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//EXAMPLE 5-33 PG NO=327 R1=15; //RESISTANCE R2=8; //RESISTANCE R3=12; //RESISTANCE R4=10; R5=5.14;; R6=7.429; R7=32.74; V=60; Ra=(R1*R2)/ (R1+R2+R3); Rb=(R3*R2)/(R1+R2+R3); Rc=(R1*R3)/(R1+R2+R3); TR=R4+R5+((R6*R7)/(R6+R7)); //TOTAL RESISTANCE I=V/TR disp('i) Resistance (Ra) is = '+string (Ra) +' ohms '); disp('i) Resistance (Rb) is = '+string (Rb) +' ohms '); disp('i) Resistance (Rc) is = '+string (Rc) +' ohms '); disp('i) Total Resistance (TR) is = '+string (TR) +' ohms '); disp('i) Current (I) is = '+string (I) +' A ');
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clc // Given That D1 = 3e-3 // diameter of nth dark fringe when liquid is absent between the lens and the plate in m D2 = 2.5e-3 // diameter of nth dark fringe when liquid is introduced between the lens and the plate in m c = 3e8 // velocity of light in vacuum in m/sed // Sample Problem 40 on page no. 1.54 printf("\n # PROBLEM 40 # \n") mu = D1^2 / D2^2// calculation for refractive index v = 3e8 / mu // calculation for velocity of light printf("\n Standard formula used \n mu = D1^2 / D2^2. \n v = 3e8 / mu. \n") printf("\n Refractive index of liquid = %f.\n velocity of light in the liquid = %e m/sec.",mu,v)
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Dataset before being written to: [ 99 99 99 99 99 99 99] [ 99 99 99 99 99 99 99] [ 99 99 99 99 99 99 99] [ 99 99 99 99 99 99 99] Dataset after being written to: [ 0 -1 -2 -3 -4 -5 -6] [ 0 0 0 0 0 0 0] [ 0 1 2 3 4 5 6] [ 0 2 4 6 8 10 12] Dataset after extension: [ 0 -1 -2 -3 -4 -5 -6 99 99 99] [ 0 0 0 0 0 0 0 99 99 99] [ 0 1 2 3 4 5 6 99 99 99] [ 0 2 4 6 8 10 12 99 99 99] [ 99 99 99 99 99 99 99 99 99 99] [ 99 99 99 99 99 99 99 99 99 99]
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clear; path = "C:/xampp2/htdocs/iniciacao/ImageProcessing/"; exec(path+"lib.sci"); foto = "crescimento-tratado.jpg"; logicalImage = toLogicalImage(foto); figure(); ShowImage(logicalImage, 'invertida'); alturas = getHeights(logicalImage); cols = size(logicalImage, 'c'); pos = 1:1:cols; plot(pos, alturas); disp(stdev(alturas));
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//Example 1.11: clc; clear; close; //given data : p=0.15;// in ohm-m mu_e=0.39;// mobility of electron in m^2/V-s e=1.6*10^-19;// in C Na=1/(e*mu_e*p); format('e',9) disp(Na,"The value of donor concentration,Na(m^-3) = ")
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function [ma]=edge_number(g) [lhs,rhs]=argn(0) if rhs<>1 then error(39), end ma=prod(size(g('tail')))
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funcprot(0) test_cases = list() test_cases($+1) = struct('input', struct('input', [0,0,3,0,2,0,6,0,0;9,0,0,3,0,5,0,0,1;0,0,1,8,0,6,4,0,0;0,0,8,1,0,2,9,0,0;7,0,0,0,0,0,0,0,8;0,0,6,7,0,8,2,0,0;0,0,2,6,0,9,5,0,0;8,0,0,2,0,3,0,0,9;0,0,5,0,1,0,3,0,0]), 'output', struct('output', [4,8,3,9,2,1,6,5,7;9,6,7,3,4,5,8,2,1;2,5,1,8,7,6,4,9,3;5,4,8,1,3,2,9,7,6;7,2,9,5,6,4,1,3,8;1,3,6,7,9,8,2,4,5;3,7,2,6,8,9,5,1,4;8,1,4,2,5,3,7,6,9;6,9,5,4,1,7,3,8,2])) function Result = test_case(index) Result = test_cases(index) endfunction function Result = test_case_count() Result = size(test_cases) endfunction function show(index) tc = test_case(index) disp('Inputs') disp('input') disp(tc.input.input) disp('Outputs') disp('output') disp(tc.output.output) endfunction function Result = check(index) tc = test_case(index) [output] = solve(tc.input.input) Result = %t Result = Result & isequal(output, tc.output.output) endfunction function Result = failures() n = test_case_count() failures = [] for index = 1:n if ~check(index) then failures = [ failures, index ] end end Result = failures endfunction function report() [temp, n] = size(failures()) disp( strcat( [ "Number of test cases: ", string(test_case_count()) ] ) ) disp( strcat( [ "Number of failures: ", string(n) ] ) ) disp( strcat( [ "Number of successes: ", string(test_case_count() - n) ] ) ) if n == 0 then disp("SUCCESS") else disp("FAIL") end endfunction
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//Example 11.11.a // impedence clc; clear; close; r1=6;//in ohms r2=3.95;//in ohms R=r1+r2;//in ohms L1=0.21;//IN HENRY L2=0.14;//in henry C1=30;// in micro farads C2=60;//in micro farads V=220;//IN VOLTS F=50;// IN HERTS Xc1=(1/(2*%pi*F*C1*10^-6));// capacitive reactance in ohms Xc2=(1/(2*%pi*F*C2*10^-6));// capacitive reactance in ohms Xc=Xc1+Xc2;//IN OHMS Xl1=2*%pi*F*L1;// inductive reactance in ohms Xl2=2*%pi*F*L2;// inductive reactance in ohms Xl=Xl1+Xl2;//in ohms Z=sqrt(R^2+(Xl-Xc)^2);// impedence in ohms disp(round(Z),"impedence in ohms is")
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//ANALOG AND DIGITAL COMMUNICATION //BY Dr.SANJAY SHARMA //CHAPTER 2 //AMPLITUDE MODULATION clear all; clc; printf("EXAMPLE 2.18(PAGENO 110)"); //given from the figure P_maxpp = 2*50//maximum peak to peak power in watts P_minpp = 2*15//minimum peak to peak power in watts //calculations m_a = (P_maxpp - P_minpp)/(P_maxpp + P_minpp)//modultaion index M = m_a*100//percentage modulation index //results printf("\n\ni.Modulation index =%.4f",m_a); printf("\n\nii.Percentage modulation index = %.2f percent",M)
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SplItTER X {} FIltEr y { cC:EB:Ed:bF:AA:cD > t OR c ( 24, +8, ) DC:ea:ae:5b:F8:cb < adA::B:BD:cB:bcDA:d or noT 4e220 << Yhnk or -36e5 > Q oR BItOR () } FILTeR y {NOT G Or not DOv or noT nUG } WOk BrAnCh gA -> T GrouPer ozcb {agGREGAte max(lvg) As Mj } ungrouPER Dx { } GRoupfILTeR q {} merGER N { EXPorT CcN }
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//force exerted per unt// pathname=get_absolute_file_path('4.06.sce') filename=pathname+filesep()+'4.06-data.sci' exec(filename) //X-component of reaction force per unit width of the gate(in N/m): Rxw=(d*(V2^2*D2-V1^2*D1))-(d*g/2*(D1^2-D2^2)) //Horizontal force exerted per unt width on the gate(in N/m): Kxw=-Rxw printf("\n\nRESULTS\n\n") printf("\n\nHorizontal force exerted per unt width on the gate: %.3f kN/m\n\n",Kxw/1000)
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// ErturkMe - Copyright 2011 - 2022 // http://erturk.me // ierturk@ieee.org // See license.txt tbx_builder_help_lang(["en_US"], .. get_absolute_file_path("builder_help.sce"));
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//Line commuted Converters// //Example 5.8// E2=415;//input voltage in volts// Emax=sqrt(2)*E2;//maximum value of dc voltage// A=%pi/6;//triggering angle in degrees// Edc=Emax*cos(A)/%pi;//dc output voltage in volts// printf('dc output voltage=Edc=%fvolts',Edc);
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//Ex 1.40.9 clc;clear;close; format('v',9); //Given : ni=10^16;//per m^3 ND=10^22;//per m^3 n=ND;//per m^3//ND>>ni disp(n,"Electron concentration(per m^3) : "); p=ni^2/n;//per m^3 disp(p,"Electron concentration(per m^3) : ");
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//scilab 5.4.1 clear; clc; printf("\t\t\tProblem Number 4.9\n\n\n"); // Chapter 4 : The Second Law Of Thermodynamics // Problem 4.9 (page no. 158) // Solution //For reversible isothermal process, //In problem 4.8, q=843.7; //Heat //Unit:Btu //at 200 psia t=381.86; //(unit:fahrenheit) //converting temperatures to absolute temperatures; T=t+460; //Unit:R" deltaS=(q/T); //Change in entropy //Btu/lbm printf("Change in entropy is %f Btu/lbm*R\n",deltaS); //1 LBm of saturated water //In problem 4.9 t1=381.86; //(unit:fahrenheit) //Source temperature t2=50; //(unit:fahrenheit) //Sink temperature //converting temperatures to absolute temperatures; T1=t1+460; //Source temperature //Unit:R T2=t2+460; //Sink temperature //Unit:R qin=q;//heat added to the cycle n=(1-(T2/T1))*100; //Efficiency printf("Efficiency is %f percentage\n",n); wbyJ=qin*n*0.01;//work output printf("Work output is %f Btu/lbm\n",wbyJ); Qr=qin-wbyJ; //heat rejected printf("Heat rejected is %f Btu/lbm\n\n",Qr); printf("As an alternative solution and refering to figure 4.12,\n") qin=T1*deltaS; //heat added //btu/lbm Qr=T2*deltaS; //Heat rejected //btu/lbm printf("Heat rejected is %f Btu/lbm\n",Qr); wbyJ=qin-Qr; //Work output //Btu/lbm printf("Work output is %f Btu/lbm\n",wbyJ); n=(wbyJ/qin)*100; //Efficiency printf("Efficiency is %f percentage\n",n);
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// Scilab Code Ex3.31:: Page-3.49 (2009) clc; clear; theta1 = 18; // Direction at which first spectral line appears, degrees theta2 = 18+5/(60*60); // Direction at which second spectral line appears, degrees d_theta = (theta2-theta1)*%pi/180; // Angular separation of two spectral lines, radians d_lambda = 50e-010; // Linear separation of two spectral lines just seen as separate, cm DP = d_theta/d_lambda; // Dispersive power of grating n = 1; // Order of diffraction // As dispersive power of grating d_theta/d_lambda = DP = n/((a_plus_b)*cosd(theta1)), solving for a_plus_b a_plus_b = n/(DP*cosd(theta1)); // Grating element, cm // But a_plus_b*sind(theta1)=n*lambda1, solving for lambda1 lambda1 = a_plus_b*sind(theta1)/n; // Wavelength of first spectral line, cm lambda2 = lambda1+d_lambda/1e-002; // Wavelength of second spectral line, cm // As resolving power of grating, lambda/d_lambda = n*N, solving for N N = lambda1/(d_lambda*n); // No. of lines required per cm on grating w = N*a_plus_b; // Minimum grating width required to resolve two wavelengths, cm printf("\nThe wavelength of first spectral line = %4.0f angstrom", lambda1/1e-008); printf("\nThe wavelength of second spectral line = %4.0f angstrom", lambda2/1e-008); printf("\nThe minimum grating width required to resolve two wavelengths = %3.1f cm", w); // Result // The wavelength of first spectral line = 6702 angstrom // The wavelength of second spectral line = 6752 angstrom // The minimum grating width required to resolve two wavelengths = 2.9 cm
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<?xml version="1.0" encoding="utf-8"?> <test> <description>Kovasznay Flow P=8</description> <executable>IncNavierStokesSolver</executable> <parameters>KovaFlow_m8.xml</parameters> <files> <file description="Session File">KovaFlow_m8.xml</file> <file description="Session File">KovaFlow_m8.rst</file> </files> <metrics> <metric type="L2" id="1"> <value variable="u" tolerance="1e-12">4.70499e-05</value> <value variable="v" tolerance="1e-12">0.000157969</value> <value variable="p" tolerance="1e-12">0.00158632</value> </metric> <metric type="Linf" id="2"> <value variable="u" tolerance="1e-12">6.85934e-05</value> <value variable="v" tolerance="1e-12">0.000191491</value> <value variable="p" tolerance="1e-12">0.00500792</value> </metric> </metrics> </test>
