0
$\begingroup$

I understand the FR of an ideal BPF is just 1 between the two cutoff frequencies and 0 everywhere else above and below them both. But with arbitrary cutoffs, how would one find the IR of this ideal basic filter type?

$\endgroup$
4
  • $\begingroup$ is this a discrete-time filter or a continuous-time ideal BPF? $\endgroup$ Apr 20, 2018 at 19:49
  • $\begingroup$ i guess it doesn't matter much. $\endgroup$ Apr 20, 2018 at 19:50
  • $\begingroup$ I mean, wouldn't it be continous by default? Since any ideal filter only exists with infinite samples? I'm still super lost on how to approach calculating the impulse though, especially since I need to then take this Impulse Response function and turn that into different FIR filters for several sample lengths. Any ideas? $\endgroup$
    – Swift142
    Apr 20, 2018 at 20:05
  • $\begingroup$ what is "default" is normally a choice. some people design digital filters directly in the discrete-time domain, or the $z$-plane. other people design digital filters first in the continuous-time domian, or the $s$-plane and covert $H(s)$ to $H(z)$ using something, either the impulse invariant method or the bilinear transform. $\endgroup$ Apr 20, 2018 at 23:46

2 Answers 2

1
$\begingroup$

Hint:

\begin{align} h(t) &= \int_{-\infty}^\infty H(f)\exp(j2\pi ft)\,\mathrm df\\ &= \int_{-f_2}^{-f_1} H(f)\exp(j2\pi ft)\,\mathrm df + \int_{f_1}^{f_2} H(f)\exp(j2\pi ft)\,\mathrm df\\ &= \int_{-f_2}^{-f_1} \exp(j2\pi ft)\,\mathrm df + \int_{f_1}^{f_2} \exp(j2\pi ft)\,\mathrm df \end{align} which you ought to be able to compute for yourself. A little algebra and trignometry might be needed to massage the answer into a nice formula involving the sinc function times a cosine.

$\endgroup$
2
  • $\begingroup$ k so I understand why it's two integrals, to represent the space between 0 and nyquest frequency in both positive and negative sides. I understand that integrals are only targeting where H(f) is 1, so that can be factored out. But I don't understand how this turns into a sinc function... $\endgroup$
    – Swift142
    Apr 20, 2018 at 20:36
  • $\begingroup$ Nothing is factored out, and if you will try to calculate the integrals (further hint: the antiderivative of $\exp(ax)$ is $\frac 1a\exp(ax)$), maybe you will get some idea of how a sinc might arise. $\endgroup$ Apr 20, 2018 at 21:03
0
$\begingroup$

Here's another way to look at Dilip's answer... just wanna get the definitions down.

continuous Fourier transform:

$$ X(f) \triangleq \mathscr{F} \Big\{ x(t) \Big\} \triangleq \int\limits_{-\infty}^{+\infty} x(t) \, e^{-j 2 \pi f t} \ \mathrm{d}t $$

and inverse:

$$ x(t) \triangleq \mathscr{F}^{-1} \Big\{ X(f) \Big\} = \int\limits_{-\infty}^{+\infty} X(f) \, e^{+j 2 \pi f t} \ \mathrm{d}f $$

It's not hard to show:

$$ \mathscr{F} \Big\{ \operatorname{rect}(t) \Big\} = \operatorname{sinc}(f) $$

where

$$ \operatorname{rect}(u) \triangleq \begin{cases} 1 \qquad & \mathrm{for} \ |u| < \tfrac12 \\ \tfrac12 \qquad & \mathrm{for} \ |u| = \tfrac12 \\ 0 \qquad & \mathrm{for} \ |u| > \tfrac12 \\ \end{cases} $$

and

$$ \operatorname{sinc}(u) \triangleq \begin{cases} 1 \qquad & \mathrm{for} \ u = 0 \\ \frac{\sin(\pi u)}{\pi u} \qquad & \mathrm{for} \ u \ne 0 \\ \end{cases} $$

.

And duality tells us if $X(f) = \mathscr{F} \Big\{ x(t) \Big\}$, then $x(-f) = \mathscr{F} \Big\{ X(t) \Big\}$.

We know, for the ideal LPF

$$ H_\mathrm{LP}(f) = \operatorname{rect}\left( \tfrac{f}{f_B} \right) $$

and impulse response

$$ h_\mathrm{LP}(t) = f_B\operatorname{sinc}(f_B \, t) $$

Frequency translation tells us that

$$\begin{align} \mathscr{F} \Big\{ 2 \cos(2 \pi f_0 t ) \ h_\mathrm{LP}(t) \Big\} &= \mathscr{F} \Big\{ \big( e^{j 2 \pi f_0 t} + e^{-j 2 \pi f_0 t} \big) \ h_\mathrm{LP}(t) \Big\} \\ \\ &= H_\mathrm{LP}(f-f_0) + H_\mathrm{LP}(f+f_0) \\ \\ &= \operatorname{rect}\left( \tfrac{f-f_0}{f_B} \right) + \operatorname{rect}\left( \tfrac{f+f_0}{f_B} \right) \\ \end{align}$$

Now, with all of those explicit definitions above, can you tell us how $f_0$ and $f_B$ relate to Dilip's $f_1$ and $f_2$? And then how to specify your ideal BPF in terms of these frequencies?

$\endgroup$
1
  • $\begingroup$ and, so far, there is no Nyquist frequency. we're in continuous-time mode for the time being. $\endgroup$ Apr 21, 2018 at 2:58

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.