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FibonacciElementVME.tst
// This file is part of the materials accompanying the book // "The Elements of Computing Systems" by Nisan and Schocken, // MIT Press. Book site: www.idc.ac.il/tecs // File name: projects/08/FunctionCalls/FibonacciElement/FibonacciElementVME.tst load, // Load all the VM files from the current directory output-file FibonacciElement.out, compare-to FibonacciElement.cmp, output-list RAM[0]%D1.6.1 RAM[261]%D1.6.1; set sp 261, repeat 110 { vmstep; } output;
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// Calculation of maximum punch force clc L = 30 // diameter of punching in mm t = 3 // thickness of sheet in mm UTS = 1e3 // Tensile strength in MN printf("\n Example 7.1") F = 0.7*UTS*t*1e-3*L*1e-3*%pi printf("\n Maximum required punching force is %.3f MN.",F) // Answer in book is 0.197 MN
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clc //initialisation of variables h= 8 //ft h1= 10 //ft //CALCULATIONS A= h X= (h1/2) Ig= h^3/12 I0= Ig+A*X^2 h2= I0/(A*X) //RESULTS printf ('depth at which the hinge of the shutter= %.2f ft ',h2)
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clf; plot2d(0,0,0,rect=[0,0,10,10],frameflag=3) xgrid(4) // grid xfrect(0,4,2,4) // first rectangle E=gce(); E.background=1; // black inside xrect(2,4,2,2) // second rectangle (square) E=gce(); E.foreground=5; // red outline xrect(4,6,2,2) // third rectangle (square) E=gce(); E.foreground=2; // blue outline E.line_style=3; // dotted line E.thickness=5; // thickness xfrect(6,8,4,2) // fourth rectangle E=gce(); E.background=7; // yellow inside
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//Example 2 // intensity clc; clear; close; I=1;//assume a1=1*I;// a2=4*I;// ph1=0;//degree i1=(a1+a2)+a2*cosd(ph1);// disp("intensity where phase difference is zero is "+string(i1)+"*I") ph2=90;//degree i2=(a1+a2)+a2*cosd(ph2);// disp("intensity where phase difference is pi/2 is "+string(i2)+"*I") ph3=180;//degree i3=(a1+a2)+a2*cosd(ph3);// disp("intensity where phase difference is pi is "+string(i3)+"*I")
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clc //Initialization of variables g=9.81 //kN/m^2 hc=16.25 //m l=1.5 //m b=2.5 //m f=0.3 Pi=50 //kN //calculations P=g*hc*l*b Preq=Pi+f*P //results printf("Force required to lift the gate = %.2f kN",Preq)
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clear; clc; disp('Example 5.12'); // aim : T0 determine // (a) change in internal nergy of the air // (b) work done // (c) heat transfer // Given values m = .25;// mass, [kg] P1 = 140;// initial pressure, [kN/m^2] V1 = .15;// initial volume, [m^3] P2 = 1400;// final volume, [m^3] cp = 1.005;// [kJ/kg K] cv = .718;// [kJ/kg K] // solution // (a) // assuming ideal gas R = cp-cv;// [kJ/kg K] // also, P1*V1=m*R*T1,hence T1 = P1*V1/(m*R);// [K] // given that process is polytropic with n = 1.25; // polytropic index T2 = T1*(P2/P1)^((n-1)/n);// [K] // Hence, change in internal energy is, del_U = m*cv*(T2-T1);// [kJ] mprintf('\n (a) The change in internal energy of the air is del_U = %f kJ',del_U); if(del_U>0) disp('since del_U>0, so it is gain of internal energy to the air') else disp('since del_U<0, so it is gain of internal energy to the surrounding') end // (b) W = m*R*(T1-T2)/(n-1);// formula of work done for polytropic process,[kJ] mprintf('\n (b) The work done is W = %f kJ',W); if(W>0) disp('since W>0, so the work is done by the air') else disp('since W<0, so the work is done on the air') end // (c) Q = del_U+W;// using 1st law of thermodynamics,[kJ] mprintf('\n (c) The heat transfer is Q = %f kJ',Q); if(Q>0) disp('since Q>0, so the heat is received by the air') else disp('since Q<0, so the heat is rejected by the air') end // End
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////Chapter No 14 Air Standard Cycles ////Example No 14.6 Page No 308 ///Find standard efficiency //Input data clc; clear; T1=27+273; //Initial temp in degree celsius T2=450+273; //Final temp in degree celsius gamma1=1.4; //Calculation r=(T2/T1)^(1/(gamma1-1)); //Isentropic process eta=100*(1-(1/(r^(gamma1-1)))); //Otto cycle air standard effeciency in % //Output printf('compression ratio= %f \n ',r); printf('standard efficiency= %f percent \n',eta);
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//example 11.12 clc; funcprot(0); z1=21/2; Lg=9; Bg=6; Qg=500*1000; Cc1=0.3; Cc2=0.2; Cc3=0.25; H2=12; H3=6; H1=21; e1=0.82; e2=0.7; e3=0.75; s1=Qg/(Lg+z1)/(Bg+z1); //sigma1 s2=500*1000/(9+27)/(6+27);//sigma2 s3=500*1000/(9+36)/(6+36);//sigma3 ss1=6*105+(27+21/2)*(115-62.4);//sigmadash1 ss2=6*105+(27+21)*(115-62.4)+(120-62.4)*6;//sigmadash2 ss3=6*105+48*(115-62.4)+12*(120-62.4)+3*(122-62.4);//sigmadash3 sc1=Cc1*H1/(1+e1)*log10((ss1+s1)/ss1); sc2=Cc2*H2/(1+e2)*log10((ss2+s2)/ss2); sc3=Cc3*H3/(1+e3)*log10((ss3+s3)/ss3); sc=sc1+sc2+sc3; disp(sc*12,"total settlement in inch");
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clear //given //how many gallon does the same tank hold of 25 inches x=25. y=64 z=x/y printf("\n \n gallons does hold %.2f ",z) //volume fraction //1783 is last problem got result we are using it same one Z=0.3611 Ct=Z*1783. printf("\n \n content of tank %.2f gal",Ct)
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clear; clc; close; disp("given y[n]=x[n]+x[n-1] this can be converted to high pass filter by multiplying with (-1)^n") disp("then y[n]=x[n]-x[n-1] taking fourier transform"); disp("H(w)=1-e^-j*w"); w=-3:0.01:3; Hw=1-%e^(-%i*w); subplot(2,1,1) plot(w,abs(Hw)); xtitle('|H(w)|','w') subplot(2,1,2) plot(w(1:find(w==0)-1),phasemag(Hw(1:find(w==0)-1))*%pi/180) a=gca(); a.y_location="origin"; plot(w(find(w==0)+1:$),phasemag(Hw(find(w==0)+1:$))*%pi/180) xtitle('phase(H(w))','w')
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//CHAPTER 4- MEASURING INSTRUMENTS //Example 2 clc; disp("CHAPTER 4"); disp("EXAMPLE 2"); //VARIABLE INITIALIZATION G=10; //galvanometer resistance in Ohms S=1; //shunt resistance in Ohms r=12; //total resistance in Ohms emf=2; //emf of cell in Volts //SOLUTION I=emf/r; //current in the circuit I_g=(S*I)/(S+G); disp(sprintf("The current through the galvanometer is %f A",I_g)); //END
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//Example No. 3.3 clc; clear; close; format('v',8); //Given Data : 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 disp(Power/1000,"Final KW rating of the motor of train : "); Force=P;//down the inclined force in N u=v;//initial velocity in m/s v=0;//final velocity in m/s m=w*1000/g;//in Kg KE=1/2*m*u^2;//in Joule d=KE/P;//distance in meter disp(d,"Distance covered in meter : ");
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// Example 13_2 clc;funcprot(0); // Given data D_piston=2.00;// ft W_out=20;// hp L=4.00;// ft/stroke m_b=4000;// lbf d=15.0;// ft Duty=35.0*10^6; N=18.0;// strokes per minute // Calculation // (a) A=(%pi*D_piston^2)/4;// ft^2 W_out=20*33000;// ft.lbf/min p_avg=W_out/(A*L*N);// lbf/ft^2 p_avg=p_avg/144;// lbf/in^2 // (b) n_T=(Duty/(8.5*10^8))*100;// The actual thermal efficiency of the engine in % // (c) W_out=20;// hp Q_boiler=(W_out*2545)/(n_T/100);// Btu/h printf("\n(a)The average pressure of the cycle,p_avg=%2.1f lbf/in^2 \n(b)The actual thermal efficiency of the engine,n_T=%1.2f percentage \n(c)The heat rate produced by the boiler,Q_boiler=%1.2e Btu/h",p_avg,n_T,Q_boiler);
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function [t,d,v]=euler2(a,b,h,d0,v0) t=a:h:b d(1)=d0 v(1)=v0 n=length(t) for i=2:n d(i)=d(i-1)+h*df(t(i-1),d(i-1),v(i-1)) v(i)=v(i-1)+h*df2(t(i-1),d(i-1),v(i-1)) end endfunction function a=df(t,d,v) a=v endfunction function r=df2(t,d,v) K=5/7; g=25634.7 teta=0.0913 r=-K*g*sin(teta) endfunction
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// Exa 8.9 format('v',6) clc; clear; close; // Given data R_C= 1;// in k ohm V_CC= 5;// in V V_CEsat= 0;// in V V_BE= 0.7;// in V bita_min= 50; bita_max= 100; // For the transistor to go to saturation, I_C= (V_CC-V_CEsat)/R_C;// in mA bita= bita_min;// for driving the transistor into saturation I_Bmin= I_C/bita;//minimum base current in mA // So, (V_CC-V_BE)/R_B >= I_B or R_B= (V_CC-V_BE)/I_Bmin;// in k ohm disp(R_B,"The maximum permissible value of R_B in k ohm is : ") // For actual calculation one may take V_CEsat= 0.3 V V_CEsat= 0.3;// in V I_C= (V_CC-V_CEsat)/R_C;// in mA bita= bita_min;// for driving the transistor into saturation I_Bmin= I_C/bita;//minimum base current in mA // So, (V_CC-V_BE)/R_B >= I_B or R_B= (V_CC-V_BE)/I_Bmin;// in k ohm disp("For actual calculation at V_CE(sat) = 0.3 V, the maximum value ") disp(R_B,"of R_B in k ohm is : ")
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src= imread("../images/image_0197.jpg"); //another image gray = rgb2gray(src); loc=localMaximaFinder(gray,[1 3],4,1);
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Time = 0.000 Temperature = 0.22 Volume = 0.00 Pressure = -4.51E+08 Energy = 0.000E+00 Time = 0.002 Temperature = 28.62 Volume = 404.81 Pressure = -4.35E+08 Energy = -5.684E+05 Time = 0.004 Temperature = 110.24 Volume = 404.41 Pressure = -4.48E+08 Energy = -5.598E+05 Time = 0.006 Temperature = 132.03 Volume = 404.00 Pressure = -4.72E+08 Energy = -5.379E+05 Time = 0.008 Temperature = 141.38 Volume = 403.57 Pressure = -4.87E+08 Energy = -5.342E+05 Time = 0.010 Temperature = 147.51 Volume = 403.12 Pressure = -4.92E+08 Energy = -5.333E+05 Time = 0.012 Temperature = 151.01 Volume = 402.67 Pressure = -4.88E+08 Energy = -5.325E+05 Time = 0.014 Temperature = 152.14 Volume = 402.23 Pressure = -4.76E+08 Energy = -5.318E+05 Time = 0.016 Temperature = 152.58 Volume = 401.79 Pressure = -4.60E+08 Energy = -5.310E+05 Time = 0.018 Temperature = 154.21 Volume = 401.38 Pressure = -4.40E+08 Energy = -5.301E+05
