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// Exa 5.1 clc; clear; close; // Given data T1 = 550;// in degree C T1 = T1 + 273;// in K T2 = 27;// in degree C T2 = T2 + 273;// in K Eta = ((T1-T2)/T1)*100;// in % disp(Eta,"Maximum possible efficiency for staem turbine plant in % is"); T1 = 2500;// in degree C T1 = T1 + 273;// in K T2 = 400;// in degree C T2 = T2 + 273;// in K Eta = ((T1-T2)/T1)*100;// in % disp(Eta,"Maximum possible efficiency for internal combustion engine in % is"); T1 = 450;// in degree C T1 = T1 + 273;// in K T2 = 15;// in degree C T2 = T2 + 273;// in K Eta = ((T1-T2)/T1)*100;// in % disp(Eta,"Maximum possible efficiency for nuclear power plant in % is");
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clear //Given Ig=0.015 //A G=5 I=1 V=15 //Calculation S=(Ig*G)/(I-Ig) R=G*S/(G+S) R1=(V/Ig)-G R2=R1+G //Result printf("\n (i) Resistance of ammeter of range 0-1 A is %0.3f ohm", R) printf("\n (ii) Resistance of ammeter of range 0-15 A is %0.3f ohm", R2)
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//EX3_38 Pg-3.75 clc Vo=5; Il=20e-3; Pz=500e-3; Rl=Vo/Il; Il_min=Il;//minimum load current Il_max=Il;//maximum load current Iz_max=Pz/Vo;//maximum zener current Iz_min=5e-3;//minimum zener current V=12;//input DC voltage Vin_min=12-3;//min input voltage Vin_max=12+3;//max input voltage Rmax=(Vin_min-Vo)/(Il_max+Iz_min); printf("\n maximum resistance required is %.0f ohm \n",Rmax) Rmin=(Vin_max-Vo)/(Il_min+Iz_max); printf("\n minimum resistance required is %.2f ohm \n",Rmin) printf("\n So series resistance must be selected between %.2f ohm to %.0f ohm \n",Rmin,Rmax)
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THE OPTIMIZATION ALGORITHM HAS CHANGED TO THE EM ALGORITHM. ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 0.255103D+00 2 -0.337816D-02 0.193959D-02 3 -0.949508D-01 0.756480D-03 0.401425D+00 4 0.306227D-03 -0.737404D-03 -0.567711D-02 0.333262D-02 5 -0.269897D-03 -0.254751D-04 0.533181D-03 -0.229508D-04 0.260393D-02 6 0.125413D-03 0.222875D-04 0.175098D-03 -0.451798D-04 0.401423D-03 7 0.594951D-03 0.316171D-04 0.717580D-03 -0.134503D-04 0.227583D-03 8 -0.749739D-04 0.386614D-04 -0.224816D-05 0.134488D-03 -0.351759D-04 9 -0.227832D+00 -0.185362D-02 0.182194D+00 0.141412D-01 0.419874D-01 10 -0.216321D+00 -0.627659D-02 0.180554D+00 -0.182361D-02 0.121148D+00 11 -0.123206D+00 0.167338D-02 0.118724D+00 -0.253496D-01 0.130468D-01 12 -0.241799D+00 -0.395052D-02 0.396967D+00 0.160008D-01 0.574940D-01 13 -0.671366D-01 -0.131920D-01 -0.386493D-02 0.673655D-03 0.155726D-01 14 -0.203268D+00 0.840256D-02 0.199820D+00 0.409522D-02 -0.274249D-01 15 -0.151576D+01 0.161531D-01 -0.252775D+00 0.101468D-01 -0.108740D+00 16 -0.114369D-01 -0.748777D-02 0.171779D-01 -0.169348D-03 0.936683D-04 17 -0.447888D-03 -0.335912D-03 -0.703601D-03 0.112303D-03 -0.194062D-03 18 -0.354698D+00 0.289058D-01 -0.121659D+00 -0.253703D-01 -0.201047D-01 19 0.800963D-01 0.770574D-03 -0.815369D-02 -0.286847D-02 -0.111950D-01 20 -0.330008D-01 -0.284628D-01 0.126357D+01 0.180914D-01 0.146902D-01 21 -0.607700D-01 0.183734D-04 -0.913305D-02 0.446219D-02 0.980240D-02 22 0.199861D-02 0.539028D-04 0.187219D-02 0.248481D-03 -0.651481D-04 23 -0.959518D-02 -0.406048D-03 -0.616456D-02 0.105932D-01 0.239883D-03 24 0.243031D-02 0.219733D-03 0.328499D-02 -0.171490D-03 -0.285702D-03 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 0.814234D-03 7 0.626403D-03 0.275239D-02 8 -0.136122D-03 0.292392D-03 0.355454D-02 9 0.169267D-01 0.174281D-01 0.440739D-02 0.353783D+02 10 0.208012D-01 0.140885D-01 0.939491D-02 0.116089D+01 0.137753D+02 11 0.232116D-01 0.370970D-01 0.234293D-01 -0.227005D+01 0.174289D+01 12 -0.273609D-02 0.192828D-01 0.652543D-01 0.109478D+02 0.182363D+01 13 0.502045D-01 0.728658D-01 -0.283572D-01 0.252397D+01 0.661904D+00 14 -0.321152D-01 -0.175908D-01 0.362125D+00 0.364333D+01 0.208577D+01 15 -0.230520D-01 -0.449037D-01 0.352724D-02 -0.294043D+01 -0.864119D+01 16 0.716808D-04 0.224852D-03 -0.859379D-04 0.490762D+00 0.544276D-01 17 -0.556420D-04 -0.179011D-07 0.378379D-05 -0.753283D-01 0.552764D-02 18 -0.425315D-01 -0.609456D-01 -0.849291D-02 -0.199018D+01 -0.393477D+00 19 -0.690273D-02 0.169234D-01 -0.887762D-02 0.548902D+00 -0.112086D+01 20 0.264486D-02 0.145582D-01 -0.287216D+00 0.366485D+00 -0.234649D+01 21 0.622991D-02 -0.160754D-01 0.121171D-01 -0.920415D+00 0.108676D+01 22 -0.780075D-04 -0.184104D-03 0.221997D-03 0.137887D-01 -0.514829D-02 23 -0.455022D-03 -0.466653D-03 -0.841611D-03 0.720203D-01 -0.861615D-01 24 0.289501D-03 0.224301D-03 -0.146581D-03 -0.206561D-01 0.791425D-02 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 0.382626D+02 12 -0.156976D+02 0.105299D+03 13 -0.165284D+01 0.107338D+01 0.105067D+02 14 0.275240D+01 0.523779D+01 -0.614593D+01 0.697341D+02 15 -0.365661D+01 0.364137D+01 -0.781360D+00 -0.153695D+01 0.166372D+03 16 -0.341253D-01 0.377981D+00 0.125527D+00 0.679678D-02 0.310197D+00 17 0.533886D-01 -0.613931D-01 0.477847D-02 0.299471D-01 -0.717441D+00 18 0.241606D+01 0.113892D+02 -0.515046D+01 0.297932D+01 -0.332999D+02 19 -0.120553D+01 -0.421413D+00 -0.168924D+00 -0.127988D+01 0.686154D+00 20 0.325935D+01 -0.182719D+02 0.335832D+01 -0.482688D+02 0.172637D+02 21 0.161517D+01 0.284681D+00 0.619772D-01 0.151342D+01 -0.174253D+01 22 -0.865570D-01 0.208962D-01 -0.927457D-02 0.264129D-01 0.181073D+00 23 -0.545477D-01 0.568237D+00 -0.159631D+00 -0.270479D+00 0.365648D-01 24 -0.185238D-01 -0.613916D-01 0.195884D-01 -0.220867D-01 -0.632271D-01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 0.309859D+00 17 -0.134547D-01 0.933216D-02 18 -0.528418D+00 0.223913D+00 0.155152D+03 19 0.901279D-01 -0.752434D-02 -0.146568D+01 0.403392D+01 20 -0.148688D+00 -0.547403D-01 -0.654058D+02 0.126998D+01 0.430685D+03 21 -0.139266D+00 0.119895D-01 0.336988D+01 -0.376262D+01 -0.257853D+01 22 0.897221D-03 -0.165124D-02 -0.680741D+00 0.111689D-01 0.206346D+00 23 0.523090D-02 0.320413D-02 -0.719749D+00 -0.159618D+00 0.465639D+01 24 0.270771D-02 0.258823D-03 0.237332D+00 0.170111D-01 -0.172424D+01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 0.445880D+01 22 -0.448862D-01 0.733777D-02 23 -0.990813D-01 0.103541D-01 0.715984D+00 24 0.646189D-03 -0.260796D-02 -0.621617D-01 0.176529D-01 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 1.000 2 -0.152 1.000 3 -0.297 0.027 1.000 4 0.011 -0.290 -0.155 1.000 5 -0.010 -0.011 0.016 -0.008 1.000 6 0.009 0.018 0.010 -0.027 0.276 7 0.022 0.014 0.022 -0.004 0.085 8 -0.002 0.015 0.000 0.039 -0.012 9 -0.076 -0.007 0.048 0.041 0.138 10 -0.115 -0.038 0.077 -0.009 0.640 11 -0.039 0.006 0.030 -0.071 0.041 12 -0.047 -0.009 0.061 0.027 0.110 13 -0.041 -0.092 -0.002 0.004 0.094 14 -0.048 0.023 0.038 0.008 -0.064 15 -0.233 0.028 -0.031 0.014 -0.165 16 -0.041 -0.305 0.049 -0.005 0.003 17 -0.009 -0.079 -0.011 0.020 -0.039 18 -0.056 0.053 -0.015 -0.035 -0.032 19 0.079 0.009 -0.006 -0.025 -0.109 20 -0.003 -0.031 0.096 0.015 0.014 21 -0.057 0.000 -0.007 0.037 0.091 22 0.046 0.014 0.034 0.050 -0.015 23 -0.022 -0.011 -0.011 0.217 0.006 24 0.036 0.038 0.039 -0.022 -0.042 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 1.000 7 0.418 1.000 8 -0.080 0.093 1.000 9 0.100 0.056 0.012 1.000 10 0.196 0.072 0.042 0.053 1.000 11 0.132 0.114 0.064 -0.062 0.076 12 -0.009 0.036 0.107 0.179 0.048 13 0.543 0.428 -0.147 0.131 0.055 14 -0.135 -0.040 0.727 0.073 0.067 15 -0.063 -0.066 0.005 -0.038 -0.181 16 0.005 0.008 -0.003 0.148 0.026 17 -0.020 0.000 0.001 -0.131 0.015 18 -0.120 -0.093 -0.011 -0.027 -0.009 19 -0.120 0.161 -0.074 0.046 -0.150 20 0.004 0.013 -0.232 0.003 -0.030 21 0.103 -0.145 0.096 -0.073 0.139 22 -0.032 -0.041 0.043 0.027 -0.016 23 -0.019 -0.011 -0.017 0.014 -0.027 24 0.076 0.032 -0.019 -0.026 0.016 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 1.000 12 -0.247 1.000 13 -0.082 0.032 1.000 14 0.053 0.061 -0.227 1.000 15 -0.046 0.028 -0.019 -0.014 1.000 16 -0.010 0.066 0.070 0.001 0.043 17 0.089 -0.062 0.015 0.037 -0.576 18 0.031 0.089 -0.128 0.029 -0.207 19 -0.097 -0.020 -0.026 -0.076 0.026 20 0.025 -0.086 0.050 -0.279 0.064 21 0.124 0.013 0.009 0.086 -0.064 22 -0.163 0.024 -0.033 0.037 0.164 23 -0.010 0.065 -0.058 -0.038 0.003 24 -0.023 -0.045 0.045 -0.020 -0.037 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 1.000 17 -0.250 1.000 18 -0.076 0.186 1.000 19 0.081 -0.039 -0.059 1.000 20 -0.013 -0.027 -0.253 0.030 1.000 21 -0.118 0.059 0.128 -0.887 -0.059 22 0.019 -0.200 -0.638 0.065 0.116 23 0.011 0.039 -0.068 -0.094 0.265 24 0.037 0.020 0.143 0.064 -0.625 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 1.000 22 -0.248 1.000 23 -0.055 0.143 1.000 24 0.002 -0.229 -0.553 1.000
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clc // // // //Variable declaration lambdaa=5.46*10**-7 //Wavelength t=6.3*10**-6 //thickness //Calculations mu=((6*lambdaa)/t)+1 //Result printf("\n The refractive index is %0.3f ",mu)
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clear; clc; x=input('Enter the sequence'); k=input('Enter the nmber of times the signal repeats'); x1=x; for i=1:k x1=[x1 x]; end x=abs(fft(x)); x1=abs(fft(x1)); subplot(211);plot2d3(x); subplot(212);plot2d3(x1);
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clc; k=1.38*10^-23; //Boltzmann's constant in J/K tk=273+100; //absolute temp (in Kelvin) KE=3/2*(k*tk); //calculating average Kinetic Energy in Joule using kinetic theory of gases disp(KE,"Average Kinetic Energy in Joule = "); //displaying result
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//Introduction to Fiber Optics by A. Ghatak and K. Thyagarajan, Cambridge, New Delhi, 1999 //Example 11.2 //OS=Windows XP sp3 //Scilab version 5.5.2 clc; clear; //given //For Cr+3 ions in ruby N1=1.6e19;//Population density of E1 energy level in cm^(-3) N2=0;//Population density of E2 energy level in cm^(-3) n=1.76;//refractive index of medium Tsp=3e-3;//Spontaneous emission lifetime of atom in sec //Let g(v0) be g g=6.9e-12;//normalized lineshape function in s lambda0=694.3e-7;//wavelength at which absorption takes place in cm c=3e10;//speed of electrons in cm/s v=c/lambda0; //Let Y(v0) be Y Y=((c/n)^2)*g*(N2-N1)/(8*%pi*Tsp*(v^2));//Corresponding gain coefficient of medium mprintf("\n Absorption coefficient = %f",Y);//The answers vary due to round off error //negative sign implies absorption
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//Exa 6.1 clc; clear; close; // given data epsilon_r=2.5; epsilon_o=8.854*10^-12; d=.2*10^-3;// in m A=20*10^-4;// in m^2 omega=2*%pi*10^6;// in radians/s f=10^6; tan_delta=4*10^-4; C=epsilon_o*epsilon_r*A/d;// in F disp("Capicitance is : "+string(C*10^12)+" miu miu F"); // Formula P=V^2/R, so // R=V^2/P and P= V^2*2*%pi* f * C * tan delta, putting the value of P, we get R=1/(2*%pi*f*C*tan_delta);// in ohm disp("The element of parallel R-C circuit is : "+string(R*10^-6)+" M ohm");
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clc clear all; close; x1=[1,3,7,-2,5]; x2=[2,-1,0,3]; y=conv(x1,x2)
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//check o/p when the i/p arg is a uniformly sampled hamming window win=window('hm',1000); en=enbw(win); disp(en); //output // 1.3638074 //
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//Example No. 3_23 //Associative law //Pg No. 58 clear ; close ; clc ; x = 0.400000*10^40 y = 0.500000*10^70 z = 0.300000*10^(-30) disp('In book they have considered the maximum exponent can be only 99, since 110 is greater than 99 the result is erroneous') disp((x*y)*z,'xy_z = ','but in scilab the this value is much larger than 110 so we get a correct result ') disp(x*(y*z),'x_yz = ')
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//Example 5_5 clc(); clear; //To calculate the average frictional force developed m=2000 //units in Kg vf=20 //units in meters/sec d=100 //units in meters f=(0.5*m*vf^2)/d //units in Newtons printf("Average frictional force f=%d N",f)
