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clc;clear; //Example 3.3 //given values B=2.179*10^-16;//a constant in J h=6.625*10^-34;//plank's constant in J-s //calculation E3=-B/3^2; E2=-B/2^2; f=(E3-E2)/h; disp(f,'frequency(in Hz) of radiation')
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//Exa 5.10 clc; clear; close; //Given data : format('v',5); A=1000;//gain(unitless) Beta=1/20;//feedback ratio (unitless) //Formula : Af=A/(1+A*Beta) Af=A/(1+A*Beta);//gain with feedback(unitless) Af=20*log10(Af);//in dB disp(Af,"Gain with feedback in dB : ");
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clc //initialisation of variables m= 1 //lb cp= 0.240 //btu/lb F T2= 150 //F T1= 50 //F //CALCULATIONS S= m*cp*(log(460+T2)-log(460+T1)) //RESULTS printf ('Entropy change = %.4f Btu/Fabs',S) //This result is same as the above since change in entropy does not depend on the process involved // but only on the ...
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// Example 2.5 :Thevenin's and Norton's Equivalent clc; close; format('v',7) clear; // given : vs1=10;//voltage in volts R1=50;//resistance in ohms R2=50;//resistance in ohms R3=25;//resistance in ohms disp("(a) Applying Thevenins Theorem ") voc=(R1/(R1+R2))*vs1;//voltage in volts req=((R1*R2)/(R1+R2))+R3;//resistance ...
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plot2d1('enl',1,(1:10:10000)'); xtitle('plot2d1 log scale','t','y log scale'); // 3 = green 4 =blue xgrid(3);
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1 0 0 0 0 0 1 0 0 0 1 1 1 The Lone Charger -100 -100 -10 100 100 100 sin(z) 0 0 0 0 0 299792448 5 0
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clear; clc; disp("--------------Example 21.7----------------") // multicast IP address 230.43.14.7 multicast_IP_address=dec2bin(230,5)+dec2bin(43,7)+dec2bin(14,7)+dec2bin(7,7); s=strsplit(multicast_IP_address,length(multicast_IP_address)-23); b=strsplit(s(2),[9 16]); starting_Ethernet_addr = "01:00:5E"; // 01:0...
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clear; clc; RS=[10 12 14 30 40]; REQS=0;REQp1=0; for i=1:1:5 REQS=REQS+RS(1,i); REQp1=REQp1+(1/RS(1,i)); end REQp=1/REQp1; disp("A RESISTENCIA EQUIVALENTE EM SERIE DADA EM OHMS:",REQS); disp("A RESISTENCIA EQUIVALENTE EM PARALELO DADA EM OHMS:",REQp);
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//Chapter 1, Example 1.4, Page 23 clc clear //Density of Hydrogen atom in water p = 1 // density of water in g cm^-3 Na = 6.022*10^23 // molucules/mol A = 18 // atomic weight of water in g/mol N = (p*Na)/A NH = 2*N printf("The density of water = %e molecules/cm3",N); printf("\n The density of hydrogen atoms = %e atoms/...
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// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India. //Chapter-5,Example 9,Page 173 //Title: Change in entropy of water //================================================================================================================ clear clc //INPUT m=1;//amount of s...
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%Generated from '../examples/shi/shi_pelda.dig'. query(instances(aconcept('Gazdag')), [d, e, c]). query(instances(aconcept('Boldog')), [d, e, a]). implies(some(inv(arole(utodja)), aconcept('Gazdag')), aconcept('Gazdag')). implies(some(arole(gyereke), aconcept('Gazdag')), aconcept('Boldog')). subrole(arol...
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//Section-4,Example-1,Page no.-I.78 //To calculate the number of protons in a sample exposed to the given magnetic field. clc; T=20+273 k=1.38*10^-23 //dl_E=g1*muN*B_0 B_01=1 dl_E1=2.821*10^-26*B_01 //N=N_a/N_b(ratio of protons having a&b spins respectively) N_1=((k*T)/((k*T)-(dl_E1))) disp(N_1,'Ratio for 1.0 Tesla mag...
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****************************************************************** * * * ****** ****** ***** ***** ***** ** ******* * * ** ** ** ** ** ** ** ** ** ** ** ** * * ** ** ** ** **...
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//Book Name:Fundamentals of Electrical Engineering //Author:Rajendra Prasad //Publisher: PHI Learning Private Limited //Edition:Third ,2014 //Ex4_16.sce. clc; clear; q1=-2e-9; q2=3e-9; q3=2e-9; q4=1e-9; AB=1; //Given square side as 1 metre BC=1; epsilon_not=8.854e-12; AP=sqrt(AB^2+BC^2)/2...
