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// Grob's Basic Electronics 11e // Chapter No. 16 // Example No. 16_10 clc; clear; // For the tantalum capacitor shown in Fig. 16–21, determine the capacitance C in both pF and uF units. Also, determine the voltage rating and tolerance. disp ('Moving from top to bottom, the first two color bands are yellow and v...
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i have 18 minutes to tell you what happened over the past six million years all right we all have come from a long way here in africa and converged in this region of africa which is a place where 90 percent of our evolutionary process took place and i say that not because i am african but it s in africa that you find t...
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//Section-14,Example-1,Page no.-PC.112 //To calculate the pH values of the following. clc; C_HCl=0.001 //(M) C_1=C_HCl //Since HCl is a strong acid,[H3O+]=[HCl](M) pH_1=-log10(C_1) disp(pH_1,'pH of 0.001 M HCl') C_NaOH=0.0001 //(M) C_2=C_NaOH //Since NaOH is a strong...
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<?xml version="1.0" ?> <TestCase name="Sub" version="5"> <meta> <create version="10.1.0" buildNumber="10.1.0.283" author="admin" date="06/08/2018" host="NESJO02" /> <lastEdited version="10.1.0" buildNumber="10.1.0.283" author="admin" date="06/08/2018" host="NESJO02" /> </meta> <id>806EAC186B1211E8BEF7FAAA20524...
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--> a = 0.6; --> n = 0:10; --> x = a^n x = column 1 to 6 1. 0.6 0.36 0.216 0.1296 0.07776 column 7 to 9 0.046656 0.0279936 0.0167962 column 10 to 11 0.0100777 0.0060466 --> plot2d3(x)
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clc //initialisation of variables k= 2.63*10^-3 //min^-1 //CALCULATIONS t1= 0.693/k //RESULTS printf (' Half time= %.f min',t1+1)
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//Chapter 5, Example 5.2 clc //Initialisation' i=6 //current in amp n=500 //no of turns l=0.4 //mean circumference pi=3.14 //pi uo=4*pi*10**-7 //dielectric constant a=300*10**-6 ...
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clear // // //Initilization of Variables //Flange b=100 //mm //Width D=80 //mm //Overall Depth t=10 //mm //Thickness of web and flanges L=3000 //mm //Length of strut E=200*10**3 //N/mm**2 //Modulus of Elasticity //Calculations //Let centroid be at depth y_bar from top fibre y_bar=(b*t*t*2**-1+(D-t)*t*((D-t)*2**-1...
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function X =Resolve_com_LU(C, B, P) // C e B são como especificados no enunciado e P é a matriz de permutações n = size(C,1); m = size(B,2); //permuta os elementos dos vetores b_i B = P*B; // Y=zeros(n,m); // resolve LY=B, ou seja, resolve L y_i = b_i com 1<=i<=m for j=1:m Y(1, j)=B(1,j); for i=2:n Y(...
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//Print the alternate characters of the file output1.dat warning('off'); fstream1=mopen('output1.dat','r'); for a=1:13 ch=mfscanf(fstream1,"%c"); //Read data from file output.dat printf("%c ",ch); //Print the data mseek(1,fstream1,'cur'); //Skeep 1 character after every reading end mclose(fstrea...
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s=%s; syms eps p=s^5+s^4+2*s^3+2*s^2+3*s+5 r=coeff(p); n=length(r); routh=[r([6,4,2]);r([5,3,1])] syms eps; routh=[routh;eps,-det(routh(1:2,2:3))/routh(2,2),0]; routh=[routh;-det(routh(2:3,1:2))/routh(3,1),-det(routh(2:3,2:3))/routh(4,2),0]; routh=[routh;-det(routh(4:5,1:2))/routh(5,1),0,0]; disp(routh,"routh...
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// Exa 7.3 clc; clear; close; // Given data I_DSS = 10;// in mA V_GS = 0;// in V I_D = 0;// in mA V_P = -4;// in V V_GS= 0:-0.1:V_P;// in V I_D = I_DSS*(1-(V_GS/V_P))^2;// mA plot(V_GS,I_D); xlabel("V_gs in volts"); ylabel("I_D in mA"); title("Transfer characteristics for an n-channel depletion type MOSFE...
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//Example 3.69.b:resistance and capacitance clc; clear; close; rab=521;//ohms lab=2;//micro-H cab=550;//pF rbc=1200;//ohms lbc=5;//micro-H cbc=250;//pF rda=12.1;//ohms rda1=1.5;//m-ohms f=10;//kHz cda=0.045;//micro-F zab=1/((1/(rab+(%i*2*%pi*f*10^4*lab*10^-6)))+(%i*2*%pi*f*10^3*cab*10^-12));// zbc=1/((1/(rbc+(%i*2*%pi*...
