# Sampling of bandpass signal with bandpass filter

Trying to solve the following problem about bandpass signal sampling. I have a signal whose Fourier transform is such that $$X(w)= 0$$ if $$w>w_h$$ or $$w. The reconstruction is done using a bandpass filter rather than a lowpass and asks me for the minimum sampling frequency $$w_s$$ given that $$w_l > w_h - w_l$$.

I followed the steps of the sampling theorem and conjectured that if $$w_s> w_h-w_l$$ I should be able to reconstruct the signal with a lowpass filter. Indeed, denoting with $$B=(w_h-w_l)/2$$ I will have a copy of the signal centred at $$0$$ and with extremes $$-B/2, B/2$$ and then another copy centred at $$w_s$$ and with extremes $$w_s-B/2$$ and $$w_s+B/2$$. So, if $$w_s > B$$ I should have no aliasing. Is it correct? A lowpass filter should recover the signal centred at $$0$$.

Now, for the reconstruction with a bandpass filter. I have that the filter is such that $$H(w)= T$$ if $$w_l\leq w \leq w_h$$ or $$-w_l\leq w \leq -w_h$$. The question is: does this filter allow me to reconstruct the original signal $$x$$? I think it does not. First, it would give me two copy of the signal (right?) and also, there is no guarantee that there exist an integer $$k$$ such that $$k w_s = (w_h+w_l)/2$$.

The last question is then: assuming that $$w_l>w_h-w_l$$ what is the smallest sampling frequency $$w_s$$ and largest sampling interval $$T$$ that allow me to reconstruct the signal? Honestly I do not understand, the issue to me remains the same, why would that assumption be enough to reconstruct the signal?

• Does reading the answers to this question help? – Matt L. Jul 14 at 11:11
• I think I covered that part with my $w_s>B$. My signal is a one-sided signal so it should be enough. The question is more regarding the reconstruction via a bandpass filter which I'm not sure how to use... – EngineeringStudent Jul 14 at 11:15
• Are you sure that the signal is complex-valued? It is likely that the problem assumes a real-valued signal, and that the negative frequency content is implied. – Matt L. Jul 14 at 11:17

Assuming that we're really talking about an analytic complex-valued signal with no negative frequency components, then a sampling frequency $$\omega_s>\omega_h-\omega_l$$ will guarantee that the shifted spectra don't overlap, i.e., there will be no aliasing. However, it's not guaranteed that you'll have an image of the spectrum centered at DC. This is only the case if the sampling frequency satisfies

$$k\omega_s=\frac{\omega_l+\omega_h}{2},\qquad k\in\mathbb{Z}^+\tag{1}$$

Eq. $$(1)$$ means that the sampling frequency must be an integer multiple of the signal's center frequency.

So in the case of an analytic signal, reconstruction with a bandpass filter is possible if there is no aliasing, i.e., if $$\omega_s>\omega_h-\omega_l$$ holds.

But since you're given the extra condition $$\omega_l>\omega_h-\omega_l$$ I strongly suspect that the signal is actually real-valued with a (conjugate) symmetric spectrum. In this case, this answer contains the relevant information. Since $$\omega_l>\omega_h-\omega_l$$, we have

$$n_{max}=\left\lfloor{\frac{\omega_l}{\omega_h-\omega_l}}\right\rfloor\ge 1\tag{2}$$

And, from Eq. $$(1)$$ in the quoted answer, the lowest possible sampling frequency must satisfy

$$\frac{2\omega_h}{n_{max}+1}<\omega_s<\frac{2\omega_l}{n_{max}}\tag{3}$$