Frequency response of a sampled filter

Figure 15: Comparing the calculation of the frequency response in the continuous case and sampled case.

Reminder: for analogue Signals we had $H(s)$ with $s=j \omega$ for the frequency response. Let's substitute $s$ by $z$: $z^{-1} = e^{-sT}$ which gives us in general a mapping between the continuous domain and the sampled domain:

  $$z = e^{sT}$$ (98)

With $s=j \omega$ this gives us $z = e^{j \omega}$ and therefore the frequency response of the digital filter is:

  $$H(e^{j \omega})$$ (99)

which then leads to the amplitude and phase response:

• Amplitude/Magnitude:
    $$\vert H(e^{i \omega})\vert$$ (100)
• Phase
    $$\arg H(e^{i \omega})$$ (101)

Fig. 15 shows the difference between continuous and sampled signal. While in the continuous case the frequency response is evaluated along the imaginary axis, in the sampled case it happens along the unit circle which makes the response periodic! This is a subtle implication of the sampling theorem.

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