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Thread: Mathematics of the DFT; Rewriting complex exponential as a product

  1. #1
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    Mathematics of the DFT; Rewriting complex exponential as a product

    I'm reading a book about the Discrete Fourier Transform. In one of the chapters (available here) discussing Spectral Leakage, I have difficulties following a derivation step. Specifically, I don't understand how:

    $\displaystyle \frac{1-e^{j(\omega_x - \omega_k)NT}}{1-e^{j(\omega_x - \omega_k)T}} = e^{j(\omega_x-\omega_k)(N-1)T/2}\frac{sin[(\omega_x-\omega_k)NT/2]}{sin[(\omega_x-\omega_k)T/2]}$

    where $\displaystyle j$ is the imaginary unit $\displaystyle \sqrt{-1}$,
    $\displaystyle T$ is a positive real value (sampling period),
    $\displaystyle N$ is a positive integer (the number of samples),
    $\displaystyle \omega_k = 2\pi\frac{k}{N}f_s$ for $\displaystyle k \in [0, N-1]$ and $\displaystyle f_s = \frac{1}{T}$ is the sampling frequency,
    $\displaystyle \omega_x$ is a positive real value (AFAIK, is an angle in radians but is not any particular $\displaystyle \omega_k$, as defined above -- please correct me if I'm wrong)

    My algebra is a bit weak so a hint is appreciated!

    Mathematics of the DFT; Rewriting complex exponential as a product-file1.jpg
    Last edited by Elusive1324; Aug 26th 2019 at 05:57 AM.
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    Re: Mathematics of the DFT; Rewriting complex exponential as a product

    Quote Originally Posted by Elusive1324 View Post
    $\displaystyle \frac{1-e^{j(\omega_x - \omega_k)NT}}{1-e^{j(\omega_x - \omega_k)T}} = e^{j(\omega_x-\omega_k)(N-1)T/2}\frac{sin[(\omega_x-\omega_k)NT/2]}{sin[(\omega_x-\omega_k)T/2]}$
    I worked backwards and figured it out. To reduce the clutter, let some variable $\displaystyle \theta = \omega_x - \omega_k$. First rewrite $\frac{sin(\theta NT/2)}{sin(\theta T/2)}$ as a quotient of exponential terms using the identity $sin(\alpha) = \frac{e^{j\alpha} - e^{-j\alpha}}{2j}$. Some substitution and algebra later, you'll find that the left-hand term was multiplied by 1, written in the form $\frac{-e^{-j\theta T/2}}{-e^{-j\theta T/2}}$.

    $\displaystyle \frac{1-e^{j\theta NT}}{1-e^{j\theta T}} \left(\frac{-e^{-j\theta T/2}}{-e^{-j\theta T/2}}\right) = e^{j(\omega_x-\omega_k)(N-1)T/2}\frac{sin[(\omega_x-\omega_k)NT/2]}{sin[(\omega_x-\omega_k)T/2]}$
    Last edited by Elusive1324; Aug 26th 2019 at 07:10 AM.
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  3. #3
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    Re: Mathematics of the DFT; Rewriting complex exponential as a product

    use the identity

    $$1-e^{j a}=e^{j a/2}\left(e^{-j a/2}-e^{j a/2}\right)$$

    for any real number $a$
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