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// 08.09.13 // 10.02.16 Eps => [Eps, Epsmag] function Out5=Sfbdparadata(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=[0.05,1]; // if Nargs>=3 Eps=varargin(3); end; if length(Eps)==1 // Eps=[Eps,1]; // end; // Eps2=0.2; if Nargs>=4 Eps2=varargin(4); end; Ts=timer(); [Zval,Xval,Yval]=Evlptablepara(Mix(Fd,Np)); Out3=Implicitplot(Zval,Xval,Yval); IMPLICITDATA=Out3; Mixdisp(Mix('ImplicitData obtained ',timer())); Out4=Cuspsplitpara(Out3,Fd,Eps(1)); // CUSPDATA=Out4; CUSPPT=CUSPSPLITPT; Mixdisp(Mix('CuspData obtained ',timer())); Out5=Borderparadata(Out4,Fd,Np,Eps,Eps2); Mixdisp(Mix('BorderData obtained ',timer())); endfunction;
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clear ; close ; clc ; A=[2 3;4 0]; disp(A,'A='); B=[1 2 0;5 -1 0]; disp(B,'B'); disp(A*B,'AB=') //end
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function [x,y,typ]=REGISTER_f(job,arg1,arg2) x=[];y=[];typ=[] select job case 'plot' then standard_draw(arg1) [graphics,model]=arg1(2:3); [orig,sz,orient,label]=graphics(1:4) dly=model(8) xstringb(orig(1),orig(2),['Shift';'Register';string(dly)],sz(1),sz(2),'fill') case 'getinputs' then [x,y,typ]=standard_inputs(arg1) case 'getoutputs' then [x,y,typ]=standard_outputs(arg1) case 'getorigin' then [x,y]=standard_origin(arg1) case 'set' then x=arg1; graphics=arg1(2);label=graphics(4) model=arg1(3);z0=model(7) while %t do [ok,label,z0]=getvalue('Set delay parameters',.. ['Block label';'Register initial condition'],.. list('str',1,'vec',-1),[label;strcat(string(z0),';')]) if ~ok then break,end if prod(size(z0))<2 then x_message('Register length must be at least 2') ok=%f end if ok then graphics(4)=label; model(7)=z0 x(2)=graphics;x(3)=model break end end case 'define' then z0=zeros(10,1) model=list('delay',1,1,1,0,[],z0,[],[],'d',%f,[%f %f]) x=standard_define([2.5 2.5],model) end
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// example 5.29 // evaluate the given double integral using the simpsons rule; // I= double integral f(x)=1/(x+y) in the range x=[1,2],y=[1,1.5]; h=.5; k=.25; deff('[w]=f(x,y)','w=1/(x+y)') I=(.125/9)*[{f(1,1)+f(2,1)+f(1,1.5)+f(2,1.5)}+4*{f(1.5,1)+f(1,1.25)+f(1.5,1.5)+f(2,1.25)}+16*f(1.5,1.25)]; disp(I);
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// Example 2.52:Ai,Ri,Av clc; clear; close; Rs=0.5;//Internal resistance in killo ohms Rl=5;//Load resistance in killo ohms //H Paramters are Hie=1;//in killo ohms Hfe=50; Hoe=25*10^-6;// in ampere per volt Ai= (1+Hfe)/(1+Hoe*Rl*10^3);// Current gain Ri= Hie+(Ai*Rl);// Input resistance in killo ohms Av= Ai*(Rl/Ri);// Voltage Gain disp(Ai,"Current gain is") disp(Ri,"Input resistance in killo ohms is") disp(Av,"Voltage gain is")
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//Pressure at entry(in kPa): p1=350; //Temperature at entry(in K) T1=333; //Velocity at entry(in m/s): V1=183; //Mach no. at exit: M2=1.3; //Stagnation pressure at exit(in kPa): p02=385; //Stagnation temperature at exit(in K): T02=350; //Value of k: k=1.4; //Gas constant(in N-m/kg-K) R=287; //Specific heat at constant pressure(kJ/(kg-K): cp=1;
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//Problem 2.03: //initializing the variables: mg = 100; // in lb Pg = 35; // in psig A = 3; // in in2 gc = 1; // in lb/lbf Pa = 14.7; // in psi //calculation: F = mg/gc Pli = F/A // in lbf/in2 Plf = Pli*144 // in lbf/ft2 P = Pg + Pa printf("\n\nResult\n\n") printf("\n pressure at the base is %.0f lbf/ft2\n",Plf) printf("\n absolute pressure is %.1f psia\n",P)
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A109345 post 0 5;n;1;-;n;*;2;/;^ 0
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clc; clear; //Example 5.13 mh_dot=1.25 //[kg/s] Cpw=4.187*10^3 //Heat capacity of water in [J/kg.K] lambda=315 //[kJ/kg] Q=mh_dot*lambda //Rate of heat transfer from vapour [kJ/s] Q=Q*10^3 //[W] Ts=345 //Temperature of condensing vapour[K] t1=290 //Inlet temperature of water [K] t2=310 //Outlet temperature of water[K] dT1=Ts-t1 //[K] dT2=Ts-t2 //[K] dTlm=(dT1-dT2)/log(dT1/dT2) //[K] //Heat removed from vapour = Heat gained mw_dot=Q/(Cpw*(t2-t1)) //[kg/s] hi=2.5 //[kW/sq m.K] hi=hi*1000 //[W/sq m.K] Do=0.025 //[m] Di=0.020 //[m] hio=hi*(Di/Do) //Inside heat transfer cosfficient referred to outside dia in [W/sq m.K] ho=0.8 //Outside heat tranbsfer coefficient in [kW/sq m.K] ho=ho*1000 //[W/sq m.K] Uo=1/(1/ho+1/hio) //[W/sq m.K] //Ud is 80% of Uc Ud=(80/100)*Uo //[W/sq m.K] Ao=Q/(Ud*dTlm) //[sq m] L=1 //[m] A=%pi*Do*L //Outside area of pipe per m length of pipe len=Ao/A //Total length of piping required. rho=1000 //[kg/m^3] V=mw_dot/rho //[m^3/s] v=0.6 //[m/s] a=V/v //Cross-sectional area for flow pass [sq m] a1=(%pi*Di^2)/4 //[sq m] //for single pass on tube side fluid(water) n=round(a/a1) //No. of tubes per pass l=len/n //Length of each tube in [m] //For two passes on water side: tn=2*n //Total no of tubes l2=len/tn //Length of each tube in [m] //For four passes on water side/tube side tn2=4*n //Total no. of tubes l3=len/tn2 //Length of each tube in [m] printf("\nNo. of tubes=%d ,\nLength of tube=%f m",tn2,l3);
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// Example 9.2 // Calculation of the total power at the fiber output. // Page no 393 clc; clear; close; //Given data p=0; // Power per channel fl=0.2; // Fiber loss f=50; // Wavelength // The total power at the fiber output. pc=10^(0.1*p); tp=pc*11; tp1=10*log10(tp); tfl=fl*f; to=tp1-tfl; //Displaying results in the command window printf("\n The total power at the fiber output = %0.3f dBm ",to);
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// Chapter 2 example 3 clc; clear; // Variable declaration h = 6.63*10^-34; // plancks constant in J.s m = 9.1*10^-31; // mass of electron in kg a = 5*10^-10; // width of infinite potential well in m e = 1.6*10^-19; // charge of electron in coulombs n1 = 1; // energy level constant n2 = 2; // energy level constant n3 = 3; // energy level constant // Calculations E1 = ((n1^2)*(h^2))/(8*m*(a^2)*e); // first energy level in eV E2 = ((n2^2)*(h^2))/(8*m*(a^2)*e); // second energy level in eV E3 = ((n3^2)*(h^2))/(8*m*(a^2)*e); // third energy level in eV // Result mprintf('First Three Energy levels are \n E1 = %3.2f eV\n E2 = %3.2f eV\n E3 = %3.2f eV',E1,E2,E3); mprintf('\n Above calculation shows that the energy of the bound electron cannot be continuous')
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load Divide.hack, output-file Divide.out, compare-to Divide.cmp, output-list RAM[15]%D2.6.2; set RAM[13] 10, // 10/2 = 5 set RAM[14] 2, repeat 1000000 { ticktock; } output; set RAM[13] 1, set RAM[14] 1, repeat 1000000 { ticktock; } output; set RAM[13] 1000, set RAM[14] 1000, repeat 1000000 { ticktock; } output; set RAM[13] 555, set RAM[14] 789, repeat 1000000 { ticktock; } output; set RAM[13] 17555, set RAM[14] 14, repeat 1000000 { ticktock; } output;
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// Example 9.8 // Calculation of the number of subcarriers required to transmit information. // Page no 413 clc; clear; close; //Given data M=4; np=2; // No of polarization nc=24; // No of channels bs=10*10^9; // Symbol rate per polarization d=5000*10^3; // Transmission distance b=22*10^-27; ts= 49.3*10^-9; // The total data rate B=bs*log2(M); T=d*b*%pi*bs; //L=T/(b*2*%pi*N*bs); N=(bs*ts)/2; //Displaying results in the command window printf("\n The number of subcarriers required to transmit information = %0.0f ",N); // The answers vary due to round off error
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// Constants g = 9.81; u0 = 0; v0 = 0; b = 0; h0 = 5030; consts=struct('g',g,'u0',u0,'v0',v0,'b',b,'h0',h0); //Domain definition // Define the x domain ni = 51; //ni=41; xmax = 100000; dx = xmax/(ni-1); x = [0:dx:xmax]; // Define the y domain nj = 51; //nj=41; ymax = 100000; dy = ymax/(nj-1); y = [0:dy:ymax]; nt=10; step=0; tmax = 200; steeringenabled=0; finishsteering=0; domain=struct('ni',ni,'xmax',xmax,'nj',nj,'ymax',ymax,'nt',nt,'tmax',tmax,'step',step,'steeringenabled',steeringenabled,'finishsteering',finishsteering); sf=10;//source frequency sa=5;//source amplitude sx=20;//source x location sy=30;//source y location source=struct('sf',sf,'sa',sa,'sx',sx,'sy',sy); metadata.directory='out'; metadata.author='MikeG'; metadata.sdate=date(); metadata.platform='felix'; metadata.desc='A simple test of SAAS'; metadata.name='intsaas1'; elist=list(2); //elist(1)='192.168.1.104'; elist(1)='localhost'; elist(2)=8081; elist(3)=0;
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// Example 1.1 A conductor material has a free- electron density of 10^24 electrons per metre^3.When a voltage is applied, a constant drift velocity of 1.5x10^-2 metre/second is attained by the electrons. If the cross- sectional area of the material is 1 cm^2, calculate the magnitude of the current. Electronic charge is 1.6x10^-19. // 1 metre = 100 centimetre n = 10^24;// charge density (e/m^3) Vd = 1.5*10^-2; //drift velocity attained by electrons(m/s) A = 10^-4; // crossectional area of the material (m^2) e = 1.6*10^-19; // charge of an electron (coulombs) // let i be the magnitude of the current // FORMULA : i = nAeVd i = prod([n,A,e,Vd]) // calculation disp(i,"magnitude of the current(in ampere)= ")
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clc;clear; //Example 21.3 //calculation of relative permeability //given values X=3.7*10^-3;//susceptibility at 300k T=300;//temp in K T1=200;//temp in K T2=500;//temp in K //calculation C=X*T;//curie constant XT1=C/T1; disp(XT1,'relative permeability at T1 is '); XT2=C/T2; disp(XT2,'relative permeability at T2 is')
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//Ex11_1 clc VCC = 20//collector voltage RL = 12//load resistance disp("VCC = "+string(VCC)+"V") disp("RL = "+string(RL)+"ohm") Pi_dc = (VCC^2)/(2*RL)//input power disp("Pi(dc) = (VCC^2)/(2*RL) = "+string(Pi_dc)+"W") Po_ac = (VCC^2)/(8*RL)//output power disp("Po_ac = (VCC^2)/(8*RL) = "+string(Po_ac)+"W") eta = Po_ac/Pi_dc//efficiency disp("eta = Po_ac/Pi_dc = "+string(eta*100)+"%") // note : has modifed variables: // using Po_ac instead of Po(ac) // and Pi_dc instead of Pi(dc). // note: there is a misprinting in the above problem given in the textbook // author want to ask for efficiency instead of frequency.
