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//Find the triangular factors L and U for the matrix A = [1,2,4;0.5,6,3;0.2,-2,9] clc;clear; A=[1,2,4;0.5,6,3;0.2,-2,9]; U=A; disp(A,'The given matrix is A=') m = det(U(1,1)); n = det(U(2,1)); a=n/m; U(2,:) = U(2,:) - U(1,:)/(m/n); n = det(U(3,1)); b= n/m; U(3,:) = U(3,:) - U(1,:)/(m/n); m = det(U(2,2)); n...
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//Bore of the engine(in cm) D=10; //Length of stroke for square engine(in cm) L=10; //Number of cylinders k=4; //Volumetric effciency nv=0.75; //Speed(in rev/s) N=40; //Density of air Pa=1.15; //Coefficient of air flow Cd=0.75; //Area of orifice(in m^2) A2=0.25*%pi*0.03^2;
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//Chapter 13 : Thin Film Preparation Techniques and their Applications clear; //Variable declaration delV1=2*10**-3 //milivolts to volts delI1=4*10**-6 //microAmpere to Ampere //Calculations Rs=delV1/delI1 //Result mprintf("Series Resistance = %d V/m",Rs)
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${ using Typewriter.Extensions.Types; Template(Settings settings) { settings.IncludeProject("Host"); settings.OutputFilenameFactory = (file) => { return file.Name.Replace(".cs", ".ts"); }; } }namespace Api {$Enums(x => x.Namespace == "Imglib.Host.Controller.Model")[ export enum $Name {$Values[ $Name ...
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clc // Given that lambda = 5893 // mean wavelength in angstrom n = 2 // order N = 5000 // Grating lines per cm theta = 2.5 // Separation in second // Sample Problem 25 on page no. 166 printf("\n # PROBLEM 25 # \n") d_theta = %pi/180*theta/60 // Angle in radian d_lambda = d_theta*sqrt((1/(n*N)^2)-(lambda*1e-...
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//Caption:In a single phase transformer Calculate (a)The efficiency at full load, unity power factor. (b)The efficiency at full load, 0.8 lagging power factor. (c)The efficiency at full load, 0.8 leading power factor. //Exam:3.27 clc; clear; close; P_f1=1;//power factor unity P_f2=0.8;//power factor 0.8 lagging o...
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/* ============================================================================= Escola Politécnica da USP PME3402 - Laboratório de Medição e Controle Discreto -------------------------------------------------------------- ATIVIDADE 5 - CONTRO...
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clear // // //Initilization of Variables sigma1=30 //N/mm**2 //Stress in tension d=20 //mm //Diameter sigma2=90 //N/mm**2 //Max compressive stress sigma3=25 //N/mm**2 //Calculations //In TEnsion //Corresponding stress in shear P=sigma1*2**-1 //N/mm**2 //Tensile force F=%pi*4**-1*d**2*sigma1 //In Compression //...
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// This file is released under the 3-clause BSD license. See COPYING-BSD. // Generated by builder.sce: Please, do not edit this file try getversion("scilab"); catch error("Scilab 5.4 or more is required."); end; fileQuit = get_absolute_file_path("unloader.sce") + "etc\" + "arduino.quit"; if isfile(fileQui...
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//=========================================================================== //chapter 5 example 20 clc; clear all; //variable declaration L = 170; //length of the wire in mm dL = 0.2; //increase in length in mm L1 =100; //length of the second wire in mm //calculations S = sqrt(...
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//Parameters model of pendulum M=0.696 //Masa del carro (Kg) m=0.017 //masa del pendulo (Kg) l=0.3 //longitud de la barra (m) g=9.8 //aceleracion gravitacional (m/s2) b=0.001 //coeficiente de friccion (Ns/m) I=0.0011 //Inercia del pendulo (Kgm2)
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//clear// //Example 4.4 // Continuous Time Fourier Transform //and Frequency Response of a Square Waveform // x(t)= A, from -T1 to T1 clear; clc; close; // CTS Signal A =1; //Amplitude Dt = 0.005; T1 = 4; //Time in seconds t = -T1/2:Dt:T1/2; for i = 1:length(t) xt(i) = A; end // // Continuous-tim...