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clc; //e.g 22.9 gm=2.5*10**-3; rd=500*10**3; RD=10*10**3; rL=(RD*rd)/(rd+RD); disp('10^3 ohm',rL*10**-3,"rL="); AV=-gm*rL; disp(AV);
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// Scilab code Exa13.4 : : Page-600(2011) clc; clear; a_v = 14.0; // Volume binding energy constant, mega electron volts a_s = 13.0; // Surface binding energy constant, mega electron volts a_c = 0.583; // Coulomb constant, mega electron volts a_a = 19.3; // Asymmetric constant, mega electron volts a_p = 33.5; // Pairing energy constant, mega electron volts Z = 92; // Atomic number // For U-236 A = 235; // Mass number E_exc_236 = a_v*(A+1-A)-a_s*((A+1)^(2/3)-A^(2/3))-a_c*(Z^2/(A+1)^(1/3)-Z^2/A^(1/3))-a_a*((A+1-2*Z)^2/(A+1)-(A-2*Z)^2/A)+a_p*(A+1)^(-3/4); // Excitation energy for uranium 236, mega electron volts // For U-239 A = 238; // Mass number E_exc_239 = a_v*(A+1-A)-a_s*((A+1)^(2/3)-A^(2/3))-a_c*(Z^2/(A+1)^(1/3)-Z^2/A^(1/3))-a_a*((A+1-2*Z)^2/(A+1)-(A-2*Z)^2/A)+a_p*((A+1)^(-3/4)-A^(-3/4)); // Excitation energy for uranium 239 // Now calculate the rate of spontaneous fissioning for U-235 N_0 = 6.02214e+23; // Avogadro's constant, per mole M = 235; // Mass number t_half = 3e+17*3.15e+7; // Half life, years lambda = 0.693/t_half; // Decay constant, per year N = N_0/M; // Mass of uranium 235, Kg dN_dt = N*lambda*3600; // Rate of spontaneous fissioning of uranium 235, per hour printf("\nThe excitation energy for uranium 236 = %3.1f MeV\nThe excitation energy for uranium 239 = %3.1f MeV\nThe rate of spontaneous fissioning of uranium 235 = %4.2f per hour", E_exc_236, E_exc_239, dN_dt); // Result // The excitation energy for uranium 236 = 6.8 MeV // The excitation energy for uranium 239 = 5.9 MeV // The rate of spontaneous fissioning of uranium 235 = 0.68 per hour
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clc // Given that d = 4e-4 // separation between slits in meter b = 2e-4 // slit-width in meter fringe_width = 2.5e-3 // fringe width in meter D = 1.6 // distance between screen and slits // Sample Problem 24 on page no. 2.47 printf("\n # PROBLEM 24 # \n") lambda = (fringe_width * d) / D // calculation for wavelength of light r = (b + d) / b // calculation for ratio of n with m m1 = 1 n1 = r * m1 // calculation for missing order m2 = 2 n2 = r * m2 // calculation for missing order m3 = 3 n3 = r * m3 // calculation for missing order printf("\n Standard formula used \n lambda = (fringe_width * d) / D. \n r = (b + d) / b. \n n = r * m. \n") printf("\n Wavelength of light = %e meter. \n Missing order = %d,%d,%d....",lambda,n1,n2,n3)
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i = imread('test2.jpg'); corners = detectMinEigenFeatures(i,'ROI',2); disp(corners);
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//caption:check_for_contrallability_of_system //example 9.10.20 //page 410 A=[0 1 0;0 0 1;0 -2 -3] B=[0 1;0 0;1 1] P=cont_mat(A,B); disp(P,"Controllability Matrix="); S=[P(1) P(4) P(7);P(2) P(5) P(8);P(3) P(6) P(9)];//collecting columns from P to form a square matrix (3*3) d=det(S); if d==0 printf("matrix is singular, so system is uncontrollable"); else printf("system is controllable"); end;
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//determine the fukux density F=0.5e-3;//webers A=4*10^-4;//meter^2 B=F/A; disp('flux density is = '+string(B)+' Wb/m^2');
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//Chapter 8 //Example 8_20 //Page 191 clear;clc; l=275; d=1.96; us=8060; sf=2; ice_t=1.27; w=0.865; pr=2; wcc=0.91; wp=3.9; t=us/sf; vol=%pi*ice_t*(d+ice_t)*100; wi=wcc*vol/1000; ww=wp*(d+2*ice_t)*100/1000; wt=sqrt((w+wi)^2+ww^2); sag=wt*l^2/8/t; printf("Working tension = %.0f kg \n\n", t); printf("Volume of ice per metre length of the conductor = %.0f cm^3 \n\n", vol); printf("Weight of ice per metre length of conductor is %.2f kg \n\n", wi); printf("Wind force/m length of conductor is %.3f kg \n\n", ww); printf("Total weight of conductor per metre length of conductor is %.3f kg \n\n", wt); printf("Sag = %.2f m \n\n", sag);
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Exa_1_15.sce
// Exa_1_15 // Graphical representation of beats A = 1; w = 20; delta = 1; for i = 1: 1001 //making t and x matrix for various points t(i) = 15 * (i-1)/1000; x(i) = 2 * A * cos(delta*t(i)/2) * cos((w + delta/2)*t(i)); end plot(t,x); //plotting xlabel('t'); ylabel('x(t)'); title('Phenomenon of beats');
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Example_6_6.sce
clc; clear; printf("\n Example 6.6\n"); G=15; //Mass flow rate of organic liquid printf("\n Given:\n Mass flow rate of organic liquid = %d kg/s",G) L_ow=2;//Length of the weir printf("\n Length of the weir = %.1f m",L_ow); rho_l=650; printf("\n Density of liquid = %d kg/m^3",rho_l); Q=G/rho_l; //Use is made of the Francis formula (equation 6.43), h_ow=(2/3)*(Q/L_ow)^(2/3); printf("\n\n Calculations:\n Height of liquid flowing over the weir = %.2f mm",h_ow*1e3);
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example_5_6.sce
clear; clc; disp("--------------Example 5.6---------------") n=3; // number of bits per sample N=3; // bit rate = 3 MHz Fc=10; // carrier f requency = 10 MHz L=2^n; // number of levels S=N/n; // baud rate two_df=S; // 2df = 1 MHz B=L*S; // bandwidth printf("\nThe number of levels is %d , the signal rate is %d Mbaud and the bandwidth is %d MHz.",L,S,B);// display result // display the figure clf(); xname("--------------Example 5.6----------------"); xarrows([0 1],[.2 .2],.5); xset("font size",5); xstring(1,.1,"Frequency"); xpoly([.1 .9],[.55 .55]); xpoly([.1 .1],[.57 .53]); xpoly([.9 .9],[.57 .53]); xstring(.4,.6,"Bandwidth = 8 MHz"); x=linspace(.15,.85,8); for i=1:8 xpoly([x(i) x(i)],[.18 .22]); end x=linspace(.1,.9,9); for i=1:9 xpoly([x(i) x(i)],[.2 .3]); end for i=0:7 xarc(.17+(i/10),.31,.03,.03,0,90*64); xarc(.1+(i/10),.31,.03,.03,90*64,91*64); end for i=0:7 xpoly([.11+(i/10) .19+(i/10)],[.31 .31]); end xset("thickness",2); xpoly([.5 .5],[.2 .35]) xset("font size",3); x=linspace(.15,.85,8); for i=1:8 s=6.5+i-1; xstring(x(i),.14,"f"+string(i)); xstring(x(i),.1,string(s)); xstring(x(i),.06,"MHz"); end xstring(.5,.14,"fc"); xstring(.5,.1,"10"); xstring(.5,.06,"MHz");
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ex12_2.sce
// Exa 12.2 format('v',6) clc; clear; close; // Given data // (a) For 1024 number of bits No_of_bits= 1024;// bits Req_add_bits= log(No_of_bits)/log(2); disp(Req_add_bits,"Address bits required for a memory that has 1024 number of bits"); // (b) For 256 number of bits No_of_bits= 256;// bits Req_add_bits= log(No_of_bits)/log(2); disp(Req_add_bits,"Address bits required for a memory that has 256 number of bits"); // (c) For 4098 number of bits No_of_bits= 4096;// bits // 2^12= 4096, 2^13= 8192, where 4096<4098<8192 or 2^12<4098<2^13, hence Req_add_bits= 13; disp(Req_add_bits,"Address bits required for a memory that has 4098 number of bits"); // d) For 16384 number of bits No_of_bits= 16384;// bits Req_add_bits= log(No_of_bits)/log(2); disp(Req_add_bits,"Address bits required for a memory that has 16384 number of bits");
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Ex2_14.sce
//Ex2_14 clc I = 5 L = 5 WL = L*(I^2)/2 disp("I = "+string(I)+"A")//current flow disp("L = "+string(L)+"H")//inductance disp("WL= "+string(WL)+"joules")//energy stored
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example3_14.sce
clc // Given that lambda = 5.89e-7 // wavelength of light in meter mu_e1 = 1.5 // refractive index for extraordinary light in first case mu_o1 = 1.55 // refractive index for ordinary light in first case mu_e2 = 1.57 // refractive index for extraordinary light in second case mu_o2 = 1.55 // refractive index for ordinary light in second case // Sample Problem 14 on page no. 3.28 printf("\n # PROBLEM 14 # \n") t1 = lambda / (4 * (mu_o1 - mu_e1)) t2 = lambda / (4 * (mu_e2 - mu_o2)) // calculation for thickness of plate of quartz printf("\n Standard formula used \n t = lambda / (4 * (mu_o - mu_e)) ") printf("\n Thickness of plate of quartz in first case = %e meter,\n And thickness of plate of quartz in second case = %e meter",t1,t2)
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Ex9_1_iii.sce
clc // given data G=39.0 // temperature gradient in K/km. h2=10.0 // depth in km rhor=2700.0 // kg/m^3 cr=820.0 // in J/kg-K h1=120/G // T1-T0=120 K is given h21=h2-h1 // in km E0byA=(rhor*(1000**3)*G*cr*h21**2)/2 // in J/km^2 Heat content per square km thetao=G*h21/2.0 // in degree K tau=rhor*cr*h21*(1000**3)/(QbyA*rhow*cw) // in seconds tau=tau/(2*60*60*24*365) // in years theta=thetao*exp(-t/tau) // in degree Kelvin Heatinitial=E0byA/(60*60*365*24*tau)/1000000 // intial heat extraction rate in MW /km^2 Heat25=Heatinitial*exp(-t/tau) // heat extraction rate after 25 years in MW /km^2 printf( "Initial Heat extraction rate is %.2f MW/km^2",Heatinitial) printf(" \n Final Heat extraction rate is %.2f MW/km^2",Heat25)
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Ex4_14.sce
//problem 14 pagenumber 4.44 //given n=7;format(6); vmax=25.4;//volt r=1/(2^n-1); disp('Change in voltage = '+string(r*vmax)+' volt');
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Ch05Ex06.sce
// Scilab code Ex5.6: Pg.209 (2008) clc; clear; // Part (1) m = 1e-09; // Mass of microscopic particle, kg L = 1e-06; // Length of box, m h_cross = 1.055e-034; // Reduced plank's constant, J-s e = 1.6e-19; // Charge on electron, C E_bar = h_cross^2/(2*m*L^2); // Minimum Kinetic energy, J E = E_bar/e ; // Minimum kinetic energy, eV printf("\nThe minimum kinetic energy of particle = %4.2e J or %4.2e eV", E_bar, E); // Part (2) v = sqrt((2*E_bar)/m); // Corresponding speed, m/s printf("\nThe speed corresponding to the obtained kinetic energy = %4.2e m/s",v); // Result // The minimum kinetic energy of particle = 5.57e-048 J or 3.48e-029 eV // The speed corresponding to the obtained kinetic energy = 1.05e-019 m/s
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Test_PROC_delete_student_enrollments.tst
PL/SQL Developer Test script 3.0 7 declare begin -- successful test case --student_registration_system.DELETE_STUDENT_ENROLLMENT('b001', 'c0002'); -- prerequisite test case student_registration_system.DELETE_STUDENT_ENROLLMENT('b001', 'c0001'); end; 0 0
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Example4_8.sci