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//Electric Power Generation, Transmission and Distribution by S.N.Singh //Publisher:PHI Learning Private Limited //Year: 2012 ; Edition - 2 //Example 10.8 //Scilab Version : 6.0.0 ; OS : Windows clc; clear; d=0.25; //Diameter of conductor in m r=0.0125; //Radius of conductor in m Dab=5; //Distance between conductors a & b in m Dbc=4; //Distance between conductors b & c in m Dac=6; //Distance between conductors a & c in m Deq=nthroot((Dab*Dbc*Dac),3); //Diameter equivalent of line in m Can=(2*%pi*10^(-9)/(36*%pi))/log(Deq/r); //Capacitance between phase a & neutral in pF/m printf("\nThe capacitance of the transmissin line is %.2f pF/m",Can/(10^-12));
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//Ex:2.17 clc; clear; close; r_o=40;//resis at 0 degree r_t=44;//at 100 degree t=100;//temperature diff. temp_coeff=(1/t)*((r_t/r_o)-1); printf("Temperature Coefficient = %f per degree centigrade",temp_coeff);
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exec("evalpoly.sce", 0) exec("deflpoly.sce", 0) exec("my_laguerre.sce", 0) exec("polyroots.sce", 0) a = [1, 3, 4, 5, 6, 7, 8]; r0 = polyroots(a) r0_sci = roots(a) n = length(r0) printf("\nResult of built-in roots function of Scilab:\n") for i = 1:n printf("Root %d = %18.10f %18.10f\n", i, real(r0_sci(i)), imag(r0_sci(i))) end if getscilabmode() ~= "STD" quit() end
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//clear// clear; clc; //Example 24.3 //Given Tw = 80; //[F] Tdb = 120; //[F] v = 3.5; //[ft/s] rho = 120; //[lb/ft^3] Xe = 0; Xc = 0.09; lambda = 1049; //[Btu/lb] M = 29; B = 24; //[in.] D = 2; //[in.] Dc = 2; //[ft] X2 = 0.20; X1 = 0.10; Dcyl = 1/4; //[in.] L = 4; //[in.] Vbar = 3.5; //[ft/s] Thb = 120; //Solution //Since the Xc is less than 10 percent, all drying takes place //in the constant-rate period and the vaporrization temperature, //as before, is 80 F. //From Exapmle 24.1, mass of water to be evaporated mdot = 8*(X2-X1); //[lb] //The quantity of heat to be transferred QT = mdot*lambda; //[Btu] //mass of the dry soild in one cylinder is mp = %pi/4*(Dcyl/12)^2*(L/12)*rho; //[lb] //surface area of one cylinder is Ap = %pi*(Dcyl/12)*(L/12); //[ft^2] //Total area exposed by 8 lb solids A = 8/mp*Ap; //[ft^2] //The heat transfer coefficient is found from the //equivalent form of Eq.(21.62) //hDbyk = 1.17*Nre^0.585*Npr^(1/3) //For air at 1 atm and 120F, the properties are rho_a = M/359*492/580; //[lb/ft^3] mu_a = 0.019; //[cP], from Appendix 8 k_a = 0.0162; //[Btu/ft-h-F], from Appendix 12 Cp_a = 0.25; //[Btu/lb-F], from Appendix 15 Nre = 1/48*Vbar*rho_a/(mu_a*6.72*10^-4); Npr = mu_a*2.42*Cp_a/k_a; //Form Eq.(21.62) h = (k_a*1.17*Nre^0.585*Npr^(1/3))/(1/48); //[Btu/ft^2-h-F] mdot_g = v*3600*rho_a; //[lb] //From Fig. 23.2 cs = 0.25; delta_Thb = Thb-Tw; //[F] delta_Tha = 8.24; //[F] //The heat transferred form the gas to a thin section of the bed delta_TL = (delta_Thb-delta_Tha)/log(delta_Thb/delta_Tha); //[F] //rate of heat transfer qT = h*A*delta_TL; //[Btu/h] //drying time tT = QT/qT; //[h] disp('h',tT,'Required drying time is')
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4_2.sce
clc,clear printf('Example 4.2\n\n') Slots=120 Pole=8 Phase=3 //number of phases n=Slots/Pole //Slots per Pole m=Slots/Pole/Phase //Slots per Pole per Phase beeta=180/n //Slot angle in degree K_d=sind(m*beeta/2) /(m*sind(beeta/2)) //Distribution Factor printf('Distribution Factor:\nK_d=%.3f',K_d)
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Example_a_3_2.sce
//Example_a_3_2 page no:129 clc; R=(((8+1.07)*1)/(8+1.07+1))+1; I1=10/R; It=5/5.8; I5=(It*10)/18.5; I2=(I5*1)/2; Ir3=(10*6.07)/(6.07+3.5); I3=(Ir3*(1/2)); I=I1-I2-I3; disp(I,"the current passing through the circuit is (in A)");
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10_1.sce
clear; clc; printf("\t\t\tExample Number 10.1\n\n\n"); // overall heat transfer coefficient for pipe in air // Example 10.1 (page no.-520-522) // solution Tw = 98;// [degree celsius] temperature of hot water k_p = 54;// [W/m degree celsius] heat transfer coefficient of pipe Ta = 20;// [degree celsius] atmospheric air temperature u = 0.25;// [m/s] water velocity // from appendix A the dimensions of 2-in schedule 40 pipe are ID = 0.0525;// [m] OD = 0.06033;// [m] // the properties of water at 98 degree celsius are rho = 960;// [kg/cubic meter] mu = 2.82*10^(-4);// [kg/m s] k_w = 0.68;// [W/m degree celsius] Pr = 1.76;// prandtl number // the reynolds number is Re = rho*u*ID/mu; // and since turbulent flow is encountered, we may use equation(6-4): Nu = 0.023*Re^(0.8)*Pr^(0.4); hi = Nu*k_w/ID;// [W/square meter degree celsius] // for unit length of pipe the thermal resistance of the steel is Rs = log(OD/ID)/(2*%pi*k_p); // again, on a unit length basis the thermal resistance on the inside is Ai = %pi*ID;// [square meter] Ri = 1/(hi*Ai); Ao = %pi*OD;// [square meter] // the thermal resistance for outer surface is as yet unknown but is written, for unit lengths, is Ro = 1/(ho*Ao) (a) // from table 7-2(page no.-339), for laminar flow, the simplified relation for ho is // ho = 1.32*(dT/d)^(1/4) = 1.32*((To-Ta)/OD)^(1/4) (b) // where To is the unknown outside pipe surface temperature. we designate the inner pipe surface as Ti and the water temperature as Tw; then the energy balance requires // (Tw-Ti)/Ri = (Ti-To)/Rs = (To-Ta)/Ro (c) // combining equations (a) and (b) gives // (To-Ta)/Ro = %pi*OD*1.32*(To-Ta)^(5/4)/OD^(1/4) // this relation may be introduced into equation (c) to yield two equations with the two unknowns Ti and To: // (Tw-Ti)/Ri = (Ti-To)/Rs (1) // (Ti-To)/Rs = %pi*OD*1.32*(To-Ta)^(5/4)/OD^(1/4) (2) // this is a non-linear equation which can be solved as for Ti = 50:0.001:100 Q = ((Ti-(Ti-(Tw-Ti)*(Rs/Ri)))/Rs)-(%pi*OD*1.32*((Ti-(Tw-Ti)*(Rs/Ri))-Ta)^(5/4)/OD^(1/4)); if Q>0 & Q<6 then Tinew = Ti; else Ti = Ti; end end Ti = Tinew;// [degree celsius] To = (Ti-(Tw-Ti)*(Rs/Ri));// [Degree celsius] // as a result, the outside heat transfer coefficient and thermal resistance are ho = 1.32*((To-Ta)/OD)^(1/4);// [W/square meter degree celsius] Ro = 1/(OD*7.91*%pi);// // the overall heat transfer coefficient based on the outer area is written in terms of these resistances as Uo = 1/(Ao*(Ri+Ro+Rs));// [W/area degree celsius] // in this calculation we used the outside area for 1.0 m length as Ao // so Uo = Uo;// [W/square meter degree celsius] printf("overall heat transfer coefficient is %f W/square meter degree celsius",Uo);
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eg2_3.sce
clear; clc; s=5*10^6; l=50*10^3; pf=0.8; eff=.9; v=33*10^3; rho=2.85*10^(-8); pl=0.1*s*pf; i=s/v; a1=2*i*i*rho*l/pl; vol=2*l*a1; printf("the volume of theconductor required is:%.2f cubic meter",vol); //b) il=s/(sqrt(3)*v); a2=3*il*il*rho*l/pl; vol=3*l*a2 printf("\n the volume of theconductor required is:%.2f cubic meter",vol);
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Chapter1_example3.sce
clc clear //Input data m=0.2//Mass of the ball in kg r=1.5//Radius of vertical circle in m q=35//Angle made by the ball in degrees v=6//Velocity of the ball in m/s //Calculations T=(m*((v^2/r)+(9.8*cosd(q))))//Tension in the string in N at=9.8*sind(q)//Tangential acceleration in m/s^2 ar=(v^2/r)//Radial acceleration in m/s^2 a=sqrt(at^2+ar^2)//Acceleration in m/s^2 //Output printf('Tension in the string is %3.1f N \n Tangential acceleration is %3.2f m/s^2 \n Radial acceleration is %i m/s^2',T,at,ar)
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Example10_14.sce
/////////Chapter 10 Properties Of Steam ///Example 10.14 Page No:197 //// Find Degree of superheat ///Input Data clc; clear; P=7; //Absolute pressure in bar t=200; //Absolute temperature ts=165; //In degree celsius from steam table //Calculation dos1=t-ts; //Degree of superheat in degree celcius //Output printf('Degree of superheat=%f degree celsius \n ',dos1);
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Exa9_4.sce
//Exa 9.4 clc; clear; close; //given data : HT=100;//in meter d=60;//in Km //Formula : d=4.12*(sqrt(HT)+sqrt(HR));//in Km HR=(d/4.12-sqrt(HT))^2;//in meter disp(HR,"Height of receiving antenna in meter : ");
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7_1.sce
RL=100 Zo=50 PM=0.05 c=3*10^8 f=900*10^6 lambda=c/f Z1=sqrt(RL*Zo) l=lambda/4 fractional_bandwidth=2-4/%pi*acos(abs(2*PM*sqrt(Zo*RL)/(RL-Zo)/sqrt(1-PM^2))) printf("\nZ1=%f ohm\nl=%.4f m\nfractional bandwidth=%.7f",Z1,l,fractional_bandwidth)
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Exa11_1.sce
//Exa 11.1 clc; clear; close; //Given data R1=15;//in kohm R2=15;//in kohm C1=0.005;//in uF C2=0.005;//in uF R=R1;//in Kohm C=C1;//in uF T=0.69*(R*10^3*C*10^-6+R*10^3*C*10^-6);//in second f=1/T;//in Hz disp(f*10^-3,"Frequency of oscillators in KHz : ");
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Ex2_8.sce
