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clc,clear printf('Example 6.4\n\n') V_l=10000 V_ph=V_l/sqrt(3) VA=10*10^6 I_FL=VA/(V_l*sqrt(3)) //Current at full laod IX_s=(20/100)*V_ph //product of I and X_s X_s=IX_s/I_FL N_s=1500 f=50 P=120*f/N_s //poles delta_dash_mech=%pi/180 //phase displacement in degree mechanical delta_dash_elec=delta_dash_...
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clc; disp("Example 4.14") h= 5 // in m f=0.005 Q=(18200/3600)*0.001 // flow rate in m^3/s l=50 // in m //from the formulae used in the last problem as well d=(((2*f*l*Q*4*Q*4)/(%pi*%pi*h*9.81))^0.2) disp("The diameter is ") disp(d)
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//Example 4-2 //To Find out the order of a Low Pass Butterworth Filter clc ; ap =3; //db as =40; //db fp =500; //Hz fs =1000; //Hz op =2* %pi *fp; os =2* %pi *fs; N= log( sqrt ((10^(0.1* as) -1) /(10^(0.1* ap) -1)))/log(os/op); disp ( ceil (N), ' Order o f the f i l t e r , N =' );
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clear;lines(0); A=rand(5,2)*rand(2,4); // 4 col. vectors, 2 independent. [X,dim]=rowcomp(A);Xp=X'; svd([Xp(:,1:dim),A]) //span(A) = span(Xp(:,1:dim) x=A*rand(4,1); //x belongs to span(A) y=X*x norm(y(dim+1:$))/norm(y(1:dim)) // small
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// Example 10.5 clc; clear; close; // Given data format('v',6); V_REF= 1.25;// in V R1= 2.5*10^3;// in Ω R2= 1*10^3;// in Ω I= V_REF/R2;// in A // The output voltage, Vout= I*(R1+R2);// in V disp(Vout,"The regulated output voltage in volts is : ")
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lya V;PROG;PL;3;PST lya V;PL;1;PST+RCT lya V;PROG;2;PRS lya V;SG;1;PST+RCT lya V;PL;3;PST+RCT lya V;PROG;PL;3;PRS lya V;PROG;PL;1;PRS lya V;PL;2;PST+RMT lya V;PL;2;FUT+IMMED lya V;PROG;PL;2;PRS lya V;PROG;3;PST lya V;PL;1;PRS lya V;INTEN;1;FUT lya V;1;PRS lya V;PROG;1;PRS lya V;PL;3;FUT+IMMED lya V;SG;2;PST+RCT lya V;P...
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clear; clc; printf("\t\t\tExample Number 5.4\n\n\n"); // isothermal flat plate heated over entire length // illustration5.4 // solution // total heat transfer over a certain length of the plate is desired, so we wish to calculate average heat transfer coefficients. // for this purpose we use equations (5-44...
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clc// // // //Variable declaration mew=9.4*10^-24; H=2; //magnetic field(weber/m^2) k=1.38*10^-23; //boltzmann constant //Calculation T=2*mew*H/(log(2)*k); //temperature(K) //Result printf("\n temperature is %0.1f K",T)
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//EXAMPLE 8-34 PG NO-551 N=2.985; Ro=175; R1=Ro*[(N-1)/(N+1)]; R2=Ro*[(2*N)/(N^2-1)]; disp('ii) Resistance (R1) is = '+string (R1) +' ohm '); disp('ii) Resistance (R2) is = '+string (R2) +' ohm ');
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errcatch(-1,"stop");mode(2);//Example 2.7://resistance ; ; i=20;//amperes v=100;//volts pf=1;//power factor p=v*i*pf;//watts rp=5500;//ohms th=150;//angle kd=((rp*th)/p);//constant wd=20;//watts per degree rp1=wd*kd;//in ohms adr=rp1-rp;// disp(adr,"addition resistance in ohm is") exit();
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//Exa 1.2 clc; clear; close; // given : A=[2 5 6] // vector A B=[1 -3 6] // vector B A+B // summation of two vectors A-B // subtraction of two vectors disp(A+B,"summation of two vectors:") disp(A-B,"subtraction of two vectors:")
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clc //initialisation of variables v= 5 //ft/sec Q= 500 //cuses w= 25 //ft g= 32.2 //ft/sec^2 //CALCULATIONS h= (Q/v)/w E= h+(v^2/(2*g)) //RESULTS printf ('Specific energy = %.2f ft ',E)
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function [ok,bllst]=adjust(bllst,inpptr,outptr,inplnk,outlnk) // [outoin,outoinptr]=connmat(inpptr,outptr,inplnk,outlnk) // for hh=1:length(bllst) ok=%t for blkout=1:length(bllst) for portout=1:outptr(blkout+1)-outptr(blkout) jj=outptr(blkout)+portout-1 for kk=outoinptr(jj):outoinptr(jj+1)-1 blkin=...
