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//Page Number: 5.31 //Example 5.15 clc; mp=1; //Assume peak amplitude is unity //Given del=0.02*mp; L=(mp*2)/del; for (i=0:10) j=2^i; if(j>=L) L1=j; break; end end n=log2(L1);// bits per sample disp(n,'Number of bits');
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// Scilab code Exa7.1: : Page-292 (2011) clc; clear; h = 6.6261e-034; // Planck's constant, joule sec C = 2.998e+08; // Velocity of light, metre per sec f = 2; // Radius of focal circle, metre d = 1.18e-010; // Interplaner spacing for quartz crystal, metre E_1 = 1.17*1.6022e-013; // Energy of the gamma rays, joule E_2 = 1.33*1.6022e-013; // Energy of the gamma rays, joule D = h*C*f*(1/E_1-1/E_2)*1/(2*d); //Distance to be moved for obtaining first order reflection for two different energies, metre printf("\nThe distance to be moved for obtaining first order Bragg reflection = %4.2e metre", D); // Result // The distance to be moved for obtaining first order Bragg reflection = 1.08e-003 metre
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//Chapter-9,Example9_30,pg 9_85 Po=8.952*10^3 V=440 Ra=1.1 Rsh=650 Rint=0.4 Rreg=50 Ml=450 Vbr=2//brush drop Il=24 Rat=Ra+Rint//series connection Rsht=Rsh+Rreg//series connection Ish=V/Rsht Ia=Il-Ish Acl=(Ia^2)*Rat//armature copper loss Fcl=(Ish^2)*Rsht//feild copper loss Bdl=Vbr*Ia//brush drop loss TL=Acl+Fcl+Bdl+Ml n=Po*100/(Po+TL) printf("efficiency of motor\n") printf("n=%.2f ",n)
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//CHAPTER 2 ILLUSRTATION 1 PAGE NO 57 //TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS clc clear //=========================================================================================== //INPUT DATA Na=300;//driving shaft running speed in rpm Nb=400;//driven shaft running speed in rpm Da=60;//diameter of driving shaft in mm t=.8;//belt thickness in mm s=.05;//slip in percentage(5%) //========================================================================================== //calculation Db=(Da*Na)/Nb;//finding out the diameter of driven shaft without considering the thickness of belt Db1=(((Da+t)*Na)/Nb)-t///considering the thickness Db2=(1-s)*(Da+t)*(Na/Nb)-t//considering slip also //========================================================================================= //output printf('the value of Db is %3.0f cm',Db) printf('\nthe value of Db1 is %f cm',Db1) printf('\nthe value of Db2 is %f cm',Db2)
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-- Fuzzy Logix, LLC: Functional Testing Script for DB Lytix functions on Netezza -- -- Copyright (c): 2014 Fuzzy Logix, LLC -- -- NOTICE: All information contained herein is, and remains the property of Fuzzy Logix, LLC. -- The intellectual and technical concepts contained herein are proprietary to Fuzzy Logix, LLC. -- and may be covered by U.S. and Foreign Patents, patents in process, and are protected by trade -- secret or copyright law. Dissemination of this information or reproduction of this material is -- strictly forbidden unless prior written permission is obtained from Fuzzy Logix, LLC. -- Functional Test Specifications: -- -- Test Category: Basic Statistics -- -- Test Unit Number: FL-Netezza-01 -- -- Name(s): FLPercentWin -- -- Description: Aggregate which calculates the percentage value of each observation in the data series. -- -- Applications: -- -- Signature: FLPercentWin(D DOUBLE PRECISION) -- -- Parameters: See Documentation -- -- Return value: FLOAT -- -- Last Updated: 07-04-2017 -- -- Author: Kamlesh Meena -- -- BEGIN: TEST SCRIPT \time --.run file=../PulsarLogOn.sql --.set width 2500 -- BEGIN: POSITIVE TEST(s) ---- Positive Test 1: Returns expected result --- Return expected results, Good select FLPercentWin(ClosePrice) over (partition by TickerId ) from finstockprice order by TickerId LIMIT 10; ---- Positive Test 2: Returns expected result; parameter is Integer --- Return expected results, Good select FLPercentWin(Volume) over (partition by TickerId ) from finstockprice order by TickerId LIMIT 10; ---- Positive Test 3: Returns expected result; partition parameter is changed --- Return expected results, Good select FLPercentWin(ClosePrice) over (partition by TickerSymbol ) from finstockprice order by TickerId LIMIT 10; -- END: POSITIVE TEST(s) -- BEGIN: NEGATIVE TEST(s) ---- Negative Test 1: First parameter in FLPercentWin has to be double precision select FLPercentWin(TickerSymbol) over (partition by TickerSymbol ) from finstockprice order by TickerId LIMIT 10 -- END: NEGATIVE TEST(s) \time -- END: TEST SCRIPT
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clear; clc; l=100;R=10;G=10^-5;Vs=40;Zr=0; Zo=sqrt(R/G); P=sqrt(R*G); Zin=Zo*(Zr+(Zo*tanh(P*l)))/(Zo+(Zr*tanh(P*l))); Is=Vs/Zin; V=(Vs*(cosh(P*l)))-(Is*Zo*(sinh(P*l))); Vm=2*V; printf("Potential at mid point = %f volts",Vm); //the difference in result is due to erroneous value in textbook. disp("The difference in result is due to erroneous value in textbook")
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clc; clear; close; //46ex //nth term in the sequence 2,4,6,8,10...is 2n. 5th term is? term5=2*5 //nth term in the sequence 1,4,9,16,25 is n^2. 5th term is? ex2_term5=5^2
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function mdaq_led_write(link_id, led, state) if link_id < 0 then disp("Wrong link ID!") return; end if led > 2 | led < 1 then disp("Wrong LED number!") return; end if state <> 0 then state = 1; end result = call("sci_mlink_led_set",.. link_id, 1, "i",.. led, 2, "i",.. state, 3, "i",.. "out",.. [1, 1], 4, "i"); if result < 0 then mdaq_error(result); end endfunction
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@relation led7digit @attribute Led1 real[0.0,1.0] @attribute Led2 real[0.0,1.0] @attribute Led3 real[0.0,1.0] @attribute Led4 real[0.0,1.0] @attribute Led5 real[0.0,1.0] @attribute Led6 real[0.0,1.0] @attribute Led7 real[0.0,1.0] @attribute number{0,1,2,3,4,5,6,7,8,9} @inputs Led1,Led2,Led3,Led4,Led5,Led6,Led7 @outputs number @data 0 9 4 9 7 7 8 8 8 8 4 4 4 4 9 9 3 3 8 8 8 8 0 0 1 7 3 7 4 4 4 9 5 5 6 6 3 3 7 7 7 7 0 0 3 3 4 4 5 5 5 5 7 1 2 6 3 ? 3 9 6 ? 7 ? 1 1 2 2 7 7 8 8 8 8 9 9 2 2 2 2 2 ? 7 7 9 8 0 0 2 2 2 8 6 8 7 1 7 7 9 ?
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clc clear //Initialization of variables Hi=55 Pi=11.8 xi=0.514 H2=18.1 H3=26.9 Pi2=17.4 //calculations ai=Pi/Hi gam=ai/xi a2=Pi/H2 gam2=a2/xi a3=Pi2/H3 gam3=a3/(1-xi) //results disp("part a") printf("Activity of acetic acid = %.4f ",ai) printf("\n Activity coefficient = %.4f ",gam) disp("part b") printf("Activity of acetic acid = %.4f ",a2) printf("\n Activity coefficient = %.4f ",gam2) disp("part c") printf("Activity of toluene = %.4f ",a3) printf("\n Activity coefficient = %.4f ",gam3)
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//Example 5.8// cx=0.01;// distance of x c0=0;////for initially pure A c=cx-c0 mprintf("c = %f ",c) a=1;//given e=a-c mprintf("\ne = %f ",e) b=0.9928;//As z= 1.90 erf(z)=0.9928 //Interpolating table 5.1 gives d=0.99;//Interpolating table 5.1 gives f=0.9891;//As z=1.80 erf(z)=0.9891 //Interpolating table 5.1 gives h=1.90;//given i=1.80;//given z=-((((b-d)/(b-f))*(h-i))-h) mprintf("\nz = %f ",z) D=1*10^-10;//m^2/s// grain boundary D1=1*10^-14;//m^2/s // volume of bulk grain t=1;//h //hour //time t1=3.6*10^3;//s/h //time x=2*z*sqrt(D*t*t1) mprintf("\nx = %e m ",x) a1=10^3;//(As 1milli = 10^-3) a2=a1*x mprintf(" = %f mm",a2) //(b) For comparison x1=2*z*sqrt(D1*t*t1) mprintf("\nx1 = %e m ",x1) b1=10^6;//(As mew = 10^-6) b2=b1*x1 mprintf(" = %f mew m",b2)
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clc; t=(225990+3360)/2769; //calculating temperature disp(t,"Temperature in celcius = "); //displaying result
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//Chapter 15: Environmental Pollution and Control //Problem: 2 clc; //Declaration of Variables v0 = 30 // cm cube, effluent v1 = 9.8 // cm cube, K2Cr2O7 M = 0.001 // M, K2Cr2O7 // Solution Oeff = 6 * 8 * v1 * M mprintf("30 cm cube of effluent contains =:%.4f mg of O2\n",Oeff) cod = Oeff * 1000 / 30. mprintf(" 1l of the effluent requires %.2f mg of O2\n",cod) mprintf(" COD of the effluent sample=%.2f ppm",cod)
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clc //calc pressuer at different heights considering on density change in air p_atm=14.7;//psia g=9.81;//m/s^2 //P2=P1*[1-(acc. due to gravity)*(mass of air)*(height)/(univ. gas const.)/(temp.)] T=289;//K R=8314//N.m^2/Kmol/K //for height of 1000ft=304.8m h=304.8//m p_1000=p_atm*[1-g*29*h/R/T]; disp("pressure at 1000ft is") disp(p_1000) disp("psia") //for height of 10000ft=3048m h=3048//m p_10000=p_atm*[1-g*29*h/R/T]; disp("pressure at 10000ft is") disp(p_10000) disp("psia") //for height of 100000ft=30480m h=30480//m p_100000=p_atm*[1-g*29*h/R/T]; disp("pressure at 100000ft is") disp(p_100000) disp("psia") //NOTE that the pressure comes out to be negative at 100000ft justifying that density of air changes with altitude
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//Length of belt N2=80 N1=200 d1=240 d2=d1*N1/N2 //mm r1=120 //mm r2=300 //mm l=2500 //mm //length of crossbelt L=%pi*(r1+r2)+2*l+((r1+r2)^2)/l //mm printf("The length of crossbelt L=%.2f mm",L)
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clc //initialisation of variables clear wcb= 2 //ton wc= 100 //ton wa= 6.5 //ton wca= 20 r= 0.8 r1= 1.2 //CALCULATIONS wca1= wc/wa wca2= wcb*(wca1/wca)^1.5 Wca= wcb*r^(9/4)*(1/r1)^(9/4)*(wca1/wca)^1.5 //RESULTS printf ('(Wc/W)a = %.2f ',wca1) printf ('\n Wc,a = %.2f ton',wca2) printf ('\n Wc,a = %.2f ton',Wca)
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errcatch(-1,"stop");mode(2);//ex4.3 disp('cant be shown') exit();
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// Calculate the time constant ess=5; A=0.1; tc=ess/A; disp(tc,'time constant(s)')
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clear; clc; Po=20000; N=1440; w_m=2*%pi*N/60; T_e=Po/w_m; f1=120; P=4; w_s=4*%pi*f1/P; r2=.4; x2=1.6; f2=50; Z1=r2+%i*x2*f1/f2; Z=abs(Z1); ph=3; V_s=400; s=(ph/w_s)*(V_s/(Z*sqrt(3)))^2*(r2/T_e); N=w_s*f1/(4*%pi)*(1-s); printf("motor speed at rated laod=%.2f rpm",N); s_m=r2/imag(Z1); printf("\nslip at which max torque occurs=%.4f",s_m); T_em=(3/w_s)*(V_s/sqrt(3))^2/(2*imag(Z1)); printf("\nmax torque=%.3f Nm",T_em);
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function [mm] = stat(x) mm = 0 x_mean= mean(x) standard_deviation = st_deviation(x) y = x - mean(x) medx = median(x) //modx = mode(x) medy = median(y) //mody = mode(y) y_mean = mean(y) printf('\nmean of data is: %f', x_mean) printf('\nmedian of data is: %f', medx) //printf('\nmode of data is: %f', modx) printf('\nstandard deviation of data is : %f', standard_deviation) printf('\nmean of noise is :%f', y_mean) printf('\nmedian of noise is: %f', medy) //printf('\nmode of noise is: %f', mody) result = [x_mean,standard_deviation,y_mean] //printf('\n stat result is : %f ', result) // 2*sigma^2 term den = standard_deviation; den = 2 * den * den; // sqrt(2*pi*sigma^2) term coeff = standard_deviation; coeff = 2 * %pi * coeff * coeff; coeff = sqrt(coeff); num = y; for i=1:100 num(i) = y(i)*y(i); end printf("\nSize %f",size(num,1)); // gaussian distribution function gaussian = (1/coeff) * exp(0-(num/den)); mm = mm + gaussian; //printf('total prob is %f',mm) endfunction
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clc //initialisation of variables sp=1400 //speed in revolutions per min ma=15 //mass in kgs r=0.287 p1=1 //pressure in bar t1=303 //temparature in k p2=7 //pressure in bar c=0.05 //clearance volume/stoke volume pi=(22/7) n=1.2 m1=15 meff=0.85 //mechanical efficinecy //CALCULATIONS rp=(p2/p1) m=ma/sp va=(m1*r*t1)/(p1*100) eff1=(1+c-c*(rp)^(1/n)) vs=va/eff1 d1=((4*vs)/pi)^(1/3) pr=((n/(n-1))*m1*r*t1*((rp)^((n-1)/n)-1))/60 prs=pr/meff d2=((prs*4)/(7*100*pi*700))^0.333 //RESULTS printf('volumetric efficiency is %2f',eff1) printf('\nlengh of the stroke is %2fm',d1) printf('\nindicated power is %2fkw',pr) printf('\npower required at the shaft of the compressor is %2fkw',prs) printf('\ndiameter of the piston is %2fm',d2)
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// exp 6 function f = fourier(x,n) f=0 for i=0:(n+1) f=f+((sin(2*(2*n+1)*%pi*x))/(2*n+1)) end f=f*(4/%pi) endfunction n = input('Entrer le nombre de termes : ') x=(-1/2):0.01:(1/2) plot2d(x, fourier(x,n), style=5) xtitle("Graphe série de Fourier avec n=" + string(n)) xlabel("x"); ylabel("f(x)")
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//Variable declaration: k = 9.1 //Thermal coductivity of steel rod (Btu/h.ft.°F) p = 0.29*1728 //Density of steel rod (lb/ft^3) Cp = 0.12 //Heat capacity of steel rod (Btu/lb.°F) P = 15+14.7 //Absolute pressure (psia) Ta = 71.0 //Initial temperature (°F) L = 20.0/12.0 //Length of rod (ft) t = 30.0/60.0 //Time taken (h) x = 0.875/12.0 //Length from one of end (ft) pi = %pi e = %e //From assumption: n = 1.0 //First term //From tables in Appendix: Ts = 249.7 //Saturated steam temperature (°F) //Calculation: a = k/(p*Cp) //Thermal diffusivity (ft^2/s) T = Ts+(Ta-Ts)*(((n+1)*(-1)**2 + 1 )/pi)*e**((-a*((n*pi)/L)**2)*t)*sin((n*pi*x)/L) //Temperature 0.875 inches from one of the ends after 30 minutes (°F) //Result: printf ("The temperature 0.875 inches from one of the ends after 30 minutes is : %.0f °F.",T)
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// Exa 8.10 clc; clear; close; // Given data f_x= 1000;// in Hz Y= 2;// points of tangency to vertical line X= 5;// points of tangency to horizontal line f_y= f_x*X/Y;// in Hz disp(f_y,"Frequency of vertical input in Hz")
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//chapter12 //example12.5 //page242 Vcc=12 // V gain_beta=100 Vbe=0.3 // V Ic=1 // mA // since gain_beta=Ic/Ib Ib=Ic/gain_beta // since Vcc=Ib*Rb+Vbe we get Rb=(Vcc-Vbe)/Ib gain_beta2=50 // since Vcc=Ib*Rb+Vbe we get Ib2=(Vcc-Vbe)/Rb Ic2=Ib2*gain_beta2 printf("for beta = 100, base resistor = %.3f kilo ohm \n",Rb) printf("for beta = 50, zero signal collector current for same Rb is = %.3f mA \n",Ic2)
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clc; clear; printf("\t\t\tChapter6_example1\n\n\n"); // Determination of the fluid outlet tetnperature and the tube-wall temperature at the outlet. // properties of ethylene glycol at 20 degree celsius from appendix table C5 Cp_20=2382; rou_20=1.116*1000; v_20=19.18e-6; kf_20=.249; a_20=.939e-7; Pr_20=204; // specifications of 1/2 standard type M seamless copper water tubing from appendix table F2 OD=1.588/100; ID=1.446/100; A=1.642e-4; Q=3.25e-6; V=Q/A; printf("\nThe average flow velocity is %.1f m/s",V*100); // calculation of Reynold's Number to check flow regime Re=V*ID/v_20; printf("\nThe Reynolds Number is %.1f",Re); // since Re>he 2100, the flow regime is laminar and the hydrodynamic length can be calculated as Z_h=0.05*ID*Re; printf("\nThe hydrodynamic length is %.1f cm",Z_h*100); Tbi=20; // bulk-fluid inlet temperature in degree celsius qw=2200; // incident heat flux in W/m^2 L=3; // Length of copper tube in m R=ID/2; // inner radius in m Tbo=Tbi+(2*qw*a_20*L)/(V*kf_20*R); printf("\nThe bulk-fluid outlet temperature is %.1f degree celsius",Tbo); // This result is based on fluid properties evaluated at 20°C. taken as a first approximation Z_t=0.05*ID*Re*Pr_20; printf("\nThe thermal entry length is %.1f m",Z_t); Two=Tbo+(11*qw*ID)/(48*kf_20); // The wall temperature at outlet in degree celsius printf("\nThe wall temperature at outlet is %.1f degree celsius",Two); //The result is based on first approximation based on flow properties evaluated at the fluid inlet temperature.
