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clc //initialisation of variables v=400//Btu per lb t=1698//R c=0.1715//ft p=0.24//ft v1=15//ft v2=1.414//V p1=560//R q=800//ft h=401600//ft-lb per lb s=13.63//ft //CALCULATIONS T=(v/(1*c))//R T1=t+T//R T2=v/(1*p)//R T3=T1+T2//R V=v1/v2//ft T4=T3/(V)^0.4//R Q=1*c*(T4-p1)//Btu per lb Q1=q-Q//Btu per...
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//3.7 clc; V=20; A=20*10^-3; Rv=10*10^3*20; Rx=(V/(A-(V/Rv)))/1000; printf("The resistance=%.2f Kohm",Rx) E_total=2.5+2.5; printf("Maximum possible error=%.0f percent",E_total)
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//clear// //Example3.8:Fourier Series Representation of Periodic Impulse Train clear; clc; close; T =4; T1 = T/4; t = [-T,0,T]; xt = [1,1,1]; //Generation of Periodic train of Impulses t1 = -T1:T1/100:T1; gt = ones(1,length(t1));//Generation of periodic square wave t2 = [-T1,0,T1]; qt = [1,0,-1];//Derivativ...
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clc //initialisation of variables dH= 6896 //cal mole^-1 T= 68.7 //C //CALCULATIONS dS= dH/(273.1+T) //RESULTS printf ('entropy change per mole= %.2f cal deg^-1 mole^-1',dS)
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errcatch(-1,"stop");mode(2);//Example 5.10.3: resolution ; ; format('v',8) //given data : n=4 R=1/10^n; disp(R,"resolution,R = ") exit();
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chdir('/home/mike/proj/math/scilab/quant'); exec('/home/mike/proj/math/scilab/quant/psi.sce');disp('exec done'); phi=0:0.05:%pi; theta=0:0.05:2*%pi; npoints=126; l=1; m=0; xval=xmat(npoints, theta, phi); yval=ymat(npoints, theta, phi); yres=ypol(npoints,theta,phi,l,m); plot3d(xval, yval, yres);
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load Not.hdl, output-file Not.out, compare-to Not.cmp, output-list in%B2.1.1 out%B2.1.2; set in %B0, eval, output; set in %B1, eval, output;
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10 258 740 156 244 458 680 390 694 844 817 ~~~~~~~~~~~~~~~~~~~~~ 102
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// Example 8_1 clc;funcprot(0); // Given data P_1=10;// kPa P_2=2000;// kPa T_3=400;// °C h_f=191.8;// kJ/kg h_1=h_f;// kJ/kg h_3=3248;// kJ/kg s_3=7.1279;// kJ/kg.K // Calculation v_1=0.001;// m^3/kg w_P=v_1*(P_2-P_1);// The pump work in kJ/kg h_2=h_1+w_P;// kJ/kg q_B=h_3-h_2;// The heat input in kJ/kg...
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clc; u=14; //object distance in cm f=-21; //focal distance in cm v=(-5/42); //simplifying(1/f)=(1/v)-(1/u) I=(3*-8.4)/(-14); //using m=(1/0)=(v/u); disp(v,"Image distance in cm = "); //displaying result disp(I,"I in cm = "); //displaying result
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exec("swigtest.start", -1); try Spam = new_Spam() catch swigtesterror(); end if Foo_blah(Spam)<>0 then swigtesterror; end exec("swigtest.quit", -1);
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//Example 11.10 clc; clear; close; format('v',7); //Given data : Cd=0.62;//Coeff of discharge H=250/1000;//meter L=400/1000;//meter g=9.81;//gravity acceleration Q=2/3*Cd*sqrt(2*g)*L*H^(3/2);//m^3/s or cumec disp(Q,"Discharge in cumec : ");
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mode(-1); lines(0); my_handle = scf(0); clf(my_handle,"reset"); demo_viewCode("aggloms.dem.sce"); // DEMO START Data = read(SCI+'/contrib/OpenPR-0.0.2/etc/data/aggloms_data',2,1810); Label = read(SCI+'/contrib/OpenPR-0.0.2/etc/data/aggloms_label',1,1810); sigma = 0.4; // kernel bandwidth ite_num = 60; // iteration ...
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//Expression for the sum of energy stored by inductor and capacitor connected in parallel at resonance clc; clear; printf('v = Vm * cos(w0*t)\n') printf('The energy stored by the capacitor = C*(Vm^2)*(cos(w0*t)^2)\n') printf('The energy stored by the inductor = L*(i^2)/2\n\n') printf('v = L *(di/dt)\n di = ...