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clear; clc; // Illustration 1.9 // Page: 27 printf('Illustration 1.9 - Page:27 \n\n'); // Solution //*****Data*****// // A-acetic acid(solute) B-acetone(solvent) T = 313; // [K] // The following data are available (Reid, et al., 1987): // Data for acetic acid T_bA = 390.4; // [K] T_cA = 594.8; // [K] P_cA = 57.9; // [bar] V_cA = 171; // [cubic cm/mole] M_A = 60; // [gram/mole] // Data for acetone T_bB = 329.2; // [K] T_cB = 508; // [K] P_cB = 47; // [bar] V_cB = 209; // [cubic cm/mole] u_bB = 0.264; // [cP] M_B = 58; // [gram/mole] phi = 1; printf('Illustration 1.9 (a) - Page:27 \n\n'); // Solution (a) // Using equation 1.48 V_bA = 0.285*(V_cA)^1.048; // [cubic cm/mole] // Using the Wilke-Chang correlation , equation 1.52 D_abo1 = (7.4*10^-8)*(sqrt(phi*M_B))*T/(u_bB*(V_bA)^.6); printf("Diffusivity of acetic acid in a dilute solution in acetone at 313 K using the Wilke-Chang correlation is %e square cm/s\n",D_abo1); // From Appendix A, the experimental value is 4.04*10^-5 square cm/s D_aboexp = 4.04*10^-5; // [square cm/s] percent_error1 = ((D_abo1-D_aboexp)/D_aboexp)*100; // [%] printf("The percent error of the estimate, compared to the experimental value is %f\n\n ",percent_error1); printf('Illustration 1.9 (b) - Page:28 \n\n'); // Solution (b) // Using the Hayduk and Minhas correlation for nonaqueous solutions V_bA = V_bA*2; // [cubic cm/mole] V_bB = 0.285*(V_cB)^1.048; // [cubic cm/mole] // For acetic acid (A) T_brA = T_bA/T_cA; // [K] // Using equation 1.55 alpha_cA = 0.9076*(1+((T_brA)*log(P_cA/1.013))/(1-T_brA)); sigma_cA = (P_cA^(2/3))*(T_cA^(1/3))*(0.132*alpha_cA-0.278)*(1-T_brA)^(11/9); // [dyn/cm] // For acetone (B) T_brB = T_bB/T_cB; // [K] // Using equation 1.55 alpha_cB = 0.9076*(1+((T_brB*log(P_cB/1.013))/(1-T_brB))); sigma_cB = (P_cB^(2/3))*(T_cB^(1/3))*(0.132*alpha_cB-0.278)*(1-T_brB)^(11/9); // [dyn/cm] // Substituting in equation 1.54 D_abo2 = (1.55*10^-8)*(V_bB^0.27)*(T^1.29)*(sigma_cB^0.125)/((V_bA^0.42)*(u_bB^0.92)*(sigma_cA^0.105)); printf("Diffusivity of acetic acid in a dilute solution in acetone at 313 K using the Hayduk and Minhas correlation is %e square cm/s\n",D_abo2); percent_error2 = ((D_abo2-D_aboexp)/D_aboexp)*100; // [%] printf("The percent error of the estimate, compared to the experimental value is %f\n\n ",percent_error2);
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//Ex 2.11.4 clc;clear;close; format('v',8); //Given : T=300;//K m_Si=1.5;//for Si m_Ge=1.5;//for Ge EGO_Si=1.21;//volt EGO_Ge=0.785;//volt Eta_Si=2; Eta_Ge=1; VT=26/1000;//V disp("Part(i) : "); d_logIoBYdt_Ge=m_Ge/T+EGO_Ge/(Eta_Ge*T*VT);//per degree C disp(d_logIoBYdt_Ge,"d(log(Io))/dt for Ge (per degree C) : "); d_logIoBYdt_Si=m_Si/T+EGO_Si/(Eta_Si*T*VT);//per degree C disp(d_logIoBYdt_Si,"d(log(Io))/dt for Si (per degree C) : "); disp("Part(ii) : "); V=0.2;//volt dVBYdt_Ge=V/T-Eta_Ge*VT*d_logIoBYdt_Ge disp(dVBYdt_Ge*1000,"dV/dt for Si (mV per degree C) : "); V=0.6;//volt dVBYdt_Si=V/T-Eta_Si*VT*d_logIoBYdt_Si disp(dVBYdt_Si*1000,"dV/dt for Si (mV per degree C) : ");
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clc // initialization of variables clear SF=1.75 p1=300 //MPa S_rr=-SF*p1 S_th=SF*325 Y=1/sqrt(2)*sqrt((S_th-S_rr)^2+S_rr^2+S_th^2) printf(' Y = %.1f MPa',Y)
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results = read(get_absolute_file_path("LoadData.sce") + "..\Data\log" + currIdx + ".txt", -1, 2); angle = results(:, 1)*%pi/180; time = results(:, 2)/1000; plot2d(time, angle, 2);
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clc clear x=8 z=21 pop=10 iter=10000 a=zeros(pop,z) rag=zeros(iter) rag=zeros(iter) bd=10000000*ones(1,pop) gppd=zeros(pop,x) cap=[288 95 115 133 107 22 34 28 186 190 33 56 100 90 82 143 68 166 44 73 72 60 68 8 20 ] tim=[0 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1 0 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1 0 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1 0 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1 0 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1 0 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1 0 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1 0 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1 0 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1 0 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 0 1 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 0 1 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 0 1 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 0 1 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 0 1 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 0 1 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 0 1 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 0 1 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 0 1 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 0 1; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 0; ] dib=[1 1 1 1 1 1 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1 1 1 1 1 1 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1 1 1 1 1 1 1 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1000 1000 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1000 1000 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1000 1000 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1000 1000; 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1; ] cad=rand(pop,z) for i=1:pop for j=1:z for k=1:z if cad(i,k)==min(cad(i,:)) a(i,j)=k cad(i,k)=10000 break end end end end disp (a) best=a function [bkop,grp,sis,kip] =finddis(a,bkom) temp=zeros(pop,x) tempo=zeros(pop,x) dis=zeros(pop,x) dis1=zeros(pop,x) rep=zeros(pop,z) rep1=zeros(pop,z) for ka=1:pop po=zeros(x,z) po1=zeros(x,z) prob=rand(1,1) kop=1+round((z-1)*rand(1,(x-1))) if prob<0.4 kom=kop end if ka==1 kom=kop end i=1 for k=1:x r=1 while i<min(kom) & i<=z po(k,r)=a(ka,i) r=r+1 if i<=z i=i+1 end end for g=1:x-1 if kom(g)==min(kom) kom(g)=100 break end end end hkom=bkom(ka,:) i=1 for k=1:x r=1 while i<min(hkom) & i<=z po1(k,r)=a(ka,i) r=r+1 if i<=z i=i+1 end end for g=1:x-1 if hkom(g)==min(hkom) hkom(g)=100 break end end end kom=kop mik=1 for k=1:x for i=1:z if po(k,i+1)~=0 dis(ka,k)=dis(ka,k)+tim(po(k,i),po(k,i+1)) if tim(po(k,i),po(k,i+1))>10 rep(ka,mik)=po(k,i) end mik=mik+1 else break end end end mik=1 for k=1:x for i=1:z if po1(k,i+1)~=0 dis1(ka,k)=dis1(ka,k)+tim(po1(k,i),po1(k,i+1)) if tim(po1(k,i),po1(k,i+1))>10 rep1(ka,mik)=po1(k,i) end mik=mik+1 else break end end end for k=1:x for i=1:z if(po(k,i)~=0) temp(ka,k)=temp(ka,k)+1 end end if temp(ka,k)>0 dis(ka,k)=dis(ka,k)+dib(k,po(k,1))+dib(k,po(k,temp(ka,k))) end end for k=1:x for i=1:z if(po1(k,i)~=0) tempo(ka,k)=tempo(ka,k)+1 end end if tempo(ka,k)>0 dis1(ka,k)=dis1(ka,k)+dib(k,po1(k,1))+dib(k,po1(k,tempo(ka,k))) end end if sum(dis1(ka,:))>sum(dis(ka,:)) dis1(ka,:)=dis(ka,:) bkom(ka,:)=kom tempo(ka,:)=temp(ka,:) rep1(ka,:)=rep(ka,:) end end bkop=bkom grp=tempo sis=dis1 kip=rep1 endfunction function [n] = mutate(b,grd,kip) gnd=ones(pop,x) for k=1:pop for i=1:x if i==1 gnd(k,i)=0 else gnd(k,i)=gnd(k,i-1)+grd(k,i-1) end end oll=rand(1,1) if oll<(y/iter) then xio=rand(1,1) if xio<0.5 r=1 for i=1:x poll=rand(1,1) if i==1 mut1=1+round((grd(k,i)-1)*rand(1,1)) mut2=1+round((grd(k,i)-1)*rand(1,1)) if poll<0.2 & mut1>0 & mut2>0 & mut1~=mut2 temper=b(k,mut2) b(k,mut2)=b(k,mut1) b(k,mut1)=temper r=r+1 end else mut1=gnd(k,i)+round((grd(k,i)-1)*rand(1,1)) mut2=gnd(k,i)+round((grd(k,i)-1)*rand(1,1)) if poll<0.2 & mut1>0 & mut2>0 & mut1~=mut2 temper=b(k,mut2) b(k,mut2)=b(k,mut1) b(k,mut1)=temper r=r+1 end end if r~=1 break; end end else for mi=1:x if kip(k,mi)~=0 mut1=1+round((z-1)*rand(1,1)) temper=b(k,mut1) b(k,mut1)=b(k,mi) b(k,mi)=temper end end end else mut1=1+round((z-1)*rand(1,1)) mut2=1+round((z-1)*rand(1,1)) temper=b(k,mut2) b(k,mut2)=b(k,mut1) b(k,mut1)=temper end end n=b endfunction function [rkom,glg,gog,kip] =roulewheel(dis,a,gpd,bkom,kip) tdis=zeros(1,pop) pdis=zeros(1,pop) cdis=zeros(1,pop) calm=a calp=gpd gkom=bkom jip=kip for i=1:pop tdis(i)=sum(dis(i,:)) tdis(i)=1/(1+tdis(i)) end for i=1:pop pdis(i)=tdis(i)/(sum(tdis)) end for i=1:pop if i==1 cdis(i)=cdis(i)+pdis(i) else cdis(i)=cdis(i-1)+pdis(i) end end for i=1:pop jin=rand(1,1) for j=1:pop if j==1 if jin<cdis(j) calm(i,:)=a(j,:) calp(i,:)=gpd(j,:) gkom(i,:)=bkom(j,:) jip(i,:)=kip(j,:) break end else if cdis(j-1)<jin & jin<=cdis(j) calm(i,:)=a(j,:) calp(i,:)=gpd(j,:) gkom(i,:)=bkom(j,:) jip(i,:)=kip(j,:) break end end end end rkom=gkom gog=calm glg=calp kip=jip endfunction toper=1000000 ygh=1 kip=zeros(pop,z) esup=zeros(1,z) egd=zeros(1,x) bkom=1+round((z-1)*rand(pop,(x-1))) bbom=zeros(pop,(x-1)) for y=1:iter [bkom,gp,dis,kip]=finddis(a,bkom) mog=zeros(1,pop) for i=1:pop mog(i)=sum(dis(i,:)) if mog(i)<bd(i) then best(i,:)=a(i,:) bd(i)=mog(i) gppd(i,:)=gp(i,:) bbom(i,:)=bkom(i,:) end if toper>bd(i) then toper=bd(i) esup=best(i,:) egd=gppd(i,:) end if round(sum(bd)/pop)-round(min(bd))<15 & y>1000 if mog(i)<1.5*bd(i) then best(i,:)=a(i,:) bd(i)=mog(i) gppd(i,:)=gp(i,:) bbom(i,:)=bkom(i,:) end end end rag(y)=min(bd) ryg(y)=y a=best gpd=gppd [bkom,gpd,a,kip]=roulewheel(dis,a,gpd,bkom,kip) [a]=mutate(a,gpd,kip) end plot(ryg,rag) tomp=zeros(x) sup=esup gd=egd for j=1:x if(j==1) tomp(j)=gd(j) else tomp(j)=tomp(j-1)+gd(j) end end fine=zeros(x,z) for j=1:x if j==1 then for k=1:tomp(j) fine(j,k)=sup(k) end else i=1 for k=tomp(j-1)+1:tomp(j) fine(j,i)=sup(k) i=i+1 end end end namer=['CHEDIKULAM' 'URUPUMKUTTY' 'EDAPUZHA' 'EDOOR' 'KOLAYAD' 'VELLARVALLY' 'ARYAPARAMBA' 'PERUVA' 'KAPPAD' 'ATTENCHERY' 'PERAVOOR' 'MALOOR' 'THRIKADARIPOIL' 'THODEEKKALAM' 'EDATHOTTY' 'PALAPPUZHA' 'THALIPPOYIL' 'VATTIARA' 'PERUMPARAMBU' 'PADIKACHAL' 'PUNNAD' 'KODOLIPRAM' 'MARUTHAYI' 'VELLIYAMPARAMBA KSS LTD' 'KANHILERI' ] bmc=["KEEZHPALLY" "KEEZHPALLY" "ODEMTHODE" "ODEMTHODE" "THOLUMBRA" "THILLANKERI" "PAZHASSI RAJA NAGAR" "PAZHASSI RAJA NAGAR" ] yum=0 for i=1:1:x for j=1:z if fine(i,j)>0 if j==1 disp(bmc(i)) disp(dib(i,fine(i,j))) yum=yum+dib(i,fine(i,j)) end disp(namer(fine(i,j))) if j~=gd(i) then disp(tim(fine(i,j),fine(i,j+1))) yum=yum+tim(fine(i,j),fine(i,j+1)) end if j==gd(i) disp(dib(i,fine(i,j))) yum=yum+dib(i,fine(i,j)) disp(bmc(i)) end end end end disp(yum)
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clc clear printf("Example 7.5 | Page number 192 \n\n"); //Part(a) Find the COP of reversible heat engine as refrigerator. //Part(b) Find the COP of reversible heat engine as heat pump. //Given Data T1 = 273 + 37 //K T2 = 273 - 13 //K //Part(a) printf("Part (a)\n"); COP_ref = T2/(T1-T2) //COP of reversible heat engine as refrigerator. printf("COP of reversible heat engine as refrigerator = %.1f\n\n",COP_ref) //Part(b) printf("Part (b)\n"); COP_hp = T1/(T1-T2) //COP of reversible heat engine as heat pump. printf("COP of reversible heat engine as heat pump = %.1f",COP_hp)
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clc //Initialization of variables G=100 //lb/s g=32.2 V2=300 //fps V1=250 //fps //calculations Qh= (V2^2 -V1^2)/(2*g) Q=Qh*G //results printf("Thermal energy added = %d ft lb/s",Q)
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//Fluid System By Shiv Kumar //Chapter 6 - Kaplan and Propeller Turbines //Example 6.7 //To Find (a)Diameter of Runner (b)Speed (c)Specific Speed clc clear //Given: H=32; //Head, m P=16000; //Shaft Power, KW D_per=190; //Percentage by which Diameter of Runner(D)is Larger than diameter of Boss(d) eta_0=91/100; //Overall Efficiency Ku=2; //Speed Ratio Kf=0.64; //Flow Ratio //Data Required: rho=1000; //Density of Water, Kg/m^3 g=9.81; //Acceleration due to gravity, m/s^2 //Computations Vfi=Kf*sqrt(2*g*H); //Velocity of Flow at Inlet, m/s ui= Ku*sqrt(2*g*H); //Velocity of Runner at Inlet, m/s Q=P*10^3/(rho*g*H*eta_0); //Discharge, m^3/s d=sqrt(Q/((%pi/4)*Kf*sqrt(2*g*H)*((D_per/100+1)^2-1))); // Diameter of Hub ,m //(a) Diameter of Runner ,D D=d+(D_per/100)*d; //m //(b) Speed,N N=ui*60/(%pi*D); // rpm //(iii) Specific Speed of Turbine, Ns Ns=N*P^(1/2)/(H^(5/4)); // SI Units //Results printf("(a)Diameter of Runner , D=%.3f m\n",D) printf(" (b)Speed, N =%.2f rpm\n",N) //The answer vary due to round off error printf(" (c)Specific Speed, Ns =%.2f (SI Units)\n",Ns) //The answer provided in the textbook is wrong.