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clc clear printf("example 9.4 page number 387\n\n") printf("this is a theoretical question, book shall be referred for solution")
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clc // Given that x1 = 1 // coordinate on x axis in first case y1 = 2 // coordinate on y axis in first case z1 = 3 // coordinate on z axis in first case x2 = 1 y2 = 1 z2 = 0 // coordinate of first plane in second case x3 = 1 y3= 1 z3 = 1 // coordinate of second plane in second case // Sample Problem 8 on page no. 13....
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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 IV : UTILIZATION AND TRACTION // CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS // EXAMPLE : 1.7 : // Page number 685-686 clear ; clc ; close ; // Clear the wo...
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clc; nC=0.120;//kmol nO=0.115;//kmol nN=0.765;//kmol m_C=44;//kg/kmol m_O=32;//kg/kmol m_N=28;//kg/kmol miC=m_C*nC;//kg miO=m_O*nO;//kg miN=m_N*nN;//kg m=miC+miO+miN; cpC=1.271;//kJ/kgK cpO=1.110;//kJ/kgK cpN=1.196;//kJ/kgK cp=cpC*(miC/m)+cpO*(miO/m)+cpN*(miN/m); R_=8.3145;//kJ/kg K R=(miC/m)*(R_/m_C)+(miO/m)*(R...
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clc clear //Input data t1=30;//Normal temperature of black body in degree centigrade t2=100;//Heated temperature of black body in degree centigrade s=20.52*10^-8;//Stefan Boltzmann constant in kJ/hrK^4 A=1;//Assume area in m^2 //Calculations T1=273+t1;//Black body temperatures in kelvin K T2=273+t2;//Heated ...
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//CHAPTER 5 ILLUSRTATION 5 PAGE NO 163 //TITLE:Inertia Force Analysis in Machines //Figure 5.3 clc clear pi=3.141 N=1800// speed of the petrol engine in rpm r=.06// radius of crank in m l=.240// length of connecting rod in m D=.1// diameter of the piston in m mR=1// ...
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//Example No.4.11. //Page No.138. clc;clear; n = 1;//For the lowest energy value n=1. h = 6.626*10^(-34);//Planck's constant. L = 1*10^(-10);//Width of the potential well -[m]. m = 9.1*10^(-31);//Mass of the electron. E = ((n^(2)*h^(2))/(8*m*L^(2))); E = ((h^(2))/(8*m*L^(2)));// For the lowest energy value n=...
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//Section-14,Example-5,Page no.-PC.17 //To calculate the temperature at which v_rms(He)=v_rms(H2). clc; //v_rms=sqrt((3*K*T)/m) //K=1(let) K=1 T_He=(3*K*200*4)/(3*K*2) disp(T_He,'Required temperature(K)')
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//Example 1.8 clc(); clear; //To calculate the distance from the edge of wedge alpha=0.01 //units in radians n=10 lamda=6000 //units in armstrongs lamda=lamda*10^-10 //units in mts x=((2*n-1)*lamda)/(4*alpha) //units in mts printf("Distance from the edge of the wedge is %.6fmts",x)
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clc clear //INPUT DATA Tc=9.15//critical temperature of Nb in K t=6//temperature of critical field in K Ho=0.196//The critical field AT 0K in T //CALCULATION Hc=(Ho*(1-(t/Tc)^2))//The critical field at 6K in T //OUTPUT printf('The critical field at %iK is %3.4f T',t,Hc)
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clc; cj=2700; // The effective jet velocity from jet engine in m/s ci=1350; // Flight velocity in m/s ma=78.6; // Air flow rate in m/s a=ci/cj; F=ma*(cj-ci); // Thrust P=F*ci; // Thrust power eff_P=2*a/(a+1); // Propulsive efficiency disp ("N",F,"(i).Thrust = "); disp ("MN",P/10^6,"(ii). Thrust power = ");...
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function Mawarikomi(varargin) FL='default'; haba='10cm'; Nargs=length(varargin); if Nargs>0 haba=varargin(1); if Nargs>=2 FL=varargin(2); if FL=='' FL='tmp.tex' end end end StrM=[ '\begin{mawarikomi}%',... '%<1>[5](0,0)%',... '{'+haba+'}{%',... ...