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//Exa 3.2 clc; clear; close; // given data AF=100;//unitless A=2*10^5;//unitless Ri=1;//in Mohm Ro=75;//in ohm //let R1 =1 ohm R1=1;//in ohm //formula : AF=1+RF/R1 RF=(AF-1)*R1;//in kohm B=1/AF;//unitless RiF=(1+A*B)*Ri*10^6;//in ohm RoF=Ro/(1+A*B);//in ohm disp(RF,"Value of RiF in kohm is : "); disp(R...
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// Problem 1.8,Page no.12 clc;clear; close; alpha=%pi/2 //degree //In case of semicircle //Semicircle-1 r_1=20 //cm //radius of semicircle y_1=4*r_1*(3*%pi)**-1 //cm //distance from the base a_1=(%pi*r_1**2)*2**-1 //cm**2 //area of semicircle //Semicircle-2 r_2=16 //cm //radius of semicircle y_2=4*r_2*(3*%pi)*...
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errcatch(-1,"stop");mode(2);function [y]=series(sys1,sys2) y=sys1*sys2 endfunction exit();
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clc //initialization of variables l = 0.07 // flim thickness in cm v = 3 // water flow in cm/sec D = 1.8*10^-5 // diffusion coefficient in cm^2/sec crat = 0.1 // Ratio of c1 and c1(sat) //Calculations z = (((l^2)*v)/(1.38*D))*((log(1-crat))^2) //Column length //Results printf("the column length needed is %.1...
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//binary to decimal conversion// //example 3// clc //clears the command window// clear // clears // p =1; // initialising // q =1; z =0; b =0; w =0; f =0; //bin= input ( Enter the binary no to be converted to its decimal equivalent : ) //accepting the binary input from user// bin =1100.11; d =modulo(bin...
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// Electric Machinery and Transformers // Irving L kosow // Prentice Hall of India // 2nd editiom // Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS // Example 9-8 clear; clc; close; // Clear the work space and console. // Given data (Exs.9-5 through 9-7) P = 8 ; // Number of poles in the SCIM f ...
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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_...
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//Graphical// //Figure 8.9 and 8.10 //PROGRAM TO DESIGN AND OBTAIN THE FREQUENCY RESPONSE OF FIR FILTER //LOW PASS FILTER clear; clc; close; M = 61 //Filter length = 61 Wc = %pi/5; //Digital Cutoff frequency Tuo = (M-1)/2 //Center Value for n = 1:M if (n == Tuo+1) hd(n) = Wc...
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example_4_14.sce
//Scilab Code for Example 4.14 of Signals and systems by //P.Ramakrishna Rao clear; clc; close; a=.5; A=1/(sqrt(2)*%pi); t=-10:0.1:10; x=A*exp(-a*t.*t); disp("Guassian pulse signal x(t)=(1/sqrt(2)*%pi)*exp(-a*t^2)"); disp("X(w)=integral(exp(-a*t^2)*exp(-%i*w*t)) w.r.t dt"); disp("d(X(w))/dw=-%i*w/(2*a)*inte...
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//Ch25_Ex21 clc; clear; close; r=7; h=24; w=1.25; l=sqrt(h^2+r^2); area=%pi*r*l; lengthCanvas=area/w; mprintf("The length of canvas is %.0f meter",lengthCanvas);
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rows = 3; cols = 3; A = zeros(rows, cols); disp("Enter the 3x3 matrix A"); for i = 1:rows for j = 1:cols A(i,j) = input("value for A:") end end n = length(A(1,:)); aug = [A, eye(n,n)]; for j = 1:n-1 for i = j+1:n aug(i, j:2*n) = aug(i, j:2*n) - aug(i, j) / aug(j, j) * aug(j, j:2*n); en...
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clc //given ne=31 na=25 nb=90 nc=83 Ta=10 //lbft //Ne-Nf/(Nc-Nf)=-83/31 k=114/83//k=Nc/Nf As Ne = 0, on simplification we get Nc/Nf= 114/83 j=-90/25//j=Na/Nb //Nc=Nb, Thus Na/Nc=-90/25 //Na/Nf=(Na/Nc)*(Nc/Nf) ie Na/Nf=k*j //Tf*Nf=Ta*Na Tf=Ta*k*j printf("\nTorque exerted on driven shaft = %.1f lb.ft\n",T...
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clc clear //INPUT DATA t2=50;//dry bulb temperature in Degree c t1=30;//dry bulb temperature in Degree c t11=25;//wet bulb temperature in Degree c V=300;//volume in m^3 Ra1=287.3;//rate of flow p=760;//pressure in mm of Hg pva=23.74;//Saturation pressure in mm Hg cp=1.005;//specific pressure ps2=92.54;//Satu...