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<?xml version="1.0" encoding="utf-8"?> <test> <description> Process npart to tecplot file </description> <executable>FieldConvert</executable> <parameters> -f -e --nparts 2 chan3D_xml:xml chan3D.fld chan3D.plt </parameters> <files> <file description="Session File Directory">chan3D_xml</file> <f...
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//Example 2.7 clc; n=-8:1:8; for i=1:length(n) x(i)=exp(-n(i)^2); h(i)=3.*n(i)^2; end y=convol(x,h); figure plot2d3(n,h); title('Impulse responce'); figure plot2d3(n,x); title('Input signal'); figure n1=-16:1:16 plot2d3(n1,y); title('Output signal');
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///////////////////////////////////////// //   適応フィルタ(LMSアルゴリズム)       //   Adaptive Filter(LMS Algorithm)     //                //           M.Tsutsui //////////////////////////////////////// clear all; funcprot(0) function [y_opt]=lms_v(myu,update);//myu:ステップサイズ,update:更新回数 R=x*x'; //E[...
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// Example 6_9 clc;funcprot(0); // Given data a=1.5;// Radius in cm W=3;// Length in cm hbar=5*10^-5;// Clearance in m mu=2*10^-2;// Viscosity of lubricating oil in Pa s rho=9*10^2;// Density of lubricating oil in kg/m^3 N=3600;// rpm n=0.5;// The eccentricity // Calculation // (a) omega=(2*%pi*N)/60;// ...
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clc; clear; close; x=poly(0,'x'); p1=x^4; p2=p1; ex1_ans=p1*p2 a=poly(0,'a'); p1=2*a^7; p2=a^3; ex2_ans=p1*p2 b=poly(0,'b'); p1=5*b^2; p2=3*b^5; ex3_ans=p1*p2 //a^2*ab^2 ex4_ans=string('a^3b^3')
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// Get the scalar part of a quaternion // // This function ensures that the correct index is used // to retrieve the scalar part of the quaternion. // // INTPUT // - q: input quaternion // // OUTPUT // - scal: scalar part of the input quaternion // // USAGE // scal = quat_getScal(q); // // HISTORY // 28/03/2014: T. Par...
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//Exa:1.6 clc; clear; close; i_amp=0.075;//given i_milamp=i_amp*1000;//current in milliamp. printf("%f amp current is %f mA",i_amp,i_milamp);
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//Eg-15.2 //pg-608 clear clc // Using D in the place of greek alphabet delta // Taking square grid => Dx = Dy = 0.25 // Applying central difference approximation to the second derivatives, we obtain // [T(i+1,j) - 2*T(i,j) + T(i-1,j)]/(Dx)^2 + [T(i,j+1) - 2*T(i,j) + T(i,j-1)]/(Dy)^2 = 0. i = 1,2,3; ...
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// fonction tfdd // Calcule la transformée de Fourier directe discrète, en 2 dimensions. // @param Img Image à partir de laquelle la transformée de Fourier est calculée. L'image doit être en niveau de gris. // @return ImgTrans Résultat de la transformée de Fourier appliquée à l'image Img. function [retour] = fourie...
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//To calculate the maximum radius of the sphere //a = 4*r/sqrt(2) and R = (a/2) - r //R = (4*r/(2*sqrt(2))) - r = (2*r/sqrt(2)) - r //R = r*(sqrt(2)-1) = 0.414*r printf("maximum radius of the sphere that can fit into the void is 0.414*r");
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//Example 16.3. clc format(6) amp = 15; vi_t=3; // transition voltage t=0:0.1:2*%pi; vi=amp*sin(t); vo=vi+3; // output voltage disp (vi_t,'transition voltage:'); for i=1:length(t) if(vo(i)<=0) vo(i)=0; end end subplot(2,2,1) plot2d1(t,vo,2,'011','',[0,0,7,18]); xtitle('Ouptut voltage in sin wave','t','v...
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//Problem 35.04: In the network shown in Figure 35.4 the load consists of a fixed capacitive reactance of 7 ohm and a variable resistance R. Determine (a) the value of R for which the power transferred to the load is a maximum, and (b) the value of the maximum power. //initializing the variables: rv = 60; // in vo...