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//PI tuning x=0; function [temp,CO,et,setpoint] = pid(disturbance) global temp heat_in fan_in et SP CO eti et1 CObias x if x<=600; setpoint = 36; elseif (x>600 & x<=1200); setpoint = 40; elseif x>1200; setpoint = 31.25; end L1 = 10; tau1 = 90; k1 = 0.62; Ts = 0.4; R1 = k1/tau1; a1 = R1*L1; //Proportional controller //kp = 1/a1; kc = kp; //et = setpoint - temp; //CO = CObias + kc*et //PI controller kpi = 0.9/a1; taui = 3*L1; kc = kpi; et = setpoint - temp; eti = eti + et; CO = CObias + kc*(et + (eti*Ts)/taui); //PID controller //kpid = 1.2/a1; taui = 2*L1; taud = 5*L1; //kpid = 1.2/a1; taui = 1.2*L1; taud = 0.25*L1; //kpid = 1.2/a1; taui = 3.5*L1; taud = 0.05*L1; //kpid = 1.2/a1; taui = 3.5*L1; taud = 0.001*L1; //et = setpoint - temp; etd = et - et1; //eti = eti + et; kc = kpid; //CO = CObias + kc*(et + ((eti*Ts)/taui) + ((etd*taud)/Ts)); if CO>40 CO = 40; elseif CO<0 CO = 0; end; heat_in = CO; fan_in = disturbance; ok = writebincom(handl,[254]); //heater ok = writebincom(handl,[heat_in]); ok = writebincom(handl,[253]); ok = writebincom(handl,[fan_in]); ok = writebincom(handl,[255]); [temp3,ok,nbytes] = readbincom(handl,2); //upper byte sleep(1); ok = writebincom(handl,[255]); [temp4,ok,nbytes] = readbincom(handl,2); sleep(1); ok = writebincom(handl,[255]); [temp5,ok,nbytes] = readbincom(handl,2); temp6 = [temp3 temp4 temp5]; for i=1:6 if temp6(i) >10 temp1 = temp6(i); else temp2 = temp6(i); end end temp7 = temp; temp = temp1 + 0.1*temp2; // if temp < 10 // temp = temp7; // end; x=x+1; //disp(x) endfunction;
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function x=EliminaçãoGauss(A,b) [N N]=size(A); //Saída- x(N) com solução para A(N,N)*x(N)=b(N). C=[A b]; printf("Matriz Aumentada [C=A|b]") disp(C) for p=1:N-1 [max_linha_p,index]=max(abs(C(p:N,p))); // pivotamento if(index<>1) then printf("Trocando linhas %d e %d",p,index+p-1) C([ p, (index+p-1)],:) = C([ (index+p-1) , p ],:) //troca linhas disp(C) end pivot=C(p,p) if pivot==0 then printf("Não há solução única pois matriz A é singular") return end printf("Eliminando coluna %d com Pivô %f\n",p,pivot) for lin=p+1:N //eliminação progressiva printf("(L%d)=(L%d)-(%f)/(%f)*L(%d)",lin,lin,C(lin,p),pivot,p) m=C(lin,p)/pivot; C(lin,p:N+1)=C(lin,p:N+1)-m*C(p,p:N+1); if lin<N then printf("\n") end end disp(C) end printf("Substituição regressiva"); x(N)=C(N,N+1)/C(N,N); for lin=N-1:-1:1 x(lin)=(C(lin,N+1)-C(lin,lin+1:N)*x(lin+1:N))/C(lin,lin); end disp(x) endfunction function x=EliminaçãoGauss1(A,b) //Saída- x(N) com solução para A(N,N)*x(N)=b(N). [N N]=size(A); C=[A b]; printf("Matriz Aumentada [C=A|b]") disp(C) for p=1:N-1 pivot=C(p,p) if pivot==0 then printf("Não é possível continuar eliminação pois pivot=0") return end printf("Eliminando coluna %d com Pivô %f\n",p,pivot) for lin=p+1:N //eliminação progressiva printf("(L%d)=(L%d)-(%f)/(%f)*L(%d)",lin,lin,C(lin,p),pivot,p) m=C(lin,p)/pivot; C(lin,p:N+1)=C(lin,p:N+1)-m*C(p,p:N+1); if lin<N then printf("\n") end end disp(C) end printf("Substituição regressiva"); x(N)=C(N,N+1)/C(N,N); for lin=N-1:-1:1 x(lin)=(C(lin,N+1)-C(lin,lin+1:N)*x(lin+1:N))/C(lin,lin); end disp(x) endfunction function x=EliminaçãoGauss2(A,b) [N N]=size(A); C=[A b]; for p=1:N-1 pivot=C(p,p) if pivot==0 then printf("Não é possível continuar eliminação pois pivot=0") return end for lin=p+1:N m=C(lin,p)/pivot; C(lin,p:N+1)=C(lin,p:N+1)-m*C(p,p:N+1); end end x(N)=C(N,N+1)/C(N,N); for lin=N-1:-1:1 x(lin)=(C(lin,N+1)-C(lin,lin+1:N)*x(lin+1:N))/C(lin,lin); end endfunction
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clear all; close; clf(); //----------------------------------------------------------------------------- s = chdir('C:\Users\work\OneDrive\Documents\SciLab\lab_v6') exec('CLIP_F.sce') exec('DIST_F.sci') // Our recorded IRC [signal, Fs, s_b] = wavread("C:\Users\work\OneDrive\Documents\SciLab\lab_v6\guitar.wav"); signal = signal(1, :) // Before applying filter res frequinces = (0:length(signal)-1)/length(signal) * Fs; figure(0) subplot(3,1,1) plot(signal) xlabel("Time", 'fontsize', 2) ylabel("Amplitude", 'fontsize', 2) title("Time domain original signal", 'fontsize', 3) subplot(3,1,2) plot2d("nl", frequinces, abs(fft(signal)),2) xlabel("Frequency, Hz", 'fontsize', 2) ylabel("Freq amplitude", 'fontsize', 2) title("Frequency response of signal", 'fontsize', 3) subplot(3,1,3) s = abs(fft(signal)) s(s>0.1) = 0 plot2d("nl", frequinces, s,2) xlabel("Frequency, Hz", 'fontsize', 2) ylabel("Freq amplitude", 'fontsize', 2) title("Frequency response of signal with treshold 0.1", 'fontsize', 3) //Applying CLIP filter signal_clip = CLIP_F(signal, 0.1) frequinces = (0:length(signal_clip)-1)/length(signal_clip) * Fs; figure(1) subplot(3,1,1) plot(signal_clip) xlabel("Time", 'fontsize', 2) ylabel("Amplitude", 'fontsize', 2) title("Time domain clipped signal", 'fontsize', 3) subplot(3,1,2) plot2d("nl", frequinces, abs(fft(signal_clip)), 2) xlabel("Frequency, Hz", 'fontsize', 2) ylabel("Freq amplitude", 'fontsize', 2) title("Frequency response of signal", 'fontsize', 3) subplot(3,1,3) s = abs(fft(signal_clip)) s(s>0.1)=0 plot2d("nl", frequinces, s, 2) xlabel("Frequency, Hz", 'fontsize', 2) ylabel("Freq amplitude", 'fontsize', 2) title("Frequency response of signal", 'fontsize', 3) //Play clipped sound savewave('clipped.wav', signal_clip, Fs) //Applying DISTORTION filter signal_dist = DIST_F(signal, 3, 5) frequinces = (0:length(signal_dist)-1)/length(signal_dist) * Fs; figure(2) subplot(3,1,1) plot(signal_dist) xlabel("Time", 'fontsize', 2) ylabel("Amplitude", 'fontsize', 2) title("Time domain distortion effect", 'fontsize', 3) subplot(3,1,2) plot2d("nl", frequinces, abs(fft(signal_dist)), 2) xlabel("Frequency, Hz", 'fontsize', 2) ylabel("Freq amplitude", 'fontsize', 2) title("Frequency response of signal", 'fontsize', 3) subplot(3,1,3) s = abs(fft(signal_dist)) s(s>3) = 0 plot2d("nl", frequinces, s, 2) xlabel("Frequency, Hz", 'fontsize', 2) ylabel("Freq amplitude", 'fontsize', 2) title("Frequency response of signal", 'fontsize', 3) //Play disted sound savewave('distortion.wav', signal_dist, Fs)
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ex4_2.sce
errcatch(-1,"stop");mode(2); disp('the probability of getting an even no. 1/2=') 1/2 exit();
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19_6.sce
clear; clc; V=220e3; Zl=complex(.8, .2); Xline=.2; Xt=.05; Sb=100e6; Vb=220e3; v=V/Vb; X=Xline+ Xt; I=conj(Zl/v) phi1=atand(imag(I)/real(I)) Vbus=1+ I * X*exp(%i * %pi/2) phi2=atand(imag(Vbus)/real(Vbus)) vbus=abs(Vbus) vbus=round(vbus *1000)/1000 vbus=vbus*Vb*1e-3; pf=cosd(-phi1+phi2) mprintf("Voltage at bus = %.2f Kv, pf= %.3f lagging", vbus, pf)
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Ex14_23_29.sce
//Section-14,Example-1,Page no.-PC.128 clc; Mo_SA=1*(0.3154)*(0.02192) disp(Mo_SA,'Moles of salicylic acid') Ma_SA=2.968*10^-3*138.12 disp(Ma_SA,'Mass of salicylic acid') pr_SA=(0.4099/0.4208)*100 disp(pr_SA,'Percentage by weight of sample that is salicylic acid')
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function pythonEnd() call("py_end") endfunction
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Ex7_21.sce
clc //initialization of new variables clear a=30 //cm t=2.5 //cm S=15 //cm s=5 //Tonne // calculations I=a*a^3-25*25^3 I=I/12 tau_zx=s*1000*27.5*t*25/(4*35000*t) FA=S*t*tau_zx tau_xy=s*1000*a*t*27.5/(4*35000*t) FB=tau_xy*t*S //Results printf('case A \n F = %d kg',FA) printf('\n case B \n F= %d kg',FB)
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set_infos.sci
function set_infos(t,level) global('m2sci_infos') select level case 0 then txt=[txt;' ';'// '+t] case 1 then m2sci_infos(level)=%t txt=[txt;' ';'//! '+t] case 2 then m2sci_infos(level)=%t if logfile>0 then write(logfile,t,'(a)'),end txt=[txt;' ';'//!! '+t] end txt=resume(txt)
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ex13_17.sce
clc; v=120; //potential diff in volt r=240; //resistance in ohm i=v/r; //current in Ampere using Ohm's law disp(i,"(a)Current in each bulb in Ampere = "); //displaying result p=i*i*r; //power in Watt disp(p,"Power dissipated in each bulb in Watt = "); //displaying result
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function [F,G,ind] = OraclePG(qc,ind) //AdTInv = inv(AdT); //B = [-AdTInv*AdC;eye(n-md,n-md)] //q0 = [AdTInv*fd;zeros(n-md,1)] v = q0+B*qc; u = r.*v.*abs(v); if ind==2 then F = 1/3*u'*v+pr'*(Ar*v); G = %nan; elseif ind==3 then G = B'*(r.*v.*abs(v))+(Ar*B)'*pr; F = %nan; elseif ind==4 then F = 1/3*u'*v+pr'*(Ar*v); G = B'*(r.*v.*abs(v))+(Ar*B)'*pr; else F = %nan; G = %nan; end endfunction
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Ex3_5.sce
clear //Given m=9*10**9 q=4*10**-6 //Calculation V=2*q*m //Result printf("\n Electric potential is %0.3f *10**3 V", V*10**-3)
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Ex3_17.sce
//example-3.17 //page no-101 //given //atomic radii of polonium,rhodium and chromium are rPo=1.7*10^(-10) //m rRh=1.34*10^(-10) //m rCr=1.25*10^(-10) //m //latice strucrure of polonium , rhodiun and chromium are SC, FCC and BCC resp //so lattice constants are aPo=2*rPo //m aRh=2*sqrt(2)*rRh //m aCr=4/sqrt(3)*rCr //m //planer density on (100) in polonium is given by rhoPo=1/(aPo)^2 //per m^2 //planer denity on (110) in rhodium rhoRh=sqrt(2)/(aRh)^2 //per m^2 //planer density on (111) in chromium rhoCr=sqrt(3)/(aCr)^2 //per m^2 printf ("the planer density of polonium in (100) is %f ,rhodium in (110) is %f per m^2 and chromium in (111) is %e per m^2",rhoPo,rhoRh,rhoCr)