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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 10: POWER SYSTEM STABILITY // EXAMPLE : 10.2 : // Page number 270 clear ; clc ; close ; // Clear the work space and cons...
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//Problem 25.07: (a) For the network diagram of Figure 25.8, determine the value of impedance Z1 (b) If the supply frequency is 5 kHz, determine the value of the components comprising impedance Z1. //initializing the variables: RL = %i*6; // in ohm R2 = 8; // in ohm Z3 = 10; // in ohm rv = 50; // in volts theta...
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//Example 12.4 //Probability to find the required sample size. disp('Let A be the event of choosing a sample size of 6 containing two red, one green , two blue and one white blue ball.'); funcprot(0) function c = combination ( n , r ) c = prod ( n : -1 : n-r+1 )/ prod (1:r) endfunction disp('The number of combin...
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clear //variable declaration A1=150.0*12.0 //Area of 1 ,mm^2 A2=(200.0-12.0)*12.0 //Area of 2,mm^2 X1=75 X2=6 Y1=6 Y2=12+(200-12)/2 A=A1+A2 xc=(A1*X1+A2*X2)/A printf("\n xc= %0.2f ",xc) yc=(A1*Y1+A2*Y2)/A printf("\n yc= %0.2f mm",yc) printf("\nThus, the centroid is at x = 36.62 mm and y = 61.6...
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//Part A Chapter 3 Example 3 clc; clear; close; format('v',6); t_ice=0;//degree C E0=0.003*t_ice-5*10^-7*t_ice^2+0.5*10^-3;//V t_steam=100;//degree C E100=0.003*t_steam-5*10^-7*t_steam^2+0.5*10^-3;//V t=30;//degree C E30=0.003*t-5*10^-7*t^2+0.5*10^-3;//V t=((E30-E0)/(E100-E0))*(t_steam-t_ice);//degree C dis...
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-- Fuzzy Logix, LLC: Functional Testing Script for DB Lytix functions on Netezza -- -- Copyright (c): 2014 Fuzzy Logix, LLC -- -- NOTICE: All information contained herein is, and remains the property of Fuzzy Logix, LLC. -- The intellectual and technical concepts contained herein are proprietary to Fuzzy Logix, LLC. -...
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clc clear //Input data Tu=645;//The temperature at the end of compression process in K usu=310;//The internal energy at the end of compression process in kJ/kg air pu=(15.4*1.013);//The pressure at the end of the compression process in bar Vu=0.124;//The volume at the end of the compression process in m^3/kg ai...
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- 4*a - 3*a^2 - 2*a^3 - a^4 + w - 2; a + 3*a^2 + 2*a^3 + a^4 + x + 2; a^3 + y - 2; 3*a + 3*a^2 + 2*a^3 + z + 2; w^3 + x^3 + y^3 + z^3 isolated Signature: /z.01 isolated variable: z with Coefficient 1 remaining RelationSet: - 4*a - 3*a^2 - 2*a^3 - a^4 + w - 2; a + 3*a^2 + 2*a^3 + a^4 + x + 2; a^3 + y - 2; w^3 + x^3...
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kp=xget("pixmap");xset("pixmap",1); xset("wwpc"); // clean pixmap t=%pi*(-5:5)/5; //first plot, to fix boundaries plot3d1(t,t,sin(t)'*cos(t),35,45," ",[1,2,4]); xset("wshow"); // show pixmap if driver()=='Pos' then st=4;else st=2;end; for i=35:st:80, // loop on theta angle xset("wwpc"); plot3d1(t,t,sin(t)...
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//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504 //Chapter-2, illustration 4, Page 36 //Title: Expressing weight percent into mole percent //=================================================...
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//system// s=%s; sys=syslin('c',12/(s*(s+1)*(s+2))) nyquist(sys) show_margins(sys,'nyquist') gm=g_margin(sys) if (gm<=0) printf("system is unstable") else printf("system is stable");end;
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// PG (6) // Taylor series for e^(-x^2) upto first four terms deff('[y]=f(x)','y=exp(-x^2)') funcprot(0) deff('[y]=fp(x)','y=-2*x*exp(-x^2)') funcprot(0) deff('[y]=fpp(x)','y=(1-2*x^2)*(-2*exp(-2*x^2))') funcprot(0) deff('[y]=g(x)','y=4*x*exp(-x^2)*(3-2*x^2)') funcprot(0) deff('[y]=gp(x)','y=(32*x^4*exp...
