# Terminology question on band-shifting

I am trying to figure out if there is anything more to the phrase 'complex band shifting' than just, a simple band shift. Disclaimer, this is NOT a communications problem, and I only say this so that some answers do get carried away with bit rates, etc.

Basically, let us say that I am interested in a signal, that exists in the passband, from frequencies $f_1$ through $f_2$. Let us further assume that I band pass filter this signal with a brick wall BPF. Then, I multiply it with a sine wave of frequency $f_1$. This operation shifts my signal 'down to baseband', and at base band, it now has a (single-sided) bandwidth of $f_2$ - $f_1$, (as it did before). (I can now LPF to get rid of the double frequency component). That is, my signal now exists from $0$ to $f_2$-$f_1$.

To me, this is all there is to band shifting. That is, the key component is a multiplication with a real sinusoid.

Why then, and what does, the phrase 'complex band shift' entail exactly?

I understand that a multiplication with a complex sinusoid is done in comm systems because we then try to decipher phase from the I and Q, so I get that, but if I simply want to shift bands of a signal, (even to an arbitrary frequency), what is 'complex band shifting' and is that always necessary? Cant I just use a real sinusoid for general band shifting?

Thanks.

You are correct that you can move your signal to 0 to $f_2-f_1$ Hz in the method that you describe. You could do the same thing in one step by multiplying the pass band signal by a complex sinusoid with a frequency of $-f_1$.
• Thanks Jim - some qs, 1) So it seems in order to multiply my signal with a complex sinusoid of -$f_1$, I have to compute my entire signals' hilbert transform first, so that I now have real and imag of my original signal, is this correct? 2) When you say there is no aliasing - can you expand on why that is the case here? You mean since there are no frequency products created that can fold back? Thanks Jun 21, 2012 at 16:23
• Ah yes. The convolution in the frequency domain of the complex exponential's delta sitting at -$f_1$ with my orignal signal's spectra will simply (circularly) shift my original signal's spectra hence giving me the band shifting. I assume at this point one could then BPF where the signal now sits, and resample from there... Jun 21, 2012 at 16:55
• I assume you meant LPF, not BPF. Yes, you can do exactly that. I realized after I answered that if your original signal is real if you do the complex band shift you will still have the spectrally inverted signal at $-2f_1$, so you would need to LPF to get rid of it. Jun 21, 2012 at 17:00