clc(); clear; // to compute the temperatures at different points a=0.02; // thermal diffusivity in ft^2/hr M=4; // the value of 4 is selected for M x=9/12; // thickness of wall in ft delx=1.5/12; delr=delx^2/(a*M); // at time interval the heat transfeered will change the temperature of sink from tb2 to tb2o printf("The time interval is to be of %.3f hr \n",delr); t1o=370; t2o=435; t3o=480; t4o=485; t5o=440; t6o=360; t7o=250; // tempetaures at different positions at wall in degF initially // we know qo=Z*delx*dely*rho*Cp(tb2'-tb2)/delr So on solving equations we get tb2'=(tb1+tb3+ta2+tc2)/4 // using above formula, temperaures at different positions as shown below can be calculated in degF ta=[370 430 470 473 431 352 250]; tb=[370 425 461 462 422 346 250]; tc=[370 420 452 452 413 341 250]; td=[370 415 444 442 404 336 250]; printf(" The temperatures at different positions 0.78 hr after, are as follows \n"); for i=1:7 printf(" The temperature at point %d is %d degF \n",i,td(i)); end
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Ex9_3.sce
//============================================================================= // chapter 9 example 3 clc clear // Variable declaration P = 400; // tensile force in newtons d = 6*10^-3; // diameter of steel rod m // Calculations r =d/2; E_stress = P/((%pi/4)*r*r); //e_stress in N/m^2 // Result mprintf('Engineering stress = %3.2f MPa',E_stress/10^6); //===========================================================================
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Ex14_22.sce
//example 14.22 PG-14.41 clc clear printf("\n a) Given=> \n\n") printf(" ((A+B'').(A''+B))'' = (A+B'')''+(A''+B)'' ....Since (AB)''= A''+B''\n") printf(" DeMorgan''s Therem\n\n") printf(" = A''.B''''+A''''.B'' ....Since A''+B''=(AB)''\n") printf(" DeMorgan''s Therem\n\n") printf(" = A''.B + A.B'' .......Since (A'')''=A\n") printf(" Therefore \n") printf(" ((A+B'').(A''+B))'' = A''.B + A.B''") printf("\n\n b) Given=> \n\n") printf(" (((A.B)'')C)''D)'' = ((AB)'')C)''''+D'' ....Since (AB)''= A''+B''\n") printf(" DeMorgan''s Therem\n\n") printf(" = (AB)''.C + D'' ....Since (A'')''=A\n\n") printf(" = (A''+B'')C + D'' ....Since (AB)''=A''+ B''\n") printf(" DeMorgan''s Therem\n\n") printf(" Therefore \n") printf(" (((A.B)'')C)''D)'' = (A''+B'')C + D'' ")
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quickSort.sce
function y = quickSort(n,x) y = []; if x == [] then y = []; elseif ((size(x,'r') > 1) | (length(n) > 1)) then error ('quickSort(n,x): n deve ser escalar, e x deve ser matriz-linha') else select n case 1 then [lt,gt] = partition(x(1),x) y = [quickSort(1,lt) x(1) quickSort(1,gt)] else if n > length (x) y = quickSort(n-1,x) else ps = quickSort(1,x(1:n)); xs = x((n+1):$); for i = 1:n if i == 1 then [lt,gt] = partition(ps(i),xs); y = [quickSort(n,lt) ps(i)]; elseif i < n then [lt,gt] = partition(ps(i),gt); y = [y quickSort(n,lt) ps(i)]; else [lt,gt] = partition(ps(i),gt); y = [y quickSort(n,lt) ps(i) quickSort(n,gt)] end end end end end endfunction function varargout = partition(x,xs) lt=[]; gt=[]; for i = 1:length(xs) if xs(i)<x lt = [lt xs(i)]; elseif xs(i)>x gt = [gt xs(i)]; end end varargout = list(lt,gt) endfunction function b = isOrd(x) b = %T; for i = 2 : length(x) if x(i)<x(i-1) then b = %F; break end end endfunction
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Example4_12.sce
// Redox Titrations clear; clc; printf("\t Example 4.12\n"); MKMnO4=0.1327;//molarity of KMnO4, M VKMnO4=16.42;//volume of KMnO4, mL nKMnO4=MKMnO4*VKMnO4/1000; nFeSO4=5*nKMnO4; VFeSO4=25;//volume of FeSO4, mL MFeSO4=nFeSO4/VFeSO4*1000; printf("\t the molarity of FeSO4 solution is : %4.4f M\n",MFeSO4); //End
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Ex21_9.sce
clc; VCC=10; R1=30*10**3; R2=20*10**3; RE=1.5*10**3; B1=150; B2=100; VBE=0.7; Ai=B1*B2; disp(Ai); VR2=VCC*(R2/(R1+R2)); VB2=VR2-VBE; VE2=VB2-VBE; IE2=VE2/RE; re2=25/(IE2*10**3); disp('ohm',re2*1,"re2="); Ib2=IE2/B2; IE1=Ib2; re1=25/(IE1*10**3); disp('ohm',re1*1,"re1="); Ri1=(R1*R2)/(R1+R2); disp('Kohm',Ri1*10**-3,"Ri1="); Av=RE/((re1/B2)+(re2+RE)); disp(Av);
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//Exa 3.9 // To calculate probability of exceeding signal beyond the receiver sensitivity. clc; clear all; SSmean=-100; //signal strength(dBm) Sr=-110; //receiver sensitivity(dBm) sd=10; //standard deviation(dB) //solution P_Smin=(0.5-0.5*erf((Sr-SSmean)/(sqrt(2)*sd))); printf('probability of exceeding signal beyond the receiver sensitivity is %.2f \n',P_Smin);
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clc; clear; h=4.136*10^-15 //Plancks constant in eV c=3*10^8 //velocity of light in m/s R=1.097*10^7 //Rydberg constant m^-1 lambda1= 900 //wavelength in nm T1_by_T2=1/3 //Ratio of temperature T1 to T2 n1=2 //energy level of atom n2=3 //energy level of atom //calculation lambda2=(lambda1*T1_by_T2)//wavelength in nm E=(h*c)/(lambda2*10^-9) //Energy of incident photon in eV Ex=R*h*c*((1/n1^2)-(1/n2^2)) //Excitation energy in eV W=E-Ex mprintf("The work function of the metal is = %1.2f eV",W)
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//Ex2_6 clc C = 12* 10^ -6 f = 1.0*10^3 Xc = 1/(2*%pi*f*C) disp("C = "+string(C)+"F")//capacitance disp("at... f = "+string(f)+"Hz")//frequency disp("Xc = 1/(2*pi*f*C) = "+string(1/(2*%pi*f*C))+"ohm")//calculation for capacitive reactance
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//find power of periodic signal x(t)=3cos(5*w0*t) clc; //To plot signal*************************// t=-5:0.01:5; w0=0.25*%pi; x=3*cos(5*w0.*t);//given signal plot(t,x); //***************************************// t=-50:0.01:50; x=3*cos(5*w0.*t); T=length(t); //To calculate Energy xsq=x.^2;//adds squares of all the 'x' values(integrates 'x^2' terms) v=sum(xsq);//energy //To calculate Power xsq=x.^2; P=1/T*v;//divide by 2T,to take the average rate of energy(gives power) disp('watts',P,'Power=');
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/587/CH11/EX11.3/example11_3.sce
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example11_3.sce
clear; clc; //Example11.3[The Condensation of Steam in a Condenser] //Given:- Th_in=30,Th_out=30,Tc_in=14,Tc_out=22;//Inlet and Outlet temperatures of hot and cold liquids [degree Celcius] A=45;//[m^2] U=2100;//[W/m^2.degree Celcius] h_fg=2431;//Heat of vapourisation of water at Th_i[kJ/kg] Cp=4184;//Specific heat of cold water [J/kg] //Solution:- del_T1=Th_in-Tc_out;//[degree Celcius] del_T2=Th_out-Tc_in;//[degree Celcius] del_T_lm=(del_T1-del_T2)/(log(del_T1/del_T2));//[degree Celcius] disp("degree Celcius",del_T_lm,"The logrithmic Mean temperature difference is") Q=U*A*del_T_lm;//[W] disp("W",Q,"The heat transfer rate in the condenser is") mw=Q/(Cp*(Tc_out-Tc_in));//[kg/s] disp("kg/s",mw,"The mass flow rate of the cooling water is") ms=(Q/(1000*h_fg));//[kg/s] disp("kg/s",ms,"The rate of condensation of steam is")
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/866/CH16/EX16.18/16_18.sce
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//CLC x1= 7 //m x2= 7 //m x3= 4 //m x4= 4 //m x5= 4 //m x6= 12 //m x7= 4 //m Fab= 12 //KN Fbc1= 7 //KN Fbc2= 7 //KN Fcd= 22 //KN Fe= 5 //KN //CALCULATIONS MfAB= -Fab*(x1+x2)/(x3+x4) MfBC= -(Fbc1*x3*(x4+x5)^2/(x6)^2)-(Fbc2*x3^2*(x4+x5)/(x6)^2) MfCD= -Fcd*x6/x6 MfDE= -Fe*x7 DFba= (3/(x1+x2))/((3/(x1+x2))+(4/(x3+x4+x5))) DFbc= 1-DFba DFcb= (4/(x3+x4+x5))/((3/(x6))+(4/(x3+x4+x5))) DFcd= 1-DFcb //RESULTS printf("DFba = %.2f",DFba) printf("DFbc = %.2f",DFbc) printf("DFcb = %.2f",DFcb) printf("DFcd = %.2f",DFcd)
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Ex15_12.sce
clc; AoL=150000; av=0.005; AcL=AoL/(1+(av*AoL)); disp('',AcL,"AcL=");//The answers vary due to round off error
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/615/CH8/EX8.5/8_5.sce
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//Fuels and Combustion// //Example 8.5// WBaSO4=0.0482;//weight of BaSO4 in grams// W=0.5248;//weight of sample in grams// PS=32*WBaSO4*100/(233*W);//percentage of sulphur in the sample// printf('percentage of sulphur in the sample=PS=%f',PS);
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Am=5; Ac=10; m=Am/Ac; fc=50; fm=10; t=0:0.001:.5; Sm=Am*cos(2*%pi*fm*t); subplot(2,2,1); plot(t,Sm); Sc=Ac*cos(2*%pi*fc*t); subplot(2,2,2); plot(t,Sc); Sam=(1+(m*cos(2*%pi*fm*t))); Sam1=Ac*Sam; Sam2=cos(2*%pi*fc*t).*Sam1; subplot(2,2,3); plot(t,Sam2);
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// let q be the function of t q=f(t) deff('q=f(t)','q=5*t*sin(4*%pi*t)'); //i is the current at t=0.5seconds in Amperes derivative(f,0.5); i=ans; disp("i=") disp(i) units='Amperes A' i=[string(i) units]; disp(i) // in amperes A // the current i is 31.415 Amperes
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function x = STI(A,b) // Autores: Jorge Zavaleta // Funcion que resuelve el sistema triangular inferior (STI) de ecuaciones // lineales Ax = b (sustitucion hacia adelante) //***************************************************************************** // -> Entrada // A (Matriz Real) - Matriz triangular inferior de tamaño n x n con entradas // reales. // // b (Vector Real) - Vector de tamaño n con entradas reales. // // -> Salida // x (Vector Real) - Vector de tamaño n con entradas reales que representan la // solucion al sistema de ecuaciones. //***************************************************************************** n = size(A,'r'); x = zeros(n,1); for i = 1:n //(*) x(i) = (b(i) - A(i,1:i-1)*x(1:i-1))/A(i,i); end endfunction function x = STS(A,b) // Autores: Jorge Zavaleta // Funcion que resuelve el sistema triangular superior (STS) de ecuaciones // lineales Ax = b (sustitucion hacia atras) //***************************************************************************** // -> Entrada // A (Matriz Real) - Matriz triangular superior de tamaño n x n con entradas // reales. // // b (Vector Real) - Vector de tamaño n con entradas reales. // // -> Salida // x (Vector Real) - Vector de tamaño n con entradas reales que representan la // solucion al sistema de ecuaciones. //***************************************************************************** n = size(A,'r'); x = zeros(n,1); for i = n:-1:1 //(*) x(i) = (b(i) - A(i,i+1:n)*x(i+1:n))/A(i,i); end endfunction function [L,U,signo] = faclupp(A) // Autores: Jorge Zavaleta, Hector E. Gomez Morales // Funcion que realiza la factorizacion LU de la matriz A (A = L*U) usando pivoteo parcial //***************************************************************************** //->Entrada // A (Matriz Real) - Matriz de tamaño n x n con entradas reales. // //->Salida // L (Matriz Real) - Matriz triangular inferior de tamaño n x n. // // U (Matriz Real) - Matriz triangular superior de tamaño n x n. // // signo (Real) - Indica el signo de la determinante de la