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//Марчук Л.Б. 5307 подгруппа 3 //Данный модуль принимает на вход контурную матрицу //и матрицу-столбец источников напряжений ветвей; //возвращает матрицу-столбец сумм напряжений источников напряжений контуров. function [result] = FindLoopEMF(LoopMatrix, EMFBranches) [rows columns] = size(EMFBranches); ...
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function [x,y,typ]=REGISTER(job,arg1,arg2) // Copyright INRIA x=[];y=[];typ=[] select job case 'plot' then standard_draw(arg1) case 'getinputs' then [x,y,typ]=standard_inputs(arg1) case 'getoutputs' then [x,y,typ]=standard_outputs(arg1) case 'getorigin' then [x,y]=standard_origin(arg1) case 'set' then x=arg1;...
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//Transport Processes and Seperation Process Principles //Chapter 2 //Example 2.7-5 //Principles of Momentum Transfer and Overall Balances //given data //cgs units V=69.1;//volumetric flow rate in gallon/min fr1=V*(1/60)*(1/7.481);//converting flow rate in ft3/s A1=0.0233;//cross section area in ft2 of the pipe...
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clear; clc; h = 80;// feet p = 28;// lb. per sq.foot rho = 126;// lb. per cubic foot p_0 = rho*h/2240;// tons-ft^2 p_max = 7;// tons/ft^2 d = 4;// feet p_b = p_max - p_0;// tons/ft^2 D = sqrt(3*p*h^2 /(2*p_b*2240) +sqrt(d^4 + (3*p*h^2 /(2*p_b*2240))^2));// feet t = 0.5*(D-d);// feet printf('The necessary thi...
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//A Textbook of Chemical Engineering Thermodynamics //Chapter 5 //Some Applications of the Laws of Thermodynamics //Example 2 clear; clc; //Given: P = 1000; //pressure of saturated steam (kPa) T = 398; //temperature of escaping steam (K) //Referring steam tables H_vap = 2778; //enthalpy of saturated...
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clear ; clc; // Example 23.5 printf('Example 23.5\n\n'); //page no. 695 // Solution // Basis : 1 g mol of gas //Given T1 = 550 ;// Initial temperature - [degree F] T2 = 200 ;// Final temperature - [degree F] CO2 = 9.2/100 ;// Mole fraction CO = 1.5/100 ;// Mole fraction O2 = 7.3/100 ;// Mole fraction N2 = 82.0/100...
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// 08.09.21 // 08.10.11 // 08.10.15 // 10.02.16 Eps function Out=Crv3onsfpersdata(varargin) global PARTITIONPT HIDDENDATA CRV3ONSFHIDDENDATA Nargs=length(varargin); Fk=varargin(1); Fbdy=Projpers(varargin(2)); Fd=varargin(3); N=4; FdL=Fullformfunc(Fd); Fxy=Mixop(1,FdL); Np=[50,50]; if Nargs>=N Np=...
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//To find the torque required clc //Given: d=22,p=3 //mm funcprot(0) beta=60/2 //degrees W=40*1000 //N mu=0.15 //Solution: //Calculating the helix angle alpha=atan(p/(%pi*d)) //radians //Calculating the virtual coefficient of friction mu1=mu/cosd(beta) //Calculating the force required at the circumference ...
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//EX9_3 PG-9.14 clc f=1e3;//frequency of oscillation C=0.1e-6;//We choose the value of each capacitance of the RC phase shift oscillator R=1/(2*%pi*sqrt(6)*f*C); printf("\n each resistance of the RC phase shift oscillator is %.3f ohm \n",R) disp(" The standard value of R=680 ohm")
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//clc() PC1 = 85;//% ( Percent carbon in coal ) PA1 = 15;//% ( Percent ash in coal ) PA2 = 80;//% ( Percent ash in cinder ) PC2 = 20;//% ( Percent carbon in cinder ) m = 100;//kg (weight of coal ) mash = PA1 * m / 100; w = mash * 100 / PA2;// weight of cinder mcarbon = w - mash; Plost = mcarbon * 100 / ( m - m...
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function r=polyval2(x,c) [l,n]=size(c); i=0:n-1; u=x^i; r=u*c'; endfunction
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clc //initialisation of variables H1= 2 //m H2= 3 //m H3= 4 //m k1= 1e-4 //cm/sec k2= 3.2e-2 //cm/sec k3= 4.1e-5 //cm/sec //calculations H= H1+H2+H3 Kh= (1/H)*((k1*H1)+(k2*H2)+(k3*H3)) Kv= H/((H1/k1)+(H2/k2)+(H3/k3)) P= Kh/Kv //results printf ('ration of equivalent hydraulic conductivity = % 2f ',P)
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//Ex 3.4 clc;clear;close; format('v',5); V=110;//V(Supply voltage) R=50;//ohm Vm=V*sqrt(2);//V(maximum voltage) Im=Vm/R;//A(maximum current) Iav1=0.637*Im;//A(average current Over +ve half cycle) Iav2=0;//(average current Over -ve half cycle) Iav=(Iav1+Iav2)/2;//A(average current Over whole cycle) disp(Iav,"(...
