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I was asked which impulse response $h_1(t)$ or $h_2(t)$ has a better decoding performance for two filter lengths: $l_1 = 50$ and $l_2 = 500$.

$$h_1(n) = \cfrac{1}{\sqrt{l_1}}\cos{(0.2\pi n f/f_s)}$$ $$h_2(n) = \cfrac{1}{\sqrt{l_2}}\cos{(0.2\pi n f/f_s)}$$ with $0\leq n <l_k$ and $f \in$ [697 Hz , 1477 Hz]

The signal that I'm trying to decode is a discrete-time signal: $$x(t) = \cos{(2 \pi f_1 t)} + \cos{(2 \pi f_2 t)}$$ with sample length of 0.2s, sampling frequency $f_s = 4000$ Hz, $f_1$ and $f_2 \in$ [697 Hz , 1477 Hz] (you may recall that is the telephone number signal)

Q1) I would like to know if the sampling frequency of $x(t)$ is the same as the $f_s$ in $h_1(n)$ and $h_2(n)$.

Q2) I would like to know which of $l_1$ or $l_2$ values is better for encoding.

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  • $\begingroup$ Hi sss. I've edited your answer for formatting reasons, but you need to also edit it to reflect the claims you make. Specifically: - your impulse responses have no dependency on $t$ even though you write them $h(t)$ (they have a dependency on $n$ in your equations). - your $x(t)$ as written is a continuous-time signal, NOT discrete (where is $f_s$?). Keep in mind continuous signals are usually written as being dependent on time $t$, and discrete time signals are usually written as being dependent on sample number $n$. Once that's done I'm sure someone will be glad to answer! $\endgroup$
    – Jdip
    Sep 13, 2022 at 6:31
  • $\begingroup$ Thanks for pointing it out. h(t) should be h(n) instead. $\endgroup$
    – sss
    Sep 13, 2022 at 6:59

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