# Development of Hilbert transform relationship

In the development of Hilbert transform relationships, Prof. Oppenheim has chosen $$\int_{-\pi}^{\pi}X_R\left(e^{j\theta}\right)\sum_{k=-\infty}^{\infty}\delta(\omega-\theta-2\pi{}k)d\theta =X_R\left(e^{j\omega}\right)$$ 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:

# 1:

\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*}

OR

# 2:

\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*}

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
$$\int_{-\pi}^{\pi}X_R\left(e^{j\theta}\right) \sum_{k=-\infty}^{\infty}\delta(\omega-\theta-2\pi{}k)d\theta=X_R(e^{j(\omega-2\pi l)})=X_R(e^{j\omega}\cdot e^{-j2\pi l})=X_R(e^{j\omega})$$
• 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: $X_R(e^{j\omega})$. Jun 7 '14 at 10:48