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clear; clc; //Example13.10[Heat Transfer through a Tubular Solar Collector] k=0.02588;//[W/m.degree Celcius] Pr1=0.7282,Pr2=0.7255;//Prandtl no nu1=1.608*(10^(-5)),nu2=1.702*10^(-5);//[m^2/s] T1=20,T2=40;//[degree Celcius] Tavg=((T1+T2)/2)+273;//[K] Do=0.1,L=1;//Dimensions of glass tube[m] Di=0.05;//Inner diameter of tube[m] Q_glass=30;//Rate of heat transfer from the outer surface of the glass cover[W] g=9.81;//[m^2/s] eo=0.9,ei=0.95;//Emissivity //Solution:- Ao=%pi*Do*L;//Heat transfer surface area of the glass cover[m^2] disp(Ao,Tavg) Ra_Do=g*Tavg*(T2-T1)*(Do^3)*Pr1/(nu1); disp(Ra_Do,"The Rayleigh number is") Nu=((0.6+((0.387*(Ra_Do^(1/6)))/((1+((0.559/Pr1)^(9/16)))^(8/27))))^2); disp(Nu,"The nusselt number is") ho=k*Nu/Do;//[W/m^2.degree Celcius] Qo_conv=ho*Ao*(T2-T1);//[W] Qo_rad=eo*5.67*10^(-8)*Ao*(((T2+273)^4)-((T1+273)^4));//[W] Qo_total=Qo_conv+Qo_rad;//[W] disp("W",Qo_total,"The total rate of heat loss from the glass cover Lc=(Do-Di)/2;//The characteristic length Ai=%pi*Di*L;//[m^2] //Assuming T_tube=54,T_cover=26;//Temperature of tube and glass cover[degree Celcius] T_avg=((T_tube+T_cover)/2)+273;//[K] Ra_L=g*T_avg*(T_tube-T_cover)*(Lc^3)*Pr2/(nu2); disp(Ra_L,"The Rayleigh number in this case is") F_cyl=((log(Do/Di))^4)/((Lc^3)*(((Di^(-3/5))+(Do^(-3/5)))^5)); k_eff=0.386*k*((Pr2/(0.861+Pr2))^(1/4))*((F_cyl*Ra_L)^(1/4)); disp("W/m.degree Celcius",k_eff,"The effective thermal conductivity is") QL_conv=2*%pi*k_eff*(T_tube-T_cover)/(log(Do/Di)); disp("W",QL_conv,"The rate of heat transfer between the cylinders by convection is") QL_rad=((5.67*10^(-8))*Ai*(((T_tube+273)^4)-((T_cover+273)^4)))/((1/ei)+(((1-eo)/eo)*(Di/Do))); disp("W",QL_rad,"The radiation rate of heat transfer is") QL_total=QL_conv+QL_rad;//[W] disp("W",QL_total,"The total rate of heat loss from the glass cover is")
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//Caption:Calculate (i)-repeller voltage V_r ,(ii)-beam current necessary to give gap voltage of 200V //Exa:8.16 clc; clear; close; e_m_ratio=1.759*10^11;//(e/m) V_o=300;//in volts R_sh=20*10^3;//in ohms f=8*10^9;//inHz w=2*%pi*f; n=2;//mode L=0.001;//spacing between repeller & cavity (in m) x=(e_m_ratio)*(2*%pi*n-%pi/2)^2/(8*w^2*L^2); volt_diff=sqrt(V_o/(x)); V_r=(volt_diff)+V_o;//repeller volatge J=0.582; V_1=200;//given (in volts) I_o=V_1/(R_sh*2*J); disp(V_r,'Repeller voltage(in volts) ='); disp(I_o*10^3,'Necessary beam current (in milliAmp.s) =');
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int f(float x, int y), g(int z) /* ошибка */ { }
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\begin{document} The second part of the semester project focuses on gaining experience with lex (flex) and yacc (bison), coupled with continued design (and redesign) skills for context free grammars (CFGs). The second part of the project is divided into two major tasks, and is worth a total of 50 points: \begin{enumerate} \item Revising the common grammar to remove the shift or reduce and rule-not-reduced conflicts. These corrections are made to the yacc or bison file, and the result must be tested extensively for different input latex documents. This task is worth 35 points. \item Using your result grammar of the first task, redesign the common grammar to support the recognition of verbatim blocks and also nested blocks. That is, recall that the different blocks (itemize, enumerate, single, etc.) can also be nested to yield different combinations of formatted text. Again, test your result using yacc.bison to demonstrate that the grammar revisions for these two changes work correctly. This task is worth 15 points. \end{enumerate} To serve as a common base for the project, there is a directory that contains a number of important files: \begin{verbatim} latex.in : A sample input file. latex.l : A sample latex lex file. latexp2.y : A yacc specification with fprintf's for debugging. latexp2clean.y: Equivalent specification without the fprintfs. y2conflicts : The shift.reduce conflicts generated by YACC. b2conflicts : The shift.reduce conflicts generated by BISON. \end{verbatim} Note that intentional errors have been placed in the files. Note also that the grammar files contain numerous shift reduce errors and other problems, as will be discussed in class. The second part of the project is due on April 6, 1994. Please hand in the following: \begin{enumerate} \item The revised yacc.bison specification for task 1. DO NOT HAND IN C files! \item A log file that documents the changes made to the grammar to eliminate the shift.reduce errors and other problems for task 1. Make sure that you provide both the original and revised grammar segments for each change that you make! Also include any remaining shift.reduce or reduce.reduce errors, but NOT the entire output file. \item The revised yacc.bison specification for task 2. DO NOT HAND IN C files! \item A log file that documents the changes made to your grammar of task 1 in support of verbatim and nested blocks. These changes may occur in both the lex (flex) and yacc (bison) files! Again, provide original.revised segments for each change, and remaining S.R or R.R errors. \item Test cases and test results for both tasks, clearly marked and organized. \item The directory location for your files and detailed compilation instructions. \end{enumerate} {\it IMPORTANT:} Do not open your directory for access by the world until 7:00 p.m. on April 6. \end{document}
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//i/p args are x and p clc; clear; exec('/home/debdeep/Desktop/TEST NOW!!/rooteig/rooteig.sci'); x=[1 2 3 4 5 67 8 9]; p=2; y=rooteig(x,p); disp(y); //output // 0. // 2.1797329
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//Example No.2.8. // Page No.62. clc;clear; Op = 5*10^(-3);//Output power -[W]. I = 10*10^(-3);//Current -[A]. V = 3*10^(3);//Voltage -[V]. Ip = (10*10^(-3)*3*10^(3));//Input power. Eff = (((Op)/(Ip))*(100));//Efficiency of the laser. printf("\nThe efficiency of the laser is %.6f percent",Eff);
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// Tester name: Usama A.F. Khalil // usamafoad@gmail.com // Before run this testes do not forget to execute ammod function in Scilab, ex. // exec('<Drive Letter><Path to ammod.sci file>\ammod.sci',-1) // then you can run code for each block by selecting it then Ctrl + E // // The aim of current test(s) is to check the three status for ammod function, i.e., // I) Y = ammod(x, Fc, Fs) // II) Y = ammod(x, Fc, Fs, INI_PHASE) // III) Y = ammod(x, Fc, Fs, INI_PHASE, CARRAMP) //==========(Case I)============================================================= // I) Y = ammod(x, Fc, Fs) // Test the amplitude modulation where the modulated signal has zero initial // phase and zero carrier amplitude (suppressed-carrier modulation). I'll use // the function to calculate the modulation and compare the results to manual // calculation to the same data. The expected result is equality between the // results from this two different method - if the function work correctly. // The data will satisfy that Fs > 2*(Fc+BW) // // results: Passed // fc = 10; fs = 100; t = [0:1/fs:fs/10]; x = sin(2*%pi*t); y_fun = ammod(x,fc,fs); x=x(:); t = (0:1/fs:(size(x,1)-1)/fs)'; t = t(:,ones(1,size(x,2))); y_man = (0 + x).*cos(2*%pi*fc*t+ 0); y_man = y_man'; figure subplot(3,1,1); plot(y_fun); title('Modulated signal calculated by function'); subplot(3,1,2); plot(y_man); title('Modulated signal calculated manualy'); subplot(3,1,3); plot(y_fun, 'g'), plot(y_man, 'r') ; title('Both (only one line will show if they identical)'); if y_fun == y_man then disp("Passed"); else disp("Failed"); end //==========(Case II)============================================================ // II) Y = ammod(x, Fc, Fs, INI_PHASE) // Test the amplitude modulation with initial phase in radians. // I'll use the function to calculate the modulation and compare the results to: //(A) manual calculation to the same data with initial phase //(B) manual calculation to the same data WITHOUT specifying initial phase // The expected results/output is equality between the results from the function // and the first case (A) and difference between the function results and the // second case (B). i.e., we need to know if adding the ini_phase to the function // make any difference or not. // The data will satisfy that Fs > 2*(Fc+BW) // // results: Passed // fc = 10; fs = 100; t = [0:1/fs:fs/10]; x = sin(2*%pi*t); ini_phase = 5; y_fun = ammod(x,fc,fs,ini_phase); x=x(:); t = (0:1/fs:(size(x,1)-1)/fs)'; t = t(:,ones(1,size(x,2))); //(A) y with ini_phase y_man_A = (0 + x).*cos(2*%pi*fc*t+ ini_phase); y_man_A = y_man_A'; //(B) y without ini_phase y_man_B = (0 + x).*cos(2*%pi*fc*t+ 0); y_manB = y_man_B'; title_A=["Both Modulated signal y_fun & y_man_A";"(only one line will show if they identical)"]; title_B=["Both Modulated signal y_fun & y_man_B";"(only one line will show if they identical)"]; figure subplot(3,1,1); plot(y_fun); title('Modulated signal calculated by function'); subplot(3,1,2); plot(y_fun, 'r'), plot(y_man_A, 'g'); title(title_A); subplot(3,1,3); plot(y_fun, 'r'), plot(y_man_B, 'g') ; title(title_B); if y_fun == y_man_A then disp("Passed"); else disp("Failed"); end if y_fun == y_man_B then disp("Failed"); else disp("Passed"); end //==========(Case III)=========================================================== // III) Y = ammod(x, Fc, Fs, INI_PHASE, CARRAMP) // Test the amplitude modulation with initial phase carrier amplitude // (transmitted-carrier modulation) I'll use the function to calculate the // modulation and compare the results to: //(A) manual calculation to the same data WITH carrier amplitude //(B) manual calculation to the same data WITHOUT specifying carrier amplitude // The expected results/output is equality between the results from the function // and the first case (A) and difference between the function results and the // second case (B). i.e., we need to know if adding the CARRAMP to the function // make any difference or not. // The data will satisfy that Fs > 2*(Fc+BW) // // results: FAILED // fc = 10; fs = 100; t = [0:1/fs:fs/10]; x = sin(2*%pi*t); ini_phase = 5; carr_amp = 9; y_fun = ammod(x,fc,fs,ini_phase, carr_amp); x=x(:); t = (0:1/fs:(size(x,1)-1)/fs)'; t = t(:,ones(1,size(x,2))); //(A) y with carr_amp y_man_A = (carr_amp + x).*cos(2*%pi*fc*t+ ini_phase); y_man_A = y_man_A'; //(B) y without carr_amp y_man_B = (0 + x).*cos(2*%pi*fc*t+ ini_phase); y_manB = y_man_B'; title_A=["Both Modulated signal y_fun & y_man_A";"(only one line will show if they identical)"]; title_B=["Both Modulated signal y_fun & y_man_B";"(only one line will show if they identical)"]; figure subplot(3,1,1); plot(y_fun); title('Modulated signal calculated by function'); subplot(3,1,2); plot(y_fun, 'g'), plot(y_man_A, 'r'); title(title_A); subplot(3,1,3); plot(y_fun, 'r'), plot(y_man_B, 'g') ; title(title_B); if y_fun == y_man_A then disp("Passed"); else disp("Failed"); end if y_fun == y_man_B then disp("Failed"); else disp("Passed"); end // The results show that the function always ignore carrier amplitude value and // consider it as zero. The reason for this is the 'else' statemnt in lines 47-48 // this allows evaluate to true and set carr_amp=0!