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//Example 16-3,Page No- 623 clear clc printf('The c band is approximately 4 to 6 Ghz since a waveguide \n acts as a high pass filter with cut off frequency of \n 9.08 Ghz it will not pass c band signal')
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clc clear //INPUT DATA Eg=1.2*1.6*10^-19//The energy gap of intrinsic semiconductor in J T1=600//Temperature in K T2=300//Temperature in K e=1.6*10^-19//charge of electron in coulombs kb=1.38*10^-23//Boltzmann's constant m^2 Kg s^-2 k^-1 //CALCULATION x=exp((-Eg/(2*kb))*((1/T1)-(1/T2)))/10^5//The ratio of co...
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clc; clear all; disp("Required tube length") m=50/60;//kg/sec di=2.5/100;//m inner diameter ts=100;// degree C ti=25;// degree C to=55;// degree C cp=4187;//J/kg.C t1=100-25; t2=100-55; tm=(t1-t2)/log(t1/t2); Q=m*cp*(to-ti);//W As=%pi/4*di^2; ;tw=(ts+(ti+to)/2)/2;// degree C temperature at which the prope...
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//Example 2.9 clear; clc; //Given R=8.314;// gas constant in J K^-1 mol^-1 t1=298;// temperature in K p1=303975;// initial pressure in Pa v1=1.43;//initial volume in dm^3 v2=2.86;// final volume in dm^3 Cp=29.1;// heat capacity at constant pressure in J mol^-1 K^-1 // To determine t2,p2,w,delH and DelE C...
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clc clear //Input data T1=450;//The temperature of the source in k H1=1000;//The amount of heat taken by the engine at T1 in calories T2=350;//The temperature of the sink in K //Calculations H2=(T2/T1)*H1;//The amount of heat rejected to the sink in each cycle in calories n=(1-(T2/T1))*100;//The efficien...
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// Example 1.12.1 clc; clear; Bandwidth = 2d6; //Bandwidth of channel Signal_to_Noise_ratio = 1; //Signal to Noise ratio of channel Capacity = Bandwidth * log2(1 + Signal_to_Noise_ratio); //computing capacity Capacity=Capacity/10^6; printf("Maximum capacity of channel is %d Mb/sec.",Capacity);
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clc //initialisation of variables dH293= -115595.8 //cal T1= 1500 //K T2= 298 //K k1= -5.6146 //cal deg^-1 mole^-1 k2= 1.8931*10^-3 //cal deg^-2 mole^-1 k3= 4.723*10^-7 //cal deg^-3 mole^-1 //CALCULATIONS dH=dH293+ k1*(T1-T2)+(k2*(T1^2-T2^2)/2)+(k3*(T1^3-T2^3)/3) //RESULTS printf ('dH1500= %.f cal ',dH)
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//2.11 clc; RAB=125; Rtotal=5000; R2=75/125*Rtotal; R4=2500; ei=5; eo=[(R2/Rtotal)-(R4/Rtotal)]*ei; printf("Output voltage = %.1f V",eo)
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<cmd> ./main_test/lexer_tst 'echo "we are the \"best ones\" yeah"; echo "Share our \\\"SHELL\\\" yeah"; echo "Share\\" our \\" shell";'</cmd> <ref> cat ./lexer/test3.ref</ref> <ret> 0</ret>
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//Example 1_37 clc; clear;close; //Given data: V=300;//V RL=10;//ohm L=0;//H Ith=100;//A f=2000;//Hz dv_by_dt=100*10^6;//V/s //dv/dt=(vth(tau)-vth(0))/tau //dv/dt=RL*(1-0.368)*V/(R+RL)/((R+RL)*C) R=V/Ith;//ohm C=RL*(1-0.368)*V/(R+RL)/(R+RL)/dv_by_dt disp(R,"Value of R(ohm)"); disp(C,"Value of C(F)"); ...
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20_08.sce
//Problem 20.08: A 100 kVA, 4000 V/200 V, 50 Hz single-phase transformer has 100 secondary turns. Determine (a) the primary and secondary current, (b) the number of primary turns, and (c) the maximum value of the flux. //initializing the variables: S = 100000; // in VA V1 = 4000; // in Volts V2 = 200; // in Volts...
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ex_13_8.sce
clc //soltuion //given P=90*10^3//W N=250//rpm ts=40//N/mm^2 q=0.0175 tb=30//N/mm^2 //let d be dia T=(P*60*1000)/(2*%pi*N)//N-mm //T/J=ts/(d/2) //T/(%pi*d^4/32)=ts/(d/2)//considering strength iof shaft d1=(35*10^6/80)^(1/3)//mm //considering rigidity //T/J=(C*q/l) //T/(%pi*d^4/32)=84000*0.0175/(20*d) d2=(35*10^6/73.5)...
