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// Ex8_2 clc; // Given: N=10^5;// electron multiplication v=10^-6;// in V e=1.6*10^-19;// electron charge // Solution: e1=N*e; C=e1/(2*v); C1=C*10^9; printf("The capacitance that would be required is = %f nF",C1)
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// Exa 3.10 clc; clear; close; // (i) Given amplifier is an inverting amplifier, where // V_out= -R_f/R_in*V_in = 1Mohm/1Mohm*V_in = -V_in, So // Av= V_out/V_in Av=-1; disp(Av,"Input impedence :"); // (ii) Because it is a unity gain inverter, So I_in= I_out // A_in = I_out/I_in A_in = 1; disp(A_in,"Voltage ...
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// Example 3.1.a:fracture stress clc; clear; close; la=0.16;//bond length in nm st=2.6*10^6;//psi psi=6894.76;//Nm^-2 e=9*10^10;//NM^-2 yp=((4*la*10^-9*(st*psi)^2)/(e));//in joules c=10^-8;// sf=sqrt((2*e*yp)/(%pi*c));//N/m^2 sf1=sf/(psi);//psi disp(sf1,"fracture stress in psi is")
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// Scilab code Exa4.10 : : Page 180 (2011) clc; clear; E = 4e+006; // Energy lost in the scintillator, eV N_pe = E/10^2*0.5*0.1; // Number of photoelectrons emitted G = 10^6; // Gain of photomultiplier tube e = 1.6e-019; // Charge of the electron, C Q = N_pe*G*e; ...
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PL/SQL Developer Test script 3.0 14 begin -- Call the procedure pkg$xftp_messages.getoperationhistory(pdate => :pdate, pxml => :pxml); open :cur for with rowset as (SELECT extractValue(VALUE(t), 'OPER/CREDITWALLETNO') AS creditwalletno, extractVal...
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//chapter 12 //example 12.23 //page 528 printf("\n") printf("given") hie=1*10^3;hfe=50;hoe=10*10^-6;Cc=5*10^-12;Cp=330*10^-12;Lp=75*10^-6;Rw=1;Rl=5*10^3;fo=1*10^6;zP=224*10^3;rC=100*10^3;K=.015;Ls=50*10^-6; RL=(Zp*Rc)/(Rc+Zp) disp("voltage gain from the input to the primary memory winding") Avp=(hfe*RL)/hie Vs...
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funcprot(0) function [polar] = rect2polar(x,y) //Function to convert rectangular coordinates to polar coordinates polar=ones(1,2) polar(1)=sqrt((x^2)+(y^2)) polar(2)=atan(y/x) polar(2)=(polar(2)*180)/%pi endfunction function [rect] = polar2rect(r,theta) //Function to convert polar coordinates to rectang...
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//This scilab script is to compute the solution of a linear equation clear clc //Example of Ax + b = 0 where unique solution exists A = [1 2 3;3 2 1;2 4 5]; b = [7;7;12]; [x,kerA] = linsolve(A,b); disp(x,'Solution of Ax + b',kerA,'kernel of A'); //Example of Py+q = 0 where solution does not exist P = [1 2 3;3 2 1;4 ...
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//Chapter-3,Example3_6_4,pg 3-19 Pin=8.6 //Input power in mW Pout=7.5 //output power in mW Pl=(-10)*log10(Pout/Pin) //Power loss or attenuation L=0.5 ...
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//************************* Half Wave Rectifier ****************************** if (blk_name.entries(bl) =='h_rect') then mputl("# h_rect",fd_w); for ss=1:scs_m.objs(bl).model.ipar(1) //.subckt h_rect in[0]=in1 in[1]=in2 out=out #h_rect_bias[0] =1e-6 & h_rect_fg[0] =0 cap_str= ".subckt h_rect i...
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// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal // Chapter 2 - Pressure and its measurements // Problem 3.38 //Given Data Set in the Problem dens=1000 g=9.81 h=1.5 L=4 b=2 a=4 alpha=30 //calculations //1) a_x=a*cos(alpha/180*%pi) a_y=a*sin(alpha/180*%pi) theta=(atan(a_x/(a_y+g...
