Frequency response design procedure

Frequency-response design of a discrete-time system is done in the w-plane. The bilinear transformation maps the unit circle of the z-plane onto the imaginary axis of the w-plane, so the usual Bode-diagram methods can be used.

Procedure

1. Find the pulse transfer function of the plant with the zero-order hold: $G(z) = (1-z^{-1})\,\mathcal{Z}[G_p(s)/s]$.
2. Transform to the w-plane with the bilinear transformation: $$z = \frac{1 + (T/2)w}{1 - (T/2)w}, \qquad G(w) = G(z)\Big|_{z=\frac{1+Tw/2}{1-Tw/2}}$$ Write $G(w)$ in time-constant form. The fictitious frequency $\nu$ ($w=j\nu$) is related to the actual frequency by $\nu = \frac{2}{T}\tan\frac{\omega T}{2}$.
3. Choose the controller gain $K$ to meet the static error constant ($K_p$, $K_v$ or $K_a$), using $\lim_{w\to0}$ in the w-plane, which gives the same constants as in the s-plane.
4. Draw the Bode diagram of $KG(j\nu)$ and read the uncompensated gain crossover frequency, phase margin and gain margin.
5. Select the compensator $G_D(w) = \dfrac{1+\tau w}{1+\alpha\tau w}$:
   • lead ($\alpha<1$): needed phase $\phi_m = \text{PM}_{\text{required}} - \text{PM}_{\text{existing}} + 5^\circ\text{ to }12^\circ$; $\sin\phi_m = \frac{1-\alpha}{1+\alpha}$; place $\phi_m$ at the new crossover where $|KG(j\nu)| = \sqrt{\alpha}$; $\tau = 1/(\nu_c\sqrt{\alpha})$;
   • lag ($\alpha>1$ in this form): put the new crossover where the uncompensated phase gives the required PM (plus about $5^\circ$), choose $\alpha$ so that the gain there drops to 0 dB, and place the zero corner $1/\tau$ about one decade below the new crossover.
6. Check the compensated Bode diagram for phase margin, gain margin and bandwidth. Repeat step 5 if needed.
7. Transform back to the z-plane with $$w = \frac{2}{T}\,\frac{z-1}{z+1}$$ to get $G_D(z)$, which is then realised as a difference equation in the computer.
8. Verify with the closed-loop step response (simulation), and check that the sampling period is small compared with the closed-loop bandwidth (8–10 samples per cycle).

 G(s) --ZOH--> G(z) --bilinear--> G(w) --Bode--> GD(w)
                                                   |
       GD(z)  <---------- w = (2/T)(z-1)/(z+1) ----+

Points to remember

• Unlike $G(s)$, $G(w)$ usually has as many zeros as poles, so its magnitude does not roll off to zero as $\nu \to \infty$. It usually has a right-half-plane zero at $w = 2/T$ (non-minimum phase), which adds phase lag at high frequency.
• Since $\nu \approx \omega$ for $\omega T \ll 1$, the design reads like a continuous-time design when sampling is fast.