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//Example 3.7 //Program to Compute circular convolution of following sequences //x[n]=[1,2,2,1,0] //Y[k]=exp(-j*4*pi*k/5).X[k] clear; clc ; close ; x=[1,2,2,1,0]; X=fft(x,-1); k=0:1:4; j=sqrt(-1); pi=22/7; H=exp(-j*4*pi*k/5); Y=H.*X; //IDFT Computation y=fft(Y,1); //Display sequence y[n] in command window disp(round(y),"y[n]="); //Plots n=0:1:4; a = gca (); a.y_location ="origin"; a.x_location ="origin"; plot2d3(n,round(y),5); poly1=a.children(1).children (1); poly1.thickness=2; xtitle('Plot of sequence y[n]','n','y[n]');
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// find output current and maximum load resistance // Electronic Principles // By Albert Malvino , David Bates // Seventh Edition // The McGraw-Hill Companies // Example 20-9, page 768 clear; clc; close; // Given data R=15*10^3;// in ohms Vin=3;// input voltage in volts Vcc=15;// in volts // Calculations iout=-Vin/R;// output current in amperes Rlmax=(R/2)*((Vcc/Vin)-1);// maximum load resistance in ohms disp("Amperes",iout,"output current=") disp("ohms",Rlmax,"Maximum load resistance=") // Result // Output current is -0.2 mAmperes // Maximum load resistance is 30 Kohms
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// Scilab Code Ex2.52:: Page-2.36(2009) clc; clear; lambda = 5893e-010; // Wavelength of light used, m mu = 1; // Refractive index of the glass b = 1; // Assume fringe width to be unity, cm // As b = l/20, solving for l l = b*20; // Length of the film, m // As b = lambda/(2*mu*theta) and theta = t/l, solving for t t = lambda*l/(2*mu); // Thickness of the wire separating two glass surfaces, m printf("\nThe thickness of the wire separating two glass surfaces = %4.2e m", t); // Result // The thickness of the wire separating two glass surfaces = 5.89e-06 m
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fwWnls = read(get_absolute_file_path("ReadOmegaNls.sce") + "..\Data\FwWnls.txt", 10, 1) rvWnls = read(get_absolute_file_path("ReadOmegaNls.sce") + "..\Data\RvWnls.txt", 10, 1)
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//Limpa a tela e os dados anteriores clc clear nomes = ['WR975','WR650','WR430','WR284','WR187','WR137','WR90','WR62'] guias = [ [9.75, 4.875], [6.5, 3.25], [4.3,2.15], [2.84,1.34], [1.872,0.872], [1.372,0.622], [0.9,0.4], [0.622,0.311] ]; sg=size(guias) //Modo TMmn m=1;n=1; //Numero de partes pa=20;pb=20; //Entradas printf("Você deseja plotar um:\n\t1-Guia de Onda da Tabela\n\t2-Guia personalizado") op = input("Selecione uma opção:") if(op==1) clc printf('Guias disponíveis:\n') for i=1:sg(1) printf("\t%i-%s\n",i,nomes(i)) end op = input("Selecione uma opção:") g=guias(op,:); a=g(1); b=g(2); elseif(op==2) a = input("Insira a:") b = input("Insira b: ") end //Calculo dos Betas betax=m*%pi/a; betay=n*%pi/b; //Dados do gráfico de contorno .: Gráfico 1 for i=0:1:a*pa x(i+1)=i/pa; for j=0:b*pb y(j+1)=j/pb; Ez(i+1,j+1)=sin(betay*y(j+1))*sin(betax*x(i+1)); end end //Dados do gráfico convencional 1, com y=b/2 .: Gráfico 2 yc1=b/2; xc1=0:a/pa:a; Ezc1=20*sin(betax*xc1)*sin(betay*yc1); //Dados do gráfico convencional 2, com x=a/2 .: Gráfico 3 xc2=a/2; yc2=0:b/pb:b; Ezc2=20*sin(betax*xc2)*sin(betay*yc2); //Dados do gráfico surf (3D) .: Gráfico 4 [X,Y]=meshgrid(0:a/pa:a,0:b/pb:b); Ezs=10*sin(betax.*X).*sin(betay.*Y); //Plotar gráfico 1 subplot(2,2,3) contour2d(x,y,Ez,6,rect=[0,0,a,b]) title("Gráfico de curvas de nivel") xlabel("x") ylabel("y") //Plotar gráfico 2 subplot(2,2,2) plot2d(xc1,Ezc1) title("Intensidade de Ez ao longo de a em y=b/2") xlabel("x") ylabel("20*Ez em y = b/2") //Plotar gráfico 3 subplot(2,2,1) plot2d(yc2,Ezc2) title("Intensidade de Ez ao longo de b em x=a/2") xlabel("y") ylabel("20*Ez em x = a/2") //Plotar gráfico 4 subplot(2,2,4) surf(X,Y,Ezs) title("Gráfico em 3d da intensidade de 10*Ez") xlabel("x") ylabel("y") zlabel("10*Ez")
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// This file is part of www.nand2tetris.org // and the book "The Elements of Computing Systems" // by Nisan and Schocken, MIT Press. // File name: projects/02/Zero16.tst load Zero16.hdl, output-file Zero16.out, compare-to Zero16.cmp, output-list in%B1.16.1 sel%B3.1.3 out%B1.16.1; set in %B0000000000000000, set sel 1, eval, output; set in %B1111111111111111, set sel 1, eval, output; set in %B1010101010101010, set sel 1, eval, output; set in %B0000000000000000, set sel 0, eval, output; set in %B1111111111111111, set sel 0, eval, output; set in %B1010101010101010, set sel 0, eval, output;
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MANAGER.tst
-- VectorCAST 18.sp3 (10/18/18) -- Test Case Script -- -- Environment : MANAGER -- Unit(s) Under Test: manager -- -- 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: manager -- Subprogram: Add_Included_Dessert -- Test Case: COND_1_ROW_1_PAIR_a_TTT TEST.UNIT:manager TEST.SUBPROGRAM:Add_Included_Dessert TEST.NEW TEST.NAME:COND_1_ROW_1_PAIR_a_TTT TEST.MCDC_BASIS_PATH:2 of 8 TEST.NOTES: This is an automatically generated test case. Test Path 2 (1) if ((Order->Entree == (STEAK) && Order->Salad == (CAESAR)) && Order->Beverage == (MIXED_DRINK)) ==> TRUE Row number 1 forms a pair with Row 5 for Condition #1, subcondition "a". Condition a ==> TRUE Condition b ==> TRUE Condition c ==> TRUE Test Case Generation Notes: TEST.END_NOTES: TEST.VALUE:manager.Add_Included_Dessert.Order:<<malloc 1>> TEST.VALUE:manager.Add_Included_Dessert.Order[0].Salad:CAESAR TEST.VALUE:manager.Add_Included_Dessert.Order[0].Entree:STEAK TEST.VALUE:manager.Add_Included_Dessert.Order[0].Beverage:MIXED_DRINK TEST.END -- Test Case: COND_1_ROW_2_PAIR_c_TTF TEST.UNIT:manager TEST.SUBPROGRAM:Add_Included_Dessert TEST.NEW TEST.NAME:COND_1_ROW_2_PAIR_c_TTF TEST.MCDC_BASIS_PATH:4 of 8 TEST.NOTES: This is an automatically generated test case. Test Path 4 (1) if ((Order->Entree == (STEAK) && Order->Salad == (CAESAR)) && Order->Beverage == (MIXED_DRINK)) ==> FALSE Row number 2 forms a pair with Row 1 for Condition #1, subcondition "c". Condition a ==> TRUE Condition b ==> TRUE Condition c ==> FALSE Test Case Generation Notes: TEST.END_NOTES: TEST.VALUE:manager.Add_Included_Dessert.Order:<<malloc 1>> TEST.VALUE:manager.Add_Included_Dessert.Order[0].Salad:CAESAR TEST.VALUE:manager.Add_Included_Dessert.Order[0].Entree:STEAK TEST.VALUE:manager.Add_Included_Dessert.Order[0].Beverage:NO_BEVERAGE TEST.END -- Test Case: COND_1_ROW_3_PAIR_b_TFT TEST.UNIT:manager TEST.SUBPROGRAM:Add_Included_Dessert TEST.NEW TEST.NAME:COND_1_ROW_3_PAIR_b_TFT TEST.MCDC_BASIS_PATH:3 of 8 TEST.NOTES: This is an automatically generated test case. Test Path 3 (1) if ((Order->Entree == (STEAK) && Order->Salad == (CAESAR)) && Order->Beverage == (MIXED_DRINK)) ==> FALSE Row number 3 forms a pair with Row 1 for Condition #1, subcondition "b". Condition a ==> TRUE Condition b ==> FALSE Condition c ==> TRUE Test Case Generation Notes: TEST.END_NOTES: TEST.VALUE:manager.Add_Included_Dessert.Order:<<malloc 1>> TEST.VALUE:manager.Add_Included_Dessert.Order[0].Salad:NO_SALAD TEST.VALUE:manager.Add_Included_Dessert.Order[0].Entree:STEAK TEST.VALUE:manager.Add_Included_Dessert.Order[0].Beverage:MIXED_DRINK TEST.END -- Test Case: COND_1_ROW_5_PAIR_a_FTT TEST.UNIT:manager TEST.SUBPROGRAM:Add_Included_Dessert TEST.NEW TEST.NAME:COND_1_ROW_5_PAIR_a_FTT TEST.MCDC_BASIS_PATH:1 of 8 TEST.NOTES: This is an automatically generated test case. Test Path 1 (1) if ((Order->Entree == (STEAK) && Order->Salad == (CAESAR)) && Order->Beverage == (MIXED_DRINK)) ==> FALSE Row number 5 forms a pair with Row 1 for Condition #1, subcondition "a". Condition a ==> FALSE Condition b ==> TRUE Condition c ==> TRUE Test Case Generation Notes: TEST.END_NOTES: TEST.VALUE:manager.Add_Included_Dessert.Order:<<malloc 1>> TEST.VALUE:manager.Add_Included_Dessert.Order[0].Salad:CAESAR TEST.VALUE:manager.Add_Included_Dessert.Order[0].Entree:NO_ENTREE TEST.VALUE:manager.Add_Included_Dessert.Order[0].Beverage:MIXED_DRINK TEST.END -- Test Case: COND_2_ROW_1_PAIR_a_TTT TEST.UNIT:manager TEST.SUBPROGRAM:Add_Included_Dessert TEST.NEW TEST.NAME:COND_2_ROW_1_PAIR_a_TTT TEST.MCDC_BASIS_PATH:6 of 8 TEST.NOTES: This is an automatically generated test case. Test Path 6 (1) if ((Order->Entree == (STEAK) && Order->Salad == (CAESAR)) && Order->Beverage == (MIXED_DRINK)) ==> FALSE (2) if ((Order->Entree == (LOBSTER) && Order->Salad == (GREEN)) && Order->Beverage == (WINE)) ==> TRUE Row number 1 forms a pair with Row 5 for Condition #2, subcondition "a". Condition a ==> TRUE Condition b ==> TRUE Condition c ==> TRUE Test Case Generation Notes: TEST.END_NOTES: TEST.VALUE:manager.Add_Included_Dessert.Order:<<malloc 1>> TEST.VALUE:manager.Add_Included_Dessert.Order[0].Salad:GREEN TEST.VALUE:manager.Add_Included_Dessert.Order[0].Entree:LOBSTER TEST.VALUE:manager.Add_Included_Dessert.Order[0].Beverage:WINE TEST.END -- Test Case: COND_2_ROW_2_PAIR_c_TTF TEST.UNIT:manager TEST.SUBPROGRAM:Add_Included_Dessert TEST.NEW TEST.NAME:COND_2_ROW_2_PAIR_c_TTF TEST.MCDC_BASIS_PATH:8 of 8 TEST.NOTES: This is an automatically generated test case. Test Path 8 (1) if ((Order->Entree == (STEAK) && Order->Salad == (CAESAR)) && Order->Beverage == (MIXED_DRINK)) ==> FALSE (2) if ((Order->Entree == (LOBSTER) && Order->Salad == (GREEN)) && Order->Beverage == (WINE)) ==> FALSE Row number 2 forms a pair with Row 1 for Condition #2, subcondition "c". Condition a ==> TRUE Condition b ==> TRUE Condition c ==> FALSE Test Case Generation Notes: TEST.END_NOTES: TEST.VALUE:manager.Add_Included_Dessert.Order:<<malloc 1>> TEST.VALUE:manager.Add_Included_Dessert.Order[0].Salad:GREEN TEST.VALUE:manager.Add_Included_Dessert.Order[0].Entree:LOBSTER TEST.VALUE:manager.Add_Included_Dessert.Order[0].Beverage:NO_BEVERAGE TEST.END -- Test Case: COND_2_ROW_3_PAIR_b_TFT TEST.UNIT:manager TEST.SUBPROGRAM:Add_Included_Dessert TEST.NEW TEST.NAME:COND_2_ROW_3_PAIR_b_TFT TEST.MCDC_BASIS_PATH:7 of 8 TEST.NOTES: This is an automatically generated test case. Test Path 7 (1) if ((Order->Entree == (STEAK) && Order->Salad == (CAESAR)) && Order->Beverage == (MIXED_DRINK)) ==> FALSE (2) if ((Order->Entree == (LOBSTER) && Order->Salad == (GREEN)) && Order->Beverage == (WINE)) ==> FALSE Row number 3 forms a pair with Row 1 for Condition #2, subcondition "b". Condition a ==> TRUE Condition b ==> FALSE Condition c ==> TRUE Test Case Generation Notes: TEST.END_NOTES: TEST.VALUE:manager.Add_Included_Dessert.Order:<<malloc 1>> TEST.VALUE:manager.Add_Included_Dessert.Order[0].Salad:NO_SALAD TEST.VALUE:manager.Add_Included_Dessert.Order[0].Entree:LOBSTER TEST.VALUE:manager.Add_Included_Dessert.Order[0].Beverage:WINE TEST.END -- Test Case: COND_2_ROW_5_PAIR_a_FTT TEST.UNIT:manager TEST.SUBPROGRAM:Add_Included_Dessert TEST.NEW TEST.NAME:COND_2_ROW_5_PAIR_a_FTT TEST.MCDC_BASIS_PATH:5 of 8 TEST.NOTES: This is an automatically generated test case. Test Path 5 (1) if ((Order->Entree == (STEAK) && Order->Salad == (CAESAR)) && Order->Beverage == (MIXED_DRINK)) ==> FALSE (2) if ((Order->Entree == (LOBSTER) && Order->Salad == (GREEN)) && Order->Beverage == (WINE)) ==> FALSE Row number 5 forms a pair with Row 1 for Condition #2, subcondition "a". Condition a ==> FALSE Condition b ==> TRUE Condition c ==> TRUE Test Case Generation Notes: TEST.END_NOTES: TEST.VALUE:manager.Add_Included_Dessert.Order:<<malloc 1>> TEST.VALUE:manager.Add_Included_Dessert.Order[0].Salad:GREEN TEST.VALUE:manager.Add_Included_Dessert.Order[0].Entree:NO_ENTREE TEST.VALUE:manager.Add_Included_Dessert.Order[0].Beverage:WINE TEST.END -- Subprogram: Add_Party_To_Waiting_List -- Test Case: Add_Party_To_Waiting_List.001 TEST.UNIT:manager TEST.SUBPROGRAM:Add_Party_To_Waiting_List TEST.NEW TEST.NAME:Add_Party_To_Waiting_List.001 TEST.COMPOUND_ONLY TEST.VALUE:manager.Add_Party_To_Waiting_List.Name:<<malloc 4>> TEST.VALUE:manager.Add_Party_To_Waiting_List.Name:"Bob" TEST.END -- Subprogram: Get_Next_Party_To_Be_Seated -- Test Case: Get_Next_Party_To_Be_Seated.001 TEST.UNIT:manager TEST.SUBPROGRAM:Get_Next_Party_To_Be_Seated TEST.NEW TEST.NAME:Get_Next_Party_To_Be_Seated.001 TEST.COMPOUND_ONLY TEST.EXPECTED:manager.Get_Next_Party_To_Be_Seated.return:"Bob" TEST.END -- Subprogram: Place_Order -- Test Case: Lobster TEST.UNIT:manager TEST.SUBPROGRAM:Place_Order TEST.NEW TEST.NAME:Lobster TEST.REQUIREMENT_KEY:FR17 TEST.VALUE:manager.Place_Order.Order.Entree:LOBSTER TEST.EXPECTED:uut_prototype_stubs.Update_Table_Record.Data.Check_Total:18 TEST.END -- Test Case: Place_Order.001 TEST.UNIT:manager TEST.SUBPROGRAM:Place_Order TEST.NEW TEST.NAME:Place_Order.001 TEST.REQUIREMENT_KEY:FR17 TEST.VALUE:manager.Place_Order.Order.Entree:STEAK TEST.EXPECTED:uut_prototype_stubs.Update_Table_Record.Data.Check_Total:14 TEST.END -- Test Case: Place_Order.002 TEST.UNIT:manager TEST.SUBPROGRAM:Place_Order TEST.NEW TEST.NAME:Place_Order.002 TEST.VALUE:manager.Place_Order.Table:<<MIN-1>> TEST.END -- Test Case: Place_Order.003 TEST.UNIT:manager TEST.SUBPROGRAM:Place_Order TEST.NEW TEST.NAME:Place_Order.003 TEST.END -- Test Case: chicken TEST.UNIT:manager TEST.SUBPROGRAM:Place_Order TEST.NEW TEST.NAME:chicken TEST.REQUIREMENT_KEY:FR17 TEST.VALUE:manager.Place_Order.Order.Entree:CHICKEN TEST.EXPECTED:uut_prototype_stubs.Update_Table_Record.Data.Check_Total:10 TEST.END -- Test Case: chicken.001 TEST.UNIT:manager TEST.SUBPROGRAM:Place_Order TEST.NEW TEST.NAME:chicken.001 TEST.REQUIREMENT_KEY:FR17 TEST.VALUE:manager.Place_Order.Order.Entree:CHICKEN TEST.EXPECTED:uut_prototype_stubs.Update_Table_Record.Data.Check_Total:10 TEST.END -- COMPOUND TESTS TEST.SUBPROGRAM:<<COMPOUND>> TEST.NEW TEST.NAME:Waiting_llist TEST.SLOT: "1", "manager", "Add_Party_To_Waiting_List", "1", "Add_Party_To_Waiting_List.001" TEST.SLOT: "2", "manager", "Get_Next_Party_To_Be_Seated", "1", "Get_Next_Party_To_Be_Seated.001" TEST.END --
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//Transport Processes and Seperation Process Principles //Chapter 11 //Example 11.1-1 //Vapour Liquid Seperation Processes //given data pa=[116.9 135.5 155.7 179.2 204.2 240] pb=[46 54 63.3 74.3 86 101.32] P=101.32 //eq: paxa+pb(1-xa)=P xa=[]; ya=[]; for i=1:6 xa(i)=(P-pb(1,i))/(pa(1,i)-pb(1,i)) ya(i)=((pa(1,i))*xa(i))/P end xa(7)=1; ya(7)=1; m=linspace(0,1,10) n=linspace(0,1,10) plot(m,n) plot2d(xa,ya,rect=[0 0 1 1]) xtitle("raolts law","xa","ya") mprintf("xa=%f",xa); mprintf("ya=%f",ya);
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clc //initialization of varaibles h1=1399.2 //B/lb h2s=976 //B/lb wt=8 //lb /hp hr //calculations Wt=2545/wt etaT=Wt/(h1-h2s) h2=h1-Wt //results printf("Engine efficiency = %.3f",etaT)
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d1=17.5//mm(Diameter at vena contracta) d2=20//mm(diameter of orifice)
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clc //initialisation of variables g=32.2//ft/sec^2 p0=8.56//lb/in^2 A=2.082*10^-4//ft^2 W=0.0212//lb/sec v0=1057.6//ft/sec T1=213.7//R T2=206//R //CALCULATIONS b=2*g*778*0.24 k1=sqrt(b) c=144*A*g/W k=p0+v0/c v1=k1*sqrt(T1-T2) p1=k-v1/c //RESULTS printf ('\n velocity= %.f ft/sec',v1 ) printf ('\n pressure= %.2f lb/in^2',p1 )
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//MÉTODO DE JACOBI //A = matriz de coeficientes do sistema linear //b = vetor de termos independentes do sistema //xo = solução inicial, pois se trata de um método iterativo //e = erro de aceitação estipulado //n = número máximo de iterações que algoritmo deve realizar //iter = número de iterações realizadas para a convergência do sistema //x = vetor de soluções do sistema //xa = x anterior, vetor de soluções atualizado a cada iteração function [x,iter] = jacobi(A,b,xo,e,n) //pegando o tamanho da matriz [l,c] = size(A); //inicializando o erro, solução inicial e iterações realizadas erro = 1; x = xo; iter = 0; //repetição do método enquanto o erro estipulado não é satisfeito while erro > e & iter < n xa = x; //percorrendo as linhas e isolando as incógnitas for i = 1:l soma = 0; for j = 1:l if j ~= i //utilizando o valor inicial da solução (xa) //para calular os elementos de x soma = soma + A(i,j)*xa(j); end end //cálculo das incógnitas x(i) = (b(i) - soma) / A(i,i); end iter = iter + 1; //atualizando o valor do erro a cada iteração erro = max(abs(x-xa)) / max(abs(x)); end endfunction
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//Ex:1.14 clc; clear; close; r=12;//in ohms v=6;//in volts i=(v/r); printf("Current = %f Amp",i);
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//4.16 clc; Vav=250; V=150; Toff=1*10^-3; Ton=(Vav/V)*Toff-Toff; printf("Period of conduction = %.6f sec", Ton)
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// Copyright (C) 2015 - IIT Bombay - FOSSEE // // This file must be used under the terms of the CeCILL. // This source file is licensed as described in the file COPYING, which // you should have received as part of this distribution. The terms // are also available at // http://www.cecill.info/licences/Licence_CeCILL_V2-en.txt // Author:Gursimar Singh & Suraj Prakash // Organization: FOSSEE, IIT Bombay // Email: toolbox@scilab.in function bboxes = peopleDetector(image, varargin) // Detects people in an image // // Calling Sequence // [bboxes] = peopleDetector(image) // [bboxes] = peopleDetector(image, ["Parameter1", value1, ["Parameter2", value2 ... ]]) // // Parameters // image : input image // hitThreshold : threshold for distance between features and SVM classifying plane. Default value is 0. // winStride : Window stride. Multiple of block stride. Default value (8, 8) // padding : Default value (16, 16). // scale : Coefficient of the detection window increase. Default value 1.05 // finalThreshold : Coefficient to regulate the similarity threshold. Some people may cover more boxex. To regulate that it is used. Default value 2. // useMeanshiftGrouping : Default value false. // bboxes : M * 4 matrix denoting M bounding boxes for detected people // // Description // The peopleDetector function detects standing people in the image using the Histogram of Oriented Gradient (HOG) features and a trained Support Vector Machine // (SVM) classifier. It detects standing peole in the image. It returns M * 4 matrix having M detected people in the image. Each row of matrix contains // [x, y, width, height] field. x, y represent the upper left corner point of the bounding box. width and height represent the width and height of the bounding box // respectively. // // Examples // I = imread('images/peopletest.jpg'); // [bboxes] = peopleDetector(I); //sz=size(x); //for i=1:sz(1) // im=rectangle(im,x(i,1),x(i,2),x(i)+x(i,3),x(i,2)+x(i,4),0,255,0,1,4,0); //end //imshow(im); // // Examples // I = imread('images/peopletest2.jpeg'); // [bboxes] = peopleDetector(I, "scale", 1.02); //sz=size(x); //for i=1:sz(1) // im=rectangle(im,x(i,1),x(i,2),x(i)+x(i,3),x(i,2)+x(i,4),0,255,0,1,4,0); //end //imshow(im); // // Authors // Suraj Prakash // Gursimar Singh // //See also //rectangle //facePredict [ lhs rhs ] = argn(0) if lhs > 1 then error(msprintf("Too many output arguments")) end if rhs > 13 then error(msprintf("Too many input arguments")) end if modulo(rhs, 2) <> 1 then error(msprintf("Wrong number of input arguments")) end image_list = mattolist(image) msprintf("Too many input arguments"); bboxes=[]; if rhs == 1 then bboxes = raw_peopleDetector(image_list); elseif rhs == 3 then bboxes = raw_peopleDetecotr(image_list, varargin(1), varargin(2)); elseif rhs == 5 then bboxes = raw_peopleDetector(image_list, varargin(1), varargin(2), varargin(3), varargin(4)); elseif rhs == 7 then bboxes = raw_peopleDetector(image_list, varargin(1), varargin(2), varargin(3), varargin(4), varargin(5), varargin(6)); elseif rhs == 9 then bboxes = raw_peopleDetector(image_list, varargin(1), varargin(2), varargin(3), varargin(4), varargin(5), varargin(6), varagin(7), varargin(8)); elseif rhs == 11 then bboxes = raw_peopleDetector(image_list, varargin(1), varargin(2), varargin(3), varargin(4), varargin(5), varargin(6), varargin(7), varargin(8), varargin(9), varargin(10)); elseif rhs == 13 then bboxes = raw_peopleDetector(image_list, varargin(1), varargin(2), varargin(3), varargin(4), varargin(5), varargin(6), varargin(7), varargin(8), varargin(9), varargin(10), varargin(11), varargin(12)); end endfunction