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//Example 15.34 //find bandwidth of the transfer function clear;clc; xdel(winsid()); s=%s A=1 B=(s+1) tf=A/B disp("when A/B(jw)=1/sqrt(2), w=w1") w1=(1/0.707)^2-1 //w1=bandwidth of the transfer function disp("Hence the bandwidth is 1 rad/sec")
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clc; clear; format('v',6); f=10; d=1.6*50/f^(1/3); disp(d,"d(in km)=");
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// Scilab Code Ex2.24:: Page-2.19 (2009) clc; clear; t = 9.21e-05; // Thickness of the mica sheet, cm mu = 1.5; // Refractive index of material of sheet n = 1; // Order of interference fringes // As path difference, (mu - 1)*t = n*lambda, solving for lambda lambda = (mu - 1)*t/n; // Wavelength of lig...
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errcatch(-1,"stop");mode(2);//Initilization of variables P=[4.82, -2.33, 5.47] //N Q=[-2.81,-6.09,1.12 ] //m //Calculations M=P*Q' //Nm //Results printf('Result is:%f N.m',M) //N-m exit();
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clc; clear; v=1400;//rms voltage f=60;//frequency kva1=125; pf1=0.28; //inductive load and lagging power factor kw2=10; kvar2=-40; //active and reactive power of a capacitive load kw3=15;//resistive load theta1=acos(pf1); s1=complex(125*cos(theta1),125*sin(theta1)); s2=complex(kw2,kvar2); s3=complex(kw3,0); ...
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//example 7 //work potential of compressed air in tank clear clc To=300 //in K T1=To R=0.287 //kPa-m^3/kg-K V=200 //in m^3 P1=1000 //kPa m1=P1*V/(R*T1) //in kg Po=100 //in kPa o1=R*To*(log(P1/Po)+Po/P1-1) //kJ/kg X1=m1*o1 //exergy content of compressed air in kJ printf("\n Hence, the exergy content of comp...
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Ex7_13.sce
clc //Chapter7 //Ex_13 //Given fs=1 //in MHz k=0.1 fa=fs/(sqrt(1-k^2)) disp(fa,"fa value in MHz for given fs is") printf("thus fa-fs is only %f kHz, which means they are very close ",(fa-fs)*10^3)
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// ################## 1 :: i1=imread("C:\Users\co250\Desktop\home_image.jpg"); i2=imread("C:\Users\co250\Desktop\mandrill.jpg"); i3=imread("C:\Users\co250\Desktop\RGB2.png"); imshow(i1); imshow(i2); imshow(i3); i4=imwrite(i3,'RGB3.jpg','.1'); // png to jpeg // ################## 2 :: i1=imread("C:\Users\...
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// example 6.32 // caption: solving the IVP by numerov method // u''=(1+t^2)*u // u(0)=1, u'(0)=0 ,[0,1] // h=0.2, // expression for numerov method is //u(j+1)-2*u(j)+u(j-1)=(h^2/12)*(u''(j+1)+10*u''(j)+u''(j-1)); // observing the IVP we can reduce the numerov method to //u(2)=2*u(1)-u(0)+(.2^2/12)*(1.16...
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Ex16_13.sce
//chapter16 //example16.13 //page354 Zin=15// kilo ohm Ai=240 mi=0.015 Zin_dash=Zin/(1+mi*Ai) printf("input impedence with negative feedback = %.3f kilo ohm \n",Zin_dash)
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clear; clc; disp('Example 4.14'); // aim : To estimate // the dryness fraction // Given values M = 1.8;// mass of condensate, [kg] m = .2;// water collected, [kg] // solution x = M/(M+m);// formula for calculation of dryness fraction using seprating calorimeter mprintf(' \n The dryness fractio...
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EX1_6.sce
//Finding of increase of Pressure //Given k=2.07*10^6; // Bulk Modulus in KN/m^2 dv=0.01; //Change in Volume //To Find p=k*(dv); // Change in pressure disp(" Increase in Pressure ="+string(p)+" KN/m^2");
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//Finding of Depth of Water //Given p=100.5525*10^4; //pressure intensity in N/m^2 spgr=1.025; //Specific gravity rho=1000; //Density of water in kg/m^3 g=9.81; //Gravitational force due to acceleration in m/sec^2 w=rho*g; //To Find h=p/w; disp(...