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Name=WCP #4 PlayerCharacters=pistol Launchman BotCharacters=crocbot.bot IsChallenge=true Timelimit=150.0 PlayerProfile=pistol Launchman AddedBots=crocbot.bot;crocbot.bot;crocbot.bot;crocbot.bot;crocbot.bot;crocbot.bot;crocbot.bot;crocbot.bot PlayerMaxLives=0 BotMaxLives=0;0;0;0;0;0;0;0 PlayerTeam=1 BotTeams=2;2;2;2;2;2...
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//Exa 4.7 clc; clear; close; //Given data : P=12000;//in Rs A1=10000;//in Rs G=1000;//in Rs i=18;//in % per annum n=10;//in years //Formula : (P/A,i,n)=(((1+i/100)^n)-1)/((i/100)*(1+i/100)^n) //Formula : (A/G,i,n) :(((1+i/100)^n)-i*n/100-1)/(((i/100)*(1+i/100)^n)-i/100) PW=-P+(A1+G*(((1+i/100)^n)-i*n/100-1...
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## Copyright (C) 2016-2017 Rik Wehbring ## ## This file is part of Octave. ## ## Octave is free software; you can redistribute it and/or modify it ## under the terms of the GNU General Public License as published by ## the Free Software Foundation; either version 3 of the License, or (at ## your option) any later versi...
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a = 0.8 l = 10 T = 60 n = 2000 delta_x = 2*l / (n + 1) n_t = 3000 delta_t = T / n_t t_inter = 2 * T / 3 t_fin = T mu = delta_t * (n+1)**2 / (2 * l)**2 F_cible = [-0.1, -0.18] function c=C(x, x_d) c = 1 - a * exp(-(x - x_d)**2 / 4) endfunction function c=Ci(i, x_d) c = C(i * delta_x - l, x_d) endfunction functi...
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clc; clear; //x(t)=2sin(4000*pi*t)+3sin(5000*pi*t)+4sin(8000*pi*t) fh=8000/2; fl=4000/2; disp(fh,"a) Highest Frequency component(in Hz)"); disp(fl,"Lowest Frequency component(in Hz)"); fs=2*fh; disp(fs," Minimum Sampling frequency(in Hz)"); Bw=fh-fl; disp(Bw," b)Bandwidth(in Hz) is"); n=fh/Bw; d...
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//chapter 24 Ex 22 clc; clear; close; side=20; diagonal1=24; x=sqrt(side^2-(diagonal1/2)^2); diagonal2=2*x; area=(1/2)*(diagonal1*diagonal2); mprintf("The area of rhombus is %d square cm",area);
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// Commands, syntax: // STATUS: = get status message // S_MSG: = setMsg <sip_id>|<a_msg_id>|<prio>|<ttl>|text| // S_MSG_FILE: = setMsg <sip_id>|<a_msg_id>|<file>| setMsg from <file> // D_MSG: = delMsg <sip_id>|<a_msg_id> // POS_BLE: = ble position req <sip_id> // POS_BLE: = dect position req <sip_i...
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clc //initialisation of variables V= 12 //km/L //CALCULATIONS MPG= V*3.7854/1.609 //RESULTS printf ('car mileage = %.2f MPG',MPG)
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//Network Theorem 1 //page no-3.48 //example3.42 //calculation of Vth disp("Removing the variable resistor RL from the network:"); disp("I1=50");....//equation 1 disp("Applying KVL to mesh 2:"); disp("5*I1-10*I2=0");....//equation 2 A=[1 0;5 -10]; B=[50 0]' X=inv(A)*B; disp(X); disp("I2 = 25 A"); disp("Wr...
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//Initilization of variables F=[20;-10;30] //N //co-ordinates in meters a=2 //m b=4 //m c=7 //m d=3 //m e=2 //m f=4 //m //Calculations R=F(1,1)+F(2,1)+F(3,1) //N M_o=F(1,1)*a+F(2,1)*b+F(3,1)*c //N-m x=M_o/R //m M_x=-F(3,1)*e-F(1,1)*d+F(2,1)*f //N-m z=-M_x/R //m //Result clc printf('The resultant is %f ...
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//Obtain path of solution file path = get_absolute_file_path('solution4_3.sce') //Obtain path of data file datapath = path + filesep() + 'data4_3.sci' //Clear all clc //Execute the data file exec(datapath) //Calculate permissible stresses for cotter (N/mm2) //Tensile stress sigma sigma = Syt/fs //Yield str...