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// Macro for single perceptron -- Scilab // Function to train a single perceptron model with given // rate and iterations to converge function theta = perceptronTrain(x, y, rate, iter) // Making an initial guess theta = (1.0/length(x(1, :)))*ones(1, length(x(1, :))); // Training for i = 1:iter sigm = x*theta'; theta = theta + rate.*((y - sigm)'*x); error = 0.5*norm(y - x*theta'); end endfunction // Function to predict the target given features and model parameter theta function ypred = perceptronPredict(x, theta) ypred = 1.0./(1.0 + exp(-x*theta')); endfunction
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//Variable declaration k=0.0002 //device parameter Vt=4 //thevinin voltage(V) Vdd=24 //drain voltage(V) Id0=3 //drain current(mA) //Calculations Vgs=(sqrt(Id0/k))+4 //as Id=k(Vgs-Vt)^2 Rd=-(Vgs-Vdd)/Id0 //as Vds=Vdd-IdRd and Vgs=Vds=7.87 k=0.0003 //device parameter syms Id expr = solve([Id**2-7.5*Id+13.7],[Id]) printf ("equation has 2 solutions") disp(expr) // putting value of k=0.0003 in eq of Id, Id1=3.15 // we get Vgs=Vds=24-5.4Id and putting Vgs again in Id we get, // Id^2-7.5Id+13.7=0 Idchange=((Id1-Id0)/Id0)*100 //changed Id(mA) //Result printf ("change in Id is %.1f %% increase",Idchange)
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Datatype tag for DS1 is: "Character array" DS1[0]: OPAQUE0 DS1[1]: OPAQUE1 DS1[2]: OPAQUE2 DS1[3]: OPAQUE3
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clc;clear;close; A=[] disp("Rayleigh Power Method") printf("\nEnter a 3x3 matrix:\n\n") for i=1:3 for j=1:3 printf("Enter element A(%d,%d):",i,j) A(i,j)=input("") end end disp(A,"The matrix is:") x0=[] disp("*****************************************************************") printf("\nEnter the initial Eigen Vector:\n") for i=1:3 x0(i,1)=input("") end disp(x0,"Initial Eigen Vector") a=max(x0) disp(a,"Initial Largest Eigen Value") disp("****************************************************************") v=A*x0 i=1 while abs(max(v)-a)>0.002 then printf("Iteration Number = %d\n",i) i=i+1 a=max(v) disp(a,"Current Eigen Value:") x1=v/a disp(x1,"Current Eigen Vector:") v=A*x1 disp("************************************************************") end format('v',5) disp("*************************************************************") printf("Iteration Number = %d \n",i) printf("(Equal Eigen Vectors in iteration number %d and %d)\n",i-1,i) disp("The largest Eigen Value:") disp(max(v)) disp("The largest Eigen Vector:") disp(v/a)
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// Chapter6 // Page.No-238 // Example_6_16_a // Design of differentiator // Given clear;clc; C1=0.1*10^-6; // Assume fa=1*10^3; // Freq at which gain is 0 dB Rf=1/(2*%pi*fa*C1); // Using fa=1/(2*%pi*Rf*C1) printf("\n Feedback resistance is = %.1f ohm \n",Rf) Rf=1.5*10^3; // Approximation fb=20*10^3; // Gain limiting freq R1=1/(2*%pi*fb*C1); printf("\n Resistance,R1 is = %.1f ohm \n",R1) R1=82; // Approximation Cf=R1*C1/Rf; printf("\n Capacitance,Cf is = %.10f farad \n",Cf) Cf=0.005*10^-6; // Approximation
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clc //initialisation of variables a=1.025//in^2 h=18//in h1=24//in x=8.2//percent v=15//in v2=6.9//ft^3 p=0.74//lb/in^2 p1=50//lb/in^2 p2=83//lb/in^2 P3=48.0//lb/in P1=29.8//lb/in^2 P2=14.6//lb/in^2 h2=29.8//in D=(%pi/4)*(3/2)^2*2//ft^3 v1=23400//ft.lb W=a*v1//ft.lb V=0.082*D//ft^3 Q=1.530//ft^3 //CALCULATIONS I=V+Q//ft^3 P=P3+P2//lb/in^2 V1=p*v2//ft^3 W1=I/V1//lb S=p2+P2///l/in^2 H=659.06//C.H.U/lb T=W/(H*W1*1400)*100//percent //RESULTS printf('The thermal efficiency of the engine=% f percent',T)
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clc clear //Input data Pt=100;//Pressure of air when the bulb is placed in hot water in cm of Hg P100=109.3;//Pressure of air in a constant volume thermometer at 100 degree centigrade in cm of Hg P0=80;//Pressure of air in a constant volume thermometer at 0 degree centigrade in cm of Hg //Calculations t=((Pt-P0)/(P100-P0))*100;//The temperature of the hot water in degree centigrade //Output data printf('The temperature of the hot water is %3.2f degree centigrade',t)
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f0=1E6 Cmax=500E-12 C=450E-12 w0=2*%pi*f L=1/(w0*w0*Cmax) w=1/sqrt(L*C) f=w/(2*%pi) wb=2*2*%pi*(f-f0) r=wb*L Q0=2*%pi*f*L/r disp(Q0,L,r) ////////calculation mistakes in book
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clc; funcprot(0); //Example 16.3 Horsepower required at sea level // Initialisation of variables W = 11200; S = 365; rho = 0.002378; // Calculations Cl = poly(0,'Cl'); Cd = 0.023 + 0.0445*Cl^2; Coeff_Velo = 19.77*sqrt(W/S); //Using equtaion 8.5.2 Coeff_HP = W^1.5/(550*sqrt(rho*S/2)); //Using equation 8.16.1 Cl = [0.2 0.3 0.4 0.6 0.8 1.0 1.2 1.4]'; Result = zeros(8,8); Result(:,1) = Cl; Result(:,2) = Cl^2; Result(:,3) = sqrt(Cl); Result(:,4) = Cl^1.5; Result(:,5) = horner(Cd,Cl); Result(:,6) = diag(diag(Result(:,5))*inv(diag(Result(:,4)))); Result(:,7) = Coeff_HP*Result(:,6); Result(:,8) = Coeff_Velo*diag(inv(diag(Result(:,3)))); disp(Result,"!!Cl Cl^2 Cl^0.5 Cl^1.5 Cd Cd/Cl^1.5 Hp req V(mph) !!") ; clf(); plot2d(Result(:,8),Result(:,7)); xlabel("Miles Per Hour"); ylabel("HorsePower"); title("Power Curves "); set(gca(),"grid",[1 1])
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// Scilab code Ex7.5: Pg:288 (2008) clc;clear; m = 3.34e-027; // Mass of deutron, gm q = 1.6e-019; // Charge, coulomb r = 0.2; // Radius of the path of deutron, meter B = 1.5; // Magnetic field, weber per meter square v = q*B*r/m; // velocity of the deutron, m/s E = 1/2*m*v^2/1.6e-013; // Energy of the deutron, MeV printf("\nThe velocity of deutron = %5.3e m/s ", v); printf("\nThe energy of deutron = %5.3f MeV ", E); // Result // The velocity of deutron = 1.437e+007 m/s // The energy of deutron = 2.156 MeV
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clear all; b = chdir('C:\Users\work\OneDrive\Documents\SciLab\lab_v5') exec('DFT.sci') exec('FFT.sci') figure(0) f = 5; Fs = 64; dt=1/Fs; T = 4; t =0:dt:T-dt; signal = cos(2*%pi*f*t); plot(signal); xlabel("Time") ylabel("Amplitude") figure(1) subplot(2,1,1) a = DFT(signal) frequinces = (0:length(signal)-1)/length(signal)*Fs; plot(frequinces, abs(a)) xlabel("Frequency, Hz", 'fontsize', 2) ylabel("Freq amplitude", 'fontsize', 2) title("Frequency response of final signal (my dft)", 'fontsize', 3) subplot(2,1,2) plot(frequinces, abs(fft(signal))) xlabel("Frequency, Hz", 'fontsize', 2) ylabel("Freq amplitude", 'fontsize', 2) title("Frequency response of final signal (build-in function)", 'fontsize', 3) figure(2) subplot(2,1,1) a = DFT(signal) a = fftshift(a) frequinces = linspace(-Fs/2, Fs/2, length(a)); plot(frequinces, abs(a)) xlabel("Frequency, Hz", 'fontsize', 2) ylabel("Freq amplitude", 'fontsize', 2) title("Frequency response of final signal (my dft)", 'fontsize', 3) subplot(2,1,2) a = fft(signal) a = fftshift(a) frequinces = linspace(-Fs/2, Fs/2, length(a)); plot(frequinces, abs(a)) xlabel("Frequency, Hz", 'fontsize', 2) ylabel("Freq amplitude", 'fontsize', 2) title("Frequency response of final signal (build-in function)", 'fontsize', 3) figure(3) b = FFT(signal) b = fftshift(b) plot(frequinces, abs(b)) xlabel("Frequency, Hz", 'fontsize', 2) ylabel("Freq amplitude", 'fontsize', 2) title("Frequency response of final signal (my fft)", 'fontsize', 3)
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clc pathname=get_absolute_file_path('6_5_2.sce') filename=pathname+filesep()+'652.sci' exec(filename) printf(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook") outputx=S/(S+100) printf("x=%f Kg KNO3/Kg",outputx) disp("Water balance") m1=basis*(1-inputx)/(1-outputx) printf(" \n m1=%f Kg",m1) disp("Mass balance") m2=basis-m1 printf(" \n m2=%f kg",m2) percentage=m2*100/(basis*inputx) printf(" \n Percentage of KNO3 in the feed that crystallizes is %f",percentage)
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function [image] = cv_rgb2gray_path(path) pyImport rgb2gray_file image=rgb2gray_file.rgb2gray(path) endfunction
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// Example 2.14.b:Maximum Core Readius clc; clear; close; n1=1.48;//Waveguide Refractive Index d= 0.01;// Cange in core-cladding refractive index a=2;// parabolic refractive index h=1.3;//wavelngth in micro meters v= 2.4*sqrt(1+(2/a));//maximum value of normalised frequence a= (v*h)/(2*%pi*n1*sqrt(2*d));//Core Radius disp(a,"maximum core radius in micro meters")
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//Example 1.14<c> // Find wheather the following signal is periodic or not x3(n)=e^(i*7*pi*n) clc; n=-21:21; x=exp(%i *7* %pi *n); f=(7*%pi)/(2*%pi); N=1/f; disp(N,'the given signal is periodic');
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//Solutions to Problems In applied mechanics //A N Gobby clear all; clc //initialisation of variables d=60//rev/min s=5//in v=5//in/s a=25.2//in/s x=2.23//in b=4.59//in z=20.0//in //CALCULATIONS U=x*v//in/s V=b*v//in/s B=V/z//rad/s //RESULTS printf('the angular velocity=% f rad/s',B)
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T0 = 300; T = 1500; Q = -8.5; W = 8.5; // Case (a) I = Q*(1-T0/T) + W; R = Q*(1-T0/T); disp("kW",I,"and",R,"Rate of availability transfer with heat and the irreversibility rate are") // Case (b) T1 = 500; Ib = - Q*(1-T0/T) + Q*(1-T0/T1); disp("kW",Ib,"Rate of availability in case b is")
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// Example 18_9 clc;funcprot(0); // Given data T=20.0+273.15;// K // Calculation theta_v=2740;// K c_vbyR=(5/2)+((((theta_v/T)^2)*exp((theta_v/T)))/(exp(theta_v/T)-1)^2); Y=8.3143;// kJ/kg.K M_NO=30.01;// The molecular mass of nitrous oxide R_NO=Y/M_NO;// kJ/kg.K c_v_NO=R_NO*c_vbyR;// kJ/kg.K printf("\nThe value of c_v/R for nitrous oxide is %1.2f.",c_vbyR);
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<?xml version="1.0" encoding="utf-8" ?> <test> <description>3D channel flow, Pyramidic elements, using SVV</description> <executable>IncNavierStokesSolver</executable> <parameters>Pyr_channel_SVV.xml</parameters> <files> <file description="Session File">Pyr_channel_SVV.xml</file> </files> <metrics> <metric type="L2" id="1"> <value variable="u" tolerance="1e-10">2.6682e-10</value> <value variable="v" tolerance="1e-10">2.67015e-10</value> <value variable="w" tolerance="1e-9">1.15296e-09</value> <value variable="p" tolerance="1e-8">2.8591e-08</value> </metric> <metric type="Linf" id="2"> <value variable="u" tolerance="1e-9">1.838e-09</value> <value variable="v" tolerance="1e-9">1.90318e-09</value> <value variable="w" tolerance="1e-8">1.42728e-08</value> <value variable="p" tolerance="1e-6">1.0353e-06</value> </metric> </metrics> </test>