//exa 2.8 clc;clear;close; format('v',8); //dC1/dP1=0.2*P1+22;//Rs./MWh //dC2/dP2=0.15*P2+30;//Rs./MWh B22=0;B12=0;//Because Loss is independent wrt P2 P1=100;//MW PL=15;//MW B11=PL/P1^2;//MW^-1 L1=1/[1-0.003*P1];//Penalty Factor plant 1 L2=1;//Penalty Factor of plant 2 lambda=60; //lambda=dC1/dP1*L1=dC2/dP2*L2 //dC1/dP1*L1=dC2/dP2*L2 P2=((0.2*P1+22)*L1-30)/0.15;//MW P=P1+P2-B11*P1^2;//MW//Total Load disp(P1,"Required generation at plant1(MW)"); disp(P2,"Required generation at plant2(MW)"); disp(P,"Total Load(MW)");
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Ex12_19.sce
clear //Given L=500*10**-3 I1=20*10**-3 //A I2=10*10**-3 //A //Calculation U1=0.5*L*I1**2 U2=0.5*L*I2**2 //Result printf("\n Magnetic energy stored in the coil is %0.3f *10**-4 J",U1*10**6) printf("\n New value of energy is %0.3f J",U2)
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sadiku_11_5.sce
clear; clc; format('v',6); Zl=100+150*%i; Zo=75; zl=Zl/Zo; T=(Zl-Zo)/(Zl+Zo); disp(T,'T ='); s=(1+abs(T))/(1-abs(T)); disp(s,'s =') format('v',5); Yl=1/Zl; disp(Yl*1000,'Load admittance in mS'); B=2*%pi,l=.4; Zin=Zo*(Zl+Zo*tan(B*l)*%i)/(Zo+Zl*tan(B*l)*%i); format('v',6); disp(Zin,'Zin at .4 l from load')//for .4l B=2*%pi,l=.6; Zin=Zo*(Zl+Zo*tan(B*l)*%i)/(Zo+Zl*tan(B*l)*%i); format('v',6); disp(Zin,'Zin at .6 l from load')//for .6l
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Ex4_5.sce
clc //initialisation of variables p=8000*10^-3//liters/min r=15*10^-2//cm v=2.5//m //CALCULATIONS V1=(4*(p)*(1/60))/(%pi*(r)^2)//m/s D=sqrt(4*(p)*(1/60)/(%pi*v))*100//cm //RESULTS printf('The diameter of the suction line is=% f cm',D)
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14Q1.sce
clc G=15.7 a=0 b=15 c=30 H=3 Kp=4.977 // from table 13.9 Pp=Kp*G*H^2/2 printf('a)the passive force = %f kN/m\n',Pp) // for part b Kp=4.53 Pp=Kp*G*H^2/2 printf(' b)the passive force = %f kN/m\n',Pp) // for part c Kp=4.13 Pp=Kp*G*H^2/2 printf(' c)the passive force = %f kN/m\n',Pp) //for part d Kp=4.56 Pp=Kp*G*H^2/2 printf(' d)the passive force = %f kN/m\n',Pp)
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ex_3_26.sce
//Example 3.26 : density clc; clear; close; //given data : format('v',5) n=4; N=6.023*10^23; // avogadro's number r=1.278*10^-8;// in cm A=63.5; a=(r*4)/sqrt(2);// in cm b=(A*n)/(a^3*N); disp(b,"density of copper,b(g/cc) = ")
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Chapter9_Example13.sce
clc clear //INPUT t1=915;//temperature at the beggining in K t2=2040;//temperature at the end in K d=12.6;//adiabatic expansion ratio y=1.39;//coefficent of expansion //CALCULATIONS x=t2/t1;//ratio temparatures n=1-(1/d)^(y-1)*((x^y)-1)/(y*(x-1));//efficiency of the engine //OUTPUT mprintf('the efficiency of the engine is %3.3f',n)
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q=170//L/s(Discharge) t=60//degrees g=9.81//m/s^2(Acceleration due to gravity)
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minimum_spanning_tree.sce
//// minimum_spanning_tree // Construct minimum spanning tree on the mesh. Replace Matlab's // graphminspantree function. // // Basically this is a classical implementation of minimum spanning tree // algorithm, using adjacency matrix. Speed is about 2(large mesh)-4(smaller mesh) // times slower comparing with Matlab's built-in graphminspantree, which is // implemented via mex function graphalgs. // //// Syntax // [tree,previous] = minimum_spanning_tree(graph) // [tree,previous] = minimum_spanning_tree(graph,source) // //// Description // graph : sparse matrix, nv x nv, adjacency matrix of graph (or triangle // mesh), elements are weights of adjacent path // source: integer scaler, optional, source node of spanning tree. If not // provided, will search for smallest node in the graph. // // tree: sparse matrix, nv x nv, minimum spanning tree, if there are k // nodes in the graph, then there are k+1 nonzeros elements in tree. // previous: double array, n x 1, predecessor nodes of the minimal spanning // tree, predecessor of source node is 0 // //// Contribution // Author : Wen Cheng Feng // Created: 2014/03/21 // Revised: 2014/03/24 by Wen, add doc // // Copyright 2014 Computational Geometry Group // Department of Mathematics, CUHK // http://www.lokminglui.com function [tree,previous] = minimum_spanning_tree(graph,source) if ~exist('source','var') [I,J] = find(graph); source = J(1); end nv = size(graph,1); previous = nan(nv,1); node = source; rem = (1:nv)'; // rem(source) = []; ind = false(nv,1); ind(I) = true; ind(source) = false; rem(~ind) = []; previous(source) = 0; nvc = sum(ind); TI = zeros(nvc,1); TJ = zeros(nvc,1); TV = zeros(nvc,1); k = 1; // most time is spent on accessing submatrix, overall speed is about two // times slower than Matlab's build-in function graphminspantree. For smaller // graph, it's even slower, say, four times. while ~isempty(rem) [I,J,V] = find(graph(node,rem)); // time consuming, need to improve [v,ind] = min(V); i = node(I(ind)); j = rem(J(ind)); TI(k) = i; TJ(k) = j; TV(k) = v; k = k+1; previous(j) = i; // node(k) = j; node = [node;j]; rem(J(ind)) = []; end tree = sparse(TI,TJ,TV,nv,nv); tree = tril(tree+tree');
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// Chapter 3 example 21 //------------------------------------------------------------------------------ clc; clear; // Given data a = 6; // width of waveguide in cm b = 3; // narrow dimension of waveguide in cm lamda = 4; // operating wavelength in cm c = 3*10^8; // velocity of EM wave in cm/s // Calculations lamda_c = 2*a; // cut-off wavelength in dominant mode lamda_g = lamda/(sqrt(1 - (lamda/lamda_c)^2)) // guide wavelength Vp = (lamda_g/lamda)*c b = (2*%pi)/lamda_g; // phase shift constant // Output mprintf('Guide wavelength = %3.2f cm\n Phase velocity = %3.2e m/s\n Phase shift constant = %3.2f radians/cm',lamda_g,Vp,b) //------------------------------------------------------------------------------
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function [f]=fwhmjlt(y,varargin) //This function Computes peak full-width at half maximum //calling sequence //f = fwhm (y) //f = fwhm (x, y) //f = fwhm (…, "zero") //f = fwhm (…, "min") //f = fwhm (…, "alevel", level) //f = fwhm (…, "rlevel", level) //Description //Compute peak full-width at half maximum (FWHM) or at another level of peak maximum for vector or matrix data y, optionally sampled as y(x). If y is a matrix, return FWHM for each column as a row vector. //The default option "zero" computes fwhm at half maximum, i.e. 0.5*max(y). The option "min" computes fwhm at the middle curve, i.e. 0.5*(min(y)+max(y)). //The option "rlevel" computes full-width at the given relative level of peak profile //The option "alevel" computes full-width at the given absolute level of y. //Example //t=-50:0.01:50; //y=(1/(2*sqrt(2*%pi)))*exp(-(t.^2)/8); //z=fwhmjlt(y) //Output: 470.96442 rhs = argn(2) if(rhs<1 | rhs>5) error("Wrong number of input arguments.") end select(rhs) case 1 then f = callOctave("fwhm",y) case 2 then f = callOctave("fwhm",y,varargin(1)) case 3 then f = callOctave("fwhm",y,varargin(1),varargin(2)) case 4 then f = callOctave("fwhm",y,varargin(1),varargin(2),varargin(3)) case 5 then f = callOctave("fwhm",y,varargin(1),varargin(2),varargin(3),varargin(4)) end endfunction
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clc; clear; rho=975 //[kg/m^3] k=0.671 //[W/(m.K)] mu=3.8*10^-4 //[N.s/m^2] dT=10 //[K] lambda=2300*10^3 //[J/kg] L=1 //[m] g=9.81 //[m/s^2] h=0.943*((rho^2*lambda*g*k^3)/(mu*L*dT))^(1/4) //W/(sq m.K) //[W/sq m.K] printf("\n (i)- Average heat transfer coefficient is %d W/(m^2.K)\n",round(h)); //Local heat transfer coefficient //at x=0.5 //[m] x=0.5 //[m] h=((rho^2*lambda*g*k^3)/(4*mu*dT*x))^(1/4) //[W/sq m.K] printf("\n (ii)-Local heat transfer coefficient at 0.5 m height is %d W/(sq m.K)\n",round(h)); delta=((4*mu*dT*k*x)/(lambda*rho^2*g))^(1/4) //[m] delta=delta*10^3 //[mm] printf("\n (iii)-Film thickness is %f mm",delta);
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// To find the value of measured Resistance R // example 7-1 in page 165 clc; // Given data I=0.5;//measured current in amps V=500;// voltmeter indication in volts Ra=10;//ammeter resistance in ohms //calculation R=(V/I)-Ra;// measured resistance printf("The value of R=%d ohm",R); //result // The value of R=990 ohm
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function xl=newton_raphson(x0,f,df,e) xl = x0 - f(x0)/df(x0) while abs(f(xl)) > e xl = xl - f(xl)/df(xl) end endfunction function y=func(x) y = -88.9*exp(-0.0472*x) + (17.3 + 0.527*x) endfunction function y=dfunc(x) y = 0.527+4.19608*exp(-0.0472*x) endfunction xm = newton_raphson(0,func,dfunc,0.1) x = linspace(-50,50) plot(x,func(x),'b-') plot(xm,func(xm),'ro')
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//Caption: Sampling Rate //Example 5.1 //page no 220 //Find Sampling Rate clear; clc; f1=4*10^3; f2=4.5*10^3; fsmin=2*f2; disp("kHz",fsmin/1000,"Sampling rate");
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clc; // page no 683 // prob no 18.8 // refer ex 18.7 fb=40*10^6;// bit rate in bps Pr_dBm=-62;//power at the receiver in dBm Pr=10^(Pr_dBm/10)*10^-3;// power at the receiver in W Eb=Pr/fb;// the energy per bit in J k=1.38*10^-23;//Boltzmann constant T=350; // the noise power density is No=k*T; // Energy per bit to noise density ratio in dB is Eb_No=10*log10(Eb/No); disp('dB',Eb_No,'Energy per bit to noise density ratio is ');