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//Variable declaration: MC = 2000.0 mc = 1000.0 U = 2000.0 A = 10.0 T1 = 300.0 t1 = 60.0 e = %e //Calculation: B = 1.0/mc b = 1.0/MC x = B/b y = U*(B-b) T2 = ((x-y)*T1 + x*(e-y)*t1)/(2*e-1) t2 = t1+(T1-T2)/x //Result: printf("T2 = : %.0f ",T2) printf("t2 = : %.0f ",t2)
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//Chapter-4,Example 11,Page 96 clc; close; R= 8.31 //gas constant T= 273+25 // temperature in Kelvin P1= 2 //pressure in atm P2= 1 //pressure in atm w= 2.303 *R*T*log10(P1/P2) //maximum work printf('maximum work done is %.f J', w) //mistake in textbook
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//Example 2_10 clc(); clear; //To calculate the width of the central maxima d=2 //units in meters lemda=500*10^-9 //units in meters a=1.5*10^-3 //units in meters x=((2*d*lemda)/a)*10^3 printf("width of central maximum is %.2f mm",x)
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clear() x0= input("valor de x0: ") x1= input("valor de x1: ") function y=f(x) y=(x^7)-1000 endfunction x2 = 0 for i = 0:12 if x1 == x0 then disp("el valor de x1 y x2 son iguales :" ) abort else x2 = x1 - (f(x1)/((f(x1)-f(x0))/(x1-x0))) disp(x2,"el valor de"+ string(i)+ " :" ) x0 = x1 x1 = x2 end end
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//QUESTION 1 //GRAM - SCHMIDT ORTHOGONALIZATION IN R3 clc;clear;close A=input("enter 3x3 matrix") disp(A,'A='); [m,n]=size(A); for k=1:n V(:,k)=A(:,k); for j=1:k-1 R(j,k)=V(:,j)'*A(:,k); V(:,k)=V(:,k)-R(j,k)*V(:,j); end R(k,k)=norm(V(:,k)); V(:,k)=V(:,k)/R(k,k); end disp(V,'Q='); //QUESTION 2 //EIGEN VALUES AND EIGEN VECTORS FOR 3x3 MATRIX clc; close(); clear; A =input("enter 3x3 matrix") lam = poly(0,'lam') lam = lam charMat = A-lam*eye(3,3) disp(charMat,'The characteristic Matrix is') charPoly = poly(A,'lam') disp(charPoly,'The characteristic polynomial is') lam = spec(A) disp(lam,'The eigen values of A are') function[x,lam] = eigenvectors(A) [n,m] = size(A); lam = spec(A)'; x = []; for k=1:3 b = A-lam(k)*eye(3,3); //characteristic matrix c = b(1:n-1,1:n-1); //coeff mat for the reduced system b1 =-b(1:n-1,n); //rhs vector for the reduced system y = c\b1; //solution for the reduced system y = [y:1]; //complete eigen vector y = y/norm(y); //make unit eigen vector x = [x y]; end endfunction get f('eigenvectors') [x,lam] = eigenvectors(A) disp(x,'The eigen vectors of A are') //QUESTION 3 //NUMERICALLY LARGEST EIGEN VALUE USING RAYLEIGH POWER METHOD clear; clc; close(); a = input("enter 3x3 matrix") disp(a,'A = ') //initial vector u0 = [1 1 1]'; disp(u0,'The initial vector is') v = a*u0 a1 = max(u0) disp(a,'First approximation to eigen value is ') while abs(max(v)-a1)>0.002 disp(v,'Current eigen vector is ') a1 = max(v) disp(a1,'Current eigen value is ') u0 = v/max(v) v = a*u0 end format('v',4) disp(max(v),'The largest eigen value is: ') format('v',5) disp(u0,'The corresponding eigen vector is: ')
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disp("σ=e*n*μe+e*p*μh=e*ni*(μe+μh)"); b=1350; //say b=μe c=450; //say c=μh e=1.6*10^-19; ni=1.45*10^10; a=e*ni*(b+c); printf('\n The value of σ is %1.2f*10^-6',a*10^6); L=1;A=1; g=4.18*10^-6; //rounding off σ R=L/(g*A); printf('\n The value of R is %f',R); Nsi=5*10^22; Nd=Nsi/10^9; n=5*10^13; h=5*10^13; printf('\n The value of Nd is %f',Nd); p=(ni^2)/Nd; printf('\n Tha value of p is %f',p); a1=e*n*b; //say σ=a1 printf('\n The value of σ1 is %f/ohm/cm',a1); R1=L/(a1*A); printf('\n The value of R1 is %f ohm',R1); a2=e*h*c; //say a2=σ printf('\n The value of σ2 is %f/ohm/cm',a1); R2=L/(a2*A); printf('\n The value of R2 is %f ohm',R2);
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//Example 4.13 clc; clear; close; format('v',4); //Given data : S1=1.25;//sp. gravity S2=1.05;//sp. gravity S3=0.79;//sp. gravity h=30/1000;//m w=1000;//kg/m^3 //pA=pB h=(0.15*w*S2-S1*w*0.15)/(S3*w-w*S2);//m h=h*1000;//mm disp(h,"Reading of manometer in mm : ");
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//ex5.5 R1=68*10^3; R2=47*10^3; R_C=1.8*10^3; R_E=2.2*10^3; V_CC=-6; V_BE=0.7; B_DC=75; R_IN_base=B_DC*R_E; disp('input resistance as seen from base is not greater than 10 times R2 so it should be taken into account') //R_IN_base in parallel with R2 V_B=((R2*R_IN_base)/(R2+R_IN_base)/(R1+(R2*R_IN_base)/(R2+R_IN_base)))*V_CC; V_E=V_B+V_BE; I_E=V_E/R_E; I_C=I_E; V_C=V_CC-I_C*R_C; V_CE=V_C-V_E; disp(I_C,'collector current in amperes') disp(V_CE,'collector emitter voltage in volts')
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//clc; //clear; pi=22/7; j=input("CHOOSE linear,cubic or quad INTERPOLATION") select(j) case(1) disp('enter the value of w0,wt,T') w0=input('enter w0 ') wt=input(' enter wt ') T=input(' input T ') b=w0 a=(wt - w0)/(T); t=0:0.1:T w=(a.*t)+(b); figure(1) grid on; plot(t,w,'o') title('DISPLACEMENT CURVE') xlabel('time') ylabel('velocity') t=0:0.1:T a1=0; figure (3) grid on title('ACCELERATION CURVE ') xlabel('time') ylabel('acceleration') case(2) disp('enter the value of w0, wt, T') w0=input('enter w0') wt=input('enter wt') T= input ('input T') c=w0 a=((wt-c)/T.^2) b=0 t=0:0.1:T w=(a.*t.*t)+(b.*t)+(c); figure (1) grid on; plot (t,w,'o') title ('DISPLACEMENT CURVE') xlabel('time') ylabel('position in degrees') t=0:0.1:T v=(2*a.*t)+(b); figure(2) grid on; plot(t,v,'o') title('VELOCITY CURVE ') xlabel('time') ylabel (' velocity') t=0:0.1:T a1=(2*a); figure (3) grid on ; plot (t,a1,'o') title ('ACCELERATION CURVE ') xlabel ('time') ylabel ('acceleration') plot(t,a1,'o'); case(3) disp ('enter the value of w0,wt,T') w0=input('enter w0') wt=input('enter wt') T=input('input T') D=w0 C=0 a=((2*d-2*wt)/T.^3) b=((3*(wt-d))/(T.^2)) t=0:0.1:T w=(a.*t.*t.*t)+(b.*t.*t)+(c.*t)+d; figure (1) grid on; plot(t,w,'o') title ('DISPLACEMENT CURVE ') xlabel('time') ylabel('position in degrees') T=0:0.1:T v=(3*a.*t.*t)+(2*b.*t)+(c); figure (2) grid on; plot (t,v,'o') title ('VELOCITY CURVE') xlabel('time') ylabel('velocity') t=0:0.1:T a1=(6*a.*t)+(2*b); figure(3) grid on; plot (t,a1,'o') title('ACCELERATION CURVE ') xlabel ('time') ylabel ('acceleration ') end
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clear; clc; close; disp("Example 10.3") M0=0 po=101.33 //in kPa T0=288.2 gmc=1.4 Cpc=1004 pd=0.95 pc=20 ec=0.9 mc2=33 Nc2=7120 Mz2=0.6 Qr=428000000 pb=0.98 eb=0.97 Tt4=1850 gmt=1.33 Cpt=1156 et=0.8 em=0.995 QrAB=4280000 pAB=0.95 eAB=0.98 Tt7=2450 pAB=1.3 CpcAB=1243 pn=0.93 p=1 //p=p9/p0 Mo0=2 po0=20 To0=223 gm0=1.4 Cpc0=1004 pdo=0.8 ec0=0.9 Qr=42800000 pb0=0.98 ebo=0.97 Tt4o=1850 gmto=1.33 cpto=1156 eto=0.8 emo=0.995 QrABo=42800000 pABo=0.95 eab=0.98 Tt7o=2450 gmABo=1.3 Cpco=1243 pno=0.93 po=1 a0=276.4 Tt2=T0 tc=pc^((gmc-1)/(ec*gmc)) Tt3=tc*Tt2 f=(Cpt*Tt4-Cpc*Tt3)/(Qr*eb-Cpt*Tt4) tt=1-(1/((1+f)*em))*(Cpc*Tt2/(Cpt*Tt4))*(tc-1) disp(tt,"Turbine expansion parameter at on and off design :") //Off-design analysis: Tt2o=To0*(1+(gmc-1)/2*(Mo0^2)) tcOD=1+(1.036)*0.995*(1156*1850/(1004*401.4))*(1-0.7915) pcOD=tcOD^((gmc)*ec/((gmc-1))) disp(pcOD,"New compressor pressure ratio :") mc2D=pcOD/pc*((Tt4o/Tt2)/(Tt4o/Tt2o))^(1/2) mc2OD=mc2*mc2D disp(mc2OD,"Off-line mc2 rate in Kg/s :") Nc2r=((Tt4o/Tt2o)/(Tt4/Tt2))^(1/2) Nc2OD=Nc2r*Nc2 disp(Nc2OD,"Off-design Nc2,O-D in rpm:") pref=101.33 //in kPa pt0=po0*(1+(gmc-1)/2*Mo0^2)^((gmc)/(gmc-1)) pt2=pdo*pt0 del2=pt2/pref Tref=288.2 the2=Tt2o/Tref m2=mc2OD*del2/(the2)^(1/2) disp(m2,"Off-design mass flow in kg/s") Tt3=859.2 Tt4=1850 fOD=0.03305 tcr=(1+fOD)/(1+f) pt5=413.7// kPa pt7=393.04 fAB=0.0367 pt9=365.52 M9=2.524 T9=1253 V9=1725 ndst=(1+f+fAB)*V9/a0-M9 disp(ndst,"Nondimensional specific thrust :") TSFC=55.94 //in mg/s/N disp(TSFC,"Thrust specific fuel consumption(TSFC) in mg/s/N :")
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errcatch(-1,"stop");mode(2);//page 238 ; ; disp('The eigen values of a projection matrix are 1 or 0.') P=[1/2 1/2;1/2 1/2]; eig=spec(P); [V,Val]=spec(P); disp(eig,'Eigen values:') x1=V(:,1); x2=V(:,2); disp(x1,x2,'Eigen vectors:'); //end exit();
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//Caption: calculate the total torque in Nw-m //Exam:2.14 clc; clear; close; P=6;//poles A=P;//number of parallel paths S=60;//slots in motor C_s=12;//conductor per slot Z=S*C_s;//total conductor in machine I_a=50;//armature current(in Amp) F_1=20//flux per pole(in m Wb) F_2=F_1*10^-3;//flux per pole)(in Wb) T=0.15924*F_2*Z*P*I_a/A;//total torque (in Nw-m) disp(T,'total torque by motor (in Nw-m)=');
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clc //initialisation of variables m= 14 //gms M= 28 //gms S= 6.94 //cal per mole T= 127 //C T1= 27 //C S1= 4.94 //cal per mole //CALCULATIONS dS= (m/M)*S*log((273+T)/(273+T1)) dS1= (m/M)*S1*log((273+T)/(273+T1)) //RESULTS printf (' Entropy change = %.2f E.U',dS-0.01) printf (' \n Entropy change = %.2f E.U',dS1)
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clc clear printf("Example 12.8 | Page number 425 \n\n"); //Find actual air and excess air //Given Data xCO2 = 9.7 //mole percent CO2 xCO = 1.1 //mole percent CO xO2 = 4 //mole percent O2 xN2 = 85.2 //mole percent N2 //Solution //by C balance b = 2 //by H2 balance d = 2 //by O2 balance a = b+d*.5 //by N2 balance c = 3.76*a Stoichiometric_air = a*(32+3.76*28)/28 //kg/kg ethylene //by C balance x = (xCO2+xCO)/2 //kmol of ehtylene be burnt //by H2 balance q = 2*x //by O2 balance p = xCO2 + xCO/2 + xO2 + q/2 actual_air = p*(32+3.76*28)/(x*28) //kg/kg ethylene excess_air = (actual_air - Stoichiometric_air)/Stoichiometric_air*100 printf("Actual air = %.1f kg/kg ethylene \n",actual_air) printf("Excess air = %.1f%%",excess_air)
