--- title: "Frequency response design procedure" subject: "Digital Control System" chapter: 4 key: "semester_6/digital_control_system:frequency-response-design-procedure" source: https://www.bctengineer.com/q/semester_6/digital_control_system/frequency-response-design-procedure site: BCT Engineering --- # Frequency response design procedure Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System - Asked in: 2072 Asoj, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/frequency-response-design-procedure - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4 ## Question Write down the general procedures for frequency response design in discrete time control system. ## Short answer Obtain the pulse transfer function G(z) with ZOH. Transform to the w-plane via bilinear substitution z=(1+(T/2)w)/(1-(T/2)w). Choose gain K for the required static error constant. Draw the Bode plot of KG(jν) and select lead/lag compensator GD(w) to meet phase and gain margins. Transform GD(w) back to GD(z) and verify by simulation. ## Answer 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). ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is frequency-response design of a discrete system done in the w-plane? A. The bilinear transformation maps the unit circle onto the imaginary axis, so Bode methods can be used B. The w-plane removes the effect of the ZOH C. The w-plane makes the system continuous in time D. It removes the need for sampling Answer: A. The bilinear transformation turns the unit circle into the imaginary axis, so s-plane style Bode design applies. (Key point: Why the w-plane) 2. Which transformation takes G(z) to G(w)? A. z = e^(wT) B. z = (1 + (T/2)w)/(1 - (T/2)w) C. z = 1 + Tw D. z = (1 - (T/2)w)/(1 + (T/2)w) squared Answer: B. The bilinear transformation z = (1 + Tw/2)/(1 - Tw/2) is used; the inverse is w = (2/T)(z-1)/(z+1). (Key point: Bilinear transformation) 3. In lead design, the maximum phase phi_m is related to alpha by: A. sin phi_m = (1 + alpha)/(1 - alpha) B. tan phi_m = alpha C. sin phi_m = (1 - alpha)/(1 + alpha) D. cos phi_m = alpha Answer: C. For (1 + tau w)/(1 + alpha tau w), the maximum lead satisfies sin phi_m = (1 - alpha)/(1 + alpha). (Key point: Lead compensator design) 4. How is the required lead phi_m chosen? A. Required PM minus existing PM, plus a safety margin of about 5 to 12 degrees B. Required PM plus existing PM C. Equal to the gain margin D. Exactly 90 degrees Answer: A. The crossover shifts after adding the lead, so extra degrees are added to the deficit. (Key point: Extra phase allowance) 5. After the w-plane compensator is designed and checked, what is done next? A. The result is used directly in the w-plane B. It is transformed back to G_D(z) with w = (2/T)(z-1)/(z+1) and then verified by simulation C. T is doubled D. The ZOH is removed Answer: B. The controller must be realised in z as a difference equation, then verified with a step response. 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