MATLAB: Matlab FFT interpretation issue

amfftfourierfourier transformMATLAB and Simulink Student Suite

I've got an issue interpreting the frequency spectrum produced using the Matlab FFT on some discrete data.

Below is a probability distribution for a simulation I'm running that varies with time (each point on the plot represents a time step). To me, it clearly resembles a modulated wave with a carrier freq (fc) of 2 cycles per time step and a modulation freq (fm) 1/320 cycles per second.

However, I find the following frequency spectrum.(Normalised scales)

Which shows 2 clear peaks in the center. I believe the peak at 0 frequency can be attributed to the mean of the data and ignored. However, for an AM signal I think I should have 3 peaks, with the central peak having the greatest amplitude (fc) and two peaks either side at fc + fm and fc – fm.

I also found that if I take the difference of the two peaks and divide by 2, I find the expected fm (in accordance with the upper and lower bands of an AM signal).

I'm not sure how to interpret this and any light would be appreciated.

Best Answer

‘However, for an AM signal I think I should have 3 peaks, with the central peak having the greatest amplitude (fc) and two peaks either side at fc + fm and fc - fm.’

Not necessarily.

In an analogue process, you would get a carrier and two sidebands, unless you suppressed the carrier.

In a discrete ‘modulation’, it works differently. If you multiply two different frequency sinusoids (with the same sampling frequency), you get a double-sideband suppressed-carrier (DSB-SC) signal, described by Product-to-sum and sum-to-product identities (link). (The Signal Processing Toolbox modulate (link) function, for example, adds the carrier frequency back into the DSB-SC signal to create a DSB-TC signal.

Related Solutions

MATLAB: DSB SC DEMODULATION in matlab

Hi,

Here, message signal amplitude is ‘Am’,carrier signal amplitude is ‘Ac’,

1)Modulated signal is product of message signal and carrier signal ,amplitude of modulated signal is given by Am*Ac.

2)Demodulated signal is product of modulated signal and carrier signal,amplitude of demodulated signal is Am*Ac*Ac.

3)To get this demodulated signal amplitude equal to message signal it needs to be divided by Ac*Ac

Refer to below attached code for better understanding:

t = 0:0.001:5;
fH = 15;
fL = 2;
Ah=5;
Al=10;
xH = Ah*sin(2*pi*fH.*t);
xL = Al*sin(2*pi*fL.*t);
% Modulation
y =  xL.*xH;
% De-Modulation By Synchoronous Method
m = (y.*xH)./(Ah*Ah);
% Filtering High Frequencies
[n,w] = buttord(2/1000,4/1000,.5,5);
[a,b] = butter(n,w,'low');
dem = filter(a,b,m);
subplot(2,2,1);
plot(t,xH,'b',t,xL,'r');
title('m(t) & c(t)');
grid;
subplot(2,2,2);
plot(t,y,'k');
title('DSBSC');
grid;
subplot(2,2,3);
plot(t,m);
title('De-Modulated');
grid;

MATLAB: How to specify ‘nsamp’ for an oversampled Orthogonal Frequency Division Multiplexing (OFDM) signal

In the time domain, four symbols are modulated by four different cosine frequencies (or OFDM tones); they are then added together to make an OFDM signal. This process can be accomplished by using an IFFT since the symbols are essentially the coefficients x(k)'s in the Fourier Transform since they are multiplied times a cosine. In order to "recover" the corresponding time signal, the IFFT is taken of the symbols.

There are a number of considerations such as inserting zeros to make a frequency guard band or reorganizing the symbols to make them symmetric (to produce a real time signal).

In any case, 'nsamp' should be set to Nfft/Np, where Nfft is the number of OFDM tones (e.g., 64) and Np is the number of non-zero tones (usually smaller than Nfft). This assumes that the output of each modulator (before OFDM frame assembly) is not oversampled. Note that this interpretation of nsamp is not the one given in the documentation, but it is the right formulation for OFDM.

Best Answer

Related Solutions

MATLAB: DSB SC DEMODULATION in matlab

MATLAB: How to specify ‘nsamp’ for an oversampled Orthogonal Frequency Division Multiplexing (OFDM) signal

Related Question