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clc; Vcc=20; //volt R2=1000; //ohm R1=6800; //ohm Vb=(R2/(R1+R2))*Vcc; //volt//voltage divider rule Ve=Vb-0.7; //volt Re=1000; //ohm Ie=Ve/Re; //Ampere Hfe=50; Ib=Ie/(Hfe+1); //Ampere disp('Amperes',Ib,"Ib=");//The answers vary due to round off error
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function out=moduloluyurt(x,m) l=length(x); n=1; for i=1:1:l if modulo(x(i),e)==0 v(n)=x(i); n=n+1; end end out=v; endfunction
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T1=300;//initial temperature in kelvin// T2=600;//final temperature in kelvin// T3=373;//initial temperature in kelvin// T4=746;//final temperature in kelvin// Cv=6.09;//molar heat capacity in cal per deg// dS2=Cv*2.303*log10(T2/T1);//change in entropy for temperature change between 300k to 600k// printf('Change ...
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// Scilab Code Ex5.19: Page-294 (2008) clc; clear; h = 6.62e-034; // Planck's constant, Js e = 1.602e-019; // Energy equivalent of 1 eV, J m = 1.67e-027; // Rest mass of a proton, kg r = 5e-015; // Radius of the nucleus, m delta_x = 2*r; // Minimum uncertainty in position of the proton, m delta_p ...
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clear(); // effacer les variables M=fscanfMat('data.txt'); // lire les donnees x=M(:,1); // recuperation de la 1ere colonne : x y=M(:,2); // recuperation de la 2ieme colonne : y [a,b,sig]=reglin(x',y') // regression lineaire ylin=a*x+b plot(x,[y ylin]) // trace
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// Calculating the ceil. y = [1.2, 1, 1.9; 4, 2.6, 5; 2.3, 8, 7]; ceilres = armaMat("ceil",y)
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//Solutions to Problems In applied mechanics //A N Gobby clear all; clc //initialisation of variables w=20//lbf p=12//ft/s v1=15//ft/s g=32.2//ft v2=10//ft/s d1=6//in d2=9//in a=10.82//in //CALCULATIONS Um=(v2*p)/sqrt(a^2-d2^2)//sec^-1 P=2*%pi/Um//sec V=w*a//in/s M=w^2*a/p//ft/s F=(w/g)*M//lbf //RESULTS...
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function [arr,n] = str2arr(str,nn) //function [arr,n] = str2arr(str) // Ouput variables initialisation (not found in input variables) arr=[]; n=[]; // Number of arguments in function call [%nargout,%nargin] = argn(0) // Display mode mode(0); // Display warning for floating point exception ieee(1); // Form a string...
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clc clear d=0.25//Diameter of the cylinder in m L=0.35//Stroke in m Cv=1500//Clearance volume in c.c s=5//cut off ratio takes place at 5 percent of stroke a=1.4//Explosion ratio g=1.4//Ratio of specific heats for air //Calculations Vs=(3.14/4)*d^2*L//Stroke volume in m^3 r=(Vs*10^6+Cv)/Cv//Compression ratio...
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m3=1150 //Kg/h H3=2676 //KJ/Kg H2=3074 //KJ/Kg H1=3278 //KJ/Kg
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// Exa 9.30 clc; clear; close; // Given data I_DD= 20;// in mA R2 = 10;// in k ohm R1 = 30;// in k ohm R_S= 1.2;// in k ohm R_D= 500*10^-3;// in k ohm V_DD = 12;// in V Vp= -6;// in V V_G = (R2/(R2+R1))*V_DD;// in V I_D= poly(0,'I_D') V_GS= V_G-I_D*R_S;// in V I_D=I_D-I_DD*(1-V_GS/Vp)^2; I_D= roots(I_D)...
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// Copyright INRIA scifuncs=['modstr','stacc']; //Scilab functions files=G_make(['/tmp/ex6fi.o','/tmp/ex6f.o'],'ex6f.dll'); addinter(strcat(files,' '),'intex6f',scifuncs); //a's to o's x=modstr('gaad'); if x<>'good' then pause,end //variable read in Scilab stack param=1:10; z=stacc(); if norm(z-param) > %eps then ...
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// Demo for decision tree -- Scilab getd('../macros') // Data preparation M = csvRead('Datasets/forestfires.csv') x = M(:,[5,6,7,8,9]); y = M(:, 13); y(or(isnan(x),'c'),:) = [] x(or(isnan(x),'c'),:) = [] n = length(y(:, 1)) for i = 1:n if(y(i)>0) y(i) = 1 end end [questions,flag] = decisionTreeFit(x, y); pred...
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/* ============================================================== Escola Politécnica da USP PME3402 - Laboratório de Medição e Controle Discreto -------------------------------------------------------------- ATIVIDADE 3 ------------------------------------------------------------...
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# ATWM1 MEG Experiment scenario = "ATWM1_Working_Memory_MEG_salient_cued_run2"; #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_monito...