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[EXPLICITE] paramètres.sce
//------------------------------------------------------------------------------ // PARAMETRES DU MODELE //------------------------------------------------------------------------------ //Température Temperature 20; T_opt_L = 37.5; // température optimale de croissance des légionnelles T_i_L = 55; // paramètre d'aju...
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bow.1_18.tst
1 10:0.5 19:0.05263157894736842 30:0.1 66:0.25 149:1.0 203:1.0 1 4:0.09090909090909091 11:1.0 49:0.08333333333333333 249:0.125 345:1.0 1 3:0.3333333333333333 13:0.3333333333333333 14:0.07407407407407407 17:0.03225806451612903 28:1.0 30:0.1 84:0.2 187:1.0 298:1.0 883:1.0 1 4:0.09090909090909091 6:0.5 10:0.5 14:0.0740740...
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Ch014Ex12.sce
// Scilab code Ex14.12 : Pg:725(2011) clc;clear; I_E = 2; // Emitter current, mA alpha = 0.88; // Emitter current amplification factor I_C = alpha*I_E; // Collector current, mA I_B = I_E - I_C; // Base current of BJT in CB mode, mA printf("\nThe base current of BJT in CB mode = %4.2f mA", I_B); ...
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ex5_6.sce
// Exa 5.6 clc; clear; close; // Given data f_c = 2;// in kHz f_c = f_c * 10^3;// in Hz C2 = 0.033;// in µF C2 = C2 * 10^-6;// in F C3 = C2;// in F C = C2;// in F R2 = 1/(2*%pi*f_c*C);// in ohm R2 = R2 * 10^-3;// in k ohm R3=R2;// in kohm disp(R2,"The value of R2 and R3 in kΩ is : "); //R_F= 0.586*R1 R1...
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Ex1_6.sce
clc; clear; V=2000 //potential in V e=1.602*10^-19 //electronic charge in eV m=9.1*10^-31 //mass of electron in kg //Calculation u=sqrt((2*V*e)/m) mprintf("velocity with which electron beam will travel= %.2e m/s",u)
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example_8_5.sce
//example 8.5 clc; clear; disp('Here what happens at each point in time' ); disp('Time t0: S = 0, R = 0, no change in Q (Q remains 0)'); disp('Time t1: S = 1, R = 0, Q changes from 0 to 1'); disp('Time t2: S = 0, R = 1, Q resets to 0'); disp('Time t3: S = 1, R = 0, Q sets to 1 '); disp('Time t4: S = 0, R = 0, ...
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example10_3.sce
//clc() Nflue = 100;//kmol NCO2 = 9; NCO = 2; NO2 = 3; NN2 = 86; NCflue = NCO2 + NCO ; MC = 12; mC = MC * NCflue ; //let A kmol air supplied, taking N2 balance, Nair = NN2 * 100/79; NO2supplied = Nair - NN2; // if CO in the flue gas was to be completely converted to CO2, then, the moles of oxygen present in...
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Ex5_8.sce
// FUNDAMENTALS OF ELECTICAL MACHINES // M.A.SALAM // NAROSA PUBLISHING HOUSE // SECOND EDITION // Chapter 5 : DIRECT CURRENT MOTORS // Example : 5.8 clc;clear; // clears the console and command history // Given data I_L1 = 5 // dc shunt motor current in A V_t = 230 // supply voltage in V R_a...
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ko.tst
#------------------------------------------------------------------------------ # Ko # # Computer Go Test Collection http://www.cs.ualberta.ca/~games/go/ # # $Source: /usr/cvsroot/project_cgtc/ko.tst,v $ # $Id: ko.tst,v 1.3 2004/04/28 17:10:39 emarkus Exp $ #-------------------------------------------------------------...
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Ex8_17.sce
clc// // // //Variable declaration RH=6.85*10^-5; //hall coefficient(m^3/coulomb) e=1.6*10^-19; sigma=250; //conductivity(m-1 ohm-1) //Calculation n=1/(RH*e); //density of charge carriers(m^3) mew=sigma/(n*e); //mobility of charge carriers(m^2/Vs) //Result printf("\n density of charge ...
