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clear;lines(0); s=poly(0,"s");p=1+s+2*s^2; A=rand(2,2);poly(A,"x")
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//example 4.43 //calculate average depth of precipitation clc;funcprot(0); //given A=[90 140 125 140 85 40 20]; //area of isohytes I=[13:-2:1]; //average isohytel interval s=0;t=0; for i=1:7 s=s+A(i)*I(i); t=t+A(i); end Pavg=s/t; Pavg=round(Pavg*10)/10; mprintf(" average depth...
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Name=QL MidAir Rocket PlayerCharacters=Rocketman BotCharacters=Pigeon.bot;Quaker Bot.bot IsChallenge=true Timelimit=60.0 PlayerProfile=Rocketman AddedBots=Pigeon.bot;Pigeon.bot;Pigeon.bot;Quaker Bot.bot;Quaker Bot.bot;Quaker Bot.bot PlayerMaxLives=0 BotMaxLives=0;0;0;0;0;0 PlayerTeam=1 BotTeams=2;2;2;0;0;0 MapName=boxe...
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//Interpolação Lagrange clear close clc x = [0; 0.2; 0.4; 0.5] y = [0; 2.008; 4.064; 5.125] p1 = 0.3 p2 = poly(0, 'p') n = size(x,1) L = zeros(n,1) function [s] = main(p) for i=1 :(n) aux1 = 1 aux2 = 1 for j=1 : (n) if j<>i aux1 = aux1 * (p - x(j, 1)) ...
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// sum 10-4 clc; clear; p=1.25; D=200; nt=0.75; C=9; sigta=20; t=(p*D)/(2*sigta*nt)+C; t=18; D1=D+(2*t); dr=D1+10; sigp=310; sigba=sigp/4; db=16; Db=dr+32+5; Do=Db+(2*db); P=%pi*(251+db)^2*1.25/4; n=6; Y=(Db-dr)/2; M=P/n*Y; Z=dr*tand(30)/6; tf=sqrt(M/(sigta*Z)); tf=22; Deff=dr+db+5; // prin...
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function [stk,txt,top]=sci_struct() // Copyright INRIA txt=[] names=['struct'];vals for k=1:2:rhs-1 names=[names;stk(top-rhs+k)(1)] end names='['+strcat(names,';')+']' for k=2:2:rhs vals=[vals;stk(top-rhs+k)(1)] end stk=list('tlist('+makeargs([names;vals])+')','0',string(rhs+1),'?','16','?')
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// Data Reconciliation Benchmark Problems From Literature Review // Author: Edson Cordeiro do Valle // Contact - edsoncv@{gmail.com}{vrtech.com.br} // Skype: edson.cv // aux functions to ipopt solver //*************************************************************** //This function analyses the structure of the problem ...
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//Ex 6.3 clc;clear;close; format('v',5) n=8;//no. of bits Range=0:10;//range LSB=max(Range)/2^n;//V MSB=max(Range)/2^0;//V VFS=MSB-LSB;//V disp(LSB*1000,"LSB(mV)"); disp(MSB,"MSB(V)"); disp(VFS,"VFS(V)");
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clc // Given that V = 1 // Volume of the flask in litre p = 1 // Pressure in atm t = 300 // Temperature in K r = 1.8e-10 // Radius of oxygen gas molecule in m m = 5.31e-26 // Mass of oxygen molecule in kg printf("\n Example 22.7 \n") n = (p*(1.013e5))/((1.38e-23)*(t)*1000) sigma = 4*%pi*(r^2) v = ((8*(1.38e-2...
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Pv-P -PvP (PvQvR)&(-P&-Q&-R) (-Pv-Q)<->-(P&Q) (-P&-Q)<->-(PvQ) -(P&-P)
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clear; clc; close; x=poly(0,'x'); p1=3*x/5+x/2; p2=5*x/4-3; p3=p1-p2; x=roots(p3) //by the law of signs
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clc //Example 14.10 //Install Symbolic toolbox //Determine the transform of rectangular pulse syms t s v=integ(exp(-s*t),t,2,%inf)-integ(exp(-s*t),t,5,%inf) disp(v,'V(s)=')
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//modified Impulse invariant Design //(c)H(s)=4s+7/s^2+5s+4 s=%s;z=%z; HS=(4*s+7)/(s^2+5*s+4); [d1]=degree(numer(HS)); [d2]=degree(denom(HS)); HZ=((3*z)/(z-%e^-2))+(z/(z-%e^-0.5)) if (d2-d1==1) then h=(4+7/%inf)/(1+5/%inf+4/%inf) HMZ=HZ-0.5*h else HMZ=HZ end HS1=4/((s+1)*(s^2+4*s+5)) HZ1=(0.21...
