0
$\begingroup$

enter image description here

Looking at the picture above, how did the author get from point A) to B)?

My Approach: Multiply A) by $e^{j\omega/2}/e^{j\omega/2}$. Now I am stuck with simplying the numerator.

$\endgroup$
0
$\begingroup$

There's a mistake in the denominator of B. This is the way it should be done:

$$\begin{align}R(\omega)&=\frac{1-e^{-j\omega (M+1)}}{1-e^{-j\omega}}\\&=\frac{e^{-j\omega(M+1)/2}}{e^{-j\omega/2}}\cdot\frac{e^{j\omega(M+1)/2}-e^{-j\omega(M+1)/2}}{e^{j\omega/2}-e^{-j\omega/2}}\\&=e^{-j\omega M/2}\frac{\sin\left(\frac{M+1}{2}\omega\right)}{\sin(\omega/2)}\end{align}$$

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.