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ex5_5.sce
// Exa 5.5 format('v',7) clc; clear; close; // Given data Ri = 1;// in k ohm Ro = 80;// in k ohm Aiint = 150; R_S = 5;// in k ohm R_L = 1;// in k ohm //Aiext = i1/i2 = (R_S/(R_S+Ri)) * (Ro/(Ro+R_L))*Aiint; Aiext = (R_S/(R_S+Ri)) * (Ro/(Ro+R_L))*Aiint; disp(Aiext,"The external current gain is");
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ex_1_20_2.sce
//Example 1.20.2: Peak Current and turn off time clc; clear; close; format('v',6) //given data : Vs=200;//in volts R1=10;// in ohm R2=R1; Vc=200;//in volts C=10;// in micro-farad I1=Vs/R1; I2=(Vs+Vc)/R2; It1=I1+I2; disp(It1,"Peak Current,It1(A) = ") Tc=(R1*C)/1.44; disp(Tc,"The Circuit Turn Off Time,Tc(micro-sec) = ")
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eg8_3b.sce
noise_var = 4; noise_mean= 0; num = 5; Xbar = 8.5; u = 8; statistic = sqrt(num/noise_var)*(Xbar - u); prob = 2*cdfnor("PQ", -1*statistic , 0,1 ); disp(prob, "P-value is")
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2_4.sce
clc //initialisation of variables a= 4.086 //A //CALCULATIONS d= a*sqrt(2) r= d/4 //RESULTS printf (' radius of silver atom= %.3f A ',r)
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clc clear //DATA GIVEN T1=1990; //Temperature of the heat Source in K T2=850; //Temperature of the heat Sink in K Qs=32.5; //heat supplied in kJ/min P=0.4; //power developed by the engine in kW ETAcarnot=(T1-T2)/T1; //Also ETAth=wo...
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clc vl=415 //assigning values to the parameters r=15 l=0.1 c=0.000000177 f=50 vph=vl/sqrt(3) xl=2*%pi*f*l xc=1/(2*%pi*f*c) zph=r+%i*(xl-xc) [r1,t]=polar(zph) iph=vph/zph il=iph p=sqrt(3)*vl*il*cos(t) q=sqrt(3)*vl*il*sin(t) s=sqrt(3)*vl*il disp("Amperes",polar(iph),"The phase current is") disp...
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int main(void) { int (**p)(int)[3]; /* ошибка: p -- указатель на функцию, возвращающую массив */ }
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//clc(); clear; //To determine the lattice constant n=2; //number of atoms per unit cell M=55.85; //atomic weight in a.m.u N=6.02*10^26; //avagadro number in kg/mol rho=7860; //density in kg/m^3; a=((n*M)/(N*rho))^(1/3); printf("lattice constant in m is "); disp(a); /...
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function GPIO = GPIO(time,single) [path,fname,extension] = fileparts(file_name); hid_dir = path + '.' + fname; if single ==[] then [a1,b1]=unix_g("sudo ~/rasp30/prog_assembly/libs/sh/asm2ihex.sh ./sftreg ~/rasp30/prog_assembly/libs/asm_code/sftreg_GPIO.s43 16384 16384 16384") if b1==1 disp('There is an error in you...
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clc clear // Características do sistema: pi = %pi ro = 0.1 // kg/m g = 9.80 // m/s² l = 0.5 // m comp = 2*l // m L = 10 // m m = comp*ro // Condições iniciais: //E=[-pi/6,pi/4,L/5,0,comp] // alpha, beta, sigma zero, sigma_ponto_zero, lambda_zero //E=[-pi/6,pi/6,L/5,0,comp] //E=[-pi/6,pi/4,L/5,0,comp] //E=[-pi/4,-pi...
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recursionlimit(110); function [y] = integral(x0, x1, y0, N) printf("recursion: %d N: %d\n", recursionlimit("current") - 1, N) if(N == 0) y = y0; else y = integral(x0, x1, y0, N-1); printf("y: %f x0: %f x1: %f y0: %f N: %f\n", y, x0, x1, y0, N) y = y0 + x...
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pathname=get_absolute_file_path('4_3.sce') filename=pathname+filesep()+'4_3_data.sci' exec(filename) //Question 4.3 //New Pressure(in bar) p2=p1*(rc^n) //New temperature(in kelvin) t2=(t1)*(p2/p1)*(1/rc) //Average temperature(in kelvin) //t=(t3+t2)/2 //Heat transfer(in mJ) Q23=CV/AF //Finding t3 using formu...
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function [x,y,typ]=EVENTSCOPE_f(job,arg1,arg2) x=[];y=[];typ=[] select job case 'plot' then standard_draw(arg1) graphics=arg1(2); [orig,sz]=graphics(1:2) thick=xget('thickness');xset('thickness',2) xrect(orig(1)+sz(1)/10,orig(2)+(1-1/10)*sz(2),sz(1)*8/10,sz(2)*8/10) xarrows([orig(1)+sz(1)/5,orig(1)+sz(1)/5; ...