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.ig XMERGE1.TST VERS:- 01.00 DATE:- 09/26/86 TIME:- 09:37:37 PM .. Test file consisting of several CAS abstracts for use in trials .. of xmerger (CAS abstracts are the default option): .. Usage: xmerger <xmerge1.tst xmerge2.tst +results_file * XMERGE1.TST ANSWER 1 OF 8 AN CA103(23):193943...
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function scs_m=changeports(scs_m,k,o_n) // Move change number of ports of block k and modify connected links if any //! //look at connected links // Copyright INRIA connected=[];dx=[];dy=[] o=scs_m(k) [nin_n,nout_n,ncin_n,ncout_n]=o_n(3)(2:5) [sz,orient,ip,op,cip,cop]=o(2)([2:3,5:8]) [nin,nout,ncin,ncout]=o(3)(2:5) ...
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//Chapter 26 Ex5 clc; clear; close; lagBA=40; lagCA=64; //distance B and C are lagging from A //assuming A covers 1000 m A=1000; B=A-lagBA; //from given condition C=A-lagCA; lagCB=A*(C/B); //Distance C is lagging from B mprintf("B should give C a start of %.0f meter",A-lagCB);
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MoveRelative 300 0 1 Sinusoid RotateTo 8 0.5 Sinusoid2 Wait 0.5 RotateTo 0 0.5 Sinusoid Wait 0.5 MoveRelative -300 0 1 Sinusoid RotateTo -8 0.5 Sinusoid2 Wait 0.5 RotateTo 0 0.5 Sinusoid Wait 0.5 SendEvent 1 Wait 0.01 Loop
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clear clc disp('total no.of possible outcomes=10*9=') 10*10 disp('no. of favourable outcomes=5*5+5*5=') 5*5+5*5 disp('p=') 50/100
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// Example 1.59 clear; clc; close; format('v',6); // Given data Zouter=0.05+%i*0.11;//in ohm Zinner=0.015+%i*0.5;//in ohm //Calculations R2odash=real(Zouter);//in ohm X2odash=imag(Zouter);//in ohm R2idash=real(Zinner);//in ohm X2idash=imag(Zinner);//in ohm TouterByTinner=R2odash/(R2odash^2+X2odash^2)*...
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load arg.asm, output-file arg.out, compare-to arg.cmp, output-list RAM[401]%D1.6.1 RAM[402]%D1.6.1; repeat 75 { ticktock; } output;
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clear clc //initialisation of variables h= 80 //ft f= 0.008 l= 3000 //ft r1= 6.07 r2= 377.5 r3= 4733 r4= 0.0466 r5= 3220 r6= 51.5 //CALCULATIONS Q= sqrt(h*10/(f*l)) Q1= sqrt(r2+sqrt(r2^2-4*r1*r3)/(2*r1))/3 Q2= Q1-r4*sqrt(r5-r6*Q1^2) //RESULTS printf ('rate discharge when valve B is closed= %.2f cusecs...
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// determine the absolute pressure in the tank clc patom=47.2 // pressure of an atom pg=40 // pressure at 40kpa from table pa=patom-pg mprintf('\n absoulte pressure in the tank is %f kPa',pa)
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function o = holding(i) //Datos de entrada Mini = i(5) T = i(7) // ºC //Parámetros D1211m = 0.05; //s Zm = 10; //ºC //Resolución del modelo xini = Mini; tfin = 300; dt = 0.01; // s t=0:dt:tfin; x = ode(xini,0,t,odeholding); M = x; Mfin = ...
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clc //initialisation of variables m= 0.35 //kg u2= 211.785 //kJ/kg u1= 182.267 //kJ/kg p2= 300 //kPa v3= 0.085566 //kJ/kg v2= 0.076218 //kJ/kg h3= 260.391 //kJ/kg h2= 234.650 //kJ/kg u4= 199.460 //kJ/kg u3= 234.721 //kJ/kg p4= 250 //kPa v1= 0.076218 //kJ/kg v4= 0.085566 //kJ/kg h1= 201.322 //kJ/kg h4= 2...