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clear; clc; clear function [I_TAV]=theta(th) n=360/th; I=1; //supposition I_av=I/n; I_rms=I/sqrt(n); FF=I_rms/I_av; I_rms=35; I_TAV=I_rms/FF; endfunction disp("when conduction angle=180"); th=180; I_TAV=theta(th); printf("avg on current rating=%.3f A",I_TAV); disp("when conduction angle=90"); th=90; I_TAV=theta(th); printf("avg on current rating=%.1f A",I_TAV); disp("when conduction angle=30"); th=30; I_TAV=theta(th); printf("avg on current rating=%.4f A",I_TAV);
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//Example 2_4 clc(); clear; //To find the lever arms and torques for the forces printf("For F1 it is Zero\n") printf("For F2 it is a*F2 Counter clockwise\n") printf("For F3 it is a*F3 Clock Wise\n") printf("For F4 it is b*F4 Counter Clock wise")
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// Given:- P1 = 1.00 // initial pressure in bar T1 = 300.00 // initial temperature in kelvin T2 = 650.00 // final temperature in kelvin // Part(a) // From table A-22 pr2 = 21.86 pr1 = 1.3860 k = 1.39 // From table A-20 // Calculations p2 = P1*(pr2/pr1) p2a = P1*((T2/T1)**(k/(k-1))) // Results printf( ' P2 = %f bar.',p2) printf( ' Part(b) IT software problem'); printf( ' P2a = %f bar',p2a);
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clc e=3.795; hcut_per_c=1973; m=0.511*(10^6); a=((hcut_per_c)^2)/(m*(e^2)); n=input("enter the energy state n :"); function zprim=f(r, z) zprim(1)=z(2) zprim(2)=-(-(1/(a*n)^2)+(2/(a*r)))*z(1) endfunction i=input("Enter the scale pan along r axis : ") r0=0.00001;rmax=i*a; r=r0:0.01*a:rmax; z0=0;zprim0=2/((n*a)^(3/2)); y=ode([z0;zprim0],r0,r,f); u=y(1,:); R=u./r; p=R.*R; D=(r.^2).*p; plot(r,R,"r"); plot(r,D,"b"); xlabel("r(CGS)"); ylabel("Wave function and probability"); xtitle("Plot of s-wave for Hydrogen atom"); legend("Wave funcion","Probability");
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//Example 3.71.b:resistance and capacitance clc; clear; close; cl=10^-4;//micro-F c2=0.004;//micro-F c3=0.001;//micro-F r3=10;//killo ohms r4=5;//killo ohms f=1;//kHz rx=((c3+cl)/c2)*r4;//killo ohms cx=(r3/r4)*c2;//micro-F disp(rx,"resistance is ,(k-ohm)=") disp(cx,"capacitance is,(micro-F)=")
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exec("signal_noise.sci",-1) function[]=plotsignalnoise(a,N,Variance) nvec=linspace(0,N,N+1) signnoise=signalnoise(a,N,Variance) xset('window',67) xtitle('Signal with gaussian noise', 'X-axis', 'Y-axis') plot(nvec,signnoise) endfunction plotsignalnoise(0.01,1000,0.1)
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/* This experiment is a replication from the article "Errors Are Aversive: Defensive Motivation and the Error-Related Negativity" by Greg Hajcak and Dan Foti (2008). doi: 10.1111/j.1467-9280.2008.02053.x I'm running it as part of the Reproducibility Project, which you can learn more about at http://openscienceframework.org/. */ # ---------- begin header ---------- # scenario = "mlewis_replication.sce"; scenario_type = trials; active_buttons = 2; #specifies how many buttons can be used button_codes = 1,2; #specifies button codes default_background_color = 0,0,0; #black default_text_color = 255,255,255; #white default_font = "Palatino Linotype"; default_font_size = 36; write_codes = true; # ---------- end header ---------- # begin; ellipse_graphic { ellipse_height = 150; ellipse_width = 150; height = 100; width = 100; color = 255, 255, 255; } flash; picture { text { caption = "+"; font_size = 48; }; x=0; y=0; # centers text } default; # sets fixation cross as default image wavefile { filename = "startle3amplified.wav"; } startle; #name of stimulus is "startle" bitmap { filename = "LC.jpg"; } bmLC; #name of stimulus is "LC" for "left(-facing center) congruent" bitmap { filename = "LI.jpg"; } bmLI; #name of stimulus is "LI" for "left(-facing center) incongruent" bitmap { filename = "RC.jpg"; } bmRC; #name of stimulus is "RC" for "right(-facing center) congruent" bitmap { filename = "RI.jpg"; } bmRI; #name of stimulus is "RI" for "right(-facing center) incongruent" picture { bitmap bmLC; x = 0; y = 0; } LC; picture { bitmap bmLI; x = 0; y = 0; } LI; picture { bitmap bmRC; x = 0; y = 0; } RC; picture { bitmap bmRI; x = 0; y = 0; } RI; trial { trial_type = first_response; trial_duration = 800; picture { bitmap bmLC; x = 0; y = 0; ellipse_graphic flash; x = 920; y = -500; }; time = 0; duration = 200; target_button = 1; code = 10; port_code = 10; } trial_LC; # name is "trial_LC" and means center arrow left, congruent w/flankers trial { trial_type = first_response; trial_duration = 800; picture { bitmap bmRC; x = 0; y = 0; ellipse_graphic flash; x = 920; y = -500; }; time = 0; duration = 200; target_button = 2; code = 11; port_code = 11; } trial_RC; # name is "trial_RC" and means center arrow right, congruent w/flankers trial { trial_type = first_response; trial_duration = forever; picture { bitmap bmLI; x = 0; y = 0; ellipse_graphic flash; x = 920; y = -500; }; time = 0; duration = 800; target_button = 1; code = 20; port_code = 20; } trial_LI; # name is "trial_LI" and means center arrow left, incongruent w/flankers trial { trial_type = first_response; trial_duration = 800; picture { bitmap bmRI; x = 0; y = 0; ellipse_graphic flash; x = 920; y = -500; }; time = 0; duration = 200; target_button = 2; code = 21; port_code = 21; } trial_RI; # name is "trial_RI" and means center arrow right, incongruent w/flankers trial { trial_duration = 350; stimulus_event { sound { wavefile startle; }; deltat = 300; duration = 50; }; nothing {}; # delays code corresponding to auditory stimulus to be more accurate; may need to be readjusted deltat = 20; code = 150; port_code = 150; } trial_auditory_error; #name is "trial_auditory_error" and is a white noise to be played at 108 db following an error trial { trial_duration = 350; stimulus_event { sound { wavefile startle; }; deltat = 300; duration = 50; }; nothing {}; # delays code corresponding to auditory stimulus to be more accurate; may need to be readjusted deltat = 20; code = 151; port_code = 151; } trial_auditory_predictable; #name is "trial_auditory_predictable" and is a white noise to be played at 108 db trial { trial_duration = 350; stimulus_event { sound { wavefile startle; }; deltat = 300; duration = 50; }; nothing {}; # delays code corresponding to auditory stimulus to be more accurate; may need to be readjusted deltat = 20; code = 152; port_code = 152; } trial_auditory_unpredictable; #name is "trial_auditory_unpredictable" and is a white noise to be played at 108 db trial { trial_duration = 1; #this is set to a random integer in PCL below picture { text { caption = "+"; font_size = 48; }; x=0; y=0; # centers text }; } trial_ISI; #name is "trial_ISI" trial { trial_duration = 1; #this is set to a random integer in PCL below picture { text { caption = "+"; font_size = 48; }; x=0; y=0; # centers text }; code = 101; port_code = 101; } short_ISI; #name is "short_ISI" trial { trial_duration = 3000; picture { text { caption = "Please try to be more accurate."; }; x=0; y=0; # centers text }; } trial_accurate; trial { trial_duration = 3000; picture { text { caption = "Please try to respond faster."; }; x=0; y=0; # centers text }; } trial_faster; trial { trial_duration = 3000; picture { text { caption = "You're doing a great job."; }; x=0; y=0; # centers text }; } trial_greatjob; # -------------PCL -------------# begin_pcl; loop #block loop int block_hits = 0; int j = 0; until j == 8 #number of blocks begin array <int> porkchops[30] = {1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4}; porkchops.shuffle(); loop #trial loop int i = 1; int ISI_times = 0; # variable is an integer for duration of ISI without startle int short_time = 0; # variable is an integer for duration of ISI following startle bool current = true; bool previous = true; until i > 30 begin ISI_times = random(500,1000); trial_ISI.set_duration(ISI_times); short_time = random(150,650); short_ISI.set_duration(short_time); if (porkchops[i] == 1) then trial_LC.present(); elseif (porkchops[i] == 2) then trial_RC.present(); elseif (porkchops[i] == 3) then trial_LI.present(); elseif (porkchops[i] == 4) then trial_RI.present(); end; #ends if/elseif regarding stimulus presentation if (response_manager.hits() == 1) then current = true; else current = false; end; /*ends if/else specifying state that determines probability of playing white noise, where if current = true it remains at 4% and if current = false probability goes to 50%*/ block_hits = block_hits + response_manager.hits(); #this is how I keep track of accuracy within a trial to guide the feedback at the end if (previous && current) then int auditory = random(0,24); if auditory == 0 then trial_auditory_unpredictable.present(); short_ISI.present(); # this ISI only ranges from 150-650ms because startle already took 350ms else trial_ISI.present(); end; #if both the last and the trial before it are accurate, white noise probability is 4% elseif (!previous && current) then int auditory = random(0,1); if auditory == 0 then trial_auditory_predictable.present(); short_ISI.present(); # this ISI only ranges from 150-650ms because startle already took 350ms else trial_ISI.present(); end; #if the current trial was correct but the last trial was not, white noise probability is 50% else int auditory = random(0,1); if auditory == 0 then trial_auditory_error.present(); short_ISI.present(); # this ISI only ranges from 150-650ms because startle already took 350ms else trial_ISI.present(); end; #otherwise, the probability should be 50% end; previous = current; i = i+1; end; #ends trial loop if block_hits < 22 then trial_accurate.present(); trial_ISI.present(); elseif block_hits > 27 then trial_faster.present(); trial_ISI.present(); else trial_greatjob.present(); trial_ISI.present(); end; j = j+1; # increments count of block end; #ends block loop
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// Scilab Code Ex1.22 Effective mass and speed of electron: Pg: 29 (2008) c = 3e+08; // Speed of light, m/s e = 1.6e-019; // Electron-volt equivalent of 1 joule, eV/joule U = 2*1e+06*e; // Total energy of electron, J // As E = (m - m0)*c^2, solving for m m = U/c^2; // Effective mass of electron, kg m0 = 0.511*1e+06*e/c^2; // Rest mass of the electron, kg // As m = m0/sqrt(1 - (v/c)^2), Relativistic mass of electron, kg, solving for v, we have v = sqrt(1 - (m0/m)^2)*c; // Velocity of moving electron, m/s printf("\nThe effective mass of electron = %4.1e kg", m); printf("\nThe relativistic speed of electron = %4.2fc m", v/c); // Result // The effective mass of electron = 3.6e-030 kg // The relativistic speed of electron = 0.97c m
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function out=formeQ(x,A,b) out=.5*x'*A*x-b'*x; endfunction function out=gradient(x,A,b) out=A*x-b; endfunction function trace(xk) c=formeQ(xk)-formeQ(xc); a=sqrt(2*c/d(1)); b=sqrt(2*c/d(2)); N=128; t=linspace(0,2*%pi,N); x=P*[a*cos(t);b*sin(t)]+xc(:,ones(1,N)); plot(x(1,:),x(2,:),'linewidth',2) endfunction A=[2 -1;-1 2]; b=[1 1]'; xc=A\b; // Point qui minimise le gradient (donc annule la fonction) [D,P]=bdiag(A); // Matrice reelle (blocs de diagonalisation) d=diag(D); x=[1;0]; //point de depart arbitraire rho=0.5; //rho fixe clf; set(gca(),"isoview","on"); trace(x); for i=1:50 last=x; x=last-rho*gradient(last,A,b); plot([last(1) x(1)],[last(2) x(2)],"r"); trace(x); end title("Algorithme du gradient a pas fixe",'fontsize',3);