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// Example 5.2 clc; clear; close; // Given data format('v',6); C= 1;// in µF C=C*10^-6;// in F R_A= 4.7;// in kΩ R_A=R_A*10^3;// in Ω R_B= 1;// in kΩ R_B=R_B*10^3;// in Ω T_on= 0.693*(R_A+R_B)*C;//positive pulse width in second T_on= T_on*10^3;// in ms T_off= 0.693*R_B*C;// pulse width in second T_off= T...
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// This file was generated by generateLibraryHelp.sci on 06-01-2020 12:06 // This file is released under the 3-clause BSD license. See COPYING-BSD. here = get_absolute_file_path("builder_help.sce"); d = dir(here); d = d.name(d.isdir); la_LA = d(grep(d, "/[a-z]{2}_[A-Z]{2}/", "r")); // filtering out ./images/ ./mml/, ...
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// Convolution spatiale avec le filtre gradient function image_out = filtreDerive(image) SizeX = size(image, 1); SizeY = size(image, 2); image_out = zeros(SizeX, SizeY); for X = 2 : SizeX-1, for Y = 2 : SizeY-1, // Utilisation de la formule sqrt(Ix²+Iy²) et avec les pixels Ix et...
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clc clear //Input data H=18;//The length of furnace wall riser in m O=76.2;//The outer diameter of the furnace wall riser in mm T=6.1;//The thickness of the furnace wall riser in mm P=80;//Pressure at which saturated water is recieved in bar V=1.5;//The velocity of the saturated water in m/s CR=12.5;//Assuming...
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function [y] = HadamardCode(len,index,opsamples) y=[]; // Display mode mode(0); // Display warning for floating point exception ieee(1); //HadamardCode Generate hadamard code //Y = HadamardCode(len,opsamples)) outputs a frame of the hadamard // code in column vector Y. It g...
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//Transport Processes and Seperation Process Principles //Chapter 2 //Example 2.2-1 //Principles of Momentum Transfer and Overall Balances //given data //a) Fl=3*32.174*(1/32.174);// lbm*(ft/s^2)*(1/(ft/s^2))=m*g/gc //b) Fdyne=3*453.59*980.665;//lbm*(g/lbm)8(cm/s^2) //c) Fnewt=3*(1/2.2046)*9.80665;//(lbm*(kg/...
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//Chapter 9 //Example 9_15 //Page 226 clear;clc; f=50; v=132*1e3; d1=4; d2=4; d3=8; r=1e-2; deq=(d1*d2*d3)^(1/3); printf("Deq = %.2f m \n\n", deq); e0=8.854*1e-12; c=2*%pi*e0/log(deq/r); printf("Capacitance of each conductor to nuetral = %.4f uF/km \n\n", c*1e9); cn=c*1e9*100; printf("Capacitance/phase for 100km lin...
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//Example 12.5 r1=(6.40*10^-2)/2;//Radius of the hose (m) A1=%pi*r1^2;//Cross-sectional area of the hose (m^2) r2=(3*10^-2)/2;//Radius of the nozzle (m) A2=%pi*r2^2;//Cross-sectional area of the nozzle (m^2) Q=40;//Flow rate (L/s) Q=Q/10^3;//Flow rate (m^3/s) v1=Q/A1;//Speed of water in hose (m/s) v2=Q/A2;//Spe...
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function y = idct2(x,varargin) //This function computes the inverse 2-D discrete cosine transform of input matrix. //Calling Sequence //Y = idct2(X) //Y = idct2(X, M, N) //Y = idct2(X, [M, N]) //Parameters //X: Matrix or integer //M, N: If specified Matrix X is padded with M rows and N columns. //Description // This fu...
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//Network Theorem 2 //pg no 3.18 //example 3.17 disp("removing the 3 Ohm resistor from the network"); disp("Applying KVL to mesh 1"); disp("11*I1-9*I2=50");....//equation 1 disp("Applying KVL to mesh 2"); disp("-9*I1+18*I2=0");....//equation 2 A=[11 -9;-9 18];//solving the equations in matrix form B=[50 0]' ...
