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// Example No.6.11. // Page No.190. clc;clear; l1 = 6*(1/2); l2 = 6*(1/3); l3 = (6*1/6); disp('For the plane (231) the intercepts are (a/2),(b/3),(c/1)'); disp('Ratio of the intercepts made by (231) plane in simple cubic crystal is as follows :'); disp('l1:l2:l3 = 3:2:6'); //As there are no numerical steps...
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// exa 6.1 Pg 168 clc;clear;close; // Given Data Sut=650;// MPa Syt=380;// MPa F1BYF2 = 2.5;// ratio of tensions Fmax=2.5;// kN da=200;// mm db=400;// mm L=1*1000;//mm Km=1.5;// fatigue factor Kt=1;// shock factor tau_d1=0.30*Syt;// MPa tau_d2=0.18*Sut;// MPa tau_d=min(tau_d1, tau_d2);// MPa (taking minimum value) t...
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// Test # 9 : Valid input test case #1 exec('./allpasslp2bsc.sci',-1); [n,d]=allpasslp2bsc(0.65,[0.34,0.56]); disp(d); disp(n); // //Scilab Output //d=1. -0.0870805-0.5498046i //n=0.5566580 -0.1564345-0.9876883i // //Matlab Output //n= 0.5567 + 0.0000i -0.1564 - 0.9877i //d= 1.0000 +...
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clc; clear; Ey=3*10^4 //electric field in y-axis in N/C Ex=0 //electric field in x-axis in N/C q=1.6*10^-19 //electric charge in C me=9.1*10^-31 //in kg //Calculation //F=q*E Fy=-q*Ey //Force in y direction ay=Fy/me format("e",8) disp(ay,"Acceleration of the electron is =") //The negative sign tel...
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//Eg-7.8 //pg-354 clear clc y = [3.21 3.25 4 3.62 3.76 4.55 5.32 4.39 4.59 5 3.68 3.18 5 0 3.7 3.4 0 2.33]; x(1,1:18) = [0.12 0.12 0.17 0.24 0.1 0.11 0.10 0.10 0.17 0.17 0.15 0.23 0.21 0.37 0.28 0.32 0.28 0.22]; x(2,1:18) = [3.2 2.7 2.7 2.8 2.6 2.0 2.0 2.0 2.2 2.4 2.4 2.2 1.9 2.3 2.4 3.3 3.5 3.0]; x(3,...
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clc; clear; mprintf('MACHINE DESIGN\n Timothy H. Wentzell, P.E.\n Example 4.4 Page no 71') P=20; //[hp] Power transmitted by chain drive n=500; //[rpm] speed d=8; //[in] Pitch diameter of sprocket fos=2; D=1.25; //[in] Diameter of shaft L=12; //[in] Distance b...
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//let resistance and inductance of the coil be R and L respectively V=220//voltage applied to coil f=50//frequency of supply I=60//current indicated by ammeter Z1=V/I //when the frequency is increased to 100 Hz I=40//current indicated by ammeter Z2=V/I //on solving for L L=sqrt((Z2^2-Z1^2)/3)/(100*%pi) R=sq...
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x=[0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100 102 104 106 108 110 112 114 116 118 120 122 124 126 128 130 132 134 136 138 140 142 144 146 148 150 152 154 156 158 160 162 164 166 168 170 172 174 176 178 180]; y=[234...
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// Variable declaration alpha = 0.01 // level of significance n1 = 10 n2 = 17 U = 6 // Calculation z_thr = 2.575 MU = (2*n1*n2)/(n1+n2) + 1 SigmaU = sqrt(((2*n1*n2)*(2*n1*n2-n1-n2)) / (((n1+n2)^2)*(n1+n2-1))) z_prt = (U - MU)/SigmaU // Result printf ( "practical Z: %.2f" ,z_prt) if(z_prt <...
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; ; TEMPLOS 1 - Vocabulary ; ; 11-10-90 ; ======== ;________________________________________________________________________ /VOC ;______________________________________________________________________________ ; ; -Movements ie verbs and nouns < 14 ; -if it is a direccion you can order to a charac...
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f=500*10^6 W=2*%pi*f Rp=50 Rs=10 Q=sqrt(Rp/Rs-1) Xs=Q*Rs Xp=Rp/Q Ls=(Xs-10)/W Cp=1/W/Xp Cs=1/W/(Xs+10) Lp=Xp/W printf("\nQ=%.0f\nXs=%.0f ohm\nXp=%.0f ohm\nLs=%.4e H\nCp=%e F\nCs=%e F\nLp=%e H",Q,Xs,Xp,Ls,Cp,Cs,Lp)
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//============================================================================= // chpter 6 example 16 clc; clear; //input data e0 = 40000; //dielectric strength in volts/m d = 33000; //thickness in kV t = d/e0; //required thickness of insulation in mm //result ...