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//chapter 6 //example 6.12 //page 277 printf("\n") printf("given") Rc=5.6*10^3;Rl=33*10^3;rs=600; hfe=100;hie=1.5*10^3;vs=50*10^-3; disp(" CE circuit operation with vs at transistor base and Re bypassed") Av=(hfe*((Rc*Rl)/(Rc+Rl)))/hie Zb=hie Rb=(R1*R2)/(R1+R2); Zi=(Rb*Zb)/(Rb+Zb) vi=(vs*Zi)/(rs+Zi) vo=Av*...
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clear clc CAo=0.1; eA=3; rA=[3.4;5.4;7.6;9.1]; CA=[0.039;0.0575;0.075;0.092]; XA=zeros(4,1); inv_rA=zeros(4,1); for i=1:4 XA(i)=(1-CA(i)/CAo)/(1+eA*CA(i)/CAo); inv_rA(i)=1/rA(i); end //W=FAo*integral(dXA/-rA) from 0 to 0.35 //Using Trapezoidal rule to find area,XA must be in increasing order //Sorting XA ...
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//page no 187 //example no 6.9 //TURN OFF THE AIR CONDITIONER. //to turn OFF the air conditioner, reset bit D7 //Assuming the same input as earlier as it is a continuation of previous example. clc; A=[1 0 1 0 1 0 1 0]; B=[0 1 1 1 1 1 1 1]; Y=bitand(A,B); //ANDing input (A) with B to keep the D4 bit always set ...
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// WRD frame TopFr = uicontrol(frame_addWRD, 'style', 'frame','constraints', createConstraints('border', 'top', [0, 140]),.. 'layout', 'grid','backgroundcolor', [1 1 1],'layout_options', createLayoutOptions('grid', [1,2])); BotFr = uicontrol(frame_addWRD, 'style', 'frame','constraints', createConstraint...
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//points A,E,F,G are at the same potential Rab=20 Reb=50 R1=Rab*Reb/(Rab+Reb) //equivalent resistance of Rab and Reb Rbc=25 R2=R1+Rbc //equivalent resistance of R1 and Rbc Rfc=50 R3=Rfc*R2/(Rfc+R2) //equivalent resistance of R2 and Rfc Rcd=30 R4=R3+Rcd //equivalent resistance of R3 and Rcd R=R4*50/(50+R...
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function[] = plot_format() //Get the handle of current axes g = gca() //Give labels and set label properties g.labels_font_color=5 g.font_size=3 g.grid=[1,1] g.box="off" endfunction //Obtain path of solution file path = get_absolute_file_path('solution5_20.sce') //Obtain path ...
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2_15.sce
clear clc disp('Exa-2.15'); r=1.5*10^11; I=1.4*10^3; //radius and intensity of sun s=4*%pi*r^2 //surface area of the sun Pr=s*I // Power radiated in J/sec c=3*10^8; //velocity of light m=Pr/c^2 //rate od decrease of mass printf('The rat...
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example8_3.sce
//example8.3 clc disp("Given : I_B=20 uA, I_E=6.4 mA") disp("I_E=I_B+I_C=I_B+[(I_B)*(beta_dc)]=(I_B)*(1+(beta_dc))") b=(6.4*10^-3)/(20*20^-6) disp(b,"(beta_dc)+1=(I_E)/(I_B)=") b=320-1 disp(b,"Therefore, (beta_dc)=") a=319/320 format(7) disp(a,"(alpha_dc)=(beta_dc)/(1+(beta_dc))=319/(1+319)=") i=319*20 disp...
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6_7.sce
// example:-6.7,page no.-312. // program to design three section binomial transformer. Zl=50;Zo=100;N=3;taom=0.05; A=(2^-N)*abs((Zl-Zo)/(Zl+Zo)); frac_bw=2-(4/%pi)*acos(0.5*(taom/A)^2); for c=1 Z1=Zo*((Zl/Zo)^((2^-N)*(c^N))); disp(Z1,'Z1 = ') end for c=3^(1/3) Z2=Z1*((Zl/Zo)^((2^-N)*(c^N))); disp(Z2,'Z...
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Ex13_7.sce
// Variable Declaration I_a = 10.0*exp(%i*90*%pi/180) //Line current(A) I_b = 10.0*exp(%i*-90*%pi/180) //Line current(A) I_c = 10.0*exp(%i*0*%pi/180) //Line current(A) // Calculation Section a = 1.0*exp(%i*120*%pi/180) //Operator I_a0 = 1.0/3*(I_a+I_b+I_c) //Zero-sequence component(A) I...