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//[f,g,newm]=tangent(nl_sys,xe,ue) //[f,g,newm]=tangent(ff,xe,[ue]) // lin\'earise autour du point d'\'equilibre (xe,ue) // le champ du syst\`eme dynamique d\'efinit par xdot=ff(t,x,[u]) // (suppos\'e en fait autonome) // Arguments : // ff : chaine de caract\`ere donnant le nom du syt\`eme a lin\'eriser // x0 : v...
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// problem 7.12 b=10 d=4 i=1/1000 N=0.03 A=b*d P=b+(2*d) m=A/P z1=23+(0.00155/i)+(1/N) z2=1+((23+(0.00155/i))*(N/(m^0.5))) C=z1/z2 Q=A*C*((m*i)^0.5) disp(Q*1000,"discharge through the rectangular channel in litres/sec")
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clc //initialisation of variables w=82//kg p=45//kg //CALCULATIONS F=w-p//kg //RESULTS printf('The force is he exerting on rope =% f kg',F)
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function z = levy_function(x) z = (4*(x(1)^2)) - (2 * 1 * (x(1)^4)) + ((1/3)*(x(1)^6)) + (x(1) * x(2)) - (4*(x(2)^2)) - 1 + (x(2)^4) plot ( x(1) , x(2) , "g." ) endfunction function z = levy_function_countour(x, y) z = (4*(x^2)) - (2 * 1 * (x^4)) + ((1/3)*(x^6)) + (x * y) - (4*(y^2)) - 1 + (y^4) ...
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clc; P=10; //Ohm Q=3; //Ohm R=12; //Ohm S=6; //Ohm G=20; //Ohm disp("-12I+22I1+IgG=0 (1)"); //KVL disp("6I-9I1+29Ig=0 (2)"); //KVL disp("13I1-3Ig=2 (3)"); //KVL Ig=7.797*10^-3; //from (1)(2)(3) disp(Ig,"Current through Galvanometer in Ampere = "); //displaying result
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disp('chapter 5 ex5.1') disp('given') disp('maximum signal voltage Vs=.5volt') disp('voltage gain Av=10') disp('IBmax=1.5*10^(-6)A and I1=100*IBmax') Vs=.5 Av=10 IBmax=1.5*10^(-6) I1=100*IBmax disp('amperes',I1) disp('R1=Vs/I1') R1=Vs/I1 disp('ohms',R1) disp('R2=Av*R1') R2=Av*R1 disp('ohms',R2) disp('R3...
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clear; clc; vg=11e3; sg=90e6; xg=.25; st1=100e6; vt1a=10e3; vt1b=132e3; nt1=vt1a/vt1b; xt1=.06; st2=30e6 *3; vt2a=66e3 * sqrt(3); vt2b=10e3; nt2=vt2a/vt2b; xt2=.05; sm1=50e6; vm1=10e3; xm1=.2; sm2=40e6; vm2=10e3; xm2=.2; xl=100; Sb=sg; Vbg=vg; Xg=xg * (vg/Vbg)^2 * (Sb/sg); Xt1=x...
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[[i= partials/header ]] <link rel="stylesheet" href="/assets/css/index.css"> <div class="container" style="align-items: center;"> <div class="header"> <img class="logo" src="/assets/img/logo.png" alt="Quick Mark Logo"> </div> <div class="app"> <div id="welcome" class="homeBlock"> ...
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getInformation.sci
// function to read a character string of interest // Logic : we need to know the measure of interest, to find corresponding index // measure = what we want to read (could be a matrix) // namefile= Matrix of character string // apply "getInformation" function //endfunction function [value]=getInformation(measure,name...
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clc //initialisation of variables m= 0.3 //kg R= 8.314 //J/mol K M= 28 //gms T1= 500 //C p1= 500 //kPa k= 1.4 V3= 0.3 //m^2 //CALCULATIONS V1= m*R*(273.15+T1)/(M*p1) T3= k*(273.15+T1) p3= m*R*T3*100/(M*V1) //RESULTS printf (' final pressure = %.1f kPa',p3)
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//ques10 //Mass of Air in a Room clc l=4;//length in metres b=5;//breadth in metres h=6;//height in metres V=l*b*h;//volume in m^3 P=100;//Pressure in kPa R=0.287;//Gas constant for a given gas in kPa.m^3/Kg.k T=298;//Temp in K m=P*V/(R*T);//mass in Kg printf('Mass =P*V/(R*T)= %.1f kg',m);
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clear; clc; R=28;C=0.04*(10^-6);G=0;L=0;f=1600;s=2000;r=3.7;l=88*(10^-3); Rc=R+(r/(s*(10^-3))); Lc=L+(l/(s*(10^-3))); Fc=1/(%pi*(sqrt(Lc*C))); printf("-Cutoff frequency = %f Hz\n",round(Fc)); w=2*%pi*f; a=sqrt(w*C*R/2); al=((Rc/2)*sqrt(C/Lc))+((G/2)*sqrt(Lc/C)); A=(a-al)*8.686; printf("-Decrease in attenuati...
