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Homework Help: Inverse Discrete Time Fourier Transform (DTFT) Question

  1. Mar 25, 2014 #1
    1. Given: The DTFT over the interval [itex] |ω|≤\pi, X\left ( e^{jω}\right )= cos\left ( \frac{ω}{2}\right ) [/itex]
    Find: [itex] x(n) [/itex]

    2. Necessary Equations: IDTFT synthesis equation: [itex] x(n)=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}X\left ( e^{jω} \right ) e^{j\omega n}d\omega[/itex]
    Euler's Identity: [itex] cos\left ( \omega \right ) = \frac{e^{j\omega n} + e^{-j\omega n}}{2}[/itex]

    b]3. Summary: My intuition tells me that a sinusoid in the frequency domain should pop out an impulse in the time domain.

    BUT when running the synthesis equation where I normally end up with an orthogonality situation landing me a pair of delayed impulses, I end up with a delay of (n-1/2) and (n+1/2). As 'n' is a discrete time integer sample there is no data for (n-1/2) and (n+1/2) so I would expect the result to fall between samples and be zero.

    Running all the way through the synthesis results in the following expression:

    [itex] x(n) = \frac{1}{\pi}\left [ \frac{2sin\left ( \pi \left ( n+\frac{1}{2} \right ) \right )}{n+\frac{1}{2}} + \frac{2sin\left ( \pi \left ( n-\frac{1}{2} \right ) \right )}{n-\frac{1}{2}} \right ] [/itex]

    Which is great (I guess) - except that this expression results in a sinc function centered on zero.

    Any guidance is appreciated!

    I have a faint suspicion that the ω/2 would smear the time domain signal - but I don't quite have grasp enough of the theory to prove it.

    Thanks for your help.

  2. jcsd
  3. Mar 28, 2014 #2


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    For what it's worth, your calculations differ from mine by a factor of 4. Other then that, they agree in terms of the sinc() functions and all.

    The frequency domain of DTFT is assumed to be periodic with a period equal to 1 over the sample rate (the sample period). In this problem, the sample rate is normalized, such that the period is equal to 2π.

    The reason you don't get the pair of impulses is that although the frequency domain signal is periodic, it is not sinusoidal. Recall the function cos(ω/2) defined from -π < ω < π. The function is always positive [Edit: well, technically non-negative]. When you repeat the function for higher or lower values of ω, it stays positive. You'll end up with a periodic function that's really the absolute value of a sinusoidal function.

    If you'd like to analyze it further (perhaps to help build up your intuition), think of the sinusoidal function being multiplied by a rectangular "box" function. Then realize that multiplication in one domain is equivalent to convolution in the other domain. What is the transform of a rectangular "box" function? What is the result if you convolve that with a pair of impulses? :wink:
    Last edited: Mar 28, 2014
  4. Mar 31, 2014 #3

    Thank you for the response - the bit about being periodic but not sinusoidal was very helpful. I did not recognize that before (my fault in not sketching the magnitude plot - I need to remember to do that).

    As for the example at the bottom - I understand the relationship between convolution and multiplication, and I see the connection to the problem, good insight!
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