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// given data clear clc rho=1.226 // air density in kG/m^3 alpha =0.14 H=10.0 // height at which wind speed is given in m uH=12.0 // speed in m/s z=100.0 // tower height in m D=80.0 // diameter in m effigen=0.85 // efficiency og generator A=%pi*(D**2)/4 // area in m^3 u0=uH*(z/H)**alpha // velocity at 10...
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clc //Intitalisation of variables clear T= 18 //C n1= 7.5 n2= 3 n3= 6 R= 2*10^-3 //kcal dH= -783.4 //kcal //CALCULATIONS dE= dH+R*(273+T)*(n2+n3-n1) //RESULTS printf ('Heat of the reaction = %.1f kcal',dE)
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clear; clc; //Example - 12.4 //Page number - 424 printf("Example - 12.4 and Page number - 424\n\n"); //Given //component 1 = formic acid //component 2 = water T = 20 + 273.15;//[K] - Temperature Mol_form = 46.027;//Molecular weight of formic acid Mol_water = 18.015;// Molecular weight of water Wt_perc=...
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clc;close;clear; [x,Fs,bits]=wavread("machali.wav"); Fs=8000; bits=16 fm=3000 //freq of noise signal given t=0.0001:1/Fs:length(x)*1/Fs; xn=sin(2*%pi*fm*t); y=x+xn; h=[1 -2*cos(2*%pi*(3000/Fs)) 1]; //impulse response of filter //overlap and save method //getting values of x1,x2,x3..... i=1; j=1; r(1,1)=0; r(1,2)=0; j...
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//Example 8_2<c> //determine the nyquist rate of x(t)=sinc(200*pi*t)+sinc2(200*pi*t) //here,sinc(400t)=0.5cos(400t)/400t+ clc; clear all; wq=400; wp=200; wf=0; if wp>=wq then wf=wp; else wf=wq; end F1=wf/2; Fs=2*F1; disp('Nyquist Rate='); disp(Fs);
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clear //variable declaration Ea=70*1000 //Young's modulus of aluminium,N/mm^2 Es=200*1000 //Young's modulus of steel,N/mm^2 alphaa=(0.000011) //expansion coefficient,/°C alphas=(0.000012) //expansion coefficient,/°C Aa=600 //Area of aluminium portion,mm^2 As=400 ...
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clear;lines(0); x=sin(2*%pi*(0:5)/5); y=cos(2*%pi*(0:5)/5); plot2d(0,0,-1,"010"," ",[-2,-2,2,2]) xset("pattern",5) xfpoly(x,y) xset("default")
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errcatch(-1,"stop");mode(2);//ex4.3 disp('cant be shown') exit();
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errcatch(-1,"stop");mode(2); //input m=140//mass v=8//speed r=5//radius g=9.8//acceleration due to gravity //calculation t=((m*v^2/5)^2)+(140*9.8)^2 //applying parallelogram of vectors t1=sqrt(t) //output printf("the tension in arm is %3.3f N",t1) exit();
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function y=testa(W) //carrega o treinamento. load ('ANN_treina.sod', 'W', 'NeuralNetwork'); //entrada dos valores de teste x=[ 0 0 1 0 0 1 1 1 1 0 0 1 0 0 0 1 0 1 1 0 0 ; 0 0 1 0 0 1 1 1 1 0 0 1 0 0 0 1 0 0 1 0 0 ; 0 0 1 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 1 1 ; 1 0 1 1 0 1 1 0 1 1 0 0 1 0 0 1 0 1 0 0 1 ; 1 0 0 ...
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[1/2,2/3,-1/5] * -1/2 3 = [0,0,0,-1/4,-1/3,1/10]
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// 08.09.13 function Out=BorderHiddenData() global BORDERHIDDENDATA Out=BORDERHIDDENDATA; endfunction;
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clear;lines(0); fd=mopen(TMPDIR+'/Mat','w'); mfprintf(fd,'Some text.....\n'); mfprintf(fd,'Some text again\n'); a=rand(6,6); for i=1:6 , for j=1:6, mfprintf(fd,'%5.2f ',a(i,j));end; mfprintf(fd,'\n'); end mclose(fd); a1=fscanfMat(TMPDIR+'/Mat')
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clear all; clc; disp("We have h4=104 Btu/lbm,h2=h1=53 Btu/lbm,p4=p1=20psia,s4=s-s3=0.226Btu/lbm-R and Hs3=122 Btu/lbm") h4=104 hs3=122 Eta_c=0.75 h_3dash=h4+(hs3-h4)/Eta_c printf("h_3dash=%0.0f Btu/lbm",h_3dash) w_i=h_3dash-h4 printf("\n The compressor work required per unit mass is w_i =%0.0f Btu/lbm",w_i)...
