# 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.

## 1 Answer

Complex band shifting is a little more flexible than real band shifting and filtering, and may be a little more efficient too, though if you change the sample rates when moving a signal to baseband that may not be true.

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 $-f_1$.

Also, when you band shift with a real sinusoid you have to watch out for aliasing. Band shifting with a complex sinusoid never causes aliasing (unless you reduce the sample rate at the same time, of course).

• 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 – Spacey Jun 21 '12 at 16:23
• @Mohammad 1) No, you don't have to do any transforms. You can generate a time-domain sinusoid of the appropriate frequency that is the same length as your signal, and then just multiply the two. This works whether your band pass signal is real or complex. 2) There is no aliasing because in the frequency plane a complex sinusoid only has one spike. Multiplying a signal by it thus becomes a "barrel shift" in the frequency plane. Real sinusoids, on the other hand, have two spikes in the frequency plane, which creates multiple images that can overlap (i.e. aliasing). – Jim Clay Jun 21 '12 at 16:33
• 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... – Spacey Jun 21 '12 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. – Jim Clay Jun 21 '12 at 17:00
• Jim, LPF in general, yes. I said BPF only because I am imagining a general band shift to another frequency, close to DC, but yes, we are on the same page. :-) Now, regarding the complex band shift, (Im still digesting it), but it then seems to be that the main advantage here is that we do not need to worry about aliasing, or even BPF the signal in the original pass band, mix with a real sinusoid, then LPF. Instead, we can just start off my multiplying by a complex sinusoid, then LPF/BPF, and be done with it. This would be then, the main advantage from what I understand thus far... – Spacey Jun 21 '12 at 17:25