Transfer function given as z-Domain fraction: gain calculation

Question

I am stuck on a question regarding transfer functions. I have the answer and have attempted the question but am having trouble solving it.

Q: Given the transfer function of a digital filter is (5z-5)/z, compute the gain (in dB) of the digital filter at the cutoff frequency (10 rad/s)

The answer given is 14.65dB. (or a gain of 5.401)

I am unsure how to do this:

I see that the transfer function can be rewritten as 5(z-1)/z leading me to believe the gain would be 5 which is 13.98 dB, more than half a dB off.

What am I doing wrong here?

Answer 1

Obviously, you might need to use the given frequency somehow to solve the problem. Guessing alone will usually not suffice. Since this is a homework type problem, and since you still need to learn a lot of basics, I'll give you a few hints to get you started.

First, the frequency domain behavior can be obtained from a (stable) transfer function by substituting $z=e^{j\omega T}$, where $T$ is the sampling period, which equals the inverse of the sampling frequency $f_s$ ($T=1/f_s$). So you won't be able to solve the problem without knowing the sampling frequency (unless the frequency is given as a normalized value, which is not the case in your exercise).

The gain at the given frequency $\omega_c$ is then given by the magnitude of the transfer function evaluated at $z=e^{j\omega_cT}$:

$$g=\big|H(e^{j\omega_cT})\big|$$

I trust that you know how to convert that linear gain to the corresponding value in dB.