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clear ; clc; //Page No.367 // Example 12.5 printf('Example 12.5\n\n'); // Solution fig.E12.5 // Given F = 100 ;// Overall feed/basis - [kg] F_n_C5H12 = 0.80 ;// Fraction of n_C5H12 in overall feed F_i_C5H12 = 0.20 ;// Fraction of i_C5H12in overall feed S_i_C5H12 = 1 ;// Fraction of i_C5H12 in isopentane stream P_n_C5H12 = .90 ;// Fraction of n_C5H12 in overall product P_i_C5H12 = .10 ;// Fraction of i_C5H12 in overall product // Overall Balances P = (F*F_n_C5H12)/P_n_C5H12 ;//Product Material Balance of n_C5H12 -[kg] S = F - P ;// Isopentane stream (S) from overall material balance - [kg] // Balance around isopentane tower // Let x be kg of butane free gas going to isopentane tower , y be the n-C5H12 stream leaving the isopentane tower // Solve following Equations by Matrix method // x = S + y - By Total materal balance // x*F_n_C5H12 = y a = [1 -1;F_n_C5H12 -1] ;// Matrix of coefficients of unknown b = [S;0] ;// Matrix of constants x = a\b ;// Matrix of solutions, x(1) = x , x(2) = y xf = x(1)/F ;// Fraction of butane-free gas going to isopentane tower printf('Fraction of butane-free gas going to isopentane tower is %.3f .\n',xf);
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//example 1.7 clc; funcprot(0); //parta e1=0.92; e2=0.86; Hc=2.8; s2=212;//sigma2dash s1=140;//sigma1dash Cc=(e1-e2)/log10(s2/s1); Sc=Cc*Hc/(1+e1)*log10(s2/s1); disp(Sc*1000,"consolidated depth in mm"); //part b Sct=40; T50=0.197; t=4.5; Cr=T50*12.7^2/t; U=Sct/Sc*100/1000; H=Hc/2; Tv=%pi/4*U^2/100^2; t=Tv*H^2/Cr*1000^2/60/24; disp(t,"time required in days");
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///Chapter No 7 Fluid Mechanics ///Example 7.1 Page No:113 ////Find mass density of liquid //Input data clc; clear; V=5; //volume of the liquid in m**3 W=45*10^3; //weight of the liquid in KN g=9.81; //acceleration due to gravity in m/s**2 rhow=1000; //constant value ////Calculation m=W/g; //mass in Kg rho=m/V; //Mass density in kg/m**3 w=W/V; //Weight Density in N/m**3 v=V/m; //Specific volume in m**3/kg S=rho/rhow; //Specific gravity //Output printf('mass=%f kg \n',m); printf('Mass density= %f kg/m^3 \n ',rho); printf('Weight Density= %f N/m^3\n ',w); printf('Specific volume=%f m^3/kg \n',v); printf('Specific gravity= %f \n',S);
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//ramanujan's method //example 2.23 //page 47 clc;clear;close; deff('y=f(x)','1-2*((3/2)*x+(1/4)*x^2-(1/48)*x^4+x^6/1440-x^8/80640)'); a1=3/2,a2=1/4,a3=0,a4=1/48,a5=0,a6=1/1440,a7=0,a8=-1/80640; b1=1; b2=a1; b3=a1*b2+a2*b1; b4=a1*b3+a2*b2+a3*b1; b5=a1*b4+a2*b3+a3*b2; b6=a1*b5+a2*b4+a3*b3; b7=a1*b6+a2*b5+a3*b4; b8=a1*b7+a2*b6+a3*b5; b9=a1*b8+a2*b7+a3*b6; printf('\n%f',b1/b2); printf('\n%f',b2/b3); printf('\n%f',b3/b4); printf('\n%f',b4/b5); printf('\n%f',b5/b6); printf('\n%f',b6/b7); printf('\n%f',b7/b8); printf('\n it appears as if the roots are converging at around %f',b7/b8)
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clc; //e.g 27.12 AV=1000; f1=50; f2=200*10**3; D=0.05; beta=0.01; AV1=AV/(1+beta*AV); disp(AV1); fl1=f1/(1+beta*AV); disp('HZ',fl1,"fl1="); fu2=(1+beta*AV)*f2; disp('MHZ',fu2*10**-6,"fu2="); D1=D/(1+beta*AV); disp('%',D1*100,"D1=");
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-- VectorCAST 6.4c (02/03/16) -- Test Case Script -- -- Environment : LIBC -- Unit(s) Under Test: abort1 abs atof atoi atol bLib memchr memcmp memcpy memmove memset ns16550 qsort rand random random_r strcat strchr strcmp strcpy strlcat strlcpy strlen strncat strncmp strncpy strpbrk strspn strtod strtok strtok_r strtol strtoul -- -- Script Features TEST.SCRIPT_FEATURE:C_DIRECT_ARRAY_INDEXING TEST.SCRIPT_FEATURE:CPP_CLASS_OBJECT_REVISION TEST.SCRIPT_FEATURE:MULTIPLE_UUT_SUPPORT TEST.SCRIPT_FEATURE:MIXED_CASE_NAMES TEST.SCRIPT_FEATURE:STATIC_HEADER_FUNCS_IN_UUTS -- -- Unit: strtok -- Subprogram: strtok -- Test Case: strtok TEST.UNIT:strtok TEST.SUBPROGRAM:strtok TEST.NEW TEST.NAME:strtok TEST.BASIS_PATH:1 of 1 TEST.NOTES: No branches in subprogram TEST.END_NOTES: TEST.VALUE:strtok.strtok.separator:<<malloc 1>> TEST.VALUE:strtok.strtok.return:<<malloc 3>> TEST.VALUE_USER_CODE:strtok.strtok.string <<strtok.strtok.string>> = ( "a,b,c" ); TEST.END_VALUE_USER_CODE: TEST.VALUE_USER_CODE:strtok.strtok.separator.separator.separator[0] <<strtok.strtok.separator>>[0] = ( "," ); TEST.END_VALUE_USER_CODE: TEST.END -- Test Case: strtok.001 TEST.UNIT:strtok TEST.SUBPROGRAM:strtok TEST.NEW TEST.NAME:strtok.001 TEST.VALUE:strtok.strtok.string:<<malloc 9>> TEST.VALUE:strtok.strtok.string:<<null>> TEST.VALUE:strtok.strtok.separator:<<malloc 2>> TEST.VALUE:strtok.strtok.separator:"1" TEST.EXPECTED:strtok.strtok.return:<<null>> TEST.END -- Test Case: strtok1 TEST.UNIT:strtok TEST.SUBPROGRAM:strtok TEST.NEW TEST.NAME:strtok1 TEST.NOTES: No branches in subprogram TEST.END_NOTES: TEST.VALUE:strtok.strtok.string:<<malloc 6>> TEST.VALUE:strtok.strtok.string:"a b c" TEST.VALUE:strtok.strtok.separator:<<malloc 1>> TEST.VALUE:strtok.strtok.return:<<malloc 3>> TEST.VALUE_USER_CODE:strtok.strtok.separator.separator.separator[0] <<strtok.strtok.separator>>[0] = ( " " ); TEST.END_VALUE_USER_CODE: TEST.END
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//ナイキスト線図 s=%s; H=1; G=(s+2)^2/((s+1)*(s^2-2*s+4)); GH=G*H; sys=syslin("c",GH); clf(); nyquist(sys) CharEq=denom(GH)+numer(GH) roots(CharEq)
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//chapter 12 //example 12.1 // page 413 n=8;// number of bits Vofs=2.55;//in volts R=2^n;//resolution disp(R) Resolution=Vofs/(2^8-1); disp(Resolution)// an input change of 1LSB cause the output to change by 10mV
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clc // Given that p1 = 0.56 // Ambient pressure in bar t1 = 260 // Ambient temperature in K r_p = 6 // Pressure ratio of compressor n_c = 0.85 // Efficiency of compressor v = 360 // Speed of aircraft in km/h d = 3 // Propeller diameter in m n_p = 0.8 // Propeller efficiency n_g = 0.95 // Gear reduction efficiency r_e = 5 // Expansion ratio n_t = 0.88 // Turbine efficiency t3 = 1100 // Temperature at the entrance of turbine in K n_n = 0.9 // Nozzle efficiency cv = 40 // Calorific value in MJ/kg printf("\n Example 21.6\n") gama = 1.4 // Heat capacities ratio for air Vo = v*(5/18) p2 = p1*r_p t2_s = t1*((r_p)^(0.286)) t2 = t1+((t2_s-t1)/n_c) Cp = 1.005 // The value of heat capacity of air as given in the book in kJ/kgK Wc = Cp*(t2-t1) m_f = (t3-t2)/((cv*1000/Cp)-t3) m_a = 1/m_f p3=p2 p4 = p3/r_e t4_s = t3/((r_e)^(0.286)) t4 = t3-((t3-t4_s)*n_t) Wt = (1+m_f)*(t3-t4)*Cp Pp = Wt-Wc p5 = p1 t5_s = t4/((p4/p5)^((gama-1)/gama)) Vj = sqrt(2*Cp*1000*(t4-t5_s)*n_n) Ft = (1+m_f)*Vj-1*Vo V = Vo/n_p V4 = 2*V-Vo Q = (%pi/4)*(d^2)*V Pt = (1/2)*(p1*(10^5)/(287*t1))*Q*((V4^2)-(Vo^2))/1000 PT = Pt/n_g ma_c = PT/Pp Fp = Pt*n_p/V printf("\n Air-fuel ratio = %f,\n Thrust power of the propeller = %f kJ/s ,\n Thrust by the propeller = %f kN,\n Mass flow rate of air flowing through the compressor = %f kg/s,",m_a,Pt,Fp,ma_c) // The answers are given in the book contain calculation error.
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// SAMPLE PROBLEM 5/1 clc;clear;funcprot(0); // Given data n_1=1800;// rev/min t_0=0;// s // alpha=4t; n_2=900;// rev/min // Calculation // (a) omega_1=(-2*%pi*n_1)/60;// rad/s // omega=-(60*%pi)+2t^2 omega_2=(-2*%pi*n_2)/60;// rad/s t=sqrt((omega_2-omega_1)/2);// s // (b) // The flywheel changes direction when its angular velocity is momentarily zero. Thus, t_b=sqrt((0-omega_1)/2);// s // (c) t_0=0;// s t_1=t_b;// s theta_1=integrate('omega_1+(2*t^2)','t',t_0,t_1);// rad N_1=abs(-theta_1/(2*%pi));// rev(clockwise) t_1=t_b;// s t_2=14;// s theta_2=integrate('omega_1+(2*t^2)','t',t_1,t_2);// rad N_2=theta_2/(2*%pi);// rev N=N_1+N_2;// rev printf("\n(a)The time required for the flywheel to reduce its clockwise angular speed,t=%1.2f s \n(b)The time required for the flywheel to reverse its direction of rotation,t=%1.2f s \n(c)The total number of revolutions,N=%3.0f rev",t,t_b,N);
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//chapter 4 //example 4.6 //page 169 printf("\n") printf("given") Vbe=.7;Vce=-6; Ib=20*10^-6 Ic=2.5*10^-3//from output characteristics Bdc=Ic/Ib
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//variable initialization x=0:0.1:9999; x0=0; x1=9999; //calculation I=integrate('x^2*exp(-x)','x',x0,x1); A=sqrt(1/(I*(%pi/2))); r=(1/4)*integrate('x^3*exp(-x)','x',x0,x1); printf("\n A = %f*a0^-1.5\n r = %.1f*a0",A,r);
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// clc; clear ; getd('../lib'); function exemplo_Gauss(label, A, b, prec) if ~exists("prec", "local") then prec = 3; end tamanho = prec + 5; pausa = %F; mprintf("\n###################################################################################\n"); mprintf("Exemplo %s de solução de sistema linear por método de Gauss\n", label) mprintf("-----------------------------------------------------------\n"); try [x, r, Det] = gauss_dp(A, b, %F, tamanho, prec, pausa) catch mprintf("!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!\n"); mprintf('%s\n',lasterror()); mprintf("!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!\n"); end mprintf("\n===================================================================================\n"); mprintf("Exemplo %s de solução de sistemas lineares por método de Gauss com pivotação parcial\n", label); mprintf("------------------------------------------------------------------------------------\n"); try [x, r, Det] = gauss_dp(A, b, %T, tamanho, prec, pausa) catch mprintf("!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!\n"); mprintf('%s\n',lasterror()); mprintf("!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!\n"); end endfunction //------------------------------------------ // exemplo 1 de sistema linerar (pág. 51) // Matriz de coeficientes A = [ 1 -3 2; -2 8 -1; 4 -6 5]; // Vetor de termos_independentes b = [ 11; -15; 29]; exemplo_Gauss('1', A, b) //------------------------------------------ // exemplo 2 de sistema linerar (pág 53) // Matriz de coeficientes A = [1, 6, 2, 4; 3, 19, 4, 15; 1, 4, 8, -12; 5, 33, 9, 3]; // Vetor de termos_independentes b = [ 8; 25; 18; 72 ]; exemplo_Gauss('2', A, b)
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// calculo de erro de interpolaçao // x | x1 | x2 // f(x) | f(x1) | f(x2) // n = 2 // Estimativa do erro // Er(x) <= |(x-x1)(x-x2)| / 2! * max λ∈[x1, x2] * |f^(n)(λ)| // Sendo f(x) = 1/x^2 determine o pol. interpolador para x1 = 3 e x2 = 4 usando o método de Lagrange // f(x1)*(x-x2) / (x1-x2) + f(x2)*(x-x1)/(x2-x1) function y=f(x) y = x.^(-2) endfunction // derivada de segunda ordem function y=df(x) y = 6.*x.^(-4) endfunction // maximo da derivada t = 3:0.05:4 w = abs(df(t)) M = max(w) R = abs((3.5-3).*(3.5-4))./factorial(2) erro = M.*R printf("Estimativa de erro:") disp(erro)
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errcatch(-1,"stop");mode(2);//Chapter 12, Problem 6 ; hFE=125; //common-emitter current gain Ic=50*10^-3; //collector current Ib=Ic/hFE; //calculating base current printf("Base current Ib = %d microampere",Ib*10^6); exit();
