# Transfer function given as z-Domain fraction: gain calculation

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?

• You seem to miss a lot of basic knowledge concerning discrete-time filters. Please read up on the basics otherwise any new exercise will give you the same trouble as this one. Dec 8 '19 at 10:09
• Hello gunter! I think in addition to what is covered by Matts' answer, you might have missed the fact that the term $(z-1)/z$ also has a frequency-dependent gain, which in logarithmic domain is added on top of your 13.98 dB. As Matt mentioned, it would help if you could include the sampling frequency in your specific problem. Dec 8 '19 at 13:42

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