# How does sampling rate affect discrete filters?

I am just starting to learn discrete filters and I could use some help. I understand continuous signals and filters.

I am trying to understand the math behind discrete filtering. For example in the s domain, a simple low pass filter can be recognized as $$\dfrac{1}{\tau s+1}$$ How does sampling rate affect a discrete low pass filter?

• The cut-off frequency of a discrete-time low pass filter is specified relative to the sampling frequency. So by changing the sampling frequency, you also change its cut-off frequency. Is that what you're asking? – Matt L. May 22 '15 at 8:36
• I guess what I am trying to understand where if the equation above was brought to the z-domain, where would the sampling time have a role? – Adam May 22 '15 at 12:59

IF we take the sampling period as T.First I will convert the filter to time domain impulse response as $$h(s)=\frac{1}{\tau s+1}\\h(s)=\frac{1/\tau}{s+\frac{1}{\tau}}$$ this in time domain is $$h(t)=\frac{e^{-t/\tau}}{\tau}$$ IF the above signal is sampled at time T the it can be written as $$h(Tn)=\frac{e^{\frac{-Tn}{\tau}}}{\tau}\\h(n)=\frac{e^{\frac{-Tn}{\tau}}}{\tau}$$ if Z transform taken for the impulse response.Then the equation is $$H(z)=\frac{1/\tau}{1-e^{\frac{T}{\tau}}z^{-1}}$$ Here we can see the sampling time coming in denominator which could put the pole inside or outside the unit circle
• since h(n) is a geometric sequence we can find standard H(z) for h(n) as $$\frac{1}{1-a^{-1}z^{-1}}$$ here a is $$e^{T/\tau}$$ and T\/tau should be negative – Vinith May 23 '15 at 2:42