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//Section-1,Example-3,Page no.-AC.173 clc; C=750 C_1=75 H=52 H_1=5.2 O=121 O_1=12.1 S=0 Net_O2=((32/12)*C)+((16/2)*H)-O W_air=Net_O2*(100/23)*10^-3 disp(W_air,'Weight of air required(kg)') W_air40=W_air*(140/100) disp(W_air40,'Weight of air required when 40% excess air is supplied(kg)') GCV=(1/100)*((8080*C_1)+(34500*(H_1-(O_1/8)))+(2240*S)) disp(GCV,'Gross calorific value(kCal/kg)') NCV=GCV-(0.09*H_1*587) disp(NCV,'Net calorific value(kCal/kg)')
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2023-05-25T08:21:31.003675
2023-05-11T16:19:59
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sce
outhex2volt.sce
function [voltage_vector]=outhex2volt (chip_num, board_num, hex_vector) select board_num case 2 then brdtype = ''; case 3 then brdtype = '_30a'; case 4 then brdtype = '_30n'; case 5 brdtype = '_30h'; else messagebox('Please select the FPAA board that you are using.', "No Selected FPAA Board", "error"); abort end temp_size= size(hex_vector); No_of_mite=temp_size(1,2); clear voltage_vector;voltage_vector=hex_vector; exec("~/rasp30/prog_assembly/libs/scilab_code/characterization/char_miteADC.sce",-1); // Converts Output hex values to voltages based on the Measure_Voltage block calibration file if No_of_mite ~= 0 then j=1; i_start = 1; i_end= No_of_mite; for i=i_start:i_end if mite_info_array(j,2) == 977 then voltage_vector(:,i)=polyval(p_mite_977_10uA,voltage_vector(:,i),S_mite_977_10uA); end if mite_info_array(j,2) == 978 then voltage_vector(:,i)=polyval(p_mite_978_10uA,voltage_vector(:,i),S_mite_978_10uA); end if mite_info_array(j,2) == 979 then voltage_vector(:,i)=polyval(p_mite_979_10uA,voltage_vector(:,i),S_mite_979_10uA); end if mite_info_array(j,2) == 980 then voltage_vector(:,i)=polyval(p_mite_980_10uA,voltage_vector(:,i),S_mite_980_10uA); end if mite_info_array(j,2) == 981 then voltage_vector(:,i)=polyval(p_mite_981_10uA,voltage_vector(:,i),S_mite_981_10uA); end if mite_info_array(j,2) == 982 then voltage_vector(:,i)=polyval(p_mite_982_10uA,voltage_vector(:,i),S_mite_982_10uA); end if mite_info_array(j,2) == 983 then voltage_vector(:,i)=polyval(p_mite_983_10uA,voltage_vector(:,i),S_mite_983_10uA); end if mite_info_array(j,2) == 984 then voltage_vector(:,i)=polyval(p_mite_984_10uA,voltage_vector(:,i),S_mite_984_10uA); end if mite_info_array(j,2) == 985 then voltage_vector(:,i)=polyval(p_mite_985_10uA,voltage_vector(:,i),S_mite_985_10uA); end if mite_info_array(j,2) == 986 then voltage_vector(:,i)=polyval(p_mite_986_10uA,voltage_vector(:,i),S_mite_986_10uA); end if mite_info_array(j,2) == 1009 then voltage_vector(:,i)=polyval(p_mite_1009_10uA,voltage_vector(:,i),S_mite_1009_10uA); end if mite_info_array(j,2) == 1010 then voltage_vector(:,i)=polyval(p_mite_1010_10uA,voltage_vector(:,i),S_mite_1010_10uA); end if mite_info_array(j,2) == 1011 then voltage_vector(:,i)=polyval(p_mite_1011_10uA,voltage_vector(:,i),S_mite_1011_10uA); end if mite_info_array(j,2) == 1012 then voltage_vector(:,i)=polyval(p_mite_1012_10uA,voltage_vector(:,i),S_mite_1012_10uA); end if mite_info_array(j,2) == 1013 then voltage_vector(:,i)=polyval(p_mite_1013_10uA,voltage_vector(:,i),S_mite_1013_10uA); end if mite_info_array(j,2) == 1014 then voltage_vector(:,i)=polyval(p_mite_1014_10uA,voltage_vector(:,i),S_mite_1014_10uA); end if mite_info_array(j,2) == 1015 then voltage_vector(:,i)=polyval(p_mite_1015_10uA,voltage_vector(:,i),S_mite_1015_10uA); end if mite_info_array(j,2) == 1016 then voltage_vector(:,i)=polyval(p_mite_1016_10uA,voltage_vector(:,i),S_mite_1016_10uA); end if mite_info_array(j,2) == 1017 then voltage_vector(:,i)=polyval(p_mite_1017_10uA,voltage_vector(:,i),S_mite_1017_10uA); end if mite_info_array(j,2) == 1018 then voltage_vector(:,i)=polyval(p_mite_1018_10uA,voltage_vector(:,i),S_mite_1018_10uA); end end end //disp(voltage_vector) endfunction
4f372cc80ce66e397f9ff66e4d4f71fc64003fd7
fd6e45f66c41ad779a3d47c3bf8ebfa140d3d657
/P3 - Non-linear equations /Métodos/6- newton sistemas.sce
6e71b6478acd35676ad912f1e22b7bc4b869583e
[]
no_license
jere1882/Numerical-Analysis-Assignments
7f474e2020d010f9f9c3dceff5e48c03b0d38652
1074f92ca93d0a402259f92a0f61f105f25e5230
refs/heads/master
2021-09-06T20:00:36.411386
2018-02-10T18:04:38
2018-02-10T18:04:38
121,039,769
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null
null
null
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UTF-8
Scilab
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1,206
sce
6- newton sistemas.sce
// Código resolución de sistemas no lineales function [p,k,err] = NewtonSistemas(F,p,delta,max1) //F es un vector de funciones (Un campo escalar) //p es una aproximación inicial al cero (R^n) Y=F(p) for k=1:max1 J=derivative(F,p'); Q= p - (J \ Y)'; //para volvera fila ; Z = F(Q); err = norm(Q-p); //Como el error es entre vectores, hay varias maneras de calcular //la norma -distancia- entre ellos. Usamos norm p=Q; Y=Z; //condicion para que pare: if (err<delta)|(abs(Y)<delta) then break end end endfunction deff('Y=F(X)',['Y(1)= X(1)^2+X(1)*(X(2)^3)-9', ' Y(2)= 3*(X(1)^2)*X(2)-4-X(2)^3']); // BIEN ESCRITO. Fijarse si van los .^para el cucuadrado [sol,it,err]= NewtonSistemas(F,[-2 , 2.5],10^(-12),50) // por defecto norma2 es norm //RESULTADO CON LOS 4 PUNTOS: SE VA A LAS 4 DISTINTAS SOLUCIONES... //Extension del método de newton-raphson // Sea alpha en R^2 una raíz de F : R^2 -> R^2. Si X0=(x0,y0) es una aproximacion de alpha, etnonces: // F(x0,y0) + JF (x0,y0) * (alpha-X0) = 0 //despejamos alpha = X0 - [JF(X0)]^-1 * F(X0) la iteración queda asi con alpha=xn y X0=xn-1
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/set4/s_College_Physics(volume_2)_R._A._Serway_And_J._S._Faughn_2072.zip/College_Physics(volume_2)_R._A._Serway_And_J._S._Faughn_2072/CH27/EX27.8/EX27_8.sce
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[]
no_license
hohiroki/Scilab_TBC
cb11e171e47a6cf15dad6594726c14443b23d512
98e421ab71b2e8be0c70d67cca3ecb53eeef1df6
refs/heads/master
2021-01-18T02:07:29.200029
2016-04-29T07:01:39
2016-04-29T07:01:39
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EX27_8.sce
errcatch(-1,"stop");mode(2);//Chapter 27 //Example 8 //given h=6.63*10^-34 //in J.s m_e=9.11*10^-31 // in Kg v=1*10^7 //in m/s lambda=h/(m_e*v) disp(lambda,"de Broglie wavelength for an electron in meters is") exit();
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/3869/CH1/EX1.54/Ex1_54.sce
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no_license
FOSSEE/Scilab-TBC-Uploads
948e5d1126d46bdd2f89a44c54ba62b0f0a1f5e1
7bc77cb1ed33745c720952c92b3b2747c5cbf2df
refs/heads/master
2020-04-09T02:43:26.499817
2018-02-03T05:31:52
2018-02-03T05:31:52
37,975,407
3
12
null
null
null
null
UTF-8
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Ex1_54.sce
clear // // // //Variable declaration R=70 //radius of curvature of lens(cm) n=10 Dn=0.433 //diameter of 10th dark ring(cm) //Calculation lamda=Dn**2/(4*R*n) //wavelength of light(cm) //Result printf("\n wavelength of light is %0.3f *10**-5 cm",lamda*10**5) printf("\n answer given in the book varies due to rounding off errors")
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/2420/CH3/EX3.3/3_3.sce
f60a797b0642834d6ff32a074559e0528b408f7b
[]
no_license
FOSSEE/Scilab-TBC-Uploads
948e5d1126d46bdd2f89a44c54ba62b0f0a1f5e1
7bc77cb1ed33745c720952c92b3b2747c5cbf2df
refs/heads/master
2020-04-09T02:43:26.499817
2018-02-03T05:31:52
2018-02-03T05:31:52
37,975,407
3
12
null
null
null
null
UTF-8
Scilab
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false
136
sce
3_3.sce
clc clear //Initialization of variables per=20 Dp=100 //calculations r=Dp/per +1 //results printf("Compression ratio = %d ",r)
07020ffcb69ed05a51473bfc388c50c60916467d
1ffd0259451af009bc55a18827746ae10e9da8ef
/task1/norm1.sce
768dee83a4a86d32563e36d23fd27b2c5c2afc5d
[]
no_license
oborovsky/vychmeth
fb7c0f2e77249ec4fea40d7a05dac2740f8e9082
ccef228095b99798e64946af41029c7b79b505ab
refs/heads/master
2020-05-31T00:09:44.080491
2016-05-05T19:10:18
2016-05-05T19:10:18
42,015,944
0
0
null
null
null
null
UTF-8
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8,981
sce
norm1.sce
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plot(xh,yy,'m--'); plot(xh,yp,'r'); plot(x,y,'*'); xgrid(); xtitle('norm delta=0.22707','X', 'Y');
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2023-05-29T04:40:42.507269
2021-06-04T13:25:58
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NewtonRaphsonMethod.sce
clc; deff('y=f(x)','y=x^3+x^2-3*x-3') deff('y=df(x)','y=3*x^2+2*x-3') x(1)=input('Enter Initial Guess:'); e= input("Answer correct upto : "); for i = 1 : 100 x(i+1)=x(i)-f(x(i))/df(x(i)); err(i)=abs((x(i+1)-x(i))/x(i)); if err(i) < e break; end end printf('The solution is %f',x(i))
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TrapezoidalRule.sce
clc; clear; close; deff('y=f(x)','y=1/(1+x^2)') x0=0; xn=6; n=6; h=(xn-x0)/n; s=0; for i=1:n s=s+f(x0+(i-1)*h)+f(x0+i*h); end integral=(h*s)/2; printf('\nThe value of integral is=%g\n',integral)
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example_6_7.sce
//Chapter 6 //Example 6-7 //ProbOnUnipolarTriangularWaveGenerator //Page 159 clear;clc; //Given p = 2.8 ; Vsatm = -13.8 ; Ri = 28*10^3 ; C = 0.05*10^-6; Vut = - ((Vsatm+0.6)/p); f = p / (2*Ri*C); printf ( "\n\n Peak Voltage = %.4f V ", Vut ) printf ( "\n\n frequency = %.4f Hz ", f )
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iris.sce
data = csvRead('Iris.csv',[],[],'string') dataNew = data(2:151,2:6) c1 = dataNew(1:150,5) == 'Iris-setosa' c2 = dataNew(1:150,5) == 'Iris-versicolor' c3 = dataNew(1:150,5) == 'Iris-virginica' temp = csvRead('Iris.csv') temp = temp(2:151,2:5) temp(1:150,5) = c1 temp(1:150,6) = c2 temp(1:150,7) = c3 data = temp X = data(:,1:4)' y = data(:,5:7)' [X, y] = ann_pat_shuffle(X, y) split = size(X)(2)*0.7 X_train = X(:,1:split) y_train = y(:,1:split) X_test = X(:,split+1:size(X)(2)) y_test = y(:,split+1:size(X)(2)) N = [4 16 3] // [4 16 3] - 97.77 lp=0.1 epochs-20 W = ann_FF_init(N) lp = [0.3,0.000001] T = 100 t = 1 train_losses = [] test_losses = [] train_accuracies = [] test_accuracies = [] while t<=T W_updated = ann_FF_Std_online(X_train, y_train, N, W, lp, 1) //W_updated = ann_FF_Std_batch(X_train, y_train, N, W, lp, 1) W = W_updated y_train_predicted = ann_FF_run(X_train, N, W) y_test_predicted = ann_FF_run(X_test, N, W) disp(t,"Epoch: ") train_loss = ann_sum_of_sqr(y_train_predicted, y_train)/split disp(train_loss,'Training Loss: ') test_loss = ann_sum_of_sqr(y_test_predicted, y_test)/(size(X)(2) - split) disp(test_loss,'Testset Loss: ') train_accuracy = 1 - ann_sum_of_sqr(y_train_predicted>=0.5,y_train)/split test_accuracy = 1 - ann_sum_of_sqr(y_test_predicted>=0.5,y_test)/(size(X)(2) - split) disp(train_accuracy,'Training Accuracy: ') disp(test_accuracy,'Test Accuracy: ') disp("") train_losses = [train_losses train_loss] test_losses = [test_losses test_loss] train_accuracies = [train_accuracies train_accuracy] test_accuracies = [test_accuracies test_accuracy] if t>=2 then clf(1) figure(1) plot((1:t)',[train_losses;test_losses]') //plot(1:t,test_losses,c='b') xlabel('Epochs') ylabel('Loss') title('Train & Test Losses') hl=legend(['train loss';'test loss']'); clf(2) figure(2) plot((1:t)',[train_accuracies;test_accuracies]') //plot(1:t,test_accuracies,c='b') xlabel('Epochs') ylabel('Accuracy') title('Train & Test Accuracy') ha=legend(['train accuracy';'test accuracy']'); end if test_accuracy >= 0.98 break end t = t+1 end