matriz //***************************************************************************** n=size(A,'r'); signo = 1; for i=1:n-1 [val,r] = max(abs(A(i:n,i))); r = r + i - 1; // Realizamos pivoteo si es necesario, llevamos cuenta del signo if r ~= i signo = signo * -1 A([i r],1:n) = A([r i],1:n); end for k=i+1:n A(k,i)=A(k,i)/A(i,i); A(k,i+1:n)=A(k,i+1:n)-A(k,i)*A(i,i+1:n); end end // Se extrae L y U de A. L = eye(n,n) + tril(A,-1); U = triu(A); endfunction function x = lur(A,b) // Autores: Jorge Zavaleta, Hector E. Gomez Morales // Funcion que resuelve un sistema de ecuaciones usando factorizacion LU // el sistema de ecuaciones lineales Ax = b mediante factorizacion LU, STI y STS //***************************************************************************** //->Entrada // A (Matriz Real) - Matriz de tamaño n x n con entradas reales. // // b (Vector Real) - Vector de tamaño n con entradas reales. // // -> Salida // x (Vector Real) - Vector de tamaño n con entradas reales que representan la // solucion al sistema de ecuaciones. //***************************************************************************** n=size(A,'r'); for i=1:n-1 [val,r] = max(abs(A(i:n,i))); r = r + i - 1; // Realizamos pivoteo si es necesario if r ~= i A([i r],1:n) = A([r i],1:n); b([i r]) = b([r i]); end for k=i+1:n A(k,i)=A(k,i)/A(i,i); A(k,i+1:n)=A(k,i+1:n)-A(k,i)*A(i,i+1:n); end end // Se extrae L y U de A. L = eye(n,n) + tril(A,-1); U = triu(A); // Resolvemos los sistemas usando STI y STS y = STI(L,b) x = STS(U,y) endfunction function d = mdet(A) // Autores: Jorge Zavaleta, Hector E. Gomez Morales // Funcion que realiza el calculo del determinante usando factorizacion LU // y la propiedad de que el determinante de matrices escalonada es el producto // de su diagonal y ademas que det(AB) = det(A)det(B) //***************************************************************************** //->Entrada // A (Matriz Real) - Matriz de tamaño m x n con entradas reales. // //->Salida // d (Real) - Valor del determinante de A cuando se puede calcular o %nan // en el caso contrario. //***************************************************************************** nr = size(A,'r'); if nr ~= size(A,'c'); //Matriz no cuadrada disp('La matriz no es cuadrada'); d = %nan; else select nr case 1 //Escalar d = A; else //Matriz nxn [L,U,signo] = faclupp(A) dl = prod(diag(L)) du = prod(diag(U)) d = dl*du end end endfunction A = [2,4,-2;4,9,-3;-2,-1,7;] B = [4;8;-6] C = [2,4,-2;4,-1,-3;-2,9,7;] disp("Matriz A:") disp(A) disp("Vector b:") disp(B) disp("Solucion sistema Ax = b:") disp(lur(A,B)) disp("Determinante de A:") disp(mdet(A)) disp("Matriz C:") disp(mdet(C))
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function [x,y,typ]=nfet_gldn(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 'getorigin' then [x,y]=standard_origin(arg1) case 'set' then x=arg1; graphics=arg1.graphics; exprs=graphics.exprs model=arg1.model; while %t do [ok,gain,over,exprs]=getvalue('Set NFET block parameters',['Gain';'Do On Overflow(0=Nothing 1=Saturate 2=Error)'],list('mat',[-1, -1],'vec',1),exprs) if ~ok then break,end if ok then graphics.exprs=exprs x.graphics=graphics;x.model=model break end end case 'define' then over=0 gain=2 model=scicos_model() model.sim=list('ota_func',5) model.in=[-1;-1] model.in2=[-2;-3] model.intyp=[1 1] model.out=-1 model.out2=0 model.outtyp=-1 model.evtin=[] model.evtout=[] model.state=[] model.dstate=[] model.rpar=[] model.ipar=[] model.blocktype='c' model.firing=[] model.dep_ut=[%t %f] exprs=[sci2exp(gain);sci2exp(over)]; gr_i= ['text=[''Src'';'' Gate''];';'xstringb(orig(1),orig(2),text,sz(1),sz(2),''fill'');'] x=standard_define([6 3],model,exprs,gr_i) end endfunction
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// Find the value of Rs // Basic Electronics // By Debashis De // First Edition, 2010 // Dorling Kindersley Pvt. Ltd. India // Example 7-43 in page 344 clear; clc; close; // Given data Idss=10*10^-3; // Drain-source current in mA Vp=-5; // Pinch off voltage in V // Calculation Vgs = 5*(sqrt(6.4/10)-1); Rs=-Vgs/(6.4*10^-3); printf("Rs = %0.0f ohms",Rs); // Result // Rs = 156 ohms
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//example 2 //discharge of water from a tank clear clc h0=4 //height of cylindrical water tank in ft h2=2 //final water level in tank in ft g=32.2 //acc. due to gravity in ft/s^2 Dt=3*12 //diameter of tank in inches Djet=0.5 //diameter of water jet in inches t=(h0^0.5-h2^0.5)*(Dt)^2/((Djet)^2*(g/2)^0.5) //time taken for water level to fall to half of its initial value in seconds printf("\n Hence, the time taken for water level to fall to half of its initial value is = %.1f min. \n",t/60);
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clc disp("Example 5.15") printf("\n") disp("For CE amplifier shown in fig 5.5 find R1,R2,Re & Rc") printf("Given\n") Vcc=24 //load resistance RL=120*10^3 //since Rc<<RL Rc=RL/10 //select Ve & Vce Ve=5 Vce=3 Vrc=Vcc-Vce-Ve //from circuit Ic=Vrc/Rc //find Re Re=Ve/Ic R2=10*Re //Vbe for si transistor Vbe=0.7 Vb=Vbe+Ve I2=Vb/R2 R1=(Vcc-Vb)/I2 printf("The resistance values are\nR1=%f ohm\nR2=%f ohm\nRe=%f ohm\nRc=%f ohm\n",R1,R2,Re,Rc)
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function [r] = prog2(A) r = min(A) - max(A); endfunction
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// Scilab Code Ex2.34:: Page-2.26 (2009) clc; clear; t = 3.1e-05; // Thickness of the soap film, cm mu = 1.33; // Refractive index of the soap film r = 0; // Angle of refraction of the light ray on the soap film, degrees // For bright fringe in reflected pattern, // 2*mu*t*cosd(r) = (2*n+1)*lambda/2 lambda = zeros(3); for n = 1:1:3 lambda(n) = 4*mu*t*cosd(r)/(2*(n-1)+1); // Wavelengths for n = 1, 2 and 3 if lambda(n) > 4000e-008 & lambda(n) < 7500e-008 then lambda_reflected = lambda(n); end end printf("\nThe wavelength reflected strongly from the soap film = %5.3e cm", lambda_reflected); // Result // The wavelength reflected strongly from the soap film = 5.497e-05 cm
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function varargout=%hm_size(M,job) // Copyright INRIA // returns dimensions of an hyper matrix [lhs,rhs]=argn(0) dims=M('dims') if rhs==2 then if job=='*' then x1=prod(M('dims')) elseif type(job)==1 then if size(job,'*') >1 then error('Second argument is incorrect'),end if job<=0|job>size('dims') then error('Second argument is incorrect'),end x1=M('dims')(job) else error('Second argument is incorrect') end return end if lhs==1 then varargout(1)=M('dims')' else if lhs>size(M('dims'),'*') then error('Too many LHS args'),end for k=1:lhs varargout(k)=dims(k) end end
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clc; warning off; exec('../Classificacao_gaussiana/geragauss.sce',-1); exec('../Plota_Dados_E_Reta/plotdata.sce',-1); function bits = de2bi(Yd) str = dec2bin(Yd); bits = [] sz = size(str) for i=1:sz(2), if Yd(i) ~= 0 bits(1,i) = 0; bits(2,i) = strtod(part(str(i),1)); bits(3,i) = strtod(part(str(i),2)); else bits(1,i) = 1; bits(2,i) = strtod(part(str(i),1)); bits(3,i) = strtod(part(str(i),2)); end end endfunction function [Treino, Teste] = run() nc = 3; npc = [120 150 140]; mc = [1.2 3.3 6.5; 4.1 1.7 4.5]; varc = [1.89 1.15 1.51; 1.21 2.14 1.32]; [X Yd] = geragauss(nc, npc, mc, varc); [X Yd] = mixvalues(X, Yd); N = size(X); N = N(2); TR = N * 0.8; TE = N * 0.2; ind = grand(1, "prm", (1:TR)); X_TR = []; Yd_TR = []; for i = 1:TR, aux = X(:,ind(i)); X_TR = [X_TR aux]; aux = Yd(ind(i)); Yd_TR = [Yd_TR aux]; end X_TE = []; Yd_TE = []; for i = 1:N, naoEsta = 1; for j = 1:TR, if i == ind(j), naoEsta = 0; end end if naoEsta == 1, aux = X(:,i); X_TE = [X_TE aux]; aux = Yd(i); Yd_TE = [Yd_TE aux]; end end Yd_BIN = de2bi(Yd_TR); disp('Configurando parametros...'); // numero de neuronios por camada, incluindo a entrada N_ = [2, 9 , 3]; // parametros de treinamento: taxa de aprendizado lp = [0.05 , 0]; // numero maximo de epocas para treinar T = 400; disp('Inicializando a rede...'); W = ann_FF_init(N_); disp('Treinando a rede...'); W_out = ann_FF_Std_online(X_TR,Yd_BIN,N_,W,lp,T); //VetorSEQ Sem momentum disp('Calculando SEQ..'); VetorSEQ = []; VetorTESTE = []; W_error = W; for i = 1:T W_error = ann_FF_Std_online(X_TR,Yd_BIN,N_,W_error,lp,1); y = ann_FF_run(X_TR,N_,W_error); VetorTESTE = [VetorTESTE, ann_sum_of_sqr(ann_FF_run(X_TE,N_,W_error), de2bi(Yd_TE))]; VetorSEQ = [VetorSEQ, ann_sum_of_sqr(y, Yd_BIN)]; end aux = round(ann_FF_run(X_TR,N_,W_out)) == Yd_BIN; Treino = length(aux(aux == %t)); aux = round(ann_FF_run(X_TE,N_,W_out)) == de2bi(Yd_TE); Teste = length(aux(aux == %t)); endfunction Teste = []; Treino = []; for i = 1:10, [teste, treino] = run(); Teste = [Teste teste]; Treino = [Treino treino]; end // scf(1); // title("Erros Quadraticos"); // plot(VetorSEQ,'-b'); // plot(VetorTESTE,'-r'); // xlabel('Epoca'); // ylabel('VetorSEQ'); // // Grafico de dados e da reta em duas dimensões para a função 1 // scf(2); // title("Dados Originais"); // plotadc2d(X, Yd); // // Grafico de dados e da reta em duas dimensões para a função 1 // scf(3); // title("Dados Reclassificados"); // plotadc2d(X_TR, ann_FF_run(X_TR,N_,W_out)); // plotadc2d(X_TE, ann_FF_run(X_TE,N_,W_out)); // // Grafico de dados e da reta em duas dimensões para a função 1 // scf(2); // title("Dados Treinamento"); // plotadc2d(X_TR, Yd_TR); // // Grafico de dados e da reta em duas dimensões para a função 1 // scf(3); // title("Dados Teste"); // plotadc2d(X_TE, Yd_TE); // // for i = 1:nc, // // plotareta(Wout(i,:), bout(i,:), [0 5]); // // end
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function A = page_rank_matrice(M, alpha) [Q, d] = matrice_ponderations(M); taille = length(d); P = Q + ones(d')*d./taille; A = alpha.*P + (1/taille).*(1-alpha).*ones(M) endfunction
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//Eg-13.12 //pg-552 clear clc y(1) = 1; h = 1; printf('For h = 1\n') printf(' x y\n') for(i = 1:3) x(i) = i; //since h = 1 y(i+1) = y(i)/(1+2*h); printf(' %f %f\n',x(i),y(i+1)) end h = 0.5; printf('\nFor h = 0.5\n') printf(' x y\n') n = (3.0-0.5)/0.5+1; for(i = 1:n) x(i) = 0.5 + (i-1)*h; //since h = 0.5 y(i+1) = y(i)/(1+2*h); printf(' %f %f\n',x(i),y(i+1)) end printf('Observe that the implicit method is stable for h = 1, whereas the explicit method is not.')
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clc // initialization of variables TL1=-5+273 // lower temperature in kelvin for first situation TH=20+273 // higher temperature in kelvin TL2=-25+273 //lower temperature in kelvin for second situation //solution COP1=TL1/(TH-TL1) // carnot refrigerator COP for first situation // Let Heat be 100 kJ QL=100 // assumption W1=QL/COP1 // work done for situation 1 // for situation 2 COP2=TL2/(TH-TL2) // COP carnot for second situation W2=QL/COP2 // work done Per=(W2-W1)*100/W1 // percentage increase in work done printf(" The perccentage increase in work is %.1f%%",Per)