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function [] = kiks_generate_wall_patch(arena) // Display mode mode(0); // Display warning for floating point exception ieee(1); // ----------------------------------------------------- // (c) 2000-2004 Theodor Storm <theodor@tstorm.se> // http://www.tstorm.se // ----------------------------------------------------...
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// chapter14 // example14.17 // page 312 Rp=300 //ohm Rs=3 // ohm Ro=3d3 // ohm // since output resistance of transistor Ro=Rp+n^2*Rs for maximum power transfer, making n as subject we get n=((Ro-Rp)/Rs)^0.5 printf("turn ratio for maximum power transfer = %d \n",n)
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//chatper 2 //example 2_6 //page 23 clear; clc; capacity = 5*10^6; h = 200; n_overall = 75; density=1000; w = capacity*density*9.81; printf("Weight of water available is W = %.2f N \n", w); e = w*h*n_overall/3600/1000; //e - electrical energy printf("Electrical energy available = W*H*n_overall = %.2f kWh \n", e);
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//Example 3.65:resistance and capacitance clc; clear; close; rab=2000;//ohms f=1;//kHz cab=0.047;//micro-farad zab=(1/((1/rab)+(%i*2*%pi*f*10^3*cab*10^-6)));// rbc=1000;//ohms cbc=0.47;//micro-F zbc=rbc-(%i/(2*%pi*f*10^3*cbc*10^-6));// cda=0.5;//micro-F zda=(-%i/(2*%pi*f*10^3*cda*10^-6));// zx=(zda*zbc)/zab;// rx=real(...
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clear all; clc; disp("From equation 5.11 with Vu=constant,we have (Vu^2/r)dr+(Va)dVa=0 ") disp("Or Vu^2lnr+Va^2/2=C") disp("Or Vu^2*ln(r/rh)=(Vah^2-Va^2)/2") disp("Or Va^2=Vah^2-2*Vu^2*ln(r/rh)") V_ah=70 //let m=Vah^2 m=V_ah^2 printf("\n Vah^2 = %0.0f",m) V_u=500 //let n=Vu^2 n=V_u^2 printf("\n Vu^2= ...
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clc; Q=200; //heat in Btu(British Thermal Unit) m=50; //mass in lb c=0.5; //specific heat capacity inBtu/lb.F delT=Q/(m*c); //calculating change in temperatur using Q=mc(del T) disp(delT,"Change in Temperature in Farenheit = "); //displaying result. disp(25-delT,"Final Temperature in Farenheit = "); //...
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// This file is part of www.nand2tetris.org // and the book "The Elements of Computing Systems" // by Nisan and Schocken, MIT Press. // File name: projects/07/MemoryAccess/PointerTest/PointerTestVME.tst load PointerTest.vm, output-file PointerTest.out, compare-to PointerTest.cmp, output-list RAM[256]%D1.6.1 RAM[3]%D1....
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//chapter-14,Example14_6,pg 510 lam=-150//max. gauge factor si=-9.25*10^-10//resistivity change mu=0.5//max poisson's ratio E=((lam-1-(2*mu))/si)//young's modulus printf("young modulus\n") printf("E=%.2f N/m^2",E)
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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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// Exa 2.30 clc; clear; close; // given : sigma=5 // conductivity of sea water in mho/m f1=25 // frequency in kHz f1=25*10^3 // frequency in Hz omega1=2*%pi*f1 // angular frequency in Hz f2=25 // frequency in MHz f2=25*10^6 // frequency in Hz omega2=2*%pi*f2 // angular frequency in Hz epsilon_r=81 // relati...
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clc sigmayp=340 //MPa tauyp=sigmayp/2 //MPa disp(tauyp,"in MPa is=") a=0.1 //m b=0.15 //m v=0.3 //pi=4*p0 //sigmatheta=(pi*(a^2+b^2)-2*p0*b^2)/(b^2-a^2) //sigmatheta=1.7*pi //sloution a: maxi principal stress theory sigmatheta=1.7 pi=sigmayp/sigmatheta disp(pi,"in MPa is= ") //sloution b: maxi shea...