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Example_1_21.sce
//Example 1.21 A batch contains 10 articles of which 4 are defective . clc; clear; N=120; disp(N,"total no. of ways in which 3 articles are selected out of 10 ="); M= 20; disp(M,"No. of favourable cases are "); P=M/N; disp(,P,"Probability that 3 articles are chosen at random and none is defective = ");
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sci
pammod.sci
function P = pammod(x, m, phi, datatype) // //Function Description //pammod: This function modulates a sequence of integers //x into a complex baseband phase amplitude modulation signal. // //Calling sequence:- //pam = pammod(x,m) //pam = pammod(x,m,phi) //pam = pammod(x,m,phi,datatype) // //Parame...
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Example8_11.sce
//Chapter-8,Example8_11,pg 8_56 R1=1.2*10^3 R2=4.7*10^3 C1=1*10^-6 C3=1*10^-6 f=0.5*10^3 w=2*%pi*f Rx=R2*C1/C3 Cx=R1*C3/R2 D=w*Cx*Rx printf("unknown capacitance and resistance\n") printf("Rx=%.f ohm\n",Rx) printf("Cx=%.8f F\n",Cx) printf("dissipation factor\n") printf("D=%.3f",D)
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test3.tst
load Larc.hdl, set RAM4K[0] %X8101, // 1. li R1 1 set RAM4K[1] %X8202, // 2. li R2 2 set RAM4K[2] %X0000, // 3. nop set RAM4K[3] %X0000, // 4. nop set RAM4K[4] %X0000, // 5. nop R1 <-- 1 set RAM4K[5] %XD512, // 6. sw 5(R1) R2 R2 <-- 2 set RAM4K[6] %XC324...
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2009-05-11T05:43:11
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kiks_error.sci
function [] = kiks_error(errstr) // Display mode mode(0); // Display warning for floating point exception ieee(1); // ----------------------------------------------------- // (c) 2000-2004 Theodor Storm <theodor@tstorm.se> // http://www.tstorm.se // ----------------------------------------------------- // !! L.8...
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Ex1_31.sce
// Ex 31 Page 374 clc;clear;close; // Given G=37;//dB f1=50;//Hz f2=18.7*1000;//Hz BW1=f2;//Hz (f2-f1~=f2) A1=10**(G/20);//Gain A3=sqrt(A1);//Gain RL1BYRL2=A1/A3;//ratio RL2BYRL1=A3/A1;//ratio //BW=2*%pi*Cd*RL BW1BYBW2=RL2BYRL1; BW2BYBW1=RL1BYRL2; f2dash=f2*sqrt(sqrt(2)-1); BW2=BW2BYBW1*f2dash;//Hz printf("Bandwidth...
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Ex1_7.sce
//refer Fig.1.13 in the textbook //by applying KVL //for mesh ABCDA, 7.45*i1-3.25*i2=10 //for mesh EFBAE, 8.55*i2-5.3*i3-3.25*i1=10 //for mesh HGBFEAH, 11.3*i3-5.3*i2=80 a=[7.45 -3.25 0;-3.25 8.55 -5.3;0 -5.3 11.3] b=[10;10;80] i=inv(a)*b i1=i(1,1) i2=i(2,1) i3=i(3,1) mprintf("Current in 6 ohm resistor=%f ...
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/Fourier-Transform-in-Image-Formation/aperture.sce
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2020-04-26T15:40:55.463187
2019-12-06T16:25:15
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aperture.sce
r0 = imread("C:\Users\csrc-lab03\Desktop\Activity5\Aperture0.jpg"); r1 = imread("C:\Users\csrc-lab03\Desktop\Activity5\Aperture1.jpg"); r2 = imread("C:\Users\csrc-lab03\Desktop\Activity5\Aperture2.jpg"); r3 = imread("C:\Users\csrc-lab03\Desktop\Activity5\Aperture3.jpg"); r4 = imread("C:\Users\csrc-lab03\Desktop\Act...
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/1820/CH5/EX5.5/Example5_5.sce
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Example5_5.sce
// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN // TURAN GONEN // CRC PRESS // SECOND EDITION // CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES // EXAMPLE : 5.5 : clear ; clc ; close ; // Clear the work space and console // GIVEN DATA C_a = 0.45 * 1...
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7_1.sce
//type of system s=%s G1=syslin('c',(1+0.5*s)/(s*(1+s)*(1+2*s)*(1+s+s^2))) disp(G1,"G(s)=") printf("type 1 as it has one s term in denominator") G2=syslin('c',(1+2*s)/s^3) disp(G2,"G(s)=") printf("type 3 as it has 3 poles at origin")
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Ex3_7.sce
clear // // // //Variable declaration d=0.203*10**-9 //lattice spacing(m) h=1 k=1 l=0 //miller indices of (110) lamda=1.5 //wavelength of X-rays(angstrom) //Calculation a=d*sqrt(h**2+k**2+l**2) //length(m) V=a**3 //volume of unit cell(m**3) r=sqrt(3)*a/4 //radius of atom(m) //Result p...