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// Scilab code Ex17.7 : Pg:892 (2011) clc;clear; amu = 931.5; // Energy equivalent of 1 amu, MeV nucleus = cell(4,3); nucleus(1,1).entries = 'P'; nucleus(1,2).entries = 15; nucleus(1,3).entries = 31; nucleus(1,4).entries = 30.98356; nucleus(2,1).entries = 'n'; nucleus(2,2).entries = 0; nucleus(2,3).entries...
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clc clear //input data N=9000//The rotational speed in rpm dT0=20//The stagnation temperature rise in K DhDt=0.6//The hub to tip ratio l=0.94//The work donee factor ns=0.9//The isentropic efficiency of the stage C1=150//Inlet velocity in m/s P01=1//The ambient pressure in bar T01=300//The ambient temperature ...
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clc; clear; gamma1=9.8;//kN/m^3 gamma2=15.6;//kN/m^3 h1=1;//m h2=0.5;//m //pA-(gamma1)*h1-h2*(gamma2)+(gamma1)*(h1+h2)=pB //pA-pB=diffp diffp=((gamma1)*h1+h2*(gamma2)-(gamma1)*(h1+h2)); disp("kPa",diffp,"The difference in pressures at A and B =")
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//Determine the maximum range of a deep-space radar lambda1 = 30/2.5; lambda = lambda1/100; Pt = 2.5e+7; D = 64; S = 1; dF = 5e+3; F = 1.1; Rmax = 48*sqrt(sqrt(((Pt*D^4*S)/(dF*lambda^2*(F-1))))); disp(Rmax, 'Maximum range of a deep-space radar is (in Km)')
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clc;funcprot(0);//EXAMPLE 4.10 // Initialisation of Variables r=7;....................//Compression Ratio t2=715;.................//Temperature at the end of isentropic compression in Kelvin t4=1610;................//Temperature at the end of expansion in Kelvin //Calculations vr2=65.8;..................//From st...
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clear;lines(0); x=-10:10; y=-10:10;m =rand(21,21); grayplot(x,y,m,"111",[-20,-20,20,20]) t=-%pi:0.1:%pi; m=sin(t)'*cos(t); xbasc() grayplot(t,t,m)
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//calculating Kc //Example 6.10 clc clear //E'cell=0.0591*logKc/n Ecell=-0.8277 n=1 Kc=10^(n*Ecell/0.0591)//equilibrium constant printf('Thus the equilibrium constant for the reaction = %e',Kc)
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clc disp("example 2.8") disp("the chronological load curve is plotted in fig 1") a=[0 5 9 18 20 22 24] //time in matrix format b=[50 50 100 100 150 80 50]//load in matrix format for x=1:6 z(1,x)=((b(1,x)+b(1,x+1))/2)*(a(1,(x+1))-a(1,x)) end e=sum(z); printf("energy required required by the system in 24 hrs ...
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//1 Load Image img_input=readpbm("C:\Users\DimitriXPS\Documents\GitHub\Exolife\Exolife\Images\Mission 9\Gliese 581d V2.pbm"); display_gray(median(img_input,2));
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clear; clc; // A Textbook on HEAT TRANSFER by S P SUKHATME // Chapter 3 // Thermal Radiation // Example 3.4(a) // Page 123 printf("Example 3.4(a), Page 123 \n\n") e = 0.08; //emissivity T = 800; //temperature, [K] Stefan_constt = 5.67*10^(-8); //[W/m^2.K^4] // From Stefan Boltzmann law, equation...
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//Variable declaration h=6.63*10**-34; //planck's constant c=3*10**8; //velocity of light(m/s) Eg=1.44; //energy gap(eV) e=1.6*10**-19; //Calculation lamda=h*c/(Eg*e); //wavelength(m) //Result printf('wavelength is %0.3f angstrom \n',(lamda*10**10))
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clc //ex16.5 n_m_1=1200; //speed in rpm T_out_1=12; //motor torque W_m_1=n_m_1*2*%pi/60; //angular speed //As we are neglecting losses, the output torque and power are equal to the developed torque and power respectively P_out_1=W_m_1*T_out_1; //output power //For Torque=24 T_out_2=24; T_de...
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// chapter 7 , Example 7.10 , pg 214 T1=300 //temperature (in K) e=1.6*10^-19 //charge of electron (in C) k=1.38*10^-23 //Boltzmann constant (in J/K) T2=330 //temperature (in K) E1=0.3 // E1=(Ec-Ef_300) (in eV) E2=(E1*T2)/T1 //E2=(Ec-Ef_330) (in eV) printf("At 330 K the Fer...
