# When I double bit rate I do not see a difference in I_spectrum and Q_spectrum, but why?

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.

• What do you mean by a higher or faster graph? Why do you expect a different result? Sep 18, 2020 at 2:45
• I expected the graph's altitude to be different because bit rate = Frequency × bit depth × channels Sep 18, 2020 at 2:48

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: 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: 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]);

• Thank you so much for the great explanation! Sep 18, 2020 at 4:27
• @sawpythonnewbie Also I apologize for all of the mistakes in spelling, among others! Sep 18, 2020 at 18:51