How to use FFTshift to produce a 0 frequency centered plot

38 次查看(过去 30 天)
Here is my plot, but I don't know why the right plot is shifted after using FFTshift function, how can I solve this problem?
fs = 250;
t = (0:1/fs:(0.1-.5/fs)); % [0, 0.1)
g = sin(2*pi*50*t)+0.3;
plot(t, g,'-o'),
title('g(t) = sin(2 * pi * 50 * t)+0.3' );
xlabel('time/s');
ylabel('g(t) ');
grid on
n = length(g);
L = 25;
Y = fft(g);
f = fs*(0:L-1)/L;
G = fftshift(Y);
fshift = (-n/2:n/2-1)*(fs/n); % zero-centered frequency range
powershift = abs(G).^2/n; % zero-centered power
figure
plot(fshift,powershift,'-o') % zero-centered plot which have shifts
grid on
figure
plot(f, abs(Y),'-o'), % correct frequency plot
xlabel 'Frequency (Hz)';
ylabel 'Magnitude';
title('Magnitude Plot');
grid on
  2 个评论
Star Strider
Star Strider 2021-9-28
What precisely is the problem?
The right plot looks to be the appropriate result after using fftshift.
What were you expecting?
.
Yian Chen
Yian Chen 2021-9-28
编辑:Yian Chen 2021-9-28
Hi, thanks for your answer, if you observe the the right plot, you can see that the peaks are exactly not 50Hz,may be less than 50Hz, and not zero-centered. I just wanna figure out why and how to solve.

请先登录,再进行评论。

采纳的回答

Ashutosh Singh Baghel
Hi Yian,
I understand your curves and data values are true, yet the plot is shifted left by "-5". Please try to update the starting and final value of 'fshift' to,
fs = 250;
t = (0:1/fs:(0.1-.5/fs)); % [0, 0.1)
g = sin(2*pi*50*t)+0.3;
n = length(g);
L = 25;
Y = fft(g);
f = fs*(0:L-1)/L;
G1 = abs(Y);
G = fftshift(Y);
fshift = (-(n+1)/2+1:(n)/2)*(fs/n) % zero-centered frequency range
fshift = 1×25
-120 -110 -100 -90 -80 -70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90 100 110 120
powershift = abs(G).^2/n; % zero-centered power
figure
plot(fshift(1:end),powershift(1:end),'-o');% zero-centered plot which have shifts
grid on
Please find the MATLAB documentation link for 'fftshift'.

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Spectral Measurements 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by