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I wanted to compare two different I-Q spectrum plots of a basic transmitter, one in which I doubled the bit rate. I expected a different result, that of possibly a higher/faster graph. It looks sort of the same, but that the doubled bit rate has more noise? I am not understanding the results here.

Bit rate of 1MHz Bit rate of 1MHz

Bit rate of 2MHz Bit rate of 2MHz

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  • $\begingroup$ What do you mean by a higher or faster graph? Why do you expect a different result? $\endgroup$ – Envidia Sep 18 '20 at 2:45
  • $\begingroup$ I expected the graph's altitude to be different because bit rate = Frequency × bit depth × channels $\endgroup$ – sawpythonnewbie Sep 18 '20 at 2:48
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Do you know how ADS is plotting the spectrum? Plotting the spectrum without doing some kind of normalization will give you a higher magnitude. Once you determine the size of the FFT, normalizing by the length will give you the same magnitude no mater what sampling rate you choose.

For example let's take two rectangular signals, one sampled at 1 MHz and the other at 2 MHz. Below are their spectrums without normalization:

enter image description here

Since the bottom one is sampled twice as fast, it eventually produces an FFT size that is twice as long, hence the 6 dB increase in the peak.

Now compare this to the same exact signals, but now their magnitudes are normalized by their respective FFT sizes:

enter image description here

Now you can see that the peaks are the same magnitude. You can play with normalizing all day long to fit your need. It is the shape of the spectrum that is usually most important.

Here is some quick MATLAB code so you can maybe try it yourself and play around a bit.

%% Signal generation and FFT

% Sampling rates
fs1 = 1e6;
fs2 = 2e6;

% Rectangular pulse signals
t1 = 0:1/fs1:1e-5;
t2 = 0:1/fs2:1e-5;
pulseSignal1 = ones(1, numel(t1)); 
pulseSignal2 = ones(1, numel(t2)); 

% FFT setup
nfft1 = 100*numel(t1); 
f1 = fs1.*(-nfft1/2:nfft1/2-1)/nfft1; 

nfft2 = 100*numel(t2); 
f2 = fs2.*(-nfft2/2:nfft2/2-1)/nfft2; 


%% Without Normalization

figure;
subplot(2, 1, 1);
plot(f1./1e6, 20*log10(abs(fftshift(fft(pulseSignal1, nfft1)))));
xlabel("Frequency (MHz");
ylabel("Magnituide (dB)");
legend("F_s = 1 MHz");
ylim([-40 50]);

subplot(2, 1, 2);
plot(f2./1e6, 20*log10(abs(fftshift(fft(pulseSignal2, nfft2)))));
xlabel("Frequency (MHz");
ylabel("Magnituide (dB)");
legend("F_s = 2 MHz");
ylim([-40 50]);

%% With Normalization

figure;
subplot(2, 1, 1);
plot(f1./1e6, 20*log10(abs(fftshift(fft(pulseSignal1, nfft1)./nfft1))));
xlabel("Frequency (MHz");
ylabel("Magnituide (dB)");
legend("F_s = 1 MHz");
ylim([-80 -10]);

subplot(2, 1, 2);
plot(f2./1e6, 20*log10(abs(fftshift(fft(pulseSignal2, nfft2)./nfft2))));
xlabel("Frequency (MHz");
ylabel("Magnituide (dB)");
legend("F_s = 2 MHz");
ylim([-80 -10]);
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  • $\begingroup$ Thank you so much for the great explanation! $\endgroup$ – sawpythonnewbie Sep 18 '20 at 4:27
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    $\begingroup$ @sawpythonnewbie Also I apologize for all of the mistakes in spelling, among others! $\endgroup$ – Envidia Sep 18 '20 at 18:51

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