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ex1_44.sce
// Exa 1.44 clc; clear; close; format('v',6) // Given data V_T = 0.7;// in V V = 5;// in V R = 2;// in k ohm R = R * 10^3;// in ohm Vs = 0.7; Vx = Vs+V_T;// in V // The value of I1 I1 = (V-Vx)/R;// in A I1 = I1 * 10^3;// in mA disp(I1,"The value of I1 in mA is"); // The value of I2 I2 = I1;// in mA ...
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test1.tst
PL/SQL Developer Test script 3.0 10 -- Created on 20.06.2017 by OLEG declare -- Local variables here i integer; BEGIN jober.P_CREATE_JOB (p_paiment_id => 82); -- jober.P_RUN_JOB (p_paiment_id => 53); --ui.start_up (p_paiment_id => 70); end; 0 0
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26_8.sce
clear// //Variables VCC = 10.0 //Source voltage (in volts) RE = 1.5 * 10**3 //Emitter resistance (in ohm) R1 = 30.0 * 10**3 //Resistance (in ohm) R2 = 20.0 * 10**3 //Resistance (in ohm) beta1 = 150.0 //Common emitter c...
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ex12_2.sce
clear; //clc(); // Example 12.2 // Page: 319 printf("Example-12.2 Page no.-319\n\n"); //***Data***// T = 298.15;//[K] temperature P = 0.987;//[atm] pressure g_0_NO = 86.6;//[kJ/mol] Free energy of formation the NO from elements R = 8.314;//[J/(mol*K)] // And the corresponding values for the elements N2...
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Ex10_1.sce
clc; //e.g 10.1 Ic=10; Vce=10; hie=500; hoe=10**-5; hfe=100; hre=10**-4; gm=Ic/25; disp('ohm',gm*1,"gm="); rbe=hfe/gm; disp('ohm',rbe*1,"rbe="); rbb=hie-rbe; disp(rbb); gbc=hre/rbe; disp('*10^-7',gbc*10**7,"gbc="); rce=-1/((hoe-(1+hfe)*gbc)); disp('kohm',rce*10**-3,"rce=");
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Ex1_6.sce
//Variable declaration I=10 //Ge diode carries current(mA) V=0.2 //forward bias voltage(V) //Calculation //Part a Is=I/(exp(40*V)-1) //reverse current(mA) //part b I1=1*10**3 V1=(log(1/3.355*10**3 + 1))/40 //voltage(V) I2=100*10**-3 //current(mA) V2=(log(100/3.355*10**3+1))/40 ...
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Ex18_15.sce
// scilab code Exa 18.15 Impulse Steam Turbine 3000 rpm P=500; // Power Output in kW u=100; // peripheral speed of the rotor blades in m/s cy2=200; // whirl component of the absolute velocity at entry of the rotor cy3=0; // whirl component of the absolute velocity at exit of the rotor alpha2=65; // nozzle angle ...
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// EXPT NO. 2: - SCREW TRANSFORMATON //AIM: - TO STUDY THE SCREW TRANSFORMATION OF A POINT // ABOUT A SPECIFIED AXIS USING scilab AND TO PROVE THAT SCREW TM // IS CUMULATIVE [ Rot(?,k)*Trans(?,k) = Trans(?,k)* Rot(?,k) ] clear ; close ; clf clc disp('ENTER COORDINATES OF THE POINT') x=input('...
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Ch02Ex13.sce
// Scilab Code Ex2.13:: Page-2.13 (2009) clc; clear; b = 0.125; // Fringe width of the interfernce pattern due to biprism, cm d = 1; // For simplicity assume distance between sources to be unity, cm d_prime = 3/4*d; // New distance between sources, cm // As b is proportional to 1/d, so b_prime = b*d/d_prime; // ...
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Example2_4.sce
//Example 2.4 //Program to Obtain Output Voltage Vo from Given A.C. Equivalent of an Amplifier using a Transistor clear; clc ; close ; //Given Circuit Data //Input Side Vs=10*10^(-3);//i.e. 10 mV Rs=1*10^3;//i.e. 1 kOhms //Output Side Ro1=20*10^3;//i.e. 20 kOhms Ro2=2*10^3;//i.e. 2 kOhms //Calculation i=Vs...
