$H(e^{j\omega})$ represented with a limited no of samples

Question

I have a low pass filter defined as:

$$ H_{lp}(e^{j\omega})=\left\{ \begin{aligned} &1 &|\omega|>\omega_c \\ &0 &\omega_c<|\omega|<\pi \end{aligned} \right. $$

and the corresponding sequence in the discrete-time domain is

$$h[n]=\frac{1}{2\pi} \int^{\omega_c}_{-\omega_c}e^{j\omega n}d\omega=\frac{\sin(\omega_cn)}{\pi n}$$

The discrete time-domain signal $h[n]$ has infinite samples.

If we consider only a finite amount of samples:

$$H_M(e^{j\omega})=\sum^{M}_{n=-M}\frac{\sin(\omega_c n)}{\pi n}e^{-j\omega n}$$

And then the book says

$H_M(e^{j\omega})$ can also be represented as:

$$H_M(e^{j\omega})=\frac{1}{2\pi}\int^{\omega_c}_{-\omega_c} \frac{\sin[(2M+1)\frac{(\omega-\theta)}{2}]}{\sin[\frac{(\omega-\theta)}{2}]} d\theta$$

How was this expression evaluated?

This is an example from Oppenheim Discrete-Time Signal Processing book

Answer 1

HINT:

1. figure out the frequency response $W(e^{j\omega})$ of a symmetric rectangular window $w[n]$ of length $2M+1$
2. note that the truncated impulse response is $h_M[n]=h[n]w[n]$
3. from the previous equation, note that the frequency response $H_M(e^{j\omega})$ can be written as the convolution of $W(e^{j\omega})$ with a rectangle. This results in the representation given in your question.