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// Electric Machinery and Transformers // Irving L kosow // Prentice Hall of India // 2nd editiom // Chapter 14: TRANSFORMERS // Example 14-27 clear; clc; close; // Clear the work space and console. // Given data // From diagram in fig.14-23a P_L = 14400 ; // Load output power in W V_L = 120 ; // Load ...
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//sin wave clc; fm=5; t=linspace(0,1,500); sgl=0.2*sin(2*%pi*fm*t); //msg sig msg=[]; for i=1:500 if sgl(i)>0 then msg=[msg 1]; else msg=[msg 0]; end end subplot(3,1,1) plot(t,msg); title('msg signal'); xlabel('time'); ylabel('amplitube'); //carrier sig fc=30; ...
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//ANALOG AND DIGITAL COMMUNICATION //BY Dr.SANJAY SHARMA //CHAPTER 11 //Information Theory clear all; clc; printf("EXAMPLE 11.4(PAGENO 489)"); //given Px_1 = 1/4//probability wrt to binary PCM '0' Px_2 = 3/4//probability wrt to binary PCM '1' //calculations Ix_1 = log2(1/Px_1)//amount of information of zer...
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clc; clf; clear all; function [yn, yorigin]=multi(x1n, x1origin,x2n, x2origin); n=(-x1origin):(size(x1n,2)-x1origin-1); n2=(-x2origin):(size(x2n,2)-x2origin-1); if(x1origin>=x2origin) then x1=[x1n,zeros(1,x1origin-x2origin)]; x2=[zeros(1,x1origin-x2origin),x2n]; yn=x1.*x...
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//Determine bandwidth requirement for an FM signal del = 10; fm = 2; fms = 8; mf = del/fm; bw = fm*fms*2; disp(bw, 'Bandwidth requirement for an FM signal (in Khz) ');
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___________ | | | | | |____ | | | | | | ____ | | | | | | __ | | | | | |___________|
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clc R=('Rock') P=('Paper') S=('Sizzors') input hand = ('Select your hand') if hand == R then print('you win') else print('You lose') end
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//Obtain path of solution file path = get_absolute_file_path('solution7_7.sce') //Obtain path of data file datapath = path + filesep() + 'data7_7.sci' //Clear all clc //Execute the data file exec(datapath) //Calculate the tensile force induced in appropriate bolt P1 (N) if ((l1 > l2) & (l1 > l3)) then P...
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// Universidade Federal Fluminense // Departamento de Engenharia Química e de Petróleo // Codigo para solução do modelo de`Produção Gas Lift // Prof. Lizando Santos // Fontes: Alessandro Jacoud Peixoto  Diego Pereira-DiasArthur F. S. Xaud Argimiro Resende Secchi // Modelling and Extremum Seeking Contr...
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//Chapter-4, Illustration 8, Page 171 //Title: Steam Nozzles and Steam Turbines //============================================================================= clc clear //INPUT DATA P1=3.5;//Dry saturated steam in bar P2=1.1;//Exit pressure in bar At=4.4;//Throat area in cm^2 h1=2731.6;//Enthalpy at P1 in k...
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clc clear //INPUT DATA t1=15;//dry bulb temperature in Degree C t3=41;//heating coil temperature in Degree C t11=11;//wet bulb temperature in Degree C p=760;//pressure in mm of Hg x=0.4;//realtive humidity in percentage pva=9.83;//Saturation pressure in mm Hg ps2=33.68;//Saturation pressure in mm Hg cp=1.005;...
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//example 8.4(a)// clc //clears the screen// clear //clears the variables// disp('In this case of flip flop, J and K are initially 0 & 1 respectively. Thus J is active. With the first leading edge of clock input, Q and therefore J goes to logic 1 state. The second leading edge edge forces Q to go to logic 0 state ...
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// Calculation of required quantity of magnesium clc j = 15 // current density in mA m^-2 m = 0.0243 // molar mass of magnesium F = 96490 // farad charge n = 2 // charge on ion t = 10 // time in years printf("\n Example 13.2") a = m*j*1e-3*(t*365*24*3600)/(n*F) printf("\n Amount of magnesium required is %0.1f ...
