# Cutoff frequency condition of input signal so that it can be reconstructed , given frequency response (Answer explanation?)

For the following system: We have: $$r(t) = s(t) + Ks(t-\tau)$$

where $$|K|< 1$$, with the following impulse response $$h(t)$$ and frequency-response $$H(f)$$: $$h(t) = \delta_0(t) + K\delta_0(t-\tau)$$ $$H(f) = 1 + Ke^{-j2\pi f\tau}$$

For a band-limited input signal $$s(t)$$ with cutoff frequency $$f_g$$ (for the spectrum $$S(f)$$ applies $$S(f)=0$$ for $$|f| > f_g$$ ), which condition must $$f_g$$ satisfy, so that $$s(t)$$ can be completely reconstructed from $$r(t)$$?

When I attempt to answer this question, Shannon's sampling theorem comes to mind. The system-frequency should be greater than $$2\cdot f_g$$. But first, from the $$H(f)$$ of this system, how can I derive the systems frequency?

Theoretically, there are for $$|K| < 1$$ no bandwidth restrictions. ($$|K|= 1$$ would lead to complete cancellations at discrete frequencies). In practice, however, even for $$K$$ values close to one, the recovery is associated with a strong increase in noise if the bandwidth $$f_g$$ becomes so large that the pronounced attenuation maxima of the channel are in the band occupied by $$S(f)$$.

Could someone provide a more detailed explanation? Why are there in principle theoretically no restrictions?

• Hi cb_ann, please try as much as you can to not use images for equations, but rather use LaTeX formatting. I went ahead and edited your question this time, but for future questions you have try to abide by this rule!
– Jdip
Jan 24 at 22:22
• There's no sampling in this problem. The sampling theorem is irrelevant here.
– MBaz
Jan 24 at 22:38
• I've seen a question almost identical to this, perhaps stemming from the same textbook question. Alas, my ability to search stackexchange posts is severely lacking. Feb 24 at 1:24

Hint: Assume $$K = 1$$. Is there a signal $$s(t) \neq 0$$ that produces an output equal to zero? You would need a signal such that, when delayed, is equal to its negative: $$s(t) + s(t-\tau) = 0$$.

Bonus question: would such an $$s(t)$$ be periodic necessarily?

Hint 2: Why did we special-cased $$K = 1$$ above? What happens if $$K \neq 1$$?

• Hi Mbaz, trying to follow your hints, 1) If K = 1, s is a signal such as sin(t), a delay of τ=pi would cancel down the signal s(t) and the output r(t) =0. Generalizing, I would say that this cancellation of the input signal through a delay is only possible for periodic signals. If the signal is not periodic, it wouldn't be possible to cancel it down with a "delayed version of itself". Jan 25 at 0:16
• 2) If K ≠1 (either greater or smaller), I think this cancellation wouldn't be possible. I guess we are talking about cancellation because as long as there is no cancellation and the output r(t)≠0, s(t) can be reconstructed... And because K ≠1 this is guaranteed. If all of the above is correct, I understand that in principle there are not limitations for fg Jan 25 at 0:17
• And trying to understand the second part of the provided answer... When K is close to 1, The signal does not totally get cancelled, but this "destructive summation" of s(t) + s(t-τ) would result in an output r(s) a with a high frequency... So the cutoff frequency fg shoud be small enough, so that this high frequency of the resulting r(s) does not fall within the band of S(f). Correct? Jan 25 at 0:17
• The problem is not the high frequency, but the low amplitude. Say K=0.999 and s(t) = -s(t-τ); then you'd have r(t) = s(t) - 0.999s(t) = 0.001s(t). To recover s(t), you'd have to multiply by 1000. If there is any noise in the input, it will also be amplified by 1000. So, s(t) must not contain frequencies that are strongly attenuated at the output.
– MBaz
Jan 25 at 2:09
• Which frequencies are strongly attenuated at the output in this case? Lower frequencies or larger frequencies? Jan 25 at 15:11