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// @Harness: verifier // @Purpose: "Test for unresolved formats" // @Result: "UnresolvedFormat @ 7:19" architecture unr_format_02 { format F = { s[15:0] } addr-mode AM { encoding = F where { s = 1 } encoding = F2 where { s = 0 } } }
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clc; clear; format('v',11); w=2*%pi*10^7; //from inspection of the given E field. f=w/(2*%pi); c=3*10^8; //c=velocity of the wave in air. lemda=c/f; k=2*%pi/lemda; disp(lemda,"The wavelength(in meter)="); disp(k,"The propagation constant,k(in rad/m)=");
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clear clc // Given that L = 60 // Cooling load in kW p = 1 // Pressure in bar t = 20 // Temperature in degree celsius v = 900 // Speed of aircraft in km/h p1 = 0.35 // Pressure in bar T1 = 255 // Temperature in K nd = .85 // Diffuser efficiency rp = 6 // Pressure ratio of compressor nc = .85 // Copressor ef...
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//Problem 44.02:A transmission line has an inductance of 4 mH/loop km and a capacitance of 0.004 μF/km. Determine, for a frequency of operation of 1 kHz, (a) the phase delay, (b) the wavelength on the line, and (c) the velocity of propagation (in metres per second) of the signal. //initializing the variables: L = 0.00...
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getd ../common_files/ exec ../common_files/loader.sce exec ser_init.sce exec ramp_test.sci xcos ramp_test.xcos
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//Chapter 13: Fuel and Combustions //Problem: 2 clc; //Declaration of Variables C = 90 // % O = 3.0 // % S = 0.5 // % N = 0.5 // % ash = 2.5 // % LCV = 8490.5 // kcal / kg // Solution mprintf("HCV = LCV + 9 * H / 100 * 587\n") mprintf(" H...
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clc clear //INPUT DATA k=0.05;//clearance p1=0.98;//initial pressure in bar pd=6.4;//delivery pressure in bar n=1.32;//index of compression and expansion p0=1;//initial pressure t1=305;//temperature in K v0=17;//volume in m^3 t0=288;//teperature in K vs=0.02;//volume per stroke in m^3 //CLACULATIONS nv=1...
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errcatch(-1,"stop");mode(2);//Example 4.6: Resistance ; ; vr=5;//V r=10;//k-ohm x=vr*r*10^3;// R=x;// disp(R*10^-3,"resistance is ,(k-ohm)=") exit();
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//i/p arg x is a matrix x=[1 2 3; 4 5 7;8 9 8]; fc=100; fs=500; y = modulate(x,fc,fs,'am'); disp(y); ////output // 1. 2. 3. // 1.236068 1.545085 2.163119 // - 6.472136 - 7.2811529 - 6.472136 //
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clc //Initialization of variables cp=0.25 t2=3460 //R t1=946.2 //R etat=0.45 Q=-489 t3=520 //R etat2=0.384 //calculations Qa=cp*(t2-t1) w=etat*Qa eps=-w/Q I=w+Q Qa2= cp*(t2-t3) W2=etat2*Qa2 eps2=-W2/Q I2=W2+Q //results printf("In case 1, Effectiveness of cycle = %d percent",eps*100) printf("\n in ca...
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//Caption:Find the frequency of voltage generated //Exa:13.1 clc; clear; close; p=16//Number of poles n=375//Speed of alternator(in r.p.m) f=(p*n)/120 disp(f,'Frequency of voltage generated(in c/s)=')
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// ----------------------------------------------------------------------- /// \brief Calcule un terme de contrainte a partir d'une homographie. /// /// \param H: matrice 3*3 définissant l'homographie. /// \param i: premiere colonne. /// \param j: deuxieme colonne. /// \return vecteur definissant le terme de contrainte...
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//Kunii D., Levenspiel O., 1991. Fluidization Engineering(II Edition). Butterworth-Heinemann, MA, pp 491 //Chapter-6, Example 5, Page 161 //Title: Reactor Scale-up for Geldart B Catalyst //========================================================================================================== clear clc //IN...