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//Chapter 11 Thermodynamics Some Basic Concepts clc; clear; //Initialisation of Variables m= 9 //gms T= -10 //C T1= 0 //C R= 0.5 //cal per gram per degree H= 79.7 //cal per gram R1= 1 //cal per gram per degree T2= 100 //C H1= 539.7 //cal per gm R2= 8.11 //cal per gram per degree M= 18 //gms T3= 40 //C //CALCULATIONS dH= m*R*(T1-T) dH1= m*H dH2= m*R1*(T2-T1) dH3= m*H1 dH4= (m/M)*R2*(T3-T1) dH5= dH+dH1+dH2+dH3+dH4 //RESULTS mprintf("Value of dH= %.1f cal",dH5)
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//A Textbook of Chemical Engineering Thermodynamics //Chapter 1 //Introduction and Basic Concepts //Example 5 clear; clc; //Given: A = (%pi/4)*(0.1^2); //area in m^2 P = 1.01325*10^5; //pressure in N/m^2 m = 50; //mass of piston and weight in kg g = 9.81; //acceleration due to gravity (N/m^2) //To determine the force exerted pressure work done and change in potential energy //(a) Fa = P*A; //force exerted by atmosphere in N Fp = m*g; //force exerted by piston and weight in N F = Fp+Fa; //total force exerted in N mprintf('Total force exerted by the atmosphere, the piston and the weight is %f N',F); //(b) Pg = F/A; //pressure of gas in N/m^2 mprintf('\nPressure of gas is %5.4e Pa',Pg); //(c) S = 0.4; //displacement of gas in m W = F*S; //work done by gas in J mprintf('\nWork done by gas is %f J',W); //(d) PE = m*g*S; //change in potential energy in J mprintf('\nChange in potential energy is %f J',PE); //end
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//Ex6_5 clc VBB = 1 VCC = 12 IC = 12*10^-3 VCE = 4 beta = 80 VBE = 0.7 disp("VBB = "+string(VBB)+"V")//base supply voltage disp("VCC = "+string(VCC)+"V")//collector supply voltage disp("IC = "+string(IC)+"A")//collector current disp("VCE = "+string(VCE)+"V")//voltage across collector and emitter disp("beta = "+string(beta))//current gain disp("VBE = "+string(VBE)+"V")//voltage across base and emitter IB = IC/beta disp("IB = IC/beta = "+string(IB)+"A")//base current RC = (VCC - VCE)/IC disp("RC = (VCC - VCE)/IC = "+string(int(RC))+"ohm")//collector resistance
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clc //initialisation of variables clear T= 25 //C T1= 0 //C h= 79.8 //cal g^-1 j= 4.18*10^7 //ergs //CALCULATIONS Wc= (T-T1)*h/(273+T1) W= (T-T1)*h*j/(273+T1) //RESULTS printf ('Work required = %.1f cal',Wc) printf ('\n Work required = %.2e ergs',W)
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//Exa 2.13 clc; clear; close; //Given data Beta=100;//unitless IC=1;//in mA VCC=12;//in volt VBE=0.3;//in volt(For Ge) //Prt (i) IB=IC/Beta;//in mA //Formula : VCC=VBE+IB*RB RB=(VCC-VBE)/(IB*10^-3);//in Ampere disp(RB/10^3,"Resistance RB in kOhm : "); //part (ii) Beta=50;//unitless IB=(VCC-VBE)/RB;//in Ampere IC=Beta*IB;//in Ampere disp(IC*10^3,"Zero signal IC in mA:");
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@relation led7digit @attribute Led1 real[0.0,1.0] @attribute Led2 real[0.0,1.0] @attribute Led3 real[0.0,1.0] @attribute Led4 real[0.0,1.0] @attribute Led5 real[0.0,1.0] @attribute Led6 real[0.0,1.0] @attribute Led7 real[0.0,1.0] @attribute number{0,1,2,3,4,5,6,7,8,9} @inputs Led1, Led2, Led3, Led4, Led5, Led6, Led7 @outputs number 1 1 3 3 3 3 8 2 9 5 5 6 6 5 7 7 2 2 2 3 3 3 4 5 5 5 5 5 6 6 6 6 7 1 8 2 9 5 9 7 9 3 3 3 7 3 0 8 2 2 2 3 2 2 3 3 3 3 4 4 8 3 0 3 1 1 8 3 0 8 4 4 5 4 5 5 7 5 7 7 7 7 8 6 8 8 2 2 4 4 5 5 0 2 1 1 5 5 7 7
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//Example2.17 T1=523 //[K] T2=323 //[K] r1=0.05 //[m] r2=0.055 //[m] r3=0.105 //[m] r4=0.155 //[m] k1=50 //[W/(m.K)] k2=0.06 //[W/(m.K)] k3=0.12 //W/(m.K) //CASE 1 Q_by_L1=2*%pi*(T1-T2)/((log(r2/r1))/k1+(log(r3/r2))/k2+(log(r4/r3))/k3) //[W/m] printf("Heat loss=%f W/m",Q_by_L1) //Case 2 Q_by_L2=2*%pi*(T1-T2)/((log(r2/r1))/k1+(log(r3/r2))/k3+(log(r4/r3))/k2) perct=(Q_by_L2-Q_by_L1)*100/Q_by_L1 printf("If order is changed then heat loss=%f W/m",Q_by_L2) printf("\n loss of heat is increased by %f percent by putting material with higher thermal conductivity near the pipe surface",perct)
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A1 Abou Jaoude Yann TD3.sce
//EXERCICE I function derivey=y1(t,y), derivey=-y+t+1; endfunction function solution=sy1(t) solution = t+exp(-t); endfunction function [t, y] = Euler(a,b, N, alpha, f) h = (b-a)/N; t(1) = a; y(1) = alpha; i = 1; for i=1:N y(i+1)= y(i)+h*f(t(i), y(i) ); t(i+1)= t(i) + h; end endfunction function [t,y] = heun(a,b,n,alpha,f) h = (b - a) / n; halfh = h / 2; y(1) = alpha; t(1) = a; for i = 1 : n t(i+1) = t(i) + h; g = f(t(i),y(i)); z = y(i) + h * g; y(i+1) = y(i) + halfh * ( g + f(t(i+1),z) ); end; endfunction function [t,y] = rk4(a,b,n,alpha,f) h = (b - a) / n; halfh = h / 2; y(1) = alpha; t(1) = a; h6 = h/6; for i = 1 : n t(i+1) = t(i) + h; th2 = t(i) + halfh; s1 = f(t(i), y(i)); s2 = f(th2, y(i) + halfh * s1); s3 = f(th2, y(i) + halfh * s2); s4 = f(t(i+1), y(i) + h * s3); y(i+1) = y(i) + (s1 + s2+s2 + s3+s3 + s4) * h6; end; endfunction Alpha=1; A=0 B=5 //N=10 //N=100 N=1000 //Quand nous augmentons le n 10X, nous reduisonsl'erreur environ 100X [t,y2]=Euler(A,B,N,Alpha,y1) [t,y3]=heun(A,B,N,Alpha,y1); [t,y4]=rk4(A,B,N,Alpha,y1); sol(1)=sy1(t(1)); for i = 1:N+1 sol(i) = sy1(t(i)); end subplot(121); plot(t,sol,'k-'); subplot(121); plot(t,y2,'b-'); subplot(121); plot(t,y3,'g-'); subplot(121); plot(t,y4,'r-'); subplot(122); erreur = abs(y2-sol); plot(t,log(erreur),'b-'); subplot(122); erreur = abs(y3-sol); plot(t,log(erreur),'g-'); subplot(122); erreur = abs(y4-sol); plot(t,log(erreur),'r-'); legend(['Log erreur euler','Log erreur heun','Log erreur rk4'],pos = "2");
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//Chapter-4,Example4_15_15,pg 4-34 //for case1 t1=2*10^-3 //thicknesss of plate d=2.65*10^3 //density Y=8*10^10 //Young's modulus k=1 //consider 1st harmonic n1=(k/(2*t1))*sqrt(Y/d) //formula of natural frequency printf(" 1)natural frequency =") disp(n1) printf("Hz") //for case2 n2=3*10^6 //frequency t2=(k/(2*n2))*sqrt(Y/d) //arranging formula of natural frequency t=t1-t2 //change in thickness printf(" 2)change in thickness =") disp(t) printf("meter")
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PL/SQL Developer Test script 3.0 8 -- Created on 15.11.2017 by V.ZHURAVOV declare -- Local variables here i integer; begin -- Test statements here dv_sr_lspv_docs_api.build_tax_diff; end; 0 0
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/* Compare performance of repeated querying of data to caching in the PGA (packaged collection) and the new Oracle 11g Result Cache. To compile and run this test script, you will first need to run the following script. Note that to compile the my_session package and display PGA usage statistics, you will need SELECT authority on: sys.v_$session sys.v_$sesstat sys.v_$statname Author: Steven Feuerstein */ @@plvtmr.pkg @@mysess.pkg @@11g_emplu.pkg @@11g_emplu_compare.sp SET SERVEROUTPUT ON BEGIN test_emplu (100000); /* With 100000 iterations: PGA before tests are run: session PGA: 2057168 Execute query each time Elapsed: 5.65 seconds. Factored: .00006 seconds. session PGA: 1139664 Oracle 11g result cache Elapsed: .3 seconds. Factored: 0 seconds. session PGA: 1139664 Cache table in PGA memory Elapsed: .12 seconds. Factored: 0 seconds. session PGA: 1336272 */ END; / /*====================================================================== | Supplement to the fifth edition of Oracle PL/SQL Programming by Steven | Feuerstein with Bill Pribyl, Copyright (c) 1997-2009 O'Reilly Media, Inc. | To submit corrections or find more code samples visit | http://oreilly.com/catalog/9780596514464/ */
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function image_traitee = appliquerFiltre(image, filtre, coef, nom) afficherLogs("Application d''un filtre : "+ nom); [N, M] = size(image) [NMasque, MMasque]= size(filtre) NMasque_moitie = floor(NMasque/2); MMasque_moitie = floor(MMasque/2); // Matrice temporaire dans laquelle nous recopions la matrice de l'image de base // + élargissement de l'image pour pouvoir appliquer le filtre // Les contours seront initialisés par défaut à 0 xtemp = 2*NMasque_moitie + N; ytemp = 2*MMasque_moitie + M; temp = zeros(xtemp,ytemp); [Ntemp, Mtemp] = size(temp); temp(NMasque_moitie + 1 : Ntemp - NMasque_moitie, MMasque_moitie + 1 : Mtemp - MMasque_moitie) = image; // disp(temp) image_traitee = zeros(N,M) for x = 1+NMasque_moitie : N+NMasque_moitie for y = 1+MMasque_moitie : M+MMasque_moitie somme_px = 0; for i = 1 : NMasque for j = 1 : MMasque x1 = x+i-NMasque_moitie-1; y1 = y+j-MMasque_moitie-1; px_img = temp(x1, y1); px_filtre = filtre(i,j); // Gros bug ici car j'avais mis y1 = y+i-MMasque_moitie-1 au lieu de j // disp("x:"+ string(x)+ " ; y:"+ string(y)+ " | i:"+ string(i)+ " ; j:"+ string(j)) // disp("Temp x: "+string(x1)) // disp("Temp y: "+string(y1)) somme_px = somme_px + px_img * px_filtre; end end if coef ~= 0 then image_traitee(x-NMasque_moitie, y-MMasque_moitie) = floor(somme_px/coef); else image_traitee(x-NMasque_moitie, y-MMasque_moitie) = floor(somme_px); end end end // On affiche la nouvelle matrice après l'application du filtre afficherLogs(image_traitee) endfunction
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function [z] = pointmilieu(a, t0, T, N, f) h = T/N; t = t0 + h*[0:N-1]'; zn = a; z = [zn]; for i = 0:N-1 K0 = f(t(i+1),zn); K1 = f(t(i+1)+h/2,zn+h/2*K0); zn = zn + h*K1; z = [z,zn]; end endfunction
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instrinsicMatrix = [1 0 0; 1 2 0; 3 4 0]; rotationMatrix = [ 0.1417 -0.7409 0.6565; 0.9661 -0.0410 -0.2548; 0.2157 0.6703 0.7100]; translationVector = [ -29.2584 35.7824 725.5824]; camMatrix = cameraMatrix(instrinsicMatrix,rotationMatrix,translationVector) I = imread('Barrel-Distortion.jpg'); J = undistortImage(I, camMatrix);
