Getting different outputs from indexing and graphically.How do I verify this indexing value?..

Question

Yogesh 2024-5-6

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2115546-getting-different-outputs-from-indexing-and-graphically-how-do-i-verify-this-indexing-value

评论： Paul 2024-5-6

采纳的回答： Paul

在 MATLAB Online 中打开

clear all

close all

clc

L=10;

n=1.45;

Omega=1.020663297455751e+11;

GAMMAb=3.590391604102621e+08;

c=2.9979e8;

dt=6e-12; %time increment

T=10*2*L*n/c; %total time

fmax = 1e9; %maximun frequency

%fs=80*fmax;

TA=-T/2:dt:T/2; %time axis for the signal

fs=1/dt; %sampling frequency

%t = (-T/2/dt:1:T/2/dt)*dt;

Nt=round(T/dt);

vsine = 1;

phi = vsine*sin(2*pi*fmax*TA);

EL1t=1.274e7*exp(1i*phi); %signal

plot(TA,(EL1t));

Warning: Imaginary parts of complex X and/or Y arguments ignored.

%FA = ((0:Nt-1)-floor(Nt/2))/Nt*fs;

FA = (-Nt/2:Nt/2-1)/Nt*fs; %frequency axis

FP=fft(phi); %fft of the signal

%fs=1/dt/Nt;

figure;

Z=plot(FA,fftshift(abs(fft(EL1t/Nt)))); %fft plot with frequency axis

xlim([(Omega/2/pi-GAMMAb/2/pi/2) (Omega/2/pi+GAMMAb/2/pi/2)])

freqlim1=dsearchn(FA',(Omega/2/pi-GAMMAb/2/pi/2))

freqlim1 = 96299

freqlim2=dsearchn(FA',(Omega/2/pi+GAMMAb/2/pi/2))

freqlim2 = 96355

FA1=freqlim1:freqlim2;

freqlim=dsearchn(FA',Omega/2/pi)

freqlim = 96327

Y1=fftshift(abs(fft(EL1t(freqlim)/Nt))) %value of the signl at Omega

Y1 = 79.0205

I have been trying the index of the signal corresponding to frequency at Omega.Graphically it is around 197 and when i substitute the index number in the fft expression , it is showing around 79.

Is this the case of indexing gone wrong??...

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Answer 1

Paul 2024-5-6

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2115546-getting-different-outputs-from-indexing-and-graphically-how-do-i-verify-this-indexing-value#answer_1453001

在 MATLAB Online 中打开

L=10;

n=1.45;

Omega=1.020663297455751e+11;

GAMMAb=3.590391604102621e+08;

c=2.9979e8;

dt=6e-12; %time increment

T=10*2*L*n/c; %total time

fmax = 1e9; %maximun frequency

%fs=80*fmax;

TA=-T/2:dt:T/2; %time axis for the signal

fs=1/dt; %sampling frequency

%t = (-T/2/dt:1:T/2/dt)*dt;

Nt=round(T/dt);

vsine = 1;

phi = vsine*sin(2*pi*fmax*TA);

EL1t=1.274e7*exp(1i*phi); %signal

%plot(TA,(EL1t));

%FA = ((0:Nt-1)-floor(Nt/2))/Nt*fs;

FA = (-Nt/2:Nt/2-1)/Nt*fs; %frequency axis

FP=fft(phi); %fft of the signal

%fs=1/dt/Nt;

figure;

Z=plot(FA,fftshift(abs(fft(EL1t/Nt)))); %fft plot with frequency axis

xlim([(Omega/2/pi-GAMMAb/2/pi/2) (Omega/2/pi+GAMMAb/2/pi/2)])

freqlim1=dsearchn(FA',(Omega/2/pi-GAMMAb/2/pi/2))

freqlim1 = 96299

freqlim2=dsearchn(FA',(Omega/2/pi+GAMMAb/2/pi/2))

freqlim2 = 96355

FA1=freqlim1:freqlim2;

freqlim=dsearchn(FA',Omega/2/pi)

freqlim = 96327

At this point, freqlim is a single number

EL1t(freqlim)
ans = 7.1267e+06 + 1.0560e+07i

So its fft

Y1=fftshift(abs(fft(EL1t(freqlim)/Nt)))    %value of the signl at Omega 
Y1 = 79.0205

is simply

abs(EL1t(freqlim))/Nt
ans = 79.0205

Are you trying to find the value of the FFT of EL1t at the frequency closest to Omega? Or something else?

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Yogesh 2024-5-6

I am trying to find the value of fft of El1t at Omega.

That should be equal to the value shown in the plot.

The value in the plot is around 198 and the indexing value is around 79.

This is what I am not understanding.

Paul 2024-5-6

在 MATLAB Online 中打开

To find the value of the FFT of EL1t at the frequency closest to Omega, wouldn't that just be (untested)

Y1 = fftshift(abs(fft(EL1t/Nt))));
Y1(dsearchn(FA',Omega/2/pi))

If EL1t is truly finite duration (i.e., not periodic), then you can use freqz to evaluate its Discrete Time Fourier Transform at any frequencies you want (check the doc, you have to query for at least two frequencies).

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