Problem with amplitude and smoothness of FFT

Hi,

I am trying to create 10 individual signals with on-off keying modulation format, add them up to have a final signal, and then take a Fourier transform of that signal. I tried to zero padd the final signal before taking fft so that I can correct amplitude for my spectrum based on this link:

https://www.mathworks.com/help/signal/ug/amplitude-estimation-and-zero-padding.html

However, I still do not get amplitude one for frequencies of interest.

clc, clear all, close all;

NStep  = 20;        % the number of bits
R   = 20;           % Bit rate (Gbps)
Tr  = 1/R;          % bit period (nanosecond)
fc  = [195.89e3 195.79e3 195.69e3 195.59e3 195.49e3 195.39e3 195.29e3 195.19e3 195.09e3 194.99e3]; %carrier freq.(GHz)
Signal = 0;
for Iter = 1:numel(fc)
    Fc = fc(Iter);                   % Carrier frequency
    x  = randi([0,1],1,NStep);       % Binary information as stream of bits (binary signal 0 or 1)
    N  = length(x);
    nb = 1e6;                        % Digital signal per bit
    t  = Tr/nb:Tr/nb:nb*N*(Tr/nb);   % Time period (ns)
    Digit = [];
    for n = 1:1:N
        if x(n) == 1;
            sig = ones(1,nb);
        else x(n) == 0;
            sig = zeros(1,nb);
        end
        Digit = [Digit sig];
    end
    mod = Digit .* cos(2*pi*Fc*t);    
    Signal = Signal + mod;             % final signal 
end
subplot(2,1,1)
plot(t,Signal);
xlabel('Time (ns)');
ylabel('Amplitude');
title('OOK Modulated Signal');

%% fft
lpad = length(Signal); 
xdft = fft(Signal,lpad);
xdft = xdft(1:lpad/2+1);
xdft = xdft/length(Signal);
xdft(2:end-1) = 2*xdft(2:end-1);          %multiply all frequencies except 0 and the Nyquist by 2.
fs = 1/(Tr/nb);
freq = 0:fs/lpad:fs/2;
subplot(2,1,2) 
plot(freq,abs(xdft)/max(abs(xdft)))
axis([194.94e3 195.94e3 0 1]);
xlabel('Frequency (GHz)');
ylabel('Amplitude');
title('FFT of final Signal');

Answer 1

[Paul]

The time domain signal, Signal, is not a sum of pure cosines because of how Digit is calculated, so we shouldn't expect its fft to be "nice."

Change this line

% mod = Digit .* cos(2*pi*Fc*t);    % OOK Modulation
mod = cos(2*pi*Fc*t);    % OOK Modulation

and the result will be very nice, which illustrates the effect of Digit.

[Amy Lg]

Thanks Paul for your explanation. Do you have any idea about the amplitude?

[Paul]

Greetings Amy,

I don't know what you're asking about here. Amplitude of what? Can you clarify the question in your comment?

[Amy Lg]

I mean the amplitude of my spectrum. I think for all the peaks, which are my carrier frequencies (fc), I should get the amplitude one in my spectrum.

[Paul]

Let's run the code, but only for one carrier frequency, and we won't normalize the spectrum by its max value

rng(100); % for repeatability

NStep  = 20;        % the number of bits
R   = 20;           % Bit rate (Gbps)
Tr  = 1/R;          % bit period (nanosecond)
fc  = [195.89e3 195.79e3 195.69e3 195.59e3 195.49e3 195.39e3 195.29e3 195.19e3 195.09e3 194.99e3]; %carrier freq.(GHz)
Signal = 0;
for Iter = 1  %1:numel(fc)
    Fc = fc(Iter);                   % Carrier frequency
    x  = randi([0,1],1,NStep);       % Binary information as stream of bits (binary signal 0 or 1)
    N  = length(x);
    nb = 1e6;                        % Digital signal per bit
    t  = Tr/nb:Tr/nb:nb*N*(Tr/nb);   % Time period (ns)
    Digit = [];
    for n = 1:1:N
        if x(n) == 1;
            sig = ones(1,nb);
        else x(n) == 0;
            sig = zeros(1,nb);
        end
        Digit = [Digit sig];
    end
    mod = Digit .* cos(2*pi*Fc*t);    % OOK Modulation
    Signal = Signal + mod;             % final signal 
end
figure;
subplot(2,1,1)
plot(t,Signal);
xlabel('Time (ns)');
ylabel('Amplitude');
title('OOK Modulated Signal');
%% fft
lpad = length(Signal); 
xdft = fft(Signal,lpad);
xdft = xdft(1:lpad/2+1);
xdft = xdft/length(Signal);
xdft(2:end-1) = 2*xdft(2:end-1);          %multiply all frequencies except 0 and the Nyquist by 2.
fs = 1/(Tr/nb);
freq = 0:fs/lpad:fs/2;
subplot(2,1,2) 
% plot(freq,abs(xdft)/max(abs(xdft)))
plot(freq,abs(xdft))
axis([194.94e3 195.94e3 0 1]);
xlabel('Frequency (GHz)');
ylabel('Amplitude');
title('FFT of final Signal');