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// calculate voltage gain // Electronic Principles // By Albert Malvino , David Bates // Seventh Edition // The McGraw-Hill Companies // Example 16-3, page 569 clear; clc; close; // Given data f=10;// cutoff frequency in hertz Avmid=100000;// midband voltage gain f1=100;// input frequency in hertz f2=10^3;// input frequency in hertz f3=10^4;// input frequency in hertz f4=10^5;// input frequency in hertz f5=10^6;// input frequency in hertz // Calculations Av1=Avmid/((1+(f1/f)^2)^0.5)// Voltage gain for input frequency below midband Av2=Avmid/((1+(f2/f)^2)^0.5)// Voltage gain for input frequency below midband Av3=Avmid/((1+(f3/f)^2)^0.5)// Voltage gain for input frequency below midband Av4=Avmid/((1+(f4/f)^2)^0.5)// Voltage gain for input frequency below midband Av5=Avmid/((1+(f5/f)^2)^0.5)// Voltage gain for input frequency below midband disp(Av1,"Voltage gain 1=") disp(Av2,"Voltage gain 2=") disp(Av3,"Voltage gain 3=") disp(Av4,"Voltage gain 4=") disp(Av5,"Voltage gain 5=") // Result // Voltage gain for an input frequency of 100 Hertz is approximately 10000 // Voltage gain for an input frequency of 1000 Hertz is approximately 1000 // Voltage gain for an input frequency of 1000 Hertz is approximately 100 // Voltage gain for an input frequency of 10000Hertz is approximately 10 // Voltage gain for an input frequency of 100000 Hertz is approximately 1
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//<f>=%rap(f,m) // %rap(f,m) calcule la somme d'une matrice de fractions rationnelles f et //d'une matrice de polynomes p //! num=f(2)+m.*f(3) [num,den]=simp(num,f(3)) f(2)=num;f(3)=den //end
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Name=Splitgatee Pistol training PlayerCharacters=Clay Pigeon BotCharacters=Long Strafe Bot.bot IsChallenge=true Timelimit=90.0 PlayerProfile=Clay Pigeon AddedBots=Long Strafe Bot.bot PlayerMaxLives=0 BotMaxLives=0 PlayerTeam=1 BotTeams=2 MapName=kovaim1.map MapScale=3.8125 BlockProjectilePredictors=true BlockCheats=true InvinciblePlayer=false InvincibleBots=false Timescale=1.0 BlockHealthbars=false TimeRefilledByKill=0.0 ScoreToWin=1000.0 ScorePerDamage=1.0 ScorePerKill=0.0 ScorePerMidairDirect=0.0 ScorePerAnyDirect=0.0 ScorePerTime=0.0 ScoreLossPerDamageTaken=0.0 ScoreLossPerDeath=0.0 ScoreLossPerMidairDirected=0.0 ScoreLossPerAnyDirected=0.0 ScoreMultAccuracy=false ScoreMultDamageEfficiency=true ScoreMultKillEfficiency=false GameTag=Tracking WeaponHeroTag=Pistol DifficultyTag=2 AuthorsTag=Klastitis BlockHitMarkers=false BlockHitSounds=false BlockMissSounds=true BlockFCT=true Description=Shoot the target with your pistol GameVersion=1.0.7.2 ScorePerDistance=0.0 [Aim Profile] Name=At Feet MinReactionTime=0.3 MaxReactionTime=0.4 MinSelfMovementCorrectionTime=0.001 MaxSelfMovementCorrectionTime=0.05 FlickFOV=30.0 FlickSpeed=1.5 FlickError=15.0 TrackSpeed=3.5 TrackError=3.5 MaxTurnAngleFromPadCenter=75.0 MinRecenterTime=0.3 MaxRecenterTime=0.5 OptimalAimFOV=30.0 OuterAimPenalty=1.0 MaxError=40.0 ShootFOV=15.0 VerticalAimOffset=-200.0 MaxTolerableSpread=5.0 MinTolerableSpread=1.0 TolerableSpreadDist=2000.0 MaxSpreadDistFactor=2.0 [Aim Profile] Name=Low Skill At Feet MinReactionTime=0.35 MaxReactionTime=0.45 MinSelfMovementCorrectionTime=0.001 MaxSelfMovementCorrectionTime=0.05 FlickFOV=30.0 FlickSpeed=1.5 FlickError=20.0 TrackSpeed=3.0 TrackError=5.0 MaxTurnAngleFromPadCenter=75.0 MinRecenterTime=0.3 MaxRecenterTime=0.5 OptimalAimFOV=30.0 OuterAimPenalty=1.0 MaxError=60.0 ShootFOV=25.0 VerticalAimOffset=-200.0 MaxTolerableSpread=5.0 MinTolerableSpread=1.0 TolerableSpreadDist=2000.0 MaxSpreadDistFactor=2.0 [Aim Profile] Name=Low Skill MinReactionTime=0.35 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x=%e*%pi,y=%i, // two variables save('myvar.sod','x','y') // saving to the file myvar.sod ls('*.sod') // file created in the current directory clear x // erase x y=y+1 // modify y load('myvar.sod') // reload x and y x,y
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clc; close; clear; x=poly(0,'x'); p=(x+9)*(x-9) p1=string('(ab+10)(ab-10)=a^2*b^2-100') p2=(4*x+5)*(4*x-5) p3=(1+x)*(1-x) a=poly(0,'a'); p4=(a-1/2)*(a+1/2) p5=string('(a/3 + b/4)*(a/3 - b/4)=a^2/9 - b^2/16') p6=string('{(a+b)+c}*{(a+b)-c}=(a+b)^2-c^2') p7=string('{a+(b+c)}*{a-(b+c)}=a^2-(b+c)^2')
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Ex6_1.sce
// Variable declaration n = 10 // Sample size N = 1000 // population size // Calculation // as we know correction factor = (N-n)/(N-1) corr_fact = ((N-n))/(N-1) // correction factor // Result printf ( "Correction Factor: %.3f",corr_fact)
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//Example 7.3.a: terminal voltage clc; clear; close; // given data: W=10;// output of the generator in k-w V=250; // voltage in volts R=0.07; // in ohm Il=(W*1000)/V;// load current in A Vf=Il*R;// voltage drop in feeder Vt=V+Vf; disp(Vt,"terminal voltage,Vt(V) = ")
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main { int a; a := 2; if (5 < a) then { a := 1; } else { a := 0; } fi print(a); return a; }
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//scilab 5.4.1 //Windows 7 operating system //chapter 9 Basic Voltage and Power Amplifiers clc clear //For two identical transistors employed by an RC-coupled amplifier hfe=100//hfe=current gain hie=2*10^3//hie=input impedance in ohm Cob=2*10^-12//Cob=capacitance in farad quoted by the transistor manufacturers C=0.4*10^-6//C=coupling capacitance in farad RL=8*10^3//RL=load resistance in ohms for each transistor CW=10*10^-12//CW=wiring capacitance in farad fl=1/(2*%pi*C*(hie+RL))//fl=lower half power frequency format("v",5) disp("Hz",fl,"The lower half-power frequency is =") hfb=-hfe/(1+hfe)//hfb=current gain for common base transistor Coc=Cob/(1+hfb)//Coc=transistor collector capacitance in farad Cs=Coc+CW//Cs=shunt capacitance in farad Ro=(hie*RL)/(hie+RL)//Ro=equivalent resistance of the parallel combination of hie and RL fh=1/(2*%pi*Cs*Ro)//fh=upper half power frequency format("v",5) disp("kHz",fh/10^3,"The upper half-power frequency is =")//fh is converted in terms of kHz
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sce
Ex9_2.sce
clear; clc; funcprot(0); //given data P = 4.0;//in MW N = 375;//in rev/min H_eps = 200;//in m KN = 0.98;//nozzle velocity coefficient d = 1.5;//in m k = 0.15;//decrease in relative flow velocity across the buckets alpha = 165;//in deg g = 9.81;//in m/s^2 rho = 1000;//in kg/m^3 //Calculations U = N*%pi*d*0.5/30; c1 = KN*sqrt(2*g*H_eps); nu = U/c1; eff = 2*nu*(1-nu)*(1-(1-k)*cos(alpha*%pi/180)); Q = (P*10^6 /eff)/(rho*g*H_eps); Aj = Q/(2*c1); dj = sqrt(4*Aj/%pi); omega_sp = (N*%pi/30)*sqrt((P*10^6)/rho)/((g*H_eps)^(5/4)); //Results printf('(i)The runner efficiency = %.4f',eff); printf('\n (ii)The diameter of each jet = %.4f m',dj); printf('\n (iii)The power specific speed = %.3f rad',omega_sp);
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no_license
grenkin/compiler
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2020-06-20T12:44:17.903582
2016-11-27T03:08:20
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49.tst
int main(int argc, char* argv[]) { int x; x++; return 0; }
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/3809/CH4/EX4.6/EX4_6.sce
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[]
no_license
FOSSEE/Scilab-TBC-Uploads
948e5d1126d46bdd2f89a44c54ba62b0f0a1f5e1
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refs/heads/master
2020-04-09T02:43:26.499817
2018-02-03T05:31:52
2018-02-03T05:31:52
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EX4_6.sce
//Chapter 4, Example 4.6 clc //Initialisation' c1=10*10**-6 //capacitance in farad c2=25*10**-6 //capacitance in farad //Calculation c=(c1*c2)/(c1+c2) //equivalent parallel capacitance in farad //Results printf("Total Capacitance, C = %.2f uF",c*10**6)
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089894a36ef33cb3d0f697541716c9b6cd8dcc43
/NLP_Project/test/tweet/bow/bow.20_11.tst
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no_license
mandar15/NLP_Project
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2020-05-20T13:36:05.842840
2013-07-31T06:53:59
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tst
bow.20_11.tst
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// ******************************************************************* // Lagrange Interpolation for the given data: **** // (x1, f1), (x2, f2), (x3, f3), ......(xn, fn) **** // NPDE-TCA UG-Level workshop at IMA Bhubaneshwar **** // By Manas,FOSSEE,IITB **** //******************************************************************** function y = Lagrange(x0, x,f, n) m = n + 1; N = ones(1,m); D = N; C = N; y = 0.0; for j = 1:m for k = 1:m if (k<>j) then N(j) = N(j)*(x0 - x(k)) D(j) = D(j)*(x(j) - x(k)) end end L(j) = N(j)/D(j); y = y + L(j)*f(j); end disp(L','L') disp(f,'f(x)') endfunction
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// Display mode mode(0); // Display warning for floating point exception ieee(1); //clearxdel(winsid()) clc //% Load full image disp("Loading full image...") // !! L.5: Matlab function imread not yet converted, original calling sequence used. A = imread("C:\Users\Gavin\Desktop\Code\Python\other_projects\flatten\flatten\samples\images\blue_flower_very_small","jpeg"); // !! L.6: Matlab function figure not yet converted, original calling sequence used. // L.6: (Warning name conflict: function name changed from figure to %figure). %figure(3); // !! L.7: Matlab toolbox(es) function imshow not converted, original calling sequence used imshow(A) //% Make image black and white // !! L.10: Matlab toolbox(es) function rgb2gray not converted, original calling sequence used Abw2 = rgb2gray(A); [nx,ny] = size(Abw2); // !! L.12: Matlab function figure not yet converted, original calling sequence used. // L.12: (Warning name conflict: function name changed from figure to %figure). %figure(1) subplot(2,2,1)// !! L.12: Matlab toolbox(es) function imshow not converted, original calling sequence used imshow(Abw2) title("Original image","FontSize",18) //% Compute the FFT of our image using fft2 disp("Doing FFT analysis for sparsity check...") tic; At = fft2(Abw2); F = log(mtlb_a(abs(fftshift(At)),1)); // !! L.21: Matlab toolbox(es) function mat2gray not converted, original calling sequence used F = mat2gray(F);// Use mat2gray to scale the image between 0 and 1 // !! L.22: Matlab function figure not yet converted, original calling sequence used. // L.22: (Warning name conflict: function name changed from figure to %figure). %figure(4) // !! L.23: Matlab toolbox(es) function imshow not converted, original calling sequence used imshow(F,[]);// Display the result //% Zero out all small coefficients and inverse transform disp("Zeroing out small Fourier coefficients...") tic; count_pic = 2; %v0 = abs(At);%v1 = max(%v0,firstnonsingleton(%v0)); for thresh = (0.1*[0.001,0.005,0.01])*max(%v1,firstnonsingleton(%v1)) ind = mtlb_logic(abs(At),">",thresh); count = mtlb_s(nx*ny,mtlb_sum(mtlb_sum(ind))); Atlow = At .