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clear; close; clc; A=[1 0;0 1;1 1]; disp(A,'A='); B=[1;1;0]; disp(B,'B='); x=(A'*A)\(A'*B); disp(x,'x=');; C=x(1,1); D=x(2,1); disp(C,'C='); disp(D,'D='); disp('The line of best fit is B=C+Dt');
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//clc(); clear; //To determine the wavelength Eg=1.43*1.602*10^-19; //band gap energy in J T=300; //temperature in K h=6.626*10^-34; c=3*10^8; lambda=(h*c)/Eg; disp(lambda); lambda=lambda*10^6; //converting into micrometre printf("the GaAs photodetector will cease to operat...
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clc;clear; a=imread('gtw.jpg'); d=double(mtlb_double(a)); me=d(:,:,1); hi=d(:,:,2); bi=d(:,:,3); bme=me; bhi=hi; bbi=bi; m=(1/9)*ones(3,3); [r1,c1]=size(mtlb_double(a)); for i=2:r1-1 for j=2:c1-1 a1me=1/me(i-1,j-1)+1/me(i-1,j)+1/me(i-1,j+1)+1/me(i,j-1)+1/me(i,j)+1/me(i,j+1)+1/me(i+1,j-1)+1/me(i+1,j)+1/me(i+...
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function [found, coordinates] = findChessboardCorners(image1,pts_row,pts_col,flags) img1 = mattolist(image1); [found coordinates] = opencv_findChessboardCorners(img1,pts_row,pts_col,flags); endfunction
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function obj=scan_mdaq_blocks(scs_m) global %microdaq; obj = []; mdaq_sim_blocks = ["mdaq_adc_sim","mdaq_dac_sim","mdaq_dio_config_sim",.. "mdaq_dio_get_sim","mdaq_dio_set_sim","mdaq_encoder_sim",.. "mdaq_func_key_sim","mdaq_led_sim","mdaq_pru_reg_get_sim",.. "mdaq_pru_reg_set_sim","mdaq_pw...
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// Exa 7.6 clc; clear; // Given data // A notch filter fo=50; // cutoff frequency for notch filter(Hz) //Solution printf('As Given fo=50 Hz. Let C=0.1 μF.'); C=0.1*10^-6; // Farads // since fo=1/(2*%pi*R*C); // Therefore R - R=1/(2*%pi*fo*C); printf(' \n For R/2, take two resistors of 31.8 k Ohms ...
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exec("chi2probln.sci",-1); exec("chi2cdf.sci",-1); exec("nonzero.sci",-1); function y=chi2conj(x,k) [df,nsamps]=size(x); lnp=chi2probln(x,k); y0=exp((1/(k/2-1))*(lnp+(k/2)*log(2)+gammaln(k/2))); y0=max(y0,max(0,2*sqrt(k-2)-sqrt(x)) .^ 2); y=y0; iterate=%T; i=0; cutoff=sqrt(%eps); while (iterate) ...
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//minimum size of the particle in the mixture of quartz and galena(mm) clear all; clc; printf("\n Example 1.4"); //maximum size of the particle(mm) d_max=0.065; //minimum size of the particle(mm) d_min=0.015; //density of quartz(kg/m^3) p_quartz=2650; //density of galena (kg/m^3) p_galena=7500; //minimum ...
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clc //Chapter 4:Frequency selective networks and transformers //example 4.2 //given //Forty decibles corresponds to a voltage ratio of 100:1 therefore since A(jwo)=1 Ajwo=0.01 n=5//no. of harmonics Q=n/(Ajwo*(n^2-1))//quality point mprintf('the minimum circuit Q is =Qmin = %f ',Q)
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function [stk,txt,top]=sci_meshgrid() // Copyright INRIA txt=[] if rhs==2 then X=stk(top-1)(1) Y=stk(top)(1) txt='// translation of meshgrid('+makeargs([X,Y])+')' if ~isname(X) then X=gettempvar(1) txt=[txt; X+'='+stk(top-1)(1)] end txt=[txt; X+'='+X+'(:)'''] if ~isname(Y) then Y=gett...
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//Initilization of variables Wc=100 //lb r= 1 //ft F=80 //lb k=50 //lb/ft s=6 //in g=32.2 //ft/s^2 //Calculations //Work done on the system U=-0.5*k*(1)+F*(s/12) //ft-lb //Initial KE is zero Vo=sqrt(U/(0.5*(Wc/g+0.5*(Wc/g)*r))) //ft/s //Result clc printf('The initial speed is %f ft/s',Vo)
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//pathname=get_absolute_file_path('1.03.sce') //filename=pathname+filesep()+'1.03-data.sci' //exec(filename) //Difference in mercury column(in m): h=30*10^-2 //Atmospheric Pressure(in kPa): pa=101 //Acceleration due to gravity(in m/s^2): g=9.78 //Guage pressure(in kPa): gp=13550*g*h*10^-3 //Actual pressure: ...