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eg4_7.sce
Dn = 30; Dp =15; Na = 5*10^16; Nd = 5*10^17; q = 1.6*10^-19; tn = 10^-8; tp = 10^-7; ni = 1.84*10^6; np = ni^2/Na; pn = ni^2/Nd; Ln = (Dn*tn)^0.5; Lp = (Dp*tp)^0.5; disp(Ln,"diffusion length, Ln (in cm) = ") disp(Lp,"diffusion length, Lp (in cm) = ") n = (q*Dn*np/Ln)/((q*Dn*np/Ln)+(q*Dp*pn/Lp)); disp(n,"...
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Ex4_15.sce
//Ex4_15 clc RON=100 disp("RON= "+string(RON)+ " ohm") //ON resistance of analog series switch ROFF=10^(10) disp("ROFF= "+string(ROFF)+ " ohm") //OFF resistance analog series switch Vip=1 disp("Vip= "+string(Vip)+" volts")// Peak amplitude of analog voltage Rs=100 disp("Rs= "+string(Rs)+ " ohm") //Voltage ...
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solution4_18.sce
//Function to round-up a value such that it is divisible by 5 function[v] = round_five(w) v = ceil(w) rem = pmodulo(v,5) if (rem ~= 0) then v = v + (5 - rem) end endfunction //Obtain path of solution file path = get_absolute_file_path('solution4_18.sce') //Obtain path of data file dat...
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clc clear //INPUT DATA Eg=1.43*1.6*10^-19//The energy gap of intrinsic GaAs in J xe=0.85//The electron mobility in m^2 V^-1 s^-1 xh=0.04//The hole mobility in m^2 V^-1 s^-1 me=0.068*9.11*10^-31//effective mass of electron in m mh=0.5*9.11*10^-31//effective mass of hole in m h=6.625*10^-34//Planck's constant in ...
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Ex3_3.sce
////Caption Nquist Rate //Example 3.3 //page no 104 //Find Nquist Rate //given clc; clear; f=100; fs=2*f;//Nyquist rate disp(" Hz",fs,"(i)To avoid aliasing Nquist Rate is ");
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eg4_4.sce
Na = 10^19; Nd = 10^16; E1 = 4*10^5; E2 = 10^7; eps0 = 8.84*10^-14; //in F/m eps = 11.9*eps0; q = 1.6*10^-19; V1 = eps*E1^2/(2*Nd*q); disp(V1,"breakdown voltage for Si = ") V2 = eps*E2^2/(2*Nd*q); disp(V2,"breakdown voltage for diamond = ")
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clc; //e.g 3.22 rho=9*10**-3; mup=0.003; sigma=1/rho; disp('S/m',sigma*1,"sigma="); RH= mup/sigma; disp('m^3*C',RH*1,"RH=");
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7_5.sce
clc; //page no 295 // prob no 7.5 // in the given problem fs=40; m=14; // the minimum data rate needed to transmit audio is given by D=fs*m; disp('Kb/s',D,'The minimum data rate needed to transmit audio is ');
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Example12_1.sce
clear; clc; // Example: 12.1 // Page: 471 printf("Example: 12.1 - Page: 471\n\n"); // This problem involves proving a relation in which no mathematics and no calculations are involved. // For prove refer to this example 12.1 on page number 471 of the book. printf(" This problem involves proving a relatio...
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Example_8_22.sce
//Example 8.22 clear; clc; //Given T=298;//Temperature in K R=8.314;//gas constant in J K^-1 mol^-1 k=4.814-(2059/T);//k=log(K),where K is the equillibrium constant //To determine the values of delGo,delHo and delSo delSo=4.814*R;//entropy change in J K^-1 mol^-1 delGo=-R*T*k;//free energy change in J mol^...
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// Given:- p1 = 3*(10**5) // initial pressure in pascal v1 = 0.1 // initial volume in m3 v2 = 0.2 // final volume m = 4.0 // mass of the gas in kg deltau = -4.6 ...
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clc //Variable Initialisation V=230//Input Voltage of motor in volts Ra=1.5//Armature resistance in ohm La=1e-3//Armature inductance in ohm Ia=15//Armature Current in Ampere k=0.05//Voltage constant //Solution Eb=0 //when d=0 Ea=Eb+(Ia*Ra) d=Ea/V Eb1=V-(Ia*Ra) //when d1=1 N=Eb1/k printf('\n\n Range of spe...
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* * This file was put into the public domain 2016-11-29 * by John P. Hartmann. You can use it for anything you like, * as long as this notice remains. * * Null problem state program for education *Testcase problem sysclear archmode z loadcore "$(testpath)/problem.core" runtest .1 *Compare r 88.4 *Want 00020000 *Done...