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THE OPTIMIZATION ALGORITHM HAS CHANGED TO THE EM ALGORITHM. ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 0.354061D+00 2 -0.283621D-02 0.303013D-02 3 -0.105206D+00 0.117462D-02 0.441608D+00 4 0.778745D-03 -0.757411D-03 -0.589612D-02 0.379278D-02 5 -0.363013D-04 0.171110D-03 -0.508299D-03 0.145094D-03 0.273233D-02 6 0.564229D-04 0.941603D-04 0.236403D-03 -0.109579D-03 -0.239512D-03 7 0.146188D-02 0.441788D-04 -0.154084D-02 -0.260273D-04 -0.668685D-03 8 -0.392328D-03 0.142947D-04 -0.738252D-03 0.469629D-04 -0.205284D-03 9 -0.940586D-01 0.195763D-01 -0.137999D+00 0.302407D-01 0.110247D+00 10 -0.114125D+00 0.179401D-01 0.101666D+00 0.134533D-01 0.191820D+00 11 -0.931157D-01 -0.106934D-02 -0.164560D+00 0.243534D-01 -0.783471D-01 12 -0.618056D+00 0.158550D-01 0.696342D+00 0.222496D-01 0.374117D-01 13 0.156295D+00 0.199751D-01 -0.622483D-02 -0.968875D-02 -0.152333D-01 14 -0.242231D+00 0.160532D-01 0.736288D-01 0.230437D-02 0.779403D-02 15 -0.834537D+00 0.812334D-01 -0.472794D+00 -0.116037D-01 -0.129071D+00 16 -0.350750D-01 -0.127444D-01 0.270714D-01 0.823272D-03 0.252565D-03 17 -0.126833D-01 -0.206129D-02 0.402240D-02 0.652342D-03 -0.692002D-03 18 -0.495044D+00 -0.545076D-02 -0.540285D+00 0.117071D-01 0.704507D-01 19 0.272218D-01 -0.675518D-03 0.145964D+00 -0.141234D-01 -0.549101D-02 20 -0.922104D+00 -0.137262D-02 0.174376D+01 0.633145D-02 -0.499680D-01 21 0.266031D-01 0.260297D-02 -0.179563D+00 0.788993D-02 0.205309D-02 22 0.311035D-02 0.594525D-03 -0.119643D-02 -0.567149D-03 0.875085D-04 23 -0.368003D-01 -0.268935D-02 -0.238845D-01 0.124167D-01 0.128197D-02 24 0.419237D-02 -0.278515D-03 0.329033D-02 -0.390520D-03 0.111698D-03 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 0.828535D-03 7 0.795832D-03 0.343981D-02 8 0.313900D-04 0.363908D-03 0.310993D-02 9 -0.181057D-01 -0.211841D-01 -0.104948D-01 0.812900D+02 10 -0.263181D-01 -0.812779D-01 -0.412930D-01 0.503932D+01 0.320531D+02 11 0.180674D-01 0.366517D-01 0.570959D-01 -0.106130D+02 -0.566648D+01 12 -0.262854D-01 -0.350387D-01 0.800634D-01 -0.233021D+00 0.760773D+01 13 0.648534D-01 0.152357D+00 -0.246992D-01 -0.145660D+01 -0.366492D+01 14 -0.343439D-02 0.138294D-01 0.393745D+00 0.356026D+01 0.278037D+01 15 0.442200D-01 0.582067D-01 0.312386D-01 -0.116845D+02 -0.201198D+02 16 -0.793585D-03 -0.251092D-02 0.111863D-02 0.626389D+00 -0.896896D-01 17 -0.260151D-03 -0.537728D-04 0.412996D-03 -0.209944D+00 -0.307175D-01 18 -0.802998D-01 -0.135815D+00 -0.479765D-01 0.305742D+01 0.128541D+02 19 -0.971543D-02 0.144059D-01 0.116341D-01 0.266935D+01 -0.610279D+00 20 0.285646D-01 -0.580981D-02 -0.341772D+00 -0.732122D+01 0.715340D+01 21 0.112303D-01 -0.125699D-01 -0.135075D-01 -0.285447D+01 0.465021D+00 22 0.117727D-03 -0.125846D-03 0.441237D-03 0.295902D-01 -0.209120D-01 23 -0.310052D-03 -0.183396D-02 -0.591895D-03 0.280668D-01 0.149171D+00 24 -0.450988D-04 0.134408D-03 0.132296D-03 0.311927D-01 -0.399108D-01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 0.519913D+02 12 -0.273098D+01 0.168260D+03 13 -0.376398D+01 -0.133348D+01 0.202633D+02 14 0.708103D+01 0.960353D+01 -0.807806D+01 0.113560D+03 15 0.132585D+02 0.205456D+02 0.435211D+01 -0.225660D+01 0.372947D+03 16 0.147055D+00 0.566289D-01 -0.989638D-01 0.124822D+00 0.120520D+01 17 -0.333022D-01 -0.471390D-01 -0.319094D-01 0.121993D+00 -0.181911D+01 18 -0.637874D+01 0.886851D+01 -0.860200D+01 -0.502968D+00 -0.912629D+02 19 -0.990208D-01 0.369122D+01 0.119143D+00 0.746158D+00 -0.875628D+00 20 -0.122653D+02 -0.233716D+02 0.666595D+01 -0.718835D+02 0.476039D+02 21 0.387260D+00 -0.353138D+01 -0.127515D+00 -0.999247D+00 0.980909D-01 22 -0.541493D-01 0.397359D-01 0.419108D-02 0.651386D-01 0.432112D+00 23 -0.174144D-01 0.123770D+01 -0.845989D-01 -0.293903D+00 0.545363D+00 24 0.114729D-02 -0.163904D+00 0.210750D-02 -0.294356D-01 -0.192580D+00 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 0.527007D+00 17 -0.208536D-01 0.213707D-01 18 -0.374648D+00 0.425702D+00 0.280504D+03 19 -0.521081D-01 0.130941D-01 0.426083D+00 0.733248D+01 20 0.154713D+00 -0.209751D+00 -0.110175D+03 0.992944D+00 0.552885D+03 21 -0.165879D+00 -0.119592D-01 0.434259D+01 -0.683332D+01 -0.243753D+01 22 0.105287D-02 -0.455768D-02 -0.127840D+01 -0.164361D-01 0.396117D+00 23 0.467948D-01 0.625417D-03 -0.532301D+00 -0.154363D+00 0.556389D+01 24 0.140843D-02 0.112897D-02 0.443171D+00 -0.177560D-01 -0.242480D+01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 0.801262D+01 22 -0.432466D-01 0.131895D-01 23 -0.128543D+00 0.605328D-02 0.873066D+00 24 0.358859D-01 -0.383943D-02 -0.795550D-01 0.257971D-01 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 1.000 2 -0.087 1.000 3 -0.266 0.032 1.000 4 0.021 -0.223 -0.144 1.000 5 -0.001 0.059 -0.015 0.045 1.000 6 0.003 0.059 0.012 -0.062 -0.159 7 0.042 0.014 -0.040 -0.007 -0.218 8 -0.012 0.005 -0.020 0.014 -0.070 9 -0.018 0.039 -0.023 0.054 0.234 10 -0.034 0.058 0.027 0.039 0.648 11 -0.022 -0.003 -0.034 0.055 -0.208 12 -0.080 0.022 0.081 0.028 0.055 13 0.058 0.081 -0.002 -0.035 -0.065 14 -0.038 0.027 0.010 0.004 0.014 15 -0.073 0.076 -0.037 -0.010 -0.128 16 -0.081 -0.319 0.056 0.018 0.007 17 -0.146 -0.256 0.041 0.072 -0.091 18 -0.050 -0.006 -0.049 0.011 0.080 19 0.017 -0.005 0.081 -0.085 -0.039 20 -0.066 -0.001 0.112 0.004 -0.041 21 0.016 0.017 -0.095 0.045 0.014 22 0.046 0.094 -0.016 -0.080 0.015 23 -0.066 -0.052 -0.038 0.216 0.026 24 0.044 -0.032 0.031 -0.039 0.013 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 1.000 7 0.471 1.000 8 0.020 0.111 1.000 9 -0.070 -0.040 -0.021 1.000 10 -0.161 -0.245 -0.131 0.099 1.000 11 0.087 0.087 0.142 -0.163 -0.139 12 -0.070 -0.046 0.111 -0.002 0.104 13 0.501 0.577 -0.098 -0.036 -0.144 14 -0.011 0.022 0.663 0.037 0.046 15 0.080 0.051 0.029 -0.067 -0.184 16 -0.038 -0.059 0.028 0.096 -0.022 17 -0.062 -0.006 0.051 -0.159 -0.037 18 -0.167 -0.138 -0.051 0.020 0.136 19 -0.125 0.091 0.077 0.109 -0.040 20 0.042 -0.004 -0.261 -0.035 0.054 21 0.138 -0.076 -0.086 -0.112 0.029 22 0.036 -0.019 0.069 0.029 -0.032 23 -0.012 -0.033 -0.011 0.003 0.028 24 -0.010 0.014 0.015 0.022 -0.044 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 1.000 12 -0.029 1.000 13 -0.116 -0.023 1.000 14 0.092 0.069 -0.168 1.000 15 0.095 0.082 0.050 -0.011 1.000 16 0.028 0.006 -0.030 0.016 0.086 17 -0.032 -0.025 -0.048 0.078 -0.644 18 -0.053 0.041 -0.114 -0.003 -0.282 19 -0.005 0.105 0.010 0.026 -0.017 20 -0.072 -0.077 0.063 -0.287 0.105 21 0.019 -0.096 -0.010 -0.033 0.002 22 -0.065 0.027 0.008 0.053 0.195 23 -0.003 0.102 -0.020 -0.030 0.030 24 0.001 -0.079 0.003 -0.017 -0.062 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 1.000 17 -0.197 1.000 18 -0.031 0.174 1.000 19 -0.027 0.033 0.009 1.000 20 0.009 -0.061 -0.280 0.016 1.000 21 -0.081 -0.029 0.092 -0.891 -0.037 22 0.013 -0.271 -0.665 -0.053 0.147 23 0.069 0.005 -0.034 -0.061 0.253 24 0.012 0.048 0.165 -0.041 -0.642 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 1.000 22 -0.133 1.000 23 -0.049 0.056 1.000 24 0.079 -0.208 -0.530 1.000
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//Example 10.5 //Newton Raphshon Method //Page no. 320 clc;clear;close; deff('y=f1(x,y)','y=x^3-3*x*y^2+1') deff('y=f2(x,y)','y=3*x^2*y-y^3') deff('y=f11(x,y)','y=3*x^2-6*y^2') deff('y=f12(x,y)','y=-6*x*y') deff('y=f21(x,y)','y=6*x*y') deff('y=f22(x,y)','y=3*x^2-3*y^2') x=[0;1]; printf('\nx(0) = %g\ny(0) = %g\n',x(1),x(2)) for i=1:3 fx=[f1(x(1),x(2));f2(x(1),x(2))] printf('\n fx(%i) = \n',i) disp(fx) J=[f11(x(1),x(2)),f12(x(1),x(2));f21(x(1),x(2)),f22(x(1),x(2)),] disp(J,'J = ') d=det(J); if d==0 then dx1=0;dx2=0; else dx1=(fx(1)*J(2,2)-fx(2)*J(1,2))/d; dx2=(fx(2)*J(1,1)-fx(1)*J(2,1))/d; end x(1)=x(1)+dx1; x(2)=x(2)+dx2; printf('\nx(%i) = %g\ny(%i) = %g\n',i,x(1),i,x(2)) end
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// Heat generation in resistance spot welding clc I = 5500 // current in ampere R = 250 // resistance in micro ohm T = 0.15 // time in sec d = 6 // diameter in mm t = 3 // thickness in mm rho = 7850 // density in kg/m^3 E = 1400 // energy required per gram mass printf("\n Example 12.5") Heat = I^2*R*1e-6*T V = %pi/4*d^2*t m = V*rho*1e-6 E_tot = m*E H_r = Heat - E_tot H_per = H_r/Heat*100 printf("\n Amount of heat generated is %d J.", Heat) printf("\n Amount of heat in weld zone is %d J or %d%%.", H_r, H_per) // Answer in book is 196 J
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function melt() //Unpivot a DataFrame from wide to long format, optionally leaving identifiers set. // Syntax //pd.melt(df, id_vars=['A'], value_vars=['B']) // // Parameters //id_vars: tuple, list, or ndarray, optional //value_vars: tuple, list, or ndarray, optional //var_name: scalar //value_name: scalar, default ‘value’ // // For additional information on parameters, See https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.melt.html // Returns : DataFrame // // Examples // If we want to join using the key columns, we need to set key to be the index in both df and other. The joined DataFrame will have key as its index. // df.join(other, lsuffix='_caller', rsuffix='_other') // Authors // Aditya Dhinavahi // Sundeep Akella endfunction
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clear; clc; disp('Example 11.9'); // aim : To determine // (a) the actual and hypothetical mean effective pressures referred to the low-pressure cylinder // (b) the overall diagram factor // (c) the pecentage of the total indicated power developed in each cylinder // given values P = 1400;// steam supply pressure, [kN/m^2] Pb = 20;// back pressure, [kN/m^2] Chp = .6;// cut-off in HP cylinder, [stroke] dh = 300*10^-3;// HP diameter, [m] di = 500*10^-3;// IP diameter, [m] dl = 900*10^-3;// LP diameter, [m] Pm1 = 590;// actual pressure of HP cylinder, [kN/m^2] Pm2 = 214;// actual pressure of IP cylinder, [kN/m^2] Pm3 = 88;// actual pressure of LP cylinder, [kN/m^2] // solution // (a) // for HP cylinder PmH = Pm1*(dh/dl)^2;// PmH referred to LP cylinder, [kN/m^2] // for IP cylinder PmI = Pm2*(di/dl)^2;// PmI referred to LP cylinder, [kN/m^2] PmA = PmH+PmI+Pm3;// actual mean effective pressure referred to LP cylinder, [kN/m^2] R = dl^2/(dh^2*Chp);// expansion ratio Pm = P/R*(1+log(R))-Pb;// hypothetical mean effective pressure referred to LP cylinder, [kN/m^2] mprintf('\n (a) The actual mean effective pressure referred to LP cylinder is = %f kN/m^2\n',PmA); mprintf('\n The hypothetical mean effective pressure referred to LP cylinder is = %f kN/m^2\n',Pm); // (b) ko = PmA/Pm;// overall diagram factor mprintf('\n (b) The overall diagram factor is = %f\n',ko); // (c) HP = PmH/PmA*100;// %age of indicated power developed in HP IP = PmI/PmA*100; // %age of indicated power developed in IP LP = Pm3/PmA*100; // %age of indicated power developed in LP mprintf('\n (c) The pecentage of the total indicated power developed in HP cylinder is = %f percent\n',HP); mprintf('\n The pecentage of the total indicated power developed in IP cylinder is = %f percent\n',IP); mprintf('\n The pecentage of the total indicated power developed in LP cylinder is = %f percent\n',LP); // End