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//Exemplo: Ajuste de Modelo por Mínimos Quadrados //Caso com 2 Funções-base Ortogonais //%Programa:MQF1.sce clear; N = 9; xp = [-4,-3,-2,-1,0,1,2,3,4]; fp = 4*(xp.^2) + 20*xp; plot(xp,fp,'or'); //Funções Base Ortogonais g1 = xp.^2; g2 = xp; a1 = sum(fp.*g1)/sum(g1.*g1); a2 = sum(fp.*g2)/sum(g2.*g2); M = 100; xc = ...
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// Scilab code Exa3.2.7 : To calculate the mass of Ra-226 :Page no. 127 (2011) t_h = 1620*31536000; // Half life of Ra-226, S D = 0.6931/t_h; // Decay constant, S^-1 A_Ci = 3.7e+010; // Activity, Ci N_Ci = A_Ci/D; // Number of atoms decayed m = 0.226; // Mass of 6.023e+023 atoms, kg M_Ci = m*N_Ci/6.023e+023; // ...
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//check o/p when the i/p is a char matrix x=['a' 'b' 'c']; a=lsf2poly(x); //output // !--error 10000 //Input arguments must be double. //at line 21 of function lsf2poly called by : //a=lsf2poly(x); //at line 3 of exec file called by : //poly/lsf2poly4.sce', -1
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//clear// //Caption:Cut-off wavelength of photodiode //Example6.1 //page224 clear; clc; close; h = 6.625*(10^-34); //planks constant C = 3*(10^8); //free space velocity Eg = 1.43*1.6*(10^-19);//joules LambdaC = h*C/Eg; disp(LambdaC,'Cut-off Wavelength of photodiode in meters =') //Result //Cut-off Waveleng...
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clc //initialisation of variables A= 10000 //ft^2 H1= 50 //ft H2= 40 //ft l= 1500 //ft d= 6 //in f= 0.0075 g= 32.2 //f/sec^2 //CALCULATIONS t= 2*A*sqrt((1.5+(4*f*l/(d/12)))/(2*g))*(sqrt(H1)-sqrt(H2))/(%pi*(d/12)^2/4) //RESULTS printf ('Time taken to lower the level of water = %.f sec ',t)
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clc //Initialization of variables s=2.7 gamw=9810 //N/m^3 mu=0.001 //Ns/m^2 d=0.15*10^-3 //m rho=1000 //kg/m^3 //calculations gams=s*gamw U= d^2 *(gams-gamw)/(18*mu) RN= U*d*rho/mu Cd = (1+ 3/16 *RN)^0.5 *(24/RN) U22 = 4/3 *d*(gams-gamw) /(Cd*rho) U2=sqrt(U22) //results printf("Settling velocity of san...
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# Response parameters active_buttons = 1; button_codes = 1; response_matching = simple_matching; # Trigger parameters write_codes = true; pulse_width = 1; ########### ### SDL ### ########### begin; polygon_graphic { sides = 100; radius = 5; line_color = 255, 0, 0; fill_color = 255, 0, 0; } fixpoly; picture { ...
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// Scilab code Ex11.18: Pg.525 (2008) clc; clear; V = 2.00e-03*%pi/4; // Volume of the sample, cm^3 M_Zr = 91.22; // Molecular weight if Zr, g/mol rho_Zr = 6.506; // Density of Zr, g/cm^3 N_A = 6.02e+23; // Avagrado's number N_Zr = N_A*V*rho_Zr/M_Zr*0.1127; // No. of atoms of Zr sigma = 900e-003*1...
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function [stk,nwrk,txt,top]=%log2for(nwrk) // Copyright INRIA txt=[] iop=evstr(op(2)) s2=stk(top);s1=stk(top-1);top=top-1 if s1(4)=='1'&s1(5)=='1'&s2(4)=='1'&s2(5)=='1' then if s2(2)=='2' then s2(1)='('+s2(1)+')',end if s1(2)=='2' then s1(1)='('+s1(1)+')',end stk=list(s1(1)+ops(iop,1)+s2(1),'1','0','1','1') else ...
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// Exa 1.22 clc; clear; close; // Given data V_S = 10;// in V R1 = 1.5*10^3;// in ohm R2 = 1.8*10^3;// in ohm I_T = V_S/(R1+R2);// in A disp(I_T*10^3,"Using the ideal diode, the total current in mA is "); V_D1 = 0.7;// in V V_D2 = 0.7;// in V I_T = (V_S-V_D1-V_D2)/(R1+R2);// in A disp(I_T*10^3,"Using the p...