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// Example 17_17 clc;funcprot(0); //Given data m_s=250;// tons/hr T_s=40;// °C T_wi=30;//°C T_wo=36;//°C U_o=2.5;//kW/m^2°C P_t=0.078;// bar v=1.8;// m/s d_i=23;// mm d_o=25;// mm rho_w=1000;// kg/m^3 moisture=12;// Percentage x_2=(100-12)/100;// Dryness fraction p_t=0.078;// bar C_pw=4.2;// kJ/kg.°C R...
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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 10: POWER SYSTEM STABILITY // EXAMPLE : 10.11 : // Page number 303 clear ; clc ; close ; // Clear the work space and con...
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clc // given that m_capella = 0.05 // magnitude of brightness of capella at 14 parsecs m_sun = 4.8 // absolute magnitude of brightness of sun d = 14 // distance of capella in parsecs D = 10 // distance of capella considerd for observation // sample problem 2b page No. 333 printf("\n # Problem 2a # \n") printf...
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function [y]=test2p(n) //! // Copyright INRIA [o,i]=argn(0); if i < 1 then error(58); end; if type(n) <>1 then error(53,1); end; if size(n) <> [1 1] then error(89,1); end; // if n < 1 then y=1; else p=log(n)/log(2); rp=1 - p + int(p); if rp < 1e-7 then p=int(p)+1; else p=int(p); end; ...
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clc clear disp("falsa posição") disp("") function [s] = f(x) s = (2*cos(x)-(%e^x)/2) endfunction a=-1; b=2; x = 0; e = 0.01; k =1; x = ((a*f(b))-(b*f(a)))/(f(b)-f(a)) printf(' interaçoes: %d\n', k); printf(' a: %f\n', a); printf(' b: %f\n', b); printf(' x: %f\n', x); disp("-----------------------------------...
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clear all; clc; disp("Ex 2_11") a1=60 a=a1*%pi/180 b1=45 b=b1*%pi/180 c1=120 c=c1*%pi/180 f1=300 f2=700 f1_x=f1*cos(b) f1_y=f1*cos(a) f1_z=f1*cos(c) printf('\n\nF1 = (%.1fi+%.0fj%.0fk) N',f1_x,f1_y,f1_z) FR=800 f2_x=0-f1_x f2_y=800-f1_y f2_z=0-f1_z printf('\n\nF_2x = %.1f N',f2_x) printf('\n\nF_2y = %.0f N',f2_y) print...
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ATWM1_Working_Memory_MEG_Salient_Cued_Run1.sce
# ATWM1 MEG Experiment scenario = "ATWM1_Working_Memory_MEG_salient_cued_run1"; #scenario_type = fMRI; # Fuer Scanner #scenario_type = fMRI_emulation; # Zum Testen scenario_type = trials; # for MEG #scan_period = 2000; # TR #pulses_per_scan = 1; #pulse_code = 1; pulse_width=6; default_monitor...
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load Inc8.hdl, output-file Inc8.out, compare-to Inc8.cmp, output-list in%B1.8.1 out%B1.8.1; set in %B00000000, // in = 0 eval, output; set in %B11111111, // in = -1 eval, output; set in %B00000101, // in = 5 eval, output; set in %B11111011, // in = -5 eval, output;
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//<s1>=%rlr(s1,s2) // %rlr(s1,s2) calcule la division a gauche de la matrice de fractions //rationnelles s1 et de la matrice de fractions rationnelles s2 (s1\s2) //! [s1,s2]=sysconv(s1,s2) [n,m]=size(s1(2)) if n<>m then error(43),end if m*n=1 then s1=%rmr(tlist('r',s1(3),s1(2),s1(4)),s2) else // reduction de ...
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clc; r=5; //resistance in ohm p=1000; //power in Watt va=100; //potential diff in Volt for a vb=100000; //potential diff in volt for b ia=p/va; //calculating current ib=p/vb; //calculating current ha=ia*ia*r; //heat in Watt hb=ib*ib*r; //heat in Watt disp(ha,"Heat produ...