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A = [3 -5 7 1 -2 4 -8 1 -7] disp(norm(A,1)) disp(norm(A,2)) disp(norm(A,%inf)) x = [1 2 3] disp(norm(x,1)) disp(norm(x,2)) disp(norm(x,%inf))
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function x=g_imag(a) // only to be called by function imag //! select type(a) case 2 then x=imag(a) //-compat next case retained for list/tlist compatibility case 15 then if a(1)=='r' then error(43) else error(43) end case 16 then if a(1)=='r' then error(43) else error(43) end case 5 [ij,v,mn]=spget(a) x=sparse(ij,imag(v),mn) else error(43) end
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clc apsilen = 11.9*8.85*10^-14 disp("apsilen = "+string(apsilen)+"F/cm") //initializing value of relative permitivity GL= 10^22 disp("GL= "+string(GL)+"cm^-3/s") //initializing value of rate of optical signal A= 10^-4 disp("A= "+string(A)+"cm^2") //initializing value of diode area Vr= 15 disp("Vr= "+string(Vr)+"V") //initializing value of reverse bias voltage Na=2*10^16 disp("Na = "+string(Na)+"cm^-3") //initializing value of acceptor atoms Nd=10^16 disp("Nd = "+string(Nd)+"cm^-3") //initializing value of donor atoms Dp = 12 disp("Dp= "+string(Dp)+"cm^2/s")//initializing value of hole diffusion coefficient Dn = 20 disp("Dn= "+string(Dn)+"cm^2/s")//initializing value of electron diffusion coefficient Tn = 10^-8 disp("Tn= "+string(Tn)+"s")//inializing value of electron minority carrier lifetime Tp = 10^-8 disp("Tp= "+string(Tp)+"s")//inializing value of hole minority carrier lifetime e = 1.6*10^-19 disp("e= "+string(e)+"C")//initializing value of charge of electron kbT = 0.026 disp("kbT = "+string(kbT)+"eV") //initializing value of kbT at 300K ni = 1.5*10^10 disp("ni= "+string(ni)+"cm^-3")//initializing value of intrinsic carrier concentration Ln = sqrt(Dn*Tn) disp("The electron diffusion length is ,Ln = sqrt(Dn*Tn)= "+string(Ln)+"cm")//calculation Lp = sqrt(Dp*Tp) disp("The hole diffusion length is ,Lp = sqrt(Dp*Tp)= "+string(Lp)+"cm")//calculation Vbi = kbT*log((Na*Nd)/ni^2) disp("The built in voltage is ,Vbi = kbT*log((Na*Nd)/ni^2)= "+string(Vbi)+"V")//calculation W = sqrt(((2*apsilen)/e)*((Na+Nd)/(Na*Nd))*(Vbi+Vr)) disp("The depletion width is ,W = sqrt(((2*apsilen)/e)*((Na+Nd)/(Na*Nd))*(Vbi+Vr))= "+string(W)+"cm")//calculation IL = e*A*GL*(W+Ln+Lp) disp("The photo current is IL = e*A*GL*(W+Ln+Lp)= "+string(IL)+"A")//calculation
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//CHAPTER 3 ILLUSRTATION 3 PAGE NO 104 //TITLE:FRICTION //FIRURE 3.18 clc clear //=========================================================================================== //INPUT DATA W=500// WEGHT IN NEWTONS THETA=30// ANGLE OF INCLINATION IN DEGRESS U=0.2// COEFFICIENT FRICTION S=15// DISTANCE IN metres //============================================================================================ Rn=W*cosd(THETA)// NORMAL REACTION IN NEWTONS P=W*sind(THETA)+U*Rn// PUSHING FORCE ALONG THE DIRECTION OF MOTION w=P*S //============================================================================================ //OUTPUT printf('WORK DONE BY THE FORCE=%3.3f N-m',w)
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function plOutput = OffsetPolyline(plInput, mOffestRegimeBoundaries, mOffsetComponents) //plInput = FwdSelfIntSmoothing(plInput, 10, .05); iPointCount = size(plInput, 1); iSegmentCount = iPointCount - 1; iIntersectionCount = iSegmentCount - 1; lnOffsetSegmentBuffer = []; plOffsetPolylineBuffer = []; plOutputBuffer = []; iOutputPolylineCount = 1 vOffsetArray = BuildOffsetMatrix(plInput, mOffestRegimeBoundaries, mOffsetComponents); //Offset each polyline segment according to vOffsetArray for i = 1:iSegmentCount lnA = plInput(i:i+1, :); lnOffsetSegmentBuffer(i*2-1:i*2,1:2) = lnA + [1 1]' * vOffsetArray(i, :); end //Find intersection point of each adjacent segment for i=1:iSegmentCount - 1 [pIntersectionBuffer(i, 1), pIntersectionBuffer(i, 2)] = IntersectionPoint(lnOffsetSegmentBuffer(i*2 - 1:i*2,:), lnOffsetSegmentBuffer(i*2 + 1:i*2 + 2, :)); end plOffsetPolylineBuffer(1, :) = lnOffsetSegmentBuffer(1,:); plOffsetPolylineBuffer(iPointCount, :) = lnOffsetSegmentBuffer($, :); for i=1:iIntersectionCount plOffsetPolylineBuffer(i + 1, :) = pIntersectionBuffer(i, :); end // if AngleReversal(plInput, plOffsetPolylineBuffer) then // plOutput = [,]; // disp("Angle Reversal Detected."); //// pause // else // plOutput = plOffsetPolylineBuffer; // end // plOffsetPolylineBuffer = RemoveKnots(plInput, plOffsetPolylineBuffer); // plOutput = plOffsetPolylineBuffer; ////////////// plOffsetPolylineBuffer = FwdSelfIntSmoothing(plOffsetPolylineBuffer, 8, .030); ////////////// plOffsetPolylineBuffer = FwdSelfIntSmoothing(plOffsetPolylineBuffer, 8, .030); ////////////// plOffsetPolylineBuffer = FwdSelfIntSmoothing(plOffsetPolylineBuffer, 18, .075); ////////////// // plOffsetPolylineBuffer = plOffsetPolylineBuffer($:-1:1,1:2); // plOffsetPolylineBuffer = FwdSelfIntSmoothing(plOffsetPolylineBuffer, 6, .05); //// plOffsetPolylineBuffer = plOffsetPolylineBuffer($:-1:1,1:2); // plOffsetPolylineBuffer = FwdSelfIntSmoothing(plOffsetPolylineBuffer, 8, .05); // plOffsetPolylineBuffer = RemoveInvertedCircles(plOffsetPolylineBuffer, 24); // plOffsetPolylineBuffer = RemoveInvertedCircles(plOffsetPolylineBuffer, 23); // plOffsetPolylineBuffer = RemoveInvertedCircles(plOffsetPolylineBuffer, 22); // plOffsetPolylineBuffer = RemoveInvertedCircles(plOffsetPolylineBuffer, 21); // plOffsetPolylineBuffer = RemoveInvertedCircles(plOffsetPolylineBuffer, 20); //plOffsetPolylineBuffer = RemoveSharpCorners(plOffsetPolylineBuffer, 30 * (%pi/180)); plOutput = plOffsetPolylineBuffer; endfunction
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clear //given // //find the final pressure gage a=240. b=15. IP=14.7 FP=IP*(a/b) Fpg=FP-IP printf("\n \n final pressure gage is %.2f ",Fpg)
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errcatch(-1,"stop");mode(2); disp("All the values have to be read from the given graph") exit();
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errcatch(-1,"stop");mode(2);//Ex10_3 b='1001000'; disp("Binary number="+string(b))//Binary value d=bin2dec(b)// Binary to decimal value o=dec2oct(d)// Decimal to octal disp("Eqivalent Octal number="+string(o)) exit();
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sce1.0 # camera eyepos 1 -1.5 0.5 eyedir 0 1 -0.10 eyeup 0.0 0.0 1.0 wdist 1.0 fovy_deg 50 nx 400 ny 300 #options max_recursion 4 aasample 4 # scene background 0 0 0.6 ca 0.1 0.1 0.1 rotate -15 0 0 1 { #ground cr 0.4 0.5 0.4 cp 0.4 0.4 0.4 triangle -3 -10 0 3 -10 0 3 10 0 triangle -3 -10 0 3 10 0 -3 10 0 } { cr 0.6 0.6 0.6 cp 0.4 0.4 0.4 translate 1 2.9 -0.3 #rotate 25 0 1 0 scale 0.5 0.5 2 ball 1 0 0 0 } { translate 0.4 0.3 -0.56 rotate 90 1 0 0 scale 0.7 0.7 0.7 cr 0.5 0.4 0.9 cp 0.0 0.0 0.0 object_phong feline100.obj } { translate -1 -1 5 pointlight 3 0 4 0.8 0.8 0.6 } end
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//Given that M = 1.4 //in kg R = 8.5*10^-2 //in meter v = 15*10^-2 //in meter //Sample Problem 12-1 printf("**Sample Problem 12-1**\n") I = 0.5*M*R^2 w = v/R K = 0.5*M*v^2 + 0.5*I*w^2 printf("The total kinetic energy of the disk is %fJ", K)
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function[img_ret]=cv_dilate(image,iterations) pyImport morphological_file img_ret=morphological_file.dilate(image,iterations) endfunction
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<?xml version="1.0" encoding="UTF-8"?> <Project Name="map1307" Width="13" Height="13" CellSize="40" BackgroundSize="1" Background="11plus.png"> <Cell Name="雪灌木" X="1" Y="1" /> <Cell Name="木箱" X="2" Y="1" /> <Cell Name="雪灌木" X="5" Y="1" /> <Cell Name="冰块" X="7" Y="1" /> <Cell Name="企鹅(怪)" X="8" Y="1" arg0="15" /> <Cell Name="冰块" X="2" Y="2" /> <Cell Name="出生点" X="3" Y="2" /> <Cell Name="冰块" X="4" Y="2" /> <Cell Name="雪灌木" X="5" Y="2" /> <Cell Name="木箱" X="6" Y="2" /> <Cell Name="冰墙" X="7" Y="2" /> <Cell Name="木箱" X="8" Y="2" /> <Cell Name="雪灌木" X="9" Y="2" /> <Cell Name="雪灌木" X="10" Y="2" /> <Cell Name="冰墙" X="1" Y="3" /> <Cell Name="冰块" X="3" Y="3" /> <Cell Name="雪树" X="5" Y="3" /> <Cell Name="冰块" X="6" Y="3" /> <Cell Name="木箱" X="7" Y="3" /> <Cell Name="冰川帐篷" X="8" Y="3" arg0="1" arg1="1,1,0,0,0" arg2="14" /> <Cell Name="冰块" X="10" Y="3" /> <Cell Name="雪灌木" X="11" Y="3" /> <Cell Name="雪树" X="2" Y="4" /> <Cell Name="雪灌木" X="3" Y="4" /> <Cell Name="海豹(怪)" X="5" Y="4" arg0="16" /> <Cell Name="雪树" X="7" Y="4" /> <Cell Name="雪树" X="8" Y="4" /> <Cell Name="企鹅(怪)" X="9" Y="4" arg0="15" /> <Cell Name="鱼人(怪)" X="2" Y="5" arg0="17" /> <Cell Name="冰块" X="3" Y="5" /> <Cell Name="冰块" X="4" Y="5" /> <Cell Name="冰块" X="6" Y="5" /> <Cell Name="雪树" X="8" Y="5" /> <Cell Name="雪树" X="9" Y="5" /> <Cell Name="冰墙" X="10" Y="5" /> <Cell Name="木箱" X="11" Y="5" /> <Cell Name="雪树" X="1" Y="6" /> <Cell Name="雪树" X="2" Y="6" /> <Cell Name="冰川帐篷" X="4" Y="6" arg0="1" arg1="1,1,0,0,0" arg2="14" /> <Cell Name="雪灌木" X="5" Y="6" /> <Cell Name="冰墙" X="6" Y="6" /> <Cell Name="雪树" X="10" Y="6" /> <Cell Name="雪树" X="11" Y="6" /> <Cell Name="冰川帐篷" X="1" Y="7" arg0="1" arg1="1,1,0,0,0" arg2="14" /> <Cell Name="木箱" X="2" Y="7" /> <Cell Name="海豹(怪)" X="3" Y="7" arg0="16" /> <Cell Name="雪树" X="8" Y="7" /> <Cell Name="冰块" X="10" Y="7" /> <Cell Name="雪树" X="2" Y="8" /> <Cell Name="鱼人(怪)" X="5" Y="8" arg0="17" /> <Cell Name="雪灌木" X="6" Y="8" /> <Cell Name="雪树" X="7" Y="8" /> <Cell Name="鱼人(怪)" X="9" Y="8" arg0="17" /> <Cell Name="冰墙" X="10" Y="8" /> <Cell Name="鱼人(怪)" X="11" Y="8" arg0="17" /> <Cell Name="冰川帐篷" X="4" Y="9" arg0="1" arg1="1,1,0,0,0" arg2="14" /> <Cell Name="雪树" X="6" Y="9" /> <Cell Name="冰川帐篷" X="8" Y="9" arg0="1" arg1="1,1,0,0,0" arg2="14" /> <Cell Name="冰块" X="10" Y="9" /> <Cell Name="雪灌木" X="4" Y="10" /> <Cell Name="雪树" X="5" Y="10" /> <Cell Name="通关点-1" X="6" Y="10" /> <Cell Name="木箱" X="9" Y="10" /> <Cell Name="雪块" X="10" Y="10" /> <Cell Name="木屋1(大型建筑)" X="2" Y="11" arg0="3" arg1="2" arg2="0,2" /> <Cell Name="冰块" X="5" Y="11" /> <Cell Name="木箱" X="6" Y="11" /> <Cell Name="雪灌木" X="8" Y="11" /> <Cell Name="冰块" X="9" Y="11" /> <Cell Name="木箱" X="10" Y="11" /> </Project>
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function seno = seno_taylor(angulo, n) seno = 0; for i:1:n seno = seno + ((angulo)^(2*n + 1) / factorial(2*n +1)); end endfunction
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//CHAPTER 8 ILLUSRTATION 4 PAGE NO 225 //TITLE:BALANCING OF ROTATING MASSES pi=3.141 clc clear mB=30// mass of B in kg mC=50// mass of C in kg mD=40// mass of D in kg rA=18// radius of A in cm rB=24// radius of B in cm rC=12// radius of C in cm rD=15// radius of D in cm //============================= mA=3.6/.18// mass of A by measurement in kg theta=124// angle with mass B in degrees by measurement in degrees y=3.6/(.18*20)// position of A from B printf('mass of A=%i kg\n angle with mass B=%i degrees\n position of A from B=%i m towards right of plane B',mA,theta,y)