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function logistic_regression(data) //currently designed to classify inputs with only 2 features and 2 classes b0 = 5; t = b0 * rand(100,2); t = [t 0.5+0.5*sign(t(:,2)+t(:,1)-b0)]; b = 1; flip = find(abs(t(:,2)+t(:,1)-b0)<b); t(flip,$)=grand(length(t(flip,$)),1,"uin",0,1); c0 = t(find(t(:,$)==0),:); ...
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clear // // // //Variable declaration N=10000/2*10**2 //number of lines/m m=1 //order lamda1=5890*10**-10 //wavelength(m) lamda2=5896*10**-10 //wavelength(m) //Calculation sintheta1=m*N*lamda1 theta1=asin(sintheta1)*180/%pi //angle(degrees) sintheta2=m*N*lamda2...
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[[m11,m12,m13],[m21,m22,m23],[m31,m32,m33]] * [a^2-b^2,2*a*b,a^2+b^2] = [a^2*m11-b^2*m11+2*a*b*m12+a^2*m13+b^2*m13,a^2*m21-b^2*m21+2*a*b*m22+a^2*m23+b^2*m23,a^2*m31-b^2*m31+2*a*b*m32+a^2*m33+b^2*m33]
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clc //initialisation of variables HfHcl= -22.063 //kcal H298= -17.74 //kcal //CALCULATIONS HfHcl200H2O= HfHcl+H298 //RESULTS printf ('enthalpy of formation= %.2f kcal mole^-1',HfHcl200H2O)
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[[i= partials/header ]] <h1 style="text-align: center">Coming soon!</h1> [[i= partials/footer ]]
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//example 18.1 //design Sarda type fall clc;funcprot(0); //given Q=40; //full supply discharge sl_u=218.3; //supply level at upstream sl_d=216.8; //supply level at downstream D=1.8; //suplly depth L=26; //bed width bl_u=216.5; //bed level ...
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/* Determina os coeficientes do filtro FIR pelo método das janelas. fc = freq. de corte (em Hz)); fs = freq. de amostragem (em Hz)); M = ordem do filtro (inteiro positivo). filtro = 'pb' (passa-baixas) ou 'pa' (passa-altas). */ function [h,w,coef] = firr(fc,fs,M,filtro) omegac = 2*%pi*(fc/fs)...
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//********************************************************************************************************** //*************** Algoritmo para calcular la frecuencia fundamental de una señal de audio ****************** //******************************* Las señales de audio de prueba son: ******************************...
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//Example 4_16 clc(); clear; //To find the wavelength and energy of X-ray beam theta=27.5 //units in degrees n=1 h=1 k=1 l=1 H=6.625*10^-34 //planks constant c=3*10^10 //velocity of light units in meters a=5.63*10^-10 //units in met...
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//Example13.9 // Design a monostable circuit with frequency f = 25 KHz clc; clear; close; f =25*10^3 ; // Hz // The output frequency of monostable multivibrator is defined as // f = 1/(0.69*R*C); C = 0.1*10^-6 ; R = 1/(0.69*f*C); disp('The value of resistance R is = '+string(R)+ ' ohm '); // In the ...
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// Example 16_5 clc;funcprot(0); // Given data T=-20.0+273.15;// K p=0.500;// atm M=0.850;// The Mach number k=1.40;// The specific heat ratio R=286;// J/kg.K g_c=1;// The gravitational constant // Solution V=M*sqrt(k*g_c*R*T);// m/s T_os=T*(1+(((k-1)*M^2)/2));// K T_os=T_os-273.15;// °C p_os=p*(1+(((k-1...
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str = get(get('edit_blockList'),"String"); for i = 1:size(str,2)-1 str(i) = stripblanks(part(str(i),1:$-1)); end str($) = stripblanks(str($)); addblk_msgStr = []; for i = 1:size(str,2) if isempty(str(i)) then continue; end if ~isempty(find(blocks(1)==str(i))) then addblk_msgStr($+1) = st...
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//Example 3.20 : interplaner spacing d_220 clc; clear; close; format('v',6) //given data : a=0.316;// in nm h=2; k=2; l=0; d=a/sqrt(h^2+k^2+l^2); disp(d,"inter planer spacing d_220,d(nm) = ") // answer is wrong in book
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//// //Variable Declaration R = 8.314 //Ideal Gas Constant, J/(mol.K) T = 298 //Temperature of Gas, K M = 0.040 //Molecular wt of Ar, kg/mol P = 101325 //Pressure, N/m2 NA = 6.022e23 //Number of particles per mol V = 1.0 //Volume of Container, L //Calculations Z...