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clear ; clc; // Example 3.3 printf('Example 3.3\n\n'); //Page No. 59 // given P = 60000;/// Principal Amount in Pound i = 0.18;// Interest Rate n = 10;//years R = P*((i*(1+i)^n)/((1+i)^n -1));//Rate of Capital Recovery printf('The annual investment required is %.1f Pound\n',R)
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//Example 4.8: design clc; clear; close; vab=1;//V vr=2-vab;//V i=50;//mA R=vr/(i*10^-3);//ohm R1=(vr/10)/(i*10^-3);//ohm n=10;// tr1=n*R1;//ohm r2=2*R1;// l=100;//cm x=R1/l;// disp(R,"resistance R is,(ohm)=") disp(R1,"resistance R1 is,(ohm)=") disp(r2,"resistance R2 is,(ohm)=") disp(x,"resistance per cm of slide wire is ,(ohm/cm)=")
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gauss_p.sci
function x= gauss_p(A,b)//x=A\b para conferir resultado //resolução de sistemas com fatoração LU com pivotamento parcial // onde: x é o vetor solução // A é a matriz de coeficientes // b é o vetor coluna de estímulos [m,n]=size(A); //dimensão da matriz que deve ser quadrada if m~=n then error("A matriz deve ser quadrada"); end m = length(b);//comprimento de b if m~=n then error("Erro na dimensão do vetor b"); end //1° passo: aumentar a matriz A P = eye(n,n); //matriz de pivotamento //Algoritmo de eliminação progressiva for i=1:n-1 //i linha e j coluna //PIVOTAMENTO PARCIAL //i é linha [maior,k]=max(abs(A(i:n,i)))//maior valor dos elementos da matriz l = k+i-1; if l~=i then A([l,i],:)= A([i,l],:)//inverte as linhas P([l,i],:)= P([i,l],:)//inverte as linhas end //--- for j=i+1:n A(j,i) = A(j,i)/A(i,i); //multiplicador guardado onde fica 0 // for k =i+1:n //k=i+1 pois vai ser 0 em k=i // A(j,k)= A(j,k)-A(i,k)*A(j,i); //end //ou FOR IMPLÍCITO------- A(j,i+1:n)= A(j,i+1:n)-A(i,i+1:n)*A(j,i); disp(A,'A'); disp(P,'P'); end end //algoritmo de substituição progressiva b= P*b; //d=zeros(n,1) para o for implicito d(1)=b(1); for i=2:n //for j=1:i-1 //d(i)=d(i)-A(i,j)*d(j); //end d(i)= b(i)-A(i,1:i-1)*d(1:i-1); end disp(d,'d'); //substituição regressiva x = zeros(n,1); x(n)= d(n)/A(n,n); for i= n-1:-1:1 //for j=i+1:n // x(i)= x(i)+ A(i,j)*x(j); //end //Com FOR IMPLÍCITO x(i)= (d(i)-A(i,i+1:n)*x(i+1:n))/A(i,i); end endfunction
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errcatch(-1,"stop");mode(2);// Example 1.4, page no-54 temp=2.022 millivolt_cor=37.325 op=millivolt_cor-temp printf("Millivolt output available=% .3f",op) exit();
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/1853/CH4/EX4.37/Ex4_37.sce
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2020-04-09T02:43:26.499817
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Ex4_37.sce
//determine the current also power nd power factor V=200+%i*0 f=50 R1=30 L1=0.2 C1=10e-6 X1=2*%pi*f*L1 Z1=R1+%i*X1 R2=40 L2=0.12 X2=2*%pi*f*L2 Z2=R2+%i*X2 Z=(Z1*Z2)/(Z1+Z2) I=V/Z R=18.858//calculatimg Z and I we get R and Z,I Z=31.06 coso=R/Z I=6.44 P=I^2*R I1=(I*Z1)/(Z1+Z2) I2=(I*Z1)/(Z1+Z2) coso1=R1/Z1 P1=I1^2*R1 coso2=R2/Z2 P2=(I2)^2*R2 disp('P2 ='+string(P2)+ 'watt' ,'P1 ='+string(P1)+ 'watt ' , 'Total power factr='+string(coso)+'' , 'Total power='+string(P)+'watt' , 'total current ='+string(I)+'amps' , 'total impedance='+string(Z)+'ohms' )
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/projects/02/Or16WaY.tst
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Metaphor-pump/ecs
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2021-01-19T18:32:30.973439
2017-08-09T09:57:02
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73,225,947
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287
tst
Or16WaY.tst
load Or16Way.hdl, output-file Or16Way.out, output-list in%B2.16.2 out%B2.1.2; set in %B0000000000000000, eval, output; set in %B1111111111111111, eval, output; set in %B1000000000000000, eval, output; set in %B0000000100000001, eval, output; set in %B0010011000100110, eval, output;
88950f13d501a73f44d9a460b8c1651e915e4aa1
daa5574081d4c7f04bf8db4a39ee6f79d5da55b8
/MMN/MMN.sce
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[]
no_license
smasa1112/presentation_practice
fec8a78d3c3c7260d1815174169ff47b1a5a3527
599664d4fae6783c52c46d7cb540bbcf35a12b27
refs/heads/main
2023-07-18T22:46:30.056399
2021-09-23T05:39:07
2021-09-23T05:39:07
405,818,206
0
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null
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UTF-8
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1,971
sce
MMN.sce
## default settings------------------------------------------------ scenario = "MMN"; # scenario name no_logfile = false; # true -> do not make presentation log scenario_type = trials; # normal default_font_size = 50; # font size(pt) default_text_color = 255, 255, 255; # font color R, G, B (0~255) default_background_color = 0,0,0; # background color R, G, B (0~255) write_codes = true; ## scenario settings------------------------------------------------ begin; array{ sound{ wavefile{ filename="9kHz.wav"; preload = false;};}snd1; sound{ wavefile{ filename="12kHz.wav"; preload = false;};}snd2; }snd_ary; sound{ wavefile{filename="no_sound.wav";};}ns; #to stimulus trial{ stimulus_event{ sound snd1; code = "stim"; }pres_evt; }pres_trl; #reset用 #-pcl------------------------------------------------ begin_pcl; int i; # 単純なカウンター int j; double t; preset int standard_tone = 1; #出現トレイン数 int max = 12000; int devient = max / 20; # 5% double ISI = 500.0; #ms 0.5s array<int> sd[2]; if (standard_tone < 1 || 2 < standard_tone) then standard_tone = 1; end; if(standard_tone == 1)then sd[1] = 1; sd[2] = 2; else sd[1] = 2; sd[2] = 1; end; #ならし運転 ns.get_wavefile().load(); # メモリーに展開 ns.present(); #音声の準備 snd1.get_wavefile().load(); # メモリーに展開 snd2.get_wavefile().load(); # メモリーに展開 system_keyboard.set_log_keypresses(true); system_keyboard.get_input(); #刺激開始 t = clock.time_double() + 100; loop until t < clock.time_double() begin end; ns.present(); t = t + 100.0; loop i = 1; j = 1; until i > max begin loop until t < clock.time_double() begin end; if(j < 20) then snd_ary[sd[1]].present(); j = j + 1; else snd_ary[sd[2]].present(); j = 1; end; t = t + ISI; i = i + 1; end; snd1.get_wavefile().unload(); snd2.get_wavefile().unload(); ns.get_wavefile().unload();
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/2969/CH7/EX7.5/Ex7_5.sce
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[]
no_license
FOSSEE/Scilab-TBC-Uploads
948e5d1126d46bdd2f89a44c54ba62b0f0a1f5e1
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refs/heads/master
2020-04-09T02:43:26.499817
2018-02-03T05:31:52
2018-02-03T05:31:52
37,975,407
3
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null
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920
sce
Ex7_5.sce
clc clear //DATA GIVEN D=0.3; //bore of engine cylinder in m L=0.45; //stroke length in m N=300; //engine speed in R.P.M. Pmi=6; //mean effective pressure in bar NBL=1.5; //Net brake load (W-S) in kN Db=1.8; //diameter of brake drum d=0.02; //brake rope diameter n=1; //no. of cylinders k=0.5; //for 4-stroke cylinder //INDICTED POWER ,I.P.=(n*PMI*l*A*N*k*10)/6 kW A=(%pi/4)*(D^2); IP=(n*Pmi*L*A*N*k*10)/6; BP=NBL*(%pi)*(Db+d)*N/(60); eta=BP/IP; //mechanical efficiency printf(' (i) The Indicted Power, I.P. is: %5.2f kW. \n',IP); printf(' (ii) The Brake Power, B.P. is: %5.2f kW. \n',BP); printf('(iii) Mechanical efficiency is: %5.4f or %5.2f percent.\n',eta,(eta*100));
fcb6b3caca8028cfc91c611485f5332a92b25b67
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/1529/CH16/EX16.2/16_02.sce
eed1784f2e18496306a3756da2d085a6850fc6fe
[]
no_license
FOSSEE/Scilab-TBC-Uploads
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2020-04-09T02:43:26.499817
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sce
16_02.sce
//Chapter 16, Problem 2 clc; V=240; //voltage R=80; //resistance in ohm f=50; //frequency in hertz C=30e-6; //capacitance in farad Ir=V/R; //current flowing in the resistor Xc=1/(2*%pi*f*C); //capacitive reactance Ic=V/Xc; //current flowing in the capacitor I=sqrt(Ir^2+Ic^2); //supply current phi=atan(Ic/Ir); Z=V/I; //impedance P=V*I*cos(phi); //power consumed S=V*I; //apparent power, printf("(a) Current flowing in the resistor = %d A\n\tCurrent flowing in the capacitor = %.3f A\n\n",Ir,Ic); printf("(b) Supply current = %.3f A\n\n",I); printf("(c) Circuit phase angle = %.2f deg (leading)\n\n",phi*(180/%pi)); printf("(d) Circuit impedance = %.2f ohm\n\n",Z); printf("(e) Power consumed = %d W\n\n",P); printf("(f) Apparent power = %.1f VA",S);
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FOSSEE/Scilab-TBC-Uploads
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2020-04-09T02:43:26.499817
2018-02-03T05:31:52
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sce
16_1.sce
clc //initialisation of variables x= 0.02 //m y= 0.4 //m R= 0.0592 e= -0.771 //V e1= -1.520 //v n= 5 //electrons z= 0.80 //m z1= 0.5 //m //CALCULATIONS E= e-R*log10(x/y) E1= e1-(R/n)*log10(z1*z^8/x) E2= E-E1 //RESULTS printf (' Redox potential of sample= %.3f v',E) printf (' \n Redox potential of sample= %.3f v',E1) printf (' \n Redox potential of sample= %.3f v',E2)
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449d555969bfd7befe906877abab098c6e63a0e8