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# -*- coding: utf-8 -*- # vim: ft=sls {%- set service_name = 'crond' %} {%- if grains.os_family in ['Debian', 'Suse'] %} {%- set service_name = 'cron' %} {%- elif grains.os_family in ['Arch'] %} {%- set service_name = 'cronie' %} {%- endif %} verify_cron.service_available: module_and_function: service.available...
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exec("model_defaults.sce", -1) // PARAM - geometry D=0.905 B0=2.1E-3 BC=1 ALPH=.20944 XV1=8.640 LGTHC=10 AREAC=2 LGTHS=15 AREAS=3 LGTHV=10 AREAV=1.95 SPL=10 LGTHR=27.75 // PARAM - supply P0=10 L=0 CS=0.85 // PARAM - control PRE1=1 PRE2=1 POST1=0 POST2=1 P1=0.1 P2=0 TRISE=1.5E-3 exec("...
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clc; T1=15+273; // Inlet temperature of air at compressor inlet in kelvin rp=6; // Compressor pressure ratio T3=750+273; // Maximum permissible temperature in kelvin T5=T3; // After reheat Cp=1.005; // Specific heat at constant pressure in kJ/kg K r=1.4; // Specific heat ratio c=rp^((r-1)/r); T2=T1*c; // Tem...
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//example 3.5(a)// clc //clears the window// clear //clears already existing variables// disp('when V(i)=0, the transistor is cutoff because the voltage between the gate and the source is below the threshold voltage. Correspondingly the output voltage V(0)=5V(point (N) as in figure)') //answer according to the cu...
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ListE(x,y) * ListO(y,z) |- ListO(x,z)
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//Using thevenin theorem determine current through 2 ohm resistor clc; clear; // Characteristic equation to find Vth // 14i1+12i2 = 2 // 12i1+16i2 = 4 // Resistors in the circuit in order from the 2V side R1=2; R2=12; R3=1; R4=3; // Voltage Sources V1=2; V2=4; Z=[14 12; 12 16]; // Resistance Mat...
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// Prob 1.1 clc; clear; close; format('v',5); // Given data s=36;//no. of slots p=4;//no. of poles ph=3;//no. of phase s1=s/ph;//no. of slots pe phase m=s1/p;//no. of slots per pole per phase alfa=180*p/s;//slot angle in degree Kd=sind(m*alfa/2)/(m*sind(alfa/2));//distribution factor disp(Kd,"Distribution ...
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//Example 2.8 //Z transform of the signal x(n)=(0.5)^n*u(n) clear;clc; xdel(winsid()); //u(n)is unit step input n=2; x=(0.5)^n; m=1; w=1; phi=tand(0); a=1; theta=tand(45); [cxz]=czt(x,m,w,phi,a,theta)
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//Example No. 6_03 //Evaluating Polynomial using Horner's rule //Pg No. clear ; close ; clc ; //Coefficients of x terms in the increasing order of power A = [ 6 ; 1 ; -4 ; 1]; x = 2 [n,c] = size(A) ; p(n) = A(n) disp(p(n), 'p(4) = ') for i = 1:n-1 p(n-i) = p(n-i+1)*x + A(n-i) printf('\n p(%i)= %i...
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//pathname=get_absolute_file_path('12.15.sce') //filename=pathname+filesep()+'12.15-data.sci' //exec(filename) //Duration of trial(in min): t=15 //Bore diameter(in m): d=0.25 //Stroke length(in m): L=0.30 //Brake diameter(in m): bd=1.5 //Net brake load(in N): bl=300 //Speed of engine: N=240 //Steam press...
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load Comparator.hdl, output-file Comparator.out, compare-to Comparator.cmp, output-list in%B1.16.1 eqz%B2.1.2 ltz%B2.1.2; set in %B0000000000000000, eval, output; set in %B0111111111111111, eval, output; set in %B1000000000000000, eval, output; set in %B1111111111111111, eval, output;
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//Chapter 21, Problem 24 clc; v1=220; //primary voltage v2=1760; //secondary voltage R=2; //cable resistance Rl=1.28e3; //load across secondary winding N=v1/v2; //turns ratio ...