*ind; percent = mtlb_s(100,(count/(nx*ny))*100); // !! L.34: Matlab function ifft2 not yet converted, original calling sequence used. // !! L.34: Scilab uint8() does not work with Complex values: uint8() call IGNORED. // ! L.34: ifft2(Atlow) may be replaced by: // ! --> uint8(ifft2(Atlow)) if ifft2(Atlow) is Real. Alow = ifft2(Atlow); // !! L.35: Matlab function figure not yet converted, original calling sequence used. // L.35: (Warning name conflict: function name changed from figure to %figure). %figure(1) subplot(2,2,count_pic) // !! L.35: Matlab toolbox(es) function imshow not converted, original calling sequence used imshow(Alow); count_pic = count_pic+1; // L.36: Drawing events are not queued in Scilab. //drawnow // !! L.37: string output can be different from Matlab num2str output. title(string(percent)+"% of FFT basis","FontSize",18) end; // !! L.39: string output can be different from Matlab num2str output. disp(" done. ("+string(toc())+"s)") //% // !! L.42: Matlab function figure not yet converted. mtlb(figure) // !! L.43: Matlab toolbox(es) function imresize not converted, original calling sequence used Anew = imresize(Abw2,0.1); surf(double(Anew));
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// Exa 7.15 clc; clear; close; format('v',5) // Given data R1 = 20;// in k ohm R1 = R1 * 10^3;// in ohm R2 = R1;// in ohm R = R1;// in ohm C1 = 1000;// in pF C1 = C1 * 10^-12;// in F C2 = C1;// in F C = C1;// in F f = 1/(2*%pi*R*C);// in Hz f= f*10^-3;// in kHz disp(f,"The frequency of oscillations in kHz is");
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gL=0.62425 gc=0.9662 m=3 Wc=2*%pi*100*10^6 CHP=1/(Wc*gL) LHP=1/(Wc*gc) printf("\nCHP=%.3e F\nLHP=%.3e H",CHP,LHP) C1=2.5495/75*10^3 L2=75*1.6472 printf("\nC1=C3=%.0f pF\nL2=%.1f nH",C1,L2)
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//Chapter 10, Problem 18 clc; BC=100; //resistance between point B and C DA=400; //resistance between point D and A CD=10; //resistance between point C and D Rx=BC*DA/CD; //calculating unknown resistance using balance equation printf("unknown resistance = %f K ohms",Rx/1000);
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x=[0 1 2 3 4 5]; y=[1 0 1 0 1 0]; xx=-3:.01:3; v=pchip(x, y,xx,1); disp(v); //output // !--error 58 //Wrong number of input arguments.at line 5 of exec file called by : //chip/pchip10.sce', -1
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//Finding core loss //Example 6.6(pg 214) clc clear v=76300//volume in c.c P=8// no of poles N=375//rpm f=P*N/120//freqency in c/s Bmax=12000//max. flux density in lines/cm^2 n=0.002//(assumed) d=7.8//densityin gm/c.c l=1.7//loss in watts per kg Hl=n*v*f*(Bmax^1.6)*(10^-7)//Hysteresis loss in Watts Al=v*d*l/1000//Additional loss under particular running conditions Tl=Hl+Al//total core loss printf('Thus the total core loss is %4.0f Watts',Tl)
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clc; clear; D=60;//mm pdiff=4;//kPa Q=0.003;//(m^3)/sec d=789;//kg/(m^3) vis=1.19*(10^(-3));//N*sec/(m^2) Re=d*4*Q/(%pi*D*vis); //assuming B=dia/D=0.577, where dia=diameter of nozzle, and obtaining Cn from Re as 0.972 Cn=0.972; B=0.577; dia=((4*Q/(Cn*%pi))/((2*pdiff*1000/(d*(1-(B^4))))^0.5))^0.5; disp("mm",dia*1000,"Diameter of the nozzle=")
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// 08.09.21 function Out5=Sfbdrawpersdata(varargin) global IMPLICITDATA CUSPDATA CUSPPT CUSPSPLITPT; Nargs=length(varargin); Fd=varargin(1); FdL=Fullformfunc(Fd); Np=[50,50]; if Nargs>=2 Np=varargin(2); if type(Np)==1 & length(Np)==1 Np=[Np,Np]; end; end; Eps=0.05; if Nargs>=3 Eps=varargin(3); end; Ts=timer(); [Zval,Xval,Yval]=Evlptablepers(Mix(Fd,Np)); Out3=Implicitplot(Zval,Xval,Yval); BdyL=Mixop(8,FdL); if BdyL~=[] Out3=Clipindomain(Out3,BdyL) end; IMPLICITDATA=Out3; Out4=Cuspsplitpers(Out3,Fd,Eps); CUSPDATA=Out4; CCUSPPT=CUSPSPLITPT; Out5=Borderrawdata(Out4,Fd,Np,Eps); endfunction;
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java -ea trip.Main -m trip-tests/autograder_map05 <<EOF Schenectady, Battery_Park EOF
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clc;//clears the command window clear;//clears all the variables format('v',8);//making the default precision to 8 significant digits i=1; dec=548;//given decimal number which should be expressed in base 8 temp=dec; i=1; d=8; while(temp>0)//storing each integer digit in vector for convenience p(i)=(modulo(floor(temp),d)) temp=floor(temp)/d; i=i+1; end temp2=0; for j=1:length(p) //multipliying bits of integer part with their position values and adding temp2=temp2+(p(j)*10^(j-1)); end disp(temp2,"Octal number"); dec=345; //given decimal number which should be expressed in base 8 temp=dec; i=1; d=6; while(temp>0)//storing each integer digit in vector for convenience p(i)=(modulo(floor(temp),d)) temp=floor(temp)/d; i=i+1; end temp2=0; for j=1:length(p) //multipliying bits of integer part with their position values and adding temp2=temp2+(p(j)*10^(j-1)); end disp(temp2,"Base 6");
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//Caption: Probability density function //Example 2.12 //Find find the value of k clc; clear; function y=FX1(x) //for -infinte<x<=0 y=0 endfunction function y=FX2(x) //for 0<x<=10 y=k*x^2 endfunction function y=FX3(x) //for 10<x<infine y=100*k endfunction k=poly(0,"k"); //from the expression for CDF is given y=100*k //for 10<x<infine y==1; k=1/100; //k=y/100 disp(k,"i) k = "); //CDF function can be expressed // FX(x)=P(X<=x) P5=FX2(5); //x=5 disp(P5,"ii) P(X<=5) = "); //now differentiating with respect tox we ,have //PDF fX=0 for -infinte<x<=0,10<x<infine x=poly(0,"x"); m=x^2/100; df=derivat(m); // for 0<x<=10 disp(" for 0<x<=10",df,"iii)PDF a) fX(x) ="); disp(" -infinte<x<=0,10<x<infine",0," b)fX(x) = "); x1=5,x2=7; function y=z(x), y=x/50; endfunction P=intg(x1,x2,z); disp(P,"iv) P(5<X<=7) =");
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clc; //Example 26.7 //page no 393 printf("Example 26.7 page no 393\n\n"); //refer to example 26.6 D=4//diameter of bed ,ft d_p=0.00137//particle diameter ,ft rho_s=84//coal particle density ,lb/ft^3 rho_f=55//oil density,lb/ft^3 meu_f=3.13e-4//viscosity of oil e_mf=0.38//void fraction L_mf=8//bed height at minimum fluidization,ft L_f=10//bed height,ft e=1-L_mf*(1-e_mf)/L_f//bed voidage g=32.174//grav acc v_s=(d_p^2)*g*(e^3)*(rho_s-rho_f)/(150*meu_f*(1-e)) //superficial velocity printf("\n superficial velocity v_s=%f ft/s",v_s); q=(%pi/4)*D^2*v_s//volumetric flow rate printf("\n vol. floe rate q=%f ft^3/s",q); //check on the laminar flow assumption meu_f=0.01 R_e=d_p*v_s*rho_f/(meu_f*(1-e)) printf("\n reynolds no R_e=%f",R_e); printf("\n since R_e is less than 10 ,flow is laminar");
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var ident < - 4; var ident2 <= - 5; var ident34543tendi == >--- css;
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//Dikshita Kambri 118A2044 //IPMV -EXPERIMENT 1 //Spatial RESOLUTION clear all clc; a = imread("C:\Users\hp\Documents\Image Processing-Scilab\Images\coins.png"); [row , col] = size(a); i = 1; j =1; for x = 1:2 :row for y = 1:2:col c(i,j)= a(x,y); j = j+1; end j=1; i=i+1; end disp('size of input image'); disp(size(a)); disp('size of output image'); disp(size(c)); figure(2) imshow(c)
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//Caption:transfer_function // example 3.2.1 //page 32 // we have defined parallel and series function which we are going to use here //exec parallel.sce; //exec series.sce; syms G1 G2 G3 H1; // shifting take off point after block G2 to a position before block G2 a= G2*H1; b=parallel(G2,G3); //shifting take off point before (G2+G3) to After (G2+G3) c=a/b; m=1; d=b/(1+m*b); e=series(G1,d); y=(e/(1+c*e)); y=simple (y); disp (y,"C(s)/R(s)=");
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clc // Given that Bp = 40 // Brake power when all cylinder operating in kW N = 2000 // Speed in rpm Bp1 = 32.2 // Brake power with cylinder no. 1 cut out in kW Bp2 = 32 // Brake power with cylinder no. 2 cut out in kW Bp3 = 32.5 // Brake power with cylinder no. 3 cut out in kW Bp4 = 32.4 // Brake power with cylinder no. 4 cut out in kW Bp5 = 32.1 // Brake power with cylinder no. 5 cut out in kW Bp6 = 32.3 // Brake power with cylinder no. 6 cut out in kW d = 100 // Diameter of cylinder in mm L = 125 // Stroke of cylinder in mm Vc = 0.000123 // Clearance volume in m^3 m_f = 9 // Fuel consumption in kg/h cv = 40 // Heating value in MJ/kg printf("\n Example 20.11\n") Ip1 = Bp-Bp1 Ip2 = Bp-Bp2 Ip3 = Bp-Bp3 Ip4 = Bp-Bp4 Ip5 = Bp-Bp5 Ip6 = Bp-Bp6 Ip = Ip1+Ip2+Ip3+Ip4+Ip5+Ip6 n_m = Bp/Ip bmep = Bp*2*60/(L*(10^-3)*((d*(10^-3))^2)*(%pi/4)*N) Vs = (%pi/4)*((d*(10^-3))^2)*(L*(10^-3)) r_k = (Vs+Vc)/Vc n_ase = 1- (1/(r_k^(1.4-1))) n_th = Ip*3600/(m_f*cv*1000) R_e = n_th/n_ase printf("\n Mechanical efficiency = %d percent,\n Brake mean effective pressure = %f bar\n Air standard ratio = %f percent,\n Brake thermal efficiency is %f percent,\n Relative efficiency = %f percent",n_m*100,bmep*(10^-2),n_ase*100,n_th*100,R_e*100) //The value of answer for air standard efficiency is different because of round off error // Answer given in the book for bmep is 3.055 bar which is wrong. // Answer given in the book for brake thermal efficiency is 40 percent which is wrong. // Answer given in the book for relative efficiency is 68.6 percent which is wrong.