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ex8_4.sce
//Example 8.4, page no-508 clear clc //(a) T=200 T0=300 Ti=70 t=3 x=(T-T0)/(Ti-T0) tow=-t/log(x) printf("(a)\nTime constant tow=%.1f s",tow) //(b) t1=5 T5=T0+((Ti-T0)*%e^(-t1/tow)) printf("\n(b)\nTemperature after 5 seconds T5 = %.2f°C",T5)
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Ex19_1.sce
clc; //e.g 19.1 Vcc=10; Rc=10*10**3; Rb=1*10**6; beta=100; Vbe=0.7; Ib=(Vcc-Vbe)/Rb; disp('microA',Ib*10**6,"Ib="); Ic=beta*Ib; disp('mA',Ic*10**3,"Ic="); Ie=Ic; re=25/(Ie*10**3); disp('ohm',re*1,"re="); Ri=beta*re; disp('kohm',Ri*10**-3,"Ri="); Ris=(Rb*beta*re)/(Rb+beta*re); disp('kohm',Ris*10**-3,"Ris="); R0=Rc; disp('kOhm',R0*10**-3,"R0="); Av=Rc/re; disp(Av);
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style.fontSize=14; style.displayedLabel="<b>Digital In</b><br>%1$s"; pal1_2 = xcosPalAddBlock(pal1_2,"gpio_in",[],style); pal6 = xcosPalAddBlock(pal6,"gpio_in",[],style);
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clc //initialization of variables clear L=2.5 //m A=6.02 //cm^2 Q1=105 s=796.5 //kg/cm^2 // calculations P=2*A*s printf('The safe load is %d kg',P) // Results // wrong calculations in the text
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# y^4 = m^2 + n^2 ? 0 0 1 1 2 2^4 3 3^4 4 2^8 5 5^4 = (7)^2 + (2^3*3)^2 proper representation by 7^2 + 24^2 = (3*5)^2 + (2^2*5)^2 [2] 6 2^4*3^4 7 7^4 8 2^12 9 3^8 10 2^4*5^4 = (2^2*7)^2 + (2^5*3)^2 = (2^2*3*5)^2 + (2^4*5)^2 [2] 11 11^4 12 2^8*3^4 13 13^4 = (5*13)^2 + (2^2*3*13)^2 = (7*17)^2 + (2^3*3*5)^2 proper representation by 119^2 + 120^2 [2] 14 2^4*7^4 15 3^4*5^4 = (3^2*7)^2 + (2^3*3^3)^2 = (3^3*5)^2 + (2^2*3^2*5)^2 [2] 16 2^16 17 17^4 = (2^3*17)^2 + (3*5*17)^2 = (7*23)^2 + (2^4*3*5)^2 proper representation by 161^2 + 240^2 [2] 18 2^4*3^8 19 19^4 20 2^8*5^4 = (2^4*7)^2 + (2^7*3)^2 = (2^4*3*5)^2 + (2^6*5)^2 [2] 21 3^4*7^4 22 2^4*11^4 23 23^4 24 2^12*3^4 25 5^8 = (5^2*7)^2 + (2^3*3*5^2)^2 = (2^2*5*11)^2 + (3^2*5*13)^2 = (2^4*3*7)^2 + (17*31)^2 proper representation by 336^2 + 527^2 = (3*5^3)^2 + (2^2*5^3)^2 [4] 26 2^4*13^4 = (2^2*5*13)^2 + (2^4*3*13)^2 = (2^2*7*17)^2 + (2^5*3*5)^2 [2] 27 3^12 28 2^8*7^4 29 29^4 = (41)^2 + (2^3*3*5*7)^2 proper representation by 41^2 + 840^2 = (2^2*5*29)^2 + (3*7*29)^2 [2] 30 2^4*3^4*5^4 = (2^2*3^2*7)^2 + (2^5*3^3)^2 = (2^2*3^3*5)^2 + (2^4*3^2*5)^2 [2] 31 31^4 32 2^20 33 3^4*11^4 34 2^4*17^4 = (2^5*17)^2 + (2^2*3*5*17)^2 = (2^2*7*23)^2 + (2^6*3*5)^2 [2] 35 5^4*7^4 = (7^3)^2 + (2^3*3*7^2)^2 = (3*5*7^2)^2 + (2^2*5*7^2)^2 [2] 36 2^8*3^8 37 37^4 = (2^2*3*37)^2 + (5*7*37)^2 = (2^3*3*5*7)^2 + (23*47)^2 proper representation by 840^2 + 1081^2 [2] 38 2^4*19^4 39 3^4*13^4 = (3^2*5*13)^2 + (2^2*3^3*13)^2 = (3^2*7*17)^2 + (2^3*3^3*5)^2 [2] 40 2^12*5^4 = (2^6*7)^2 + (2^9*3)^2 = (2^6*3*5)^2 + (2^8*5)^2 [2] 41 41^4 = (3^2*41)^2 + (2^3*5*41)^2 = (2^4*3^2*5)^2 + (7^2*31)^2 proper representation by 720^2 + 1519^2 [2] 42 2^4*3^4*7^4 43 43^4 44 2^8*11^4 45 3^8*5^4 = (3^4*7)^2 + (2^3*3^5)^2 = (3^5*5)^2 + (2^2*3^4*5)^2 [2] 46 2^4*23^4 47 47^4 48 2^16*3^4 49 7^8 50 2^4*5^8 = (2^2*5^2*7)^2 + (2^5*3*5^2)^2 = (2^4*5*11)^2 + (2^2*3^2*5*13)^2 = (2^6*3*7)^2 + (2^2*17*31)^2 = (2^2*3*5^3)^2 + (2^4*5^3)^2 [4] 51 3^4*17^4 = (2^3*3^2*17)^2 + (3^3*5*17)^2 = (3^2*7*23)^2 + (2^4*3^3*5)^2 [2] 52 2^8*13^4 = (2^4*5*13)^2 + (2^6*3*13)^2 = (2^4*7*17)^2 + (2^7*3*5)^2 [2] 53 53^4 = (17*73)^2 + (2^3*3^2*5*7)^2 proper representation by 1241^2 + 2520^2 = (2^2*7*53)^2 + (3^2*5*53)^2 [2] 54 2^4*3^12 55 5^4*11^4 = (7*11^2)^2 + (2^3*3*11^2)^2 = (3*5*11^2)^2 + (2^2*5*11^2)^2 [2] 56 2^12*7^4 57 3^4*19^4 58 2^4*29^4 = (2^2*41)^2 + (2^5*3*5*7)^2 = (2^4*5*29)^2 + (2^2*3*7*29)^2 [2] 59 59^4 60 2^8*3^4*5^4 = (2^4*3^2*7)^2 + (2^7*3^3)^2 = (2^4*3^3*5)^2 + (2^6*3^2*5)^2 [2] 61 61^4 = (11*61)^2 + (2^2*3*5*61)^2 = (2^3*3*5*11)^2 + (7^2*71)^2 proper representation by 1320^2 + 3479^2 [2] 62 2^4*31^4 63 3^8*7^4
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clear ; clc ; close ; // CTS Signal A =1; // Amplitude Dt = 0.005; T1 = 2; //Time in seconds t = 0: Dt:T1 /2; for i = 1: length (t) xt(i) = A; end // Continuous time Fourier Transform Wmax= 2*%pi*1; // Analog Frequency = 1Hz K =4; k=0:(K/1000):K; W =k*Wmax/K; xt=xt'; XW =xt*exp(-sqrt(-1)*t'*W)*Dt; XW_Mag =real(XW); W =[-mtlb_fliplr(W), W(2:1001)]; // Omega from Wmax to Wmax XW_Mag =[mtlb_fliplr( XW_Mag ), XW_Mag(2:1001)]; // displaying the given function subplot(2 ,1 ,1); a =gca(); a.data_bounds =[ -4 ,0;4 ,2]; a.y_location ="origin"; plot(t,xt); xlabel('t in msec .'); title(' Contiuous Time Signal x(t) {Gate Function} ') // displaying the fourier Transform of the given function subplot(2 ,1 ,2); a=gca(); a.y_location ="origin"; plot(W, XW_Mag); xlabel('Frequency in Radians / Seconds '); title('Continuous time Fourier Transform X(jW)' ) mprintf('Hence Fourier transform of given Gate function is:\n A*delta*Sa[w*delta/2]/ exp(-j*w*delta/2)')
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E = 2*10^7; //in V/m m0 = 0.91 * 10^-30; //in kg q = 1.6*10^-19; h = 1.05*10^-34; //in J.s m1 = 0.065*m0; //for GaAs m2 = 0.02*m0; // for InAs E1 = 1.5; //in eV E2 = 0.4; //in eV p1 = -4*(2*m1)^0.5*(E1*q)^1.5/(3*q*h*E); disp(p1,"Tunneling probability is exponent to the power") tp1 = %e^p1; disp(tp1,"Tunneling probability = ") p2 = -4*(2*m2)^0.5*(E2*q)^1.5/(3*q*h*E); disp(p2,"Tunneling probability is exponent to the power") tp2 = %e^p2; disp(tp2,"Tunneling probability = ") disp("In InAs the band-to-band tunneling will start becoming very important if the field is ∼ 2 × 105 V/cm.")
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clear; clc; printf("\nEx-6.24\n"); //page no.-195 //given rho=1.73*10^-8;......//resistivity in ohm m M=63.5;...........//atomic weight d=8.92*10^3;......//density in Kg/m^3 N=6.023*10^23;......//avagadro no. e=1.6*10^-19;.......//charge m=9.11*10^-31;......//mass of e no=(N*d)/M........//no of electrons per unit volume printf("\nno. of electrons/ unit volume 8.463*10^25 /m^3\n"); mu=1/(rho*no*e).........//mobility printf("\nmobility is 4.1145 m^2/Vs\n"); tau=m/(no*e^2*rho)..........//relaxation time printf("\nrelaxation time is 2.25*10^-11 s");
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errcatch(-1,"stop");mode(2);//Example 8_12 ; ; //To find the fundamental frequency l=3*10^-3 //units in meters d=3.5*10^3 //units in kg/m^3 Y=8*10^10 //units in N/m^2 v=1/(2*l)*sqrt(Y/d) v=v*10^-6 //units in Hz printf("Fundamental Frequency v=%.3f Hz",v) exit();
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clc; T2=15+273; T1=800+273; eta=1-(T2/T1); p4=210;//bar p2=1;//bar R=0.218; sA_s4=R*log(p4/p2); cp=1.005; sA_s2=cp*log(T1/T2); output=(T1-T2)*(sA_s4-sA_s2); W41=T1*(sA_s4-sA_s2); cv=0.718; W21=cv*(T1-T2); gross=W41+W21; disp(W41) work=output/gross; disp("work ratio is"); disp(work)
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//Superheated water vapor enters a valve at 3.0 MPa, 320C and exits at a pressure of 0.5 MPa. The expansion is a throttling process. Determine the specific flow exergy at the inlet and exit and the exergy destruction per unit of mass flowing, each in kJ/kg. Let T0 = 25C, p0=  1 atm. //solution //variable initialization p1 = 3 //entry pressure in Mpa p2 = .5 //exit pressure in Mpa T1 = 320 //entry temperature in degree celcius T0 = 25 //in degree celcius p0 = 1 //in atm //from table A-4 h1 = 3043.4 //in kj/kg s1 = 6.6245 //in kj/kg.k h2 = h1 //from reduction of the steady-state mass and energy rate balances s2 = 7.4223 //Interpolating at a pressure of 0.5 MPa with h2 = h1, units in kj/kg.k //from table A-2 h0 = 104.89 //in kj/kg s0 = 0.3674 //in kj/kg.k ef1 = h1-h0-(T0+273)*(s1-s0) //flow exergy at the inlet ef2 = h2-h0-(T0+273)*(s2-s0) //flow exergy at the exit //from the steady-state form of the exergy rate balance Ed = ef1-ef2 //the exergy destruction per unit of mass flowing is printf(' the specific flow exergy at the inlet in kj/kg is :\n\t ef1 =%f',ef1) printf('\nthe specific flow exergy at the exit in kj/kg is:\n\t ef2 = %f', ef2) printf('\nthe exergy destruction per unit of mass flowing in kj/kg is:\n\t = %f',Ed)
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//y=x^2-5*x+5 clear; clc; clf; close; x=linspace(-2,7,10); y=x^2-5*x+5; plot2d(x,y,3); plot2d3(x,y,7); x=poly(0,'x'); y=x^2-5*x+5; x=roots(y) for x=0:5 for y=5:20 plot(x,y,'r.pentagram'); //y>0 region end end xtitle("Using quadratic inequalities to describe regions","x axis","y axis"); xgrid(); legend("y=x^2-5*x+5","y<x^2-5*x+5 region","y>x^2-5*x+5 region",4);
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//Example 5_5 clc; clear; close; format('v',6); //Part(a) Derivation //Part(b) //given data : mu_p=500;//cm^2/V-s q=1.6*10^-19;//C/electron rho=3;//ohm-cm V0=0.4;//V//Barrier Height Vd=4.5;//V//Reverse Voltage D=40;//mils D=D*10^-3;//inch D=D*2.54;//cm/in A=%pi/4*D^2;//cm^2 NA=1/rho/mu_p/q;//cm^-3 W=sqrt((V0+Vd)/(14.13*10^10));//m^2 Vj=V0+Vd;//V CT=2.9*10^-4*sqrt(NA/Vj)*A;///pF disp(CT,"CT(pF) : "); //Answer given in the textbook is not accurate.