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// Example 3_7 clc;funcprot(0); // Given data T_1=240;// °F T_2=80;// °F p_1=150;// psia p_2=14.7;// psia c_p=0.240;// Btu/lbm · R c_v=0.172;// Btu/lbm · R // Solution // (a) deltau=c_v*((T_2+459.67)-(T_1+459.67));// Btu/lbm deltah=c_p*(T_2-T_1);// Btu/lbm printf('\n(a)The change in specific internal ene...
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//Chapter 4 //page no 114 //given clc; R1=0.7; R2=0.99; ad=0.1; //compute Ld Ld=1-R1*R2*%e^-(2*ad); printf("\n Decay Loss %0.4f \n",Ld); trt=40;//fs tph=trt/Ld; printf("\n Photon lifetime %0.2f fs\n",tph); BW=1/tph; printf("\n Bandwidth %0.1f Thz\n",BW*1000);//Answer in Thz
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// Initilization of variables W1=50 // N // weight of the first block W2=50 // N // weight of the second block mu_1=0.3 // coefficient of friction between the inclined plane and W1 mu_2=0.2 // coefficient of friction between the inclined plane and W2 // Calculations // On adding eq'ns 1&3 and substuting the value...
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//Problem 2.08: //initializing the variables: tc = 0.0512 // in cal/m.s.°C //calculation: k = tc*0.3048*3600/(252*1.8) // in Btu/ft.h.°F printf("\n\nResult\n\n") printf("\n thermal conductivity in english units is %.3f Btu/ft.h.°F\n",k)
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//Example number 12.5, Page number 265 clc;clear; close; //Variable declaration n1=1.53; //Core refractive index n2=1.42; //Cladding refractive index //Calculation thetac=asin(n2/n1); //critical angle(radian) thetac=thetac*180/%pi; //critical angle(degrees) //Resul" printf("critical angle is ...
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function [x,y,typ]=DELAYV_f(job,arg1,arg2) x=[];y=[];typ=[]; select job case 'plot' then //normal position standard_draw(arg1) case 'getinputs' then [x,y,typ]=standard_inputs(arg1) case 'getoutputs' then [x,y,typ]=standard_outputs(arg1) case 'getorigin' then [x,y]=standard_origin(arg1) case 'set' then x=ar...
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// Dado un sitema de ecuaciones F y un punto inicial x, // obtiene una solución aproximada al sistema F(v) = 0 con el método de newton multivariable // con una tolerancia de epsilon y un máximo iter de iteraciones function v = metodo_newton_multivariable(F, x, eps, iter) for i = 1: iter Jinv = inv(numderiva...
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//Finding resistance //Example 15.14(pg. 400) clc clear R1=0.12//old conductor resistance in ohm d1=15//diameter of old conductor in cm d2=0.4*d1//diameter of new conductor in cm a1=%pi*(d1^2)/4//area of cross section of old conductor a2=%pi*(d2^2)/4//area of cross section of new conductor //R=rho*l/a=rho*V/a^...
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//Program for example 4 chapter 3 clear clc disp("Example 4:Display a message or the value of a variable using the command printf") disp("********************************************************************") disp("Answer : ") disp('') halt('Continue ...?? ') if (getos()~='Linux') then printf("\nTo display ...
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clc // Fundamental of Electric Circuit // Charles K. Alexander and Matthew N.O Sadiku // Mc Graw Hill of New York // 5th Edition // Part 1 : AC Circuits // Chapter 3: Magnetically Couple Circuits // Example 13 - 1 clear; clc; close; // // Given data Z1 = complex(4.0000,-1.0000); Z2 = complex...
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//problem 20 pagenumber 2.103 //given v1=2;//volt v2=3;//volt v3=6;//volt v4=8;//volt rf1=50e3;//ohm r1=40e3;//ohm r2=25e3;//ohm r3=10e3;//ohm r4=20e3;//ohm r5=30e3;//ohm //determine output voltage v0x=-(v1*rf1/r1)-(v2*rf1/r2);format(5); req=r5*r4/(r5+r4);//combination of r4 and r5 re1=(r3*r5)/(r3+r5);//...