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clc(); clear; //Given : u = 5*10^5; //horizontal velocity in m/s alpha = 35; // in degrees E = 200 ;// Electric field in V/m e = 1.6*10^-19; // electron charge in C m = 9.12*10^-31; // electron mass in kg a = (-e*E)/m; // horizontal range in m/s^2 //(a); z_max = (-(u^2)*(sind(alpha))^2)/(2*a); // maximum pene...
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clear// //Variables IE = 6.28 //Emitter current (in milli-Ampere) IC = 6.20 //Collector current (in milli-Ampere) //Calculation alpha = IC / IE //Common base current gain //Result printf("\n Common-Base current gain is %0.3f ." ,alpha)
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// Chapter 3_The Semiconductor in Equilibrium //Caption_Statistics of acceptors and donors //Ex_7//page 108 T=300; Nd=10^16 // donor concentration per cm cube kT=0.0259 Ecd=0.045 //Ec-Ed Nc=2.8*(10^19); x=1/(1+(Nc/(2*Nd))*exp(-(Ecd)/kT)) printf('Fraction of total electrons still in the donor state is %fd ...
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clear; clc; printf("\t\t\tExample Number 1.4\n\n\n"); // heat source and convection // illustration1.4 // solution d = 1*10^(-3);//[m] diameter of wire l = 10*10^(-2);//[m] length of wire Sarea = 22*d*l/7;//[square meter] surface area of wire h = 5000;//[W/square meter] heat transfer coefficient Twall = 114...
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clc; clear all; disp("cylindrical drum") d=0.35;// m diameter ts=80;// degree C rhol=956.4;//kg/m^3 k=68.23*10^(-2);//W/m.C mu=283*10^(-6);// kg/ms hfg=2201.6*10^3;// J/kg vg=0.885;// m^3/kg rhov=1/vg;//kg/m^3 g=9.81;// m/s m=70/3600;// kg/s tsat=120.2;// degree C disp("delL=(4*k*mu*(tsat-ts)*L/(g*rhol*(r...
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//Example 7.13 //Zeros for the Thermal System from a State Description xdel(winsid())//close all graphics Windows clear; clc; //------------------------------------------------------------------ // State space model of the given system F=[-7 -12; 1 0]; G=[1;0]; H=[1 2]; J=0; sysG=syslin('c',F,G,H,J) //---...
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clc //Example 13.5 printf("Given") disp('L1=30 mH L2=60 mH M=40 mH') L1=30*10^-3; L2=60*10^-3; M=40*10^-3; //The equivalent T network is UL=L1-M UR=L2-M CS=M printf("The T network has \n") printf("%d mH in the upper left arm\n",UL*10^3) printf("%3.0f mH in the upper right arm\n",UR*10^3) printf("%d mH in t...
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//Scilab Code for Example 9.11 of Signals and systems by //P.Ramakrishna Rao //Power Spectral Density clear; clc; function [y]=delta(t) if t==0 y=1 else y=0 end endfunction fo=2; n=1; RC=1/(2*%pi*10^3); w=4*%pi*10^3 H=1/(1+%i*w*RC); disp(H,'H(f)'); X=(abs(H))^2; for f=-5:5 Pxx(n)=25*[delta(f-fo)+delta(...
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exec("alaqiltest.start", -1); try foo = new_Foo(); catch alaqiltesterror(); end try foo2 = Foo_blah(foo); catch alaqiltesterror(); end try delete_Foo(foo); catch alaqiltesterror(); end try delete_Foo(foo2); catch alaqiltesterror(); end exec("alaqiltest.quit", -1);
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clear //find the work done by exhaust gases //given // w=500. v1=9000. v2=5400. t=60. V1=v1/t V2=v2/t hp=32.2 KE1=(1/2.)*(500/32.2)*(150**2) KE2=(1/2.)*(500/32.2)*(90**2) W=KE1-KE2 printf("\n \n workdone %.2f ft-lb",W)
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clear clc E=10 x=1 e=1e-5 while (E>e) f=(2*x) - (.43429 * log(x))-7 df=2-(.43429 /x) x1=x-(f/df) E=abs(x1-x) x=x1 end mprintf("x=%f",x)
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function [stable] = isStable(tf) // isStable function for SISO continuous time transfer function // Example // s = %s ; // sys = syslin('c',(s+8)/(s^3+8*s^2+7*s+9)) ; // isStable(sys) stable = %T //True ; if (degree(tf.num)>degree(tf.den)) stable = %F ; end root...