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clc clear //Page number 492 //Input data l=1.876*10^-7;//The mean free path of the gas in m v=511;//The average speed of the molecule in m/s //Calculations f=v/l;//The collision frequency in per second //Output printf('The collision frequency is %3.4g per second ',f)
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// problem 6.7 d=0.3 g=9.81 l=400 Q=0.3 f=0.032 a=3.142*d*d/4 V=Q/a Lentrance=(0.5*V*V)/(2*g) Hf=(4*f*l*V*V)/(d*2*g) Lexit=(V*V)/(2*g) Totalloss=Lentrance+Hf+Lexit disp(Totalloss,"diffrenc in elevation in m")
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//Example 17_8 page no:846 clc; R0=100; D=20; N=10^(D/20); R1=R0*(N^2-1)/(2*N); R2=R0*(N+1)/(N-1); disp(R1,"the resistance R1 is (in ohm)"); disp(R2,"the resistance R2 is (in ohm)");
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//MissionA1 function missionA1(imgA1) disp("Résultat de la mission A1 :") //on cherche les pixels qui se raprochant le plus de 255 greaterShadeofgray = max(imgA1); disp("Valeur de gris(via max): " + string(greaterShadeofgray)) //Recherche et affichage de la case concernée numeroCase =find(imgA1==gr...
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// Part (a) h1 = 2758; h2 = 1817; h3 = 192; h4 = 200; Wt = h1-h2; Wp = h4-h3; Q1 = h1-h4; Wnet = Wt-Wp; cpg = 1.1; n1 = Wnet/Wt; WR = Wnet/Wp; Q1_ = 100; PO = n1*Q1_; cp = 1000; wg = (Q1_/(833-450)); EIR = wg*cpg*((833-300)-300*(log(833/300))); n2 = PO/EIR ; disp("Part (a)") disp("%",n1*100,"n1 is") disp("%",n2*100,"n2...
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clc //Initialization of variables cp=0.24 //B/lbm F R=53.3 //ft-lb/lbm F //calculations cv=cp-R/778 //results printf("Specific heat at constant volume = %.3f B/lbm F",cv)
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//Chapter 13 //Example 13_24 //Page 336 clear;clc; va=250; rab=0.02; rbc=0.018; rcd=0.025; rda=0.02; ib=150; ic=300; id=250; //interconnector resistance icr=0.02; I=(rbc*ib+rcd*(ib+ic)+rda*(ib+ic+id))/(rab+rbc+rcd+rda); printf("I = %.2f A \n", I); //from fig(ii) vab=I*rab; vbc=186.75*rbc; vcd=113.25*rcd; vda=363.25...
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// Scilab Code Ex14.5: Magnetisation-Magnetic flux density-relative permeability of diamagnetic material : Page-307 (2010) mu_0 = 4*%pi*1e-07; // Magnetic permeability of free space, tesla metre per ampere H = 1.2e+005; // Field strength to which the diamagnetic material is subjected, ampere per metre chi = -4...
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disp('chapter 5 ex5.5') disp('given') disp('709 op-amp is used to design an noninverting amplifier') disp('voltage gain Av=50') Av=50 disp('voltage gain Av=50=34dB') disp('compensation components are listed for Av=20dB and for Av=40dB') disp('for over compensation use components for Av=20dB') disp('C1=500pF R1=...
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function s2=%p_a_lss(d1,s2) //s=%p_a_lss(d1,s2) <=> s=d1+s2 polynomail mat. + syslin list //! // Copyright INRIA s2(5)=d1+s2(5)
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//Exa 10.6 clc; clear; close; //given data : n=0.7;//refractive index N=400;//in cm^-3 //Formula : n=sqrt(1-81*N/f^2) f=sqrt(81*N/(1-n^2));//in KHz disp(f,"Frequency of wave propagation in KHz : "); //Note : Unit of Answer in the book is MHz. It is written by mistake. It is accurately calculated by scilab in ...
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// Calculation // (A) x1 limit- (1,2) x2 limit- (2,3) function ans= f1(x1,x2) ans = 6*exp(-2*x1-3*x2) endfunction //p = int2d(2,3,f1) //p = dblquad(lambda x1, x2: , 2, 3, lambda x1: 1, lambda x1: 2) a = feval([2,3],[1,2],f1) p1 = a(3) function ans = f2(x1,x2) ans = 6*exp(-2*x1-3*x2) endfunction // (B) ...
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//chapter8,Example8_3,pg 181 n=10 D10=0.5 lam=5000*10^-8 R=(D10^2)/(4*n*lam) D50=sqrt(4*50*R*lam) r50=D50/2 printf("radius of 50th dark ring\n") printf("r50=%.2f cm",r50)
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clc //Intialisation of variables s=1/4 //degree LC= 0.005 // degree n= s/LC x=n-1 //Results printf (' %f',x) printf (' such primary divisions should be taken for the length of the vernier scale and then divided into %2f parts for a direct vernier',n )
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//************************** VDD Out (Macro block) ***************************** if(blk_name.entries(bl)=='vdd_o') then for ss=1:scs_m.objs(bl).model.ipar(1) mputl("#VDD_OUT "+string(bl),fd_w); mputl(".subckt vdd_out in[0]=fb_vddout_net"+string(blk(blk_objs(bl),2))+"_"+string(ss)+" in[1]=net"+strin...