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function t = taylor(x0 , x , n) t = exp(x0) for i = 2: n t = t+exp(x0)*(x - x0)^(i-1)/factorial(i-1) end endfunction t = taylor(-4.63,3.94,17)
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// Exa 1.17 clc; clear; close; // Given data ni= 1.8*10^16;// in /m^3 q= 1.6*10^-19;// in C em=0.14;// electron mobility in m^2/v-sec hm=0.05;// hole mobility in m^2/v-sec resistivity= 1.2;// in Ωm n= 1/(q*em*resistivity);// in /m^3 disp(n,"The electron concenration in /m^3 is :") p= ni^2/n;// in /m^3 disp...
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//Calculate efficiency of transformer //Chapter 3 //Example 3.27 //page 235 clear; clc; disp("Example 3.27") kVA=40; //rating of the transformer coreloss=450; //core-loss in watts Culoss=800; //copper loss in watt pf=0.8; //power factor of th...
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//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 11.5w //calculation of the work done by an external agent //given data //E = (10 N/kg)(i + j).....given gravitational field Ex=10//value of X-component of gravitational field(in N/kg) Ey=10//value of Y-component of g...
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exec('PMatrix.sci') exec('Doolittle.sci') P = PMatrix() lu = Doolittle(P) disp('Ejercicio h:') disp("L:") disp(lu.L) disp("U:") disp(lu.U)
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//clear// //Example9.3:Lapalce Transform x(t) = 3exp(-2t)u(t)-2exp(-t)u(t) syms t s; y = laplace('3*%e^(-2*t)-2*%e^(-t)',t,s); disp(y) //Result //(3/(s+2))-(2/(s+1))
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//Chapter-3,Example 4,Page 57 clc; close; m=234 // atomic mass of uranium M_0 = 4 // initial mass of uranium t_half= 2.48*10^5 // half life of uranium t= 62000*365*24*3600 // time period lamda=8.88*10^-14 M= M_0*exp(-lamda*t) printf('Mass of uranium left unchanged is %.3f mg', M) N= (M*6...
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//Network Theorem 1 //page no-2.39 //example2.37 disp("Applying KCL to node 1:"); disp("2*V1+17*V2 = 0");....//equation 1 disp("Applying KCL to node 2:"); disp("V1+6V2 = 0");...//equation 2 disp("Solving equations 1 and 2");...//solving equations in matrix form A=[2 17;1 6]; B=[0 0]' X=inv(A)*B; disp(X); di...
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function[bet] = backward(sequence, transitionMatrix, emissionMatrix) // Author: Benjamin Fradet n = length(sequence); numberStates = size(transitionMatrix, 1); bet = zeros(n, numberStates); // we start from state 1 bet(n, :) = [1, 1]; for i = n-1:-1:1 for j = 1:numberStates ...
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b=[1 2 3 2 3 4]; a=[1 2 3; 4 5 6]; y=filtord(b,a); disp(y); //output //!--error 10000 //check input dimension //at line 36 of function filtord called by : //y=filtord(b,a);
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nyaa nyaanee V;PL;1;PST nyaa nyaachaa jiraa V;SG;1;PRS nyaa nyaattee V;SG;2;PST nyaa ni nyaadhaa V;SG;1;FUT nyaa nyaachaa jiraa V;PL;1;PRS nyaa nyaattee V;FEM;SG;3;PST nyaa ni nyaattii V;FEM;SG;3;FUT nyaa ni nyaattuu V;PL;2;FUT nyaa nyaachaa jirtii V;FEM;SG;3;PRS nyaa nyaachaa jirtaa V;SG;2;PRS nyaa ni nyaatanii V;PL;3...
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clc;funcprot(0);//EXAMPLE 4.8 // Initialisation of Variables r=14;......................//Compression ratio t1=87+273;....................//Temperature of the charge at the end of the stroke in Kelvin p1=1;......................//Pressure of the charge at the end of the stroke in bar hsupa=1700;......................
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clc //Initialization of variables p=50 //atm pc= 73 //atm t= 459.7+212 //R tc=459.7+87.9 //R R=0.73 M=44 v=0.193 //ft^3/lbm //calculations pr=p/pc tr=t/tc z=0.88 //from compressibility charts p2= z*R*t/v/M //results printf("pressure = %.1f atm",p2)
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//ques-5.19 //Calculating pH of Calcium hydroxide clc M=0.005;//molarity of calcium hydroxide c1=2*M;//content of hydroxide ion (in mol/L) c2=10^-14/c1;//content of hydrogen ion (in mol/L) p=-log10(c2); printf("the pH required is %.0f.",p);
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function y=compound_residuals(x, Ainput,Ncomp ) //****************************************************************************** // Data Reconciliation Benchmark Problems From Literature Review // Author: Edson Cordeiro do Valle // Contact - edsoncv@{gmail.com}{vrtech.com.br} // Skype: edson.cv //**********************...