/779/CH15/EX15.7/15_7.sce
d6726bf761301fc2022e18f7d6f70cfc01f23ec5
[]
no_license
FOSSEE/Scilab-TBC-Uploads
948e5d1126d46bdd2f89a44c54ba62b0f0a1f5e1
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refs/heads/master
2020-04-09T02:43:26.499817
2018-02-03T05:31:52
2018-02-03T05:31:52
37,975,407
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sce
15_7.sce
h1 = 57; h2 = h1; h3 = 42; W1 = 0.0065; W2 = 0.0088; W3 = W2; t2 = 34.5; v1 = 0.896; n = 1500; // seating capacity of hall a = 0.3; // amount of out door air suplied G = (n*a)/0.896 ; // Amount of dry air suplied CC = (G*(h2-h3)*60)/14000; // in tonns R = G*(W2-W1)*60; disp("tonnes",CC,"Capacity of the cooling coil in tonnes") disp("kg/h",R,"Capacity of humidifier")
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/sim/scripts/ejector.tst
69d011cf4dfccd6f445dd8f0656c2a9973fb5b21
[]
no_license
psy007/NNPC-CHEMICAL-SIM-
4bddfc1012e0bc60c5ec6307149174bcd04398f9
8fb4c90180dc96be66f7ca05a30e59a8735fc072
refs/heads/master
2020-04-12T15:37:04.174834
2019-02-06T10:10:20
2019-02-06T10:10:20
162,587,144
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tst
ejector.tst
units PureSI $th = VirtualMaterials.Peng-Robinson / -> $th th + WATER ejector = Ejector.EjectorOp() cd /ejector.Process P = 20270.58558 Pa T = 333.3333336 K Fraction = 1.0 MoleFlow = 4.5359244127 kgmole/h cd /ejector.Motive P = 689475.7 Pa T = 333.3333336 K Fraction = 1.0 MoleFlow = 45.359244127 kgmole/h cd /ejector.Discharge #P = 81107.4142866 Pa #T = 333.3333354 K #Fraction = 1.0 #MoleFlow = 49.8951685397 kgmole/h cd / ejector.NozzleDiameter = 0.48639 ejector.ThroatDiameter = 1.613175 ejector.Process ejector.Motive ejector.Discharge ejector.NozzleDiameter ejector.ThroatDiameter ejector.NozzleDiameter = None ejector.Discharge.P = 81107.2733997 ejector.Process ejector.Motive ejector.Discharge ejector.NozzleDiameter ejector.ThroatDiameter copy /ejector paste / ejectorClone.Process ejectorClone.Motive ejectorClone.Discharge ejectorClone.NozzleDiameter ejectorClone.ThroatDiameter
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/3754/CH14/EX14.3/14_3.sce
117bbd8c0687e7f86cd2501348027cc2125b68a3
[]
no_license
FOSSEE/Scilab-TBC-Uploads
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7bc77cb1ed33745c720952c92b3b2747c5cbf2df
refs/heads/master
2020-04-09T02:43:26.499817
2018-02-03T05:31:52
2018-02-03T05:31:52
37,975,407
3
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UTF-8
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367
sce
14_3.sce
clear// //Variables alpha = 0.967 //common base current gain IE = 10 //Emitter current (in milli-Ampere) //Calculation IC = alpha * IE //Collector current (in milli-Ampere) IB = IE - IC //Base current (in milli-Ampere) //Result printf("\n Base current is %0.3f mA." ,IB)
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/APLICACAO_MATRIZ.sce
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ivybin/PROGRAMAS-SCE
82296ac10f4fb02d6e283f64e910666c6103063a
e61918ceae020bd33fcf223908f7decdec1c41b4
refs/heads/master
2023-07-23T12:55:08.711868
2021-09-02T17:22:23
2021-09-02T17:22:23
400,595,765
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sce
APLICACAO_MATRIZ.sce
/*código que gere três matrizes aleatórias - A, B e C de dimensão 11x11 E compute os seguintes valores: a) O somatório dos elementos da diagonal principal. b) O somatório dos elementos das colunas pares. c) O somatório dos elementos das colunas ímpares. d) O logaritmo natural dos resultados dos items a), b) e c).*/ clear; clc; A= 10*rand(11,11); B= 10*rand(11,11); C= 10*rand(11,11); M=0;U=0;U2=0; for i=1:1:11 M=M+A(i,i)+B(i,i)+C(i,i); end disp(M); //B) for T=2:2:10 for K=1:1:11 U=U+A(K,T)+B(K,T)+C(K,T); end end disp(U); //C) for T2=1:2:11 for K2=1:1:11 U2=U2+A(K2,T2)+B(K2,T2)+C(K2,T2); end end disp(U2); //D) DT= [log(M);log(U);log(U2)]; disp("OS LOGARITMOS NATURAIS DOS ITENS A B E C SE APRESENTAM DE FORMA RESPECTIVA",DT);
3dd44b9de68ad894547ae3c36a7b5113e27660ba
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/macros/cvtColor.sci
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no_license
gursimarsingh/FOSSEE_Image_Processing_Toolbox
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2021-01-22T02:08:45.870957
2017-01-15T21:26:17
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sci
cvtColor.sci
function new_image = cvtColor(image, code, varargin) [ lhs, rhs ] = argn(0) image_list = mattolist(image) select rhs case 2 then out = opencv_cvtColor(image_list, code) case 3 then out = opencv_cvtColor(image_list, code, varargin(1)) end sz = size(out) for i=1:sz new_image(:, :, i) = out(i) end endfunction
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FOSSEE/Scilab-TBC-Uploads
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//voltage standing wave ratio //given clc LEMg=4.82//cm d1_d2=0.7//cm VSWR=LEMg/(%pi*d1_d2)//VSWR VSWR=round(VSWR*1000)/1000///rounding off decimals disp(VSWR,'the voltage standing wave ratio:')
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cvdhengel/MarioCards
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2021-01-23T00:48:31.597526
2017-05-31T18:52:29
2017-05-31T18:52:29
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Mario_v8.sce
#Headers active_buttons=5; default_background_color= 255,255,255; default_text_color=0,0,0; default_font="calibri"; #SDL begin; TEMPLATE "Bitmaps.tem"; TEMPLATE "Trials.tem"; #PCL begin_pcl; #include "MarioPCL.pcl"# #Response data & output response_data Rdata; output_file output = new output_file; string ParticipantNumber = logfile.subject(); if ParticipantNumber.count() > 0 then output.open (ParticipantNumber + "_MarioCards.txt"); else output.open( "999_MarioCards.txt"); end; output.print( "date/time: " + date_time("dd mmmm yyyy' 't")+"\n"); output.print("Trial\tStimuli\tPrincple\tButton\tSorting\tFeedback\tErrortype\n"); #include "Start.pcl"; #Show introduction #intro.present(); #Sorting: 1=form, 2=color, 3=amount array <int> sort[6]={1,2,3,1,2,3}; #Main Trial Start stimuli.shuffle(); #sort.shuffle(); stimulus_data laststim = stimulus_manager.last_stimulus_data(); int c=0; #correct int f=0; #false int tc=0; #total correct int tf=0; #total incorrect int p=0; #perseveration int np=0; #nonperseveration #Main Trial Loop loop int i=1 until i > stimuli.count() begin; loop int s=1 until s > sort.count() begin; VB.set_part(5, stimuli[i]); MainEvent.set_event_code(stimuli[i].description()); MainTrial.set_duration(forever); MainTrial.set_type( first_response ); MainTrial.present(); output.print(i); output.print("\t"); output.print(stimuli[i].description()); output.print("\t"); response_data respdat = response_manager.last_response_data(); int BP = respdat.button(); output.print(BP); output.print("\t"); #Matching the cards on the screen if BP == 1 then VB.set_part_x(5,-700); VB.set_part_y(5, 0); end; if BP == 2 then VB.set_part_x(5,-250); VB.set_part_y(5, 0); end; if BP == 3 then VB.set_part_x(5,250); VB.set_part_y(5, 0); end; if BP == 4 then VB.set_part_x(5,700); VB.set_part_y(5, 0); end; MainTrial.set_duration(1000); MainTrial.present(); #Putting it back for the next trial VB.set_part_x(5,0); VB.set_part_y(5, -300); #Showing Feedback if sort[s]==1 then output.print("charac"); output.print("\t"); if (stimuli[i].description().find("M") == 1 && BP == 1) || (stimuli[i].description().find("S") == 1 && BP == 2) || (stimuli[i].description().find("L") == 1 && BP == 3) || (stimuli[i].description().find("P") == 1 && BP == 4) then feedbackcorrect.present(); output.print("correct"); output.print("\t"); c=c+1; f=0; tc=tc+1; else feedbackwrong.present(); output.print("incorrect"); output.print("\t"); if (stimuli[i].description().find("1") > 0 && BP == 1) || (stimuli[i].description().find("2") > 0 && BP == 2) || (stimuli[i].description().find("3") > 0 && BP == 3) || (stimuli[i].description().find("4") > 0 && BP == 4) then p=p+1; output.print("P"); output.print("\t"); else np=np+1; output.print("NP"); output.print("\t"); end; f=f+1; c=0; tf=tf+1; end; end; if sort[s]==2 then output.print("color"); output.print("\t"); if (stimuli[i].description().find("R") == 2 && BP == 1) || (stimuli[i].description().find("G") == 2 && BP == 2) || (stimuli[i].description().find("B") == 2 && BP == 3) || (stimuli[i].description().find("V") == 2 && BP == 4) then feedbackcorrect.present(); output.print("correct"); output.print("\t"); c=c+1; f=0; tc=tc+1; else feedbackwrong.present(); output.print("incorrect"); output.print("\t"); if (stimuli[i].description().find("M") == 1 && BP == 1) || (stimuli[i].description().find("S") == 1 && BP == 2) || (stimuli[i].description().find("L") == 1 && BP == 3) || (stimuli[i].description().find("P") == 1 && BP == 4) then p=p+1; output.print("P"); output.print("\t"); else np=np+1; output.print("NP"); output.print("\t"); end; f=f+1; c=0; tf=tf+1; end; end; if sort[s]==3 then output.print("amount"); output.print("\t"); if (stimuli[i].description().find("1") > 0 && BP == 1) || (stimuli[i].description().find("2") > 0 && BP == 2) || (stimuli[i].description().find("3") > 0 && BP == 3) || (stimuli[i].description().find("4") > 0 && BP == 4) then feedbackcorrect.present(); output.print("correct"); output.print("\t"); c=c+1; f=0; tc=tc+1; else feedbackwrong.present(); output.print("incorrect"); output.print("\t"); if (stimuli[i].description().find("R") == 2 && BP == 1) || (stimuli[i].description().find("G") == 2 && BP == 2) || (stimuli[i].description().find("B") == 2 && BP == 3) || (stimuli[i].description().find("P") == 2 && BP == 4) then p=p+1; output.print("P"); output.print("\t"); else np=np+1; output.print("NP"); output.print("\t"); end; f=f+1; c=0; tf=tf+1; end; end; ########################################## #Knowing when to continue or when to end # ########################################## if c>5 then s=s+1; c=0; end; #If 6 cards (in a row) were correct, the participant continues to the next sorting principle without letting know if f>5 then s=sort.count(); i=stimuli.count(); end; #If participant misses 6 in a row, the task will end. if tc>2 then s=sort.count(); i=stimuli.count(); end; #If all cards are sorted, the task will end. end; i=i+1; end; EndTrial.present(); output.print("\n\n\Total Correct\tTotalFalse\tTotalP\tTotalNP\n"); output.print(tc); output.print("\t"); output.print(tf); output.print("\t"); output.print(p); output.print("\t"); output.print(np); output.print("\t");
