# The baseband sampling frequency when the negative spectrum is considered

Suppose I have a signal $$x(t)$$ of bandpass bandwidth of $$2W$$ Hz. The sampling frequency must be at least $$2W$$ in baseband, because the largest frequency component is $$W$$ Hz. I understand that, when we shift a signal to the baseband, we ignore the negative frequencies, and double the power in the positive half. What if we didn't do that, and kept the whole frequency components as they are, would that change the sampling frequency? I guess what I'm asking is, is the maximum frequency component in baseband is the largest positive frequency, or we need to consider the difference between the lower and upper frequencies when considering the negative part of the spectrum?

• Is $x(t)$ real? I.e. $x(t) \in \mathbb{R}$ Jan 5, 2023 at 13:54

Consider a real baseband signal $$r(t)$$ with bandwidth $$B$$. It must be sampled at rate $$2B$$. Frequency-translating to a large carrier frequency $$f_c$$ changes the signal bandwidth to $$W=2B$$, but nothing fundamental has changed. Bringing the signal back to baseband, the signal is again $$r(t)$$ with bandwidth $$B$$ and requires $$2B$$ samples per second.
Now consider the case where the baseband signal $$s(t) = s_\textrm{I}(t) + j s_\textrm{Q}(t)$$ is complex with bandwidth $$B$$. The signal must still be sampled at rate $$2B$$, but now the samples are complex. One complex sample requires two real samples (one for $$s_\textrm{I}(t)$$ and one for $$s_\textrm{Q}(t)$$). Converting to passband, the signal has bandwidth $$W=2B$$, and converting to baseband again recovers $$s(t)$$, which must obviously still be sampled at $$2B$$ complex samples per second.
As to how to measure the bandwidth: a real signal has symmetric magnitude spectrum, so its largest positive frequency $$B$$ is always equal in magnitude to its largest negative frequency $$-B$$. If the signal is complex, its spectrum is not necessarily symmetric. Say the spectrum goes from $$-f_0$$ to $$f_1$$; then $$B = \textrm{max}(f_0, f_1)$$. But note that in this case the signal could be frequency-shifed so that $$f_0 = f_1$$ to minimize both the bandwidth and the sampling rate required.
• Indeed, in my case $f_0=f_1$, and $x(t)$ is the received signal to a real signal with complex AWGN. Jan 6, 2023 at 9:48