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//recuperation de l'image originale n=512; funcprot(0); l=read('/home/marlow/scilab/TD7/lena.csv',512,512); lena=l'; x=[1:512]; y=[512:-1:1]; xset('colormap',graycolormap(256)); //resolution du SVD [U,S,V]=svd(lena); sig=diag(S); clf; plot(sig,'o'); xlabel("k",'fontsize',4); ylabel("$$\\sigma_k'+'$$",'fontsize',4); ep...
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clc //Initialization of variables d=1.2 //m w=1 //m U=60*1000/3600 //m/s nu=1.5e-5 //m^2/s Cd=0.4 rho=1.22 //kg/m^3 //calculations Rn=U*d/nu A=d*w Fd= Cd*0.5*rho*U^2 *A M= 0.5*Fd //results printf("Bending moment = %.2f h^2 N m",M)
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// Examle 3.3 // Given L1= 2L2 // From the Diagram Leq= 0.5+ (L1*L2)/(L1+L2) // there for (L1*L2)/(L1+L2)= 0.2 ,( where Leq= 0.7) // i.e (2*L2*L2)/3L2= 0.2; /...
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clear; clc; Ro=600;R1=240; R2=((Ro*Ro)-(4*R1*R1))/(4*R1); d=acosh(1+(2*R1/R2)); printf("Value of shunt resistance = %d ohm\n",R2); printf(" Attenuation = %f db",round(d*8.686*10)/10);
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//Introduction to Fiber Optics by A. Ghatak and K. Thyagarajan, Cambridge, New Delhi, 1999 //Example 21.1 //OS=Windows XP sp3 //Scilab version 5.5.2 clc; clear; //given nf=1.51;//refractive index of film ns=1.50;//refractive index of substrate nc=1.0;//refractive index of cover d=4e-6;//thickness of film in m...
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//To Compare the results the percent voltage drop ratio for different loading //Page 204 clc; clear; //Voltage Drops in Percentage VDlumped=5; VDuniform=2.5; VDincreasing=3.333; //Ratio of the percent voltage drops Rlu=VDlumped/VDuniform; Rli=VDlumped/VDincreasing; Riu=VDincreasing/VDuniform; printf(...
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clc clear //Input data C=3.5//Capacity in litres P=13.1//Indicated power in kW/m^3 N=3600//Speed in rpm ve=82//Volumetric efficiency in percent p1=1.013//Pressure in bar T1=25+273//Temperature in K rp=1.75//Pressure ratio ie=70//Isentropic efficiency in percent me=80//Mechanical efficiency in percent g=1.4...
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//To calculate the conductivity, equilibrium hole concentration and position of Fermi level ni = 1.5*10^16; ////intrinsic charge carriers per m^3 e = 1.6*10^-19; mew_e = 0.135; //electron mobility, m^2/Vs mew_h = 0.048; //hole mobility, m^2/Vs sigma = ni*e*(mew_e+mew_h); //conductivity, ohm-1 m-...
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clc mode(1) //Zadatak za samostalni 2 scicv_Init() img_gray = imread(getSampleImage("lena.jpg"), CV_LOAD_IMAGE_GRAYSCALE); figure matplot(img_gray); title('Original'); img_mat=img_gray(:,:); A=double(img_mat) n = size(A,1); [U,S,V] = svd(A); k = 5; figure subplot(2,3,1) Ak = U(:,1:k)*S(1:k,1:k)*V(:,1:k)'; matplot(Ak)...
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// Test #8 : When output arguments are less than 5 exec('./zpklp2bpc.sci',-1); [z,p,k]=zpklp2bpc(2.3*%i,[4.2,%i],3,0.3,[0.12,0.34]); disp(k); disp(p); disp(z); // //Scilab Output //k=-1.310438 - 0.3284784i //p=0.4889351 + 0.4310543i // 0.9980440 + 0.0625157i //z=0.6269358 + 0.1312146i // //Matlab Output //z=0...