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//Example 5.3. frequency of induced emf in the rotor clc disp("The given values are,") disp("P = 4, f = 50 Hz, N = 1470 r.p.m") ns=(120*50)/4 format(5) disp(ns,"N_s(in r.p.m) = 120f/P =") s=(1500-1470)/1500 disp(s,"s = N_s-N / N_s =") f=0.02*50 disp(f,"Therefore, f_r(in Hz) = s*f =")
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//Chapter 7_Operational Amplifier Characteristics //Caption : Largest Amplitude //Example7.16: An amplifier has a 10 kHz sinewave input signal. Find the largest amplitude that the output of the amplifier can be,without distortion owing to slew rate limiting. Given slew rate=0.5V/u sec. //Solution: clear; clc; Fma...
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clear; clc; disp('Example 1.9'); // Given values m_dot = 20.4; // mass flowrate of petrol, [kg/h] c = 43; // calorific value of petrol, [MJ/kg] n = .2; // Thermal efficiency of engine // solution m_dot = 20.4/3600; // [kg/s] c = 43*10^6; // [J/kg] // power output P_out = n*m_dot*c; // [W] mprintf('\n The ...
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clear //Given f=18 //cm u=1.5 //Calculation R=(u-1)*f //Result printf("\n Radius of the curvature is %0.3f cm", R)
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clc //initialisation of variables i= 1/6400 b= 40 //ft d= 5 //ft C= 140 h= 6 //ft g= 32.2 //ft/sec^2 //CALCULATIONS A= b*d P= b+2*d m= A/P v= C*sqrt(m*i) V= v*(d/h) Q= v*b*d x= h-(Q/(3.09*(b/2)))^(2/3)-(V^2/(2*g)) //RESULTS printf ('height of pump= %.2f ft',x)
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// Test # 9 : Invalid flag test exec('./allpasslp2mb.sci',-1); [n,d]=allpasslp2mb(0.4,[0.3,0.4],'j'); //!--error 10000 //Invalid option,input should be either pass or stop //at line 68 of function allpasslp2mb called by : //[n,d]=allpasslp2mb(0.4,[0.3,0.4],'j')
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// Scilab Code Ex 1.22 Page-33 (2006) clc; clear; r = 1.746e-010; // Atomic radius of lead atom, angstrom a = 4*r/sqrt(2); // Interatomic spacing, m h = 1; k = 0; l = 0; // Miller Indices for planes in a cubic crystal d_100 = a/(h^2+k^2+l^2)^(1/2); // The interplanar spacing for cubic crystals, m printf("\nThe ...
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//Example 3.1.A clc; Syms s t; A=3 laplace(A,t,s)
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clear // // //Initilization of Variables //Portion AB L_AB=600 //mm //Length of AB A_AB=40*40 //mm**2 //Cross-section Area of AB //Portion BC L_BC=800 //mm //Length of BC A_BC=30*30 //mm //Length of BC //Portion CD L_CD=1000 //mm //Length of CD A_CD=20*20 //mm //Area of CD P1=80*10**3 //N //Load1 P2=60*10**3 //N /...
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function ret = snr(power, f, n) s = zeros(1, n); for i = 1:length(s); s(1, i) = power(1, f * i + 1); power(1,f * i + 1) = 0; end ret = 10 * log10(sum(s) / sum(power)); endfunction
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clear; lines(0); np=200; q=2; for w=2.08 s0 = '~/imc/StarBrush/Neutral/n'+ string(np) + '/q' + string(q) + 'den/chi0/mn' + string(w) + '/' ; nf=66;//number of the latest file //nf=10; i=1; N=np*(q+1);//mass DEN VITAL!!! for j=1:1:nf sig=0.01*j; // v=j; s = s0 + 'CSBrush_mn' + string(w) + '_' + string(j) + '.pro';...
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clear /////////////////////////////////////////////////////////////////////////// // ProyectoFinal_1.sce // // Programa final para la clase de métodos numéricos. Este programa // provee una compilación de soluciones para todos los métodos vistos // en clase: // TODO FUNCIONA PROFE // // Autores: // Jorge Vazquez...