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ceconelli/Metodos-Computacionais
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Bissecao.sci
function[y]= f(x) y = x^2-3*x-10; endfunction function[Raiz,Iter,CondErro]=Bissecao(a,b,Toler,IterMax) Fa = f(a); Fb = f(b); if(Fa*Fb>0) then disp("Função nao muda de sinal nos extremos do intervalo dado"); return; end DeltaX = abs(b-a)/2; Iter = 0; while 1 x = (a+b)/2; Fx = f(x); disp(Iter,a,Fa,b,Fb,x,Fx,DeltaX); if((DeltaX<=Toler & abs(Fx)<=Toler) | Iter>=IterMax) then break; end if(Fa*Fx>0) then a = x; Fa = Fx; else b = x; end DeltaX = DeltaX/2; Iter = Iter+1; end Raiz = x; if(DeltaX<=Toler & abs(Fx)<=Toler) then CondErro = 0; else CondErro = 1; end endfunction
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//Determine ABS/BH switch calling rate and CCS for a switch RL = 12000; n = 80000; BL = 64000; HL = 4000; CRr = 2; CRb = 3; CRh = 10; HTr = 140; HTb = 160; HTh = 200; RLp = RL/n; BLp = BL/n; HLp = HL/n; CCSrl = CRr * (HTr/100); CCSbl = CRb * (HTb/100); CCShl = CRh * (HTh/100); SCR = (CRr*RLp) + (CRb*BLp) + (CRh*HLp) ; Sccs = (CCSrl*RLp) + (CCSbl*BLp) + (CCShl*HLp) ; Aht = (Sccs/SCR)*100; ABSc = SCR*n; ABSu = (Sccs*n)/36; Dcc = 1.5*ABSc; De = 1.5*ABSu; disp(Dcc, 'Design call capacity based on HD') disp(De, 'Design erlangs based on HD')
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clc clear printf("example 5.5 page number 173\n\n") //to find the rate of heat loss through pipeline //resistance by pipeline d1=0.15 //in m d2=0.16 //in m l=1 //in m A1=3.14*d1*l; A2=3.14*d2*l Am1=(A2-A1)/log (A2/A1); x1=(d2-d1)/2; k1=50 //in W/mK R1=x1/(k1*Am1); //resistance by insulation d2=0.16 //in m d3=0.26 //in m l=1 //in m A2=3.14*d2*l; A3=3.14*d3*l Am2=(A3-A2)/log (A3/A2); x2=(d3-d2)/2; k2=0.08 //in W/mK R2=x2/(k2*Am2); R=R1+R2; printf("total resistance = %f K/W",R) T1=120; //in K T2=40; //in K delta_T=T1-T2; Q=delta_T/R; printf("\n\nheat loss = %f W/m",Q)
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CS16B019HalfAdder.tst
load CS16B019HalfAdder.hdl, output-file CS16B019HalfAdder.out, compare-to CS16B019HalfAdder.cmp, output-list a%B3.1.3 b%B3.1.3 sum%B3.1.3 carry%B3.1.3; set a 0, set b 0, eval, output; set a 0, set b 1, eval, output; set a 1, set b 0, eval, output; set a 1, set b 1, eval, output;
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/plus_tau_cont-Omega.sci
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2020-03-16T18:48:14.182188
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sci
plus_tau_cont-Omega.sci
clear; // function that determines p_i function y=fun1(x) y=sin(5*x*%pi)+1; endfunction // function that determines r_i function y=fun2(x) //y=cos(2*x*%pi)+1; y=(-0.2+x)^2; endfunction cp=intg(0,1,fun1); cr=intg(0,1,fun2); function y=fun11(x) y=fun1(x)/cp; endfunction function y=fun21(x) y=fun2(x)/cr; endfunction d=0.01;//the length of the intervals of partitioning at the graphic drawn with dash line m=1+1/d; x=0; for i=1:m p(i)=fun11(x); r(i)=fun21(x); T(i)=x; x=x+d; end p0=p; r0=r; po1=0.0/d; po2=0.3/d; ro1=0.4/d; ro2=0.6/d; po3=0.7/d; po4=0.8/d; ro3=0.9/d; ro4=1/d; N=1000;// number of steps (graphic drawn with dash line) for k=1:N // Theta = sum(sqrt(p.*r)*d); Theta1=0; Theta2=0; for i=1:m if (0<=i)&(i<po1) then tau(i)=0 end if (po1<=i)&(i<po2) then tau(i)=p(i); Theta1=Theta1*d+p(i)*d; end if (po2<=i)&(i<ro1) then tau(i)=0 end if (ro1<=i)&(i<ro2) then tau(i)=r(i); Theta2=Theta2*d+r(i)*d; end if (ro2<=i)&(i<po3) then tau(i)=0 end if (po3<=i)&(i<po4) then tau(i)=p(i); Theta1=Theta1*d+p(i)*d; end if (po4<i)&(i<ro3) then tau(i)=0 end if (ro3<=i)&(i<ro4) then tau(i)=r(i); Theta2=Theta2*d+r(i)*d; end if ro4<=i then tau(i)=0; end end Theta=Theta1*Theta2; W = sum(tau.*d); z = 1+W+Theta; p = (p.*(1+Theta)+tau)./z; r = (r.*(1+Theta)+tau)./z; end if 1==1 then mu0(1)=p0(1)*d; nu0(1)=r0(1)*d; mu(1)=p(1)*d; nu(1)=r(1)*d; mtau(1)=tau(1)*d; for i=2:m mtau(i)=mtau(i-1)+tau(i)*d; mu0(i)=mu0(i-1)+p0(i)*d; nu0(i)=nu0(i-1)+r0(i)*d; mu(i)=mu(i-1)+p(i)*d; nu(i)=nu(i-1)+r(i)*d; end plot(T,mu0,'b--'); plot(T,mu,'b.-'); plot(T,nu0,'black--'); plot(T,nu,'black'); plot(T,mtau,'r'); end if 2==1 then plot(T,p,'b.-'); plot(T,p0,'b--'); plot(T,r0,'black--'); plot(T,r,'black'); //plot(T,tau1,'*'); // legend('p(x)','r(x)','limit'); end
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// Part (a) T1 = 273; T2 = 373; m = 1 ; cv = 4.187; Ss = m*cv*log(T2/T1); // S = S2-S1 Q = m*cv*(T2-T1); Sr = -(Q/T2); S = Ss+Sr; disp("kJ/K",S,"The entropy change of the universe is") // Part (b) T3 = 323; Sw = m*cv*(log(T3/T1)+log(T2/T3)); Sr1 = -m*cv*(T3-T1)/T3; Sr2 = -m*cv*(T2-T3)/T2; Su = Sw+Sr1+Sr2; disp("kJ/K",Su,"The entropy change of the universe is")
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//filter result //here in test case we are comparing the outputs of idfilt and filt function defined //here for a given input ,if the output of these filters will be same for given same //input the orresponding result of rbs function will also be same as rest all part of //rbs function just adjusts the outputs from the filter with their corresponding sign //and levels. //test case input signals names u1 to u5,their corresponding bands band11,band21,band31..band51 and no.of channels nu1,nu2..nu5. loadmatfile('testfile2.mat') function[u]=filt(data,band1,nu) //data:-input signal to filter,band:-band,nu:-no.of channels of input signal [lhs,rhs]=argn(0) band=band1/2 //dividing the band by 2 as we are taking data from matlab and here band is defined in between [0 0.5].while in matlab it is defined in bw [0 1] if ~and(band==[0 0.5]) then if(band(1)==0) then [hz]=iir(8,'lp','butt',[band(2) band(1)],[0 0]); //8th order butterwoth filter num=hz(2); den=hz(3); for i=1:1:nu y(:,i)=filter(num,den,data(:,i)); end elseif(band(2)==0.5) then [hz]=iir(8,'hp','butt',[band(1) band(2)],[0 0]); //8th order butterwoth filter num=hz(2); den=hz(3); for i=1:1:nu y(:,i)=filter(num,den,data(:,i)); end else [hz]=iir(8,'bp','butt',band,[0 0]); //8th order butterwoth filter num=hz(2); den=hz(3); for i=1:1:nu y(:,i)=filter(num,den,data(:,i)); end end u=y else u=data; end endfunction //testu1=filt(u1,band11,nu1); //testu2=filt(u2,band21,nu2); //testu3=filt(u3,band31,nu3); //testu4=filt(u4,band41,nu4); //testu5=filt(u5,band51,nu5);
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//Scilab Code Ex2.9 :Page:82 (2011) clc;clear; n = 1; m0 = 9.1e-031;....// Mass of the electron, kg a = 1e-10;....// Width of the box, m h = 6.63e-034;....// Planck's constant, J-s E = n^2*h^2/(8*m0*a^2); printf("\n The energy of the electron moving in 1D infinetly high potential box = %5.2e J", E); // Result // The energy of the electron moving in 1D infinetly high potential box = 6.04e-18 J
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// Chapter9 // Page.No-402, Figure.No-9.16(a) // Example_9_5 // Value of capacitor // Given clear;clc; Ra=10*10^3; // Resistance in ohm tp=10*10^-3; // Output pulse width C=tp/(1.1*Ra); printf("\n Capacitance C is = %.9f farad \n",C) // Approximately 1uF
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errcatch(-1,"stop");mode(2);//Example 7_3 ; ; //To compare the acceptance angle NA=0.3 thetaa=asin(NA)*180/%pi //units in degrees theta1=asin(NA/sin(45*%pi/180))*180/%pi //units in degrees printf("for meridional rays theta=%.2f degrees",thetaa) printf("\n for skew rays theta=%.2f degrees",theta1) exit();
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clear clc //// Retorna la representación binaria de un número en Scilab, como //// dos listas (parte entera y decimal) de tamaño fijo. function [E, D] = my_dec2bin (x) ent = floor(x); dec = x - ent; step = 30; while(step >= 1) E(step) = modulo(ent, 2); ent = floor(ent / 2); step = step - 1; end step = 30; while(step >= 1) dec = dec * 2; D(step) = floor(dec); dec = dec - floor(dec); step = step - 1; end E = E' D = D' endfunction //// Inversa de my_dec2bin function x = my_bin2dec(E, D) x = 0; for i = 1:30 x = x + E(i) * 2^(30 - i) end for i = 1:30 x = x + D(i) * 2^(i - 31) end endfunction
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// example:-3.2,page no.-87. // program to find out load impedence. Zo=100; // characteristic impedence. tao=0.560+0.215*%i; // reflection coefficient. z=(1+tao)/(1-tao); // normalized impedence(normalized w.r.t Zo) Zl=z*Zo; // result disp(Zl,'load impedence = ') // by smith chart. smith_chart(tao) // when analyse with the help of smith chart.see the angle from x=0 axis i.e Tao_real axis.if it is above this axis take angle anticlockwise and if it is below this axis.take angle clockwise from Tao_real axis below.
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-- The intellectual and technical concepts contained herein are proprietary to Fuzzy Logix, LLC. -- and may be covered by U.S. and Foreign Patents, patents in process, and are protected by trade -- secret or copyright law. Dissemination of this information or reproduction of this material is -- strictly forbidden unless prior written permission is obtained from Fuzzy Logix, LLC. -- Functional Test Specifications: -- -- Test Category: Date Functions -- -- Test Unit Number: FLIntToDate-TD-01 -- -- Name(s): FLIntToDate -- -- Description: Scalar function which converts an integer to date. The integer could be positive or negative and represents the difference in days from Jan1, 1990. The scalar function return the date Jan 1, 1990 for the integer value zero. -- -- Applications: -- -- Signature: FLIntToDate(pNumOfDays INTEGER) -- -- Parameters: See Documentation -- -- Return value: Date -- -- Last Updated: 05-11-2017 -- -- Author: <Zhi.Wang@fuzzyl.com, Joydeep.Das@fuzzyl.com>, Sam Sharma -- -- BEGIN: TEST SCRIPT -- .run file=../PulsarLogOn.sql -- .set width 2500 -- set session dateform = ANSIDATE ; -- BEGIN: POSITIVE TEST(s) ---- Positive Test 1: Manual Example --- Same Output, Good SELECT FLIntToDate(-726467) AS FLIntToDate1, FLIntToDate(-726468) AS FLIntToDate2, FLIntToDate(0) AS FLIntToDate3, FLIntToDate(5678) AS FLIntToDate4, FLIntToDate(2925591) AS FLIntToDate5, FLIntToDate(2925592) AS FLIntToDate6; ---- Positive Test 2: And more for additional coverage SELECT FLIntToDate(-72646) AS FLIntToDate1, FLIntToDate(-42646) AS FLIntToDate2, FLIntToDate(992559) AS FLIntToDate3; ---- Positive Test 3: Additional coverage, multiples of 10x SELECT FLIntToDate(10000 ); SELECT FLIntToDate(100000 ); SELECT FLIntToDate(1000000 ); SELECT FLIntToDate(10000000); -- END: POSITIVE TEST(s) -- BEGIN: NEGATIVE TEST(s) ---- Negative Test 1: Invalid Data Type --- Return expected error msg, Good SELECT FLIntToDate(NULL) AS FLIntToDate1; SELECT FLIntToDate(1.2) AS FLIntToDate1; -- SELECT FLIntToDate('2010: type') AS FLIntToDate1; -- SELECT FLIntToDate(CAST ('01/01/0001' AS DATE)) AS FLIntToDate1; -- SELECT FLIntToDate(CAST ('0001-01-01 00:00:00.000000' AS TIMESTAMP)) AS FLIntToDate1; -- END: NEGATIVE TEST(s) -- END: TEST SCRIPT