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clc //Initialization of variables E0=-0.11 //V H=10^-7 //calculations pH=-log10(H) E=E0-29.59*pH*10^-3 //results printf("Biological standard potential = %.2f V",E)
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clear// //Variable Declaration e_x=800*10**-6 //Strain in x e_y=200*10**-6 //Strain in y y_xy=-600*10**-6 //Strain in xy v=0.30 //Poissons Ratio E=200 //Youngs Modulus in GPa R_e=424.3*10**-6 //Strain e_bar=500*10**-6 //Strain //Calculations //Part 1 R_sigma=10**-6*R_e*(E*10**9/(1+v)) //Stress in MPa s...
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//Variable declaration sample1 = [0.27,0.35,0.37] // Copper content-1 sample2 = [0.23,0.15,0.25,0.24,0.30,0.33,0.26] // Copper content-2 Yvalue1 = [1,1,1] Yvalue2 = [1,1,1,1,1,1,1] //Results plot(sample1,Yvalue1,"bo") plot(sample2,Yvalue2,"ro") title("Dot Diagram") xlabel("$Copper Content$")
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function [u,x,t] = heat_1d_euler_exp(a,xf,T,initialTemp,bx0,bxf,Nx,Nt) // solve 1d heat equation using explicit Euler method // // a u_xx = u_t // // for: 0<=x<=xf, 0<=t<=T // // Initial condition: u(x,0) = it0(x) // // Boundary condition: u(0,t) = bx0(t), u(xf,t) = bxf(t) // // Nx = no. of subintervals along x-axis...
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a b a b i l א ב א ב י ל a b a d a n ע ב ד א ן a b d o h ע ב ד ו a b d u h ע ב ד ו ה a b h u r א ב ח ו ר a d e ע א ד a d o l p h e א ד ו ל ף a d r i a t i c א ד ר י א ט י ק a d w a n ע ד ו א ן a f a r ע פ א ר a h m a d א ח מ ד a j r a m ע ' ג ר ם a l a s k a n א ל ס ק א ן a l e x a n d r u א ל ק ס נ ד ר ו a l l a n א ל ...
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//find IDQ and VGSQ. //Example 4.14 page no 123 clear clc Vdd=15 //v Vdsq=7 //v Rs=3 //kΩ Rd=1 //kΩ Idq=((Vdd-Vdsq)/(Rs+Rd)) printf("\n Idq=%0.2f mA" ,Idq) Vgsq=-(Idq*Rd) printf("\n Vgsq=%0.2f V" ,Vgsq)
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//finding maximum conversion time// //example 21// clc //clears the command window//; clear //clears// c=2^10;//maximum no of counts// f=2*10^-6;//counter advance rate of 1 count per second// T=c*f;//conversion time// disp('maximum conversion time:') disp(T);//displaying result//
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//Example 4.16 //Find DTFT of x[n]=(a^n)u[n],for 0<a<1 clc; syms w a n; x=a^n; X=symsum(x*exp(-%i*w*n),n,0,%inf);
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//Exa 9.7 clc; clear; close; //given data // IC 723 Id=1;//in mA Vsense=0.7;//in volts Vo=15;//in volts Im=50;//in mA Vr=7;//in volts R1=(Vo-Vr)/(Id*10^-3); R2=Vr/(Id*10^-3); R3=(R1*R2)/(R1+R2); Rcl=Vsense/(Im*10^-3); disp("various resistance values for the circuit is as follows : ") disp(R1,"R1 :") di...
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//Example 5.40 //Newton Raphson Method //Page no. 206 clc;clear;close; deff('y=f(F)','y=-10*F^3-21*F+10') deff('y=f1(F)','y=-21-30*F^2') printf('n\txn\t\t\f(xn)\t\tf1(xn)\t\tXn+1\t\tError\n') printf('-----------------------------------------------------------------------------------------------------\n') x0=1;e...
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//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 6.6w //calculation of the values of coefficient of static and kinetic friction //given data M=2.5//mass(in kg) of the block F=15//horizontal force(in N) g=10//gravitational acceleration(in m/s^2) of the earth x=10//...
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; Procedure Tests for Cthulhu Scheme ; Scot W. Stevenson <scot.stevenson@gmail.com> ; First version: 10. Mai 2020 ; This version: 10. Mai 2020 ; Formats: ; - empty lines are ignored ; - lines that start with a semicolon ';' like this one are ignored ; - lines that start with a SECTION are printed to the output ; - Al...