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clc;funcprot(0);//EXAMPLE 20.14 // Initialisation of Variables v=14;..........//Volume of air delivered in m^3 p1=1;........//Suction pressure in bar p2=7;........//Delivery pressure in bar N=310;........//Compressor rpm n=1.35;........//Compression index k=0.05;........//Clearance ratio rld=1.5;.........//Ratio of cylinder length and diameter //Calculations etav=(1+k)-(k*((p2/p1)^(1/n)));..........//Volumetric efficiency Vs=v/(etav*N);.............//Swept volume in m^3 D=((Vs)/((%pi/4)*rld))^(1/3);......//Compressor diameter in m L=rld*D;......................//Compressor stroke in m disp(D*100,"Compressor diameter in cm:") disp(L*100,"Compressor stroke in cm:")
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@relation unknow @attribute a1 real[13.0,334177.0] @attribute a2 real[0.0,256382.0] @attribute a3 real[3.0,5323.0] @attribute a4 real[1.0,893.0] @attribute a5 real[0.0,675.0] @attribute a6 real[-457916.0,575241.0] @attribute a7 real[-297172.0,238321.0] @attribute a8 real[-205895.0,179851.0] @attribute a9 real[0.0,89958.0] @attribute a10 real[57.0,495561.0] @attribute a11{g,h} @inputs a1,a2,a3,a4,a5,a6,a7,a8,a9,a10 @outputs a11 @data g g g g g g g g g h g g g g g g g g g h g g g h g g g g g g g g g g g g g g g g g g g h g g g g g g g g g g g g g h g g g g g g g g g g g h g g g g g g g h g g g h g g g g g g g g g g g g g g g g g h g g g h g g g g g h g h g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g h g g g g g g g h g g g g g g g g g g g g g g g g g g g g g g g h g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g h g g g g g g g g g g g g g g g g g g g h g g g g g g g g g h g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g h g g g g g h g g g h g g g g g g g h g g g g g g g g g g g g g g g g g g g g g h g h g g g g g h g g g g g h g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g h g g g g g h g g g g g h g g g g g g g g g g g g g g g h g g g g g g g h g g g h g h g g g g g g g g g g g g g g g g g h g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g h h h h h h h g h h h h h g h g h h h g h g h g h g h g h h h h h h h g h h h h h g h g h h h h h h h h h h h h h g h g h g h h h h h g h g h h h h h h h g h h h h h h h h h g h h h g h h h g h h h h h h h h h h h h h h h h h g h h h g h g h h h g h g h h h g h g h h h h h g h g h h h h h g h g h g h h h h h h h g h g h h h g h g h h h g h g h h h h h g h h h h h h h h h g h g h g h h h h h h h h h h h h h h h h h h h h h h h g h h h h h h h g h h h h h h h g h h h g h h h h h h h h h h h h h h h h h g h h h g h g h h h h h h
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//clc() wf = 01.5; for i = 1:41 for j = 1:41 T(i,j) = 0; if j == 1 then T(i,j) = 0;//C else if j == 41 then T(i,j) = 100;//C end end if i == 1 then T(i,j) = 75;//C else if i == 41 then T(i,j) = 50;//C end end end end e = 100; while e>1 for i=1:41 for j = 1:41 if i>1 & j>1 & i<41 & j<41 then Tn(i,j) = (T(i + 1,j) + T(i-1,j) + T(i,j+1) + T(i,j-1))/4; Tn(i,j) = wf * Tn(i,j) + (1-wf)*T(i,j); if i==2 & j==2 then e = abs((Tn(i,j) - T(i,j)) * 100/ (Tn(i,j))); end T(i,j) = Tn(i,j); end end end end disp(T,"for error < 1, the temperatures are")
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// 08.08.22 // 09.10.25 // 09.10.26 // 09.10.27 // 09.10.29 function Out=MakeveLfaceL(VfL) // Out format // VeL Edge, Face num(as numlist), VeL num // FL Face (Vertexs) Eps=10^(-4); Tmp=VfL($); Tmp1=Tmp(1); if type(Tmp1)==1 FvL=list(VfL); else FvL=VfL; end; EL=list(); FL=list(); for Nn=1:length(FvL) Tmp=FvL(Nn); VL=Tmp(1); if length(VL)>0 FnL=Tmp(2); FaceL=list(); for I=1:length(FnL) Tmp1=FnL(I); PtL=list(); for J=1:length(Tmp1) Tmp2=Tmp1(J); PtL(J)=VL(Tmp2); end; FaceL(I)=PtL; end; else FaceL=list(Tmp(2)); end; for I=1:length(FaceL) Face=FaceL(I); Face($+1)=Face(1); FL($+1)=Face; for J=1:length(Face)-1 Edge=list(Face(J),Face(J+1)); Flg=0; for K=1:length(EL) Tmp=EL(K); Tmp1=Tmp(1); Tmp2=norm(Edge(1)-Tmp1(1))+norm(Edge(2)-Tmp1(2)); Tmp3=norm(Edge(1)-Tmp1(2))+norm(Edge(2)-Tmp1(1)); if Tmp2<Eps | Tmp3<Eps Tmp=EL(K); Tmp1=Tmp(1); Tmp2=[Tmp(2),length(FL)]; EL(K)=list(Tmp1,Tmp2,K); Flg=1; break; end; end; if Flg==0 Ntmp=length(EL); EL(Ntmp+1)=list(Edge,[length(FL)],Ntmp+1); end; end; end; end; Out=list(EL,FL); endfunction
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clc;clear; a=imread('gtw.jpg'); d=double(a); me=d(:,:,1); hi=d(:,:,2); bi=d(:,:,3); bme=me; bhi=hi; bbi=bi; m=(1/9)*(ones(3,3)); [r1,c1]=size(a); for i=2:r1-1 for j=2:c1-1 a1me=[me(i-1,j-1) me(i-1,j) me(i-1,j+1) me(i,j-1) me(i,j) me(i,j+1) me(i+1,j-1) me(i+1,j) me(i+1,j+1)]; a1hi=[hi(i-1,j-1) hi(i-1,j) hi(i-1,j+1) hi(i,j-1) hi(i,j) hi(i,j+1) hi(i+1,j-1) hi(i+1,j) hi(i+1,j+1)]; a1bi=[bi(i-1,j-1) bi(i-1,j) bi(i-1,j+1) bi(i,j-1) bi(i,j) bi(i,j+1) bi(i+1,j-1) bi(i+1,j) bi(i+1,j+1)]; a2me=gsort(a1me); a2hi=gsort(a1hi); a2bi=gsort(a1bi); medme=a2me(1); medhi=a2hi(1); medbi=a2bi(1); bme(i,j)=medme; bhi(i,j)=medhi; bbi(i,j)=medbi; end end imgrgb=cat(3,bme,bhi,bbi); figure(); subplot(121);imshow(a); title('Original Image','fontsize',8); subplot(122);imshow(uint8(imgrgb)); title('Min Filtered Image','fontsize',8);
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clear //Given e=8.854*10**-12 //C**2/N/m**2 A=10**-4 //m**2 E=3*10**6 //V/ms //Calculation Id=e*A*E //Result printf("\n Displacement current is %0.1f *10**-9 A",Id*10**9)
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//Find how much loss will occur in 300 hours //Ex:20.3 clc; clear; close; x1=0.1;//in mm t1=25;//in hours t2=300;//in hours x2=x1*sqrt(t2/t1);//in mm disp(x2,"Oxidation loss in 300 hours (in mm) = ");
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clear;lines(0); g=make_graph('foo',1,3,[1,2,3,1],[2,3,1,3]); is_connex(g) g=make_graph('foo',1,4,[1,2,3,1],[2,3,1,3]); is_connex(g)
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clc //Given that c = 3e8 // speed of light in m/s v = c/20 // Speed of proton in m/s m = 1.67e-27 // Mass of proton in Kg h = 6.625e-34 // Plank constant printf("Example 1.5") lambda = h/(m*v) // calculation of de Broglie wavelength printf("\n de Broglie wavelength of proton is %e m.\n\n\n",lambda) // Answer in book is 6.645e-14m which is a calculation mistake
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write(%io(2),a)
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// SAMPLE PROBLEM 3/4 clc;clear;funcprot(0); // Given data m=10;// The mass in kg v=2;// The speed in m/s R=8;// N // Calculation k=R/v^2;// N.s^2/m^2 // SigmaF_x=ma_x; v_0=v;// m/s v=v_0/2;// m/s t=((1/v)-(1/2));// The time in s t_0=0;// s t_1=2.5;// s x=integrate('10/(5+(2*t))','t',t_0,t_1); printf("\nThe corresponding travel distance,x=%1.2f m",x);
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// Example 3.1 // To compute the number of channels available per cell for a)four-cell reuse system a)seven-cell reuse system a)12-cell reuse system // Page No.61 clc; clear; // Given data B=33*10^6; // Total bandwidth allocated to particular FDD system in Hz Bc=25*10^3; // Bandwidth per channel in Hz Nc=2; // Number of simplex channels Bc=Bc*Nc; // Channel bandwidth in Hz Ntotal=B/Bc; // Total number of channels //a) To compute the number of channels available per cell for four-cell reuse system N=4; // frequency reuse factor chpercell=Ntotal/N; // number of channels available per cell for four-cell reuse system // Displaying the result in command window printf('\n The number of channels available per cell for 4-cell reuse system = %0.0f channels',chpercell); printf('\n One control channel and 160 voice channels would be assigned to each cell.'); // b) To compute the number of channels available per cell for seven-cell reuse system N=7; // frequency reuse factor chpercell=ceil(Ntotal/N); // number of channels available per cell for seven-cell reuse system // Answer is varrying due to round-off error // Displaying the result in command window printf('\n \n The number of channels available per cell for 7-cell reuse system = %0.0f channels',chpercell); printf('\n Each cell would have one control channel, four cells would have 90 voice channels and three cells would have 91 voice channels.'); // c) To compute the number of channels available per cell for 12-cell reuse system N=12; // frequency reuse factor chpercell=Ntotal/N; // number of channels available per cell for seven-cell reuse system // Displaying the result in command window printf('\n \n The number of channels available per cell for 12-cell reuse system = %0.0f channels',chpercell); printf('\n Each cell would have one control channel, eight cells would have 53 voice channels and four cells would have 54 voice channels.');
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clear; clc; X1=6.6*%i; X2=6.3*%i; X0=12.6*%i; r=37.5e6; v=33e3; e=1; zb=v^2/r; x1=X1/zb; x2=X2/zb; x0=X0/zb; x1g=.18*%i; x2g=.12*%i; x0g=.1*%i; x1=x1+x1g; x2=x2+x2g; x0=x0+x0g; ia=3*e/(x1+x2+x0); ia1=ia/3; a=1*%e^(%i*(120*%pi/180)); b=a^2; ibase=r/(sqrt(3)*v); ian=ia*ibase; printf("fault current=%djAmp",imag(ian)); va=e-(ia1*x1g); vb=-ia1*x2g; vc=-ia1*x0g; va0=(va+vb+vc); va1=(va+(b*vb)+(a*vc)); va2=(va+(a*vb)+(b*vc)); v=v/sqrt(3); va0=va0*v; va1=va1*v; va2=va2*v; va0r=real(va0); va0i=imag(va0); va0m=sqrt((va0r^2)+(va0i^2)); va0a=atand(va0i/va0r); va1r=real(va1); va1i=imag(va1); va1m=sqrt((va1r^2)+(va1i^2)); va1a=atand(va1i/va1r)-120; va2r=real(va2); va2i=imag(va2); va2m=sqrt((va2r^2)+(va2i^2)); va2a=atand(va2i/va2r)+120; mprintf("\nthe voltage levels are \n va=%f+j%f V \tor\t %d/_%d kV",va0r/1e3,va0i/1e3,va0m/1e3,va0a); mprintf("\n vb=%f+j(%f) kV \tor\t %d/_%d kV",va1r/1e3,va1i/1e3,va1m/1e3,va1a); mprintf("\n vc=%f+j(%f) kV \tor\t %d/_%d kV",va2r/1e3,va2i/1e3,va2m/1e3,va2a);
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= Plugins Tests 1 student-plugins/robleshs+jamespur/zodiac/robleshs+jamespur_zodiac_test.py