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clc V=0.35; //m^3 m_CO=0.4; //kg m_air=1; //kg m_O2=0.233; //kg m_N2=0.767; //kg T=293; //K R0=8.314; //kJ/kg K M_O2=32; //Molecular mass of O2 M_N2=28; //Molecular mass of N2 M_CO=28; //Molecular mass of CO disp("Partial Pressures=") p_O2=m_O2*R0*10^3*T/M_O2/V/10^5; //bar disp("partial pressure for p_O2") disp(p_O2) disp("bar") p_N2=m_N2*R0*10^3*T/M_N2/V/10^5; //bar disp("partial pressure for p_N2") disp(p_N2) disp("bar") p_CO=m_CO*R0*10^3*T/M_CO/V/10^5; //bar disp("partial pressure for p_CO") disp(p_CO) disp("bar") disp("(ii) Total pressure in the vessel") p=p_O2+p_N2+p_CO; disp("p=") disp(p) disp("bar")
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//example-21.2 //page no-621 //given //relative permeability of superalloy mur=200000 mu0=4*(%pi)*10^-7 //henry/m //intensity of magnetisation M=6000 //A/m //magnetic field is given by H=M/(mur-1) //A/m //strength of magnet B=mu0*mur*H //tesla printf ("the strength of magnet is %f T",B)
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//Caption:transfer_function // example 3.2.10 //page 43 // we have defined parallel and series function which we are going to use here //exec parallel.sce; //exec series.sce; syms G1 G2 G3 G4 G5 H1 H2 H3; a=parallel(G3,G4); //shift off the take off point before block 'a' to after block 'a' b=1/a; d=1; c=G2/(1+G2*d); e=parallel(H1,b); f=series(c,a); g=series(H2,e); h=f/(1+f*g); h=simple(h); i=series(h,G1); y=i/(1+i*H3); y=simple(y); disp(y,"C(s)/R(s)=")
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//Cargamos el archivo guardado de "linearizing" load("PLANTABIPEDO.sod","X","U","sys") ////Ponemos las matrices obtenidas de Scilab/// A=sys.A B=sys.B C=sys.C D=sys.D C=C(1:2,:); D=zeros(2,2); /////Controllability and Observability///// ///Matrices Scilab/// //controlabilidad Cc = cont_mat(A,B) rankCc=rank(Cc) //observabilidad O = obsv_mat(A, C) rankO=rank(O) ///Matrices Analiticas/// //controlabilidad //Cc1 = cont_mat(A1,B1) //rankCc=rank(Cc1) //observabilidad //O1 = obsv_mat(A1,C1) //rankO=rank(O1) /////Plotear valores singulares////// ///Valores matrices Scilab/// G = syslin('c', A, B, C, D); tr = trzeros(G) w = logspace(-3,3); sv = svplot(G,w); //ploteo valores singulares ambos modelos// //primer ploteo para las matrices de Scilab scf(1); plot2d("ln", w, [20*log(sv')/log(10)]) xgrid(12) xtitle("Valores Singulares de la Planta","Frequency (rad/s)", "Amplitude (dB)"); ////Obtencion de las funciones de transferencia//// //MatricesScilab// [h]=ss2tf(sys) //MatricesAnaliticas// //[h1]=ss2tf(G1) ////Obtencion de polos y ceros///// //Matrices Scilab// scf(2) plzr(h); xtitle("Polos y zeros de la Planta") //Matrices analiticas //scf(5) //plzr(h1); //xtitle("Polos y zeros matrices analiticas") // Escalonamiento a la planta // su = diag( [0.9614, 0.2753] ) sx = diag( [3.157, 11.47, 3.157, 11.47] ) sy = diag( [3.157 3.157] ) ap_ = sx*A*inv(sx) bp_ = sx*B*inv(su) cp_ = sy*C*inv(sx) dp_ = sy*D*inv(su) Gs_= syslin("c",ap_, bp_, cp_, dp_) // Valores singulares de la planta escalonada // sv1 = svplot(Gs_,w); scf(3) plot2d("ln", w, [20*log(sv1')/log(10)]) xgrid(12) xtitle("Valores Singulares de Planta Escalonada","Frequency (rad/s)", "Amplitude (dB)"); // Planta aumentada con integradores antes del proyecto de controlador [ns,nc] = size(bp_); //ns = número de entradas; //nc = número de controles; a_1 = [ap_ bp_ ; 0*ones(nc,ns) 0*ones(nc,nc) ]; b_1 = [0*ones(ns,nc); eye(nc,nc)]; c_1 = [cp_ 0*ones(nc,nc)]; d_1 = 0*ones(nc,nc) Gs_1= syslin("c",a_1, b_1, c_1, d_1) // Valores singulares de la planta escalonada con el integrador // sv2 = svplot(Gs_1,w); scf(5) plot2d("ln", w, [20*log(sv2')/log(10)]) xgrid(12) xtitle("Valores Singulares de Planta Escalonada con Integradores","Frequency (rad/s)", "Amplitude (dB)"); // LQR controller calculation // Recuperar Target Loop resolviendo un problema de LQR barato q = c_1'*c_1; //Matriz de ponderación del estado rho = 1e-9; //Parámetro de recuperación de control barato r = rho*eye(nc,nc) //Matriz de ponderación de control //how we calculate B B=b_1*inv(r)*b_1'; A=a_1; //Solv the ricatti equation X=riccati(A,B,q,'c','eigen'); //matriz de ganacia G_1=inv(r)*b_1'*X; ////// PREGUNTA 7 ////// //calculate observer Kalman Filter ll = inv(cp_*inv(-ap_)*bp_ + dp_); lh = -inv(ap_)*bp_*ll; l = [lh //ll, lh - Para la conformación de ll]; //bucles de baja y alta frecuencia. Gs_2= syslin("c",a_1, l, c_1, d_1) // Valores singulares del filtro de bucle Abierto // sv3 = svplot(Gs_2,w); scf(6) plot2d("ln", w, [20*log(sv3')/log(10)]) xgrid(12) xtitle("Valores Singulares de Filtro Abierto","Frequency (rad/s)", "Amplitude (dB)"); // Filtro de Kalman pnint=eye(nc,nc) //Proceso de matriz de intensidad de ruido mu=0.01; //Medicion de la intensidad de ruido mnint=mu*eye(nc,nc) //Matriz de intensidad de ruido de medicion Ch=l*l'; //Forma de Ch para "riccati" segun Scilab Ah=a_1'; //Forma de Ah para "riccati" segun Scilab Bh=c_1'*inv(mnint)*c_1; Xh=riccati(Ah,Bh,Ch,'c','eigen'); //ganacia H H_1=(inv(mnint)*c_1*Xh)'; Gs_3= syslin("c",a_1, H_1, c_1, d_1) // Valores singulares del observador Filtro Kalman sv4 = svplot(Gs_3,w); scf(7) plot2d("ln", w, [20*log(sv4')/log(10)]) xgrid(12) xtitle("Valores Singulares de Filtro de Kalman","Frequency (rad/s)", "Amplitude (dB)"); //ACTIVAR ESTA PARTE PARA LOS POLOS Y CEROS DEL FILTRO DE KALMAN GANACIA H [h2]=ss2tf(Gs_3) scf(8) plzr(h2); xtitle("Polos y Zeros del Filtro Kalman, Ganancia H") /////// PREGUNTA 8 //////// // COMPENSADOR K(S) DE LA FORMA DEL PORF. RODRIGUEZ // ak = [ a_1-b_1*G_1-H_1*c_1 0*ones(ns+nc,nc) G_1 0*ones(nc,nc) ]; bk = [ H_1 0*ones(nc,nc) ]; ck = [0*ones(nc, ns+nc) eye(nc,nc) ]; dk = [0*eye(nc,nc)]; Gs_4= syslin("c",ak, bk, ck, dk) // Valores singulares del compensador "K(s)" // sv4 = svplot(Gs_4,w); scf(9) plot2d("ln", w, [20*log(sv4')/log(10)]) xgrid(12) xtitle("Valores Singulares del compensador Ks","Frequency (rad/s)", "Amplitude (dB)"); /////// PREGUNTA 9 /////// //SENSIBILIDAD "S" Y SENSIBILIDAD COMPLEMENTARIA "T" //Analisis en bucle abierto al = [ ap_ bp_*ck 0*ones(ns+nc+nc,ns) ak ]; bl = [ 0*ones(ns,nc) bk ]; cl = [ cp_ 0*ones(nc,ns+nc+nc) ]; dl = [0*eye(nc,nc)]; Gs_5= syslin("c",al, bl, cl, dl) // Valores Singulares de bucle abierto// sv5 = svplot(Gs_5,w); scf(10) plot2d("ln", w, [20*log(sv5')/log(10)]) xgrid(12) xtitle("Valores Singulares del bucle abierto","Frequency (rad/s)", "Amplitude (dB)"); // Valores singulares de sensibilidad S // Gs_6= syslin("c",al-bl*cl, bl, -cl, eye(nc,nc)) sv6 = svplot(Gs_6,w); scf(11) plot2d("ln", w, [20*log(sv6')/log(10)]) xgrid(12) xtitle("Ploteo de la Sensibilidad","Frequency (rad/s)", "Amplitude (dB)"); // Valores singulares de sensibilidad complementaria T // Gs_7= syslin('c',al-bl*cl, bl, cl, dl) sv7 = svplot(Gs_7,w); scf(12) plot2d("ln", w, [20*log(sv7')/log(10)]) xgrid(12) xtitle("Ploteo de la Sensibilidad Complementaria","Frequency (rad/s)", "Amplitude (dB)"); // Valorers Singulares de S y T juntos // scf(13) plot2d("ln", w, [20*log(sv6')/log(10)]) plot2d("ln", w, [20*log(sv7')/log(10)]) xgrid(12) xtitle("Sensibilidad y Sensibilidad Complementaria","Frequency (rad/s)", "Amplitude (dB)");
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//clear// //Caption:Program to Calcualte Optical Power Emitted from the Light source and Optical power coupled to step-index fiber //Example5.2 //page194 clear; close; clc; rs = 35e-06;//the source radius in meter a = 25e-06; //the core radii of step-index fiber meter NA = 0.20; //the numerical aperture value Bo = 150e04;// radiance in W/square meter.sr Ps = ((%pi^2)*(rs^2))*Bo;//power emitted by the source if (rs <=a) then PLED_step = Ps*(NA^2); elseif (rs>a) then PLED_step = (((a/rs)^2)*Ps)*(NA^2); end disp(Ps,'Optical power emitted by LED light source Ps =') disp(PLED_step,'Optical Power coupled into step index fiber in Watts PLED_step ='); //RESULT //Optical power emitted by LED light source Ps = 0.0181354 //Optical Power coupled into step index fiber in Watts PLED_step = 0.0003701
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// Scilab code Ex8.16: Diffraction of thermal neutrons from planes of Ni crystal Page 294 (2010) k = 1.38e-023; // Boltzmann constant, J/mol/K h = 6.626e-034; // Planck's constant, Js theta = 28.5; // Bragg's angle, degree a = 3.52e-010; // Lattice parameter of fcc structure of nickel, m m_n = 1.67e-027; // Rest mass of neutron, kg // For fcc lattice, the interplanar spacing is given by d = a/sqrt(3); // Interplanar spacing of Ni, m // Bragg's equation for first order diffraction (n = 1) is lambda = 2*d*sind(theta); // Bragg's law, m // From kinetic interpretaion of temperature, we have // (1/2)*m*v^2 = (3/2)*k*T -- (a) // Further from de-Broglie relation // lambda = h/(m*v) -- (b) // From (a) and (b), solving for T, we have T = h^2/(3*m_n*k*lambda^2); // Effective temperature of the neutrons, K printf("\nThe effective temperature of neutrons = %d K", T); // Result // The effective temperature of neutrons = 168 K
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# eta= 0.300, alfa= 0.900, epoch=750, hidden=35 310 1.4400 13.3333 2917 1250 26 424 1.2600 15.2000 2888 1625 20 403 1.2000 13.9333 2838 2166 15 517 1.1800 14.2667 2844 1710 19 490 1.2400 14.0667 2839 1710 19 641 1.3000 14.9333 2837 1911 17 668 1.1400 15.3333 2837 1911 17 705 1.2000 13.2667 2837 1625 20 525 1.1800 14.9333 2840 1805 18 348 1.2200 14.9333 2837 1911 17 Epoch: (503 129.012) 1%: ( 1.236 0.080) 15%: ( 14.420 0.716) Time: (2851 54.59) Performance: (1762 234.70)
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//Fuels and Combustion// //Example 8.1// w=1500;//quantity of water in grams// W=125;//Water equivalent of calorimeter in grams// x=1.050;//quantity of fuel carried out in combustion in grams// t1=25;//initial temperature of water in degree C// t2=27.8;//final temperature of water in degree C// Q=(w+W)*(t2-t1)/x;//calorific value of the fuel in cal per grams// printf('Calorific value of the fuel=Q=%fcal/g',Q);
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//Example 7.4 page no 209 clear clc R1=200*10^3 R2=800*10^3 Zin=(R1*R2/(R1+R2))/1000 printf("\n The value of Zin=%0.3f Kohm",Zin) Rg=160*10^3 r1=5*10^3 vgs=Rg/(Rg+r1) printf("\n The value of vgs=%0.3f vi",vgs) Av=-1.88 Rl=2*10^3 Ai=(Av*(Rg+r1))/Rl printf("\n The value of ai=%0.3f vi",Ai)
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//Example 8.6.7: resistance and capacitance clc; clear; close; //given data : format('v',8) C2=106*10^-12;// in farad C4=0.6*10^-6;// in farad R4=1000/%pi;// in ohm R3=250;// in ohm R1=(C4/C2)*R3*10^-6; disp(R1*10^6,"resistance,R1(ohm) = ") C1=(R4/R3)*C2*10^6; disp(round(C1*10^6),"capacitance,C1(micro-farad) = ")
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printf "a\nA\n8\n 8" > foo; node sort foo; rm foo 8 8 A a
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N = input('Insira o tamanho do vetor: '); pos = input ('Insira a posição de inicio do degrau: '); a = input('Insira a amplitude do degrau: '); vet = zeros([0:1:N]); //vetor de 0 até N preenchido com zeros for j = pos:1:N+1 vet(j) = a; end disp(vet)