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//Exa 6.17 clc; clear; close; // Given data T1 = 12;// in °C T2 = 92;// in °C T1 = T1 + 273;// in K T2 = T2 + 273;// in K del_T = T2 - T1;// in K m = 20;// in kg C_v = 4.187; s= 1; Q = m * s * del_T;// in cal Q = Q * 4.18;// in J H = 2;// heat given by the heater in kw H = H * 10^3;// in J/sec t = Q/H;...
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//i/p args are x and m and the method is specified x=[1 2 3 4 5 7 4 62]; m=3; X = corrmtx(x,m,'postwindowed'); disp(X); //output //4. 3. 2. 1. // 5. 4. 3. 2. // 7. 5. 4. 3. // 4. 7. 5. 4. // 62. 4. 7. 5. // 0. 62. 4. 7. ...
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//Variable declaration epsilon0=8.85*10**-12; A=650*10**-6; //area(m**2) d=4*10**-3; //seperation(m) Q=2*10**-10; //charge(C) epsilonr=3.5; //Calculation V=Q*d/(epsilon0*epsilonr*A); //voltage(V) //Result printf('voltage is %0.3f V \n',(V))
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//check o/p when the i/p arguments are n:required order of the filter, nbar: no of constant level sidelobes, and sll: max sidelobe level w = taylorwin(6,4,-40); disp(w); ////output //0.3305716 // 1.019979 // 1.6494495 // 1.6494495 // 1.019979 // 0.3305716
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clear; clc; // A Textbook on HEAT TRANSFER by S P SUKHATME // Chapter 2 // Heat Conduction in Solids // Example 2.14 // Page 74 printf("Example 2.14, Page 74 \n\n") L = 0.02 ; // [m] t = 0.002 ; // [m] b = 0.2 ; // [m] theta1 = 200 ; // [C] h = 15 ; // [W/m^2 K] k = 45 ; // [W/m K] Bi = h*(t/2)/k ...
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function r=%spqs(a,b) //r=a.\b [ij,v,mn]=spget(a) if size(v,'*')<>mn(1)*mn(2) then error(27) else r=full(a).\b end
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// Problem no 6.6,Page No.158 clc;clear; close; L=10 //m //span of beam W=10*10**3 //N //Point Load a=6 //m //Distance from left end of beam to point Load b=4 //m ////Distance from right end of beam to point Load E=210*10**9 I=10**-4 //m //M.I of beam //Calculation //slope at left end is given by theta_A=W*b*(L**...
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//Chapter-2,Example 2_13,Page 2-36 clc() //Given Data: lam=6.328*10^-7 //Wavelength of light N=1/6000*10^-2 //N=(a+b) grating element //Calculations: //We know, N*sin(theta)=m*lam theta1=asin(1*lam/N)*180/%pi //angular position in first order maxima,m=1 printf('Angular position in first ...
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function [k]=signal_op(n,str,num) select str case "doa" then k = n + num; case "fold" then k= -n; case "scale" then k = n .* num; end endfunction /* case "delay" then n = n + abs(num); case "advance" t...
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Name=Movie Battles 2 PB reflex training FAR PlayerCharacters=Clicker BotCharacters=target_decay.bot;target_nodecay.bot IsChallenge=true Timelimit=10.0 PlayerProfile=Clicker AddedBots=target_decay.bot PlayerMaxLives=0 BotMaxLives=0 PlayerTeam=1 BotTeams=2 MapName=pblive1far.map MapScale=2.0 BlockProjectilePredictors=fal...
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//Chapter 12,Example 6, page 409 //Determine the efective electrical parameters clear clc rc = 0.0875*(1+0.004*50) // conductor resistance in ohm/km Rc = 0.105*85 // ohm w = 2*%pi*50 Rsh = 23.2*10**-6*85*10**5/(%pi*(3^2-2.5^2)) // Resistance of sheath D = 8 rsh = 1/2*(2.5+3) Xm = w*2*log(D/rsh)*10**-7*85000 Ref = Rc +...