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clc;funcprot(0);//EXAMPLE 11.11 // Initialisation of Variables a=4500;.................//Altitude afr=14;...............//Air fuel ratio at sea level t1=25;...........//Temperature at sea level in Celsius p1=1.013;...........//Pressure at sea level in bar //Calculations t2=t1-(0.0064*a);............................
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10 1073:0.16666666666666666 1074:1.0 10 7:0.5 21:0.25 22:0.058823529411764705 431:1.0 486:0.5 649:1.0 10 4:0.2 7:0.5 54:0.25 72:1.0 91:0.6666666666666666 93:1.0 150:0.5 157:0.5 299:1.0 311:0.16666666666666666 419:1.0 421:1.0 798:1.0 1066:1.0 1084:1.0 1085:1.0 1087:1.0 1164:0.5 1367:1.0 1426:1.0 1436:1.0 10 22:0.0588235...
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clc; clear; printf("\n Example 6.4\n"); l=0.3;//length of tube printf("\n Given:\n length of tube = %.1f m",l); id_t=25e-3;//Top internal diameter of tube printf("\n Top internal diameter of tube = %d mm",id_t*1e3); id_b=20e-3;//Bottom internal diameter of tube printf("\n Bottom internal diameter of tube = %d...
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//Example No. 7.10.4 clc; clear; close; format('v',7); N1=1;//no. of turns in primary N2=8;//no. of turns in secondary //a=lambda/25; aBYlambda=1/25;//(temporary calculation) //A=%pi*a^2 A_BY_lambda_sqr=%pi*aBYlambda^2;//(temporary calculation) Rr1=31200*(N1*A_BY_lambda_sqr)^2;//Ω(Radiation resistance for si...
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function [r]=%l_o_l(l1,l2) //l1==l2 //! // Copyright INRIA n1=size(l1) r=n1==size(l2) if r&n1>0 then r=%f(ones(1,n1)) sel=%f(ones(1,n1)) k1=definedfields(l1) k2=definedfields(l2) for i=intersect(k1,k2), if l1(i)==l2(i) then r(i)= %t,end end s1=1:n1;s1(k1)=[]; s2=1:n1;s2(k2)=[]; k1=intersect(s1,s...
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//chapter-7 page 278 example 7.1 //============================================================================== clc; clear; a=4;//Length of Waveguide in cm b=2.5;//breadth Waveguide in cm f=10^10;//Frequency in Hz x=0.1;//distance between twice minimum power points in cm c=3*10^10;//Velocity of Light in cm/...
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[3,2,1] | [1,1] = divide: quot[1] = 1, remd = [3,1] divide: quot[0] = 1, remd = [2] gcd: [3,2,1] / [1,1] -> [1,1] rest [2] divide: quot[1] = 1/2, remd = [1] divide: quot[0] = 1/2, remd = [0] gcd: [1,1] / [2] -> [1/2,1/2] rest [0] result: [1/2,1/2]
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function o=standard_define(sz,model,label) //initialize graphic part of the block data structure [lhs,rhs]=argn(0) if rhs<3 then label=' ',end [nin,nout,ncin,ncout]=model(2:5) if nin>0 then pin(nin,1)=0,else pin=[],end if nout>0 then pout(nout,1)=0,else pout=[],end if ncin>0 then pcin(ncin,1)=0,else pcin=[],end if ncou...
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clc; clear all; C=2.4*1e-12;//given capacitance in F e0=8.854*1e-12;//permittivity of vacume a=4*1e-4;//area in m^2 d=0.5*1e-2;//thickness tandelta=0.02; er=(C*d)/(e0*a);//relative permittivity disp(er,'relative permittivity is='); lf=er*tandelta;//loss factor disp(lf,'electric loss factor is='); delta=atan(...
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//(Welded and Riveted Joints) Example 8.27 //Refer Fig.8.70 on page 324 //Number of rivets n n = 9 //Permissible shear stress tau (N/mm2) tau = 60 //Eccentric force P (kN) P = 50 //Eccentricity e (mm) e = 300 //Radial distance of rivets 2, 4, 6 and 8 from the C.G.(5) r2 (mm) r2 = 100 //Number of rivets wi...