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we we wy wy wy x1 x2 Klasa1 Klasa2 Klasa3 -9,697 9,175 1 0 0 -8,636 8,045 1 0 0 -7,991 7,259 1 0 0 -7,221 6,718 1 0 0 -6,076 5,613 1 0 0 -4,932 4,483 1 0 0 -3,912 3,034 1 0 0 -2,996 2,518 1 0 0 -2,143 1,462 1 0 0 -1,269 0,626 1 0 0 -1,748 -0,381 1 0 0 -3,371 -0,061 1 0 0 -4,62 0,872 1 0 0 -6,888 3,746 1 0 0 -8,262 3,992 1 0 0 -9,406 3,157 1 0 0 -9,385 1,265 1 0 0 -6,742 0,258 1 0 0 -7,866 0,332 1 0 0 -8,22 -2,027 1 0 0 -7,367 -1,732 1 0 0 -6,659 -1,167 1 0 0 -5,39 -0,626 1 0 0 -4,287 -2,739 1 0 0 -3,329 -1,977 1 0 0 -6,867 -2,297 1 0 0 -9,177 -0,798 1 0 0 -7,471 5,195 1 0 0 -8,532 6,767 1 0 0 -9,531 4,778 1 0 0 6,722 6,178 0 1 0 7,679 7,259 0 1 0 8,741 7,971 0 1 0 9,261 8,634 0 1 0 5,681 5,416 0 1 0 4,87 4,385 0 1 0 3,684 2,985 0 1 0 2,04 1,977 0 1 0 0,479 0,504 0 1 0 1,062 -0,135 0 1 0 1,582 -0,798 0 1 0 2,393 -1,486 0 1 0 2,997 -2,1 0 1 0 3,538 -2,69 0 1 0 4,35 -1,584 0 1 0 5,681 -0,282 0 1 0 5,14 0,577 0 1 0 6,098 1,388 0 1 0 7,658 1,363 0 1 0 8,99 1,191 0 1 0 8,574 2,862 0 1 0 8,262 4,041 0 1 0 6,16 3,279 0 1 0 4,433 2,027 0 1 0 3,33 1,167 0 1 0 2,373 0,43 0 1 0 5,827 -3,034 0 1 0 6,597 -3,009 0 1 0 6,431 -1,216 0 1 0 8,283 -1,56 0 1 0 -4,953 -4,851 0 0 1 -3,683 -3,722 0 0 1 -1,228 -3,623 0 0 1 0,541 -3,623 0 0 1 2,477 -3,623 0 0 1 3,933 -4,016 0 0 1 5,057 -5,073 0 0 1 6,035 -5,932 0 0 1 6,743 -6,866 0 0 1 7,742 -7,775 0 0 1 8,449 -8,634 0 0 1 0,562 -5,785 0 0 1 -1,748 -5,711 0 0 1 -0,603 -6,596 0 0 1 -2,83 -6,62 0 0 1 -4,162 -5,613 0 0 1 -5,119 -6,522 0 0 1 -4,162 -7,554 0 0 1 -6,68 -7,48 0 0 1 -7,7 -7,431 0 0 1 -0,041 -9,199 0 0 1 1,499 -9,101 0 0 1 2,435 -9,003 0 0 1 4,745 -7,578 0 0 1 -7,929 -8,806 0 0 1 -5,848 -8,978 0 0 1 -4,016 -8,929 0 0 1 -1,935 -7,701 0 0 1 -6,201 -6,055 0 0 1 1,166 -7,652 0 0 1
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clc clear //Input data c=50//Capacitor in micro F Vm=220//Maximum voltage in V f=50//Frequency in Hz //Calculations Xc=(1/(2*3.14*c*10^-6*f))//Reactance in ohms I=(Vm/Xc)//Maximum current in A Irms=I/sqrt(2)//rms current in A //Output printf('rms current is %3.2f A',Irms)
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// 08.09.10 // 08.09.16 // 09.12.31 (gsort) function OutL=Clipindomain(ObjL,FigL) EEps=10^(-4); Eps=0.01; Eps2=0.2; Bdy=Kyoukai(FigL); if Mixlength(Bdy)>=2 Fbdy=Joincrvs(FigL) else Fbdy=Mixop(1,FigL); end; if Mixtype(ObjL)==1 ObjL=Mix(ObjL) end; if Mixtype(FigL)==1 FigL=Mix(FigL) end; OutL=[]; for Nobj=1:Mixlength(ObjL); Obj=Mixop(Nobj,ObjL); ParL=[1,Numptcrv(Obj)]; Tmp=IntersectcrvsPp(Obj,Fbdy,Eps,Eps2); for J=1:Mixlength(Tmp) Tmp1=Mixop(J,Tmp); ParL=[ParL,Mixop(2,Tmp1)]; end; ParL=gsort(ParL); ParL=ParL(length(ParL):-1:1); Tmp=[1]; for I=1:length(ParL) Tmp1=Tmp(length(Tmp)); Tmp2=ParL(I); if Tmp2-Tmp1>EEps Tmp=[Tmp,Tmp2]; end; end; ParL=Tmp; Tmp1=ParL(length(ParL)); Tmp2=Numptcrv(Obj); if abs(Tmp1-Tmp2)<Eps ParL=[ParL(1:length(ParL)-1),Tmp2]; end; Fig=[]; for N=1:length(ParL)-1 Tmp=(ParL(N)+ParL(N+1))*0.5; Tmp=PointonCurve(Tmp,Obj); Tmp=Naigai(Tmp,Bdy); Tmp1=modulo(sum(Tmp),2); if Tmp1==0 continue; end; Fig=Mixadd(Fig,Partcrv(ParL(N),ParL(N+1),Obj)); end; OutL=Mixjoin(OutL,Fig); end; endfunction;
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//Example 19.1// t=0;//time y=100;//nm//thickness of oxide coating c4=1;//given c5=y^2-c4*t;//substituting value in the equation mprintf("c5 = %e nm^2",c5) //For t1=1;//h //hour //time y1=200;//nm //thickness of oxide coating c4=y1^2-c5 //substituting values in the equation mprintf("\nc4 = %e nm^2/h",c4) //Then t2=24;//h//hour //time y2=c4*t2+c5 mprintf("\ny2 = %e nm^2",y2) mprintf("\nor y=854nm (=0.854 mew m) ")
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clc //initialization of variables x1=1 //in x2=4 //in T1=85 //F T2=30 //F //calculations QbyA=12*(T1-T2)/(x1/0.3 + x2/0.026) //results printf("Rate of heat flow = %.1f B/r-ft^2-F",QbyA)
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clc //Example 10.5 //Calculate the pump head N=1750//rev/min //1 min 60 sec omega=2*(%pi)*N/60//radians/sec Q=100//gal/min //1 gallon = 231 in^3 //1 ft =12 in //1 min = 60 sec d_inlet = 2.067//ft A_inlet=(%pi)/4*(d_inlet^2)//ft^2 V1=(Q/A_inlet)*231/60/12//ft/s d_outlet = 1.61//ft A_outlet=(%pi)/4*(d_outlet^2)//ft^2 V2=(Q/A_outlet)*231/60/12//ft/s g=32.2//ft/s^2 d_inner=0.086//ft d_outer=0.336//ft h=(omega)^2/g*((d_outer^2)-(d_inner)^2)+(V2^2-V1^2)/2/g//ft printf("The pump head is %f ft",h);
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// Convolution of two sequences // 3.2 h = [1 2 3]; u = [4 5 6]; y = convol(u,h)
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//Igor Yoshimitsu Ide //Cíntia Bras Mesquita //Tiago Pinheiro Camargo a = imread('C:\desafio.JPG'); im = a; count = 0; for r=1:274 // 1:274 for g=1:368 // 1:368 for b = 0:0//1 - ciano-> vermelho //2 - verde -> roxo // 3 -> amarelo azul escuro if a(r, g)>=0 && a(r, g)<125 im(r, g, 1) = 20; end if a(r, g)>=125 && a(r, g)<130 im(r, g, 3) = 10; end if a(r, g)>=130 && a(r, g)<147 im(r, g, 3) = 10; end if a(r, g)>=147 && a(r, g)<155//amarelo im(r, g, 3) = 50; end if a(r, g)>=155 && a(r, g)<160// amarelo im(r, g, 3) = 80; end if a(r, g)>=160 && a(r, g)<170 //amarelado im(r, g, 3) = 80; end if a(r, g)>=170 && a(r, g)<255 im(r, g, 3) = 200; end end end end imshow(im); imwrite(im, 'desafio_feito.jpg');
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clear close t = (0:0.05:10); t = t' ifinal=size(t);ifinal=ifinal(1); F = sin(4*t); m = 40; b = 10; k = 100; s = poly(0, 's'); g = syslin('c', 1/(m*s^2 +b*s+ k)) y = csim(F', t', g)'; filtro = 1/(s+1)^3; hf = syslin('c', filtro); u_fil = csim(F', t', hf); y_fil = csim(y', t', hf); dy_fil = csim(y', t', s*hf); d2y_fil = csim(y', t', s*s*hf); P = 10000000*eye(3, 3); theta_plot = zeros(3, ifinal); p = zeros(ifinal); p(1) = norm(P, 'fro') theta = [0;0;0]; for i=2:ifinal fi = [u_fil(i) -y_fil(i) -dy_fil(i)]'; K =P*fi/(1+fi'*P*fi); P = (eye(3,3) - K*fi')*P; p(i) = norm(P, 'fro') theta = theta + K*(d2y_fil(i) - fi'*theta); theta_plot(:,i) = theta end figure(1) plot(t,theta_plot') figure(2) plot(t,p)
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//Problem 12.11: //initializing the variables: //Antoine Eq Coeff for ethanol Ae = 8.1122; Be = 1592.864; Ce = 226.184; //Antoine Eq Coeff for toulene At = 6.95805; Bt = 1346.773; Ct = 219.693; p = 760; // mm of Hg R = 1.987; //calculation: //The saturation temperatures: Tsat_e = (Be/(Ae - log10(p))) - Ce Tsat_t = (Bt/(At - log10(p))) - Ct // xe = 0.5 xt = 0.5 T = xe*Tsat_e + xt*Tsat_t // pde = 10^(Ae - (Be/(T + Ce))) pdt = 10^(At - (Bt/(T + Ct))) // a = 0.5292 bet = 713.57 bte = 1147.86 // tou_et = bet/(R*(T+273)) tou_te = bte/(R*(T + 273)) Get = %e^(-1*a*tou_et) Gte = %e^(-1*a*tou_te) r_e = %e^(0.5^2*(tou_te*(Gte/(xe + xt*Gte))^2 + Get*tou_et/(xt + xe*Get)^2)) r_t = %e^(0.5^2*(tou_et*(Get/(xt + xe*Get))^2 + Gte*tou_te/(xe + xt*Gte)^2)) // pde = p/(r_e*xe + r_t*xt*pdt/pde) // Tn = Be/(Ae - log10(pde)) - Ce // ye = xe*r_e*pde/p printf("\n\nResult\n\n") printf("\n mole fraction at T = %.2f degC, xe = %.3f and ye = %0.3f \n Return to step 2 and use a different value for xe. Continue this until an entire T-x, y diagram is formed. \n A T-x, y diagram for ethanol and toluene, employing the NRTL method can be found in Fig. 12.11\n To generate an x–y diagram, simply plot the xe as the ordinate and ye as the abscissa.",T, xe, ye)
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clc; pathname=get_absolute_file_path('3_10_soln.sce') filename=pathname+filesep()+'3_10_data.sci' exec(filename) // Solution: // Acceleration due to gravity, g=32.2; //ft/s^2 // Energy Equation between Station 1 and Station 2 is given by, // (Z1+P1+K1+Hp-Hm-Hl)=(Z2+P2+K2) // since, There is no Hydraulic motor between Station 1 and 2, // Therefore Motor Head, Hm=0; //ft // also, cross section of oil tank is very large, as a result oil is at rest, v1=0; //ft/s // Kinetic Energy Head at inlet, K1=(v1^2)/(2*g); //ft // Height of Station 1 from Datum, Z1=0; //ft // Height of Station 2 from Datum, Z2=20; //ft // Pressure Head at inlet, P1=p1/SG; //ft // Pump Head, Hp=ceil((3950*HHP)/(Q*SG)); //ft // Pump flow, Q_1=Q/449; //ft^3/s // Area of pipe, A=((%pi)*((D/12)^2))/4; //ft^2 // Therefore, velocity in pipe, v2=Q_1/A; //ft/s // Kinetic Energy head at Station 2, K2=(v2^2)/(2*g); //ft // Therefore, Pressure Head at outlet, P2=Z1+P1+K1+Hp-Hm-Hl-Z2-K2; //ft // specific weight of oil, gamma1=SG*62.4; //lb/ft^3 // Pressure available at inlet of hydraulic motor at station 2, p2=P2*gamma1; // lb/ft^2 p2=floor(p2/144); //psi // Results: printf("\n Results: ") printf("\n The Pressure available at inlet of hydraulic motor at Station 2 is %.0f psig.",p2)
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// // Autor: Jonas Vieira de Souza // // TODO: Função para ajuste de curvas por regressão linear // Exemplo de chamada: // $$ exec( caminho + 'regressao_linear.sci', -1 ); // $$ x [1 2 5 7 9 21]; // $$ y [4 5 6 7 9 20]; // $$ [ vi ] = interpolador_newton( x, y ); // // Retornos: // $$ vi $$ variavel independente do interpolado // // Argumentos: // $$ _x $$ vetor de valores da Variavel Dependente // $$ _y $$ vetor de valores da Variavel Independente // $$ _pa $$ ponto de avalição do Polinônio Interpolador // clc function v_indep = interpolador_newton( _x, _y, _pa ) [ mx, nx ] = size(_x); [ my, ny ] = size(_y); if nx ~= ny then disp("Dados incompatíveis - Tamanho dos dados desiguais"); error("x e y devem ter a mesma dimensão"); end b = _y; //Encontrando os termos do polinomio de Newtom // f(n-1) = b1+b2(x-x1)+...+bn(x-x1)(x-x2)...(x-xn) for i = 2:nx for j = nx:-1:i b(j) = (b(j)-b(j-1))/(_x(j)-_x(j-(i-1))); // end end disp(b,"Termos de b -->"); //Avaliando o ponto no polinomio interpolador v_indep = 0; for i = nx:-1:1 jota = 1; for j = 1:i-1 jota = jota*(_pa-_x(j)); end v_indep = v_indep + jota*b(i); end endfunction
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//Chapter-11 example 26 //============================================================================= clc; clear; //input data mprintf('(PRF1) = 2(PRF2)\n'); mprintf(' Vb3 = 4Vb5\n'); mprintf(' (3Vo/2F1)(PRF1)) = 4(5Vo/2F2)(2PRF2)\n'); mprintf(' 3/2F1 = 20/F2\n'); mprintf(' Ratio of operating frequencies is F2/F1 = 40/3\n'); //=================end of the program===========================================
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// Example 8.10 clear all; clc; // Using the data from Example 8.3 to 8.8 P = 2000; // Pressure in psi v = 15.6; // Coolant velocity in ft/sec D_e = 0.0427; // Equivalent diameter in ft d = 0.42; // Diameter of the fuel rod in inches b = 0.024; // Thickness of Zircaloy-4 clad in inches a = (d/2)+b; // Radius of fuel rods in inches T_b = 600; // Bulk temeperature in F // 1. // Using Bernath correlation // Calculation T_wc = 102.6*log(P)-((97.2*P)/(P+15))-(0.45*v)+32; // Result printf(" \n Cladding temeperature = %d F\n",T_wc); // 2. D_i = (2*%pi*a)/(%pi*12); // Heated perimeter is (2*%pi*a)/12 in feet // Calculation h_c = 10890*((D_e)/(D_e+D_i))+((48*v)/D_e^0.6); // Result printf(" \n Heat transfer coefficient = %d Btu/hr-ft^2-F\n",h_c); // 3. // Calculation q_c = h_c*(T_wc-T_b); // Result printf(" \n Critical heat flux = %.2E Btu/hr-ft^2\n",q_c); // In the textbook, the unit of critical heat flux is wrong.