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//laplace// s=%s; G=syslin('c',(5*(s+2))/((s+3)*(s+4))); disp(G,"G(s)=") x=denom(G); disp(x,"Characteristics Polynomial=") y=roots(x); disp(y,"Poles of a system=")
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clear; clc; disp("--------------Example 11.11---------------") // example explaination printf("This example shows an exchange in which a frame is lost. The sequence of events that occur is as follows :\n\n* Node B sends three data frames (0, 1,and 2), but frame 1 is lost.\n\n* When node A receives frame 2, it discards it and sends a REJ frame for frame 1 since the protocol being used is Go-Back-N\nwith the special use of an REJ frame as a NAK frame.\n\n* The NAK frame does two things here: It confirms the receipt of frame 0 and declares that frame 1 and any following frames must be resent.\n\n* Node B, after receiving the REJ frame, resends frames 1 and 2.\n\n* Node A acknowledges the receipt by sending an RR frame (ACK) with acknowledgment number 3.");
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Example1.sce
clear pwd curr=ans flag=1 mode(-1) clc disp("INSTRUCTIONS : ") printf("\nHere all instructions are preloaded in the form of a demo\n\nInitially the whole perl script is displaying and then \n the result of the same can be seen in the command line interpreter.\nPLEASE MAKE SURE THAT THE PERLSCRIPT INTERPRETER\nEXISTS IN THE SYSTEM\nOR THE COMMAND WOULD NOT WORK \n\n\nPRESS ENTER AFTER EACH COMMAND to see its RESULT\nPRESS ENTER AFTER EACH RESULT TO GO TO THE NEXT COMMAND\n") halt('.............Press [ENTER] to continue.....') halt("") clc printf("\tUNIX SHELL SIMULATOR(DEMO VERSION WITH PRELOADED COMMANDS)\n\n\n") printf("\n# Enter the name of the perlscript file whichever you desire \n\n") nam=input('$ cat ','s') halt(' ') clc li(1)='#!/usr/bin/perl' li(2)='# Script: '+nam+'.pl - Shows the use of variables' li(3)="#" li(4)='print('+ascii(34)+'Enter your name: '+ascii(34)+') ;' li(5)='$name = <STDIN> ; #Input from the keyboard' li(6)='print('+ascii(34)+'Enter a temperature in Centigrade: '+ascii(34)+') ;' li(7)='$centigrade=<STDIN>; #Whitespace unimportant' li(8)='$fahrenheit=$centigrade*9/5 + 32 ; #Here too ' li(9)='print '+ascii(34)+'The temperature $name in Fahrenheit is $fahrenheit\n'+ascii(34)+' ;' li(10)='print('+ascii(34)+'\n\nType exit to go back to console\n\n'+ascii(34)+')' halt(' ') v=mopen(nam+'.pl','wt') for i=1:10 mfprintf(v,"%s\n",li(i)) if i~=10 then printf("%s\n",li(i)) end end mclose(v) if getos()=='Linux' then printf("\n\nPlease open a new terminal window and then go to the directory %s and execute the following instruction\n\nperl %s.pl [Command line parameters if any]\n\nThank You \n\n",curr,nam) halt(' ') exit end printf("\n# type the following command in the command line interpreter as soon as it appears") printf(" \n %c perl %s.pl %c[ENTER]\n\n",ascii(34),nam,ascii(34)) printf("\n$ perl %s.pl #to execute the perlscript",nam) halt(' ') dos('start') printf("\n\n\n") halt(' ---------------->Executing PerlScript in Command Line Prompt<-------------- ') printf("\n\n\n$ exit #To exit the current simulation terminal and return to Scilab console\n\n") halt("........# (hit [ENTER] for result)") //clc() printf("\n\n\t\t\tBACK TO SCILAB CONSOLE...\nLoading initial environment') sleep(1000) mdelete(nam+'.pl')
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function erro = eqm(y,d) [Input_Size,p]=size(y); erro=0; for k=1:p erro=erro+(d(k)-y(k)).^2; end erro=erro/p; endfunction
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function result=dialog(labels,valueini) //interface to x_dialog primitive to allow simple overloading for live demo // Copyright INRIA result=x_dialog(labels,valueini)
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//Calculation of Strain-Hardening Exponent clear; clc; printf("\tExample 6.5\n"); sig_t=415; //True stress in MPa et=0.1; //True strain K=1035; // In MPa n=log(sig_t/K)/log(et); printf("\nStrain - hardening coefficient is %.2f",n); //End
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//Example 1// Ch 12 clc; clear; close; // given data r1=2;//inner coaxial cable radius r2=5;//sheath radius over the insulation Em1=40;//max stress in the insulation in kV/cm Em2=25;//max stress in the insulation in kV/cm epsilon1=6; epsilon2=4; x=Em1/Em2; r=x*((epsilon1*r1)/(epsilon2));//radial thickness of the dielectric printf("radial thickness of the dielectric %f cm",r) inner=r-r1;//inner thickness of dielectric outer=r2-r;//outer thickness of dielectric printf("inner thickness of dielectric %f cm",inner) printf("outer thickness of dielectric %f cm",outer) V1=Em1*r1*log(r/r1);//voltage drop across dielectric in kV V2=Em2*r*log(r2/r);//voltage drop across outer dielectric printf("voltage drop across dielectric %f kV",V1) printf("voltage drop across outer dielectric %f kV",V2) pv = V1+V2;//peak voltage of cable printf("peak voltage of cable %f kV",pv) pvrms=pv/sqrt(2); printf("peak voltage in rms %f kV",pvrms)
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A=[-6,3;3,-10.5] //Matrix of I1,I2 Coeffecients by Mesh analysis B=[-12.5;0]; [I]=inv(A)*B; disp("Amperes",I(1,1),"Current in 1 Ohm resistor");
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//=========================================================================== //chapter 3 example 24 clc;clear all; //variable declaration x1 = 1.570; //voltage in V x2 = 1.597; //voltage in V x3 = 1.591; //voltage in V x4 =1.562; //voltage in V x5 =1.577; //voltage in V x6 = 1.580; //voltage in V x7 = 1.564; //voltage in V x8 = 1.586; //voltage in V x9 = 1.550; //voltage in V x10 = 1.575; //voltage in V n =10; //ccalculations x =(x1+x2+x3+x4+x5+x6+x7+x8+x9+x10)/(10); //arthimetic mean d1 =x1-x; //deviation d2 =x2-x; //deviation d3 =x3-x; //deviation d4 =x4-x; //deviation d5 =x5-x; //deviation d6 =x6-x; //deviation d7 =x7-x; //deviation d8 =x8-x; //deviation d9 =x9-x; //deviation d10 =x10-x; //deviation D =(abs(d1)+abs(d2)+abs(d3)+abs(d4)+abs(d5)+abs(d6)+abs(d7)+abs(d8)+abs(d9)+abs(d10))/(n); d = ((d1^2)+(d2^2)+(d3^2)+(d4^2)+(d5^2)+(d6^2)+(d7^2)+(d8^2)+(d9^2)+(d10^2)); sigma = sqrt(d/(n-1)); //standard devation r = 0.6745*sigma; //probable error of one reading v = sigma^2; rm = r/(sqrt(n-1)); //probable error of mean in V //result mprintf("arthimetic mean = %3.3f",x); mprintf("\naverage deviation = %3.3f gramme",D); mprintf("\nstandard deviation = %3.5f gramme*2",sigma); mprintf("\nprobable error of one reading = %3.5f gramme",r); mprintf("\n variance= %3.3e gramme^2",v); mprintf("\nprobable error of mean = %3.4f gramme",rm);
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// Multlayer Perceptron (backpropagatin com gradiente decrescente) // Usando as funcoes internas do Scilab // Simulação do Procesamento de polimeros // Autor: Carlos Affonso ; Renato Sassi ; Ricardo Ferreira // Data: 05/10/2010 // // X = Vetor de entrada // d = saida desejada (escalar) // W = Matriz de pesos Entrada -> Camada Oculta // M = Matriz de Pesos Camada Oculta -> Camada saida // eta = taxa de aprendizagem // alfa = fator de momento clear; clc; //===================================================================== // Dados de entrada //===================================================================== loadmatfile('-ascii','Polymer_dados.txt','f'); loadmatfile('-ascii','Polymer_alvos.txt'); dados = Polymer_dados; // Vetores de entrada alvos = Polymer_alvos; // Saidas desejadas correspondentes dados=dados' alvos=alvos' // Número de nós da camada de saída No=1 // Dimensão dos dados de entrada [LinD,ColD] = size(dados); //==================================================================== // Embaralha vetores de entrada e saidas desejadas // Normaliza componentes para media zero e variancia unitaria mi = mean(dados,2); // Media das ao longo das colunas di = stdev(dados,2); // desvio-padrao das colunas for j = 1:ColD dados(:,j) = (dados(:,j)-mi)./di; end; Dn = dados; // Define tamanho dos conjuntos de treinamento/teste (hold out) ptrn = 0.8; // Porcentagem usada para treino ptst = 1-ptrn; // Porcentagem usada para teste J = floor(ptrn*ColD); // Vetores para treinamento e saidas desejadas correspondentes P = Dn(:,1:J); T1 = alvos(:,1:J); [lP,cP] = size(P); // Tamanho da matriz de vetores de treinamento // Vetores para teste e saidas desejadas correspondentes Q = Dn(:,J+1:$); T2 = alvos(:,J+1:$); [lQ,cQ] = size(Q); // Tamanho da matriz de vetores de teste // DEFINE ARQUITETURA DA REDE //=========================== Ne = 500; // No. de epocas de treinamento Nr = 1; // No. de rodadas de treinamento/teste Nh = 8; // No. de neuronios na camada oculta eta = 0.01; // Passo de aprendizagem mom = 0.75; // Fator de momento for r=1:Nr, // Inicio do loop de rodadas de treinamento rodada=r, // Inicia matrizes de pesos WW = 0.1*(2*rand(Nh,lP+1)-1); // Pesos entrada -> camada oculta WW_old = WW; // Necessario para termo de momento MM = 0.1*(2*rand(No,Nh+1)-1); // Pesos camada oculta -> camada de saida MM_old = MM; // Necessario para termo de momento // ETAPA DE TREINAMENTO for t = 1:Ne, Epoca = t; [s,I]=gsort(rand(1,cP)); //I é uma permutação randômica de 1:ColD P = P(:,I); T1 = T1(:,I); // Embaralha vetores de treinamento e saidas desejadas EQ = 0; for tt = 1:cP, // Inicia LOOP de epocas de treinamento // CAMADA OCULTA X = [-1; P(:,tt)]; // Constroi vetor de entrada com adicao da entrada x0=-1 Ui = WW*X; // Ativacao (net) dos neuronios da camada oculta Yi = tanh(Ui); // Saida entre [-1,1] (função tanh) // CAMADA DE SAIDA Y = [-1;Yi]; // Constroi vetor de entrada DESTA CAMADA Uk = MM*Y; // Ativacao (net) dos neuronios da camada de saida Ok = tanh(Uk); // Saida entre [-1,1] (função logistica) // CALCULO DO ERRO Ek = T1(:,tt)-Ok; // erro entre a saida desejada e a saida da rede EQ = EQ + 0.5*sum(Ek^2); // soma do erro quadratico de todos os neuronios // CALCULO DOS GRADIENTES LOCAIS Dk = 0.5*(1-Ok^2); // derivada da sigmoide logistica (camada de saida) DDk = Ek.*Dk; // gradiente local (camada de saida) Di = 0.5*(1-Yi^2); // derivada da sigmoide logistica (camada oculta) DDi = Di.*(MM(:,2:$)'*DDk); // gradiente local (camada oculta) // AJUSTE DOS PESOS - CAMADA DE SAIDA MM_aux = MM; MM = MM + eta*DDk*Y' + mom*(MM-MM_old); MM_old = MM_aux; // AJUSTE DOS PESOS - CAMADA OCULTA WW_aux = WW; WW = WW + eta*DDi*X' + mom*(WW-WW_old); WW_old = WW_aux; end; // Fim do loop de uma epoca EQM(r,t) = EQ/cP; // MEDIA DO ERRO QUADRATICO P/ EPOCA end; // Fim do loop de treinamento // ETAPA DE GENERALIZACAO %%% EQ2=0; OUT2=[]; SAIDA=[]; for tt = 1:cQ, // Inicia LOOP de epocas de treinamento // CAMADA OCULTA X = [-1; Q(:,tt)]; // Constroi vetor de entrada com adicao da entrada x0=-1 Ui = WW*X; // Ativacao (net) dos neuronios da camada oculta Yi = tanh(Ui); // Saida entre [-1,1] (funcao logistica) // CAMADA DE SAIDA Y = [-1;Yi]; // Constroi vetor de entrada DESTA CAMADA Uk = MM*Y; // Ativacao (net) dos neuronios da camada de saida Ok = tanh(Uk); // Saida entre [-1,1] (funcao logistica) OUT2=[OUT2 Ok]; // Armazena saida da rede Ek = T2(:,tt)-Ok; // erro entre a saida desejada e a saida da rede EQ2 = EQ2 + 0.5*sum(Ek^2); // soma do erro quadratico de todos os neuronios SAIDA=[SAIDA; norm(Ek) T2(:,tt) Ok]; end; // Fim do loop de uma epoca EQM2(r)=EQ2/cQ; // MEDIA DO ERRO QUADRATICO COM REDE TREINADA end // Fim do loop de rodadas de treinamento // CALCULA ACERTO EQM_media=mean(EQM,1); // Curva de aprendizagem media (p/ Nr realizacoes) //plot(EQM_media); // Plota curva de aprendizagem // SALVA PESOS E SAÍDA savematfile('pesos.dat','WW','-ascii'); // RODAR A REDE COM OS PESOS SINAPTICOS OUT3=[]; for tt = 1:ColD, // Inicia LOOP de epocas de treinamento // CAMADA OCULTA X = [-1; dados(:,tt)]; // Constroi vetor de entrada com adicao da entrada x0=-1 Ui = WW*X; // Ativacao (net) dos neuronios da camada oculta Yi = tanh(Ui); // Saida entre [-1,1] (função tanh) // CAMADA DE SAIDA Y = [-1;Yi]; // Constroi vetor de entrada DESTA CAMADA Uk = MM*Y; // Ativacao (net) dos neuronios da camada de saida Ok = tanh(Uk); // Saida entre [-1,1] (função logistica) OUT3=[OUT3 Ok]; // Armazena saida da rede // PLOTAR SAIDAS plot(alvos) plot(OUT3,'r--d') end
