Captain1024 said:
Homework Statement
Consider a LTI system defined by the following difference equation: ##\mathrm{y}[n]=-2x[n]+4x[n-1]-2x[n-2]##
a) Determine the impulse response of the system
b) Determine the frequency response of the system
Homework Equations
DTFT: ##\mathrm{x}[n]=\frac{1}{2\pi}\int_{\pi}^{-\pi} \mathrm{X}(e^{j\omega})e^{j\omega n} d\omega \ \longleftrightarrow \ \mathrm{X}(e^{j\omega})=\sum_{n=-\infty}^{\infty} \mathrm{x}[n]e^{-j\omega n}##
The Attempt at a Solution
I started by finding the frequency response:
##\mathrm{Y}(e^{j\omega})=-2\mathrm{X}(e^{j\omega})+4e^{-j\omega}\mathrm{X}(e^{j\omega})-2e^{-2j\omega}\mathrm{X}(e^{j\omega})##
##=\mathrm{X}(e^{j\omega})(-2+4e^{-j\omega}-2e^{-2j\omega})##
Frequency response ##\mathrm{H}(e^{j\omega})=-2+4e^{-j\omega}-2e^{-2j\omega}##
My work on impulse response:
##\mathrm{h}[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}(-2+4e^{-j\omega}-2e^{-2j\omega})e^{j\omega n}d\omega##
##=\frac{1}{2\pi}\int_{-\pi}^{\pi}-2e^{j\omega n}+4e^{-j\omega}e^{j\omega n}-2e^{-2j\omega}e^{j\omega n}d\omega##
##=\frac{1}{2\pi}\int_{-\pi}^{\pi}-2e^{j\omega n}+4e^{j\omega (n-1)}-2e^{j\omega (n-2)}d\omega##
I believe the answer is ##\mathrm{h}[n]=\{-2, 4, -2\}##. Am i overthinking how to arrive at the answer or am I on the right track?
-Captain1024
Yes, I believe you are correct.
Note: It's been awhile since I've done this. That said, your answers seem to be correct, as far as I can tell.
Regarding the impulse response: It's exceedingly easy to find the impulse response of a finite impulse response (FIR) filter. You merely send an impulse through and record what y is at each offset. I assume that is how you obtained your result of \mathrm{h}[n]=\{-2, 4, -2\}, which is correct.
You could also find the impulse response by taking the inverse DTFT of the frequency response. You've already started that approach. That method is a bit tricky (and perhaps unnecessary since there are easier ways to do it), but is certainly possible.
It's a fun idea to do that at least once, if for no other reason to see how the math works out, and even to see where that \frac{1}{2 \pi} comes from in the inverse-DTFT formula.
Here are some hints to help you evaluate that integral, and obtain the impulse response by taking the inverse-DTFT of the frequency response:
- Each term will be 0 for all values of n expect for a single value of n, which is specific to that term.
- That particular value of n will result in a 0 in the numerator, and also a 0 in the denominator, resulting in an undefined number.
- You can find what that number is by using L'Hôpital's[/PLAIN] rule.
- And since that value is specific to a specific value of n, you can tack on a \delta[n - \nu] on that term, where \nu is a specific number 0, 1, 2, 3, ... etc., specific to that particular term. [Edit: In other words, you'll be tacking on a \delta[n], \delta[n - 1], or \delta[n - 2], etc., on each term.]
- You might find the following relationships useful: \frac{d}{dn}\{ (-1)^n \} = j \pi (-1)^n and \frac{d}{dn} \{ (-1)^{-n} \} = - j \pi (-1)^{-n}
[Edit: at least that's the way I did it.]
[Another edit: it also may be useful to recognize that e^{j \pi n} = (-1)^n, where n = ... -3, -2, -1, 0, 1, 2, 3, ...]
[Still another edit: Rather than converting the exponentionals to the form of (-1)^n you might instead prefer to take a different approach by recognizing that \frac{e^{j \pi n} - e^{-j \pi n}}{2 j} = \sin(\pi n). This latter approach is perhaps technically better. (It's more rigorous, and doesn't require that
n be an integer.)]