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clear// //Variables I = 5 //Current (in Ampere) V = 230 //Voltage (in volts) //Calculation P = V*I //Power consumed (in watt) //Result printf("\n The power consumed by the toaster is: %0.3f watt.",P)
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//Strength Of Material By G.H.Ryder //Chapter 1 //Example 8 // To Find the Maximum Stress l=1; //lenght of steel rod, Unit in m N=1000; //rpm of rod, Unit in rmp rho=7.8; //density of the material, Unit in g/cm^3 Omega=%pi*2*N/60; //Angular Velocity, Unit in rad/sec //sigma a=-rhox^2*Omega*2/2+c, formula...
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//Exa 2.60 clc; clear; close; //Given data : format('v',6); f=50;//in Hz P=4;//no. of poles phase=3;//no. of phase S=4;//in % S=4/100;//in fraction Ns=120*f/P;//in rpm N1=Ns-Ns*S;//in rpm //When speed reduced to 10% N2=N1*85/100;//in rpm(NewSpeed) disp(N2,"New speed(in rpm) :"); //New speed is reduced b...
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//Compute the Loss per Kg at a particular frequency. clc; clear; hy_ls=4.9; f1=50; maxflux=1; density=7.5; d=density*(10^6)/(10^3); hy_ls_cycle= hy_ls*d/f1; n=hy_ls_cycle/((maxflux)^1.7); disp(n,'i) The value of the Co-Efficient= ' ) mflux2=1.8; f2=25; hy_ls2=hy_ls*(f2/f1)*((1.8)^1.7); ...
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ex_3.sce
//Example 3 // velocity and direction clc; clear; close; //y=1.2*sin(3.5*t+0.5*x);//equation w=3.5;//from equation k=0.5;//from equation v=w/k;//m/s disp("wave velocity is "+string(v)+" m/s and direction of the wave is along negative X-axis")
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// Exa 4.12.4 clc; clear; close; // Given data I_DSS = 12;// in mA I_DSS = I_DSS * 10^-3;// in A V_P = -(6);// in V V_GS = -(1);// in V g_mo = (-2 * I_DSS)/V_P;// in A/V g_m = g_mo * (1 - (V_GS/V_P));// in S disp(g_m*10^3,"The value of transconductance in mS is");
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//Section-10,Example-1,Page no.-CT.29 //To calculate the maximum work done. clc; V_2=50 V_1=5 R=0.08206 T=298 n=10 W=-(n*R*T)*log(V_2/V_1) disp(W,'Maximum work done in(dm^3atm)') W1=W*(8.314/0.08206) disp(W1,'Maximum work done in(J)')
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F=60;P=4;S=0.05; Ns=1800;V=460;Tr=0.5; Ns=(120*F)/(P) N=(1-S)*Ns F2=S*F Sliprpm=S*Ns A=S*Tr*V/sqrt(3)
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exec('01_Initialize.sce'); exec('02_LoadPoints.sce');
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printf("\n trnsfer function of the given network ") printf("\n E0(s)=(1/(C*s))*I(s)") printf("\n E0(s/Ei(s)=(I(s)/C*s)*((C*s)/(I(s)*(s^2*LC+s*R*C+1)))") printf(" E0(s/Ei(s)= 1/(s^2*LC+s*R*C+1)is the required transfer function")
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# Graph of refinement dependencies (e_bc e_spec: spec; nt_items only; single set of items (e_cd e_spec_all_items: single set, NITM CITM SITM spec_item_t = NTITM of nt_item | CITM of citm_t | SITM of sym_item an abstract model, with all types of items; refines e_bc; single set of items; no staging; ...
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//metodo de legendre clc clear format (6) //x = [0 0.5 1 1.5]; //y = [1 1.6487 2.7182 4.4816]; //ponto = 1.3; //n = 4; //n = 3; //n = 2; //grau do polinomio //x = [0 0.2 0.4 0.5]; //y = [0 2.008 4.064 5.125]; //ponto = 0.3; //n = 4; //x = [1950 1960 1970 1980]; //y = [352.724 683.908 1235.030 1814.990]; //ponto = 1...
