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I have a length 10 Finite Impulse response as shown below, and I am trying to find the phase delay of this FIR.

$$h[n]=\begin{cases}0.1,& 0\le n \lt 10\\ 0,&\textrm{otherwise}\end{cases}$$

I tried plotting from first principles as well as using the phasedelay() function, however I am getting different results. The results are varying greatly. Any suggestions on what I am forgetting to do in my approach from first principles?

enter image description here

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The problem is the definition of the phase. The command angle() computes the phase $\phi(\omega)$ according to

$$H(e^{j\omega})=|H(e^{j\omega})|e^{j\phi(\omega)}\tag{1}$$

where $H(e^{j\omega})$ is the complex frequency response. At frequencies $\omega$ where the frequency response has zeros, the phase jumps by $\pi$. This is shown in the two left figures of the plot below. You can't compute the phase delay from $\phi(\omega)$ because of the jumps. Note that in your computation, the phase delay is correct until the first phase jump due to the first zero of the magnitude response.

The phase delay needs to be computed from a continuous phase $\varphi(\omega)$, which is implicitly defined by

$$H(e^{j\omega})=A(\omega)e^{j\varphi(\omega)}\tag{2}$$

where $A(\omega)$ is a real-valued but bi-polar smooth function. This continuous phase is shown in the top right figure below, and it is obviously linear. From this continuous phase, the phase delay can be computed by

$$\tau_{ph}=-\frac{\varphi(\omega)}{\omega}\tag{3}$$

and for the given linear-phase FIR filter the phase delay is constant:

$$\tau_{ph}=\frac{N-1}{2}\tag{4}$$

where $N$ is the filter length.

Note that the continuous phase $\varphi(\omega)$ cannot be obtained from $\phi(\omega)$ using the Matlab function unwrap(), because the latter just removes jumps by (multiples of) $2\pi$ due to phase ambiguity. Here we have to deal with actual phase jumps due to zeros of $H(e^{j\omega})$.

enter image description here

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