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I would like to compute the Inverse Fourier Transform of a given function using the DTFT definition

$$H(e^{jω})=\frac{1}{(2−e^{jω})(2−e^{−jω})}$$

But I have to use the definition of the DTFT $$H(e^{j\omega}) = \sum_{n = -\infty}^{n = \infty}{h(n) e^{j\omega n}}$$

My approach was that I assume that $x=e^{jω}$, which then gives the equation

$$H(x) = \frac{1}{(2-x)(2-x^{-1})}$$ Now I can apply partial fraction decomposition.

Is my approach correct?

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    $\begingroup$ An identical question has been posted to math.SE. $\endgroup$ – Dilip Sarwate Dec 11 '12 at 17:46
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    $\begingroup$ You're already starting out with a signal's DTFT. What exactly are you trying to compute? $\endgroup$ – Phonon Dec 11 '12 at 18:40
  • $\begingroup$ Like the question says trying to obtain the inverse, is the method of setting $ x = e^{j \omega} $ appropriate? In other words will doing that and then doing partial fraction work? $\endgroup$ – Anonymous Dec 11 '12 at 18:47
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    $\begingroup$ First of all, the statement of problem is false. You would like to find $h(n)$ from a given Fourier transform $H(e^{j\omega})$ (Instead, you wrote "I would like to compute the Fourier Transform..."). Second, make sure that the Fourier transform is exactly what you said. Is this a homework or a question from a specific reference? $\endgroup$ – Deniz Dec 11 '12 at 20:04
  • $\begingroup$ Its actually a sample test question, I am confused as to how to approach this problem. Btw I updated the question. If you look at last page of this document, I am not sure how they derived the answer goo.gl/OZPjm $\endgroup$ – Anonymous Dec 11 '12 at 22:21
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Yes, you can do that. Because we substitute $e^{j\omega}$ for $z$ to get the frequency response of a discrete system, you can go back by substituting $z$ for $e^{j\omega}$. This yields

$$H(z) = \frac{1}{(2-z)(2-z^{-1})} = \frac{1}{5-2z^{-1}-2z} = \frac{z^{-1}}{-2+5z^{-1}-2z^{-2}}$$

Therefore the system you're trying to find is

$$-2y[n]+5y[n-1]-2y[n-2] = -x[n]$$

$$y[n] = \frac{-x[n]-5y[n-1]+2y[n-2]}{-2} = \frac{1}{2}x[n] + \frac{5}{2}y[n-1] - y[n-2]$$

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