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// exa 9.11 Pg 273 clc;clear;close; // Given Data d=26;// mm L=0.25;//m F=300;// N mu=0.14;// coefficient of thread friction p=5;// mm (for normal series) dc=d-p;// mm dm=d-p/2;// mm l=2*p;// mm alfa=atand(l/%pi/dm);// degree fi=atand(mu);// degree To=F*L;// N.m (Torque applied by the operator) //Tf=W*dm/2*tand(alfa...
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// Exa 6.27 clc; clear; close; // Given data P1 = 2.1;// in MN/m^2 P1= P1*10^3;//in kN/m^2 P2 = 0.7;// in MN/m^2 P2= P2*10^3;//in kN/m^2 V1 = 0.1281;// in m^3 x = 0.9; n = 1.25; h_f1= 920;// in kJ/kg h_fg1= 1878.6;// in kJ/kg h_f2= 697.0;// in kJ/kg h_fg2= 2065.0;// in kJ/kg V_wet1 = x * 0.0949;// in m^...
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load Or8Way.hdl, output-file Or8Way.out, compare-to Or8Way.cmp, output-list ip1%B2.8.2 out%B2.1.2; set ip1 %B00000000, eval, output; set ip1 %B11111111, eval, output; set ip1 %B00010000, eval, output; set ip1 %B00000001, eval, output; set ip1 %B00100110, eval, output;
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disp("Experiment: a card is selected from a deck of 52 cards ") disp("A is the event of the selected card being a spade ") disp("B is the event of the selected card being a face card ") t=52 ; //the total number of cards s=13; //number of spades PA= s/t; disp(PA,'probability of selecting a spade'...
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//Ex:7.17 clc; clear; close; P_21=4/5;// ratio of the input available at port2 P_31=1/5;// ratio of the input available at port3 Lt=-10*log(P_21)/log(10);// throughput loss Lp=-10*log(P_31)/log(10);// tap loss Le=-10*log(P_21+P_31)/log(10);// excess loss printf("The throughput loss =%f dB", Lt); printf("\n ...
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// Example 12.5 mode(0) function residual = residual(x) residual = 0.3*E*(x/a)^2*(1-.00875*(phimax-20))*(1-.000175*a/x)-qcr endfunction a = 31.2;//m E = 19.0E9;//N/m^2, from carpet plots phimax = 22.62; //deg qcr = 11632;//N/m^2 t_guess = 0.1; t = fsolve(t_guess,residual)
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load Eq16.hdl, output-file Eq16.out, compare-to Eq16.cmp, output-list a%B3.16.3 b%B3.16.3 out%B3.1.3; set a %B0000000000000000, set b %B0000000000000000, eval, output; set a %B1000000000000000, set b %B0000000000000001, eval, output; set a %B1111111111111111, set b %B1111111111111111, eval, outpu...
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%XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX %XXXX QPSK Modulation and Demodulation without consideration of noise XXXXX %XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX clc; clear all; close all; data=[0 1 0 1 1 1 0 0 1 1]; % information %Number_of_bit=...
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clc disp("Example 5.37") printf("\n") disp("calculate the phase shift with negative feedback") printf("Given\n") //open loop phase shift Po=15 //open loop gain Av=60000 //closed loop gain Acl=300 //to calculate phase shift with feedback AvB=(Av/Acl)-1 k=((AvB*sin(Po*%pi/180))/(1+(AvB*cos(Po*%pi/180)))) Pc...
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/* Température critique : T_c théorique ~ 2.269 */ N = 100; J = ones(N,N,2); h = zeros(N,N); n = 100; T1 = 2.4; T2 = 2.3; T3 = 2.2; T4 = 2.1; fig_tcg = scf(); clf(fig_tcg); printf("Temps d''exécution :\n"); tic(); X1 = ising_MH(J/T1,h/T1,n*N^2); t1 = toc(); printf("T = "+string(T1)+" : "+string(t1)+"s\n"); subplot(2,...