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clc; clear; mprintf('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-13.4 Page No.288\n'); //Pitch diameter Ng=60; Pd=6; Dp=Ng/Pd; mprintf('\n Pitch diameter = %f in.',Dp); //Circular pitch Pc=%pi*Dp/Ng; mprintf('\n Circular pitch = %f in.',Pc); L=Pc; //Lead angle D=2; LA=atan(L/(%pi*D)); LA=LA*180/%pi; m...
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//Example 4.11 clc disp(" C_M = 2 pF") fs=(1/(2*%pi*sqrt(2*0.01*10^-12)))*10^-6 // in MHz format(6) disp(fs,"Now f_s(in MHz) = 1 / 2*pi*sqrt(L*C) =") ceq=(2*0.01*10^-24)/(2.01*10^-12) // in F format(9) disp(ceq," C_eq(in F) = C_M*C / C_M+C =") fp=(1/(2*%pi*sqrt(2*9.95*10^-15)))*10^-6 // in MHz format(...
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//A Textbook of Chemical Engineering Thermodynamics //Chapter 4 //Second Law of Thermodynamics //Example 2 clear; clc; //Given: m = 1; //mass of water(kg) T1 = 300; //temperature of surrounding(K) T2 = 273; //temperature of water(K) Hf = 334.11; //latent heat of fusion of ice(kJ/kg) //To determine ...
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//8.8 clc; Imax=10*10^-6; C=4000*10^-12; Slew_rate=Imax/C; printf("Slew rate=%.0f V/s", Slew_rate)
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//Example 2-30// // -5 in two’s complement form// clc //clears the console// clear //clears all existing variables// x=2^8 //Smallest nine bit number// y=5 z=x-y //subtraction// a=dec2bin(z) //binary conversion of the decimal number// disp(' -5 in two’s complement form is ') disp(a) //answer in binary fo...
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clc //Initialization of variables u = 0.04 // lb-s/ft^2 V = 2 //ft/s h = 1/60 // ft //Calculations T = (u*3*V)/h // results printf("The shearing stress is %.1f lb/ft^2",T)
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clear; clc; printf("\nEx3.1\n"); //page no.-120 //given h=6.63*10^-27;........//planck's constant l=10^-8;..............//length of each side of cube in m m=9.11*10^-20;.......//mass of electron p1=h*3^(0.5)/2*l..........//momentum on ground state in gm*cm/sec E1=(h^(2)*3)/(8*m*l^(2)).........//energy for ...
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Calcul de PGCD.sci
clear disp(' /// Programme de calcul du plus grand diviseur commun ///') ////////////////////////////////////////////////////////////////// // VARIABLES : // Pour les fonctions : a,b,m,M nombres // Pour l'algorithme : // a,b,m,M nombres // // ///////////////////////////////////////////////////////////...
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// Example 32_32 clc;funcprot(0); //Given data P=100;// MW T=[0 4 8 12 16 20 24];// Time in hr L_a=[20 20 80 80 20 20];// Load A in MW L_b=[30 60 60 60 60 10];// Load B in MW CV=45000;// kJ/kg C=10;// Cost in Rs./kg Sc=10;// Sale cost in Rs./kWh // Calculation t=[0 0 4 4 8 8 12 12 16 16 20 20 24 24];// Tim...
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[[-5,-5,-5,-5],[2,3,2,2],[-3,3,0,-3],[-5,3,4,-5]],det=0 [34,2,-15,-33], chain 2 => [60,-22,3,-59] => [90,-58,-69,-59] ?? [480,-250,-267,-605] [[-3,-2,-1,-4],[-1,-2,2,-2],[1,0,-1,2],[-3,1,-4,0]],det=-34 [34,2,-15,-33], chain 2 => [41,-2,-17,-40] => [58,9,-22,-57] ?? [58,-6,-34,-77] [[-3,-2,-1,-4],[-1,-2,2,-2],[1,0,-1,...
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//Example 15.36 //impulse response transfer function clear;clc; xdel(winsid()); s=%s; G=syslin('c',25,s^2+4*s+25); t=0:0.05:5; y=csim('impuls',t,G); plot(t,y) xtitle('Impulse response 25/(s^2+4*s+25)','t sec','Response');
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//Example number 6.4, Page number 117 clc;clear; close; //Variable declaration PE=1/100; //probability E_EF=0.5; //energy difference //Calculation x=log((1/PE)-1); T=E_EF/x; //temperature(K) //Result printf("temperature is %.4f K",T) //answer given in the book is wrong
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//************************************************************************** //output-> shows the error on console //-> means that output is correct and no modification needs to be done //************************************************************************** I1 = imread("left.jpg"); I2 = imread("right.jpg"); //I3...