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Ex2_13.sce
// FUNDAMENTALS OF ELECTICAL MACHINES // M.A.SALAM // NAROSA PUBLISHING HOUSE // SECOND EDITION // Chapter 2 : BESICS OF MAGNETIC CIRCUITS // Example : 2.13 clc;clear; // clears the console and command history // Given data k_h = 110 // hysteresis co-efficient in J/m^3 V_cvol = 0.005 // volume ...
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FullSubtractor.tst
load FullSubtractor.hdl, output-file FullSubtractor.out, compare-to FullSubtractor.cmp, output-list a%B3.1.3 b%B3.1.3 c%B3.1.3 diff%B3.1.3 borrow%B3.1.3; set a 0, set b 0, set c 0, eval, output; set a 0, set b 0, set c 1, eval, output; set a 0, set b 1, set c 0, eval, output; set a 0, set b ...
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test_ping.tst
# Wait a minute before starting, to give agent a chance to initialize DELAY FOR 60 WHENEVER True # Wait 6 minutes for a ping WAIT ping FOR 360 # Reset the ping and wait until the next one WHENEVER ping WAIT not ping FOR 10 # Don't want the current ping to confuse things ENSURE not ping FOR 120 # This should...
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Ex6_1.sce
clear; clc; funcprot(0); //given data dt = 1.0;//tip diameter in m dh = 0.9;//hub diameter in m alpha1 = 30;//in deg beta1 = 60;//in deg alpha2 = 60;//in deg beta2 = 30;//in deg N = 6000;//rotational speed in rev/min rhog = 1.5;//gas density in kg/m^3 Rt = 0.5;//degree of reaction at the tip //Calculat...
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//scilab 5.4.1 //Windows 7 operating system //chapter 17 Number Systems,Boolean Algebra,and Digital Circuits clc clear dec=263 base=5 s=dec2base(dec,base) disp(,s,"Equivalent of 263 in a code base 5 is ")
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clear; clc; // SCILAB-2.sce // Constantes A = 1; alpha = 0.5; omega = 2; // Funciones function y = f(t) y = A*exp(-alpha*t).*sin(omega*t); endfunction dt = 0.001; tfin = 10; t = 0:dt:tfin; y = f(t); // Gráficas scf(1); clf(1); plot(t,y); xgrid; xlabel('t'); ylabel('y'); // Objetivos yobj = 0.5; indexyobj = ...
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//Electric Drives:concepts and applications by V.subrahmanyam //Publisher:Tata McGraw-Hill //Edition:Second //Ex4_10 clc; clear; V=500;// voltage in V I=15;//Current in A t=0.6;//time in sec f=80;//frequency in Hz Vav=V*t; Vi=V-Vav; Ton=t/f; L=Vi*(Ton/I); disp(L,"The inductance in Henry is:")
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//example 8.20 //calculate streeses at heel and toe of dam clc;funcprot(0); //given c=1; hw=80; //heigth of water in reservior Bt=6; //width of top of dam H=84; //heigth of the dam Hs2=75; //heigth of slope on downstream side wb=56; //width of base of dam ...
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p=[0.1,0.7,0.8,0.85,0.9,0.95,1] function [n]=test() n=1 c(1,n)=4 while c(1,n)>0 u1=rand() i=1 while i<=7 if u1<=p(i) then c(1,n+1)=c(1,n)+i-1 break else i=i+1 end end i...
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//chapter17 //example17.3 //page379 L1=1000d-6 // H L2=100d-6 // H M=20d-6 // H C=20d-12 // F Lt=L1+L2+2*M f=1/(2*%pi*(Lt*C)^0.5) mv=L2/L1 printf("operating frequency = %.3f Hz or %.3f kHz \n",f,f/1000) printf("feedback function = %.3f",mv) //in book the answer is 1052 kHz but the accurate answer is 1054.029 kHz
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clear // //variable declaration Db=(20) //diameter of brass rod,mm Dse=(40) //external diameter of steel tube,mm Dsi=(20) //internal diameter of steel tube,mm Es=(2*100000 ) //Young's modulus steel, N/mm^2 Eb=(1*100000 ) //Young's mod...