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clear ; clc; printf("\t Example 10.12\n"); T1=291; //temp.of sky,K T2=308; //temp of air,K e1=0.9; //emissivity 0f black paint h=8; //heat transfer coefficient,W/(m^2*K) P=600 ; //heat flux,W/m^2 //heat loss from the roof to the inside of the barn will lower the roof temp., since we dont have enough information to evaluate the loss, we can make an upper bound on roof temp. by assuming that no heat is transferred to the interior. x=poly([0],'x'); x=roots(8*(e1*5.67*10^-8*(x^4-T1^4)+(x-T2)-e1*P)); //for white acrylic paint, by using table, e=0.9 and absorptivity is 0.26,Troof T=poly([0],'T'); T=roots(8*(e1*5.67*10^-8*(T^4-T1^4)+(T-T2)-0.26*P)); Tn=T(2)+0.6 printf("\t temp. of the root is :%.1f C or 312 K ,the white painted roof is only a few degrees warmer than the air.\n",Tn); //end
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//clear// clear; clc; //Example 10.4 //Given k = 0.075; //[Btu/ft-h-F] rho = 56.2; //[lb/ft^3] Cp = 0.40; //[Btu/lb-F] s = 0.5/12; //[ft.] Ts = 250; //[F] Ta = 70; //[F] Tb_bar = 210; //[F] //(a) Temp_diff_ratio = (Ts-Tb_bar)/(Ts-Ta); alpha = k/(rho*Cp); // From Fig.10.6 N_Fo =0.52; tT = N_Fo*s^2/alpha //[h] //(b) //Substituting in Eq.(10.23) QTbyA = s*rho*Cp*(Tb_bar-Ta) //[Btu/ft^2]
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global G; global Kc; global C; global LpAlpha; global Tc; global Kd; global Kp; global Ki; Kd = 2; Ki = 20; Kp = 10 Tc = 0.001; LpAlpha = 0.001; s = poly(0, 's'); w_n = 28.400017; xi_n = 0.2825585; function[xi_min] = getMinimumXI(desiredOvershoot) xi_min = sqrt(log(desiredOvershoot)^2/(%pi^2 + log(desiredOvershoot)^2)); endfunction function[omega_min] = getMinimumOmega(alpha, desiredSettleTime, xi_min) N = 1/sqrt(1-xi_min^2); omega_min = (log(alpha/100) - log(N))/(-1*xi_min*desiredSettleTime); endfunction // Plant G = 1/(s^2/w_n^2 + 2*(xi_n/w_n)*s +1); //G = 1/(s*(s+4)*(s+6)); Kc = 60; z0 = -10; z1 = -30; //z2 = -25; p0 = -270; //p1 = -50; C =((s-z0)*(s-z1))/((s)*(s-p0)); xi_t = getMinimumXI(0.2); w_t = getMinimumOmega(5,0.3,xi_n); clf(); evans(C*G); sgrid(xi_t,w_t); //kc_lim = kpure(syslin('c',C*G));
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Synchroniczny.sci
clc; clear; lines(0); function [M] = unipolarna(W, ac) RozmiarKolumn = size(W, 'c'); if RozmiarKolumn == 3 then //dla macierzy SynchUni X=[ 0 0 0; 0 0 1; 0 1 0; 0 1 1; 1 0 0; 1 0 1; 1 1 0; 1 1 1]; end //dla macierzy Vjeden Vcztery i VL if RozmiarKolumn == 6 then X=[0 0 0 0 0 0; 0 0 0 0 0 1; 0 0 0 0 1 0; 0 0 0 0 1 1; 0 0 0 1 0 0; 0 0 0 1 0 1; 0 0 0 1 1 0; 0 0 0 1 1 1; 0 0 1 0 0 0; 0 0 1 0 0 1; 0 0 1 0 1 0; 0 0 1 0 1 1; 0 0 1 1 0 0; 0 0 1 1 0 1; 0 0 1 1 1 0; 0 0 1 1 1 1; 0 1 0 0 0 0; 0 1 0 0 0 1; 0 1 0 0 1 0; 0 1 0 0 1 1; 0 1 0 1 0 0; 0 1 0 1 0 1; 0 1 0 1 1 0; 0 1 0 1 1 1; 0 1 1 0 0 0; 0 1 1 0 0 1; 0 1 1 0 1 0; 0 1 1 0 1 1; 0 1 1 1 0 0; 0 1 1 1 0 1; 0 1 1 1 1 0; 0 1 1 1 1 1; 1 0 0 0 0 0; 1 0 0 0 0 1; 1 0 0 0 1 0; 1 0 0 0 1 1; 1 0 0 1 0 0; 1 0 0 1 0 1; 1 0 0 1 1 0; 1 0 0 1 1 1; 1 0 1 0 0 0; 1 0 1 0 0 1; 1 0 1 0 1 0; 1 0 1 0 1 1; 1 0 1 1 0 0; 1 0 1 1 0 1; 1 0 1 1 1 0; 1 0 1 1 1 1; 1 1 0 0 0 0; 1 1 0 0 0 1; 1 1 0 0 1 0; 1 1 0 0 1 1; 1 1 0 1 0 0; 1 1 0 1 0 1; 1 1 0 1 1 0; 1 1 0 1 1 1; 1 1 1 0 0 0; 1 1 1 0 0 1; 1 1 1 0 1 0; 1 1 1 0 1 1; 1 1 1 1 0 0; 1 1 1 1 0 1; 1 1 1 1 1 0; 1 1 1 1 1 1]; end //dla Cj if RozmiarKolumn == 9 then X=[0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 1; 0 0 0 0 0 0 0 1 0; 0 0 0 0 0 0 0 1 1; 0 0 0 0 0 0 1 0 0; 0 0 0 0 0 0 1 0 1; 0 0 0 0 0 0 1 1 0; 0 0 0 0 0 0 1 1 1; 0 0 0 0 0 1 0 0 0; 0 0 0 0 0 1 0 0 1; 0 0 0 0 0 1 0 1 0; 0 0 0 0 0 1 0 1 1; 0 0 0 0 0 1 1 0 0; 0 0 0 0 0 1 1 0 1; 0 0 0 0 0 1 1 1 0; 0 0 0 0 0 1 1 1 1; 0 0 0 0 1 0 0 0 0; 0 0 0 0 1 0 0 0 1; 0 0 0 0 1 0 0 1 0; 0 0 0 0 1 0 0 1 1; 0 0 0 0 1 0 1 0 0; 0 0 0 0 1 0 1 0 1; 0 0 0 0 1 0 1 1 0; 0 0 0 0 1 0 1 1 1; 0 0 0 0 1 1 0 0 0; 0 0 0 0 1 1 0 0 1; 0 0 0 0 1 1 0 1 0; 0 0 0 0 1 1 0 1 1; 0 0 0 0 1 1 1 0 0; 0 0 0 0 1 1 1 0 1; 0 0 0 0 1 1 1 1 0; 0 0 0 0 1 1 1 1 1; 0 0 0 1 0 0 0 0 0; 0 0 0 1 0 0 0 0 1; 0 0 0 1 0 0 0 1 0; 0 0 0 1 0 0 0 1 1; 0 0 0 1 0 0 1 0 0; 0 0 0 1 0 0 1 0 1; 0 0 0 1 0 0 1 1 0; 0 0 0 1 0 0 1 1 1; 0 0 0 1 0 1 0 0 0; 0 0 0 1 0 1 0 0 1; 0 0 0 1 0 1 0 1 0; 0 0 0 1 0 1 0 1 1; 0 0 0 1 0 1 1 0 0; 0 0 0 1 0 1 1 0 1; 0 0 0 1 0 1 1 1 0; 0 0 0 1 0 1 1 1 1; 0 0 0 1 1 0 0 0 0; 0 0 0 1 1 0 0 0 1; 0 0 0 1 1 0 0 1 0; 0 0 0 1 1 0 0 1 1; 0 0 0 1 1 0 1 0 0; 0 0 0 1 1 0 1 0 1; 0 0 0 1 1 0 1 1 0; 0 0 0 1 1 0 1 1 1; 0 0 0 1 1 1 0 0 0; 0 0 0 1 1 1 0 0 1; 0 0 0 1 1 1 0 1 0; 0 0 0 1 1 1 0 1 1; 0 0 0 1 1 1 1 0 0; 0 0 0 1 1 1 1 0 1; 0 0 0 1 1 1 1 1 0; 0 0 0 1 1 1 1 1 1; 0 0 1 0 0 0 0 0 0; 0 0 1 0 0 0 0 0 1; 0 0 1 0 0 0 0 1 0; 0 0 1 0 0 0 0 1 1; 0 0 1 0 0 0 1 0 0; 0 0 1 0 0 0 1 0 1; 0 0 1 0 0 0 1 1 0; 0 0 1 0 0 0 1 1 1; 0 0 1 0 0 1 0 0 0; 0 0 1 0 0 1 0 0 1; 0 0 1 0 0 1 0 1 0; 0 0 1 0 0 1 0 1 1; 0 0 1 0 0 1 1 0 0; 0 0 1 0 0 1 1 0 1; 0 0 1 0 0 1 1 1 0; 0 0 1 0 0 1 1 1 1; 0 0 1 0 1 0 0 0 0; 0 0 1 0 1 0 0 0 1; 0 0 1 0 1 0 0 1 0; 0 0 1 0 1 0 0 1 1; 0 0 1 0 1 0 1 0 0; 0 0 1 0 1 0 1 0 1; 0 0 1 0 1 0 1 1 0; 0 0 1 0 1 0 1 1 1; 0 0 1 0 1 1 0 0 0; 0 0 1 0 1 1 0 0 1; 0 0 1 0 1 1 0 1 0; 0 0 1 0 1 1 0 1 1; 0 0 1 0 1 1 1 0 0; 0 0 1 0 1 1 1 0 1; 0 0 1 0 1 1 1 1 0; 0 0 1 0 1 1 1 1 1; 0 0 1 1 0 0 0 0 0; 0 0 1 1 0 0 0 0 1; 0 0 1 1 0 0 0 1 0; 0 0 1 1 0 0 0 1 1; 0 0 1 1 0 0 1 0 0; 0 0 1 1 0 0 1 0 1; 0 0 1 1 0 0 1 1 0; 0 0 1 1 0 0 1 1 1; 0 0 1 1 0 1 0 0 0; 0 0 1 1 0 1 0 0 1; 0 0 1 1 0 1 0 1 0; 0 0 1 1 0 1 0 1 1; 0 0 1 1 0 1 1 0 0; 0 0 1 1 0 1 1 0 1; 0 0 1 1 0 1 1 1 0; 0 0 1 1 0 1 1 1 1; 0 0 1 1 1 0 0 0 0; 0 0 1 1 1 0 0 0 1; 0 0 1 1 1 0 0 1 0; 0 0 1 1 1 0 0 1 1; 0 0 1 1 1 0 1 0 0; 0 0 1 1 1 0 1 0 1; 0 0 1 1 1 0 1 1 0; 0 0 1 1 1 0 1 1 1; 0 0 1 1 1 1 0 0 0; 0 0 1 1 1 1 0 0 1; 0 0 1 1 1 1 0 1 0; 0 0 1 1 1 1 0 1 1; 0 0 1 1 1 1 1 0 0; 0 0 1 1 1 1 1 0 1; 0 0 1 1 1 1 1 1 0; 0 0 1 1 1 1 1 1 1; 0 1 0 0 0 0 0 0 0; 0 1 0 0 0 0 0 0 1; 0 1 0 0 0 0 0 1 0; 0 1 0 0 0 0 0 1 1; 0 1 0 0 0 0 1 0 0; 0 1 0 0 0 0 1 0 1; 0 1 0 0 0 0 1 1 0; 0 1 0 0 0 0 1 1 1; 0 1 0 0 0 1 0 0 0; 0 1 0 0 0 1 0 0 1; 0 1 0 0 0 1 0 1 0; 0 1 0 0 0 1 0 1 1; 0 1 0 0 0 1 1 0 0; 0 1 0 0 0 1 1 0 1; 0 1 0 0 0 1 1 1 0; 0 1 0 0 0 1 1 1 1; 0 1 0 0 1 0 0 0 0; 0 1 0 0 1 0 0 0 1; 0 1 0 0 1 0 0 1 0; 0 1 0 0 1 0 0 1 1; 0 1 0 0 1 0 1 0 0; 0 1 0 0 1 0 1 0 1; 0 1 0 0 1 0 1 1 0; 0 1 0 0 1 0 1 1 1; 0 1 0 0 1 1 0 0 0; 0 1 0 0 1 1 0 0 1; 0 1 0 0 1 1 0 1 0; 0 1 0 0 1 1 0 1 1; 0 1 0 0 1 1 1 0 0; 0 1 0 0 1 1 1 0 1; 0 1 0 0 1 1 1 1 0; 0 1 0 0 1 1 1 1 1; 0 1 0 1 0 0 0 0 0; 0 1 0 1 0 0 0 0 1; 0 1 0 1 0 0 0 1 0; 0 1 0 1 0 0 0 1 1; 0 1 0 1 0 0 1 0 0; 0 1 0 1 0 0 1 0 1; 0 1 0 1 0 0 1 1 0; 0 1 0 1 0 0 1 1 1; 0 1 0 1 0 1 0 0 0; 0 1 0 1 0 1 0 0 1; 0 1 0 1 0 1 0 1 0; 0 1 0 1 0 1 0 1 1; 0 1 0 1 0 1 1 0 0; 0 1 0 1 0 1 1 0 1; 0 1 0 1 0 1 1 1 0; 0 1 0 1 0 1 1 1 1; 0 1 0 1 1 0 0 0 0; 0 1 0 1 1 0 0 0 1; 0 1 0 1 1 0 0 1 0; 0 1 0 1 1 0 0 1 1; 0 1 0 1 1 0 1 0 0; 0 1 0 1 1 0 1 0 1; 0 1 0 1 1 0 1 1 0; 0 1 0 1 1 0 1 1 1; 0 1 0 1 1 1 0 0 0; 0 1 0 1 1 1 0 0 1; 0 1 0 1 1 1 0 1 0; 0 1 0 1 1 1 0 1 1; 0 1 0 1 1 1 1 0 0; 0 1 0 1 1 1 1 0 1; 0 1 0 1 1 1 1 1 0; 0 1 0 1 1 1 1 1 1; 0 1 1 0 0 0 0 0 0; 0 1 1 0 0 0 0 0 1; 0 1 1 0 0 0 0 1 0; 0 1 1 0 0 0 0 1 1; 0 1 1 0 0 0 1 0 0; 0 1 1 0 0 0 1 0 1; 0 1 1 0 0 0 1 1 0; 0 1 1 0 0 0 1 1 1; 0 1 1 0 0 1 0 0 0; 0 1 1 0 0 1 0 0 1; 0 1 1 0 0 1 0 1 0; 0 1 1 0 0 1 0 1 1; 0 1 1 0 0 1 1 0 0; 0 1 1 0 0 1 1 0 1; 0 1 1 0 0 1 1 1 0; 0 1 1 0 0 1 1 1 1; 0 1 1 0 1 0 0 0 0; 0 1 1 0 1 0 0 0 1; 0 1 1 0 1 0 0 1 0; 0 1 1 0 1 0 0 1 1; 0 1 1 0 1 0 1 0 0; 0 1 1 0 1 0 1 0 1; 0 1 1 0 1 0 1 1 0; 0 1 1 0 1 0 1 1 1; 0 1 1 0 1 1 0 0 0; 0 1 1 0 1 1 0 0 1; 0 1 1 0 1 1 0 1 0; 0 1 1 0 1 1 0 1 1; 0 1 1 0 1 1 1 0 0; 0 1 1 0 1 1 1 0 1; 0 1 1 0 1 1 1 1 0; 0 1 1 0 1 1 1 1 1; 0 1 1 1 0 0 0 0 0; 0 1 1 1 0 0 0 0 1; 0 1 1 1 0 0 0 1 0; 0 1 1 1 0 0 0 1 1; 0 1 1 1 0 0 1 0 0; 0 1 1 1 0 0 1 0 1; 0 1 1 1 0 0 1 1 0; 0 1 1 1 0 0 1 1 1; 0 1 1 1 0 1 0 0 0; 0 1 1 1 0 1 0 0 1; 0 1 1 1 0 1 0 1 0; 0 1 1 1 0 1 0 1 1; 0 1 1 1 0 1 1 0 0; 0 1 1 1 0 1 1 0 1; 0 1 1 1 0 1 1 1 0; 0 1 1 1 0 1 1 1 1; 0 1 1 1 1 0 0 0 0; 0 1 1 1 1 0 0 0 1; 0 1 1 1 1 0 0 1 0; 0 1 1 1 1 0 0 1 1; 0 1 1 1 1 0 1 0 0; 0 1 1 1 1 0 1 0 1; 0 1 1 1 1 0 1 1 0; 0 1 1 1 1 0 1 1 1; 0 1 1 1 1 1 0 0 0; 0 1 1 1 1 1 0 0 1; 0 1 1 1 1 1 0 1 0; 0 1 1 1 1 1 0 1 1; 0 1 1 1 1 1 1 0 0; 0 1 1 1 1 1 1 0 1; 0 1 1 1 1 1 1 1 0; 0 1 1 1 1 1 1 1 1; 1 0 0 0 0 0 0 0 0; 1 0 0 0 0 0 0 0 1; 1 0 0 0 0 0 0 1 0; 1 0 0 0 0 0 0 1 1; 1 0 0 0 0 0 1 0 0; 1 0 0 0 0 0 1 0 1; 1 0 0 0 0 0 1 1 0; 1 0 0 0 0 0 1 1 1; 1 0 0 0 0 1 0 0 0; 1 0 0 0 0 1 0 0 1; 1 0 0 0 0 1 0 1 0; 1 0 0 0 0 1 0 1 1; 1 0 0 0 0 1 1 0 0; 1 0 0 0 0 1 1 0 1; 1 0 0 0 0 1 1 1 0; 1 0 0 0 0 1 1 1 1; 1 0 0 0 1 0 0 0 0; 1 0 0 0 1 0 0 0 1; 1 0 0 0 1 0 0 1 0; 1 0 0 0 1 0 0 1 1; 1 0 0 0 1 0 1 0 0; 1 0 0 0 1 0 1 0 1; 1 0 0 0 1 0 1 1 0; 1 0 0 0 1 0 1 1 1; 1 0 0 0 1 1 0 0 0; 1 0 0 0 1 1 0 0 1; 1 0 0 0 1 1 0 1 0; 1 0 0 0 1 1 0 1 1; 1 0 0 0 1 1 1 0 0; 1 0 0 0 1 1 1 0 1; 1 0 0 0 1 1 1 1 0; 1 0 0 0 1 1 1 1 1; 1 0 0 1 0 0 0 0 0; 1 0 0 1 0 0 0 0 1; 1 0 0 1 0 0 0 1 0; 1 0 0 1 0 0 0 1 1; 1 0 0 1 0 0 1 0 0; 1 0 0 1 0 0 1 0 1; 1 0 0 1 0 0 1 1 0; 1 0 0 1 0 0 1 1 1; 1 0 0 1 0 1 0 0 0; 1 0 0 1 0 1 0 0 1; 1 0 0 1 0 1 0 1 0; 1 0 0 1 0 1 0 1 1; 1 0 0 1 0 1 1 0 0; 1 0 0 1 0 1 1 0 1; 1 0 0 1 0 1 1 1 0; 1 0 0 1 0 1 1 1 1; 1 0 0 1 1 0 0 0 0; 1 0 0 1 1 0 0 0 1; 1 0 0 1 1 0 0 1 0; 1 0 0 1 1 0 0 1 1; 1 0 0 1 1 0 1 0 0; 1 0 0 1 1 0 1 0 1; 1 0 0 1 1 0 1 1 0; 1 0 0 1 1 0 1 1 1; 1 0 0 1 1 1 0 0 0; 1 0 0 1 1 1 0 0 1; 1 0 0 1 1 1 0 1 0; 1 0 0 1 1 1 0 1 1; 1 0 0 1 1 1 1 0 0; 1 0 0 1 1 1 1 0 1; 1 0 0 1 1 1 1 1 0; 1 0 0 1 1 1 1 1 1; 1 0 1 0 0 0 0 0 0; 1 0 1 0 0 0 0 0 1; 1 0 1 0 0 0 0 1 0; 1 0 1 0 0 0 0 1 1; 1 0 1 0 0 0 1 0 0; 1 0 1 0 0 0 1 0 1; 1 0 1 0 0 0 1 1 0; 1 0 1 0 0 0 1 1 1; 1 0 1 0 0 1 0 0 0; 1 0 1 0 0 1 0 0 1; 1 0 1 0 0 1 0 1 0; 1 0 1 0 0 1 0 1 1; 1 0 1 0 0 1 1 0 0; 1 0 1 0 0 1 1 0 1; 1 0 1 0 0 1 1 1 0; 1 0 1 0 0 1 1 1 1; 1 0 1 0 1 0 0 0 0; 1 0 1 0 1 0 0 0 1; 1 0 1 0 1 0 0 1 0; 1 0 1 0 1 0 0 1 1; 1 0 1 0 1 0 1 0 0; 1 0 1 0 1 0 1 0 1; 1 0 1 0 1 0 1 1 0; 1 0 1 0 1 0 1 1 1; 1 0 1 0 1 1 0 0 0; 1 0 1 0 1 1 0 0 1; 1 0 1 0 1 1 0 1 0; 1 0 1 0 1 1 0 1 1; 1 0 1 0 1 1 1 0 0; 1 0 1 0 1 1 1 0 1; 1 0 1 0 1 1 1 1 0; 1 0 1 0 1 1 1 1 1; 1 0 1 1 0 0 0 0 0; 1 0 1 1 0 0 0 0 1; 1 0 1 1 0 0 0 1 0; 1 0 1 1 0 0 0 1 1; 1 0 1 1 0 0 1 0 0; 1 0 1 1 0 0 1 0 1; 1 0 1 1 0 0 1 1 0; 1 0 1 1 0 0 1 1 1; 1 0 1 1 0 1 0 0 0; 1 0 1 1 0 1 0 0 1; 1 0 1 1 0 1 0 1 0; 1 0 1 1 0 1 0 1 1; 1 0 1 1 0 1 1 0 0; 1 0 1 1 0 1 1 0 1; 1 0 1 1 0 1 1 1 0; 1 0 1 1 0 1 1 1 1; 1 0 1 1 1 0 0 0 0; 1 0 1 1 1 0 0 0 1; 1 0 1 1 1 0 0 1 0; 1 0 1 1 1 0 0 1 1; 1 0 1 1 1 0 1 0 0; 1 0 1 1 1 0 1 0 1; 1 0 1 1 1 0 1 1 0; 1 0 1 1 1 0 1 1 1; 1 0 1 1 1 1 0 0 0; 1 0 1 1 1 1 0 0 1; 1 0 1 1 1 1 0 1 0; 1 0 1 1 1 1 0 1 1; 1 0 1 1 1 1 1 0 0; 1 0 1 1 1 1 1 0 1; 1 0 1 1 1 1 1 1 0; 1 0 1 1 1 1 1 1 1; 1 1 0 0 0 0 0 0 0; 1 1 0 0 0 0 0 0 1; 1 1 0 0 0 0 0 1 0; 1 1 0 0 0 0 0 1 1; 1 1 0 0 0 0 1 0 0; 1 1 0 0 0 0 1 0 1; 1 1 0 0 0 0 1 1 0; 1 1 0 0 0 0 1 1 1; 1 1 0 0 0 1 0 0 0; 1 1 0 0 0 1 0 0 1; 1 1 0 0 0 1 0 1 0; 1 1 0 0 0 1 0 1 1; 1 1 0 0 0 1 1 0 0; 1 1 0 0 0 1 1 0 1; 1 1 0 0 0 1 1 1 0; 1 1 0 0 0 1 1 1 1; 1 1 0 0 1 0 0 0 0; 