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function y=f(x) y=x^2-2; endfunction x=2; y=1; ITMAX=1000; precision=1e-10; erreur=zeros(ITMAX,1); //initialisation for k=1:ITMAX y=x-f(x)*(x-y)/(f(x)-f(y)); //passage par une variable temporaire pour inverser les valeurs de x et y temp=x; x=y; y=temp; erreur(k)=abs(y-sqrt(2)); ...
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clear //Given b=238 c=206 d=92 e=82 //Calculation a=(b-c)/4.0 A=-d+(2*a)+e //Result printf("\n (i) The emission of alpha particle will reduce the mass number by 4a and charge number by 2a")
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clear;lines(0); p=poly([0,10,1+%i,1-%i],'x'); roots(p) A=rand(3,3);roots(poly(A,'x')) // Evals by characteristic polynomial spec(A)
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function [x,y,typ]=PROD_f(job,arg1,arg2) x=[];y=[];typ=[]; select job case 'plot' then [orig,sz]=arg1(1:2) wd=xget('wdim') graphics=arg1(2); [orig,sz,orient]=graphics(1:3) thick=xget('thickness');xset('thickness',2) p=wd(2)/wd(1) rx=sz(1)*p/2 ry=sz(2)/2 xarc(orig(1),orig(2)+sz(2),sz(1)*p,sz(2),0,23040...
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clear //Initialization q=15*10**-6 //charge in coulomb a=200*10**-6 //area //Calculation d=q/a //electric flux density //Results printf("\n D = %d mC/m^2",d*10**3)
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//<f>=%sxr(n1,f2) // %sxr(,M,r) calcule le produit element par element de la matrice de //scalaires M par la matrice de fractions rationnelles r . (M.*r) //! f=list(f2(1),n1.*f2(1),f2(3),f2(4)) //end
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//Ex 5.5 clc; clear; close; format('v',4); Ad=5:200;//Gain R1max=50;//kohm(Potentiometer) R4=10;R3=10;//kohm //Case 1st : Ad=Admin &R1=R1max R1=R1max;//kohm R2=(min(Ad)-1)/2*R1max //Case 2nd : Ad=Admax &R1=R1min R1min=2*R2/(max(Ad)-1);//kohm disp(R2,"Resistance R2(kohm)"); disp(R1min,"Minimum value of res...
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// Scilab code Ex11.2: Pg 383 (2005) clc; clear; // Part (a) f = 6.42e+13; // Frequency of absorption, Hz omega = 2*(%pi)*f; // Angular frequency of absorbed radiations, Hz mu = 1.14e-26; // Reduced mass of CO molecule, kg K = mu*(omega^2); // Effective force constant of CO molecule, N/m printf("\nTh...
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Generation Of BPSK and BFSK signal: // NRZ polar line coding technique // Generating BPSK and BFSK signals clc; clear; fc = input('Enter the carrier frequency'); //2 n = input('Input Message sequence'); //[1 0 0 1 1 1 0 1] for i=1:length(n) if n(i) == 1 then seq(i) = 1; else seq(i) = -1; e...
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<?xml version="1.0" encoding="utf-8"?> <test> <description>Linear stability with coupled solver (Arpack LI): Channel Largest real Ev = (0.00223554,+/-0.249844i)</description> <executable>IncNavierStokesSolver</executable> <parameters>ChanStability_Coupled.xml</parameters> <files> <file descripti...
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clc clear //INPUT DATA L=1;//Length of the bar in m l=0.25;//Length of the pemdulum in m //CALCULATIONS k=sqrt((L^2)/12);//Radius of gyration m T=sqrt(((k^2/l)+l)/9.8)*2*3.14;//Time period of pendulum in s //OUTPUT mprintf('Time period of the pendulum is %3.3f sec',T)
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//Example Water Jet Stinking a Stationary Plate V_1 = 20 //velocity of water jet [m/s] mdot = 10 //water mass flow rate of stiking the stationary plate [kg/s] beta1 = 1 //momentum flux correction factor
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clc,clear printf('Example 4.15\n\n') P=30*1000/3 //power per phase V_ph=400/sqrt(3) //phase voltage R=(V_ph)^2/P //resistance of strip t=0.025*10^-2 //thickness of strip S=1.03*10^-6 //specific resistance of nichrome alloy l_by_w = R*t/S //because R=specific_resistance*l/(w*t) ...