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clc //initialisation of variables k= 15 //knots w= 64 //lbf/ft^3 W= 5 //tonf l= 6 //ft U= 6080 //ft/km.hr //CALCULATIONS P= (0.5/32.2)*w*(k*U/3600)^2 Ct= (W*2240)/(P*%pi*(l/2)^2) nf= 2/(1+sqrt(1+Ct)) Pb= (W*k*2240/nf)*6080/(3600*550) //RESULTS printf (' theotrical power= %.f h.p',Pb)
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//Chapter 12 : Solutions to the Exercises //Scilab 6.0.1 //Windows 10 clear; clc; //Solution for 5.21 A=[1 -1 1 0;1 -1 1 1;0 1 1 -1;0 2 3 -3] ainv=inv(A) disp(ainv,'A^-1') mprintf('(a,b,c,d)=\n') mprintf(' l1(1,1,0,0)+l2(-1,-1,1,2)+l3(1,-1,1,3)+l4(0,1,-1,-3)')
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//clc() //f(x) = exp(-x)-x //f'(x) = -exp(-x)-1 for i = 1:5 if i == 1 then x(i) = 0; else x(i) = x(i-1) - (exp(-x(i-1))-x(i-1))/(-exp(-x(i-1))-1); et(i) = (x(i) - x(i-1)) * 100 / x(i); x(i-1) = x(i); end end disp(x) disp(et)
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100 -2.5 0 0 0 4.33013 0 2.5 0 0 -2.5 7.88925 0 0 12.2194 0 2.5 7.88925 0 -2.5 0 0.5 0 4.33013 0.5 2.5 0 0.5 -2.5 4.17849 0.5 0 8.50862 0.5 2.5 4.17849 0.5 -2.5 0 1 0 4.33013 1 2.5 0 1 -2.5 1.8691 1 0 6.19923 1 2.5 1.8691 1 -2.5 0 1.5 0 4.33013 1.5 2.5 0 1.5 -2.5 8.22943 1.5 0 12.5596 1.5 2.5 8.22943 1.5 -2.5 0...
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//[r]=%lnp(l1,l2) //%lnp(l1,l2) correspond a l'operation logique l1==l2 avec l1 une liste //et l2 une matrice de polynomes //! r=%t //end
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clear //Given //we will divide this into two equal parts and other part l = 10.0 // in - The height t = 0.1 // in - The width b = 5.0 //mm- The width of the above part A = t* b //sq.in - area of part y_net = l/2 // The com of the system y_1 = l // The position of teh com of part_2 I_1 = t*(l*...
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//Book - Power System: Analysis & Design 5th Edition //Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J. Overbye //Chapter - 12 ; Example 12.7 //Scilab Version - 6.0.0 ; OS - Windows clc; clear; C1=[8e-3 10 0] //Coefficients of cost equation for unit 1 C2=[9e-3 8 0] //C...
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//find.. clc //solution //given P=45*1000//W N=1000//rpm a=(%pi/180)*12.5 D=500//mm R=250//mm u=0.2 pn=0.1//N/mm^2 T=P*60/(2*%pi*N)*1000//N-mm //let b be face width //T=2*%pi*u*R^2*b b=T/(2*%pi*pn*u*R^2)//mm printf("face width is,%f mm\n",b) Wn=pn*2*%pi*R*b//N We=Wn*(sin(a)+0.25*u*cos(a)) printf("axia...
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//---------------------------------------------------------------// //---------------------------------------------------------------// //-------------------------- TU Berlin --------------------------// //--------- Fakultaet IV: Elektrotechnik und Informatik ---------// //------------------------------------------...
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//pagenumber 221 example 34 clear em1cur=2*10^-3;//ampere v1=12;//volt vcc=12;//volt format(12); colres=5*10^3;//ohm em1res=v1/em1cur; colcur=em1cur; voltag=colcur*colres;//ic*r v1=vcc-(colres*colcur); disp("emitter current = "+string((em1cur))+"ampere"); disp("collector current = "+string((colcur))...
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// Example 10.9, page no-273 clear clc e=1.6*10^-19//C m=9.1*10^-31//kg h=6.626*10^-34 k=1.38*10^-23 eg=1.1*e mue=0.48//m^2/V.s muh=0.013//m^2/V.s T=300//K ni=2*(2*%pi*m*k*T/(h^(2)))^(1.5) ni=ni*%e^(-eg/(2*k*T)) sig=ni*e*(mue+muh) printf("\nThe carrier concentration of an intrinsic semiconductor is = %....
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X1=[0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,1,1,1,1,1,-1; 0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,1,1,1,1,1,0,0,0,0,0,-1; 0,0,1,0,0,0,0,1,0,0,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,-1; 0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,1,0,0,0,1,1,1,0,0,-1; 0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,1,0,0,0,0,1,0,0,0,-1; 1,1,1,1,1,0,0,1,0,0,0...
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clc //initialisation of variables P= 680 //lb K= 1000 //lb/in L= 6 //ft E= 30*10^6 Ina= 1.728 //in^4 //CALCULATIONS A= [((L*12)^3/(3*E*Ina)),-(1/K);1,1] b= [0;P] c= A\b Pb= c(1,1) Ps= c(2,1) //RESULTS printf ('Force in the spring= %.2f psi',Ps)
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// Updated(1-8-07) // 9.21 // Motor control problem // Transfer function a = [-1 0; 1 0]; b = [1; 0]; c = [0 1]; d = 0; G = syslin('c',a,b,c,d); Ts = 0.25; [B,A,k] = myc2d(G,Ts); [Ds,num,den] = ss2tf(G); // Transient specifications rise = 3; epsilon = 0.05; phi = desired(Ts,rise,epsilon); // Controll...