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//Exa 3.3 clc; clear; close; //Given data : //Formula Pulse Broadning per Km : deltaTmat(per Km)=(deltaTAUs*1000/c)*(lambda*d2n/dlambda^2) deltaTAUs=45;//in nm deltaTAUs=45*10^-9;//in m lambda=0.9;//in um lambda=0.9*10^-6;//in m //let say, d^2n/dlambda^2=a a=4*10^-2;//in um^-2 a=a*(10^-6)^-2;//in m^-2 c=3*...
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//chapter 2 //example 2.2 //page 70 Ib1=18*10^-6 ;Ib2=22*10^-6;// given Ib=(Ib1+Ib2)/2 //input base current disp(Ib) //result Iios=(Ib2-Ib1) // input offset current disp(Iios)// result
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clear; clc; // Example: 10.7 // Page: 408 printf("Example: 10.7 - Page: 408\n\n"); // Solution //*****Data******// Temp = 30;// [OC] A = 0.625; //**************// P1sat = exp(13.71 - 3800/Temp);// [kPa] P2sat = exp(14.01 - 3800/Temp);// [kPa] // At azeotropic point: // P = gama1*P1sat + gama2*P2s...
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clc //initialisation of variables hi=1306.9 //lbm si=1.5894//lbm x=0.077 //lbm n=0.7 //lbm\ T0=537 //F Te=586//F se=0.1897 //Btu/lbm hes=1116.2-se*(1022.2)//Btu/lbm Ws=hi-hes//Btu/lbm Wa=n*Ws//Btu/lbm he=hi-Wa //Btu/lbm S=1.9200 //Btu/lbm //CALCULATIONS Se=S-x*(1.7451)//Btu/lbm Wrev=(hi-he)-T0*(si-Se)...
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clc; //e.g7.5 C1=5*10**-12;//min C2=5*10**-12;//min L=10*10**-3; CT=(C1*C2)/(C1+C2);//CTmax disp('F',CT*1,"CT="); fo=1/(2*%pi*sqrt(L*CT)); disp('MHZ',fo*10**-6,"fo="); C1=50*10**-12;//max C2=50*10**-12;//max CT=(C1*C2)/(C1+C2);//CTmin disp('F',CT*1,"CT="); fo=1/(2*%pi*sqrt(L*CT)); disp('kHZ',fo*10**-3,"fo...
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function R = Euler2R(A) // Euler angle -> Orientation matrix a1 = A(1); a2 = A(2); a3 = A(3); R1 = [1, 0, 0; 0, cos(a1), -sin(a1); 0, sin(a1), cos(a1)]; R2 = [cos(a2), 0, sin(a2); 0, 1, 0; -sin(a2), 0, cos(a2)]; R3 = [cos(a3), -sin(a3), 0; sin(a3), cos(a3), 0; 0, 0, 1]; R...
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//Example 1.49://ARITHEMATIC MEAN,AVERAGE DEVIATION ,STANDARD DEVIATION AND VARAIANCE clc; clear; q=[1.34,1.38,1.56,1.47,1.42,1.44,1.53,1.48,1.40,1.59];//length in mm AM= mean(q);//arithematic mean in mm for i= 1:10 qb(i)= q(i)-AM; end Q= [qb(1),qb(2),qb(3),qb(4),qb(5),qb(6),qb(7),qb(8),qb(9),qb(10)];// AV=(-qb(1)-...
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v1=1.5; v2=0.96; v3=1; v4=0.014; disp("Part a"); true=v3+v4; disp("the true reading (in V) of the voltmeter is"); disp(true); disp("Part b"); cor=true-v2; disp("the voltmeter correction (in mV) is"); disp(cor*10^3); disp("Part c"); fsd=cor*100/v1; disp("The F.S.D. accuacy (in %) of the meter is"); disp(fsd)...
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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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clc // initialization of variables clear // linked to 6_5 A=3 //cm^2 E= 2*10^6 //kg/cm^2 nu= 0.25 l= 60 //m L=150 //cm d=0.5 //cm dd=10 //cm D=180 //cm //calculations K=(l*100/(A*E))+(L*D/2*D*32*2*(1+nu)/(E*%pi*dd^4*2)) P=d/K Ts=P/A fs=dd*D*P*32/(%pi*4*dd^4) // results printf('The tensile stress is ...
