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Suppose we have an ideal low-pass filter $H(e^{j\theta})$ with a cut-off frequency $\theta_c$ in the range of $0\leq \theta_c \leq \pi$.

Low-Pass Filter

I want to know the input signal $x[n]$, as well as the cut-off frequency $\theta_c$, which produces the following output signal $y[n]$.

enter image description here

To solve this I started off with transforming $y[n]$ to the frequency domain by calculating the DTFT:

$$ Y(e^{j\theta})=\sum_{n=-\infty}^{\infty}y[n]e^{-j\theta n}=1+e^{-j\theta}+2e^{-j2\theta}+e^{-3j\theta}+e^{-j4\theta} $$

Now I can calculate $X(e^{j\theta})=\frac{Y(e^{j\theta})}{H(e^{j\theta})}$. This is only possible for $|\theta| \leq \theta_c$, since outside of this region $H(e^{j\theta}) = 0$ and $X(e^{j\theta})$ would approach infinity. But inside this region $H(e^{j\theta})$ is simply 1 and when transforming $X(e^{j\theta})$ back to $x[n]$ it is just the same as $y[n]$, a behaviour that is expected of a low-pass.

The thing I am not sure about is the cut-off frequency. If I simply plot $Y(e^{j\theta})$, I see that there are frequency components until $\pi$. So I could just say alright, $\theta_c=\pi$. But is it really like that? Because the low-pass is actually just dependent on $\theta$ and not all the multiples of it, which I find in the complex exponentials of $Y(e^{j\theta})$. So how do I calculate $\theta_c$?

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    $\begingroup$ Something's wrong here. The ideal filter you mention has an infinite impulse response, so that the output sequence $y$ can't be as simple as you said. You can also see it in the frequency domain: as you said, this $y$ has frequency components all the way through $\pi$, so that it can't be the output of a filter that cuts away completely everything from $\theta_c$ to $\pi$. $\endgroup$ – fonini Nov 28 '16 at 15:01
  • $\begingroup$ @fonini The output signal is infinite, indicated by the dots to the left and right of the graph, it's just zero where n is outside of 0 to 4. You are right, that might have been the more appropriate forum .. $\endgroup$ – Daiz Nov 28 '16 at 15:16
  • $\begingroup$ The domain is infinite, but its support is finite. The output vanishes after $n=4$, that's what matters. Anyway, you can see there's something wrong using the frequency-domain argument you mention. $\endgroup$ – fonini Nov 28 '16 at 16:22
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set $\theta_c = \pi$ and set $x[n] = y[n]$ and you're done. and you're not dividing by zero anywhere.

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