In the development of Hilbert transform relationships, Prof. Oppenheim has chosen \begin{equation} \int_{-\pi}^{\pi}X_R\left(e^{j\theta}\right)\sum_{k=-\infty}^{\infty}\delta(\omega-\theta-2\pi{}k)d\theta =X_R\left(e^{j\omega}\right) \end{equation} in his book "Discrete-Time Signal Processing, 2e" Chapter 11 pp807, Eq. 11-24. I failed to understand how he arrives at it.

Does anybody any idea how has derived, please let me know?

I tried some BUT IT DID NOT GIVE the above relationship:


\begin{align*} \int_{-\pi}^{\pi}X_R\left(e^{j\theta}\right) \sum_{k=-\infty}^{\infty}\delta(\omega-\theta-2\pi{}k)d\theta&=\frac{1}{2\pi}\int_{-\pi}^{\pi}X_R\left(e^{j\theta}\right)2\pi\sum_{k=-\infty}^{\infty} \delta(\omega-\theta-2\pi{}k)d\theta\\& =\frac{1}{2\pi}\int_{-\pi}^{\pi}X_R\left(e^{j\theta}\right)e^{j\theta{}n}d\theta\\& =x_e(n) \end{align*}



\begin{align*} \int_{-\pi}^{\pi}X_R\left(e^{j\theta}\right) \sum_{k=-\infty}^{\infty}\delta(\omega-\theta-2\pi{}k)d\theta&=\sum_{k=-\infty}^{\infty}\int_{-\pi}^{\pi}X_R\left(e^{j\theta}\right) \delta(\omega-\theta-2\pi{}k)d\theta\\& =\sum_{k=-\infty}^{\infty}X_R\left(e^{j(\omega-2\pi{}k)}\right) \end{align*}


2 Answers 2


Your second try was already close. However, you treated the integral as if its limits were $-\infty$ and $\infty$. With the given integration limits there is only one delta impulse in the integration interval, i.e. only one index of the sum is relevant, all others are outside the integration interval. So if for some given value $\omega$ this index is $k=l$, then the result is

$$\int_{-\pi}^{\pi}X_R\left(e^{j\theta}\right) \sum_{k=-\infty}^{\infty}\delta(\omega-\theta-2\pi{}k)d\theta=X_R(e^{j(\omega-2\pi l)})=X_R(e^{j\omega}\cdot e^{-j2\pi l})=X_R(e^{j\omega})$$


I searched through many books and found this way only yesterday night: [source: Encyclopedia of Mathematics and its Applications 124: Hilbert transforms Vol.1, pp659 Eq.13.122 written by Frederick W. King] \begin{align*} \int_{-\pi}^{\pi}X_R\left(e^{j\theta}\right) \sum_{k=-\infty}^{\infty}\delta(\omega-\theta-2\pi{}k)d\theta& =\sum_{k=-\infty}^{\infty}\int_{-\pi+2\pi{}k}^{\pi+2\pi{}k}X_R\left(e^{j(\theta-2\pi{}k)}\right) \delta\left(\omega-\theta\right)d\theta\\& =\sum_{k=-\infty}^{\infty}\int_{(2k-1)\pi}^{(2k+1)\pi}X_R\left(e^{j(\theta-2\pi{}k)}\right) \delta\left(\omega-\theta\right)d\theta\\& =\left\{\ldots+\int_{-3\pi}^{-\pi}X_R\left(e^{j(\theta+2\pi{})}\right) \delta\left(\omega-\theta\right)d\theta\right.\\& \quad+\int_{-\pi}^{\pi}X_R\left(e^{j\theta}\right) \delta\left(\omega-\theta\right)d\theta\\& \quad+\left.\int_{\pi}^{3\pi}X_R\left(e^{j(\theta-2\pi{})}\right) \delta\left(\omega-\theta\right)d\theta+\ldots\right\}\\& =X_R\left(e^{j\omega}\right) \end{align*}

But then, I think a similar logical conclusion given by Matt L. is still required here too.

  • $\begingroup$ Yes, it's the same thing. You need to see that only one of the integrals between the braces is non-zero. Which one depends on $\omega$, but it does not matter because the result of this integral is always the same: $X_R(e^{j\omega})$. $\endgroup$
    – Matt L.
    Jun 7, 2014 at 10:48
  • $\begingroup$ Absolutely, now I understand fully. I was having difficulty convincing others about this integrals with sum! Phew!! :) Thanks @ Matt L. $\endgroup$
    – Khaaba
    Jun 7, 2014 at 14:17

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