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function [x0,f0,iter]=dfp(x0,maxiter) clf exec('cel.sci'); exec('mapa.sci',0); [f0,df0]=cel(x0); iter=0; kryt=norm(df0); g=df0; D=eye(2,2); dd=g*D'; t=1; x1=x0-t*dd; [f1,df1]=cel(x1); iter=iter+1; while f1 > f0 t=0.25*t; x1=x0-t*dd; [f1,df1]=cel(x1); iter=iter+1; end ...
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//iir filter design // Hz = iir(N,ftype,fdesign,cfreq,delta) // [Hw w] = frmag(Hz,256) // y = flts(x,Hz) clc clear close //1.initializing given parameters fs = 44100 N = 5 ftype = 'lp' fdesign = 'butt' cfreq = [1500/fs 0] delta = [0 0] //2.determining filter function Hz = iir(N,ftype,fdesign,cfreq,delta) [Hw w] = f...
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load DMux32Way.hdl, output-file DMux32Way.out, compare-to DMux32Way.cmp, output-list in%B1.1.1 sel%B1.5.1 o0%B1.1.1 o2%B1.1.1 o4%B1.1.1 o6%B1.1.1 o8%B1.1.1 o10%B1.1.1 o12%B1.1.1 o14%B1.1.1 o16%B1.1.1 o18%B1.1.1 o20%B1.1.1 o22%B1.1.1 o24%B1.1.1 o26%B1.1.1 o28%B1.1.1 o30%B1.1.1; set in 0, set sel %B00000, eval, ...
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kalman_filter.sci
function[K] = kalman_filter(P0,A,Qw,Qv,C) // P_apost = P[n|n] // P_priori = P[n|n-1] P_apost = P0; for i =1:100 P_priori = A * P_apost * A' + Qw; K(:,i) = P_priori * C' * inv(C * P_priori * C' + Qv); P_apost = P_priori - K(:,i)*C * P_priori; end endfunct...
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// sum 31-2 clc; clear; d=16; D=1.5*d; t=20; tg=4; //Let Gasket diameter in compression zone be d1 d1=D+(2*t)+tg; lg=40; E=207*10^3; kb=%pi*d^2*E/(lg*4); Ecl=90*10^3; x=(5*(lg+(0.5*d))/(lg+(2.5*d))); kp=%pi*Ecl*d/(2*log(x)); Ag=%pi*(d1^2-d^2)/4; Eg=480; kg=Ag*Eg/tg; km=kg*kp/(kg+kp); C=kb/(kb+km); A...
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clear; clc; printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 1.2 Page 11 \n')// Example 1.2 // Find a) Emissive Power & Irradiation b)Total Heat Loss per unit length d=.07; //[m] - Outside Diameter of Pipe Ts = 200+273.15; //[K] - Surface Temperature of St...
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clc; x = [1 2 3 4 5 6 7 8 9 10 ]; y = [5 0 5 0 5 0 5 0 5 0 ]; plot2d2(x,y) xlabel('VALUES OF x'); ylabel('VALUES OF y'); title('SQUARE WAVE FUNCTION');
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example6_8.sce
syms G1 G2 G3 G4 G5 G6 G7 G8; T1=G1*G8*G7*G5*G6; T2=G1*G2*G3*G4*G8; L1=-G6*H5; L2=-G3*H3; delta=1-(L1+L2)+(L1*L2) del1=1-L2; del2=1-L1; TF=(T1*del1 + T2*del2)/delta ; disp(TF,"C/R = ")
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// Scilab ( http://www.scilab.org/ ) - This file is part of Scilab // Copyright (C) 2013 - Scilab Enterprises - Antoine ELIAS // // 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...
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pathname=get_absolute_file_path('5_10.sce') filename=pathname+filesep()+'5_10_data.sci' exec(filename) clf(); i = 1; while(i<=length(M)) Cpcr(i)=(2/(y*M(i)^2))*[[(2+(y-1)*M(i)^2)/(y+1)]^(y/(y-1))-1] Cpmin(i)=Cpomin/sqrt(1-M(i)^2); i = i+1; end xlabel("Mach Number"); ylabel("Cp"); plot2d(M,Cpcr...