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//calculating Kc //Example 6.8 clc clear //E'cell=0.0591*logKc/n Ecell=0.16 n=4 Kc=10^(n*Ecell/0.0591)//equilibrium constant printf('Thus the equilibrium constant for the reaction = %e',Kc)
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// Scilab code Exa5.6: : Page 204 (2011) clc; clear; Z_D = 82; // Atomic number of Po E_Po210 = 5.3; // Alpha-source for Po210, MeV E_Po214 = 7.7; // Alpha-source for Po214, MeV log_lambda_Po210 = -1*1.72*Z_D*E_Po210^(-1/2); log_lambda_Po214 = -1*1.72*Z_D*E_Po214^(-1/2); delta_OM_t = log_lambda_Po214 - log_lambda_Po210; // Difference in order of magnitude of life times of Po214 and Po210 printf("\nThe disintegration constant increases by a factor of some 10^%2d", delta_OM_t); // Result // The disintegration constant increases by a factor of some 10^10
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clear; clc; //Example - 3.19 //Page number - 113 printf("Example - 3.19 and Page number - 113\n\n"); //Given T_1 = 600;//[C] - Temperature at entry P_1 = 15;//[MPa] - Pressure at entry T_2 = 400;//[K] - Temperature at exit P_2 = 100;//[kPa] - Pressure at exit A_in = 0.045;//[metre square] - flow in area A_out = 0.31;//[metre square] - flow out area m = 30;//[kg/s] - mass flow rate. //At 15 MPa and 600 C,it has been reported in the book that the properties of steam are, Vol_1 = 0.02491;//[m^(3)/kg] - Specific volume H_1 = 3582.3;//[kJ/kg] - Enthalpy // m = den*vel*A = (Vel*A)/Vol, substituting the values vel_1 = (m*Vol_1)/A_in;//[m/s] - Velocity at point 1. printf(" The inlet velocity is %f m/s\n",vel_1); //At 100 MPa (saturated vapour),it has been reported in the book that the properties of steam are, T_sat = 99.63 C, and Vol_vap_2 = 1.6940;//[m^(3)/kg] - specific volume of saturated vapour. H_vap_2 = 2675.5;//[kJ/kg] - Enthalpy os saturated vapour. vel_2 = (m*Vol_vap_2)/A_out;//[m/s] - Velocity at point 2. printf(" The exit velocity is %f m/s\n",vel_2); //From first law we get, q - w =delta_H + delta_V^(2)/2 //q = 0, therefore, -w = delta_H + delta_V^(2)/2 delta_H = H_vap_2 - H_1;//[kJ/kg] - change in enthalpy. delta_V_square = (vel_2^(2) - vel_1^(2))/2;//[J/kg] delta_V_square = delta_V_square*10^(-3);//[kJ/kg] w = -(delta_H + delta_V_square);//[J/kg] W_net = w*m;//[kW] W_net = W_net*10^(-3);//[MW] - power produced. printf(" The power that can be produced by the turbine is %f MW",W_net);
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clc; y=1.4; p2!p1=3; T1=288; T2s=T1*[(p2!p1)^({y-1}/y)]; nc=0.8; T2=T1+[T2s-T1]/nc cps=1.005; Wi=cps*(T2-T1); Wo=2*(Wi)/0.98; T6=923; cps2=1.15; T7=T6-Wo/cps2 nT=0.85; T7s=T6-[(T6-T7)/nT] y2=1.333; p8!p9=[p2!p1^2]/[(T6/T7s)^{y2/(y2-1)}]; T8=T6; T9s=T8/[(p8!p9)^({y2-1}/y2)]; T9=T8-nT*(T8-T9s) N=cps2*(T8-T9)*0.98; Tr=0.75; T4=420.5; T5=T4+Tr*(T9-T4) Q=cps2*([T6-T5]+[T8-T7]); Ceff=N/Q; disp(Ceff,"cycle efficiency is:"); //part II GWo=Wo+N/0.98; Wr=N/GWo; disp(Wr,"work ratio is:") //part III m=5000/N; disp("kg/s",m,"rate of flow of air is:")
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// To find the form factor and error // Modern Electronic Instrumentation And Measurement Techniques // By Albert D. Helfrick, William D. Cooper // First Edition Second Impression, 2009 // Dorling Kindersly Pvt. Ltd. India // Example 6-1 in Page 131 clear; clc; close; // Given data //let E_m = 10; //Let the peak amplitude of the square wave be 10V T = 1; //Let the time period of the square wave be 1s function y= f(t),y=(E_m)^2 ,endfunction E_rms = sqrt(1/T * intg(0,T,f)); printf("(a) The rms value of the square wave = %d V \n",E_rms); function x = ff(t),x =(E_m) ,endfunction E_av = (2/T * intg(0,T/2,ff)); printf(" The average value of the square wave = %d V\n",E_av); k = E_rms/E_av; printf(" The form factor of the square wave =%d\n",k); k_sine = 1.11; k_square = 1; %error = (k_sine - k_square)/k_square*100; printf("(b) The percentage error in meter indication = %d %%",%error); //Result // (a) The rms value of the square wave = 10 V // The average value of the square wave = 10 V // The form factor of the square wave =1 // (b) The percentage error in meter indication = 11 %
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5_02.sce
// Problem 2, Figure 5.3 clc; R=100; //Assigning the values to variable V=25; V1=10; V2=4; V3=V-V1-V2; //Calculating the voltage across Resistor R3 printf("Potential difference across R3 = %f V\n\n\n",V3); I=V/R; //Calculating the current printf("Current flowing through each resistor = %f A\n\n\n",I); R2=V2/I; //Calculating the resistance of R2 printf("Resistance R2 = %f ohm\n\n\n",R2);
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ex3_31.sce
//Ex:3.31 clc; clear; close; n1=1.46;// core refractive index a=45/2;// max radius in um y=0.85;// operating wavelength in um NA=0.17;// numerical aperture v=(2*3.14*a*NA)/y;//normalised frequency M=v^2/2;// number of modes printf("The normalised frequency =%f", v); printf("\n The number of modes =%d", M);
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Ex6_11.sce
clear; clc; //Example 6.11 Vtnd=1; Vtnl=1; Kn=30; //let W/L=x xl=1; Vdd=5; Av=10; //Av=sqrt(xd/xl) xd=(Av)^2*xl; printf('\nwidth to length ratio of driver transistor=%0.2f\n',xd) Knd=xd*Kn*0.001/2; Knl=xl*Kn*0.001/2; printf('\nconduction parameter Knd=%.2f mA/V^2\n',Knd) printf('\nconduction parameter Knl=%.3f mA/V^2\n',Knl) //Vgsd-Vtnd=(Vdd-Vtnl)-sqrt(Knd/Knl)*(Vgsd-Vtnd) y=sqrt(Knd/Knl); Vgsd=(y+5)/(1+y); printf('\nVgsd=%.2f V\n',Vgsd) Vdsd=Vgsd-1; printf('\nVdsd=%.2f V\n',Vdsd)
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eg7_6.sce
clear; //clc(); r=0.5*3*4.75/1000; d1=3; d2=6; dac1=6; dbb1=9; dca1=6; dac=6; dc1a1=6; dab=sqrt(d1*d1 + (d1/2)*(d1/2)); dbc=dab; da1b1=dab; db1c1=dab; dab1=sqrt(d1*d1 + (dac+d1/2)*(dac+d1/2)); dbc1=dab1; dba1=dab1; db1c=dab1; da1b=dab1; daa1=sqrt(d2*d2 + d2*d2); dcc1=sqrt(d2*d2 + d2*d2); mgmd=(dab*dbc*dac*dab1*dbc1*dca1*da1b*db1c*dac1*da1b1*db1c1*dc1a1)^(1/12); sgmd=(((0.7788*r)^3)*(daa1*dbb1*dcc1))^(1/6); l=2*log([mgmd/sgmd]); xl=2*(%pi)*50*l*10^(-5); printf("\n the inductance is: %.4f Ohm/km\n ",xl);
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function unix_x(cmd) //unix_x - shell command execution, results redirected in an xless window //%Syntax // unix_x(cmd) //%Parameters // cmd - a character string //%Description // cmd instruction is passed to shell, the standard output is redirected // to a background xless window //%Examples // unix_x("ls") //%See also // host unix_g unix_s //! // Copyright INRIA if prod(size(cmd))<>1 then error(55,1),end if getenv('WIN32','NO')=='OK' & getenv('COMPILER','NO')=='VC++' then tmp=strsubst(TMPDIR,'/','\')+'\unix.out'; cmd1= cmd + ' > '+ tmp; else tmp=TMPDIR+'/unix.out'; cmd1='('+cmd+')>'+ tmp +' 2>'+TMPDIR+'/unix.err;'; end stat=host(cmd1); select stat case 0 then if getenv('WIN32','NO')=='OK' & getenv('COMPILER','NO')=='VC++' then host(""""+strsubst(SCI,'/','\')+'\bin\xless.exe"" '+ tmp); else host('$SCI/bin/xless '+tmp+' & 2>/dev/null;') end case -1 then // host failed error(85) else //sh failed if getenv('WIN32','NO')=='OK' & getenv('COMPILER','NO')=='VC++' then error('unix_x: shell error'); else msg=read(TMPDIR+'/unix.err',-1,1,'(a)') error('unix_x: '+msg(1)) end end // do not delete file because it is possible xless has not yet been // launched. CLG //if getenv('WIN32','NO')=='OK' & getenv('COMPILER','NO')=='VC++' then // host('del '+tmp); //else // host('rm -f '+tmp); //end
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ex2_8.sce
errcatch(-1,"stop");mode(2);// Example 2.8 page no-49 L=1400 E_diff=12400/L //eV del_E=2.15 L2=12400/del_E printf("\nE2-E1=%.2f eV\n1850 A° line is from 6.71 eV to 0 eV\nTherefore, second photon must be from %.2f to 6.71 eV.\nLambda=%d A°.",E_diff,E_diff,L2) exit();
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function [t,x]=cyclic_euler(A,x0,t0,tf,h) FE = eye(A) + A * h; BE = inv(eye(A) - A * h); t=[t0:h:tf]; x=zeros(length(x0), length(t)); x(:,1)=x0; for k=1:length(t)-1 if modulo(k,2) == 1 then x_kp1 = FE * x(:,k); else x_kp1 = BE * x(:,k); end x(:,k+1) = x_kp1; end endfunction u=0; B=zeros(x0); x0 = [1 ; -2]; ti = 0; tf = 25; A = [0 , 1 ; -9.01, 0.2]; h = 0.01 [t,x] = cyclic_euler(A,x0,ti,tf,h); x_a=ltisol(A,B,u,x0, t); err = norm(x_a - x)/(norm(x))
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LU_Scilab.sce
clc clear // Utilizaremos la factorización LU de Scilab para resolver el sistema A*x=b A = [0 2 3; 2 0 3; 8 16 -1] b = [7 13 -3]' [L,U,P] = lu(A) disp(P) disp(L) disp(U) // Modificamos el vector b usando la matriz de permutación c = P*b // La solución del sistema A*x=b mediante la factorización LU procede en dos etapas: y = L\c x = U\y disp(x)
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Ex17_29.sce
clear //Given u=-10.0 //cm m=-3.0 //Calculation v=m*u f=1/((1/v)-(1/u)) //Result printf("\n Image formed at %0.3f cm",v) printf("\n Focal length is %0.3f cm",f)
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// problem 4.4 Cd=0.6 H1=3 H2=4 b=2 g=9.81 Q=(2*Cd*b*((2*g)^0.5)*((H2*H2*H2)^0.5-(H1*H1*H1)^0.5))/3 q1=Q*1000 disp(q1,"dischsrge flow rate in litres/sec")