As expected, the time domain (OOK) signal is not a cosine, and the amplitude of the spectrum isn't unity. Now if you were to normalize this spectrum by its max, then of course the peak would be unity.

Now let's do the same thing for all carriers

rng(100); % for repeatability

NStep  = 20;        % the number of bits
R   = 20;           % Bit rate (Gbps)
Tr  = 1/R;          % bit period (nanosecond)
fc  = [195.89e3 195.79e3 195.69e3 195.59e3 195.49e3 195.39e3 195.29e3 195.19e3 195.09e3 194.99e3]; %carrier freq.(GHz)
Signal = 0;
for Iter = 1:numel(fc)
    Fc = fc(Iter);                   % Carrier frequency
    x  = randi([0,1],1,NStep);       % Binary information as stream of bits (binary signal 0 or 1)
    N  = length(x);
    nb = 1e6;                        % Digital signal per bit
    t  = Tr/nb:Tr/nb:nb*N*(Tr/nb);   % Time period (ns)
    Digit = [];
    for n = 1:1:N
        if x(n) == 1;
            sig = ones(1,nb);
        else x(n) == 0;
            sig = zeros(1,nb);
        end
        Digit = [Digit sig];
    end
    mod = Digit .* cos(2*pi*Fc*t);    % OOK Modulation
    Signal = Signal + mod;             % final signal 
end
figure
subplot(2,1,1)
plot(t,Signal);
xlabel('Time (ns)');
ylabel('Amplitude');
title('OOK Modulated Signal');
%% fft
lpad = length(Signal); 
xdft = fft(Signal,lpad);
xdft = xdft(1:lpad/2+1);
xdft = xdft/length(Signal);
xdft(2:end-1) = 2*xdft(2:end-1);          %multiply all frequencies except 0 and the Nyquist by 2.
fs = 1/(Tr/nb);
freq = 0:fs/lpad:fs/2;
subplot(2,1,2) 
% plot(freq,abs(xdft)/max(abs(xdft)))
plot(freq,abs(xdft))
axis([194.94e3 195.94e3 0 1]);grid
xlabel('Frequency (GHz)');
ylabel('Amplitude');
title('FFT of final Signal');

As we can see, the unnormalized peaks are not all the same becuse of the randomness in how mod is formed each time through the loop. If the randomness causes mod to have more time at zero, then the peak at the corresponding carrier will be lower, and vice versa. As a thought experiment, consider what happens when mod == 0 for all time. So when you normalize the spectrum by the max(abs(xdft)), the max peak is normalized to unity and all of the others will be smaller, which is exactly what you've shown.

[Amy Lg]

Hi Paul,

Many thanks for your explanation. So, the amplitude depends on how many bit one I have. Can I define binary stream (x) in such a way that I get more ones than zeros?

If I do so and then normalize the spectrum by the max(abs(xdft)), I can get amplitude one in all the peaks, am I right?

[Paul]

I'm sure you could do various things to get more ones than zeros, but I'm not sure why you would do that. Then again, I'm not familiar with this specific application.

My sense is that any randomness in mod will ensure that the unnormalized peaks at the carriers will not be the same, so normalizing them all by the same value won't achieve unit amplitude at each peak.

[Amy Lg]

Sometimes the amplitude of the spectrum for a peak is very low and this can make problem in my system later on.

Thanks for your help.