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Xn = input("Enter the value of Xn "); i_x= input("Enter the range of Xn "); Hn = input("Enter the value of Hn "); i_h= input("Enter the range of Hn "); exec("E:\scilab\work_17BIT009\exp_6_fun.sce"); [Yn,In] = exp_6_fun(Xn,i_x,Hn,i_h); figure(1); subplot(3,1,1); plot2d3(i_x,Xn); xlabel('n'); ylabel('Xn'); title("Graph of Xn"); subplot(3,1,2); plot2d3(i_h,Hn); xlabel('n'); ylabel('Hn'); title("Graph of Yn"); subplot(3,1,3); plot2d3(In,Yn); xlabel('n'); ylabel('Yn'); title("Convolution of Xn and Hn");
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// Display mode mode(0); // Display warning for floating point exception ieee(1); clear; clc; disp("Introduction to heat transfer by S.K.Som, Chapter 4, Example 12") //A large block of nickel steel conductivity(k=20W/(m*K)),thermal diffusivity(alpha=0.518*10-5 m^2/s) is at uniform temprature(Ti) of 30°C. Ti=30; k=20; alpha=0.518*10^-5; //One surface of the block is suddenly exposed to a constant surface heat flux(qo) of 6MW/m^2. qo=6*10^6;//in W/m^2 //To determine the temprature at a depth(x) of 100mm after a time(t) of 100 seconds. t=100; x=0.1;//in metre //Similarity parameter,eta=x/(4*alpha*t) eta=x/((4*alpha*t)^0.5) //E is gaussian error function disp("gaussian error function is" ) E=erf(eta) //The equation to determine temprature is T-Ti=((2*qo(alpha*t/%pi)^0.5)/(k))*e^((-x^2)/(4*alpha*t))-((qo*x)/(k))*erf(x/(2*(alpha*t)^0.5)) //Above equation can also be written as T=Ti+((2*qo(alpha*t/%pi)^0.5)/(k))*e^((-x^2)/(4*alpha*t))-((qo*x)/(k))*erf(x/(2*(alpha*t)^0.5)) disp("The temprature at a depth(x) of 100mm after a time(t) of 100 seconds,in °C is") T=Ti+((2*qo*(alpha*t/%pi)^0.5)/(k))*%e^((-x^2)/(4*alpha*t))-((qo*x)/(k))*erfc(x/(2*(alpha*t)^0.5))//NOTE:The answer in the book is incorrect(Calculation mistake)
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style.fontSize=16; style.displayedLabel="FG Switch"; pal2 = xcosPalAddBlock(pal2,"fgswitch",[],style);
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// Exa 1.4 clc; clear all; // Given data x1= 49.7; x2= 50.1; x3= 50.2; x4= 49.6; x5= 49.7; // solution X_mean= (x1+x2+x3+x4+x5)/5; // Arithmatic mean d1= x1-X_mean; d2= x2-X_mean; // deviation from each value d3= x3-X_mean; d4=x4-X_mean; d5=x5-X_mean; d_total= d1+d2+d3+d4+d5; //Algebraic sum of deviations printf('The arithmatic mean is %.2f \n \n',X_mean); printf(' Deviation from x1 is %.2f \n ',d1); printf('Deviation from x2 is %.2f \n ',d2); printf('Deviation from x3 is %.2f \n ',d3); printf('Deviation from x4 is %.2f \n ',d4); printf('Deviation from x5 is %.2f \n \n',d5); printf(' The algebraic sum of deviation is %d \n',d_total);
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// Scilab Code Ex4.1: Page-233 (2008) clc; clear; c = 3e+008; // Speed of light in vacuum, m/s v = 3e+004; // Speed of earth, m/s d = 7; // Effective length of each path, m lambda = 7000e-010; // Wavelength of light used, m n = 2*d*v^2/(lambda*c^2); // Fringe shift printf("\nThe expected fringe shift = %3.1f", n); // Result // The expected fringe shift = 0.2
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errcatch(-1,"stop");mode(2);//page 238 ; ; A=[1 0 1 6 2;0 1 1 0 3]; b=[8 9]'; c=[0 0 7 -1 -3]'; lb=[0 0 0 0 0]' ub=[]; [x,lagr,f]=linpro(c,A,b,lb,ub); disp(x,'New corner:'); disp(f,'Minimum cost:'); //end exit();
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errcatch(-1,"stop");mode(2);//Caption:Find (a)Load supplied by second machine and its power factor (b)Power factor of total load //Exa:14.10 ; ; P1=300//Lighting load(in KW) P2=500//Industrial load(in KW) P3=200//Industrial load(in KW) P4=100//Load(in KW) Pa=500//Power supplied by first machine(in KW) pf1=0.8 pf2=0.707 pf3=0.9 pfa=0.8 La=P1+P2+P3+P4 Lr=(P2*tand(acosd(pf1)))+(P3*tand(acosd(pf2)))+(P4*tand(acosd(pf3))) Pb=La-Pa Prl=Pa*(tand(acosd(pfa))) Pc=Lr-Prl pfb=cosd(atand(Pc/Pb)) pfl=cosd(atand(Lr/La)) disp(pfb,Pb,'(a)Load supplied by second machine(in KW) and its power factor=') disp(pfl,'(b)Power factor of load=') exit();
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ima_adpcm_dec.sci
function out_samp = ima_adpcm_dec(in_pcm) // This function decodes an IMA ADPCM compressed audio. The output is reconstructed // to 16 bits per sample. // // Author: Moti Litochevski // Date: February 17, 2010 // // step quantizer adaptation lookup table STEP_LUT = [ ... 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, 19, 21, 23, 25, 28, 31, 34, ... 37, 41, 45, 50, 55, 60, 66, 73, 80, 88, 97, 107, 118, 130, 143, 157, ... 173, 190, 209, 230, 253, 279, 307, 337, 371, 408, 449, 494, 544, 598, ... 658, 724, 796, 876, 963, 1060, 1166, 1282, 1411, 1552, 1707, 1878, 2066, ... 2272, 2499, 2749, 3024, 3327, 3660, 4026, 4428, 4871, 5358, 5894, 6484, ... 7132, 7845, 8630, 9493, 10442, 11487, 12635, 13899, 15289, 16818, 18500, ... 20350, 22385, 24623, 27086, 29794, 32767]; // index quantizer adaptation lookup table INDEX_LUT = [-1, -1, -1, -1, 2, 4, 6, 8]; // prepare loop variables predictor_samp = zeros(1, length(in_pcm)+1); qstep_index = ones(1, length(in_pcm)+1); // convert the input coded signal to binary form & calculate the PCM value (without sign) pcm_bin = de2bi(in_pcm, 4); pcm_val = pcm_bin(:,[2:4]) * [4, 2, 1]'; // decoding loop for idx = [1:length(in_pcm)], // extract the current quantizer step size qstep_size = STEP_LUT(qstep_index(idx)); // de-quantizer implementation starts from the middle of the current // quantization step (qstep_size/8) dequant_samp = qstep_size; // de-quantize bit by bit dequant_samp = dequant_samp + pcm_bin(idx, 2) * qstep_size * 8; dequant_samp = dequant_samp + pcm_bin(idx, 3) * qstep_size * 4; dequant_samp = dequant_samp + pcm_bin(idx, 4) * qstep_size * 2; // update the predictor output sample according to the sign bit if (pcm_bin(idx, 1)), predictor_samp(idx+1) = predictor_samp(idx) - dequant_samp; else predictor_samp(idx+1) = predictor_samp(idx) + dequant_samp; end // check for predictor sample saturation condition if (predictor_samp(idx+1) > (2^18-1)), predictor_samp(idx+1) = 2^18-1; elseif (predictor_samp(idx+1) < -2^18), predictor_samp(idx+1) = -2^18; end // update the step size index qstep_index(idx+1) = qstep_index(idx) + INDEX_LUT(pcm_val(idx)+1); // check index saturation conditions if (qstep_index(idx+1) < 1) qstep_index(idx+1) = 1; elseif (qstep_index(idx+1) > 89) qstep_index(idx+1) = 89; end end // output sample is just the saturated & scaled predictor output out_samp = predictor_samp(2:$)/8; // implement rounding out_samp = round(out_samp); cor_idx = find((out_samp - predictor_samp(2:$)/8) == -0.5); out_samp(cor_idx) = out_samp(cor_idx) + 1; // check for saturation out_samp(find(predictor_samp(2:$)/8 > 32767)) = 32767; out_samp(find(predictor_samp(2:$)/8 < -32768)) = -32768; endfunction
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//Calculate the Root Mean Square distance traveled by a urea molucule //Example 21.2 clc; clear; D=1.18*10^-9; //Diffusion coefficient of Urea in m^2 s^-1 t=1*60*60; //Diffusion time in second meanx=sqrt(2*D*t)*1000; //Root mean square distance in mm printf("Root mean square distance traveled = %.1f mm",meanx);
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//example 6.12 //Trapezoidal and Simpson's rule //page 230 clc;clear;close; deff('y=f(x)','y=sqrt(1-x^2)'); k=10:10:50 for i=1:length(k) T_area(i)=0,S_area(i)=0; h=1/k(i); x=0:h:1 l=length(x); for j=1:l y(j)=f(x(j)); end for j=1:l if j==1|j==l then T_area(i)=T_area(i)+y(j) else T_area(i)=T_area(i)+2*y(j) end end T_area(i)=T_area(i)*(h/2); for j=1:l if j==1|j==l then S_area(i)=S_area(i)+y(j) elseif (modulo(j,2))==0 then S_area(i)=S_area(i)+4*y(j) elseif (modulo(i,2))~=0 then S_area(i)=S_area(i)+2*y(j) end end S_area(i)=S_area(i)*(h)/3; end printf(' no of subintervals Trapezoidal Rule Simpsons Rule\t \n \n') for i=1:length(k) printf(' %0.9g %0.9g %0.9g \n ',k(i),T_area(i),S_area(i)); end
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//Finding of inclination rho=1000; d=0.03; V=16; w=125; //To Find A=(%pi/4)*d^2; P=rho*A*V^2; Q=P*(16/32); theta=asin((rho*A*V^2)/w); disp("Inclination ="+string(theta)+" degrees");
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function [pt,dx,dz,dt]=acoustic(vel,tf,fc,spos,dx,dz,dt) //[pt[,dx,dz,dt]]=acoustic(vel,tf,fc,spos[,dx,dz][,dt]) ////////////////////////////////////////////////////// // // // Scilab program to simulate forward propagation // // of acoustic waves with absorbing boundary // // conditions where p is the wavefield, s is the // // source, and v is the velocity field: // // // // 2 __ 2 // // P = v \/ P + S // // tt // // // ////////////////////////////////////////////////////// // vel :Velocity distribution (matrix which represents // :distribution of velocity as function of offset x // :and depth z. Increasing offset and depth goes // :as the increasing indices of the matrix vel). // tf :Final time (initial time is zero) // fc :Center frequency of wavelet (derivative of gaussian) // spos :source postion which is a 2-vector of integers // :where 1<=spos(1)<=nr, 1<=spos(2)<=nc // :and [nr,nc]=size(vel). // dx :Sampling step in offset // dz :Sampling step in depth // dt :Sampling step in time. // ntbl :Name table containing names of all the files // :created containing data // //The parameters dx, dz, and dt are optional and are //calculated automatically, when not specified, to satisfy //stability conditions and to impose an acceptable level //of numerical dispersion. // //! //author: C. Bunks date: 29-Oct-90 // Copyright INRIA [lhs,rhs]=argn(0); lines(0); //velocity parameters [nr,nc]=size(vel); vmax=maxi(vel);vmin=mini(vel); //default check if rhs==4 then,//auto calculation of dx, dz, and dt dx=vmin/(16*fc); dz=dx; dmin=mini([dx,dz]);dmax=maxi([dx,dz]); dt=.95*dmin/(vmax*sqrt(2));//stability condition end, if rhs==5 then,//auto calculation of dx and dz dx=vmin/(16*fc); dz=dx; dmin=mini([dx,dz]);dmax=maxi([dx,dz]); end, dmin=mini([dx,dz]);dmax=maxi([dx,dz]); if rhs==6 then,//auto calculation of dt dt=.95*dmin/(vmax*sqrt(2));//stability condition end, //inform user of default choices write(%io(2),' '), write(%io(2),' CHOICES OF DX, DZ, AND DT:'), print(6,[dx,dz,dt]), write(%io(2),' '), //ERROR CHECKING //source location eflag='on'; if 1<=spos(1) then, if spos(1)<=nr then, if 1<=spos(2) then, if spos(2)<=nc then, eflag='off'; end,end, end,end, if eflag=='on' then, write(%io(2),' '), write(%io(2),'*************ERROR*************'); write(%io(2),' '), write(%io(2),' SOURCE POSITION OUT OF BOUNDS '); write(%io(2),' '), write(%io(2),'*************ERROR*************'); write(%io(2),' '), error('sim.bas'), end, //space discretization if dmax>vmin/(15*fc) then, write(%io(2),' '), write(%io(2),'****************WARNING****************'); write(%io(2),' '), write(%io(2),' NUMERICAL DISPERSION LIKELY '), write(%io(2),' VARIABLES SHOULD SATISFY: '), write(%io(2),' '), write(%io(2),' max([dx,dz]) < vmin/(15*fc) '), write(%io(2),' '), write(%io(2),'****************WARNING****************'); write(%io(2),' '), end, //time discretization and stability check // v*dt/dx < 1/sqrt(2) if (vmax*dt/dmin)>(1/sqrt(2)) then, write(%io(2),' '), write(%io(2),'*****************ERROR*****************'); write(%io(2),' '), write(%io(2),' UNSTABLE CHOICES: vel, dt, dx, and dz '), write(%io(2),' VARIABLES MUST SATISFY: '), write(%io(2),' '), write(%io(2),' max(vel)*dt*sqrt(2) < min(dx,dz) '), write(%io(2),' '), write(%io(2),'*****************ERROR*****************'); write(%io(2),' '), error('sim.bas'), end, t=0:dt:tf; //pre-calculation v2dt=(dt*dt)*(vel.