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kjh.tst
пуғдай N;ALL;SG кӱскӱ N;GEN;SG алтын N;NOM;SG чайғы N;NOM;PL хар N;GEN;SG ағас N;NOM;PL хуча N;INS;PL кӱскӱ N;ALL;PL хара хурт N;NOM;SG тарбаған N;AT;PL иир N;ALL;SG алтынзарых N;ALL;PL інек N;ACC;PL хысхы N;GEN;PL аба N;INS;PL тигір N;ACC;PL кӱзен N;NOM;PL хысхы N;NOM;SG тағ N;GEN;SG пулут N;ABL;PL хымысха N;GEN;SG тиин N;GEN;SG алтынзарых N;INS;PL часхы N;INS;PL чылтыс N;INS;PL кӧл N;ABL;PL хамнос N;DAT;PL хозан N;NOM;PL хамнос N;NOM;PL пӧрік N;NOM;SG ӱстінзарых N;ACC;PL чар N;ALL;PL хозан N;ACC;SG тирек N;NOM;SG ӧрке N;DAT;PL сабын N;AT;PL суғ N;ACC;SG киндір N;ACC;SG кӱскӱ N;INS;PL пуғдай N;ACC;SG чар N;ALL;SG тағ N;DAT;SG пуға N;ALL;SG тиин N;NOM;PL пус N;INS;SG пуға N;NOM;PL іскер N;ABL;SG часхы N;ACC;PL порсых N;NOM;SG тас N;ALL;SG кӱн N;INS;PL палых N;INS;PL кӧрік N;INS;PL іскер N;DAT;SG тағ N;INS;SG пуға N;AT;PL пулут N;GEN;PL тимір N;GEN;SG тарбаған N;GEN;SG тас N;ABL;SG часхы N;AT;SG інек N;GEN;PL мылтых N;ABL;SG парыс N;GEN;SG хысхы N;NOM;PL кӱн N;ABL;SG мылтых N;NOM;PL азах N;NOM;SG палых N;ABL;PL тіл N;AT;SG аба N;ACC;PL тайға N;INS;PL кӱскӱ N;ALL;SG хозан N;INS;PL порсых N;DAT;PL оо N;INS;PL суғ N;ALL;PL тайға N;NOM;SG пуға N;ABL;PL сабын N;GEN;PL кӧл N;AT;PL пӧрік N;ACC;PL чис N;DAT;SG кӱскӱ N;AT;PL парыс N;AT;PL хоосха N;INS;SG тиин N;ALL;SG алабарыс N;ACC;PL чылан N;NOM;PL тиин N;INS;PL кӧл N;INS;PL хымысха N;ABL;SG хоосха N;INS;PL тӱлгӱ N;ALL;SG порсых N;GEN;PL чылтыс N;GEN;SG часхы N;INS;SG ағас N;GEN;PL азах N;ABL;PL сыын N;AT;PL кӧл N;ALL;PL порсых N;INS;SG чис N;ABL;PL кӱмӱс N;GEN;PL молат N;INS;SG ағас N;ACC;PL пӧрік N;ABL;SG адай N;GEN;SG чазы N;ABL;PL харағай N;GEN;SG кӱмӱс N;DAT;PL кидер N;INS;SG кӧрік N;ABL;SG чылан N;NOM;SG адай N;DAT;SG азах N;DAT;SG чил N;NOM;PL пулут N;ALL;PL пуға N;ABL;SG тамыр N;NOM;SG чылан N;ABL;SG тибе N;ALL;PL адай N;ACC;PL тирек N;DAT;PL тайға N;ALL;SG хысхы N;ACC;PL тамыр N;ALL;PL алабарыс N;DAT;PL тіл N;DAT;SG киндір N;ALL;PL ағас N;ACC;SG хоосха N;GEN;PL парыс N;AT;SG хозан N;AT;PL кӱн N;DAT;PL кидер N;NOM;PL хамнос N;INS;PL кӱскӱ N;GEN;PL інек N;ALL;SG тигір N;AT;PL сыын N;DAT;SG тибе N;ALL;SG алтынзарых N;GEN;PL хозан N;AT;SG ӱстінзарых N;ALL;PL чон N;ACC;SG тимір N;GEN;PL сосха N;ALL;SG тас N;INS;PL часхы N;NOM;SG тас N;ALL;PL чазы N;ALL;SG хуча N;DAT;PL хузурух N;GEN;SG наңмыр N;ALL;SG тӱн N;ALL;SG пуғдай N;AT;PL талай N;ACC;PL чар N;GEN;SG іскер N;INS;PL хуча N;NOM;PL чар N;ACC;PL хысхы N;GEN;SG харағай N;ALL;SG ағас N;AT;PL сыын N;GEN;SG тирек N;INS;SG тигір N;ALL;SG кӱзен N;INS;PL кӱзен N;ABL;SG оо N;GEN;SG тас N;AT;SG пуғдай N;INS;SG тибе N;INS;SG сосха N;ABL;PL тағ N;AT;SG тибе N;ACC;SG чон N;GEN;SG тамыр N;DAT;SG хум N;NOM;PL тарбаған N;NOM;PL хамнос N;ABL;SG тарбаған N;ABL;SG иир N;ACC;PL чил N;ACC;PL тайға N;ABL;SG парыс N;ALL;PL алтынзарых N;ACC;PL суғ N;INS;PL хул N;AT;SG чылан N;ABL;PL молат N;INS;PL тас N;NOM;SG кӱзен N;ACC;PL чазы N;ACC;SG порсых N;DAT;SG молат N;GEN;PL чазы N;ACC;PL табах N;ACC;PL аба N;DAT;PL тӱлгӱ N;DAT;SG хуча N;NOM;SG пӧрік N;ALL;PL хара хурт N;AT;PL адай N;NOM;PL тарбаған N;ALL;PL хум N;AT;SG наңмыр N;NOM;PL чайғы N;ALL;SG сӧс N;NOM;SG чылтыс N;AT;SG чил N;NOM;SG тиин N;ACC;PL сосха N;ACC;SG иир N;ABL;SG сӧс N;AT;SG хамнос N;AT;PL чил N;INS;PL харағай N;ACC;PL парыс N;ACC;SG чил N;ABL;PL хуча N;ACC;SG іскер N;GEN;PL чис N;AT;PL мылтых N;GEN;SG тимір N;ACC;SG хузурух N;DAT;PL оо N;NOM;SG талай N;GEN;SG чис N;GEN;SG тигір N;ABL;SG табах N;GEN;SG парыс N;ABL;PL хум N;DAT;SG сабын N;ALL;SG тамыр N;ALL;SG наңмыр N;AT;PL сосха N;DAT;SG хузурух N;ACC;PL чайғы N;AT;PL
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Simpson's.sce
function[ans] = simpsons(a,b,n) h1 = (b-a)/n deff('[y] = f(x)',"y=sqrt(x^2 + 1)") ans = 0 x0 = a xn = b while (x0 < xn) ans = ans + (h1*(f(x0) + 4*f(x0 + h1) + f(x0+ 2*h1)))/3 x0 = x0 + 2*h1 end endfunction
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/3733/CH32/EX32.40/Ex32_40.sce
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Ex32_40.sce
// Example 32_40 clc;funcprot(0); //Given data L_cap=150;// MW L=[20 60 30];// Load in MW T=[0 8 16 24];// Time in hours n_1=0.9; n_2=2.7; // Calculation // Considering the Consumer C_1 E_1=(L(1)*(T(2)-T(1)))+(L(2)*(T(3)-T(2)))+(L(3)*(T(4)-T(3)));// MWh L_a1=E_1/24;// Average load in MW L_max1=L(2);// Maximum load in MW LF_1=L_a1/L_max1;// Load factor // Considering the Consumer C_1 T=[0 4 12 20 24];// Time in hours L_4=30;// Load in MW t_4=4;// Time in hours L_12=80;// Load in MW t_12=12;// Time in hours L_20=20;// Load in MW t_20=20;// Time in hours E_2=(L_4*(T(2)-T(1)))+(((L_12*t_12)-(L_4*t_4))/(n_1+1))+(((L_12*t_12)-(L_20*t_20))/(n_2-1))+(L_20*(T(5)-T(4))); L_a2=E_1/24;// Average load in MW L_max2=L_12;// Maximum load in MW LF_2=L_a2/L_max2;// Load factor E_t=E_1+E_2;// Total energy supplied in MW L_ap=E_t/24;// Average load on the plant in MW L_pmax=L_max1+L_max2;// Maximum load in MW LF_p=L_ap/L_pmax;// Load factor t=[1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24];// Time in hours L_5=(L_4*(t(5)/t(4))^n_1);// MW L_6=(L_5*(t(6)/t(5))^n_1);// MW L_7=(L_6*(t(7)/t(6))^n_1);// MW L_8=(L_7*(t(8)/t(7))^n_1);// MW L_9=(L_8*(t(9)/t(8))^n_1);// MW L_10=(L_9*(t(10)/t(9))^n_1);// MW L_11=(L_10*(t(11)/t(10))^n_1);// MW L_12=(L_11*(t(12)/t(11))^n_1);// MW L_12=80;// MW L_13=(L_12*((t(12)/t(13))^n_2));// MW L_14=(L_13*(t(13)/t(14))^n_2);// MW L_15=(L_14*(t(14)/t(15))^n_2);// MW L_16=(L_15*(t(15)/t(16))^n_2);// MW L_17=(L_16*(t(16)/t(17))^n_2);// MW L_18=(L_17*(t(17)/t(18))^n_2);// MW L_19=(L_18*(t(18)/t(19))^n_2);// MW L_20=(L_19*(t(19)/t(20))^n_2);// MW P_8=L(1)+L_8;// MW P_6=L(2)+L_16;// MW printf('\nPower supplied at 8th hour=%0.2f MW \nPower supplied at 16th hour=%0.2f MW',P_8,P_6);
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/sci2blif/rasp_design_added_blocks/div_by_n.sce
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div_by_n.sce
style.fontSize=16; style.displayedLabel="<table> <tr><td align=left><b>CLK<br><br><br>RESET<b></td> <td></td> <td></td> <td>Divide by N</td> <td></td> <td></td> <td align=right><b>OUT<br>1:%2$s</b></td></tr></table>"; pal3 = xcosPalAddBlock(pal3,"div_by_n",[],style);
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/1085/CH15/EX15.4/ex15_4.sce
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ex15_4.sce
//Exam:15.4 clc; clear; close; N_a=1.1*10^20;//acceptor density in atoms/m3 n_i=2.5*10^19;//concentration of majority carrier per m3 n_p=(n_i^2)/N_a;//intrinsic density R=n_p/n_i;//Ratio of n_p and n_i disp(R,'n_p/n_i=');
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/3819/CH4/EX4.6/Ex4_6.sce
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Ex4_6.sce
// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal // Chapter 4-Buoyancy and Floatation // Problem 4.6 //Given Data Set in the Problem dens=1000 g=9.81 sg=0.8 theta=135 d=15 P=9.81 OB=50 OD=35 //calculations //Let h is the depth h=OB*sin((180-theta)*%pi/180)-(OD) //in cms //volume of oil displaced v_disp=2/3*%pi*(d/2)^3+h*%pi*(d/2)^2 F_buoy=sg*dens*g*v_disp*10^-6 //taking moment about the hinge //P*20=(F_buoy-W_float)*(OB*cos 45) function[f] = F(W) f = P*20-(F_buoy-W)*(OB*cos((180-theta)/180*%pi)) endfunction W= 10; W = fsolve(W,F) //Weight of the float mprintf("The weight of the float is %f N\n",W)
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/23/CH3/EX3.1/Example_3_1.sce
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Example_3_1.sce
clear; clc; //To find Approx Value function[A]=approx(V,n) A=round(V*10^n)/10^n;//V-Value n-To what place funcprot(0) endfunction //Example 3.1 //Caption : Program to Find Volume Change and Pressure generated for Acetone //Given Values for Acetone P1=1;//Pressure=1Bar T1=20;//Temp=293.15K(20`C) Beta=1.487*10^(-3);//vol expansivity(K^-1) k=62*10^(-6);//isothermal compressibility(bar^-1) V1=1.287*10^(-3);//Volume(m^3 kg^-1) //Solution //(a) //Find (dP/dT)v?? //Using eq.(3.4),V constant hence dV=0 ans_a=round(Beta/k); disp('K^-1',ans_a,'(a)The value of (dp/dT)v is ') //(b) //Find Pressure when acetone heated at const. Vol from T1(1bar) to T2. T2_b=30;//Temp2=303.15K(30`C) del_P=ans_a*(T2_b-T1); ans_b=P1+del_P; disp('bar',ans_b,'(b)The pressure is ') //(c) //Find vol. change when acetone changed from T1(P1) to T2(P2) T2_c=0;//Temp2=273.15K(0`C) P2=10;//pressure=10bar //solve using Eq. (3.5) ln_value=(Beta*(T2_c-T1))-(k*(P2-P1));//ln(V2/V1) ratio=exp(ln_value);//taking antilog,V2/V1 V2=ratio*V1; del_V=approx(V2-V1,6) disp('(X 10^-3) m^3 kg^-1',del_V*1000,'(c)The change in Volume is ') //End
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/881/CH8/EX8.3/exa8_3.sce
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exa8_3.sce
clc; //Example 8.3 //Page No 309 //Solution SN=29; //dB NF=4; //dB FMi=16; //dB disp("The predetection signal to noise ratio is "); pre=SN-NF; disp('dB',pre,"S/N(pre) = "); disp("The postdetection signal to noise ratio is "); pst=pre+FMi; disp('dB',pst,"S/N(post) = ");