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vo = zeros (1, 50) v1 = 10*ones (1, 50) v2 = 0:0.3:10 v5 = linspace (-3, 7, 50) v6 = 1:1:25 p = 2*ones (1, 25) p.^v6 function r = truc(x) r = (1+x) .* sin(%pi .* x) endfunction x = linspace (-2, 2, 100) a = truc(x) fenetre = figure("Figure_name", "Equations", "position", [100 50 1000 600]); fenetre.backgro...
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// problem 9.3 D=0.15 s=0.3 Hs=3 Hd=30 n=0.8 a=3.142*D*D/4 N=60/60 w=9810 Q=0.62/60 Qth=(2*a*s*N) slip=(Qth-Q)/Qth power=(w*Qth*(Hs+Hd))/(1000*n) disp(slip*100,power,"power in Kw required to drive the pump,percentage slip")
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//Example 5.12 clear; clc; //Given HfNH4NO3=-365;//enthalpy of formation of NH4OH in kJ mol^-1 HfH2=0;//enthalpy of formation of H2 in kJ mol^-1 HfH2O=-242;//enthalpy of formation of H2O in kJ mol^-1 HfN2H4=50;//enthalpy of formation of N2H4 in kJ mol^-1 SoNH4NO3=150;//Standard entropy of NH4NO3 molecule in J...
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//Page Number: 43 //Example 1.18 clc; //Given er=2.2; n0=377;//ohm n2=n0/sqrt(er);//ohm n1=377;//ohm //Reflection coefficient t=(n2-n1)/(n2+n1); disp(t,'Reflection coefficient:'); //Vswr //Taking mod of reflection coefficient t1=-t; p=(1+t1)/(1-t1); disp(p,'VSWR:');
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function [stk,txt,top]=sci_upper() // Copyright INRIA txt=[] stk=list('convstr('+stk(top)(1)+',''u'')','0',stk(top)(3),stk(top)(4),'10')
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//Function migration (image list to matrix) for: boundingRect //Generated by migrate.cpp //Author: Anirudh Katoch function res = boundingRect(varargin) select length(varargin) case 01 then res = raw_boundingRect(varargin(01)) else error(39) end endfunction
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//Transport Processes and Seperation Process Principles //Chapter 7 //Example 7.2-1 //Principles of Unsteady state and convective mass transfer //given data pa1=0.2*101325;//Pa pa2=0; P=2*101325; ya1=pa1/P; ya2=pa2/P; yb1=1-ya1; yb2=1-ya2; ybm=(yb2-yb1)/((log(yb2/yb1))/log(2.71828183)); kdy=6.78/(10^5); k...
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clear; clc(); // To find the total heat flow per foot of depth through the sction and the shape factor k=0.9; // thermal conductivity of section material in Btu/hr-ft-degF // Heat is considered to flow through fictitious rods and only half of the heat flows through symmetry axes ...
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syms Y t X1=Y; X2=diff(Y,t); X3=diff(X2,t); A=[0 1 0;0 0 1; -2 -7 -4] B=[0 ; 0 ; 5] disp("OUTPUT IS C*X + D*U WHERE ") C=[1 0 0 ]; D=0; disp(C,"C=") disp(D,"D= ")
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// Exa 7.14 clc; clear; // Given // An LVDT vo = 2.6; // Output voltage(volts) of LVDT d = 0.4; // displacement in mm // Solution printf(' The sensitivity s = RMS value of output voltage/Displacement \n'); S = vo/d; // sensitivity printf(' Therefore, s = %.1f V/mm \n',S);
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disp("En=-(13.6)*(Z^2)/(n^2)"); n1=1;n2=2;n3=3; Z=3; h=6.6*10^-34; c=3*10^8; E1=-(13.6)*(Z^2)/(n1^2); E2=-(13.6)*(Z^2)/(n2^2); E3=-(13.6)*(Z^2)/(n3^2); printf('\n The value of E1 is %f eV',E1); printf('\n The value of E2 is %f eV',E2); printf('\n The value of E3 is %f eV',E3); a1=91.8;a2=108.8;a3=114.75; //...
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void main() { int a = 10; do { print("DO_OK"); } while(a == 10) { print("DO_OK"); } }