*vel); zr=0*ones(1,nc);zc=0*ones(nr,1); //integrate forward // unit=file('open','pt.dat','unknown','unformatted'); pt=[]; pkm1=0*ones(vel);pkm2=0*ones(vel);//initial conditions for tk=t, src=shot(tk,fc); pk=integrate(tk,pkm1,pkm2,src,spos); pt=[pt;pk]; // writb(unit,pk); pkm2=pkm1;pkm1=pk; end, // file('rewind',unit); // pt=readb(unit,nr*maxi(size(t)),nc); // file('close',unit); function [utp1]=integrate(t,ut,utm1,src,spos) //[utp1]=integrate(t,ut,utm1,src,spos) //compute second order time update of acoustic wave equation //(with absorbing boundaries) // t :current time // ut :wavefield at time t // utm1 :wavefield at time t-dt // src :source value at time t // spos :source position // utp1 :wavefield at time t+dt // //! //author: C. Bunks date: 29-OCT-90 // Copyright INRIA write(%io(2),'t='+string(t)); //calculate laplacian in the interior of the medium utp1= 2*ut... -utm1... +v2dt.*(([ut(:,2:nc) zc]+[zc ut(:,1:nc-1)]-2*ut)/dx**2... +([ut(2:nr,:);zr]+[zr;ut(1:nr-1,:)]-2*ut)/dz**2); //calculate boundary conditions on edges (fix velocities) //(see paper by Reynolds, Geophysics, v. 43, 1978, pp1099-1110) //right side boundary ua=ut(:,nc-1);ub=ut(:,nc);uc=utm1(:,nc-2);ud=utm1(:,nc-1); utp1(:,nc)=(ua+ub-ud)-(dt/dx)*vel(:,nc).*(ub-ua+uc-ud); //left side boundary ua=ut(:,2);ub=ut(:,1);uc=utm1(:,3);ud=utm1(:,2); utp1(:,1)=(ua+ub-ud)-(dt/dx)*vel(:,1).*(ub-ua+uc-ud); //bottom boundary ua=ut(nr-1,:);ub=ut(nr,:);uc=utm1(nr-2,:);ud=utm1(nr-1,:); utp1(nr,:)=(ua+ub-ud)-(dt/dz)*vel(nr,:).*(ub-ua-ud+uc); //top boundary (absorbing or free) ua=ut(2,:);ub=ut(1,:);uc=utm1(3,:);ud=utm1(2,:); utp1(1,:)=(ua+ub-ud)-(dt/dz)*vel(1,:).*(ub-ua-ud+uc); //calculate boundary conditions at corners utp1(1,1)=ut(1,1)+((dt/dx)*vel(1,1))*(ut(2,2)-ut(1,1)); utp1(1,nc)=ut(1,nc)+((dt/dx)*vel(1,nc))*(ut(2,nc-1)-ut(1,nc)); utp1(nr,1)=ut(nr,1)+((dt/dx)*vel(nr,1))*(ut(nr-1,2)-ut(nr,1)); utp1(nr,nc)=ut(nr,nc)+((dt/dx)*vel(nr,nc))*(ut(nr-1,nc-1)-ut(nr,nc)); //add in source sz=spos(1);sx=spos(2); utp1(sz,sx)=utp1(sz,sx)+dt**2*src; function [dg]=shot(t,fc) //[dg]=shot(t,fc) //calculate shot values as a function of time //as the derivative of a gaussian: // // g=exp(-(t-m)**2/(2*sigma**2)) // dg=-g*(t-m)/sigma**2 // //Center frequency for the derivative of the gaussian is //at fc=1/(2*m). The ratio of the minimum velocity (vm) //to center frequency (fc) (i.e., the smallest spatial wavelength) //should be between 10 and 20 times greater than the largest //spatial discretization. Here we take the factor to be 15: // // vm/fc = 2*m*vm > 15*maxi(dx,dz) //or // m = 7.5*maxi(dx,dz)/mini(vel) // // t :current time // fc :center frequency of the wavelet // dg :derivative of a gaussian at time t // //! //author: C. Bunks date: 29-OCT-90 // Copyright INRIA m=1/(2*fc); sig=m/4; sig2=sig**2; g=exp(-(t-m)**2/(2*sig2))/sqrt(2*%pi*sig2); dg=-g*(t-m)/sig2; function [pt]=get_data(ntbl,entry) //Search for a data file written on disk // ntbl :table of file names (first two entries give // :data dimensions) // entry :integer giving the entry in the table-2 // pt :returned data file // Copyright INRIA ts=maxi(size(ntbl)); nr=evstr(ntbl(1)); nc=evstr(ntbl(2)); if entry<=ts-2 then, filename=ntbl(entry+2); unit=file('open',filename,'unknown'); pt=read(unit,nr,nc); file('close',unit); plot2d(pt,'agc',[0,%pi/4],'x'); else, write(%io(2),' '), write(%io(2),' Table Entry to Large...Max Value='+string(ts-2)), write(%io(2),' '), end,
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#include "alu.inc" .code prolog #define DIV(N, I0, I1, V) ALU(N, , div, I0, I1, V) #define UDIV(N, I0, I1, V) ALU(N, _u, div, I0, I1, V) DIV(0, 0x7fffffff, 1, 0x7fffffff) DIV(1, 1, 0x7fffffff, 0) DIV(2, 0x80000000, 1, 0x80000000) DIV(3, 1, 0x80000000, 0) DIV(4, 0x7fffffff, 2, 0x3fffffff) DIV(5, 2, 0x7fffffff, 0) DIV(6, 2, 0x80000000, 0) DIV(7, 0x7fffffff, 0x80000000, 0) DIV(8, 0, 0x7fffffff, 0) DIV(9, 0xffffffff, 0xffffffff, 1) UDIV(0, 0x7fffffff, 1, 0x7fffffff) UDIV(1, 1, 0x7fffffff, 0) UDIV(2, 0x80000000, 1, 0x80000000) UDIV(3, 1, 0x80000000, 0) UDIV(4, 0x7fffffff, 2, 0x3fffffff) UDIV(5, 2, 0x7fffffff, 0) UDIV(6, 0x80000000, 2, 0x40000000) UDIV(7, 2, 0x80000000, 0) UDIV(8, 0x7fffffff, 0x80000000, 0) UDIV(9, 0x80000000, 0x7fffffff, 1) UDIV(10,0, 0x7fffffff, 0) UDIV(11,0x7fffffff, 0xffffffff, 0) UDIV(12,0xffffffff, 0x7fffffff, 2) UDIV(13,0xffffffff, 0xffffffff, 1) #if __WORDSIZE == 32 DIV(10, 0x80000000, 2, 0xc0000000) DIV(11, 0x80000000, 0x7fffffff, 0xffffffff) DIV(12, 0x7fffffff, 0xffffffff, 0x80000001) DIV(13, 0xffffffff, 0x7fffffff, 0) #else DIV(10, 0x80000000, 2, 0x40000000) DIV(11, 0x80000000, 0x7fffffff, 1) DIV(12, 0x7fffffff, 0xffffffff, 0) DIV(13, 0xffffffff, 0x7fffffff, 2) DIV(14, 0x7fffffffffffffff, 1, 0x7fffffffffffffff) DIV(15, 1, 0x7fffffffffffffff, 0) DIV(16, 0x8000000000000000, 1, 0x8000000000000000) DIV(17, 1, 0x8000000000000000, 0) DIV(18, 0x7fffffffffffffff, 2, 0x3fffffffffffffff) DIV(19, 2, 0x7fffffffffffffff, 0) DIV(20, 0x8000000000000000, 2, 0xc000000000000000) DIV(21, 2, 0x8000000000000000, 0) DIV(22, 0x7fffffffffffffff, 0x8000000000000000, 0) DIV(23, 0x8000000000000000, 0x7fffffffffffffff, 0xffffffffffffffff) DIV(24, 0x7fffffffffffffff, 0xffffffffffffffff, 0x8000000000000001) DIV(25, 0xffffffffffffffff, 0x7fffffffffffffff, 0) DIV(26, 0xffffffffffffffff, 0xffffffffffffffff, 1) UDIV(14,0x7fffffffffffffff, 1, 0x7fffffffffffffff) UDIV(15,1, 0x7fffffffffffffff, 0) UDIV(16,0x8000000000000000, 1, 0x8000000000000000) UDIV(17,1, 0x8000000000000000, 0) UDIV(18,0x7fffffffffffffff, 2, 0x3fffffffffffffff) UDIV(19,2, 0x7fffffffffffffff, 0) UDIV(20,0x8000000000000000, 2, 0x4000000000000000) UDIV(21,2, 0x8000000000000000, 0) UDIV(22,0x7fffffffffffffff, 0x8000000000000000, 0) UDIV(23,0x8000000000000000, 0x7fffffffffffffff, 1) UDIV(24,0x7fffffffffffffff, 0xffffffffffffffff, 0) UDIV(25,0xffffffffffffffff, 0x7fffffffffffffff, 2) UDIV(26,0xffffffffffffffff, 0xffffffffffffffff, 1) #endif #undef DIV #define DIV(N, T, I0, I1, V) FOP(N, T, div, I0, I1, V) DIV(0, _f, -0.5, 0.5, -1.0) DIV(1, _f, 1.25, 0.5, 2.5) DIV(0, _d, -0.5, 0.5, -1.0) DIV(1, _d, 1.25, 0.5, 2.5) prepare pushargi ok ellipsis finishi @printf ret epilog
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//<f>=%pir(i,j,f2,f) // %pir(i,j,f2,f) insere la sous matrice de polynomes f2 dans la //matrice de fractions rationnelles f1 aux lignes (colonnes) // designees par i (j). Cette macro correspond a la syntaxe; f(i,j)=f2. //! [n,d]=f(2:3),[ld,cd]=size(d),l=maxi(i),c=maxi(j) if l>ld then d(ld+1:l,:)=ones(l-ld,cd),ld=l,end if c>cd then d(:,cd+1:c)=ones(ld,c-cd),end n(i,j)=f2,[l,c]=size(f2),d(i,j)=ones(l,c) f(2)=n,f(3)=d //end
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clear; clc; printf("\t Example 2.20\n"); //this is the case of equimolar counter diffusion as the latent heat of vaporisation are very close to each other T=(360); //temperature in kelvin pt=372.4/760; //total pressure in atm R=82.06; //universal gas constant Dab=0.0506; //diffusion coefficient in cm^2/s z=0.254; //gas layer thickness in cm vp=368/760; //vapour pressure of toluene in atm xtol=.3; //mole fractoin of toluene in atm pb1=xtol*vp; //partial pressure of toluene //since pb1 is .045263 bt in book it is rounded to 0.145 pb2=xtol*pt; //parial pressure of toluene in vapour phase Na=Dab*(pb1-pb2)/(z*R*T); //diffusion flux printf("\n the diffusion flux of a mixture of benzene and toluene %f*10^-8 gmol/cm^2*s\n",Na/10^-8); printf("\nthe negative sign indicates that the toluene is getting transferred from gas phase to liquid phase(hence the transfer of benzene is from liquid to gas phase)") //end
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// This file is BASED ON part of www.nand2tetris.org // and the book "The Elements of Computing Systems" // by Nisan and Schocken, MIT Press. // File name: projects/01/And8.tst load And8.hdl, output-file And8.out, compare-to And8.cmp, output-list a%B1.8.1 b%B1.8.1 out%B1.8.1; set a %B00000000, set b %B00000000, eval, output; set a %B00000000, set b %B11111111, eval, output; set a %B11111111, set b %B11111111, eval, output; set a %B10101010, set b %B01010101, eval, output; set a %B00111100, set b %B00001111, eval, output; set a %B00010010, set b %B10011000, eval, output;
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//Ex 1.13 //Check for periodicity clc; T=2*%pi/6; t=0:0.001:T*2 x=cos((6*t)+%pi/3); subplot(3,2,1),plot(x); disp('the plot shows that the above signal is periodic'); T=2*%pi/(%i*%pi); t=0:0.001:T*2 x=exp(%i*(%pi*abs(t-1)));//exp(%i*(%pi*t-1))=exp(%i*%pi*t)/exp(%i) //since the period is a complex no so non periodic disp('T cannot be complex so non periodic T=2*%pi/(%i*%pi)'); //pi=22/7 T=2*%pi/4;//calc the fundamental period z=2*T; t=0:1/100:z x=(cos(2*t+%pi/3))^2; //sinusoid function subplot(3,2,2),plot(x) disp('the plot shows that the above signal is periodic'); k=1; N=2*k*7/6; z=2*N; n=0:1/100:z x=cos((6*%pi*n/7)+1); subplot(3,2,3),plot(x);//the plot shows that the above signal is periodic disp('the plot shows that the above signal is periodic'); k=1; N=2*%pi*k*8; z=2*N; n=0:1/100:z x=sin((n/8)-%pi); subplot(3,2,4),plot(x);//the plot shows that the above signal is periodic disp('the plot shows that the above signal is periodic'); k=1; N=2*k*12;//2*cos(n*%pi/4).*cos(n*%pi/3)=cos(7*n*%pi/12)-cos(n*%pi/12) z=2*N; n=0:1/100:z x=2*cos(n*%pi/4).*cos(n*%pi/3); subplot(3,1,3),plot(x);//the plot shows that the above signal is periodic disp('the plot shows the above signal is periodic');
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exec("chi2luck.sci",-1); function chi2plots(k) clf(); xmin=(max(0,sqrt(k-2)-3))^2; xmax=(sqrt(k-2)+3)^2; x=[xmin:0.01:xmax]; [L,U]=chi2luck(x,k); [q,p]=chi2cdf(x,k); P=exp(chi2probln(x,k)); Lest = erf(abs(sqrt(x)-sqrt(k-2))); plot(x',[L',Lest']); disp(max(abs(L-Lest))); endfunction
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errcatch(-1,"stop");mode(2);// Example 10.6, Page No-439 fr=300 bw=50 ip=320 pdop=fr+ip printf('\nPhase detector output= %d kHz', pdop) difr=ip-fr printf('\nDifference Frequency= %d kHz', difr) printf('\nAs Bandwidth is greater than difference frequency,') printf('\nPLL can acquire lock') exit();
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clear // //Given //Variable declaration L=3*1000 //Length in mm W=25*1000 //Point load in N I=1e8 //Moment of Inertia in mm^4 E=2.1e5 //Youngs modulus in N/sq.mm //Calculation //case(i):Slope of the cantilever at the free end thetaB=((W*(L**2))/(2*E*I)) //case(ii):Deflection at the free end yB=((W*L**3)/(E*I*3)) //Result printf("\n Slope at the free end = %0.3f rad",thetaB) printf("\n Deflection at the free end = %0.3f mm",yB)
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clc clear //Input data d=0.09;//The diameter of the bore in m L=0.1;//The length of the stroke in m T=120;//The torque measured in Nm pi=3.141;//Mathematical constant of pi n=4;//Number of cylinders //Calculations pmb=[(4*pi*T)/(L*(pi/4)*d^2*n)]/10^5;//The brake mean effective pressure in bar //Output printf('The brake mean effective pressure = %3.2f bar',pmb)
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clc clear printf("example 4.5 page number 133\n\n") //to find variation of losses with velocity loss_ratio=3.6; //delta_P2/delta_P1=3.6 velocity_ratio=2; //u2/u1=2 n=log2(loss_ratio); //delta_P2/delta_P1=(u2/u1)^n printf("power constant = %f flow is turbulent",n)
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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 7 Example 2") m=1;//mass of air in kg Po=1*10^5;//atmospheric pressure in pa To=(15+273);//temperature of atmosphere in K Cv=0.717;//specific heat at constant volume in KJ/kg K R=0.287;//gas constant in KJ/kg K Cp=1.004;//specific heat at constant pressure in KJ/kg K T=(50+273);//temperature of tanks A and B in K disp("In these tanks the air stored is at same temperature of 50 degree celcius.Therefore,for air behaving as perfect gas the internal energy of air in tanks shall be same as it depends upon temperature alone.But the availability shall be different.") disp("BOTH THE TANKS HAVE SAME INTERNAL ENERGY") disp("availability of air in tank,A") disp("A=(E-Uo)+Po*(V-Vo)-To*(S-So)") disp("=m*{(e-uo)+Po(v-vo)-To(s-so)}") disp("m*{Cv*(T-To)+Po*(R*T/P-R*To/Po)-To(Cp*log(T/To)-R*log(P/Po))}") disp("so A=m*{Cv*(T-To)+R*(Po*T/P-To)-To*Cp*log(T/To)+To*R*log(P/Po)}") disp("for tank A,P=1*10^5 pa,so availability_A in KJ") P=1*10^5;//pressure in tank A in pa availability_A=m*{Cv*(T-To)+R*(Po*T/P-To)-To*Cp*log(T/To)+To*R*log(P/Po)} disp("for tank B,P=3*10^5 pa,so availability_B in KJ") P=3*10^5;//pressure in tank B in pa availability_B=m*{Cv*(T-To)+R*(Po*T/P-To)-To*Cp*log(T/To)+To*R*log(P/Po)} disp("so availability of air in tank B is more than that of tank A") disp("availability of air in tank A=1.98 KJ") disp("availability of air in tank B=30.98 KJ")