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//Chapter-9 example 16 //============================================================================= clc; clear; //input data //Da = 7*lamda; diameter of antenna //Calculations //HPBW = 70*(lamda/Da) //HPBW = 70*(lamda/(7*lamda)); HPBW = 70/7;//half power beamwidth NNBW = 2*HPBW;//null to null beamwid...
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clear clc loadmatfile('A.mat'); loadmatfile('B.mat'); loadmatfile('H.mat'); loadmatfile('C.mat'); loadmatfile('observer.mat'); m=rank(B); n=size(A); n=n(1,1); qw=rank(H); p=rank(C); function [LME,LMI,OBJ]=DRC(XLIST) [delta2,delta3,deltah3,betah,Thx,Thw,gama6,gama7]= XLIST(:) LME=list(delta2-delta2...
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tab= read("polystyrene",100,101) for i=1:50 clf a=string(double(i*16/100)) plot2d(tab(i,:),rect=[0,285,101,315],style=14) xtitle("Température à "+a+" sec, barre de polystyrène","x","T(K)") sleep(100) end plot2d(tab(50,:),rect=[0,285,101,315],style=14) xtitle("Température à "+string(a)+" sec...
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errcatch(-1,"stop");mode(2);// Exa 14.2 ; all; // Given data Va=2;// Volts Vb=1;// Volts Vc=3; // Volts Ra=3;// k Ohms Rb=3;// k Ohms Rc=3;// k Ohms Rf=1;// k Ohms Rom=270;// Ohms Supply=15;// Volts // Solution disp(" Assuming that the opamp is initially nulled"); // Using equation 14.8 to det...
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//(3.8) A piston–cylinder assembly contains 0.9 kg of air at a temperature of 300K and a pressure of 1 bar. The air is compressed to a state where the temperature is 470K and the pressure is 6 bars. During the compression, there is a heat transfer from the air to the surroundings equal to 20 kJ. Using the ideal gas m...
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clc; warning("off"); printf("\n\n example5.10 - pg171"); // given (from example 5.9) vol = 1 //cm^3 na=2; // moles of a nb=3; // moles of b nc=4; // moles of c mma=2; //molecular weight of a mmb=3; //molecular weight of b mmc=4; //molecular weight of c ma=na*mma; //[g] weight of a mb=nb*mmb; //[g] weight of b m...
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//Example 15.3// pcu=8.93;// g cm^-3 //Density of Copper a=63.55;//amu //atomic mass of copper c=10^6;//cm^3/m^3 //given d=1;//g.atom //given h=0.6023*10^24;//atoms/g.atom //Avogardo's Number p=pcu*c*(d/a)*(h) mprintf("p = %e atoms/m^3",p) a1=104*10^27;//m^-3 //density of free electrons in copper at room temp...
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//ques-1.16 //Calculating degree of Total and Permanent and Temporary Hardness clc V=50;//volume of water sample (in mL) v1=15;//volume of EDTA used before boiling (in mL) v2=5;//volume of EDTA used after boiling (in mL) M=0.01;//Molarity of EDTA total=v1*M*(1000/V)*50*2;//Total hardness (in ppm) perm=v2*M*(100...
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//x[n] = 1+sin(2*%pi/N)n+3cos(2*%pi/N)n+cos[(4*%pi/N)n+%pi/4] clear; close; clc; N = 10; n = 0:0.01:N; Wo = 2*%pi/N; xn =ones(1,length(n))+sin(Wo*n)+3*cos(Wo*n)+cos(2*Wo*n+%pi/4); for k =0:N-2 C(k+1,:) = exp(-sqrt(-1)*Wo*n.*k); a(k+1) = xn*C(k+1,:)'/length(n); if(abs(a(k+1))<=0.1) a(k+1)=0; en...
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function [v]=vertex(lat_a, lon_a, lat_b, lon_b) // bA, bB en Lab bepalen bA = lat_a bB = lat_b Lab = abs(lon_b - lon_a) if(Lab > 180) Lab = 360 - Lab end // hoek GrKgrc bepalen tanK = sind(Lab) / (cosd(bA)*tand(bB)-sind(bA)*cosd(Lab)) K = atand(tanK) if(K < 0 &&...
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first = [ 0 0 1; 0 3 1; 4 3 1; 2 2 2; 7 1 2; 8 0 2]; second = [ 0 0 3; 0 3 3; 4 3 3; 2 2 4; 7 1 4; 8 0 4]; [out1 ou2t] =estimateAffine3D(first,second,3,2); //output-> // !--error 999 //confidence value should be between 0 and 1.