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exec("ode_euler.sce",-1) exec("ode_euler_PC.sce",-1) exec("ode_RK4.sce",-1) // f and y are array with size 2 // y(1) = y // y(2) = y' // // f(1) = y' = y(2) // f(2) = y'' = -y = -y(1) // There is no dependence of f to t explicitly. // However, we put it here because in general there might be dependence // to t functio...
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//10.1 clc; K=0.1*10^-3; d=60; N2=200; phi2=K*d/(2*N2); a2=25*10^-6; B=phi2/a2; N=300; I=10; l=0.1; H=N*I/l; Permability_absolute=4*%pi*10^-7; Permability_relative=B/(Permability_absolute*H) printf("Relative permability of iron=%.2f",Permability_relative)
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s=%s; syms K h=syslin('c',(K/(s*(s+1)*(0.1*s+1)))) H=syslin('c',h) fmin=0.001; fmax=1000; bode(G,fmin,fmax) show_margins(G) xtitle("uncompensated system") [gm,freqGM]=g_margin(G) [pm,freqPM]=p_margin(G) disp(gm,"gain_margin=") disp((freqGM*2*%pi),"gain margin freq="); disp(pm,"phase margin=") disp((...
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A = [1.000001 2 3;4 5 6;7 8 9] x = [1;2;5] b1 = A*x b2 = b1 + [10^(-8);0;0] //Erro de 10^(-8) adicionado x1 = resolve(A,b1) x2 = resolve(A,b2) //Residuos residuo_x1 = b1 - A*x1 residuo_x2 = b2 - A*x2 //Erro Relativo erro_entrada = norm(b1-b2)/norm(b1) erro_saida = norm(x1-x2)/norm(x1) //Kappa para cada norma kappa...
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clear; clc; // Example: 6.19 // Page: 227 printf("Example: 6.19 - Page: 227\n\n"); // Solution // *****Data******// a = 3.59;// [square L atm /square mol] b = 0.043;// [L/mol] R = 0.082;// [J/mol K] //***************// // From Eqn. 6.122: Ti = 2*a/(R*b);// [K] printf("Inversion of temperature is ...
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// Example 2.8.3 clc; clear; n1=1.482; //refractive index of core n2=1.474; //refractive index of cladding lamda=820d-9; //Wavelength NA=sqrt(n1^2 - n2^2); //computing Numerical aperture theta= asind(NA); //computing acceptance angle solid_angle=%pi*(NA)^2; //computing solid ...
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//Variable declaration l = [5.57,5.76,4.18,4.64,7.02,6.62,6.33,7.24,5.57,7.89,4.67,7.24,6.43,5.59,5.39] // calculation Mean = mean(l) var = 0 for i = 1:length(l) var = var + (l(i)-Mean)^2 end var = var/length(l) coff = sqrt(var)/Mean // Results printf ( "Maximum likelihood estimates of Mean : %.3f , ...
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//Chapter-7,Example7_7,pg 7-30 Rm=500 Im=40*10^-6 V=10 Rs=(V/Im)-Rm printf("multiplier resistance\n") printf("Rs=%.2f ohm",Rs)
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//Example 25.1 R1=6;//Resistance (ohm) R2=2.5;//Resistance (ohm) R3=1.5;//Resistance (ohm) r1=0.5;//Internal resistance (ohm) r2=0.5;//Internal resistance (ohm) emf1=18;//Emf 1 (V) emf2=45;//Emf 2 (V) //A set of three equations are required since there are three unknowns-currents I1,I2 and I3 //Equation 1: I1=...
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//Chapter-6, Illustration 10, Page 309 //Title: Refrigeration cycles //============================================================================= clc clear //INPUT DATA Tg=470;//Heating temperature in K T0=290;//Cooling temperature in K TL=270;//Refrigeration temperature in K //CALCULATIONS COP=((Tg-T0...
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//Calculate the voltage gain of the FET clear; clc; //soltion //given Idss=8*10^-3;//A Vp=4;//V rd=25*10^3;//ohm Rd=2.2*10^3;//ohm //by the help of figure Vgs=-1.8;//V gmo=2*Idss/(abs(Vp)); gm=gmo*(1-(Vgs/(-Vp))); Av=-gm*(rd*Rd/(rd+Rd)); printf("The voltage gain of the FET %.2f",Av);
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clc //initialisation of variables v=10//m/sec f=20//kg g=9.81//m/sec q=12//m/sec //CALCULATIONS M=f/q//kg*m^-1 sec^2 G=M*g//kg //RESULTS printf('the acceleration due to gravity is =% f kg',G)