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(unwatch all) (clear) (dribble-on "Actual//factsfun.out") (batch "factsfun.bat") (dribble-off) (clear) (open "Results//factsfun.rsl" factsfun "w") (load "compline.clp") (printout factsfun "factsfun.bat differences are as follows:" crlf) (compare-files "Expected//factsfun.out" "Actual//factsfun.out" factsfun) (close factsfun)
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//Chapter-3, Example 3.6, Page 111 //============================================================================= clc clear //INPUT DATA N1=250;//no of turns in a coil I1=2;//current in coil in A phi1=0.3;//flux in coil in wb dt=2//time in millisec Em2=63.75//induced voltage in V K=0.75 //CALCULATIONS L1=N1*(phi1/I1);//self inductance of first coil in H M=Em2*(dt/I1);//mutual inductance of two coils in H L2=((Em2/K)^2)/(L1);//self inductance of second coil in H phi2=K*phi1;//flux in second coil in wb N2=(Em2*dt)/phi2;//no of turns in second coil //OUTPUT mprintf("Thus the self inductance of first coil is %2.1f mH \n",L1); mprintf("mutual inductance of two coils %2.2f mH \n",M); mprintf("self inductance of second coil %4.0f mH \n",L2); mprintf("no of turns in second coil %3.0f turns \n",N2); //note:the answer given for N2 in textbook is wrong .please check the calculations //=================================END OF PROGRAM==============================
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//EXAMPLE 27.6 //DC GENERATOR clc; funcprot(0); //Variable Initialisation Z=500;......//Total number of conductors Ia=200;......//Total current in Amperes P=6;.........//Total number of poles b=10;........//Angle of lead in degrees y=1.3;...........//Leakage coefficient Aw=2;...........//Number of parallel paths for wave wound generator I=Ia/Aw;.....//Current per path in Amperes ATepole=Z*I*((1/(2*P))-(b/360));...........//Cross magnetizing ampere-turns per pole r=round(ATepole);.....//Rounding of decimal places disp(r,"(a).Cross magnetizing ampere-turns per pole:"); ATdpole=Z*I*b/360;........//Demagnetizing ampere-turns per pole r1=round(ATdpole);.......//Rounding of decimal places disp(r1,"(b).Demagnetizing ampere-turns per pole:"); S=y*r1/Ia;.......//Series turns required to balance the demagnetizing ampere-turns r2=round(S);....//Rounding of decimal places disp(r2,"Series turns required to balance the demagnetizing ampere-turns:");
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function g1=makeund(g) [lhs,rhs]=argn(0), if rhs==0 then g=the_g, end if g_direct(g) == 0 then error('graph already undirected'), end m=2*g_arcnum(g), ma=g_arcnum(g), mm=m [a1,p1,s1]=compunl1(g_la1(g),g_lp1(g),g_ls1(g)) [a2,p2,s2]=compl2(a1,p1,s1,0) [he,ta]=compht(a1,p1,s1,0) g1=list(' ',0,m,g_nodnum(g),ma,mm,a1,p1,s1,a2,p2,s2,he,ta,... g_ntype(g),g_xnode(g),g_ynode(g),g_ncolor(g),g_demand(g),g_acolor(g),... alenght(g),g_acost(g),g_mincap(g),g_maxcap(g),g_qweig(g),g_qorig(g),g_aweig(g))
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//ex3.7 V_IN=24; V_Z=15; I_ZK=0.25*10^-3; I_ZT=17*10^-3; Z_ZT=14; P_D_max=1; //output voltage at I_ZK V_out_1=V_Z-(I_ZT-I_ZK)*Z_ZT; disp(V_out_1,'output voltage in volts at I_ZK') I_ZM=P_D_max/V_Z; //output voltage at I_ZM V_out_2=V_Z+(I_ZM-I_ZT)*Z_ZT; disp(V_out_2,'output voltage in volts a I_ZM') R=(V_IN-V_out_2)/I_ZM; disp(R,'value of R in ohms for maximum zener current, no load') disp('closest practical value is 130 ohms') R=130; //for minimum load resistance(max load current) zener current is minimum (I_ZK) I_T=(V_IN-V_out_1)/R; I_L=I_T-I_ZK; R_L_min=V_out_1/I_L; disp(R_L_min,'minimum load resistance in ohms')
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clear clc disp('from the principle of counting,the required no.of ways are 12*11*10*9=') 12*11*10*9
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clc; clear; //Example 3.30 k=0.02685 //W/(m.K) v=16.5*10^-6 //kg/(m.s) Npr=0.7 //Prandtl number Beta=3.25*10^-3 //K^-1 g=9.81 //m/(s^2) Tw=333; //[k] T_inf=283 //[K] dT=Tw-T_inf //[K] L=4 //Length/height of plate [m] Ngr=(g*Beta*dT*(L^3))/(v^2) //Grashoff number //Let const=Ngr*Npr const=Ngr*Npr //Sice it is >10^9 Nnu=0.10*(const^(1.0/3.0)) //Nusselt number h=Nnu*k/L //W/(sq m.K) h=4.3 //Approx in book W=7 //width in [m] A=L*W //Area of heat transfer in [sq m] Q=h*A*dT //[W] printf("\nHeat transferred is %d W\n",Q)
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Expanding for base=2, level=7, reasons+features=base,similiar invall,norm Refined variables=x,y [0+1x,0+1y]: unknown -> [1] [0,0] x²-y³+2 ---------------- level 0 expanding queue[0]^-1,meter=[2,2]: x²-y³+2 [1+2x,1+2y]: unknown -> [1] [1,1] 2x+2x²-3y-6y²-4y³+1 endexp[0] ---------------- level 1 expanding queue[1]^0,meter=[2,2]: 2x+2x²-3y-6y²-4y³+1 [1+4x,3+4y]: unknown -> [2] [0,1] 2x+4x²-27y-36y²-16y³-6 -> solution [5,3],NONTRIVIAL [3+4x,3+4y]: unknown -> [3] [1,1] 6x+4x²-27y-36y²-16y³-4 endexp[1] ---------------- level 2 expanding queue[2]^1,meter=[2,2]: 2x+4x²-27y-36y²-16y³-6 [1+8x,3+8y]: unknown -> [4] [0,0] 2x+8x²-27y-72y²-64y³-3 [5+8x,3+8y]: unknown -> [5] [1,0] 10x+8x²-27y-72y²-64y³ -> solution [5,3],NONTRIVIAL endexp[2] expanding queue[3]^1,meter=[2,2]: 6x+4x²-27y-36y²-16y³-4 [3+8x,3+8y]: unknown -> [6] [0,0] 6x+8x²-27y-72y²-64y³-2 [7+8x,3+8y]: unknown -> [7] [1,0] 14x+8x²-27y-72y²-64y³+3 endexp[3] ---------------- level 3 expanding queue[4]^2,meter=[2,2]: 2x+8x²-27y-72y²-64y³-3 [1+16x,11+16y]: unknown -> [8] [0,1] 2x+16x²-363y-528y²-256y³-83 [9+16x,11+16y]: unknown -> [9] [1,1] 18x+16x²-363y-528y²-256y³-78 endexp[4] expanding queue[5]^2,meter=[2,2]: 10x+8x²-27y-72y²-64y³ [5+16x,3+16y]: unknown -> [10] [0,0] 10x+16x²-27y-144y²-256y³ -> solution [5,3],NONTRIVIAL [13+16x,3+16y]: unknown -> [11] [1,0] 26x+16x²-27y-144y²-256y³+9 endexp[5] expanding queue[6]^3,meter=[2,2]: 6x+8x²-27y-72y²-64y³-2 [3+16x,3+16y]: unknown -> [12] [0,0] 6x+16x²-27y-144y²-256y³-1 [11+16x,3+16y]: unknown -> [13] [1,0] 22x+16x²-27y-144y²-256y³+6 endexp[6] expanding queue[7]^3,meter=[2,2]: 14x+8x²-27y-72y²-64y³+3 [7+16x,11+16y]: unknown -> [14] [0,1] 14x+16x²-363y-528y²-256y³-80 [15+16x,11+16y]: unknown -> [15] [1,1] 30x+16x²-363y-528y²-256y³-69 endexp[7] ---------------- level 4 expanding queue[8]^4,meter=[2,2]: 2x+16x²-363y-528y²-256y³-83 [1+32x,27+32y]: unknown -> [16] [0,1] 2x+32x²-2187y-2592y²-1024y³-615 [17+32x,27+32y]: unknown -> [17] [1,1] 34x+32x²-2187y-2592y²-1024y³-606 endexp[8] expanding queue[9]^4,meter=[2,2]: 18x+16x²-363y-528y²-256y³-78 [9+32x,11+32y]: unknown -> [18] [0,0] 18x+32x²-363y-1056y²-1024y³-39 [25+32x,11+32y]: unknown -> [19] [1,0] 50x+32x²-363y-1056y²-1024y³-22 endexp[9] expanding queue[10]^5,meter=[2,2]: 10x+16x²-27y-144y²-256y³ [5+32x,3+32y]: unknown -> [20] [0,0] 10x+32x²-27y-288y²-1024y³ -> solution [5,3],NONTRIVIAL [21+32x,3+32y]: unknown -> [21] [1,0] 42x+32x²-27y-288y²-1024y³+13 endexp[10] expanding queue[11]^5,meter=[2,2]: 26x+16x²-27y-144y²-256y³+9 [13+32x,19+32y]: unknown -> [22] [0,1] 26x+32x²-1083y-1824y²-1024y³-209 [29+32x,19+32y]: unknown -> [23] [1,1] 58x+32x²-1083y-1824y²-1024y³-188 endexp[11] expanding queue[12]^6,meter=[2,2]: 6x+16x²-27y-144y²-256y³-1 [3+32x,19+32y]: unknown -> [24] [0,1] 6x+32x²-1083y-1824y²-1024y³-214 [19+32x,19+32y]: unknown -> [25] [1,1] 38x+32x²-1083y-1824y²-1024y³-203 endexp[12] expanding queue[13]^6,meter=[2,2]: 22x+16x²-27y-144y²-256y³+6 [11+32x,3+32y]: unknown -> [26] [0,0] 22x+32x²-27y-288y²-1024y³+3 [27+32x,3+32y]: unknown -> [27] [1,0] 54x+32x²-27y-288y²-1024y³+22 endexp[13] expanding queue[14]^7,meter=[2,2]: 14x+16x²-363y-528y²-256y³-80 [7+32x,11+32y]: unknown -> [28] [0,0] 14x+32x²-363y-1056y²-1024y³-40 [23+32x,11+32y]: unknown -> [29] [1,0] 46x+32x²-363y-1056y²-1024y³-25 endexp[14] expanding queue[15]^7,meter=[2,2]: 30x+16x²-363y-528y²-256y³-69 [15+32x,27+32y]: unknown -> [30] [0,1] 30x+32x²-2187y-2592y²-1024y³-608 [31+32x,27+32y]: unknown -> [31] [1,1] 62x+32x²-2187y-2592y²-1024y³-585 endexp[15] ---------------- level 5 expanding queue[16]^8,meter=[2,2]: 2x+32x²-2187y-2592y²-1024y³-615 [1+64x,59+64y]: unknown -> [32] [0,1] 2x+64x²-10443y-11328y²-4096y³-3209 [33+64x,59+64y]: unknown -> [33] [1,1] 66x+64x²-10443y-11328y²-4096y³-3192 endexp[16] expanding queue[17]^8,meter=[2,2]: 34x+32x²-2187y-2592y²-1024y³-606 [17+64x,27+64y]: unknown -> [34] [0,0] 34x+64x²-2187y-5184y²-4096y³-303 [49+64x,27+64y]: unknown -> [35] [1,0] 98x+64x²-2187y-5184y²-4096y³-270 endexp[17] expanding queue[18]^9,meter=[2,2]: 18x+32x²-363y-1056y²-1024y³-39 [9+64x,43+64y]: unknown -> [36] [0,1] 18x+64x²-5547y-8256y²-4096y³-1241 [41+64x,43+64y]: unknown -> [37] [1,1] 82x+64x²-5547y-8256y²-4096y³-1216 endexp[18] expanding queue[19]^9,meter=[2,2]: 50x+32x²-363y-1056y²-1024y³-22 [25+64x,11+64y]: unknown -> [38] [0,0] 50x+64x²-363y-2112y²-4096y³-11 [57+64x,11+64y]: unknown -> [39] [1,0] 114x+64x²-363y-2112y²-4096y³+30 endexp[19] expanding queue[20]^10,meter=[2,2]: 10x+32x²-27y-288y²-1024y³ [5+64x,3+64y]: unknown -> [40] [0,0] 10x+64x²-27y-576y²-4096y³ -> solution [5,3],NONTRIVIAL [37+64x,3+64y]: unknown -> [41] [1,0] 74x+64x²-27y-576y²-4096y³+21 endexp[20] expanding queue[21]^10,meter=[2,2]: 42x+32x²-27y-288y²-1024y³+13 [21+64x,35+64y]: unknown -> [42] [0,1] 42x+64x²-3675y-6720y²-4096y³-663 [53+64x,35+64y]: unknown -> [43] [1,1] 106x+64x²-3675y-6720y²-4096y³-626 endexp[21] expanding queue[22]^11,meter=[2,2]: 26x+32x²-1083y-1824y²-1024y³-209 [13+64x,51+64y]: unknown -> [44] [0,1] 26x+64x²-7803y-9792y²-4096y³-2070 [45+64x,51+64y]: unknown -> [45] [1,1] 90x+64x²-7803y-9792y²-4096y³-2041 endexp[22] expanding queue[23]^11,meter=[2,2]: 58x+32x²-1083y-1824y²-1024y³-188 [29+64x,19+64y]: unknown -> [46] [0,0] 58x+64x²-1083y-3648y²-4096y³-94 [61+64x,19+64y]: unknown -> [47] [1,0] 122x+64x²-1083y-3648y²-4096y³-49 endexp[23] expanding queue[24]^12,meter=[2,2]: 6x+32x²-1083y-1824y²-1024y³-214 [3+64x,19+64y]: unknown -> [48] [0,0] 6x+64x²-1083y-3648y²-4096y³-107 [35+64x,19+64y]: unknown -> [49] [1,0] 70x+64x²-1083y-3648y²-4096y³-88 endexp[24] expanding queue[25]^12,meter=[2,2]: 38x+32x²-1083y-1824y²-1024y³-203 [19+64x,51+64y]: unknown -> [50] [0,1] 38x+64x²-7803y-9792y²-4096y³-2067 [51+64x,51+64y]: unknown -> [51] [1,1] 102x+64x²-7803y-9792y²-4096y³-2032 endexp[25] expanding queue[26]^13,meter=[2,2]: 22x+32x²-27y-288y²-1024y³+3 [11+64x,35+64y]: unknown -> [52] [0,1] 22x+64x²-3675y-6720y²-4096y³-668 [43+64x,35+64y]: unknown -> [53] [1,1] 86x+64x²-3675y-6720y²-4096y³-641 endexp[26] expanding queue[27]^13,meter=[2,2]: 54x+32x²-27y-288y²-1024y³+22 [27+64x,3+64y]: unknown -> [54] [0,0] 54x+64x²-27y-576y²-4096y³+11 [59+64x,3+64y]: unknown -> [55] [1,0] 118x+64x²-27y-576y²-4096y³+54 endexp[27] expanding queue[28]^14,meter=[2,2]: 14x+32x²-363y-1056y²-1024y³-40 [7+64x,11+64y]: unknown -> [56] [0,0] 14x+64x²-363y-2112y²-4096y³-20 [39+64x,11+64y]: unknown -> [57] [1,0] 78x+64x²-363y-2112y²-4096y³+3 endexp[28] expanding queue[29]^14,meter=[2,2]: 46x+32x²-363y-1056y²-1024y³-25 [23+64x,43+64y]: unknown -> [58] [0,1] 46x+64x²-5547y-8256y²-4096y³-1234 [55+64x,43+64y]: unknown -> [59] [1,1] 110x+64x²-5547y-8256y²-4096y³-1195 endexp[29] expanding queue[30]^15,meter=[2,2]: 30x+32x²-2187y-2592y²-1024y³-608 [15+64x,27+64y]: unknown -> [60] [0,0] 30x+64x²-2187y-5184y²-4096y³-304 [47+64x,27+64y]: unknown -> [61] [1,0] 94x+64x²-2187y-5184y²-4096y³-273 endexp[30] expanding queue[31]^15,meter=[2,2]: 62x+32x²-2187y-2592y²-1024y³-585 [31+64x,59+64y]: unknown -> [62] [0,1] 62x+64x²-10443y-11328y²-4096y³-3194 [63+64x,59+64y]: unknown -> [63] [1,1] 126x+64x²-10443y-11328y²-4096y³-3147 endexp[31] ---------------- level 6 expanding queue[32]^16,meter=[2,2]: 2x+64x²-10443y-11328y²-4096y³-3209 [1+128x,123+128y]: unknown -> [64] [0,1] 2x+128x²-45387y-47232y²-16384y³-14538 [65+128x,123+128y]: unknown -> [65] [1,1] 130x+128x²-45387y-47232y²-16384y³-14505 endexp[32] expanding queue[33]^16,meter=[2,2]: 66x+64x²-10443y-11328y²-4096y³-3192 [33+128x,59+128y]: unknown -> [66] [0,0] 66x+128x²-10443y-22656y²-16384y³-1596 [97+128x,59+128y]: unknown -> [67] [1,0] 194x+128x²-10443y-22656y²-16384y³-1531 endexp[33] expanding queue[34]^17,meter=[2,2]: 34x+64x²-2187y-5184y²-4096y³-303 [17+128x,91+128y]: unknown -> [68] [0,1] 34x+128x²-24843y-34944y²-16384y³-5885 [81+128x,91+128y]: unknown -> [69] [1,1] 162x+128x²-24843y-34944y²-16384y³-5836 endexp[34] expanding queue[35]^17,meter=[2,2]: 98x+64x²-2187y-5184y²-4096y³-270 [49+128x,27+128y]: unknown -> [70] [0,0] 98x+128x²-2187y-10368y²-16384y³-135 [113+128x,27+128y]: unknown -> [71] [1,0] 226x+128x²-2187y-10368y²-16384y³-54 endexp[35] expanding queue[36]^18,meter=[2,2]: 18x+64x²-5547y-8256y²-4096y³-1241 [9+128x,107+128y]: unknown -> [72] [0,1] 18x+128x²-34347y-41088y²-16384y³-9570 [73+128x,107+128y]: unknown -> [73] [1,1] 146x+128x²-34347y-41088y²-16384y³-9529 endexp[36] expanding queue[37]^18,meter=[2,2]: 82x+64x²-5547y-8256y²-4096y³-1216 [41+128x,43+128y]: unknown -> [74] [0,0] 82x+128x²-5547y-16512y²-16384y³-608 [105+128x,43+128y]: unknown -> [75] [1,0] 210x+128x²-5547y-16512y²-16384y³-535 endexp[37] expanding queue[38]^19,meter=[2,2]: 50x+64x²-363y-2112y²-4096y³-11 [25+128x,75+128y]: unknown -> [76] [0,1] 50x+128x²-16875y-28800y²-16384y³-3291 [89+128x,75+128y]: unknown -> [77] [1,1] 178x+128x²-16875y-28800y²-16384y³-3234 endexp[38] expanding queue[39]^19,meter=[2,2]: 114x+64x²-363y-2112y²-4096y³+30 [57+128x,11+128y]: unknown -> [78] [0,0] 114x+128x²-363y-4224y²-16384y³+15 [121+128x,11+128y]: unknown -> [79] [1,0] 242x+128x²-363y-4224y²-16384y³+104 endexp[39] expanding queue[40]^20,meter=[2,2]: 10x+64x²-27y-576y²-4096y³ [5+128x,3+128y]: unknown -> [80] [0,0] 10x+128x²-27y-1152y²-16384y³ -> solution [5,3],NONTRIVIAL [69+128x,3+128y]: unknown -> [81] [1,0] 138x+128x²-27y-1152y²-16384y³+37 endexp[40] expanding queue[41]^20,meter=[2,2]: 74x+64x²-27y-576y²-4096y³+21 [37+128x,67+128y]: unknown -> [82] [0,1] 74x+128x²-13467y-25728y²-16384y³-2339 [101+128x,67+128y]: unknown -> [83] [1,1] 202x+128x²-13467y-25728y²-16384y³-2270 endexp[41] expanding queue[42]^21,meter=[2,2]: 42x+64x²-3675y-6720y²-4096y³-663 [21+128x,99+128y]: unknown -> [84] [0,1] 42x+128x²-29403y-38016y²-16384y³-7577 [85+128x,99+128y]: unknown -> [85] [1,1] 170x+128x²-29403y-38016y²-16384y³-7524 endexp[42] expanding queue[43]^21,meter=[2,2]: 106x+64x²-3675y-6720y²-4096y³-626 [53+128x,35+128y]: unknown -> [86] [0,0] 106x+128x²-3675y-13440y²-16384y³-313 [117+128x,35+128y]: unknown -> [87] [1,0] 234x+128x²-3675y-13440y²-16384y³-228 endexp[43] expanding queue[44]^22,meter=[2,2]: 26x+64x²-7803y-9792y²-4096y³-2070 [13+128x,51+128y]: unknown -> [88] [0,0] 26x+128x²-7803y-19584y²-16384y³-1035 [77+128x,51+128y]: unknown -> [89] [1,0] 154x+128x²-7803y-19584y²-16384y³-990 endexp[44] expanding queue[45]^22,meter=[2,2]: 90x+64x²-7803y-9792y²-4096y³-2041 [45+128x,115+128y]: unknown -> [90] [0,1] 90x+128x²-39675y-44160y²-16384y³-11866 [109+128x,115+128y]: unknown -> [91] [1,1] 218x+128x²-39675y-44160y²-16384y³-11789 endexp[45] expanding queue[46]^23,meter=[2,2]: 58x+64x²-1083y-3648y²-4096y³-94 [29+128x,19+128y]: unknown -> [92] [0,0] 58x+128x²-1083y-7296y²-16384y³-47 [93+128x,19+128y]: unknown -> [93] [1,0] 186x+128x²-1083y-7296y²-16384y³+14 endexp[46] expanding queue[47]^23,meter=[2,2]: 122x+64x²-1083y-3648y²-4096y³-49 [61+128x,83+128y]: unknown -> [94] [0,1] 122x+128x²-20667y-31872y²-16384y³-4438 [125+128x,83+128y]: unknown -> [95] [1,1] 250x+128x²-20667y-31872y²-16384y³-4345 endexp[47] expanding queue[48]^24,meter=[2,2]: 6x+64x²-1083y-3648y²-4096y³-107 [3+128x,83+128y]: unknown -> [96] [0,1] 6x+128x²-20667y-31872y²-16384y³-4467 [67+128x,83+128y]: unknown -> [97] [1,1] 134x+128x²-20667y-31872y²-16384y³-4432 endexp[48] expanding queue[49]^24,meter=[2,2]: 70x+64x²-1083y-3648y²-4096y³-88 [35+128x,19+128y]: unknown -> [98] [0,0] 70x+128x²-1083y-7296y²-16384y³-44 [99+128x,19+128y]: unknown -> [99] [1,0] 198x+128x²-1083y-7296y²-16384y³+23 endexp[49] expanding queue[50]^25,meter=[2,2]: 38x+64x²-7803y-9792y²-4096y³-2067 [19+128x,115+128y]: unknown -> [100] [0,1] 38x+128x²-39675y-44160y²-16384y³-11879 [83+128x,115+128y]: unknown -> [101] [1,1] 166x+128x²-39675y-44160y²-16384y³-11828 endexp[50] expanding queue[51]^25,meter=[2,2]: 102x+64x²-7803y-9792y²-4096y³-2032 [51+128x,51+128y]: unknown -> [102] [0,0] 102x+128x²-7803y-19584y²-16384y³-1016 [115+128x,51+128y]: unknown -> [103] [1,0] 230x+128x²-7803y-19584y²-16384y³-933 endexp[51] expanding queue[52]^26,meter=[2,2]: 22x+64x²-3675y-6720y²-4096y³-668 [11+128x,35+128y]: unknown -> [104] [0,0] 22x+128x²-3675y-13440y²-16384y³-334 [75+128x,35+128y]: unknown -> [105] [1,0] 150x+128x²-3675y-13440y²-16384y³-291 endexp[52] expanding queue[53]^26,meter=[2,2]: 86x+64x²-3675y-6720y²-4096y³-641 [43+128x,99+128y]: unknown -> [106] [0,1] 86x+128x²-29403y-38016y²-16384y³-7566 [107+128x,99+128y]: unknown -> [107] [1,1] 214x+128x²-29403y-38016y²-16384y³-7491 endexp[53] expanding queue[54]^27,meter=[2,2]: 54x+64x²-27y-576y²-4096y³+11 [27+128x,67+128y]: unknown -> [108] [0,1] 54x+128x²-13467y-25728y²-16384y³-2344 [91+128x,67+128y]: unknown -> [109] [1,1] 182x+128x²-13467y-25728y²-16384y³-2285 endexp[54] expanding queue[55]^27,meter=[2,2]: 118x+64x²-27y-576y²-4096y³+54 [59+128x,3+128y]: unknown -> [110] [0,0] 118x+128x²-27y-1152y²-16384y³+27 [123+128x,3+128y]: unknown -> [111] [1,0] 246x+128x²-27y-1152y²-16384y³+118 endexp[55] expanding queue[56]^28,meter=[2,2]: 14x+64x²-363y-2112y²-4096y³-20 [7+128x,11+128y]: unknown -> [112] [0,0] 14x+128x²-363y-4224y²-16384y³-10 [71+128x,11+128y]: unknown -> [113] [1,0] 142x+128x²-363y-4224y²-16384y³+29 endexp[56] expanding queue[57]^28,meter=[2,2]: 78x+64x²-363y-2112y²-4096y³+3 [39+128x,75+128y]: unknown -> [114] [0,1] 78x+128x²-16875y-28800y²-16384y³-3284 [103+128x,75+128y]: unknown -> [115] [1,1] 206x+128x²-16875y-28800y²-16384y³-3213 endexp[57] expanding queue[58]^29,meter=[2,2]: 46x+64x²-5547y-8256y²-4096y³-1234 [23+128x,43+128y]: unknown -> [116] [0,0] 46x+128x²-5547y-16512y²-16384y³-617 [87+128x,43+128y]: unknown -> [117] [1,0] 174x+128x²-5547y-16512y²-16384y³-562 endexp[58] expanding queue[59]^29,meter=[2,2]: 110x+64x²-5547y-8256y²-4096y³-1195 [55+128x,107+128y]: unknown -> [118] [0,1] 110x+128x²-34347y-41088y²-16384y³-9547 [119+128x,107+128y]: unknown -> [119] [1,1] 238x+128x²-34347y-41088y²-16384y³-9460 endexp[59] expanding queue[60]^30,meter=[2,2]: 30x+64x²-2187y-5184y²-4096y³-304 [15+128x,27+128y]: unknown -> [120] [0,0] 30x+128x²-2187y-10368y²-16384y³-152 [79+128x,27+128y]: unknown -> [121] [1,0] 158x+128x²-2187y-10368y²-16384y³-105 endexp[60] expanding queue[61]^30,meter=[2,2]: 94x+64x²-2187y-5184y²-4096y³-273 [47+128x,91+128y]: unknown -> [122] [0,1] 94x+128x²-24843y-34944y²-16384y³-5870 [111+128x,91+128y]: unknown -> [123] [1,1] 222x+128x²-24843y-34944y²-16384y³-5791 endexp[61] expanding queue[62]^31,meter=[2,2]: 62x+64x²-10443y-11328y²-4096y³-3194 [31+128x,59+128y]: unknown -> [124] [0,0] 62x+128x²-10443y-22656y²-16384y³-1597 [95+128x,59+128y]: unknown -> [125] [1,0] 190x+128x²-10443y-22656y²-16384y³-1534 endexp[62] expanding queue[63]^31,meter=[2,2]: 126x+64x²-10443y-11328y²-4096y³-3147 [63+128x,123+128y]: unknown -> [126] [0,1] 126x+128x²-45387y-47232y²-16384y³-14507 [127+128x,123+128y]: unknown -> [127] [1,1] 254x+128x²-45387y-47232y²-16384y³-14412 endexp[63] ---------------- level 7 Maximum level 7 [128] mod 2: x²-y³+2
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clc(); clear; //Given : Q = 4 ;// in MeV Ex = 2; // in MeV Ey = 5 ; // in MeV mx = 4; // in u my = 1 ; // in u My =13; // in u theta = acosd(( (Ey*(1 + (my/My))) - (Ex*(1 - (mx/My))) - Q )/((2/My)*sqrt(mx*Ex*my*Ey))); // angle of ejection in degrees printf("Angle of ejection is %.0f degrees",theta);
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code2str.man.tst
clear;lines(0); code2str([-28 12 18 21 10 11])
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ex5_8.sce
clc; clear all; u0=1.658;//given u0 ue=1.486;//given ue lamda=5893*1e-8 //in cm t=lamda/(4*(u0-ue));//Thicknesss of quarter wave plate disp(+'cm',t,'Thicknesss of quarter wave plate =')
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1_7_1.sce
//Transport Processes and Seperation Process Principles //Chapter 1 //Example 1.7-1 //Introduction to engineering principles and units //given data //Hf data at 298K //Input items: sum of the enthalpies of two streams relative to 298K //calculating H of liq Hil=2000*4.06*(30-25);//inlet mass flow rate of the liquid=2000 kg/h,Cp= 4.06 kJ/kg K, final temp-initial temp= 30 deg C - 25 deg C //Hiw(enthalpy at inlet of water)=W(4.21)(95-25) where W in kg/h Cp of water is 4.21 kJ/kg K, 95-25 is the temp diff //Output items Hol=2000*4.06*(70-25);//outlet mass flow rate of liquid is 2000 kg/h, Cp= 4.06 kJ/kg K 70-25: temp diff //How= W(4.21)(85-25) //energy at inlet = energy at outlet //4.060*10^4 + 2.947*10^2 W= 3.654*10^5 + 2.526*10^2 W // solving these equations: W= ((4.060*10^4)-(3.654*10^5))/((2.526*10^2)-(2.947*10^2)) mprintf("the outlet feed rate in kg/h is %f",W) //calculating enthalpy change of liquid: delH= Hol-Hil; mprintf(" change in enthalpy in kw in kJ/h is %f",delH) //end //s=all the calculations performed are correct but there may be certein deviations.
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Exa_3_4.sce
//Example No. 3.4 clc; clear; close; format('v',6); //Given Data : MotorOutput=200;//KW v=60;//Km/hr w=400;//KN friction=5;//N/KN weight tan_theta=1/100;//inclination g=9.81;// gravity constant //Solution : sin_theta=tan_theta; W_sin_theta=w*1000*sin_theta;//N R=friction*W_sin_theta/10;//frictional resistance in N P=W_sin_theta+R;//N v=60*1000/60/60;//m/s Power=P*v;//Watt Pdash=MotorOutput*1000-Power;//Power causes acceleration in watt or N-m/s m=w*1000/g;//in Kg a=Pdash/m;//in m/s^2 disp(a,"Acceleration in m/s^2 : ");
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/3472/CH22/EX22.2/Example22_2.sce
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Example22_2.sce
// A Texbook on POWER SYSTEM ENGINEERING // A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar // DHANPAT RAI & Co. // SECOND EDITION // PART II : TRANSMISSION AND DISTRIBUTION // CHAPTER 15: INSULATION CO-ORDINATION // EXAMPLE : 15.2 : // Page number 399 clear ; clc ; close ; // Clear the work space and console // Given data V_hv = 132.0 // Voltage at the HV side of transformer(kV) V_lv = 33.0 // Voltage at the LV side of transformer(kV) V = 860.0 // Insulator allowable voltage(kV) Z = 400.0 // Line surge impedance(ohm) BIL = 550.0 // BIL(kV) // Calculations V_rating_LA = V_hv*1.1*0.8 // Voltage rating of LA(kV) E_a = 351.0 // Discharge voltage at 5 kA(kV) I_disc = (2*V-E_a)*1000/Z // Discharge current(A) L_1 = 37.7 // Separation distance in current b/w arrester tap and power transformer tap(m) dist = 11.0 // Lead length from tap point to ground level(m) de_dt = 500.0 // Maximum rate of rise of surge(kV/µ-sec) Inductance = 1.2 // Inductance(µH/metre) di_dt = 5000.0 // di/dt(A/µ-sec) lead_drop = Inductance*dist*di_dt/1000 // Drop in the lead(kV) E_d = E_a+lead_drop // (kV) V_tr_terminal = E_d+2*de_dt*L_1/300 // Voltage at transformer terminals(kV) E_t = BIL/1.2 // Highest voltage the transformer is subjected(kV) L = (E_t-E_a)/(2*de_dt)*300 // Distance at which lightning arrester located from transformer(m) L_lead = (E_t-E_a*1.1)/(2*de_dt)*300 // Distance at which lightning arrester located from transformer taken 10% lead drop(m) // Results disp("PART II - EXAMPLE : 15.2 : SOLUTION :-") printf("\nRating of L.A = %.1f kV", V_rating_LA) printf("\nLocation of L.A, L = %.f m", L) printf("\nLocation of L.A if 10 percent lead drop is considered, L = %.1f m", L_lead) printf("\nMaximum distance at which a ligtning arrester is usually connected from transformer is %.f-%.f m", L-2,L+3)