -0.648894D+01 -0.670267D+01 0.470940D+02 19 0.664716D+00 -0.126303D+01 -0.853857D+00 -0.405854D+00 0.160263D+01 20 -0.324615D+01 -0.205549D+02 0.272998D+00 -0.181666D+02 0.312462D+02 21 -0.175059D+00 0.140864D+01 0.766352D+00 0.857607D+00 -0.481160D+00 22 -0.111694D+00 -0.655429D-01 0.242627D-01 0.955021D-02 -0.185555D+00 23 -0.781892D-01 -0.460140D+00 0.241709D+00 0.224303D+00 0.542685D+00 24 -0.305090D-01 -0.996614D-01 -0.109144D-01 -0.878228D-01 -0.269978D+00 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 0.492956D+00 17 -0.222605D-01 0.145400D-01 18 0.920693D-01 -0.191347D+00 0.193085D+03 19 0.137097D+00 -0.223799D-01 0.140965D+01 0.522138D+01 20 0.273983D+00 -0.238512D+00 0.753256D+02 0.416485D+01 0.468135D+03 21 0.313742D-01 0.314777D-01 0.274473D+01 -0.475944D+01 -0.548838D+01 22 0.678407D-02 0.288322D-02 -0.103205D+01 -0.116970D-01 -0.524814D+00 23 0.504795D-01 -0.394639D-02 -0.588838D+00 -0.569231D-02 0.412895D+01 24 -0.249432D-02 0.321608D-02 -0.474113D+00 -0.744333D-02 -0.238902D+01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 0.571627D+01 22 -0.375391D-01 0.126640D-01 23 0.111901D+00 0.687238D-02 0.701940D+00 24 0.214812D-01 0.667075D-02 -0.374606D-01 0.271705D-01 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 1.000 2 -0.092 1.000 3 0.220 -0.019 1.000 4 -0.014 0.308 -0.055 1.000 5 -0.036 0.037 -0.036 -0.002 1.000 6 0.018 0.043 0.036 -0.068 -0.063 7 -0.038 0.035 -0.010 -0.031 0.186 8 0.002 -0.022 -0.048 0.064 -0.094 9 -0.095 0.036 0.035 -0.016 0.107 10 -0.077 -0.044 0.017 0.005 0.437 11 -0.048 0.046 0.023 0.055 0.240 12 -0.046 0.006 -0.171 0.080 0.008 13 0.002 0.023 0.004 -0.060 0.043 14 -0.016 -0.044 -0.109 0.057 -0.034 15 -0.259 -0.044 0.005 -0.009 -0.130 16 -0.166 -0.080 -0.051 -0.019 -0.029 17 0.154 -0.087 -0.020 -0.053 -0.070 18 0.005 -0.042 0.037 -0.005 0.020 19 -0.142 0.025 0.031 0.029 0.057 20 0.062 0.014 -0.215 -0.032 0.041 21 0.099 -0.025 -0.035 -0.028 -0.057 22 -0.002 -0.071 -0.019 -0.038 -0.050 23 0.022 -0.047 -0.070 -0.280 0.043 24 -0.057 -0.032 0.027 -0.084 -0.013 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 1.000 7 0.399 1.000 8 -0.032 -0.061 1.000 9 0.116 0.077 0.005 1.000 10 -0.075 0.079 -0.081 0.158 1.000 11 0.079 0.193 -0.008 0.258 0.144 12 0.065 0.028 0.027 0.066 0.027 13 0.501 0.530 -0.042 0.016 0.169 14 0.056 0.042 0.474 0.050 0.038 15 0.016 -0.009 0.063 -0.080 -0.176 16 0.060 0.085 0.049 0.192 -0.052 17 -0.074 -0.068 -0.002 -0.231 -0.006 18 -0.122 -0.110 -0.040 -0.038 0.048 19 -0.231 0.073 -0.094 -0.037 0.049 20 -0.018 0.005 -0.132 -0.002 0.012 21 0.228 -0.074 0.132 0.028 -0.072 22 0.049 0.008 0.019 -0.023 -0.073 23 0.087 0.048 -0.030 0.020 0.030 24 -0.037 -0.022 0.006 -0.020 -0.011 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 1.000 12 0.239 1.000 13 -0.089 -0.001 1.000 14 0.028 -0.014 0.035 1.000 15 -0.049 0.011 -0.021 0.044 1.000 16 0.059 0.072 0.037 0.054 0.160 17 -0.107 -0.023 -0.022 -0.047 -0.541 18 -0.003 -0.009 -0.128 -0.063 0.226 19 0.046 -0.050 -0.102 -0.023 0.047 20 -0.023 -0.086 0.003 -0.110 0.096 21 -0.011 0.053 0.088 0.047 -0.013 22 -0.155 -0.053 0.059 0.011 -0.110 23 -0.015 -0.050 0.079 0.035 0.043 24 -0.029 -0.055 -0.018 -0.070 -0.109 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 1.000 17 -0.263 1.000 18 0.009 -0.114 1.000 19 0.085 -0.081 0.044 1.000 20 0.018 -0.091 0.251 0.084 1.000 21 0.019 0.109 0.083 -0.871 -0.106 22 0.086 0.212 -0.660 -0.045 -0.216 23 0.086 -0.039 -0.051 -0.003 0.228 24 -0.022 0.162 -0.207 -0.020 -0.670 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 1.000 22 -0.140 1.000 23 0.056 0.073 1.000 24 0.055 0.360 -0.271 1.000
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function [p]=derivat(p) //pd=derivat(p) computes the derivative of the polynomial or rational //function marix relative to the dummy variable //! t=type(p) if t=1 then p=0*p,return,end if t=2 then [m,n]=size(p);var=varn(p); for i=1:m for j=1:n pij=p(i,j);nij=degree(pij); if nij=0 then p(i,j)=0 else pij=coeff(pij).*(0:nij),p(i,j)=poly(pij(2:nij+1),var,'c') end; end; end; return end; if t=15 then if p(1)='r' then num=p(2);den=p(3) [m,n]=size(num) for i=1:m for j=1:n num(i,j)=derivat(num(i,j))*den(i,j)... -num(i,j)*derivat(den(i,j)) den(i,j)=den(i,j)**2 end; end; p=list('r',num,den,p(4)) return end; end; error('incorrect data type')
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// PD control law from polynomial coefficients, as explained in Sec. 9.8. // 9.22 function [K,taud,N] = pd(Rc,Sc,Ts) // Both Rc and Sc have to be degree one polynomials s0 = Sc(1); s1 = Sc(2); r1 = Rc(2); K = (s0+s1)/(1+r1); N = (s1-s0*r1)/r1/(s0+s1); taudbyN = -Ts*r1/(1+r1); taud = taudbyN * N; endfunction;
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v=[5, -5, 9, 12, -1, 0, 4] v(v>0) = 10