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function x = criteriolinha(A) // Verifica se o Criterio das Linhas é satisfeito [m,n] = size(A); if m~=n, error('Matriz A deve ser quadrada'); end for k = 1:n alfaK=0; for j = 1:n if j ~= k then alfaK = alfaK + abs(A(k,j)/A(k,k)); end end if alfaK >= 1 th...
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function neighbors = bruteForceNeighbors(points,point,k) wp = size(points,2); for i=1:k lp = size(points,1); minDis = normNoSqrt(point,points(1,:)); neighbors(i,:) = zeros(1,wp); for j =2:lp curDis = normNoSqrt(point,points(j,:)); if (curDis~...
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clc printf("\n") //Given OC=9//inches CP=36//inches XC=12//inches X=40//degrees CM=6.98//from the scaled figure N1=240//rpm N2=240//rpm (instantaneous) with angular aceleration (ao) 100 rad/s^2 ao=100 //rad/s^2 w=(%pi*N1/30) a=w^2*(OC/12) printf("Centripetal acceleration = %.f ft/s^2\n",a) Wr=w*CM/CP//r...
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// 08.08.12 // 08.10.28 function CL=Projpers(varargin) Nargs=length(varargin); Flg=0; if Nargs==1 & Mixtype(varargin(1))==1 Flg=1; end; CL=[]; for N=1:Nargs Crv=varargin(N); if Mixtype(Crv)==1 Tmp=CameraCurve(Crv); CL=Mixadd(CL,Tmp); else if Mixtype(Crv)==3 ObjL=[]...
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//exa 2.15 clc;clear;close; format('v',7); P1=poly(0,'P1'); P2=poly(0,'P2'); P3=poly(0,'P3'); Q1=0.002*P1^2+0.86*P1+20;//tons/hour Q2=0.004*P2^2+1.08*P2+20;//tons/hour Q3=0.0028*P3^2+0.64*P3+36;//tons/hour Pmax=120;//MW Pmin=36;//MW P=200;//MW C=500;//Rs./ton //C1=C*Q1;C2=C*Q2;C3=C*Q3;//Rs./ton dC1bydP1=...
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global A F Tm Fd T k Signal //A = evstr(x_dialog("Введи значение Амплитуды от 1-3",'')); //F = evstr(x_dialog("Введи значение Частота сигнала (Гц)",'')); //Tm = evstr(x_dialog("Введи значение Длительность сигнала (с)",'')); //Fd = evstr(x_dialog("Введи значение Частоты дискретизации(8000) (Г...
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//Chapter 01: The Foundations: Logic and Proofs clc; clear; v1=sqrt(2) v2=(3/2) //let p be the proposition that sqrt(2) > (3/2) if v1 > v2 then //which is false z=v1**2 >v2**2 mprintf("(sqrt(2))^2 > (3/2)^2 %s ", string([%F]))//which is false and as a result will not be printed end //The conclusi...
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clear //given M=8 d=0.2 u=4*%pi*10**-7 //Calculation B=u*2*M/(4*%pi*d**3) Beqa=B/2.0 //Result printf("\n (i) Magnetic induction at axial point %0.3f *10**-4 T", B*10**4) printf("\n (ii) Magnetic induction at equatorial point is %0.3f *10**-4 T",Beqa*10**4)
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clear //Initialisation Vo=15.2 //Output Voltage Rl=50 //Load Resistance //Calculation Po=(Vo**2)/Rl //Output Power //Result printf("\n Output Power, Po = %.1f W",Po)
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8_4.sce
clc //initialisation of variables t1=261 //temp in k t3=310 //temp in k cp=1.005 //kj/kg r=5 g = 9.81; //CALCULATIONS t2=(t1*(r)^((g-1)/g)) t4=(t3/(r)^((g-1)/g)) re=cp*(t1-t4) ma=(3.5164*3600)/re woc=cp*(t2-t1) woe=cp*(t3-t4) nw=woc-woe cop1=re/nw cop2=t1/(t3-t1) reff=cop1/cop2 //RESULTS printf('temparature at states 2...
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clc R=8.314 T=300 a=422.546 b=0.0373 v1=30 v2=5 W=integrate('((R*T)/(v-b))-(a/(v.^2))','v',v1,v2) mprintf("W=%fkJ/kmol\n",W)//ans may vary due to roundoff error deltaU=a*(1/v1-1/v2) mprintf("U2-U1=%fkJ/kmol\n",deltaU)//ans vary due to roundoff error Q=deltaU+W mprintf("Q=%fkJ/kmol\n",Q)//ans vary due to rou...