# Unit delay operator as parameter in function

I'm busy trying to implement a filter from this paper. I'm a bit confused with one particular step (30e in the paper).

Basically I need to find the zero frequency gain (K) of the following whitening/comb filter:

The relevant steps in the algorithm are:

and

where $q^{-1}$ is the unit delay operator, $\omega_0$ is the fundamental freq of the comb and $\rho$ is a constant.

I'm not sure how to calculate K. Since the function isn't acting on anything, what should be delayed? The signal (at time t)? Or does passing a constant in place of the delay mean it should be replaced?

Any help is greatly appreciated!

EDIT

Some clarification on $q^{-1}$:

The right-shift operator $q^{-1}$ acts on a discrete-time signal f(k, T) as $q^{-1}$ f(k, T) = (k − 1, T). The operation $q^{-1}$ therefore shifts the discrete-time signal to the right by one sampling interval.

• i would recommend that, instead of taking pics of equations, you should use $\LaTeX$ to write out the equations so that they can be edited. that said, please lose the $q^{-1}$ and replace with $z^{-1}$. then i might read the question, otherwise i won't bother to. Mar 10, 2015 at 14:11
• $q^{-1}$ is the symbol used in the paper, if you really want me to replace it I will, but does it really matter? Mar 10, 2015 at 14:13
• i use "$q$" for zeros and "$p$" for poles. i dunno whose book this came outa but the notation is not conventional. Mar 10, 2015 at 14:17
• I'd prefer to leave it as it is. I'm trying to understand what the author intended, so I think it would be best to use their words. Sorry you won't read my question. Mar 10, 2015 at 14:56

## 1 Answer

With $A(q^{-1},t)$ as given in the formula you get

$$A(1,t)=\prod_{k=1}^n(1-2\cos(k\hat\omega_0(t))+1)\tag{1}$$

and

$$A(\rho,t)=\prod_{k=1}^n(1-2\cos(k\hat\omega_0(t))\rho+\rho^2)\tag{2}$$

So $K(t)$ is simply the quotient of (1) and (2).

• Me and me colleague came to the same conclusion, I just found it confusingly written. Thanks for the answer. Mar 10, 2015 at 15:18