1 1 0 0 1 0 0 0 1; 1 1 0 0 1 0 0 1 0; 1 1 0 0 1 0 0 1 1; 1 1 0 0 1 0 1 0 0; 1 1 0 0 1 0 1 0 1; 1 1 0 0 1 0 1 1 0; 1 1 0 0 1 0 1 1 1; 1 1 0 0 1 1 0 0 0; 1 1 0 0 1 1 0 0 1; 1 1 0 0 1 1 0 1 0; 1 1 0 0 1 1 0 1 1; 1 1 0 0 1 1 1 0 0; 1 1 0 0 1 1 1 0 1; 1 1 0 0 1 1 1 1 0; 1 1 0 0 1 1 1 1 1; 1 1 0 1 0 0 0 0 0; 1 1 0 1 0 0 0 0 1; 1 1 0 1 0 0 0 1 0; 1 1 0 1 0 0 0 1 1; 1 1 0 1 0 0 1 0 0; 1 1 0 1 0 0 1 0 1; 1 1 0 1 0 0 1 1 0; 1 1 0 1 0 0 1 1 1; 1 1 0 1 0 1 0 0 0; 1 1 0 1 0 1 0 0 1; 1 1 0 1 0 1 0 1 0; 1 1 0 1 0 1 0 1 1; 1 1 0 1 0 1 1 0 0; 1 1 0 1 0 1 1 0 1; 1 1 0 1 0 1 1 1 0; 1 1 0 1 0 1 1 1 1; 1 1 0 1 1 0 0 0 0; 1 1 0 1 1 0 0 0 1; 1 1 0 1 1 0 0 1 0; 1 1 0 1 1 0 0 1 1; 1 1 0 1 1 0 1 0 0; 1 1 0 1 1 0 1 0 1; 1 1 0 1 1 0 1 1 0; 1 1 0 1 1 0 1 1 1; 1 1 0 1 1 1 0 0 0; 1 1 0 1 1 1 0 0 1; 1 1 0 1 1 1 0 1 0; 1 1 0 1 1 1 0 1 1; 1 1 0 1 1 1 1 0 0; 1 1 0 1 1 1 1 0 1; 1 1 0 1 1 1 1 1 0; 1 1 0 1 1 1 1 1 1; 1 1 1 0 0 0 0 0 0; 1 1 1 0 0 0 0 0 1; 1 1 1 0 0 0 0 1 0; 1 1 1 0 0 0 0 1 1; 1 1 1 0 0 0 1 0 0; 1 1 1 0 0 0 1 0 1; 1 1 1 0 0 0 1 1 0; 1 1 1 0 0 0 1 1 1; 1 1 1 0 0 1 0 0 0; 1 1 1 0 0 1 0 0 1; 1 1 1 0 0 1 0 1 0; 1 1 1 0 0 1 0 1 1; 1 1 1 0 0 1 1 0 0; 1 1 1 0 0 1 1 0 1; 1 1 1 0 0 1 1 1 0; 1 1 1 0 0 1 1 1 1; 1 1 1 0 1 0 0 0 0; 1 1 1 0 1 0 0 0 1; 1 1 1 0 1 0 0 1 0; 1 1 1 0 1 0 0 1 1; 1 1 1 0 1 0 1 0 0; 1 1 1 0 1 0 1 0 1; 1 1 1 0 1 0 1 1 0; 1 1 1 0 1 0 1 1 1; 1 1 1 0 1 1 0 0 0; 1 1 1 0 1 1 0 0 1; 1 1 1 0 1 1 0 1 0; 1 1 1 0 1 1 0 1 1; 1 1 1 0 1 1 1 0 0; 1 1 1 0 1 1 1 0 1; 1 1 1 0 1 1 1 1 0; 1 1 1 0 1 1 1 1 1; 1 1 1 1 0 0 0 0 0; 1 1 1 1 0 0 0 0 1; 1 1 1 1 0 0 0 1 0; 1 1 1 1 0 0 0 1 1; 1 1 1 1 0 0 1 0 0; 1 1 1 1 0 0 1 0 1; 1 1 1 1 0 0 1 1 0; 1 1 1 1 0 0 1 1 1; 1 1 1 1 0 1 0 0 0; 1 1 1 1 0 1 0 0 1; 1 1 1 1 0 1 0 1 0; 1 1 1 1 0 1 0 1 1; 1 1 1 1 0 1 1 0 0; 1 1 1 1 0 1 1 0 1; 1 1 1 1 0 1 1 1 0; 1 1 1 1 0 1 1 1 1; 1 1 1 1 1 0 0 0 0; 1 1 1 1 1 0 0 0 1; 1 1 1 1 1 0 0 1 0; 1 1 1 1 1 0 0 1 1; 1 1 1 1 1 0 1 0 0; 1 1 1 1 1 0 1 0 1; 1 1 1 1 1 0 1 1 0; 1 1 1 1 1 0 1 1 1; 1 1 1 1 1 1 0 0 0; 1 1 1 1 1 1 0 0 1; 1 1 1 1 1 1 0 1 0; 1 1 1 1 1 1 0 1 1; 1 1 1 1 1 1 1 0 0; 1 1 1 1 1 1 1 0 1; 1 1 1 1 1 1 1 1 0; 1 1 1 1 1 1 1 1 1]; end M = []; Y=0; // zmienna pomocnicza przechowująca odkryte punkty stałe sieci Rozmiar = size(W,'r'); printf('Rozmiar wejscia: %d\n',Rozmiar); N=2^Rozmiar; // liczba wszystkiech możliwych wektorów na wejściu printf('Liczba wektorow wejsciowych: %d\n',N); printf('Tryb działania sieci: synchroniczny.\n'); printf('BADANIE ZBIEŻNOŚCI\n\n'); for j=1:N printf('BADANIE WEKTORA NR %d @@@@@@@@@@@@@@@@@@@@@',j); disp(X(j,:)); //inicjacja wartosci: Uk=zeros(Rozmiar,1); Vk=ones(Rozmiar,1); V=X(j,:)';//=[0 1 1]'; k=1; //badanie zbieżności: Vk_1=V; printf('\nWektor V(0):'); disp(Vk_1'); //petla warunek_stop=%f; // warunek stopu nie jest spełniony while ~warunek_stop, printf('KROK\t%d\n',k); printf('Potencjał wejsciowy U(%d)',k); Uk=W*Vk_1; disp(Uk'); printf('Potencjał wyjsciowy V(%d)',k); Vk= unipolar(Uk); //M = [M ; Vk' ]; disp(Vk'); k=k+1; if Vk_1 == Vk then warunek_stop=%t; printf('Zbieżny do:'); disp(Vk'); //M = [M ; Vk' ]; if size(Y,'c')==1 then Y=Vk'; else if isInMatrix(Y,Vk')==%f then Y=[Y; Vk']; end end else Vk_1=Vk; end if k==16 then printf('BRAK ZBIEZNOSCI!!!!!\n'); break; end end end M = Y; endfunction function [M] = bipolarna(W, ac) RozmiarKolumn = size(W, 'c'); if RozmiarKolumn == 3 then X=[ -1 -1 -1; -1 -1 1; -1 1 -1; -1 1 1; 1 -1 -1; 1 -1 1; 1 1 -1; 1 1 1]; end if RozmiarKolumn == 6 then X=[-1 -1 -1 -1 -1 -1; -1 -1 -1 -1 -1 1; -1 -1 -1 -1 1 -1; -1 -1 -1 -1 1 1; -1 -1 -1 1 -1 -1; -1 -1 -1 1 -1 1; -1 -1 -1 1 1 -1; -1 -1 -1 1 1 1; -1 -1 1 -1 -1 -1; -1 -1 1 -1 -1 1; -1 -1 1 -1 1 -1; -1 -1 1 -1 1 1; -1 -1 1 1 -1 -1; -1 -1 1 1 -1 1; -1 -1 1 1 1 -1; -1 -1 1 1 1 1; -1 1 -1 -1 -1 -1; -1 1 -1 -1 -1 1; -1 1 -1 -1 1 -1; -1 1 -1 -1 1 1; -1 1 -1 1 -1 -1; -1 1 -1 1 -1 1; -1 1 -1 1 1 -1; -1 1 -1 1 1 1; -1 1 1 -1 -1 -1; -1 1 1 -1 -1 1; -1 1 1 -1 1 -1; -1 1 1 -1 1 1; -1 1 1 1 -1 -1; -1 1 1 1 -1 1; -1 1 1 1 1 -1; -1 1 1 1 1 1; 1 -1 -1 -1 -1 -1; 1 -1 -1 -1 -1 1; 1 -1 -1 -1 1 -1; 1 -1 -1 -1 1 1; 1 -1 -1 1 -1 -1; 1 -1 -1 1 -1 1; 1 -1 -1 1 1 -1; 1 -1 -1 1 1 1; 1 -1 1 -1 -1 -1; 1 -1 1 -1 -1 1; 1 -1 1 -1 1 -1; 1 -1 1 -1 1 1; 1 -1 1 1 -1 -1; 1 -1 1 1 -1 1; 1 -1 1 1 1 -1; 1 -1 1 1 1 1; 1 1 -1 -1 -1 -1; 1 1 -1 -1 -1 1; 1 1 -1 -1 1 -1; 1 1 -1 -1 1 1; 1 1 -1 1 -1 -1; 1 1 -1 1 -1 1; 1 1 -1 1 1 -1; 1 1 -1 1 1 1; 1 1 1 -1 -1 -1; 1 1 1 -1 -1 1; 1 1 1 -1 1 -1; 1 1 1 -1 1 1; 1 1 1 1 -1 -1; 1 1 1 1 -1 1; 1 1 1 1 1 -1; 1 1 1 1 1 1]; end M = []; Y=0; // zmienna pomocnicza przechowująca odkryte punkty stałe sieci Rozmiar = size(W,'r'); printf('Rozmiar wejscia: %d\n',Rozmiar); N=2^Rozmiar; // liczba wszystkiech możliwych wektorów na wejściu printf('Liczba wektorow wejsciowych: %d\n',N); printf('Tryb działania sieci: synchroniczny.\n'); printf('BADANIE ZBIEŻNOŚCI\n\n'); for j=1:N printf('BADANIE WEKTORA NR %d @@@@@@@@@@@@@@@@@@@@@',j); disp(X(j,:)); //inicjacja wartosci: Uk=zeros(Rozmiar,1); Vk=ones(Rozmiar,1); V=X(j,:)';//=[0 1 1]'; k=1; //badanie zbieżności: Vk_1=V; printf('\nWektor V(0):'); disp(Vk_1'); //petla warunek_stop=%f; // warunek stopu nie jest spełniony while ~warunek_stop, printf('KROK\t%d\n',k); printf('Potencjał wejsciowy U(%d)',k); Uk=W*Vk_1; disp(Uk'); printf('Potencjał wyjsciowy V(%d)',k); Vk= bipolar(Uk); disp(Vk'); k=k+1; if Vk_1 == Vk then warunek_stop=%t; printf('Zbieżny do:'); disp(Vk'); //M = [M ; Vk']; printf('Macierz M : '); disp(M); if size(Y,'c')==1 then Y=Vk'; else if isInMatrix(Y,Vk')==%f then Y=[Y; Vk']; end end else Vk_1=Vk; end if k==16 then printf('BRAK ZBIEZNOSCI!!!!!\n'); break; end end end M = Y; endfunction function [wy]=unipolar(a) [row col]=size(a); if row*col<>1 then wy=zeros(row,col); for i = 1:row for j = 1:col wy(i,j) = unipolar(a(i,j)); end; end else if a > 0 then wy=1; else wy=0; end end endfunction function wy_log=isInMatrix(Y,X) // funkcja pomocnicza, sprawdza czy wektor X znajduje się już w macierzy Y if size(Y,'c') <> size(X,'c') | size(X,'r') <> 1 then printf('Niepoprawne dane wejściowe: wymiar kolumn dla obu macierzy nie jest zgodny lub macierz druga nie jest wektorem wierszowym.'); end wy_log=%f; for i=1:size(Y,'r') if Y(i,:) == X then wy_log=%t end end endfunction function [wy]=bipolar(a) [row col]=size(a); if row*col<>1 then wy=zeros(row,col); for i = 1:row for j = 1:col wy(i,j) = bipolar(a(i,j)); end; end else if a > 0 then wy=1; else wy=-1; end end endfunction function [Z] = RegulaHebba(X) RozmiarK = size(X, 'c'); RozmiarW = size(X, 'r'); W = zeros(RozmiarK, RozmiarK); row = size(W,'r'); col = size(W, 'c'); for i = 1: row for j = 1: col for k = 1: RozmiarW if i <> j then W(i,j) = W(i,j) + (X(k,i) * X(k,j)); else W(i,j) = 0; end end end end Z = W/RozmiarK; endfunction function printC(Z) RozmiarK = size(Z , 'c'); for i = 1 : RozmiarK if Z(1,i) == 1 then printf('#'); else printf("*"); end if modulo(i,3) == 0 then printf('\n'); end end endfunction function printMacierz(Z) RozmiarK = size(Z, 'c'); RozmiarW = size(Z, 'r'); for i = 1 : RozmiarW for j = 1: RozmiarK if Z(i,j) == 1 then printf('#'); else printf("*"); end end printf("\n"); end endfunction function [W] = pseudoinwersja(X) w = det(X); if w <> 0 then [W] = X' * (X * X')^(-1) * X; end endfunction function czyLiniowoNiezalezne(X) w = det(X); printf("Czy Liniowo Niezalezna : %d \n ",w); endfunction function [Z] = czyRozpoznane(P,V) // printf('Macierz wektorow Rozpoznanych : \n ' ); // disp(P); // printf('Macierz V : \n ' ); // disp(V); Z = []; //L = liczba wektorow do rozpoznania; L = size(V, 'r'); for i = 1: L if isInMatrix(P, V(i,:)) then //printf("jest \n"); //disp(V(i,:)); Z = [Z; V(i,:)]; else //printf("nie ma \n"); end end endfunction function [Z] = synchronicznaSH(W, pattern, V, ac) r = size (W, 'c'); if W == -1 then printf('Regula Hebba\n'); //V1 = RegulaHebba(pattern); //printf("Pseudoinwersja\n") //czyLiniowoNiezalezne(pattern); [V1] = pseudoinwersja(pattern); disp(V1); if ac == 1 then P = unipolarna(V1); printf('Macierz wektorów zbieżnych: \n'); disp(P); printf('Macierz V : \n' ); disp(V); Z = czyRozpoznane(P, V); printf("Wypisania Macierzy Z \n"); disp(Z); printf("Wektor poczatkowy \n") //printC(pattern);//dla macierzy C4 printMacierz(pattern); printf("Wektor koncowy \n"); //porownane(P, V); //printC(Z);//dla macierzy C4 printMacierz(Z); end if ac == -1 then P = bipolarna(V1); printf('Macierz wektorów zbieżnych: \n'); disp(P); printf('Macierz V : \n' ); disp(V); Z = czyRozpoznane(P, V); printf("Wypisania Macierzy Z \n"); disp(Z); printC(Z); end else if pattern == -1 then printf('Regula Heba \n'); disp(W); if ac == 1 then //P = unipolarna(W); P = unipolarna(W); printf('Macierz wektorów zbieżnych: \n'); disp(P); printf('Macierz V : \n' ); disp(V); Z = czyRozpoznane(P, V); //printf("Wypisania Macierzy Z \n"); //disp(Z); //printC(Z); //printMacierz(Z); end if ac == -1 then P = bipolarna(W); printf('Macierz wektorów zbieżnych: \n'); disp(P); printf('Macierz V : \n' ); disp(V); Z = czyRozpoznane(P, V); //printf("Wypisania Macierzy Z \n"); //disp(Z); //printC(Z); //printMacierz(Z); end end end endfunction Wv = [-1 1 1; -1 1 -1; 1 1 -1]; Wx = [0 1 1; 0 1 0; 1 1 0]; C1 = [0 1 0 1 1 0 0 1 0]; Cx1 = [0 1 1]; Cx2 = [0 1 0]; Cx3 = [1 1 0]; Ca = [1 1 1]; Cb = [0 1 0]; Cc = [1 1 1]; Cx4 = [0 0 1 0]; Cx5 = [0 1 1 0]; Cx6 = [0 0 1 0]; Cx7 = [0 0 1 0]; Cx = [Cx1;Cx2;Cx3]; Cxx = [Cx4;Cx5;Cx6]; Cabc = [Ca; Cb; Cc]; C2 = [-1 1 -1 1 1 -1]; Cv1 = [-1 1 1]; Cv2 = [-1 1 -1]; Cv3 = [ 1 1 -1]; Cv = [Cv1;Cv2;Cv3]; WSynchUni = [ 0 -1 -3; -1 0 2; -3 2 0]; WSynchBip = [0 1 2; 1 0 -1; 2 -1 0]; C4 = [1 0 1 0 1 0 1 0 1]; Vjeden = [1 0 0 1 0 0; 0 0 1 1 0 0; 0 1 0 1 0 0; 0 0 0 1 0 0; 0 0 0 1 0 1; 0 0 0 1 1 0]; Vjeden1 = [0 0 0 1 0 0; 0 0 1 1 0 0; 0 1 0 1 0 0; 1 0 0 1 0 0; 0 0 0 1 0 0; 0 0 0 1 0 0]; Vcztery = [0 0 0 1 0 0; 0 0 1 1 0 0; 0 1 0 1 0 0; 1 1 1 1 0 0; 0 0 0 1 1 0; 1 1 0 1 0 1]; Vcztery1 = [0 0 0 1 0 0; 0 0 1 1 0 0; 0 1 0 1 0 0; 1 1 1 1 0 0; 0 0 0 1 0 0; 0 0 0 1 0 0]; VL = [1 1 0 0 0 0; 0 1 0 0 0 0; 0 1 1 1 0 0; 0 1 0 1 1 0; 0 1 0 0 1 0; 0 1 1 1 0 1]; VL1 = [0 1 0 0 0 0; 0 1 0 0 0 0; 0 1 0 0 0 0; 0 1 0 0 0 0; 0 1 0 0 0 0; 0 1 1 1 0 0]; printf('PROGRAM NAPISANY NA ĆWICZENIA - ZAAWANSOWANE METODY SZTUCZNEJ INTELIGENCJI\n autorzy: Paweł Mazur, Piotr Mazur\n\n'); //Z = synchronicznaSH(Wx, -1, Cx, -1); //dla Podpunktu b //regula Hebba unipolarna //Z = synchronicznaSH(-1, C4, C4, 1); //dla pesudoinwersji unipolarna Z = synchronicznaSH(-1, Vjeden, Vjeden1, 1); //Z = asynchronicznaSH(-1, Vcztery, Vcztery1, 1); //Z = asynchronicznaSH(-1, VL, VL1, 1); //regula Hebba bipolarna //Z = synchronicznaSH(-1, C2, C2, -1); //dla pesudoinwersji bipolarna //Z = synchronicznaSH(-1, Cv, Cv, -1); //Z = synchronicznaSH2(-1, Vx1, Vx1, 1); //dla Podpunktu a //z wagami bipolarna //Z = synchronicznaSH(WSynchUni, -1, Cx, 1); //z wagami unipolarna //Z = synchronicznaSH(WSynchUni, -1, Cv, 1);
