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Homework Help: Finding Impulse And Frequency Responses

  1. Feb 12, 2016 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant 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}##

    3. The attempt at a solution
    I started by finding the frequency response:
    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?

  2. jcsd
  3. Feb 14, 2016 #2


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    Homework Helper
    Gold Member

    Yes, I believe you are correct. :smile:

    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 [itex] y [/itex] is at each offset. I assume that is how you obtained your result of [itex] \mathrm{h}[n]=\{-2, 4, -2\} [/itex], 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 [itex] \frac{1}{2 \pi} [/itex] 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] [Broken] rule.
    • And since that value is specific to a specific value of n, you can tack on a [itex] \delta[n - \nu] [/itex] on that term, where [itex] \nu [/itex] is a specific number 0, 1, 2, 3, ... etc., specific to that particular term. [Edit: In other words, you'll be tacking on a [itex] \delta[n] [/itex], [itex] \delta[n - 1] [/itex], or [itex] \delta[n - 2] [/itex], etc., on each term.]
    • You might find the following relationships useful: [itex] \frac{d}{dn}\{ (-1)^n \} = j \pi (-1)^n [/itex] and [itex] \frac{d}{dn} \{ (-1)^{-n} \} = - j \pi (-1)^{-n} [/itex]
    [Edit: at least that's the way I did it.]

    [Another edit: it also may be useful to recognize that [itex] e^{j \pi n} = (-1)^n [/itex], where [itex] n [/itex] = ... -3, -2, -1, 0, 1, 2, 3, ...]

    [Still another edit: Rather than converting the exponentionals to the form of [itex] (-1)^n [/itex] you might instead prefer to take a different approach by recognizing that [itex] \frac{e^{j \pi n} - e^{-j \pi n}}{2 j} = \sin(\pi n) [/itex]. This latter approach is perhaps technically better. (It's more rigorous, and doesn't require that n be an integer.)]
    Last edited by a moderator: May 7, 2017
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