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q=4.0*10^-3;//quantity of electricity in coulombs// e=1.6*10^-19;//charge of an electron in coulombs// N=q/e;//no. of electrons per second// printf('No. of electrons per second=N=%f=2.5*10^16',N);
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// mistakes & errors
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clear clc s = poly(0,'s'); G = 1/((s+3)*(s+4)*(s+12)); for k = 1:50:1000 T = k*G/(1+k*G); t = 0:0.1:20; T1 = syslin('c',T); ts = csim('step' ,t, T1); plot(t , ts); end
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//Part A Chapter 5 Example 4 clc; clear; close; T1=827+273;//K T2=27+273;//K T3=-13+273;//K Q1=2000;//kJ Q2=545.45;//kJ WE=Q1-Q2;//kJ Q3BYQ4=T3/T2; WE_sub_WR=300;//kJ WR=WE-WE_sub_WR;//kJ Q43=WR;//kJ(Q4-Q3=WR) Q4=WR/(1-Q3BYQ4);//kJ Q3=Q4-Q43;//kJ Qt=Q2+Q4;//kJ disp("Heat transfered to refrigerant = "+s...
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//Calculate speed of motor and maximium torque //Chapter 4 //Example 4.24 //page 321 clear; clc; disp("Example 4.24") V=400; //supply voltage in volts f=50; //frequency in hertz P=6; //number of poles ph=3; //three phase supply R2=0.03; ...
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//Example 9.4 g=9.80;//Acceleration due to gravity (m/s^2) w_a=2.50*g;//Weight of forearm, m*g, (N) w_b=4.00*g;//Weight of load, m*g, (N) r1=4*10^-2;//Distance of force exerted by biceps from elbow (m) r2=16*10^-2;//Distance of CG of forearm from elbow (m) r3=38*10^-2;//Distance of load from elbow (m) F_B=(r2*w_...
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function [x,y,typ]=meas_volt(job,arg1,arg2) // Copyright INRIA x=[];y=[];typ=[]; select job case 'plot' then standard_draw(arg1) case 'getinputs' then //** GET INPUTS [x,y,typ]=standard_inputs(arg1) case 'getoutputs' then [x,y,typ]=standard_outputs(arg1) case 'getori...
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clear; clc; printf("\t\t\tExample Number 8.9\n\n\n"); // effective emissivity of finned surface // Example 8.9 (page no.-409-410) // solution // for unit depth in the z-dimension we have A1 = 10;// [square meter] A2 = 5;// [square meter] A3 = 60;// [square meter] // the apparent emissivity of the open cav...
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printf("\t example 17.4 \n"); printf("\t approximate values are mentioned in the book \n"); // The area between the saturation line and the operating line represents the potential for heat transfer // at T=79.3F Hs=43.4; // fig 17.12 H=30.4; // fig 17.12 d1=(Hs-H); printf("\t difference is : %.1f \n",d1); //at ...
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//To find velocity, torque and acceleration clc //Given: NAO=100 //rpm OA=150/1000,AB=600/1000,BC=350/1000,CD=150/1000,DE=500/1000 //m dA=50/1000,dB=dA,rA=dA/2,rB=dB/2 //m pF=0.35 //N/mm^2 DF=250 //mm //Solution: //Refer Fig. 8.21 //Calculating the angular speed of the crank AO omegaAO=2*%pi*NAO/60 //rad/s ...
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//Caption: (a)Power alternator can supply (b)Power factor of synchronous motor (c)Load taken by motor //Exa:15.8 clc; clear; close; P=500//Load supplied by alternator(inKW) pf=0.8//Power factor e=0.9 L=P/pf Ps=L-P disp(Ps,'(a)Power alternator can supply(in KW)=') Pr=P*tand(acosd(pf)) pfm=cosd(atand(Pr/Ps))...
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//Caption: Rank Correlation //Test of Significance of rank correlation coefficient //Example10.3 //Page375 X = [1,2,3,4,5,6,7,8,9,10];//rating by judge-1 Y = [2,4,5,1,3,6,7,9,10,8];//rating by judge-2 n = length(X); rs = 0; for i = 1:n rs = rs+(X(i)-Y(i))^2; end rs = 1-6*(rs/(n*(n^2-1))); disp(rs,'The r...
