Simplifying frequency response equation

Question

I am trying to to find a magnitude and phase response of a FIR filter as in problem 5.18 Digital Signal Processing 4th Edition by John Proakis. I got the frequency response as:

$$ H(\omega ) = 1 - e^{-j4\omega} $$

I checked the manual solution and I got the same answer, but in that manual solution, the answer could be simplify to be: $$ H(\omega)=2e^{-j2\omega}e^{j\pi/2} sin2\omega $$

I don't understand what are the steps to simplify my answer into that solution, can someone derive it from my answer into the simplified solution. Thank you so much, hope that my question is clear enough

Answer 1

$$1 - e^{-4j\omega} = e^{-2j\omega}(e^{2j\omega} - e^{-2j\omega}) \tag{1}$$

Now,

$$ \sin(2\omega) = \frac{e^{2j\omega} - e^{-2j\omega}}{2j} \tag{2}$$

Equation 2 is a consequence of Euler's formula. Multiply and divide by $2j$ in (1) and use the identity (2) in equation 1 we have:

$$1 - e^{-4j\omega} = 2je^{-2j\omega}\sin(2\omega) \tag{3}$$

Now $j = e^{\frac{j\pi}{2}}$, substituting in 3 we have

$$1 - e^{-4j\omega} = 2e^{-2j\omega}e^{\frac{j\pi}{2}}\sin(2\omega) \tag{4}$$

The term $e^{-2j\omega}$ in (1) is represntative of the delay of the envelope in time domain, the term $(e^{2j\omega} - e^{-2j\omega}) $ is reprsentative of the magnitude response and actual shape of the filter in frequency domain.

A general trick in simplifying expressions of FIR responses type is to tap expressions around the mid point of the impulse response taps, you could use this strategy to pair terms in responses with more than 2 terms as well

Answer 2

$\begin{align} 1-e^{-j4\omega} &=1-(\cos(4\omega)-j\sin(4\omega))\\ &=1-\cos(4\omega)+j\sin(4\omega)\\ &=1-(1-2\sin^2(2\omega))+j2\sin(2\omega)\cos(2\omega)\\ &=2\sin(2\omega)(\sin(2\omega)+j\cos(2\omega))\\ &=2\sin(2\omega)(j)(\cos(2\omega)-j\sin(2\omega))\\ &=2\sin(2\omega)e^{j\pi /2}e^{-j2\omega} \end{align} $

Answer 3

\begin{equation} 1 - e^{-j4w} = 1 - e^{-j2w} e^{-j2w}\\=1-(cos(2w)-jsin(2w))e^{-j2w}\\=e^{-j2w}(e^{j2w} - cos(2w) + jsin(2w))\\=(cos(2w) + jsin(2w) - cos(2w) + jsin(2w))e^{-j2w}\\=2e^{-j2w} j sin(2w)\\=2e^{j\pi/2}sin(2w) \end{equation}

In the third line I basically pulled $e^{-j2w}$ from the brackets.