Calculate Inverse Discrete Time Fourier Transform of the following where \$|a| < 1\$:

$$ X(e^{j\omega}) = \frac{1-a^2}{(1-ae^{-j\omega})(1-ae^{j\omega})} $$

Plugging this directly into the IDTFT equation, I get:

\begin{align*} x[n] &= \frac{1}{2\pi} \int_{-\pi}^\pi X(e^{j\omega}) e^{j \omega n} d\omega \\ x[n] &= \frac{1}{2\pi} \int_{-\pi}^\pi \frac{(1-a^2)e^{j \omega n}}{(1-ae^{-j\omega})(1-ae^{j\omega})} d\omega \\ \end{align*}

I am having trouble getting started. I'm not sure what to try. None of the standard Fourier Transform property laws seem to directly apply to this.

(This is problem 2.57 from Oppenheim textbook on Discrete Time Signal Processing)

  • \$\begingroup\$ Try not to do the integral rather than find properties of the Fourier Transform and common transforms. \$\endgroup\$
    – OFRBG
    Sep 8, 2015 at 0:17
  • \$\begingroup\$ I went through the Fourier theorems and properties listed in my textbook and I didn't see anything that applied... \$\endgroup\$
    – clay
    Sep 8, 2015 at 0:26
  • \$\begingroup\$ I've added an answer. I hope it helps a bit. \$\endgroup\$
    – OFRBG
    Sep 8, 2015 at 1:25

2 Answers 2


We need to play a bit with the layout of the expression. We have:

$$ F=\frac{(1+a)(1-a)}{(1-ae^{\ jw})(1-ae^{\ -jw})} =\frac{(1+a)}{(1-ae^{\ -jw})}\frac{(1-a)}{(1-ae^{\ jw})} $$

We can rewrite it as:

$$ =\left( \frac{1}{(1-ae^{\ -jw})}+\frac{a}{(1-ae^{\ -jw})}\right) \left( \frac{1}{(1-ae^{\ jw})}-\frac{a}{(1-ae^{\ jw})}\right) \\ $$

We factor out a \$-ae^{\ jw}\$ from the right-most terms and do the inverse transform:

$$ = \left( \frac{1}{(1-ae^{\ -jw})}+a\frac{1}{(1-ae^{\ -jw})}\right) \left( -\frac{1}{a}\frac{e^{\ -jw}}{(1-\frac{1}{a}e^{\ -jw})}+\frac{e^{\ -jw}}{(1-\frac{1}{a}e^{\ -jw})}\right) \\ \implies \left( a^nu[n]+a(a^nu[n]) \right) * \left( -\tfrac{1}{a}({\tfrac{1}{a}}^nu[n-1])+({\tfrac{1}{a}}^nu[n-1]) \right) $$

Finally, cleaning up:

$$ = \left( a^nu[n](1+a) \right) * \left (\tfrac{1}{a}^{n-1}u[n-1](1-\tfrac{1}{a}) \right) $$

Sorry if it is too messy. Fourier tends to be a lot of writing. Tell me where you think I may have made an error or isn't clear! If someone finds a mistake, please let me know.

  • \$\begingroup\$ I messed up factor out a term, but I think it is corrected now. :/ Also, this is the reference I used: mechmat.ethz.ch/Lectures/tables.pdf \$\endgroup\$
    – OFRBG
    Sep 8, 2015 at 1:07
  • \$\begingroup\$ The second row, "we can rewrite it as:" doesn't look algebraically correct. I pasted the following into Wolfram Alpha to verify "(1/(1-ae^(jw)) + a/(1-ae^(jw))) * (1/(1-ae^(jw)) - a/(1-ae^(jw)))" \$\endgroup\$
    – clay
    Sep 8, 2015 at 3:53
  • \$\begingroup\$ @clay Sorry! I messed up the signs. I think I've corrected them throughout. \$\endgroup\$
    – OFRBG
    Sep 8, 2015 at 4:29
  • \$\begingroup\$ @OFRGB, thank you for the help. I am pretty sure the final answer is $x[n] = a^{|n|}$ though. \$\endgroup\$
    – clay
    Sep 9, 2015 at 15:28
  • \$\begingroup\$ @clay sure! I hope that my answer helped you tackle the problem and am glad you got the answer. \$\endgroup\$
    – OFRBG
    Sep 9, 2015 at 17:39

Does this check out?

\begin{align*} X(e^{j\omega}) &= X_1(e^{j\omega}) \cdot X_2(e^{j\omega}) \\ X_1(e^{j\omega}) &= \frac{1-a^2}{1-ae^{-j\omega}} \\ X_2(e^{j\omega}) &= \frac{1}{1-ae^{j\omega}} \\ x_1[n] &= (1-a^2) a^nu[n] \\ x_2[n] &= a^{-n}u[-n] \\ x[n] &= x_1[n] * x_2[n] \\ x[n] &= \sum\limits_{k=-\infty}^\infty x_1[k] x_2[n-k] \\ x[n] &= \sum\limits_{k=0}^\infty (1-a^2) a^k a^{k-n}u[k-n] \\ x[n] &= (1-a^2) a^{-n} \sum\limits_{k=0}^\infty (a^2)^k u[k-n] \\ x[n] &= \begin{cases} (1-a^2) a^{-n} \sum\limits_{k=n}^\infty (a^2)^k & n \ge 0\\ (1-a^2) a^{-n} \sum\limits_{k=0}^\infty (a^2)^k & n < 0\\ \end{cases} \\ x[n] &= \begin{cases} (1-a^2) a^{-n} \frac{a^{2n}}{1-a^2} & n \ge 0\\ (1-a^2) a^{-n} \frac{1}{1-a^2} & n < 0\\ \end{cases} \\ x[n] &= \begin{cases} a^{n} & n \ge 0\\ a^{-n} & n < 0\\ \end{cases} \\ x[n] &= a^{|n|} \\ \end{align*}


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