# Simplifying frequency response equation

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

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

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

• That's indeed the way to do it. May 1 '20 at 6:53
• Indeed. I'd suggest to provide a little more insight to the first trick: how to factor the expression into: 1) a real term that may wiggle (function of $\omega$), akin to envelopes, 2) a pure exponential term that may wiggle (function of $\omega$), akin to modulations 3) some constant. May 1 '20 at 7:06
• thank you sir, now I understand how to simplify it and what does the simplified answer means May 1 '20 at 7:40
• sorry sir, after I read your answer again, I think I got confuse with the last paragraph. What does "tap expressions around the mid point of the impulse response taps" means ? can you please elaborate ? thanks sir May 1 '20 at 12:08
• @Anthony, that means if you were to look at the impulse response of the above frequency response you have an impulse at n=0 and n =4, the mid point here is 2 similary if you have an impulse repnse with an impulse at n=1 and n=4, then mid point is n=2.5, so basically you take out a common term around the mid point of the two terms. May 1 '20 at 12:47

\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}

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

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