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//Example5.8 //determine the feedback transfer function of an op-amp for the following condition clc; clear; close; // a) When open loop gain of 10^5 and the closed loop gain of 100 A = 10^5 ; // open loop gain Af = 100 ; //closed loop gain // Feedback transfer function is beta =(1/Af)-(1/A); disp('Feedback transfer function is = '+string(beta)+''); beta = 1/beta ; disp('OR 1/Beta is = '+string(beta)+''); // For an open loop gain of -10^5 and closed loop gain of -100 A = -10^5 ; // open loop gain Af = -100 ; //closed loop gain // Feedback transfer function is beta =(1/Af)-(1/A); disp('Feedback transfer function is = '+string(beta)+''); beta = 1/beta ; disp('OR 1/Beta is = '+string(beta)+'');
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////Ex 8.1 clc; clear; close; format('v',9); VCC=6;//V VEE=6;//V RT=4;//kohm CT=330;//pF C=240;//pF fo=0.3/(RT*1000*CT*10^-12)/1000;//kHz disp(fo,"Free running frequency(kHz)"); Vplus=(VCC-(-VEE))/2;//V deltafL=8*fo/Vplus;//kHz disp(deltafL,"Lock Range(+ve & -ve in kHz)"); //For LM 565 R=3.6;//kohm deltafC=sqrt(deltafL*1000/(2*%pi*R*1000*C*10^-12))/1000;//kHz disp(deltafC,"Capture Range(+ve & -ve in kHz)"); deltafP=2*deltafC/2;//kHz disp(deltafP,"Pull-in Range(kHz)");
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int x; int main(void) { return 0; }
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// Display mode mode(0); // Display warning for floating point exception ieee(1); clear; clc; disp("Engineering Thermodynamics by Onkar Singh Chapter 9 Example 15") r=10;//pressure ratio Cp=1.0032;//specific heat of air in KJ/kg K y=1.4;//expansion constant T3=1400;//inlet temperature of gas turbine in K T1=(17+273);//ambient temperature in K P1=1*10^5;//ambient pressure in Pa Pc=15;//condensor pressure in KPa Pg=6*1000;//pressure of steam in generator in KPa T5=420;//temperature of exhaust from gas turbine in K disp("In gas turbine cycle,T2/T1=(P2/P1)^((y-1)/y)") disp("so T2=T1*(P2/P1)^((y-1)/y)in K") T2=T1*(r)^((y-1)/y) disp("T4/T3=(P4/P3)^((y-1)/y)") disp("so T4=T3*(P4/P3)^((y-1)/y) in K") T4=T3*(1/r)^((y-1)/y) disp("compressor work per kg,Wc=Cp*(T2-T1) in KJ/kg") Wc=Cp*(T2-T1) disp("turbine work per kg,Wt=Cp*(T3-T4) in KJ/kg ") Wt=Cp*(T3-T4) disp("heat added in combustion chamber per kg,q_add=Cp*(T3-T2) in KJ/kg ") q_add=Cp*(T3-T2) disp("net gas turbine output,W_net_GT=Wt-Wc in KJ/kg air") W_net_GT=Wt-Wc disp("heat recovered in HRSG for steam generation per kg of air") disp("q_HRGC=Cp*(T4-T5)in KJ/kg") q_HRGC=Cp*(T4-T5) disp("at inlet to steam in turbine,") disp("from steam table,ha=3177.2 KJ/kg,sa=6.5408 KJ/kg K") ha=3177.2; sa=6.5408; disp("for expansion in steam turbine,sa=sb") sb=sa; disp("let dryness fraction at state b be x") disp("also from steam table,at 15KPa, sf=0.7549 KJ/kg K,sfg=7.2536 KJ/kg K,hf=225.94 KJ/kg,hfg=2373.1 KJ/kg") sf=0.7549; sfg=7.2536; hf=225.94; hfg=2373.1; disp("sb=sf+x*sfg") disp("so x=(sb-sf)/sfg ") x=(sb-sf)/sfg disp("so hb=hf+x*hfg in KJ/kg K") hb=hf+x*hfg disp("at exit of condenser,hc=hf ,vc=0.001014 m^3/kg from steam table") hc=hf; vc=0.001014; disp("at exit of feed pump,hd=hd-hc") disp("hd=vc*(Pg-Pc)*100 in KJ/kg") hd=vc*(Pg-Pc)*100 disp("heat added per kg of steam =ha-hd in KJ/kg") ha-hd disp("mass of steam generated per kg of air=q_HRGC/(ha-hd)in kg steam per kg air") q_HRGC/(ha-hd) disp("net steam turbine cycle output,W_net_ST=(ha-hb)-(hd-hc)in KJ/kg") W_net_ST=(ha-hb)-(hd-hc) disp("steam cycle output per kg of air(W_net_ST)=W_net_ST*0.119 in KJ/kg air") W_net_ST=W_net_ST*0.119 disp("total combined cycle output=(W_net_GT+W_net_ST) in KJ/kg air ") (W_net_GT+W_net_ST) disp("combined cycle efficiency,n_cc=(W_net_GT+W_net_ST)/q_add") n_cc=(W_net_GT+W_net_ST)/q_add disp("in percentage") n_cc=n_cc*100 disp("In absence of steam cycle,gas turbine cycle efficiency,n_GT=W_net_GT/q_add") n_GT=W_net_GT/q_add disp("in percentage") n_GT=n_GT*100 disp("thus ,efficiency is seen to increase in combined cycle upto 57.77% as compared to gas turbine offering 48.21% efficiency.") disp("overall efficiency=57.77%") disp("steam per kg of air=0.119 kg steam per/kg air")
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//example 3.8 // this program needs kmap.sci and noof.sci clc; disp(' C''D'' C''D CD CD''');//displaying the given kmap disp('A''B'' 0 0 0 0'); disp('A''B 0 0 1 0'); disp('AB x x x x'); disp('AB'' 0 0 x x'); k=[0 0 0 0;0 0 1 0;0 0 1 0;0 0 0 0]; disp('In a Karnaugh map if don''t care condition exits, we may consider them as ones if that gives a larger group size.'); disp('The minimal expression from the given kmap is '); kmap(k); //calling the kamp function
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clc // Fundamental of Electric Circuit // Charles K. Alexander and Matthew N.O Sadiku // Mc Graw Hill of New York // 5th Edition // Part 2 : AC Circuits // Chapter 14 : Frequency Response // Example 14 - 6 clear; clc; close; // s=poly(0,'s') h=syslin('c',10000*s/(s + 1)*(s + 5)*(s + 20)) clf();bode(h,0.01,100);
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//EXAMPLE 2-108 PG NO-146 I1=(100/1.414)^2; I2=(10/1.414)^2; R.M.S=(I1+I2)^0.5; disp('R.M.S VALUE is = '+string(R.M.S)+' A');
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//clear// clear; clc; //Example 23.1 //Given T = 320; //[F] P = 1 ; //[atm] //(1)=CO2, (2)=H2O, (3)=O2, (4)=N2 y_in = [0.14,0.07,0.03,0.76]'; Tw = 80; //[F] //Solution //(a) //Basis F = 100; //[mol], of gas Ts = 120; //[F] Cps = [9.72,8.11,7.14,6.98]'; n_in = F*y_in; //[mol] nCp = n_in.*Cps; // sum_nCp = sum(nCp); sum_n_in = sum(n_in); //[mol] Tavg = (Ts+T)/2; //[F] lambda_s = 1025.8*18; //[Btu/lb mol], at Ts, from Appendix 7 //Making a heat balance for z moles of water evaporated z = sum_nCp*(T-Ts)/(lambda_s+18*(Ts-Tw)); //Total moles of water in exit gas n_out(2) = z+n_in(2); //[mole] //Partial pressure of the water in the exit gas PH2O = n_out(2)/107.76*760; //[mm Hg] //But at 120 F, PH2Oprime = 87.5 mm Hg (Appendix 7). Saturation //temperature Ts must be greater than 120 F. Trying Ts = 126; // [F] Tavg = (Ts+T)/2; //[F] lambda_s = 1022.3*18; //[Btu/lb mol], at Ts, from Appendix 7 //Making a heat balance for z moles of water evaporated z = sum_nCp*(T-Ts)/(lambda_s+18*(Ts-Tw)); //Total moles of water in exit gas n_out(2) = z+n_in(2); //[mole] //Partial pressure of the water in the exit gas PH2O = n_out(2)/107.76*760; //[mm Hg] //This is close enough to the value of PH2Oprime disp('F',Ts,'Adiabatic saturation temperature'); //(b) //for Tin = Ts, by heat balance z = sum_nCp*(T-Ts)/(lambda_s); n_out(2) = z + n_in(2); //[mole] //Partial pressure of the water in the exit gas PH2O = n_out(2)/107.85*760; //[mm Hg] //This is higher than the vapor pressure of water at 126 F, //103.2 mm Hg, and Ts>126 F. Trying Ts = 127; //[F] Tavg = (Ts+T)/2; //[F] lambda_s = 1021.7*18; //[Btu/lb mol], at Ts, from Appendix 7 //Making a heat balance for z moles of water evaporated z = sum_nCp*(T-Ts)/(lambda_s); //Total moles of water in exit gas n_out(2) = z+n_in(2); //[mole] //Partial pressure of the water in the exit gas PH2O = n_out(2)/107.76*760; //[mm Hg] //Thus 127 is too high and 126 is too low. Hence, Ts = (126+127)/2; //[F] disp('F',Ts,'Adiabatic saturation temperature');
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; @Harness: simulator ; @Format: atmel ; @Arch: avr ; @Purpose: "Test the LSL (logical shift left instruction" ; @Result: "flags.h = 0, flags.s = 1, flags.v = 1, flags.n = 0, flags.z = 1, flags.c = 1, r16 = 0" start: ldi r16, 0b10000000 lsl r16 end: break
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clc // Given that V = 9.8e3 // voltage in V i = 2e-3 // current in amp c = 3e8 // speed of light in m/sec // Sample Problem 6 on page no. 20.8 printf("\n # PROBLEM 6 # \n") printf("Standard formula used \n ") printf("h*c/lambda = eV \n") lambda = 12400 / V f = c / lambda printf("\n Highest frequency is %e Hz.\n Minimum wavelength is %f Angstrom.",f,lambda)
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stacksize('max'); w = 60*24*30*6; time = read("P:\finance\spectrumSeparator\nhel.Composite\20.1.win.8\console\outTime",-1,w); orig = read("P:\finance\spectrumSeparator\nhel.Composite\20.1.win.8\console\outOrig",-1,w); extrap = read("P:\finance\spectrumSeparator\nhel.Composite\20.1.win.8\console\outExtrap",-1,w); clf pos = 50; timeMult = 1/60/60/24; mn = min(orig(pos,:)); mx = max(orig(pos,:)); bx=[time(pos,1),time(pos,w/2),time(pos,w/2),time(pos,w)]; by=[mn,mn,mx,mx]; plot(bx*timeMult, by,'r'); plot(time(pos,:)*timeMult, orig(pos,:),'k'); plot(time(pos,:)*timeMult, extrap(pos,:)-extrap(pos,w/2)+orig(pos,w/2),'b');
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Ch05Ex03.sce
// Scilab code Ex5.3: Pg.196-197 (2008) clc; clear; E_k = 150; // Kinetic energy of a cosmic ray proton, GeV h = 6.63e-034; // Plank's constant, J-s m = 1.67e-027; // Mass of proton, kg c = 3e+08; // Velocity of light, m/s E_0 = 0.938; // Rest energy of the proton, GeV lamda_c = h/(m*c); // Compton wavelength of proton, m r = E_k/E_0; // Ratio of kinetic energy to rest energy of proton // Since this value on the curve corresponds to about 6e-03 on the axis, therefore r_w = 6e-03; // Ratio of de broglie wavelength of proton to its compton wavelength lamda = r_w*lamda_c; // de Broglie wavelength of the proton, m printf("\nThe de-Broglie wavelength of cosmic ray proton = %3.1e m = %3.1e fm",lamda, lamda*1e+15); // Result // The de-Broglie wavelength of cosmic ray proton = 7.9e-018 m = 7.9e-003 fm
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Ex2_9.sce
//To determine the most economic power factor //Page 91 clc; clear; P=200*(10^3); //Maximum Demand pf=0.707; //Power Factor Lagging a=100; //Tariff per kVA per year b=200; //Power factor improvement cost Per kVA. r=20; //Interest Depriciation, maintenance and cost of losses amount to 20% of capital cost per year // Economic PF = sqrt(1-((b1/a)^2)) b1=r*b/100;// b' term accrding to the equation above pfeco=sqrt(1-((b1/a)^2)); //Economic Power Factor printf('The Economic Power Factor is %g \n',pfeco)
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15_16.sce
errcatch(-1,"stop");mode(2);//Find Band gap energy //Ex:15.16 ; ; c=2.99*10^8;//speed of light in m/s h=6.62*10^-24;//planck's constant w=1.771*10^-6;//wavelength in J eg=h*c/w;//in J disp(eg,"Band gap energy (in J) = "); exit();