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function result_path = mdaqToolboxPath() result_path = []; path = fileparts(get_function_path('mdaqToolboxPath')); result_path = part(path,1:length(path)-length("macros") - 1 ); endfunction
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clc //initialisation of variables veff=0.8 //efficiency rp=7 n=1.2 //constant value pi=(22/7) //CALCULATIONS c=(veff-1)/(1-(rp)^(1/n)) vs=2/c d=((4*vs)/pi)^(1/3) //RESULTS printf('stroke volume is %2fm*m*m',vs) printf('\nlenght of stroke is %2fm',d)
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// Scilab Code Ex5.22: Page-295 (2008) clc; clear; h = 6.62e-034; // Planck's constant, Js m = 1e-009; // Mass of the particle, kg v = 1; // Velocity of the particle, m/s delta_v = v*0.01/100; // Minimum uncertainty in the velocity of the particle, m/s delta_x = h/(m*delta_v); // Minimum uncertaint...
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8_3.sce
clc //initialisation of variables t1=253 //temp in k t3=313 //temp in k cp=1.005 //kj/kg r=4 //bar g=1.4 //CALCULATIONS t2=(t1*(r)^((g-1)/g)) t4=(t3/(r)^((g-1)/g)) re=cp*(t1-t4) wi=cp*((t2-t3)-(t1-t4)) cop=re/wi ma=(3.5164*10)/re p=ma*wi //RESULTS printf('cop is %2f',cop) printf('\nmass of refrigeratio...
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// funcao de correlacao function Rx = comat(x,N) px = xcorr(x,N,'biased'); for i = 1:N+1 Rx(:,i) = px((N+2-i):(2*(N+1)-i)); end endfunction //para fins didaticos, a implementacao pura do LMS function [y,e,wn]=filterLMS(x,d,mi,N) xn=zeros(N+1,1); Wn=zeros(N+1,1); M=length(x); for n=1...
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vtk2ply.sci
function vtk2ply() // Convert a VTK file to PLY format. // // Syntax // PointCloud(vtkFileName,plyFileName,"vtk2ply") // // Parameters // vtkFileName : input file of vtk format // plyFileName : output file of ply format // // Description // Input file is an VTK format which is them transformed to PLY format ...
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//Example 10.1.1 Page 350 //Non-Linear Dynamics and Chaos, First Indian Edition Print 2007 //Steven H. Strogatz clear; clear; clc; close; x=poly(0,"x"); f = (x^2)-x; //Defining Polynomial--> x(dot)=x^2 -1. Let this be f(x) disp("Fixed Points are :") y = roots(f) lambda1=evstr(2*y(1)) ...
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//Ex 3.6 clc;clear;close; format('v',6); fo=1;//kHz Ap=1.586;//Band pass gain C1=0.005;C2=0.005//micro F(Assumed) R=1/(2*%pi*fo*10^3*C1*10^-6);//ohm Rf=10;//kohm(Assumed) Ri=Rf/(Ap-1);//kohm disp("Design values are :"); disp(R/1000,"Resistance in kohm, R1=R2="); disp(Ri,"Resistance Ri(kohm)"); disp(Rf,"Resi...
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function dxdt = odeholding(t,x) Dm = D1211m*10^((121.1-T)/Zm); // s km = log(10)/Dm; // s-1 dxdt = -km*x(1) endfunction
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//when i/p is a column vector v=[1;2;34;5]; w=bartlett(v); disp(w); //output //!--error 10000 //L must be a positive integer //at line 26 of function bartlett called by : //w=bartlett(v); //MATLAB o/p // 0 // 0 // 0 // 0 //
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while 1==1, [a1,b1]=unix_g("~/rasp30/prog_assembly/libs/sh/asm2ihex.sh char_Scurve_swc ~/rasp30/prog_assembly/libs/asm_code/char_Scurve_swc.s43 16384 16384 16384"); [a2,b2]=unix_g("sudo tclsh ~/rasp30/prog_assembly/libs/tcl/write_mem2_NoRelease.tcl -start_address 0x7000 -input_file_name "+hid_dir+"/target_info_...
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//------------------------------------------------------------------------------ // Détermination d'un critère de stabilité du processus par méthode graphique //------------------------------------------------------------------------------ // Paramètres lambdaMin = 0.4 lambdaMax = 0.5 step = 0.02 mu = 0.5 tmax = 4000 ...
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A=[1,-2,3;0,4,5]; B=[4,6,8;1,-3,-7]; k=A+B; disp(k,'The addition of the two matrices A and B is:') m=3*A; disp(m,'The multiplication of a vector with a scalar is:') p=2*A-3*B
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