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.

  • $\begingroup$ 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.