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//Keyword matrix for the example Fig 2.18 // Move scilab to current file directory [u,t,n] = file() n = strcat(n) file_name = basename(n)+fileext(n) file_name = strcat(file_name) ind=strindex(n,file_name) path = part(n,1:ind-1) chdir(path) exec("Chapter_2.sci",-1) key = "PLAYFAIR EXAMPLE" printf("Keywo...
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//example no. 3.8 // solving the matrix equation with partial pivoting in gauss elimination A=[2 1 1 -2;4 0 2 1;3 2 2 0;1 3 2 -1] b=[-10;8;7;-5] pivotgausselim(A,b)
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//steady state errors from closed loop transfer functions s=%s p=poly([5 1 0],'s','coeff'); q=poly([10 10 60 17 1],'s','coeff'); M=p/q disp(M,"M(s)=") H=2; R=1; b=coeff(p) a=coeff(q) //step input if (a(1,1)==b(1,1)) then printf("for step input Ess=0 \n" ) else Ess=1/H*(1-(b(1,1)*H/a(1,1)))*R ...
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clc disp("Example 15.9") printf("\n") s=%s; printf("Given") disp("L1=0.2H L2=0.1H") disp("M=0.1H R=10ohm") disp("v1=u(t) a unit step function") L1=0.2;L2=0.1 M=0.1;R=10; v1=1; w=100; //Let Input impedance be Z1 and can be calculated as //From the equations in 15.10 disp("Z1(s)=L1*s-((M*s)^2)/(R+L2*s)") ...
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clear; clc; //Example - 7.35 //Page number - 269 printf("Example - 7.35 and Page number - 269\n\n"); //Given T = 100 + 273.15;//[K] // (dT/dP)_sat = (1/27.12) K/mm dT_dP = (1/27.12);//[K/mm] dT_dP = dT_dP*(760/101325);//[K/Pa] // The clapeyron equation is given by // (dP/dT)_sat = (-delta H_LV)/(T*delt...
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clc //Intitalisation of variables clear h= -1.35 //kcal e= 44.3 //kcal n= 2 //CALCULATIONS dH= -n*h E= e-dH //RESULTS printf ('Enthalpy of reaction = %.1f kcal',dH) printf ('\n Energy of activation = %.1f kcal',E)
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//Example 7.12 clc clear function [f] = fun1(x,y) f = 1 / (x+y); endfunction x = 1:0.25:2; y = x; m = length(x); n = length(y); del = %nan*ones(m,n); for j = 1:n for i = 1:m del(i,j) = fun1(x(i),y(j)); end end hx = x(2) - x(1); for i = 1:m I = del(i,1); for j =...
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//Given that lambda = 632.8*10^-9 //in meter i = 0 //in rad dFringes = 6 bFringes = 5 //Sample Problem 36-6 printf("**Sample Problem 36-6**\n") //Assume the difference in thickness is 't' //then the path difference will be 2*L //hence deltaL = bFringes/2*lambda printf("The difference in thickness is equa...
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//find length of weld at top and bottom clc //solution //given //ref fig 10.21 P=200*10^3//n T=75//N/mm^2 s=10//mm //a+b=200//mm //let la=length of weld at top //lb=length of weld at bottom //l=la+lb//total length of belt //P=0.707*s*l*T l=P/(0.707*s*T)//mm b=[(200-10)*10*95+(150*10*5)]/(190*10+1500)//mm ...
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clc syms x; f=%e^(-x) disp('clearly,f>0 for every x in(1,2) and integrate (f,x,0,%inf )=') integrate ('%e^(-y)','y',0,%inf ) disp('required probability=p(1<=x<=2)=integrate(f,x,1,2)=') integrate('%e^(-y)','y',1,2) disp('cumulative probability function f(2)=integrate(f,x,-%inf,2)=') integrate('%e^(-y)','y',0,2)
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héc̣a hémac̣a V;SG;1;PRS héc̣a hénic̣api V;PL;2;PRS héc̣a héc̣api V;PL;3;PRS héc̣a héuŋc̣api V;PL;1;PRS héc̣a héuŋc̣a V;PL;1+INCL;PRS héc̣a héc̣a V;SG;3;PRS héc̣a hénic̣a V;SG;2;PRS ehnake uŋkehnakapi V;PL;1;PRS ehnake ehnakapi V;PL;3;PRS ehnake ewahnake V;SG;1;PRS ehnake ehnake V;SG;3;PRS ehnake eyahnakapi V;PL;2;PRS ...