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function [in,out]=get_blocks_in_rect(scs_m,ox,oy,w,h) // Copyright INRIA in=[];out=[] for k=2:size(scs_m) o=scs_m(k) if o(1)=='Block'|o(1)=='Text' then // check is block is outside rectangle orig=o(2)(1) sz=o(2)(2) x=[0 1 1 0]*sz(1)+orig(1) y=[0 0 1 1]*sz(2)+orig(2) ok=%f for kk=1:4 ...
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(unwatch all) (clear) (set-strategy depth) (open "iofnx.rsl" iofnx "w") (dribble-on "iofnx.out") (batch "iofnx.bat") (dribble-off) (load "compline.clp") (printout iofnx "iofnx.bat differences are as follows:" crlf) (compare-files iofnx.exp iofnx.out iofnx) ; close result file (close iofnx)
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disp('chapter 5 ex5.4') disp('given') disp('709 op-amp is used design an inverting amplifier') disp('Vs=50mV') Vs=50*10^(-3) disp('voltage gain Av=100') Av=100 disp('IBmax=2nA and I1=100*IBmax') IBmax=2*10^(-9) I1=100*IBmax disp('amperes',I1) disp('R1=Vs/I1') R1=Vs/I1 disp('ohms',R1) disp('standard value ...
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//to calculate synchronising power and torque coeff/deg mech shift clc; clear j = %i; V=11000; Vt=V/sqrt(3); P=6*10^6; Ia=P/(sqrt(3)*V); ohm_b=Vt/Ia; Xs=.5; Xss=Xs*ohm_b; f=50; P=8; n_s=(120*f/P)*(2*%pi/60); Ef=Vt; dl=0; Psyn=(%pi/15)*(Ef*Vt/Xss)*cosd(dl);disp(Psyn,'synchronising power(W)'); Tsyn=Psyn/n_s;disp(Tsyn,...
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// ///////////////////////////////////////////////////////////////////////////////////////////// // ############################################################################################# // begin : header // ############################################################################################# // ////////...
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//To determine the number of lamps required //Page 344 clc; clear; A=30*20; //Area of the building B=25; //Brightness in lumen/sq. metre CoR=0.25; //Co - efficient of reflection Lwatt=500; //Lamp Wattage Ll=8000; //Lamp lumens output...
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clc; // Resolving forces 125 N and 90 N f1=125;//N f2=90;//N f3=200;//N theta=45;// degree, angle made by f1 with x axis theta=theta*%pi/180;// Conversion of angle into radian alpha=30;// degree, angle made by force f2 with y axis alpha=alpha*%pi/180;// Conversion of angle into radian f1x=f1*cos(theta);//N, X...
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// calculate dc collector current,dc collector-emitter voltage,ac resistance seen by collector // Electronic Principles // By Albert Malvino , David Bates // Seventh Edition // The McGraw-Hill Companies // Example 12-1, page 384 clear;clc; close; // Given data R1=490;// in ohms R2=68;// in ohms Rc=120;// in ohms Re=2...
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//Ex21_8 Pg-1076 clc dec=253 //decimal number bin=dec2bin(dec) //binary value of 253 disp("Binary value of 253 =") disp(bin) disp("In TTL for 1s it is represented in 5V and 0s as 0V") disp("Therefore 253 in TTL voltage terms is") disp("5V,5V,5V,5V,5V,5V,0V,5V")
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{ "EGTE 150800Z 1509/1518 18005KT 9999 FEW010 TEMPO 1510/1511 BKN010 TEMPO 1511/1512 OVC010 BECMG 1512/1515 CAVOK": { "TAF base conditions cover METAR - CAVOK": { "metar": "EGTE 150850Z 18005KT CAVOK", "test time": "20200615T0900Z", "expected": "" }, "TAF ...
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clear(); function writeToFile(filePath, var, data) fd = mopen(filePath, 'wt'); for i = [1:size(var, 2)] mfprintf(fd,"%s\t",var(i)); end mfprintf(fd, "\n"); for i = [1:size(data, 1)] for j = [1:size(data, 2)] mfprintf(fd, "%f\t", data(i, j)); end mfprintf...