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Hamming Distance.sce
clear M=[0 0 0 0;0 0 0 1;0 0 1 0;0 0 1 1;0 1 0 0;0 1 0 1;0 1 1 0;0 1 1 1;1 0 0 0;1 0 0 1;1 0 1 0;1 0 1 1;1 1 0 0;1 1 0 1;1 1 1 0;1 1 1 1] P=[0 1 1;1 0 1;1 1 0;1 1 1] I=[1 0 0 0;0 1 0 0;0 0 1 0;0 0 0 1] G=[I P] for i=1:16 c(i,:)=[0 0 0 0 0 0 0] c(i,:)=M(i,:)*G c(i,:)=modulo(c(i,:),2) end for i=1:16...
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//Caption:Simple (Single) Exponential Smoothing Method //Example10.7 //Page385 clear; clc; F1 = 600;//forecast of a product for the first week of march D1 = 550; //Demand of the first week D2 = 620; //Demand of the second week alpha = 0.2; //smoothing constant F2 = F1+alpha*(D1-F1); disp(F2,'Forecast for the ...
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//Chapter 16_Bipolar and MOS Digital Gate Circuits //Caption : AC Power //Example16.14: Calculate the ac power dissipated by a CMOS inverter which drives a 20pF load.Given f=1MHz and Vdd=10V. //Solution: clear; clc; Ct=20*10^-12;// load capacitor in Farad Vdd=10;//drain voltage supply in Volt f=1*10^6;//frequen...
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//Chapter 18, Problem 11 clc; //initializing the variables: R = 15; //resistance in ohms V = 110; //supply voltage taw = 2; //time constant t1 = 3; // in secs i2 =5; // in amperes //calculation: L = taw*R t = 0:0.1:10 I = V/R ...
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function [D,G,mu]=musolve(M,KK,T,params) // musolve - Structured Singular value problem //%Syntax // [mu [,D [,G]] ]=musolve(M,KK,T [,params]) // // M - n by n matrix for which the upper bound of SSV is to be computed. // K - m by 1 vector contains the block structure. K(i), i=1:m, is the // size...
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main var a22a, b33b; { let a22a <- 0; let b33b <- 4; if a22a < b33b then let a22a <- a22a - b33b - b33b; if a22a < b33b then let a22a <- a22a - b33b - b33b fi fi; call outputnum(a22a); call outputnewline(); call outputnum(b33b); call outputnewline() }.
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function [R, U, kr, e] = rlevinson(a, efinal) //rlevinson function computes the autocorrelation coefficients using prediction polynomial. // Calling Sequence // a = rlevinson(a, efinal) // [a, U] = rlevinson(a, efinal) // [a, U, kr] = rlevinson(a, efinal) // [a, U, kr, e] = rlevinson(a, efinal) // Paramete...
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//EXAMPLE 3-9 PG NO-178-179 Z1=10+%i*0; Z2=13+%i*7.5; Z3=-13+%i&7.5; Z4=8.66-%i*5; X=[Z1+Z2 Z3;Z3 Z2+Z4]; Z5=-104+%i*180.13; Z6=280+%i*0; Y=[Z5 Z3;Z6 Z2+Z4]; I1=det(Y/X); disp('i) Current (I1) is = '+string (I1) +' A ');
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// sum 3-20 clc; clear; H=20; D=5; d=3; rho=21; sigd=rho*H; p=2; A=D*H; P=p*A; M=P*H/2; Z=%pi*(D^4-d^4)/(32*D); sigb=M/Z; sigmax=420+sigb; sigmin=420-sigb; // printing data in scilab o/p window printf("sigmax is %0.2f kN/m^2 ",sigmax); printf("\n sigmin is %0.2f kN/m^2 ",sigmin);
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clc; m=input('Enter The Number Of Rows:') n=input('Enter The Number Of Columns:') A=zeros(m,n) B=zeros(m,n) C=zeros(m,n) disp('Enter the elements of first matrix:') for i=1:m for j=1:n A(i,j)=input('') end end disp('Enter the elements of second matrix:') for i=1:m for j=1:n B(i,j)=input('') end end for i=1:m for j=1:n ...
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function converge = verificarSeidel(B) [l c] = size(B); C = abs(B); Beta = ones(l,1); converge=%T for i=1:l if (sum(C(i,:)) > 1) converge = %F break end Beta(i) = C(i,:)*Beta; end BetaMax = max(Beta); if(BetaMax>1) then converge = %...
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// *********************************************************** // * ScicosLab Pack Installer * // * Requires: Scicoslab v4.4.1 * // * Built/Tested On: XP, Windows 7 * // * Description: Convert .COS to .COSF * ...