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// // 1. Create the spline //nx = 7; //ny = 15; //data = read("/media/D/Qt/teplovay linza/saveFile.csv", -1, 40); //y = linspace(%pi/2,4*%pi,ny); //z = cos(x')*cos(y); //C = splin2d(x, y, z, "periodic"); // // 2. Evaluate the spline on a grid //mx = 50; //my = 20; //xx = linspace(%pi,1.5*%pi,mx); //yy = linspace(%pi,2*%pi,my); //[XX,YY] = ndgrid(xx,yy); //zz = interp2d(XX,YY, x, y, C); // // 3. Plot the interpolated values //scf(); //plot3d(xx, yy, zz) z = read("/media/D/Qt/teplovay linza/saveFile.csv", -1, 40)-300; y = linspace(0,4,40); x = linspace(0,20,200); C = splin2d(x, y, z, "not_a_knot"); yy = linspace(0,4,400); xx = linspace(0,20,200); [XX,YY] = ndgrid(xx,yy); zz = interp2d(XX,YY, x, y, C); scf(); plot3d(xx, yy, zz)
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//detrmine the value of inductance I=5;//amp V=200;//volt f=50;//hertz X=V/I; L=40/(2*%pi*50); disp('the value of inductive.reactance='+string (X)+'ohms' , 'value of inductors='+string(L)+'henry');
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clc clear //INPUT DATA cp=0.03;//specific heat of lead in kj/kg.k v=10000;//initial velocity of bullet in cm/sec J=4.2*10^7;//joules constant in ergs/cal //CALCULATIONS //let mass of the bullet in gm ke=(v^2)/2;//kinetic energy of the bullet per unit mass in (cm/sec)^2 //T is the rise in temperature,then heat produced is m*cp*T //95% of kinetic energy is converted to heat T=ke*95/(cp*J*100);//rise in temperature in deg.C mprintf('the rise in temperature is %3.1f deg.C',T)
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function []=plot_graph(g,orx,ory,w,h) //This function plot a graph in the Scilab graphic window //It uses the data of the list defining the graph [lhs,rhs]=argn(0) // g check_graph(g) n=g('node_number'); ma=g('edge_number'); xnodes=g('node_x');ynodes=g('node_y'); if (xnodes==[]|ynodes==[]) then error('plot_graph: coordinates of nodes needed for plotting'); return end diam=g('default_node_diam'); if diam==[] then diam=20;end; nodediam=g('node_diam');if nodediam==[] then nodediam=zeros(1,n);end; ii=find(nodediam==0);nodediam(ii)=diam*ones(ii);ray=0.5*nodediam; if rhs==1 then lim=max(nodediam); ah=min(xnodes);bh=max(xnodes); av=min(-ynodes);bv=max(-ynodes); enx=bh-ah;enx=max(enx,5.*lim); eny=bv-av;eny=max(eny,5.*lim); orx=ah+0.09*enx;w=0.82*enx; ory=av+0.05*eny;h=0.9*eny; else if rhs<>5 then error(39), end end xsetech([0,0,1.0,1.0],[orx,ory,orx+w,ory+h]); isoview(orx,orx+w,ory,ory+h); nodecolor=nodediam nodeborder=0*ones(1,n) nodefontsize=23040.*ones(1,n); metarc=[xnodes-ray;-ynodes+ray;nodecolor;nodediam;nodeborder;nodefontsize]; xset('use color',1); ncolor=g('node_color') if ncolor=[] then ncolor=0*ones(1,n);end; xarcs(metarc,ncolor-1);vtail=g('tail');vhead=g('head'); nxx=0*ones(2,ma);nyy=nxx; nxx(1,:)=xnodes(vtail); nxx(2,:)=xnodes(vhead); nyy(1,:)=-ynodes(vtail); nyy(2,:)=-ynodes(vhead); txx=0*ones(2,ma);tyy=txx; txx(1,:)=xnodes(vtail); txx(2,:)=xnodes(vhead); tyy(1,:)=-ynodes(vtail); tyy(2,:)=-ynodes(vhead); edgecolor=g('edge_color');if edgecolor==[] then edgecolor=ones(1,ma);end; ii=find(edgecolor==0);edgecolor(ii)=ones(ii); xpolys(txx,tyy,-edgecolor);
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// Scilab code Ex13.2: Pg.624 (2008) clc; clear; // Calculation of pressure at the centre of Sun M = 1.99e+30; // Mass of Sun, kg R = 6.96e+08; // Solar radius, m G = 6.67e-011; // Gravitational constant, N-m^2/kg^2 mu = M/R^2; // Mass per unit surface area of Sun, kg/m^2 g = 0.5*G*(M/R^2); // Acceleration due to gravity in the Sun, m/s^2 P_c = mu*g; // Pressure at the center of the Sun, N/m^2 printf("\nThe pressure at the center of the sun = %1.0e N/m^2", P_c); // Calculation of pressure inside the H-atom k = 9e+09; // Coulomb's cnstant, N-m^2/C^2 e = 1.6e-19; // Charge iside the atom, C a_o = 0.5e-10; // Bohr's radius for H-atom, m A = 4*%pi*a_o^2; // Surface area of Sun, m^2 F = (k*e^2)/a_o^2; // Coulomb's force of attraction, N P = F/A; // Pressure inside the H-atom, N/m^2 printf("\nThe pressure inside the H-atom = %3.1e N/m^2",P); printf("\nSince the pressure at the centre of the sun is %3.1e times greater than that of the pressure inside the H-atom so it is unlikely to exist in its interior.", P_c/P); // Result // he pressure at the center of the sun = 6e+014 N/m^2 // The pressure inside the H-atom = 2.9e+012 N/m^2 // Since the pressure at the centre of the sun is 1.9e+002 times greater than that of the pressure inside the H-atom so it is unlikely to exist in its interior.
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clc; //Example 19.2 //page no 250 printf("Example 19.2 page no 250\n\n"); //pitot tube is located at the center line of a horizontal pipe transporting air rho=0.075//density of gas ,lb/ft^2 h=0.0166667//height difference,ft g=32.2//gravitational acc. lb/ft^2 rho_m=62.4//density of medium which is air v=sqrt(2*g*h*(rho_m-rho)/rho)//velocity printf("\n velocity v=%f ft/s",v); v_max=v//because at that point where the reading was taken is the centerline printf("\n maximum veocity v_max=%f ft/s",v_max); //since the flowing fluid is air at a high velocity the flow has a high probability of being turbilent .from chapter 14,assume //v_av/v_max=0.815 v_av=v_max*0.815 printf("\n average velocity v_av=%f ft/s",v_av);
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errcatch(-1,"stop");mode(2);//chapter 14 //example 14.4 //page 441 all; ; //given Av=1;//voltage follower printf("\nc1=500 pF\nc2=2000 pF\nc3=1000 pF") exit();
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COMPORT=8; // mount library on handle "1" h=slMount(); // handle "1": check availability of specified COM port slCheck(h,COMPORT); // handle "1": configure port slConfig(h,9600,8,0,1); // handle "1": open port slOpen(h,COMPORT); while(1) tune=input(" "); if tune==0 then break; else slSendByte(h,48+tune); end end slClose(h);
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//EX13_6 PG-13.5 clc clear disp("decimal from 0 to 9 in radix 5 ") for i=0:1:9; a=i/5; b=modulo(i,5); printf(" %d=%d%d\n",i,a,b);//conversion from decimal to radix 5 end
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// Determine R,Ids,Vgs // Determine Vgs,Id,Vds,operating region // Basic Electronics // By Debashis De // First Edition, 2010 // Dorling Kindersley Pvt. Ltd. India // Example 6-8 in page 277 clear; clc; close; // Given data Vp=-3; // Peak voltage in V Vgg=5; // Gate voltage in V Ids=10*10^-3; // Drain current in mA // Calculation R=5/(10*10^-3); printf("(a)R = %0.0f ohm\n",R); Ids=5/400; Vds=(2*Ids*R)+15; printf("(b)Idss = %0.2e A\n",Ids); printf("(c)Vds = %0.0f V\n",Vds); printf("This confirms active region\n"); Rid=14/2; Vgs=Vgg-Rid; printf("(d)Vgs = %0.0f V\n",Vgs); printf("Vds=2>Vgs-Vp=-1.5+3=1.5 -> Active region"); // Result // (a) R = 500ohm, // (b) Ids = 12.5mA, // (c) Vgs = -2V
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//Program for example 2 chapter 2 clear clc disp("Example 2: Display the type of a given variable or a command ") disp('') printf("\n*******************************************************************\n") disp("Answer :") printf("THE FOLLOWING LINES OF CODE RUN \nONLY IN SCILAB INSTALLED IN UNIX ENVIRONMENT.....") printf("\nTHE CONSOLE GIVES DIFFERENT \nOUTPUT IN OTHER OPERATING SYSTEMS") if(getos() ~= "Linux" )then ctd=input("Enter the command or variable whose type is to be determined ") clc(1) pt = input("Enter the command again to confirm ","s") clc(1) printf("Continue?....\ny :Yes\nAny other key:No") st = input('','s') clc(2) if( st ~= "y") then exit else n=type(ctd) clc(1) printf("%s is a ",pt) select n case 1 then printf("a real or complex matrix of double.") case 2 then printf('a polynomial matrix.') case 4 then printf('a boolean matrix.') case 5 then printf('a sparse matrix.') case 6 then printf('a sparse boolean matrix.') case 7 then printf('Matlab sparse matrix') case 8 then printf('a matrix of integers stored on 1 (int8), 2 (int16) or 4 (int32) bytes.') case 9 then printf('a matrix of graphic handles.') case 10 then printf('a matrix of character strings.') case 11 then printf('an un-compiled function . A function created with deff with argument [n].') case 13 then printf('a compiled function .') case 14 then printf('a function library.') case 15 then printf('a list.') case 16 then printf('a typed list (tlist).') case 17 then printf('a matrix oriented typed list (mlist).') case 128 then printf('a pointer (Use case: lufact).') case 129 then printf('a size implicit polynomial used for indexing.') case 130 then printf('a built-in Scilab function, called also gateway (C, C++ or Fortran code).') case 0 then printf('a null variable. It is mainly used internally by Scilab. If a function has no declared returned argument like disp when it is called it returns a null variable. If a function is called with an omitted argument in the argument list like foo(a,,b) the missing argument is assigned to a null variable.') end end else disp("Enter the file whose type is to be found") unix_w('read xtun;type $xtun') end printf("\n*******************************************************************\n")
