The Day I Felt Good at Digital Signal Processing

Exam context

Ten minutes, one unfamiliar proof.

There are a few moments in school when a difficult subject stops feeling like a wall and starts feeling like a language you can speak. One of mine came during the final exam for EE 699: Digital Signal Processing, taught by Dr. Michael Johnson.

I was a Ph.D. student then. After finishing the previous six questions, I had about ten minutes left for the graduate-only question. I had never seen one like it, and my first reaction was that it might be beyond me. Instead of leaving the exam hall early, I decided to use the time for an honest try: go back to what you know and connect the dots.

I had always felt comfortable with mathematics, but I had not done this kind of proof in signal processing before—at least not since proof-based mathematics classes during my undergraduate degree. So I started with the definitions from class, followed the symmetry, and gave the problem an honest try.

The question asked us to show that a Type IV FIR filter—an even-length, anti-symmetric filter—has generalized linear phase. It was not a trick question, but it asked for more than recalling a definition. It asked for the structure hidden inside the Fourier transform.

Handwritten solution to a graduate digital signal processing exam question on Type IV FIR filters and generalized linear phase. — My solution to the graduate-only final-exam question from EE 699: Digital Signal Processing. Thanks to Dr. Johnson for letting me take a picture of this memory.

Why I remember it

Getting full credit felt good, of course. More than that, it was evidence that the ideas had started to connect: time-domain symmetry, frequency-domain phase, and the practical behavior of FIR filters were no longer separate facts to memorize.

It is a small academic memory, but it is one I keep. Signal processing has a way of rewarding patience: stay with the algebra long enough, and the pattern usually introduces itself.

The idea behind the proof

For a Type IV FIR filter, $N$ is even and the impulse response is anti-symmetric:

h [n] = - h [N - 1 - n]

Start with the FIR transfer function, exactly as I did in the solution. The frequency response is obtained by evaluating it on the unit circle, $z = e^{j ω}$ . Then pair each sample with its reflected partner; anti-symmetry turns the second term into a subtraction:

H (z) = \sum_{n = 0}^{N - 1} h [n] z^{- n} \Rightarrow H (e^{j ω}) = \sum_{n = 0}^{N - 1} h [n] e^{- j ω n}

= \sum_{n = 0}^{N / 2 - 1} h [n] [e^{- j ω n} - e^{- j ω (N - 1 - n)}]

= e^{- j ω (N - 1) / 2} \cdot 2 j \sum_{n = 0}^{N / 2 - 1} h [n] \sin (ω (\frac{N - 1}{2} - n))

The final line is the key insight: the paired exponentials become a sine term, while the remaining exponential is the delay about the filter midpoint. Define the real-valued amplitude term

A (ω) = 2 \sum_{n = 0}^{N / 2 - 1} h [n] \sin (ω (\frac{N - 1}{2} - n))

Because the filter coefficients and sine terms are real, $A (ω)$ is real. Since $j = e^{j π / 2}$ , the response becomes

H (e^{j ω}) = A (ω) e^{- j ω (N - 1) / 2} e^{j π / 2}

This is the generalized linear-phase form $H (e^{j ω}) = A (ω) e^{- j α ω + j β}$ , with $α = \frac{N - 1}{2}$ and $β = \frac{π}{2}$ . The delay is linear in frequency, and the anti-symmetry contributes a fixed 90° phase offset.

Put another way, the answer was not buried in a long calculation. Once the impulse response is grouped around its midpoint, its anti-symmetry forces the transform into the right shape.

The day I felt good at digital signal processing.

Ten minutes, one unfamiliar proof.

Why I remember it

The idea behind the proof