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//Caption:Find out the speed for 6 pole machine //Exam:2.7 clc; clear; close; F=60;//flux per pole(in m Wb) F_1=F*10^-3;//flux per pole(in Wb) Z=480;//Number of armature conductors P=6;//Number of poles A=2;//Number of parallel paths(Armature wave wound) E_g=320;//generated emf (in V) N=E_g*60*A/(F_1*Z*P);/...
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function M=%s_4_hm(s,M) // Copyright INRIA //s>=M if size(s,'*')<> 1 then error('>= : arguments have incompatible dimensions') end M('entries')=s>=M('entries')
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//Variable declaration k=0.0002 //device parameter Vt=4 //thevinin voltage(V) Vdd=24 //drain voltage(V) Id0=3 //drain current(mA) //Calculations Vgs=(sqrt(Id0/k))+4 //as Id=k(Vgs-Vt)^2 Rd=-(Vgs-Vdd)/Id0 //as Vds=Vdd-I...
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function datos = euler(fxy, x1, y1, h, n) datos = [] for i=1:n+1 datos(i, 1) = i-1 datos(i, 2) = x1 datos(i, 3) = y1 f2xy = fxy(x1, y1) datos(i, 4) = f2xy datos(i, 5) = h*f2xy datos(i...
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//Chapter-2, Example 2.40, Page 2.75 //============================================================================= clc clear //INPUT DATA VLP=11000;//Primary line voltage in V VLS=440;//Secondary line voltage in V Vphp=11000;//Primary phase voltage in V Vphs=(440/sqrt(3));//Secondary phase voltage in V ILP...
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syms s t V q L C R L1 L2 X1 X2 q1 R1 R2 q2 X //F-V anolagy F=V; x=q; M=L; K=1/C; B=R; V=L1*q1*s^2 + R1*s*q1 + R2*s*(X1-X2) //0=L2*s^2*q2 + (1/C)*q2 + R2*s*(q2-q1) //REPLACING I/s=Q disp("V=L1*s*I1 + R1*I1 + R2*(I1-I2)") //LOOP 1 disp("0=L2*s*I2 + 1/(s*C) + R2*(I2-I1)") //LOOP 2 //...
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W=10;//load in N/m L=20;//L in m EI=70000;//flexural rigidity,in N.m^2
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@relation vehicle @attribute COMPACTNESS integer[73,119] @attribute CIRCULARITY integer[33,59] @attribute DISTANCECIRCULARITY integer[40,112] @attribute RADIUSRATIO integer[104,333] @attribute PRAXISASPECTRATIO integer[47,138] @attribute MAXLENGTHASPECTRATIO integer[2,55] @attribute SCATTERRATIO integer[112,265] @attr...
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<div class="navbar"> <img src="/assets/img/logo_small.png" alt="z5217759 // QM" class="logo"> <nav> [[e= navItem in navList ]] <a href="[[navItem.slug]]" class="navItem [[?= navItem.active ]] navActive [[?==]] ">[[navItem.title]]</a...
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.run file=../PulsarLogOn.sql CALL FLCoxPH('tblCoxPHdeep','ObsID', 'VarID', 'Num_Val', 15, NULL,AnalysisID);
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clc //Variable Initialization Vm=320 //Input Voltage In Volts Eb=100 //Back Emf In Volts Ra=5 //Armature Resistance In Ohm af=45 // Firing Angle Of SCR In Degree N=1200 //Speed Of Motor In RPM //Solution Va=Vm* (1/%pi)* (1+cosd(af)) //Voltage Across Armature In volts Ia=(Va-Eb)/Ra //Armature Current Amp W=(...
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//chapter 8 Ex 1 clc; clear; close; //let age of younger person be x //by the given condition: equation is 3(x-6)=x+16-6; mycoeff=[-28 2]; p=poly(mycoeff,"x","coeff"); x=roots(p); mprintf("Their present ages are %d years and %d years",x,x+16);
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//Example 7.24: weight of the locomotive abd number of axles clc; clear; close; //given data : W=12*30;//tonnes we=1.04*360;//tonnes r=5*9.81;// G=1;//in percentage mu=0.2;// alpha=0.8;// x=13.882;// y=0.041;// wlo=(x)/(mu-y);//in tonnes al=20;//allowable load in tonnes na=wlo/al;// disp(wlo,"weight of the locomotive i...