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disp("Name: Pradyumna YM") disp("SRN: PES1201700986") a=input("Enter the Matrix A:") x=input("Enter the approximation: ") ax = a*x; disp(ax); l = max(abs(ax)); //find out the highest absolute value of the components ax = ax/l; previous = [0;0;0]; //since there was no previous l while 1 do previous=ax; ax = a*ax; lT = max(abs(ax)); ax = ax/lT; disp(ax); if((ax-previous)<0.0000001) //if the difference is negligible, break from the loop break; end T = ax; end disp(lT); disp(ax);
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trainLRClassifier.sci
// Copyright (C) 2015 - IIT Bombay - FOSSEE // // This file must be used under the terms of the CeCILL. // This source file is licensed as described in the file COPYING, which // you should have received as part of this distribution. The terms // are also available at // http://www.cecill.info/licences/Licence_CeCILL_V2-en.txt // Author: Siddhant Narang // Organization: FOSSEE, IIT Bombay // Email: toolbox@scilab.in function classifier = trainLRClassifier(imgSets, bag, classifierName, varargin) // This function is used to train an image classifier using the LR algorithm. // // Calling Sequence // classifier = trainLRClassifier(imgSets, bag, classifierName) // classifier = trainLRClassifier(imgSets, bag, classifierName, learningRate) // classifier = trainLRClassifier(imgSets, bag, classifierName, learningRate, iteration) // classifier = trainLRClassifier(imgSets, bag, classifierName, learningRate, iteration, regularization) // classifier = trainLRClassifier(imgSets, bag, classifierName, learningRate, iteration, regularization, trainMethod) // classifier = trainLRClassifier(imgSets, bag, classifierName, learningRate, iteration, regularization, trainMethod, minibatch) // // Parameters // classifier: Image category classifier // imgSets: Input imageSet to train the classifier on // bag: The bagOfFeatures of the imageSet provided // learningRate: Defines the rate at which the classifier will learn. // iteration: Number of iterations the training function will perform. // regularization: Controls the kind of regularization to be applied. The types are<itemizedlist><listitem>REG_DISABLE- Regularization disabled, flag value = -1.</listitem><listitem>REG_L1- L1 norm, flag value = 0.</listitem><listitem>REG_L2- L2 norm, flag value = 1.</listitem></itemizedlist> // // trainMethod: Controls the kind of training method to be applied. The types are<itemizedlist><listitem>BATCH- flag value = 1.</listitem><listitem>MINI_BATCH- flag value = 0.</listitem></itemizedlist> // // minibatch: Specifies the number of training samples taken in each step of Mini-Batch Gradient Descent. // Will only be used if trainMethod flag value is set to 0 training algorithm. // It has to take values less than the total number of training samples. // // // Description // This function trains a LR classifier which can be used to predict classes of images given to it as // input using the predictLR() function. // // Examples // imgSet = imageSet('images/trainset_2/','recursive'); // [trainingSet testSet] = partition(imgSet,[0.8]); // bag = bagOfFeatures(trainingSet); // lrclassi = trainLRClassifier(im, bag, "lrclassi", 1, 150, 0, 1, 5); // // Examples // imgSet = imageSet('images/trainset_3/','recursive'); // [trainingSet testSet] = partition(imgSet,[0.8]); // bag = bagOfFeatures(trainingSet); // lrclassi = trainLRClassifier(im, bag, "lrclassi", 1, 150, 0); // save("var.dat", "lrclassi"); // // See also // imageSet // partition // bagOfFeatures // mlPredict // save // // Authors // Siddhant Narang bag_list = bagStructToList(bag); imgSets_list = imageSetToList(imgSets); // Handling variable arguments. [lhs rhs] = argn(0) if lhs > 1 error(msprintf("Too many output arguments")); elseif rhs < 3 error(msprintf("Not enough input arguments")); elseif rhs > 8 error(msprintf("Too many input arguments")); end if rhs == 3 temp = raw_trainLRClassifier(imgSets_list, bag_list, classifierName); elseif rhs == 4 temp = raw_trainLRClassifier(imgSets_list, bag_list, classifierName, varargin(1)); elseif rhs == 5 temp = raw_trainLRClassifier(imgSets_list, bag_list, classifierName, varargin(1), varargin(2)); elseif rhs == 6 temp = raw_trainLRClassifier(imgSets_list, bag_list, classifierName, varargin(1), varargin(2), varargin(3)); msprintf("6 arguments"); elseif rhs == 7 temp = raw_trainLRClassifier(imgSets_list, bag_list, classifierName, varargin(1), varargin(2), varargin(3), varargin(4)); elseif rhs == 8 temp = raw_trainLRClassifier(imgSets_list, bag_list, classifierName, varargin(1), varargin(2), varargin(3), varargin(4), varargin(5)); end classifier = struct("ClassifierType", temp(1), "ClassifierLocation", temp(2), "BagofFeaturesLocation", temp(3), "Description", temp(4)) endfunction
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//Chapter 1, Example 1.1 clc //Initialisation v1=15.8 //voltage v2=12.3 //voltage r=220 //resistance in ohm //Calculation v=v1-v2 //voltage i=v/r //current in ampere //Results printf("Current, I = %.1f mA",(i*1000))
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// Example 7.9 N2=800; // turns of Coil 2 N1=1200; // turns of Coil 1 Q2=0.15*10^-3; // Megnetic Flux in Coil 2 Q1=0.25*10^-3; // Megnetic Flux in Coil 1 I2=5; // Current in A Coil 2 I1=5; // Current in A Coil 1 L1=N1*(Q1/I1); // Formula for (Self Inductance of Coil 1) disp('(a) Self Induction of a Coil 1 = '+string(L1)+' H'); L2=N2*(Q2/I2); // Formula for (Self Inductance of Coil 2) disp('(b) Self Induction of a Coil 2 = '+string(L2)+' H'); k=0.6; // Coefficient of Coupling Constant Q12=k*Q1; // Formula for (Megnetic Flux in 2nd Coil) M=N2*(Q2/I1); // Formula for (Mutual Inductance of Coils) disp('(c) Mutual induction of a Coil = '+string(M)+' H'); k1=M/sqrt(L1*L2); // Mutual Inductance of Coil 1 & 2 disp('(d) Coefficient of Coupling between the Coil = '+string(k1)+' H'); // p 233 7.9
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//Example 10_8 clc;clear;funcprot(0) // Given values w=2.0;// Width in mm L=35.0;// Length in cm b=2.0;// Distance in cm v_dot=0.110;// The total volume flow rate in m^3/s u_starmax=0.159;// m/s // Calculation v_dotbyL=-(v_dot/(L/100));// Strength of line source in m^2/s u_max=-(u_starmax*(v_dotbyL/(b/100)));// Maximum speed along the floor printf('\nStrength of line source=%0.3f m^2/s \nMaximum speed along the floor,u_max=%0.2f m/s',v_dotbyL,u_max);
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clc //Variable Initialisation Ea=210//Input Voltage of motor in volts Ia1=140//Armature Current in Ampere Ia2=2*Ia1 Ra=0.08//Armature resistance in ohm N1=1100//Rated Speed of Motor in rpm N2=1200//Rated Speed of Motor in rpm //Solution Eb1=Ea-(Ia1*Ra) Eb2=(N2/N1)*Eb1 Rb=((Eb2+Ea)/Ia2)-Ra W=(2*%pi*N2)/60 T1=(Eb2*Ia2)/W Ia3=Ea/(Ra+Rb) T2=T1*(Ia3/Ia2) printf('\n\n Resistance to be placed=%0.1f ohm\n\n',Rb) printf('\n\n Braking torque=%0.1f N-m\n\n',T1) printf('\n\n torque=%0.1f N-m\n\n',T2)//The answers vary due to round off error
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//To calculate the highest order for which Bragg's reflection can be seen lamda = 1.5; //wavelength, A.U d = 1.6; //interplanar spacing, A.U theta = 90; //maximum glancing angle possible, degrees n = 2*d*sind(theta)/lamda; //maximum possible diffraction order printf("maximum possible diffraction order is %d",n);
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Example27_5.sce
clear ; clc; // Example 27.5 printf('Example 27.5\n\n'); //page no. 850 // Solution // Given V1 = 5 ;// Volume of gas initially - [cubic feet] P1 = 1 ;// Initial pressure - [atm] P2 = 10 ;// Final pressure - [atm] T1 = 100 + 460 ;// initial temperature - [degree Rankine] R = 0.7302 ;// Ideal gas constant -[(cubic feet*atm)/(lb mol)*(R)] //Equation of state pV^1.4 = constant //(a) //Energy balance reduces to del_E = del_U = del_W V2 = V1*(P1/P2)^(1/1.4) ;// Final volume - [cubic feet] W1_rev = integrate('-(P1)*(V1/V)^(1.4)','V',V1,V2) ;// Reversible work done in compresion in a horizontal cylinder with piston -[cubic feet *atm] W1 = W1_rev *1.987/.7302 ;// Conversion to Btu -[Btu] printf('\n (a)Reversible work done in compression in a horizontal cylinder with piston is %.1f Btu .\n ',W1); //(b) n1 = (P1*V1)/(R*T1) ;// Number of moles of gas W2_rev = integrate('(V1)*(P1/P)^(1/1.4)','P',P1,P2) ;// Reversible work done in compresion in a rotary compressor -[cubic feet *atm] W2 = W2_rev *1.987/.7302 ;// Conversion to Btu -[Btu] printf('\n (b)Reversible work done in a rotary compressor is %.1f Btu .\n ',W2);