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clc; funcprot(0);//Example 14.9 //Initializing the variables As = 6; d = 0.02; f =0.01; L = 1.5; K = 0.9; g = 9.81; //Calculations Ap = %pi*d^2/4; function[y] = Qinv(h) y = sqrt((4*f*L/d +K+1)/(2*g*h))/Ap; endfunction //By direct integration t = -As*intg(3.5,2.25,Qinv); // Discharge is 2 m below...
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clear,clc, xdel(winsid()); pathname = get_absolute_file_path("СХЕМА.sce") chdir(pathname) winnum=1; win=string(winnum); //Создаем рабочую панель cf=figure(winnum); set(cf,'figure_name','ОКНО УПРАВЛЕНИЯ'); cf.figure_position=[200,100] cf.figure_size=[1500,600] startButton_Nazvanie = uicontrol(c...
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Vgs1=-5; Vgs2=-0.75; Gm01=0.006; Gm02=0.002; Vgsoff1=-8; Vgsoff2=-2; Gm1=Gm01*(1-(Vgs1/Vgsoff1)); Gm2=Gm02*(1-(Vgs2/Vgsoff2)); RD=8200; RL=100000; rD=(RD*RL)/(RD+RL); Avmax=rD*Gm1; Avmin=rD*Gm2; disp(' ',Avmax,"Avmax=")//The answers vary due to round off error disp(' ',Avmin,"Avmin=")//The answers vary du...
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exec('path', -1) x=[0:.1:10]; deff('z=f(x)','z=exp(x)-3*x'); deff('z=fp(x)','z=exp(x)-3'); z=newton(f,fp,0.25,10,0.001)
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//Problem 20.21: The output stage of an amplifier has an output resistance of 112 ohm. Calculate the optimum turns ratio of a transformer which would match a load resistance of 7 ohm to the output resistance of the amplifier. //initializing the variables: R1 = 112; // in Ohms RL = 7; // in Ohms //calculation:...
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//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 7.7w //calculation of the value of force exerted by the air on the plane //given data v=900*10^3/(60*60)//speed(in m/s) of the fighter plane r=2000//radius(in m)of the vertical circle M=16000//mass(in kg) g=9.8//gra...
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clear; errcatch(4,'continue','nomessage') a if iserror(4)==1 then errclear(-1),else pause,end a;b=1; errclear(-1),if b<>1 then pause,end b=[];for k=1:3,b(1,k)=k;a,end errclear(-1),if or(b<>(1:3)) then pause,end deff('foo()','x=a','n') foo();if iserror(4)==1 then errclear(-1),else pause,end deff('foo()','x=a') foo();i...
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//Example 3.21 //Determine the response of the below relaxed system. clc; close; n=-10:.01:10; for i=1:length(n) if n(i)<0 then h(i)=0;x(i)=0; else h(i)=(1/3)^n(i); x(i)=2^n(i); end end y=convol(h,x); //figure f=scf(0); plot(n,h,'black'); xtitle('input ...
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//Exemplo: Sejam os nós x [0 0.6 1]. Encontre os pesos Ai da quadratura I = A1*f(x1) + A2 * f(x2) + A3 * f(x3) tal que o erro seja o menor possível //para integrar f no intervalo 0 a 1 //Primeiro calcula A que neste caso com os valores 0, 0.6 e 1 será "A = [1 1 1; 0 0.6 1; 0 (0.6)^2 1]" //Depois de calcula b, que no i...
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function [mod,phase] = complexe(matrice) [N,M] = size(matrice) for x = 1:N for y = 1:M mod(x,y) = abs(matrice(x,y)) phase(x,y) = atan(imag(matrice(x,y)),real(matrice(x,y))) end end endfunction
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clc p1=75.882; //cm of Hg T1=286; //K V1=0.08; //m^3 p2=76; //cm of Hg T2=288; //K V2=p1*V1*T2/p2/T1; m=28; //kg c=4.18; t2=23.5; //0C t1=10; //0C Q_received=m*c*(t2-t1); HCV=Q_received/V2; disp("Higher calorific value =") disp(HCV) disp("kJ/m^3") amt=0.06/0.08; //Amount of vapour formed per m...
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//<>=zgrid() // xselect(); square(-1.1,-1.1,1.1,1.1); xtitle( ['loci with constant damping and constant frequencies';... 'in discrete plane'],' ',' '); // xsi=0:0.1:1 // // 2 2 //roots of s + 2*xsi*w0*s +w0 //given by : w0*(-xsi+-%i*sqrt(1-sxi*xsi)) raci...
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# ATWM1 MEG Experiment scenario = "ATWM1_Working_Memory_MEG_salient_uncued_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_monit...
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clc clear //INPUT DATA h=1//miller indices with respect to x axis k=1//miller indices with respect to y axis l=0//miller indices with respect to z axis d=2.86*10^-10//the distance between miller indices in m //CALCULATION a=(d*(sqrt(h^2+k^2+l^2)))/10^-10//The lattice constant in m *10^-10 //OUTPUT printf(...