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clc //initialisations m1=10//kg t1=80//c t2=20//c t3=150//c t4=90//c t=100//c a=800//cal/kg //calculations h=m1*1000*(t1-t2)/1000 H=a*(t3-t)+540000+1000*(t-t4) k=H/1000 x=h/k //results printf(' kg of steam required per hour= % 1f kg/hr',x)
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DFT Pract 4.sce
clc function [m]=dft_mtx(N) W=int(exp(-%i*2*%pi/N)) for k=0:N-1 for n=0:N-1 m(k+1,n+1)=W**(n*k) end end endfunction A=dft_mtx(4) disp(A,"DFT Matrix= ") x=input('Enter x= ') disp(x,"x= ") //DFT Transformation X=A*x*A' disp(X,"2D DFT of x= ") //Inverse DFT Transform...
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clc //initialisation of variables T= 116 //C C1= 3.74*10^-16 C2= 1.44*10^-2 //CALCULATIONS WLmax= (2893*10^-6)/(T+273) Wb= (C1*(WLmax)^(-5))/((%e^(C2/2893*10^6))-1) //RESULTS printf ('Wavelength at which the maximum monochromatic emissive power = %.2e m',WLmax) printf (' \n Coffecient of performnance= %.2e W/...
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5 56:1.0 119:0.125 251:0.25 266:1.0 308:0.3333333333333333 1185:0.5 1387:1.0 5 1:0.42857142857142855 3:0.02631578947368421 7:2.0 25:2.0 30:1.0 44:0.14285714285714285 57:0.1 134:1.0 283:1.0 430:1.0 436:1.0 508:1.0 561:1.0 598:1.0 931:1.0 1029:1.0 1105:1.0 1783:1.0 1928:1.0 5 1:0.14285714285714285 3:0.02631578947368421 1...
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C23P5.sce
clear clc //to find speed of sound in the gas //Given: //room temperature T = 20//in degree celsius //parameter gama for air gama = 1.4 //molar gas constant R = 8.31//in J/mol.K //molar mass for air M = 0.0290//in Kg/mol //Solution: //applying laws of thermodynamics //speed of sound in the gas v = sq...
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4_13.sce
clc //initialisation of variables L= 100 //ft H= 2.25 //ft Cd= 0.95 w= 120 //ft h= 2 //ft g= 32.2 //ft/sec^2 //CALCULATIONS Q= 3.087*Cd*L*H^1.5 v0= Q/(w*(h+H)) Q1= 3.087*Cd*L*((H+(v0^2/(2*g)))^1.5-(v0^2/(2*g))^1.5) //RESULTS printf ('Discharge = %.f cuses ',Q1)
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clear /////////////////////////////////////////////////////////////////////// // Autor: Juan Luis Flores Garza // Fecha: 2/26/16 // Version: 1.0 // // Algoritmos: // // Algoritmo gauss jordan // 1. RowPivote = 1 // 2. ColPivote = 1 // 3. Mientras no se llegue y procese ult...
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//Optical Fiber communication by A selvarajan //example 9.1 //OS=Windows XP sp3 //Scilab version 5.5.1 clc; clear all; lambda=1.3*1e-6//wavelength in m c=3*1e8//velocity of light in m/s SNRoutdB=30//signal to noise ratio at outputin dB SNRout=10^(SNRoutdB/10);//signal to noise ratio at output normal scale new...
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### Recall Pilot Experiment ### Last Edit: 07.03.2017 ### Author: Daniel Cortez Stevenson ### Questions or comments? Contact: [daniel.stevenson@charite.de] # ---- Header ---- pcl_file = "S2_Reca_Pilot.pcl"; ### Button Options active_buttons = 4; button_codes = 1, 2, 3, 4; ### Response Logging Options #...
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//========================================================================== // get_comment : return a C comment // for generated code // //input : typ : a string // param : a list // //output : a C comment // //16/06/07 Author : A.Layec //Copyright INRIA function [txt]=get_comment(typ,param) tx...
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// FUNDAMENTALS OF ELECTICAL MACHINES // M.A.SALAM // NAROSA PUBLISHING HOUSE // SECOND EDITION // Chapter 2 : BESICS OF MAGNETIC CIRCUITS // Example : 2.10 clc;clear; // clears the console and command history // Given data A = 2*10^-4 // cross sectional area in m^2 N = 200 // number...
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function [s] = check(str) funcprot(0); is_AKICc = (str == "AKICc") disp(is_AKICc) endfunction
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false
789
sce
Ex1_6.sce
clear; clc; //Example 1.6 Is=10^-13;//(A)reverse saturation current V_PS=5;//(V)applied voltage R=2;//(KOhm)Resistance in circuit V_T=0.026;//(Volt)terminal voltage //V_PS=Is*R*(exp(V_D/V_T)-1)+V_D //5=(10^-13)*(2000)*(exp(V_D/V_T)-1)+V_D //let right side of equation be x=(10^-13)*(2000)*(exp(V_D/V_T)-1)+V_D ...