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clc //initialisation of variables d= 3 //in d1= 1.5 //in F= 7500 //lb //CALCULATIONS A1= (%pi/4)*(d^2-d1^2) P= F/A1 //RESULTS printf ('pressure in the cylinder = %.f psi',P-1)
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//A Textbook of Chemical Engineering Thermodynamics //Chapter 7 //Properties of Solutions //Example 10 clear; clc; //Given: K = 4.4*10^4; //Henry's law constant (bar) pp = 0.25; //partial pressure of oxygen in bar M_O2 = 32; //molecular wt of oxygen M_water = 18; //molecular wt of water //To estimate the solubility of oxygen in water at 298 K //Using eq. 7.72 (Page no. 275) x_O2 = pp/K; //mole fraction of O2 mprintf('Solubility of oxygen is %5.4e moles per mole of water',x_O2); //In mass units sol_O2 = (x_O2*M_O2)/M_water; mprintf('\n Solubility of oxygen in mass units is %4.3e kg oxygen per kg water',sol_O2); //end
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clc clear //Initialization of variables Tc=647.3 //K dh=1.1 Db=-2 v2=0.234 v1=0.27 //calculations dh2=dh+Db*(v2-v1) dhh=dh2*Tc dhbar=dhh*4.18/18 disp("From steam tables,") h1=3777.5 //kJ/kg h2=3928.2 //kJ/kg dhs=h2-h1 err=abs(dhs-dhbar)/dhs //results printf("Enthalpy departure = %d kJ/kg",dhbar) printf("\n Percentage error = %.1f ",err*100)
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//Ex 1.11.1 clc;clear;close; format('v',9); //Given : ni=1.5*10^10/10^-6;//per m^3 mu_n=1800*10^-4;//m^2/V-s mu_p=500*10^-4;//m^2/V-s q=1.6*10^-19;//Coulomb sigma_i=ni*(mu_n+mu_p)*q;//(ohm-m)^-1 disp(sigma_i,"Conductivity in (ohm-m)^-1 : "); rho_i=1/sigma_i;//ohm-m disp(rho_i,"Resistivity in ohm-m : ");
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//example 6.7 clc; clear; a=0; b=0; q=0; //aa=input(" Enter the first no (in decimal) :"); //bb=input(" Enter the number from which first no has to substracted:"); aa=175; bb=118; while(aa>0) // converting the inputs in to binary numbers x=modulo(aa,2); a= a + (10^q)*x; aa=aa/2; aa=floor(aa); q=q+1; end q=0; while(bb>0) x=modulo(bb,2); b= b + (10^q)*x; bb=bb/2; bb=floor(bb); q=q+1; end printf(' \n The binary equivalent of first no is %f\n\n',a); printf(' The binary equivalent of secnd no is %f\n\n',b); for i=1:8 a1(i)=modulo(a,10); a=a/10; a=round(a); b1(i)=modulo(b,10); b=b/10; b=round(b); end car(1)=0; for i=1:8 c1(i)=car(i)+a1(i)+ b1(i);//adding the binary numbers (binary addtion) if c1(i)== 2 then car(i+1)= 1; c1(i)=0; elseif c1(i)==3 then car(i+1)= 1; c1(i)=1; else car(i+1)=0; end end c1(9)=car(9); re=0; format('v',18); for i=1:8 re=re+(c1(i)*(10^(i-1))) end printf('If only 8 bits are taken the result will be as shown below \n\n'); printf(' and the sum of given two binary numbers will be %f\n\n',re ); q=1; b=0; f=0; a=re; while(a>0) //converting the binary output to hexadecimal r=modulo(a,10); b(1,q)=r; a=a/10; a=floor(a); q=q+1; end for m=1:q-1 c=m-1; f = f + b(1,m)*(2^c); end printf(' Sum in decimal notation is %d\n\n',f); hex=dec2hex(f); printf(' The sum in hexadecimal notation is %sH \n',hex); printf(' \n with an overflow of %d\n\n',car(9));
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//Example 6_12 clc; clear; close; format('v',6); //given data : V1=230;//V N2ByN1=1/3;//turns ratio RL=200;//ohm V2=V1*N2ByN1;//V Vm=sqrt(2)*V2;//V Im=Vm/RL;//A Pmax=Im^2*RL;//W disp(Pmax,"Maximum load power(W) : "); format('v',5); Vdc=0.318*Vm;//V Idc=Vdc/RL;//A Pdc=Idc^2*RL;//W disp(Pdc,"Average value of load power(W) : "); //Answer in the textbook is not accurate.
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//Fuels and Combustion// //Example 8.10// C=810;//weight of carbon in 1kg of coal sample in grams// O=80;//weight of oxygen in 1kg of coal sample in grams// S=10;//weight of Sulphur in 1kg of coal sample in grams// N=10;//weight of nytrogen in 1kg of coal sample in grams// H=50;//weight of hydrogen in 1kg of coal sample in grams// MO=C*32/12+H*16/2+S*32/32;//minimum weight of oxygen needed in grams// printf('minimum weight of oxygen needed=MO=%fg',MO); printf('\nOxygen already available in fuel=80grams\nNet oxygen needed=2490grams'); MA=2490*100/23;//minimum weight of air needed in grams// printf('\nminimum amount of air needed=MA=%fg',MA); printf('\nProducts of combustion are CO2 and SO2'); printf('\nFrom the equations written above,44grams of CO2 is obtained 12grams of carbon\nhence,weight of CO2 obtained from 810grams of carbon=810*44/12=2970grams'); printf('\nSimilarly,weight of SO2 obtained from 10grams of sulphur=10*64/32=20grams'); NF=10+MA*0.77;//weight of nitrogen present in the products in grams// printf('\nWeight of nitrogen present in the products=NF=%fg',NF); WD=2970+20+8346;//total weight of dry products in grams// printf('\nTotal weight of dry products=WD=%fg',WD); PCO2=2970*100/WD;//percentage composition of CO2// printf('\nPercentage composition of CO2=PCO2=%f',PCO2); PSO2=20*100/WD;//percentage composition of SO2// printf('\nPercentage composition of SO2=%f',PSO2); PN2=8346*100/WD;//percentage composition of N2// printf('\nPercentage composition of N2=PN2=%f',PN2);
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clc //Given that E = 1e9 // energy of electron in eV c = 3e8 // speed of light in m/s m_0 = 9.1e-31 // mass of electron in kg // sample problem 14 page No. 227 printf("\n \n\n # Problem 14a # \n") printf("\n Standard formula used \n E = m*c^2") m = E / c^2 * 1.6e-19 // calculation of relativistic mass of particle ratio = m / m_0// calculation of Ratio of relativistic mass and rest mass of particle printf ("\n Ratio of relativistic mass and rest mass of particle is %e.",ratio )
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// Example 18.7, page no-464 clear clc c=2*10^-6//F epsr=80 v=1000 //v E1=(c*v^2)/2 c0=c/epsr E2=(c0*v^2)/2 E=E1-E2 printf("\nThe Energy stored in capacitor =%.0f J",E1) printf("\nThe energy stored in polarising the capacitor = %.4f J",E)
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clear // //variable declaration w=(100) //wide of rectangular beam,mm h=(200) //height or rectangular beam dude,mm I=w*(h**3)/12 //At point A, which is at 30 mm below top fibre y=100-30 M=(80*1000000) //sagging moment,KN-m fx=M*y/I px=-fx F=(100*1000 ) //shear force,N b=(100) A=b*30 y1=100-15 q=(F*(A*y1))/(b*I) //shearing stress,N/mm^2 py=0 p1=(px+py)/2+sqrt((((px-py)/2)**2)+(q**2)) p2=(px+py)/2-sqrt((((px-py)/2)**2)+(q**2)) printf("\n p1= %0.2f N/mm^2",p1) printf("\n p2= %0.2f N/mm^2",p2)
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function [Ack] = AutoGenSciHelpSet() global bOK; global sciobj; Ack = %f; SciDesSet(); if(bOK) then setsize = strtod(sciobj.Input); for i = 1 : setsize SciParamSet("Input",i); if ~bOK then disp("在第"+string(i)+"个输入参数时退出."); return; end end else disp("退出帮助文档自动生成流程."); return; end if(bOK) then setsize = strtod(sciobj.Output); for i = 1 : setsize SciParamSet("Output",i); if ~bOK then disp("在第"+string(i)+"个输出参数时退出."); return; end end else disp("退出帮助文档自动生成流程."); return; end GenSciHelpXml(); Ack = %t; endfunction
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function f = DIST_F(x, a, b) f = x for i = 1:length(f) f(i) = a * atan(b*f(i)) end endfunction
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// Scilab code Ex1.14: Pg.47-48 (2008) clc; clear; fo = 3; // Actual frequency of the signal, Hz Beta = 0.8; // Boost parameter // When Homer & Ulysses are receding away from each other f_ratio_recede = sqrt((1-Beta)/(1+Beta)); // Frequency ratio while approaching printf("\nHomer receives %g flashes during the first 9 years and Ulysses receives %g flash in his first 3 years!", fo, f_ratio_recede*fo); // When Homer & Ulysses are approaching towards each other f_ratio_approach = sqrt((1+Beta)/(1-Beta)); // Frequency ratio while approaching printf("\nHomer receives %d flashes during the 10th year and Ulysses receives %d flashes during his final 3 years!", fo, f_ratio_approach*fo); // Result // Homer receives 3 flashes during the first 9 years and Ulysses receives 1 flash in his first 3 years! // Homer receives 3 flashes during the 10th year and Ulysses receives 9 flashes during his final 3 years!