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//Chapter-4,Example4_15_19,pg 4-36 t1=4*10^-3 //thickness of 1st crystal n1=400*10^3 //frequency of 1st crystal n2=500*10^3 //frequency of 2nd crystal t2=n1*t1/n2 //since frquency is inversly pro...
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clc // Given that N = 6.5e25 // no. of atom per m^3 T = 300 // room temperature in K mu_ = 4 * %pi * 1e-7 // magnetic permittivity of space k = 1.38e-23 // Boltzmann's constant in J/K m = 9.1e-31 // mass of electron in kg e = 1.6e-19 // charge in an electron in C h = 6.62e-34 // Planck constant in J sec // Sample Pro...
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expected_str = "[Handles, AppData] = dummyCallback(Handles, AppData)"; callback_str = ui_prepareAppCallback("dummyCallback"); assert_checktrue(callback_str == expected_str);
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clc h1 = 3465 // Enthalpy at state 1 in kJ/kgK h2s = 3065 //Enthalpy at state 2s in kJ/kgK h3 = 3565 //Enthalpy at state 3 in kJ/kgK h4s = 2300 // Enthalpy at state 4s in kJ/kgK x4s = 0.88 // Steam quality at state 4s h5 = 191.83// Enthalpy at state 5 in kJ/kgK v = 0.001 // specific volume in m^3/kg P = 150 //...
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//problem 9-2 data //defining blank arrays blank = part(emptystr(),1:1); blank5=[blank,blank,blank,blank,blank]; blank4=[blank,blank,blank,blank]; blank3=[blank,blank,blank]; blank2=[blank,blank]; blank1=[blank]; //number of task n=8; //no of segments k=6;
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clc; v=240; // supply voltage n=1000; // speed of motor i=40; // current drawn from supply rf=0.2; // field resistance ra=0.25; // armture resistance rd=0.3; // diverter resistance // torque is constant for different speeds // when diverter is put in parallel with series resistance then some fraction of armatur...
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ATWM1_Working_Memory_MRI_Salient_Cued_Run1.sce
# ATWM1 MRI Experiment scenario = "ATWM1_Working_Memory_MRI_salient_cued_run1"; scenario_type = fMRI; # Fuer Scanner #scenario_type = fMRI_emulation; # Zum Testen #scenario_type = trials; scan_period = 2000; # TR pulses_per_scan = 1; pulse_code = 1; #pulse_width=6; default_monitor_sounds = fa...
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clc // variable initialization d=2 //distance travelled by weight in m m=50 // mass of weight in kg g=9.8 // gravitaional acceleration in m/sec^2 // calculation of work in non-quasiequilibrium process W=m*g*d;// work in joules // the work done must be transferred as heat Q=W; printf("The heat that ...
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clc //initialisation of variables v=120//rpm r=5//ft lr=2//ft l=8.2//ft p=62.4/32.2 th=25*%pi/180//degrees cl=0.7 cd1=0.04 N=4 H=25//ft vf=15//ft/sec s=2 vr=36//ft/sec//from velocity diagram //CALCULATIONS v1=2*%pi*v*r/60 A=l*lr F=p*A*vr*vr*[(cl*sin(th))-(cd1*cos(th))]/2 hp=F*62.8*N/550 W=62.4*2*%...
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load Fill.hack; echo "Make sure that 'No Animation' is selected. Then, select the keyboard, press any key for some time, and inspect the screen."; repeat { ticktock; }
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clear; clc; printf("\t\t\tExample Number 6.1\n\n\n"); // turbulent heat transfer in a tube // illustration6.1 // solution p = 2*101325;// [Pa] pressure of air Ta = 200+273.15;// [K] temperature of air d = 0.0254;// [m] diameter of tube R = 287;// [] gas constant u = 10;// [m/s] velocity of air dT = 20;...
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// Examle 3.9 Is=3; // Source current Rs=2; // Source resistance Vs=Rs*Is; // Source voltage Rl=4; // Load resistance R=(Rs*Rl)/(Rs+Rl); // Eqviualent resistance Il1=(Is*Rs)/(Rs+Rl); // Load current i...
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//Example number 4.8, Page number 69 clc;clear; close; //Variable declaration e=1.6*10**-19; //charge(c) theta=56; //angle(degree) V=854; //voltage(V) n=1; //order of diffraction m=9.1*10**-31; //mass(kg) h=6.625*10**-34; //plank constant //Calculation theta=theta*%pi/180; //angle(radian) lamda=...