# Is it possible to sketch a bode plot for a discrete-time transfer function?

For a regular s-domain transfer function, there are simple rules you can use to draw an asymptotic bode graph, basically by taking a slope of +/-20dB per decade and a phase shift of +/-90 degree for each pole/zero.

Is it possible to do the same for a z-domain transfer function?

I'm explicitly not looking to plot the exact transfer using Scipy/Matlab, but a pen-and-paper approach that gives intuition.

Basically, if I have an s-domain TF, it's fairly easy to have an intuition of the gain, phase, and corresponding stability margins. For a discrete system, I have close to no intuition about them at all.

## 2 Answers

In linear, time-invariant, continuous time systems, the transfer function $$\H(s) \$$ is a broken rational function in s. We obtain the complex frequency response through the substitution $$\s=j\cdot \omega\$$. In this case it is rather easy to find the asymptotic behaviour of the system and to sketch the bode diagram.

In linear, time-invariant, discrete time systems, the transfer function $$\H(z) \$$ is a broken rational function in z. We obtain the complex frequency response through the substitution $$\s=e^{j\cdot \omega \cdot T_S}\$$. Since this relationship is non-linear, there are no straight lines as asymptotes.

However, if one examines the zeros and poles of the time-discrete system, it is possible to predict the behavior of the system for frequencies between 0 and $$\\frac{f_S}{2}\$$ (or rather between $$\-\frac{f_S}{2}\$$ and $$\+\frac{f_S}{2}\$$) . Imagine the poles as very high mountains in the complex z-plane and the zeros as deep valleys, and then take a walk along the unit circle. This should give you a good feeling for how the system behaves.

Is it possible to do the same for a z-domain transfer function?

It certainly is but you are somewhat constrained to the upper frequency limit being the equivalent Nyquist frequency. At this point, due to sampling and aliasing, the amplitude is usually indeterminate. Above that you also get aliasing so, unless you have some special under-sampling requirement then, the Nyquist frequency is your maximum useful point on the x-axis.

You should also remember that the amplitude of the filter response is altered as you rise towards the Nyquist frequency: - Picture from here.

• Technically a correct answer to my question, but not very helpful in how I'd go about sketching the transfer function. – Pepijn Feb 3 '20 at 15:49
• @Pepijn other than suggest use a pencil so you can rub out errors I'm failing to see the difficulty that you envisage. Are you requiring the sinc function formula? – Andy aka Feb 3 '20 at 15:53
• I must be missing something obvious then?? Say I have something simple like H(z)=z/(z-1), how would a sinc function help me sketch the magnitude and phase of this transfer function? – Pepijn Feb 3 '20 at 15:58
• Maybe a z domain to laplace conversion table might help you: lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html. For instance $\dfrac{z}{z-e^{-a\tau}}$ converts to $\dfrac{1}{s+a}$? – Andy aka Feb 3 '20 at 16:03
• $\large z\rightarrow e^{j \omega T}$, therefore $\large \frac{z}{z-1}\rightarrow \frac{e^{j \omega T}}{e^{j \omega T}-1}=\frac{cos(\omega T)+j sin(\omega T)}{cos(\omega T)-1 +j sin (\omega T)}$ – Chu Feb 3 '20 at 19:11