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chi_test.sci
clear; printf('************** Chi-square Distribution Function ****************'); printf('\n'); df = input('Enter a Degree of Freedom : '); Total = cdfchi( "PQ", 1, df); printf('Total ='); disp(string(Total)); printf('\n');
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mulluckrec.sci
function el=mulluckrec(nb,mb,lpx,k,p,y,eps) [n,m]=size(lpx); el=zeros(n,m); for yk=0:nb-mb y(k)=yk; if k < length(p)-1 then el=el+mulluckrec(nb,mb+yk,lpx,k+1,p,y,eps); else y(length(p))=nb-mb-yk; lpy=mulprobln(y,p); c=0.5*bool2s(lpy > lpx-eps)+0.5*bool2s(lpy > lpx+eps); ...
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// Example 1.66 clc;clear;close; // Given data format('v',7); P=4;//no. of poles f1=50;//in Hz S=4;//in % R2=1;//in ohm/phase X2=4;//in ohm/phase //calculations Ns=120*f1/P;//in rpm S=S/100;//slip //part (a) N=(1-S)*Ns;//in rpm disp(N,"(a) Speed of the motor in rpm : "); //part (b) f2=S*f1;//in H...
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9_3.sce
clc clear //Initialization of variables V=200 //ft/s L=5 //ft B=2 //ft rho=0.00232 //slug/ft^3 mu=3.82e-7 //lb-sec/ft^2 p2=14.815 //psia pa=14.7 //psia //calculations Nr=V*L*rho/mu Cdf=0.0032 Fdf=Cdf*%pi*L*B*0.5*rho*V^2 Fd=(p2-pa)*%pi/4 *(B*12)^2 -Fdf //results printf("Drag on the model = %.2f lb",Fd)
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clc;funcprot(0);//EXAMPLE 20.26 // Initialisation of Variables p1=1;............//Intake pressure in bar p2=4;..............//Pressure after first stage in bar p3=16;............//Final pressure in bar ns=2;............//No of stages t1=300;............//Intake temperature in K n=1.3;............//Compression in...
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3_2.sce
clc clear //Initialization of variables t1=1000 //K p1=20 //Mpa p2=10 //Mpa ti=600 //K t2=700 //K v1=0.02188 vi=0.02008 v2=0.02825 Ei=2617.5 E2=2893.1 E1=3441.8 x=0.22 m=1 //kg cp=4.186 t3=639 //K H3=2409.5 H1=3879.3 //calculations Tf= ti+ (v1-vi)/(v2-vi) *(t2-ti) Ef= Ei+ x*(E2-Ei) Q1=Ef-E1 //re...
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clc clear //Input data N=700;//Engine speed in rpm D=0.6;//Diameter of brake drum in m d=0.05;//Diameter of rope in m W=35;//Dead load on the brake drum in kg S=4.5;//Spring balance reading in kg g=9.81;//Gravitational constant in N/m^2 pi=3.14;//Mathematical constant //Calculations P=(((W-S)*g*pi*(D+d))/1...
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19 82:0.06666666666666667 115:0.2 168:0.5 205:0.2 227:0.5 305:0.2857142857142857 307:0.5 510:0.25 750:0.5 824:1.0 1017:0.5 1147:1.0 2243:0.5 3201:1.0 5171:1.0 6308:1.0 19 35:0.029850746268656716 79:0.25 134:0.09090909090909091 168:0.25 227:1.0 302:0.5 371:0.125 510:0.25 889:0.125 2754:1.0 2893:1.0 3391:1.0 5195:1.0 576...
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// Example 9.2 : To determine A_v,f_t,f_P,SR and P_D of folded cascode amplifier // Consider a design of the folded-cascode op amp I=200*10^-6; // (A) I_B=250*10^-6; // (A) V_OV=0.25; // (V) k_n=100*10^-6; // k_n=k'_n (A/V^2) k_p=40*10^-6; // k_p=k'_p (A/V^2) V_A=20; // V_A=V'_A (V/um) V_DD=2.5; // (V) V_SS=2....