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//example 11.3 //calculate //capacity of siphon //head required in oggy spillway //length of oggy weir required clc;funcprot(0); //given t=6; //tail water elevation h=1; //heigth of siphon spillway w=4; //width of siphon spillway hw=1.5; //head water elevation C=0.6; //coefficient of discharge Co=2.25; //coefficient of discharge of oggy spillway lo=4; //length of oggy spillway hc=1.5; //head on weir crest g=9.81; //acceleration due to gravity //part (a) Q=C*h*w*(2*g*(t+hw))^0.5; Q=round(Q*10)/10; mprintf("capacity of siphon=%f cumecs.",Q); //part (b) h1=(Q/(Co*lo))^(2/3); h1=round(h1*100)/100; mprintf("\nhead required in oggy spillway=%f m",h1); //part (c) L=Q/(Co*(hc)^1.5); L=round(L*100)/100; mprintf("\nlength of oggy weir required=%f m.",L);
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clear; clc; printf('\n Example 14.3"); He = 2690; //He is the enthalpy of entrained steam in kJ/kg H4 = ((1*2780)+(1.6*2690))/2.60; //Again assuming isentropic compression from 101.3 to 135 kN/m2, then: H3 = 2640; //in kJ/kg (from the chart) n = (1.0+1.6)*(2725-2640)/[1.0*(2780-2375)]; printf("\n η = %.2f ",n); printf("\n This value is low, since in good design overall efficiencies approach 0.75–0.80. Obviously the higher the efficiency the greater the entrainment ratio or the higher the saving in live steam"); Pe = 101.3; //pressure of entrained vapour in kN/m^2 discharge_P = 135;//discharge pressure in kN/m^2 printf("\n the required flow of live steam = 0.5 kg/kg entrained vapour."); printf("\n In this case the ratio is (1.0/1.6) = 0.63 kg/kg");
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//Caption:Calculate (i)-maximum power gain in dBs ,(ii)-noise figure F in dBs, (iii)-bandwidth for r=0.2 //Exa:9.12 clc; clear; close; ratio=8; r=0.2; r_Q=8; T_d=300;//in Kelvin T_o=300;//in Kelvin X=8; G=(ratio)*X/(1+sqrt(1+X))^2; G_in_dB=(10*log(G))/log(10);//gain disp(G_in_dB,'Maximum Gain (in dB)='); F=[10*log(1+(2*T_d/T_o)*[(1/(r_Q))+(1/(r_Q)^2)])]/log(10);//noise figure disp(F,'Noise figure (in dB) ='); B_W=2*r*sqrt(ratio);//bandwidth disp(B_W,'bandwidth =');
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clc clear // Características do sistema: pi = %pi R = 3 c = 0.1 ro = 0.1 // kg/m g = 9.80 // m/s² l = 0.5 // m L = 2*l // m m = L*ro alpha = l/R //Vetor de Estados Iniciais theta_0 = pi/6 omega_0 = 9 E = [theta_0,omega_0] //Vetor Tempo t0 = 0 dt = 0.005 tf = 100 t = t0:dt:tf //Integração function z_dot = deriva(t,z) dk_dt = z(2) d2k_dt2 = -(g/R)*(sin(alpha)/alpha)*sin(z(1)) - L*c*z(2) z_dot = [dk_dt;d2k_dt2] endfunction //ODE X = ode([theta_0;omega_0], 0, t, deriva) ////Plots clf() scf(0) plot(t, X(2,:))
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//Chapter 24: Space Wave Propagation //Example 24-9.3 clc; //Variable Initialization tx_h = 100 //Transmitting antenna height (m) rx_h = 16 //Receiving antenna height (m) tx_p = 40e3 //Transmitting antenna power radiation (W) f = 100e6 //Frequency (Hz) d = 10e3 //Distance (m) c = 3e8 //Speed of light (m/s) E = 1e-3 //Signal strength (V/m) //Calculations los = 4.12*(sqrt(tx_h) + sqrt(rx_h)) //LOS distance (km) wave_lt = c/f //Wavelength (m) Es = (88*sqrt(tx_p)/(wave_lt*(d**2)))*tx_h*rx_h //Field strength at distance d (V/m) dsig = sqrt(88*sqrt(tx_p)*tx_h*rx_h/(wave_lt*E)) //Distance at which field strength reduces to 1mV/m //Result mprintf( "The LOS distance is %.2f km", los) mprintf( "\nThe field strength at 10km is %.5f V/m", Es) mprintf( "\nThe distance at which field strength is 1mV/m is %.d m",dsig)
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t= 0:1/10000:0.02; // declare time interval Ec = 5; // amplitude of carrier signal Em = 4.5; // amplitude of modulating signal fc = 1000; // carrier frequency fm = 300; // modulating frequency //Carrier signal Vc = Ec *cos (((2*%pi)*fc)*t); //Modulating signal Vm = Em * cos (((2*%pi)*fm)*t); m1 = 1; // modulation index Vfm = Ec*cos(((( 2*%pi)*fc)*t)+m1*sin(((2*%pi)*fm)*t)); //Frequency modulation signal // plot signal subplot (311); plot (t, Vm); title("Modulating signal"); xlabel('Time - s'); ylabel('Amplitude'); subplot (312); plot (t,Vc); title("Carrier signal"); xlabel('Time - s'); ylabel('Amplitude'); subplot (313); plot (t,Vfm); title("Modulated-wave"); xlabel('Time - s'); ylabel('Amplitude');
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n=1:100 x=5*sin(2*%pi*n/10) figure plot(x) a=fft(x) figure plot(a)
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//Smallest weight W //Refer fig. 5.9 mu=0.4 //consider equilibrium of block B //using law of friction N1=5/((0.5)+(tand(20))*(sind(20))) //kN F1=N1*tand(20) C=N1*cosd(30)-F1*cosd(60) //kN //Consider the equilibrium of block A F2=C //kN //Law of friction N2=4.196/0.4 //kN W=N2 //kN printf("\nW=%0.2f kN",W)
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var=[]; max-cong=[]; max-trans=[]; var(1)=0.1; max-cong(1)=302.5990099009901; max-trans(1)=0.9518900343642611;
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double.man.tst
clear;lines(0); x=int8([0 12 140]) double(x)
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Ex5_9_u1.sce
//Example 5_9_u1 clc(); clear; //To determine the interplanar spacing h=6.63*10^-34 //units in m^2 kg s^-1 m=9.1*10^-31 //units in Kgs e=1.6*10^-19 //units in coulombs v=844 //units in Volts lamda=h/sqrt(2*m*e*v) //units in meters n=1 theta=58 //units in degrees d=(n*lamda)/(2*sin(theta*(%pi/180))) //units in meters printf("The interplanar spacing d=") disp(d) printf("meters")
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Ex15_3.sce
clear; clc; disp('Example 15.3'); // aim : To determine // (a) the pressure, volume and temperature at each corner of the cycle // (b) the thermal efficiency of the cycle // (c) the work done per cycle // (d) the work ratio // given values m = 1;// mass of air, [kg] P1 = 1730;// initial pressure of carnot engine, [kN/m^2] T1 = 273+300;// initial temperature, [K] R = .29;// [kJ/kg K] Gama = 1.4;// heat capacity ratio // solution // taking reference Fig. 15.15 // (a) // for the isothermal process 1-2 // using ideal gas law V1 = m*R*T1/P1;// initial volume, [m^3] T2 = T1; V2 = 3*V1;// given condition // for isothermal process, P1*V1=P2*V2, so P2 = P1*(V1/V2);// [MN/m^2] // for the adiabatic process 2-3 V3 = 6*V1;// given condition T3 = T2*(V2/V3)^(Gama-1); // also for adiabatic process, P2*V2^Gama=P3*V3^Gama, so P3 = P2*(V2/V3)^Gama; // for the isothermal process 3-4 T4 = T3; // for both adiabatic processes, the temperataure ratio is same, // T1/T4 = T2/T3=(V4/V1)^(Gama-1)=(V3/V2)^(Gama-1), so V4 = 2*V1; // for isothermal process, 3-4, P3*V3=P4*V4, so P4 = P3*(V3/V4); disp('(a) At line 1'); mprintf('\n V1 = %f m^3, t1 = %f C, P1 = %f kN/m^2\n',V1,T1-273,P1); disp('At line 2'); mprintf('\n V2 = %f m^3, t2 = %f C, P2 = %f kN/m^2\n',V2,T2-273,P2); disp('At line 3'); mprintf('\n V3 = %f m^3, t3 = %f C, P3 = %f kN/m^2\n',V3,T3-273,P3); disp('At line 4'); mprintf('\n V4 = %f m^3, t4 = %f C, P4 = %f kN/m^2\n',V4,T4-273,P4); // (b) n_the = (T1-T3)/T1;// thermal efficiency mprintf('\n (b) The thermal efficiency of the cycle is = %f percent\n',n_the*100); // (c) W = m*R*T1*log(V2/V1)*n_the;// work done, [J] mprintf('\n (c) The work done per cycle is = %f kJ\n',W); // (d) wr = (T1-T3)*log(V2/V1)/(T1*log(V2/V1)+(T1-T3)/(Gama-1));// work ratio mprintf('\n (d) The work ratio is = %f\n',wr); // there is calculation mistake in the book so answer is not matching // End
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symbolic; typedef any; typedef number checked by numberp; typedef sf checked by sfpx; typedef sq checked by sqp; procedure hugo(x1,x2); x2; assert hugo: (number,any) -> number; assert_install hugo; hugo(0,0); hugo('x,0); hugo(0,'x); assert addf: (sf,sf) -> sf; assert addsq: (sq,sq) -> sq; assert_install addf,addsq; addsq(simp 'x,numr simp 'x); algebraic; assert_analyze(); assert_uninstall_all; end;
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Hooke_Jeeves(Newton).sce
// @ metodo de Hooke-Jeeves com newton @ global d; d = [1, 0; 0, 1]; global p; p = [-10, 10]; global lamb; lamb = [2,2] N = 2 function z =f(p) z = (p(1)+2)^2+(p(2)-10)^2; endfunction function y = flamb(lamb,j) global d ; y = f([p(1) + lamb*d(j,1) , p(2) + lamb*d(j,2)]); endfunction function y1 = PrimeiraDeriv(j) global lamb ; h = 1e-6 ; y1 = (flamb(lamb(j)+h,j) - flamb(lamb(j),j))/h ; endfunction function y2 = SegundaDeriv(j) global lamb ; h = 1e-3 ; y2 = (flamb(lamb(j)+h,j)-2*flamb(lamb(j),j)+flamb(lamb(j)-h,j))/h^2 ; endfunction function Newton(j) global lamb ; for(k = 1:10) lamb(j) = lamb(j) - (PrimeiraDeriv(j)/SegundaDeriv(j)) end endfunction function HookJeeves(d,p,lamb) global d; global p ; global lamb ; for(k=1:10) for(j = 1:N) Newton(j); for (w=1:N) p(w) = p(w) + lamb(j)*d(j,w); end end end mprintf("\n O ultimo ponto (x,y) é :"); mprintf("(%.7f,",p(1)); mprintf("%.7f)",p(2)); mprintf("\n valor de f no ponto é : %.6f",f(p)) endfunction //----------------------------- main ------------------------ HookJeeves(d,p,lamb);
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// Scilab code Ex3.19: Pg 101-102 (2008) clc; clear; C = 270e-12; // Capacitance, F A = 60e-04; // Cross-sectional area of plate, m^2 E = 350e03; // Dielectric strength, V/m epsilon_r = 2.1; // Relative pemittivity epsilon_o = 8.854e-12; // Permittivity of free space // Part (a) // Since formula for capacitance, C = ((epsilon_o)*(eplison_r)*A)/d, solving for d d = ((epsilon_o)*(epsilon_r)*A)/C; // Thickness of dielectric, m // Part (b) // Since E = V/d, solving for V V = E*d; // Maximum possible working voltage, V printf("\nThe thickness of Teflon sheet required = %5.4f mm", d/1e-03); printf("\nThe maximum possible working voltage for the capacitor = %5.1f V", V); // Result // The thickness of Teflon sheet required = 0.413 mm // The maximum possible working voltage for the capacitor = 144.6 V
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//clc(); clear; // To determine the wavelength of source of light beeta=0.30; // fringe spacing in centimtres d=0.04; // distance between two slits in centimtres D=180; // distance between the slit and screen in centimetres lambda=(beeta*d*10^8)/D; printf("the wavelength of source of light is %f Armstrong",lambda);
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pathname=get_absolute_file_path('26_3.sce') filename=pathname+filesep()+'26_3data.sci' exec(filename) clear Tr=T*L1/2; mu=(8*G*t/(A*E*(b+a)))^0.5; L=L1/2; k1=((T*(b-a)*10^3)/(8*a*b*G*t)); k2=1/(mu*cosh(mu*L)); k3=((4*(b-a))/(a*b*(b+a))); k4=(2*T*(10^3)/(a*b*G*t*(b+a))); function[th]=f(z) w=(k1*((k2*sinh(mu*z))-z)); F=((k1*k3 +k4)*L*L*0.5 -(k1*k2*k3/mu)*cosh(mu*L)); th=(k1*k2*k3/mu)*cosh(mu*z) -(k1*k3 +k4)*z*z*0.5 + F; endfunction funcprot(); printf("\nangle of twist at mid-span θ: %f rad",f(0));
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refs/heads/master
2021-01-19T23:53:52.068010
2017-04-22T12:39:21
2017-04-22T12:39:21
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sce
testparamfplot2d.sce
function y=f(x,t) y=exp(-(x-t).^2/2) endfunction x=[-10:0.05:10].'; t=[-5:0.2:5]; clf; paramfplot2d(f,x,t);