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clc C_x = [485 490 545 611 606 596 567 546] // x do Centro C_y = [308 312 315 335 351 352 343 344] // y do Centro P_x = [524 317 169 582 758 867 1004 515] // x do movimento pendular P_y = [1271 1254 1194 1302 1299 1274 1178 1276] // y do movimento pendular tmps = [0 106 305 644 745 815 1020 1393] // tempos d...
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//Chapter-5, Example 5.22, Page 182 //============================================================================= clc clear //INPUT DATA Z1=7+((%i)*5);//impedance of branch1 in ohms Z2=10-((%i)*8);//impedance of branch2 in ohms V=230;//supply voltage in volts f=50;//frequency in hz //CALCULATIONS Y1=1/(Z1)...
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clc //initialisation of variables t=1200//F t1=1660//F p1=39.7//lbf p2=14.7//lbf cp=0.24//Btu v1=4930000//ft //CALCULATIONS T=(t1)/(p1/p2)^(0.4/1.4)//R V=cp*(t1-T)//Btu V2=sqrt(v1)//ft per sec //RESULTS printf('the steady of flow equation reduces=% f ft per sec',V2)
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clc //initialisation of variables A= 600 //ft^2 W= 40 //lbf/ft^2 n= 75 //percent r= 10 v= 300 //miles/hour //CALCULATIONS L= W*A D= L/r P= D*v*5280/(60*33000) hp= P*100/n //RESULTS printf (' brake horse-power of the engines= %.f h.p',hp)
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clear clc //to find change in internal energy //Given: //refer to figure 23-17 from page no. 529 //refer to problem 23-4 //final volume vf = 1.0//in m^3 //initial volume vi = 4.0//in m^3 //initialvolume pi = 10//in Pa //value of constant for monoatomic gas gama = 1.66 //number of moles of ideal gas n =...
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disp("Part a"); r1=5*10^(-2); r2=7*10^(-2); n=400; i=2; f=1.5*10^(-4); d=r2-r1; a=%pi*d^2/4; b=f/a; mmf=n*i; r=(r1+r2)/2; l=2*%pi*r; h=mmf/l; mu=b/h; disp("the permeability of the iron core (in Wb/At.m) is");disp(mu); disp("Part b"); mu0=4*%pi*10^(-7); mu1=mu/mu0; disp("the relative permeability is");...
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clc P1=100*10^3 v1=1 v2=1/16 n=1.25 T1=300 P2=P1*((v1/v2)^n) mprintf("P2=%fMPa\n",P2/(10^6)) T2=(T1*P2*v2)/(P1*v1) mprintf("T2=%fK\n",T2) R=8.314 W=(R*(T1-T2))/(n-1) mprintf("W=%fkJ/mol\n",W/1000) gama=1.4 q=((R*(T2-T1))/(gama-1))+W mprintf("q=%fkJ/mol",q/1000)
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clc clear //Input data d=0.175;//The diameter of the bore in m pi=3.141;//The mathematical constant of pi L=0.32;//The length of the stroke in m p=6.5;//Mean effective pressure in bar pp=0.4;//Pumping loop mean effective pressure in bar N=510;//The speed of the engine in rpm pm=0.65;//Diagrams from the dead cy...
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function [y,q]=movavew(x,m,w) n=length(x) y=zeros(1:n-2*m) q=0 for i=1:n-2*m y(i)=0 for j=1:2*m+1 y(i)=y(i)+x(i+j-1)*w(j) end q=q+abs(x(i+m)-y(i)) end q=q/(n-2*m) endfunction
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//Problem 14.01: Determine the periodic time for frequencies of (a) 50 Hz and (b) 20 kHz //initializing the variables: f1 = 50; // in Hz f2 = 20000; // in Hz //calculation: T1 = 1/f1 T2 = 1/f2 printf("\n\n Result \n\n") printf("\n (a) Periodic time T = %.2f secs",T1) printf("\n (b) Periodic time T = %.2...
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//chapter-8 page 338 example 8.4 //============================================================================== clc; clear; //For a reflex klystron n=2;//peak mode value V0=500;//beam voltage in V Rsh=20000;//Shunt resistance in ohms L=0.001;//distance in m f=8*10^(9);////Operation frequency in Hz V1=200...
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tst
Mux16.tst
load Mux16.hdl, output-file Mux16.out, output-list a%B1.16.1 b%B1.16.1 sel%D2.1.2 out%B1.16.1; set a 0, set b 0, set sel 0, eval, output; set sel 1, eval, output; set a %B0000000000000000, set b %B0001001000110100, set sel 0, eval, output; set sel 1, eval, output; set a %B1001100001110110, set b %B0000000000000000...