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a=imread('milkeway.jpg'); b=double(a); [m,n]=size(b); for i=1:m for j=1:n c(i,j)=10*log(1+b(i,j)); end end imshow(c);
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clc Nd=10**16 //cm^-3 Na=0 ni=1.8*10**6 //cm^-3 T=300 //K k=8.617*10^-5 //eV/K E=10 //V/cm^2 e=1.6*10**-19 un=8500 n0=((Nd-Na)/2)+sqrt((((Nd-Na)/2)^2)+ni^2) disp(n0,"n0 in cm^-3 is") p0=(ni^2)/n0 disp(p0,"p0 in cm^-3") //Jdrf=e*(un*n0+up*p0)*E= e*un*Nd*E Jdrf=e*un*Nd*E disp(Jdrf,"Jdrf in A/cm^2 i...
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function [] = rockpaperscissors(n) clc; clear all; R = 1; P = 2; S = 3; n = 1; while n == 1 pause(0.5); pause on o = 1; while o == 1 player_move = upper(input('Make your move [R/P/S]: ','s')); disp(' '); pause(0.5); pause on if player_move == 'R' disp('Player chose Rock'); o=0...
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//8.17 clc; R1=50*10^3; R=10*10^3; Vs1=4.5; Vs2=5; Vo=R1/R*(Vs2-Vs1); printf("Output voltage=%.1f V", Vo)
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# 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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di = pwd() + "/Documents" num_files = 15 num_cycles = 100 file_names = ["data_traditional_AverageConnectionSetUpTime.txt", "data_traditional_AverageUtilization.txt", "data_traditional_AveragePlaybackDelayTime.txt", "data_gcp2p_AverageRTT.txt", "data_traditional_AverageReject.txt"] conn_arr = ["data_traditional_Con...
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// Example 7.5, page no-163 clear clc rho=1.54*10^-8//ohm-m E=100//V/m n=5.8*10^28//m^-3 e=1.6*10^-19//C mu=1/(rho*n*e) vd=mu*E printf("\nMobility of electron in silvetr is %.4f*10^-3 m^2/v-s\n\nThe drift velocity of the electron in silver is %.5f m/s ",mu*10^3,vd)
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border.left: 6 border.top: 6 border.right: 6 border.bottom: 7 source: input.png
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//Example 2.5://torque clc; clear; close; d=2.5;//diameter in cm n=500;//turns b=1.1;//mWb/m^2 v=100;//volts pf=0.7;//power factor rp=2000;//ohms x=((%pi*(d*10^-2)^2*n*b*10^-3*v*pf)/(4*rp));// ang1=45;//degree ang2=90;//degree td1=x*sind(ang1);// disp(td1,"torque in Nm when angle is 45 degree") td2=x*sind(ang2);// disp...
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// s: 2014.09.07 function Objname() global OBJFIGNO OBJJOIN if OBJJOIN==0 then OBJFIGNO=OBJFIGNO+1; Gname="ketfig"+string(OBJFIGNO); Printobjstr("# "+Gname); Printobjstr("g "+Gname); end; endfunction
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clc; A=[0 3/2;-1/2 2]; B=[1;0]; C=[1 -1]; D=[0]; Mc=[B A*B]; if (det(Mc)<>0) then disp("and is controllable",rank(Mc),"Mc has a rank of") else disp("and is uncontrollable",rank(Mc),"Mc has a rank of"); end Mo=[C;C*A]; if (det(Mo)<>0) then disp("and is observable",rank(Mo),"Mo has a rank of") e...
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function [x,f,lagr,info]=linpro(x0,p,c,d,ci,cs,mi,modo,imp) [lhs,rhs]=argn(0) if rhs=8 then imp=0,end, n=maxi(size(p)) [x,f,lagr,info]=quapro(x0,0*ones(n,n),p,c,d,ci,cs,mi,modo,imp)
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//Chapter 7,Example 7.5 Page 226 clc clear E = 500 Z = 350 L = 800 E1 = E*(1-exp(-(2*Z/L)*2)) printf (" E'' = %f kV \n",E1) //Answers may vary due to round off error
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//Example 6.6.6 :minimum instantaneous load current,peak instantaneous current and maximum peak to peak ripple clc; clear; close; format('v',6) v=220;//volts r=10;//in ohms l=15.5;//in mH f=5;//in kHz Eb=20;//in volts d=0.5;// x=exp((-(1-d)*r)/(f*10^3*l*10^-3));// y=(1-x)*(Eb/r);// y1=(1-x)*((v-Eb)/r);// A=[0.94 -0.94*...
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clc // Given that P = 1e9 // power required for enlighten the city in watt e = 30 // percentage efficiency of nuclear reactor E = 3.2e-11 // energy released per fission in J // Sample Problem 25 on page no. 12.40 printf("\n # PROBLEM 25 # \n") printf("Standard formula used \n") printf(" E_total = E*N_0/n \n") E_ = E *...