plotting the frequency spectrum

Hi,

I am having trouble plotting the frequency spectrum of a sine wave. For this code, i expect the main frequency component to be centered around 1/(2*pi), but they are not. Is there something I am missing in my code?

x = [0:0.1:2*pi];
y = sin(x);
z = fft(y);
z = fftshift(z);
N = length(y);
f = [-N/2:N/2-1]/N;
plot(f,abs(z))

Answer 1

[Rick Rosson]

Please try:

   %%Time specifications:
   Fs = 100;                      % samples per second
   dt = 1/Fs;                     % seconds per sample
   StopTime = 1;                  % seconds
   t = (0:dt:StopTime-dt)';
   N = size(t,1);

   %%Sine wave:
   Fc = 12;                       % hertz
   x = cos(2*pi*Fc*t);

   %%Fourier Transform:
   X = fftshift(fft(x));

   %%Frequency specifications:
   dF = Fs/N;                      % hertz
   f = -Fs/2:dF:Fs/2-dF;           % hertz

   %%Plot the spectrum:
   figure;
   plot(f,abs(X)/N);
   xlabel('Frequency (in hertz)');
   title('Magnitude Response');

HTH.

Rick

[Mere]

Could someone please explain how you know to set up the frequency vector that way?

[shalabh mishra]

 %%Fourier Transform:
   X = fftshift(fft(x));
@Rick Rosson
In the above command, Why we are not using X=fftshift(x) or X=fft(x)?

[Le Dung]

X = fftshift(fft(x)); is first to calculate fft of x, then you will shift the fft value. If you use fftshift(x), mean that you didn't have any fft value of x to shift, or more exactly, you shift values of x but not fft of values of x.

[Karan Anvekar]

Why do we take Fs = 1000? Although, I got my answer, I want to know the theoretical reason behind choosing 1000.

[Kyle Ohlschlager]

This worked great and was exactly what I needed in knowing how to create the frequency axis on my own- it's hard to find code that doesnt rely on built in functions too much- Thank you!

[Anya Getman]

Karan, looks like Rick chose 100, and he either guessed, or looked at your time samples and found them to be 0.01 apart. Take whatever your sample time step is and 1/step. If your time samples are unevenly sampled, you will need to use a function like decimate to make them evenly sampled for analysis.

Answer 2

[kittipob techawudtipat]

i'm having trouble with this >> n = 1:2:5; >> x = -3.98*(sin((pi*n)/2))/(n*pi)+32*(sin((pi*n)/2))/((n.^3)*(pi.^3)*(10.^6)); >> y = 1.996 + x; >> plot(y(1:50)) Index exceeds matrix dimensions.

how can i solve this problem. the graph i order to plot dont happen thankyou bro

Answer 3

[Frantz Bouchereau]

There are various ways in which you can compute and plot true power spectrum or power spectral density in MATLAB (when I say 'true power spectrum' I mean that the output values correspond to actual power values).

1) If you want to compute the power spectrum without having to specify many parameters and want the function to choose the best parameters for you, you can use pspectrum. Calling the function without outputs will give you a plot with the computed power spectrum.

2) If you want to compute power spectrum or power spectral density and want full control over the window size, window overlap, window type, and number of FFT points, you can use the Welch periodogram pwelch function. Calling the function without outputs will give you a plot with the computed power spectrum.

3) If you want to just visualize the power spectrum, you can use the Signal Analyzer app. The app let's you visualize your signals simultaneously in the time, frequency, and time-frequency domains. You can zoom into signal regions of interest and analyze the spectra at those zoomed regions.

4) If you have split your signals into multiple signal frames you can use the Spectrum Analyzer scope.

Finally, here is a popular MATLAB doc page that explains the relationship between FFT and true power spectra: Power Spectral Density Estimates Using FFT.

Answer 4

[Dr. Seis]

Your "f" looks off. For your configuration (which uses fftshift):

N = length(y);
dt = x(2)-x(1); % Time increment
Nyq = 1/(2*dt); % Nyquist frequency
df = 1/(N*dt);  % Frequency increment
if mod(N,2) == 0 % N is even
    f = -Nyq : df : Nyq-df;
else            % N is odd
    f = [sort(-1*(df:df:Nyq)) (0:df:Nyq)];
end

[Dr. Seis]

Forgot the "== 0" in the if statement... whoops!

Answer 5

[reddy]

Hi every one

could you please plot spectrum for the variables that are in EXCEL file which is attached.?

thanks.

Please provide me code also.

Phanindra

Answer 6

[Deepanshu Gupta]

I am trying to plot spectrum of wave for a string plucked halfway along it's length.I am not seeing any figure when I run it. c = 343; % speed of sound air% L = 1000; % string length% w0 = 200; % Displacement distance % Bn = 0; % string at rest, -ve part is zero% x = (0:L); % distance on L%

    n = 2;              %no. of modes%
    fn = (n*c)/2*L;
    omega = 2*pi*fn;          %omega%
    omegaN = n * omega;       %w is displacement%
    kn = w0/c;               % wave no.%
    t = 1/fn;                %time%
    An = ((8*w0)/n^2*pi^2)*sin((n*pi)/2);                 %incident wave part +ve%
    W = sum((An*cos(omegaN*t) + Bn*sin(omegaN*t))*sin(kn*x));  %Displacement eq. sum n=1:10%

plot(W,t); title('hinged string') xlabel('displacement') ylabel('time')

Answer 7

[Khathutshelo Mudau]

Hy, can someone help me with determining or plotting the error spectrum of a sound signal

Answer 8

[Nermin Hamdan]

it is requested that 0 0 1 0 data be transmitted through a narrow communication channel. Using the MATLAB program, find the signal received at the receiver with the effect of the band-limited channel

First plot the submitted 0 0 1 0 data in time plane. Then find and plot the frequency spectrum. I do not have an idea what to do can you help me please