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OMEGA = 7 AMPLITUDE = .2 function z = f(x, y) z = AMPLITUDE * ( cos(OMEGA * x) + cos(OMEGA * y) ) endfunction N = 100 x = linspace(-1,1,N) y = linspace(-1,1,N) [X, Y] = meshgrid(x, y) clf() surf(X, Y, f(X, Y)) // Cosmétique {{{ a = gca() a.isoview = "on" a.data_bounds = [ -1, -1, -1; 1, 1, 1] a.children.thickness = 0 f = gcf() f.color_map = .1 + .8*bonecolormap(32) // on croppe le noir et le blanc // }}}
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clc disp("Example 8.7") printf("\n") disp("convert the following binary numbers to decimal") disp("a)1011 b)110101 c)10101") //Given binary number bin=1011 i=1 //storing each integer digit in b(i) while(bin>0) b(i)=modulo(bin,10) bin=floor(bin/10) i=i+1; end //checking whether it is a binary number or not for i=1:length(b) if(b(i)>1) then disp('not a binary number') abort end end dec=0 for i=1:length(b) //multipliying bits of integer part with their position values and adding dec=dec+(b(i)*2^(i-1)) end //displaying the output printf("decimal format is") disp(dec)
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errcatch(-1,"stop");mode(2);//Example 2.19://error ; ; n=40;//revolutions rc=0.12;//registration constant err=n/rc;//energy recorded in kWh is e2=22000;//volts e1=110;//volts i2=500;//amperes i1=5;//amperes i=5.25;//amperes lv=110;//volts pf=1;// t=61;//seconds ae=((sqrt(3)*e2*lv*i*i2*pf*t)/(e1*i1*3600))*10^-3;//kWh e=((err-ae)/ae)*100;// disp(-e,"error (slow) is (%)") exit();
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clc // Given that lambda = 6000 // wavelength of first source in angstrom d = 2 // Spacing between sources in mm D = 0.1 // Distance between source and screen in meter t = 0.5 // Thickness of plate in mm shift = 5 // Shift of fringe in mm // Sample Problem 7 on page no. 95 printf("\n # PROBLEM 7 # \n") printf("\n Standard formula used \n x = D*(mu1)*n*lambda/d \n") mu = 1+ shift*1e-3*d*1e-3/(D*t*1e-3) // Refractive index of Glass plate printf("\n Refractive index of Glass plate is %f.",mu)
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clc; p1=1; // pressure at inlet in bar T1=27+273; // Temperature at inlet in kelvin T4=1200; // Maximum temperature in kelvin t=T4/T1; // Temperature ratio r=1.4; // Specific heat ratio rp=t; c=rp^((r-1)/r); x=(1-sqrt(c)/rp)/(1-c/rp); eff2_1=x; r1=sqrt(rp); r2=r1; r3=r1; r4=r1; disp (eff2_1,"Efficiency ratio of power plants = "); disp (r4,"pressure ratio of LPT = ",r3,"pressure ratio of HPT = ",r2,"pressure ratio of HPC = ",r1,"pressure ratio of LPC = ");
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clc() clear() /// PRACTICA 8 function I = Simpsonar(f, a, b, n) /// LA CANTIDAD DE INTERVALOS n DEBE SER PAR h = (b - a) / n I = f(a) + f(b) for k = 1:n-1 x = a + k * h if modulo(k,2) == 1 then I = I + 4 * f(x) else I = I + 2 * f(x) end end I = I * h/3 // El error es // -h^4 * (b-a) / 180 * max(abs(f''''(x))) para x entre a y b endfunction function I = trapeciar(f, a, b, n) h = (b - a) / n I = 0.5 * (f(a) + f(b)) for k = 1:n-1 x = a + k * h I = I + f(x) end I = I * h // El error es // h^2 (b-a) / 12 * max(abs(f''(x))) para x entre a y b endfunction // SE BANCA PASAR FUNCIONES COMO ARGUMENTOS EN Y function I = Trapeciar2D(f, a, b, cc, dd, n) /// CANT INTERVALOS N DEBE SER PAR hx = (b - a) / n for kx = 0:n x = a + kx*hx // asumo que ambas son funciones if(typeof(cc) == "function") c = cc(x) d = dd(x) else c = cc d = dd end subI = 0 hy = (d - c) / n for ky = 0:n y = c + ky*hy w = 1 if(kx == 0 | kx == n) w = w / 2 end if(ky == 0 | ky == n) w = w / 2 end subI = subI + w*f(x, y) end subI = subI * hy I = I + subI end I = I * hx endfunction /// Ej 1 //I = Simpsonar(log, 1, 2, 20) //Ireal = 2 * log(2) - 1 //disp(I) //disp(Ireal) //disp( -0.05^4 / 180 * 6) // //I = trapeciar(log, 1, 2, 20) //Ireal = 2 * log(2) - 1 //disp(I) //disp(Ireal) //disp( 0.05^2 / 12 ) function [I]=DoubleIntegralTrap(a, b, n, c, d, m, f) // TRAPECIO 2D // a < b, n: cantidad de divisiones sobre x // c < d, m: cantidad de divisiones sobre y Dx = (b - a)/n; Dy = (d - c)/m; x = zeros(1,n+1); y = zeros(1,m+1); F = zeros(n+1,m+1); for i = 1:n+1 x(1,i) = a + (i-1)*Dx; end; for j = 1:m+1 y(1,j) = c + (j-1)*Dy; end; for i = 1:n+1 for j = 1:m+1 F(i,j) = f(x(1,i),y(1,j)); end; end; I = F(1,1) + F(1,m+1) + F(n+1,1) + F(n+1,m+1); for i = 2:n I = I + 2*(F(i,1) + F(i,m+1)); end; for j = 2:m I = I + 2*(F(1,j) + F(n+1,j)); end; for i = 2:n for j = 2:m I = I + 4*F(i,j); end; end; I = I*Dx*Dy/4; //end of DoubleIntegral function endfunction function [I]=DoubleIntegralSimp(a, b, n, c, d, m, f) // SIMPSON 2D // a < b, n: cantidad de divisiones sobre x // c < d, m: cantidad de divisiones sobre y Dx = (b - a)/n; Dy = (d - c)/m; x = zeros(1,n+1); y = zeros(1,m+1); Rx = zeros(1, n+1); Ry = zeros(1, m+1); F = zeros(n+1,m+1); for i = 1:n+1 x(1,i) = a + (i-1)*Dx; end; for j = 1:m+1 y(1,j) = c + (j-1)*Dy; end; for i = 1:n+1 for j = 1:m+1 F(i,j) = f(x(1,i),y(1,j)); end; end; resto = 2 for i = 1:n+1 Rx(i) = resto resto = 6 - resto end Rx(1) = 1 Rx(n + 1) = 1 resto = 2 for i = 1:m+1 Ry(i) = resto resto = 6 - resto end Ry(1) = 1 Ry(m + 1) = 1 R = Rx' * Ry //disp(F) I = sum(F .* R)/9 * Dx * Dy //end of DoubleIntegral function endfunction // Ej 2 //deff('y = f(x)', 'y = 1/x') //I = trapeciar(f, 1, 3, 4) //disp(I) //disp(intg(1,3,f)) function z = f(x, y) // z = 1 / (%pi * sqrt(x * (1-x))) z = sin(x + y) endfunction z1 = DoubleIntegralSimp(0, 1, 2, 0, 2, 2, f) disp(z1)
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/1775/CH3/EX3.7/Chapter3_Example7.sce
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Chapter3_Example7.sce
//Chapter-3, Illustration 7, Page 145 //Title: Internal Combustion Engines //============================================================================= clc clear //INPUT DATA n=2;//No. of cylinders N=4000;//speed of engine in rpm nV=0.77;//Volumetric efficiency nM=0.75;//Mechanical efficiency m=10;//fuel consumed in lit/h g=0.73;//spcific gravity of fuel Raf=18;//air-fuel ratio Np=600;//piston speed in m/min imep=5;//Indicated mean efective pressure in bar R=281;//Universal gas constant in J/kg-K T=288;//Standard temperature in K P=1.013;//Standard pressure in bar //CALCULATIONS L=Np/(2*N);//Piston stroke in m mf=m*g;//mass of fuel in kg/h ma=mf*Raf;//mass of air required in kg/h Va=(ma*R*T)/(P*60*(10^5));//volume of air required in (m^3)/min D=sqrt((2*Va)/(nV*L*N*3.1415));//Diameter in m IP=(2*imep*100*L*3.1415*(D^2)*N)/(4*60);//Indicated Power in kW BP=nV*IP;//Brake Power in kW //OUTPUT mprintf('Piston Stroke is %3.3f m \n Bore diameter is %3.4f m \n Brake power is %3.1f kW',L,D,BP) //==============================END OF PROGRAM=================================
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ex4_12.sce
clc; g=32 //gravitational constant in ft/sec square w=128; //mass in lb F=(1/2)*(1/2)*128' //calculating F in lb m=w/g; //calculating m in slugs disp(F,"Weight at height above the earths surface of one earth radius = "); //displaying weight disp(m,"Mass of the girl in slugs= "); //displaying mass in slugs
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Ex5_2.sce
clear; clc; //page no. 153 p1 = 300;// psia T1 = 900;// degreeF p2 = 200;// psia T2 = 780;// degreeF H2 = 1414;//Btu/lb H1 = 1471;// Btu/lb V2 = sqrt(2*31.1*778*(H1-H2)); printf('T2 = %d degreeF\n V2 = %d fps',T2,V2);
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Ex2_1.sce
clc //page 13 printf("\t Example 2.1 \n"); printf("\t approximate values are mentioned in the book \n"); Tavg=900; // average temperature of the wall,F k=0.15; // Thermal conductivity at 932 F,Btu/(hr)(ft^2)(F/ft) T1=1500; // hot side temperature,F T2=300; // cold side temperature,F A=192; // surface area,ft^2 L=0.5; // thickness,ft Q=(k)*(A)*(T1-T2)/L; // formula for heat,Btu/hr printf("\t heat is : %.2e Btu/hr \n",Q); //end
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Ex4_17.sce
clear // //Given //Variable declaration L=4*10**3 //Length of bar in mm A=2000 //Area of bar in sq.mm P1=3000 //Falling weight in N(for 1st case) h1=20*10 //Height in mm(for 1st case) P2=30*1000 //Falling weight in N(for 2nd case) h2=2*10 //Height in mm(for 2nd case) E=2e5 //Youngs modulus in N/sq.mm //Calculation V=A*L //Volume of bar in mm^3 //case(i):Maximum stress when a 3000N weight falls through a height of 20cm sigma1=(((sqrt((2*E*P1*h1)/(A*L))))) //case(ii):Maximum stress when a 30kN weight falls through a height of 2cm sigma2=((P2/A)*(1+(sqrt(1+((2*E*A*h2)/(P2*L)))))) //Result printf("\n Maximum stress induced(when a weight of 3000N falls through a height of 20cm)= %0.3f N/mm^2",sigma1) printf("\n Maximum stress induced(when a weight of 30kN falls through a height of 2cm)= %0.3f N/mm^2",sigma2)