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// Exa 10.21 clc; clear; close; format('v',6) // Given data P = 4; A = 4; Turns = 100; N = 600;// in rpm Eg = 220;// in V n = 2;// no of total conductors Z = n*Turns; // Eg = (N*P*phi*Z)/(60*A); phi = (Eg*60*A)/(N*P*Z);// in Wb disp(phi,"The useful flux per mole when armature is LAP connected in Wb is"); A = 2; // Eg = (N*P*phi*Z)/(60*A); phi = (Eg*60*A)/(N*P*Z);// in Wb disp(phi,"The useful flux per mole when armature is WAVE connected in Wb is");
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clear clc T=[10 132 6.6 .15] M=[5 6.6 .3 .2 ] B=[10 6.6] T(5)= T(3)/T(2) B(3)=B(2)* T(5) B(4)= B(1)*1e6/(sqrt(3)*B(2)*1e3) M(5)=M(4) *B(1)/M(1) M(6)=M(3) *B(1)/M(1) X1=1/((1/M(5))+(1/M(5))+(1/T(4))) IF1=round(100/X1)/100 I1=IF1*B(4) mprintf("\n(a) sub transient fault current=%.0f A", I1) It=round(100/T(4))/100 Im=1/M(5) ID=It+Im iD=ID*B(4) mprintf("\n(b) current through D=%.0f A", iD) RD=iD*1.6 mprintf("\n(c) current rating of D=%.0f A", RD) X2=1/((1/M(6))+(1/T(4))) IF2=round(100/X2)/100 I2=IF2*round(B(4)*10)/10 iCB=1.1 *I2 mprintf("\n(d) current to be interrrupted by D=%.1f A", fix(iCB*10)/10)
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//Exa 11.4 clc; clear; close; IR=6;//in % per year i=18;//in % per year AFR=5000000;//in Rs. n=7;//in years AI=500000;//in Rs. disp("End of year AFR InflationFactor InflatedAmount P/F PW"); TPW=0;//Initialising format('v',10) for n=1:5 IF=(1+IR/100)^n; IA=IF*AFR;//in Rs. PF=1/((1+i/100)^n); PW=PF*IA;//in Rs. TPW=PW+TPW;//in Rs. disp(" "+string(n)+" "+string(AFR)+" "+string(IF)+" "+string(IA)+" "+string(PF)+" "+string(PW)); AFR=AFR+AI;//in Rs. end; disp(TPW,"The value of the single deposit to be made now to recieve the specified series for the next five years is Rs. : ")
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clc //to calculate distance travelled by the beam deltat0=2.5*10^-8 //proper half life of pi mesons in (s) c=3*10^8 //light speed (m/s) v=0.8*c //mesons velocity (m/s) deltat=deltat0/sqrt(1-(v/c)^2) //half life (s) //No=initial flux ,N=flux after time t //N=N0 e^(-t/T) //N=N0/e^2 (given)=No e(-t/T) //t=2 deltat d=2*deltat*v //d=vt disp("distance travelled by the beam is d="+string(d)+"m") //answer is given in the textbook=19.96 m
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clc; Vsmin=19.5; Vsmax=22.5; RL=6*10**3; Vz=18; Izmin=2*10**-6; Pzmax=60*10**-3; rz=20; Izmax=sqrt(Pzmax/rz); IL=Vz/RL; ILmax=IL; ILmin=IL; Rsmax=(Vsmin-Vz)/(Izmin+ILmax); disp('ohm',Rsmax*1,"Rsmax="); Rsmin=(Vsmax-Vz)/(Izmax+ILmin); disp('ohm',Rsmin*1,"Rsmin=");
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function [ar]=armac(a,b,d,ny,nu,sig) // Renvoit une liste Scilab qui decrit un systeme ARMACX // A(z^-1)y= B(z^-1)u + D(z^-1)sig*e(t) // a=<Id,a1,..,a_r>; matrice (ny,r*ny) // b=<b0,.....,b_s>; matrice (ny,(s+1)*nu) // d=<Id,d1,..,d_p>; matrice (ny,p*ny); // ny : dimension de l'observation y // nu : dimension de la commande u // sig : matrice (ny,ny); // //! [na,la]=size(a); if na<>ny,write(%io(2),"a(:,1) must be of dimension "+string(ny));end [nb,lb]=size(b); if nb<>ny,write(%io(2),"b(:,1) must be of dimension "+string(ny));end [nd,ld]=size(d); if nd<>ny,write(%io(2),"d(:,1) must be of dimension "+string(ny));end ar=list('ar',a,b,d,ny,nu,sig)
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A = zeros(1000, 1000) for i=1:1000 A(i, i) = i if i+1 <= 1000 A(i, i+1) = 0.5 A(i+1, i) = 0.5 end if i+2 <= 1000 A(i, i+2) = 0.5 A(i+2, i) = 0.5 end end b = ones(1000, 1) x0 = zeros(1000, 1) D = diag(diag(A)) U = triu(A, 1) L = tril(A, -1) // Jacobi xk = x0 for step=1:15 xk_next = inv(D) * (b - (L + U) * xk) xk = xk_next end ans_jcb = xk // Gauss-Seidel xk = x0 for step=1:15 xk_next = inv(L + D) * (-U * xk + b) xk = xk_next end ans_gs = xk // SOR w=1.1 w = 1.1 xk = x0 for step=1:15 xk_next = inv(w * L + D) * ((1 - w) * D * xk - w * U * xk) + w * inv(D + w * L) * b xk = xk_next end ans_sor = xk // Conjugate Gradient d0 = b - A * x0 r0 = d0 dk = d0 rk = r0 xk = x0 for step=1:15 if rk == 0 break end alphak = (rk' * rk) / (dk' * A * dk) xk_next = xk + alphak * dk rk_next = rk - alphak * A * dk betak = (rk_next' * rk_next) / (rk' * rk) dk_next = rk_next + betak * dk dk = dk_next rk = rk_next xk = xk_next end ans_cg = xk // Conjugate Gradient with jcb preconditioner M = D r0 = b - A * x0 d0 = inv(M) * r0 z0 = d0 dk = d0 zk = z0 rk = r0 xk = x0 for step=1:15 if rk == 0 break end alphak = (rk' * zk) / (dk' * A *dk) xk_next = xk + alphak * dk rk_next = rk - alphak * A * dk zk_next = inv(M) * rk_next betak = (rk_next' * zk_next) / (rk' * zk) dk_next = zk_next + betak * dk dk = dk_next zk = zk_next rk = rk_next xk = xk_next end ans_cgp = xk
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var x, y; x <- 51; y <- 2 * x; call print (x).
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//find safe working voltage of cable clear; clc; //soltion //given d=4;//cm D=10;//cm e1=5;//realtive permeabilty e2=4;//realtive permeabilty e3=3;//realtive permeabilty d1=e1*d/e2; d2=e1*d/e3; gmax=40;//kV/cm Vper=(gmax/2)*[d*log(d1/d)+d1*log(d2/d1)+d2*log(D/d2)]; Vsafe1=Vper/sqrt(2); printf("Safe working voltage(rms) of a cable= %.2f kV\n",Vsafe1); Vpeak=(gmax/2)*[d*log(D/d)]; Vsafe2=Vpeak/sqrt(2); printf("Safe working voltage(rms) of the ungraded cable= %.2f kV",Vsafe2);
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// Scilab code Exa7.7.1 : To calculate the thickness of depletion layer of silicon detector and amplitude of voltage pulse :P.no. 316 (2011) E_r = 12; // Relative permittivity E_o = 8.85e-012; // Permittivity of free space E = E_r*E_o; // Absolute dielectric constant C = 100e-012; // Capacitance of the dielectric, F A = 1.6e-04; // Area of the detector, m^2 e = 1.602e-019; // Charge of an electrin, C E_p = 3.2; // Energy required to create an ion pair, eV E_s = 12e+06; // Energy required to stopped ion pair, eV n = E_s/E_p; // Number of ion-pair produced Q = n*e; // Charge of these ion pair, C d = A*E/(C*10^-6); // The thickness of the depletion layer, micron A = Q/C*1000; // The amplitude of voltage pulse, mV printf("\n The thickness of the depletion layer = %d micron \n The amplitude of voltage pulse: = %6.4f mV ", d, A) // Result // The thickness of the depletion layer = 169 micron // The amplitude of voltage pulse: = 6.0075 mV
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PL/SQL Developer Test script 3.0 32 -- 测试游标之动态游标[弱类型游标] declare -- 声明类型 type v_cursor_type is ref cursor; -- 声明变量 v_cursor v_cursor_type; -- 行变量 v_row emp%rowtype; v_dept_row dept%rowtype; begin -- Test statements here open v_cursor for select * from emp; loop fetch v_cursor into v_row; exit when v_cursor%notfound; dbms_output.put_line(v_row.ename); end loop; close v_cursor; open v_cursor for select * from dept; loop fetch v_cursor into v_dept_row; exit when v_cursor%notfound; dbms_output.put_line(v_dept_row.dname); end loop; close v_cursor; end; 0 0
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function [io,s]=syssize(sys) //Old stuff // io=syssize(sys) // [io,ns]=syssize(sys) // // sys : syslin list // io : io=[nout,nin] // nout: nb. ouputs // nin : nb. inputs // s : nb states. // Copyright INRIA select type(sys) case 1 then io=size(sys) s=[] //-compat next case retained for list/tlist compatibility case 15 then sys1=sys(1) select sys1(1) case 'lss' then io=size(sys(5)), [s,s]=size(sys(2)) case 'r' then io=size(sys(3)) [lhs,rhs]=argn(0); if lhs==2 then sys=tf2ss(sys);[s,s]=sys(2),end else error(97,1) end; case 16 then sys1=sys(1) select sys1(1) case 'lss' then io=size(sys(5)), [s,s]=size(sys(2)) case 'r' then io=size(sys(3)) [lhs,rhs]=argn(0); if lhs==2 then sys=tf2ss(sys);[s,s]=sys(2),end else error(97,1) end; else error(97,1), end
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//POW2DB Power to dB conversion //YDB = POW2DB(Y) convert the data Y into its corresponding dB value YDB //Example: //Calculate ratio of 2000W to 2W in decibels //y1 = pow2db(2000/2) //Answer in db //Author : Debdeep Dey function [ydb]=pow2db(y) rhs = argn(2) if(rhs~=1) error("Wrong number of input arguments.") end [r,c]=size(y); if (find(real(y(:))<0))==[] then if abs(y(:))>=0 then for i=1:r for j=1:c if abs(y(i,j))>0 then ydb(i,j)=10*log10(y(i,j)); else ydb(i,j)=-%inf; end end end end else error("The power value must be non-negative") end endfunction
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//example 3 //calculating the required force clear clc Dcyl=0.1 //cylinder diameter in m Drod=0.01 //rod diameter in m Acyl=%pi*Dcyl^2/4 //cross sectional area of cylinder in m^2 Arod=%pi*Drod^2/4 //cross sectional area of rod in m^2 Pcyl=250000 //inside hydaulic pressure in Pa Po=101000 //outside atmospheric pressure in kPa g=9.81 //acc. due to gravity in m/s^2 mp=25 //mass of (rod+piston) in kg F=Pcyl*Acyl-Po*(Acyl-Arod)-mp*g //the force that rod can push within the upward direction in N printf("\n hence,the force that rod can push within the upward direction is F = %.3f N. \n",F)