--- title: "(1+aTs)/(1+Ts) as lead or lag compensator" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lead-or-lag-1-ats" source: https://www.bctengineer.com/q/semester_3/control_systems/lead-or-lag-1-ats site: BCT Engineering --- # (1+aTs)/(1+Ts) as lead or lag compensator Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2073 Shrawan, 4 marks - Video: [What are Lead Lag Compensators? An Introduction.](https://www.youtube.com/watch?v=xLhvil5sDcU) — Brian Douglas - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/lead-or-lag-1-ats - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question How can a controller with transfer function Gc(s) = (1 + aTs)/(1 + Ts) be used as lead or lag compensator? Explain. ## Short answer For Gc(s)=(1+aTs)/(1+Ts), if a>1 the zero is closer to origin: lead compensator, adding positive phase and high-frequency gain boost. If a<1 the pole is closer to origin: lag compensator, attenuating high frequencies to improve steady-state accuracy. ## Answer $$ G_c(s)=\frac{1+aTs}{1+Ts} $$ It has a **zero at $s=-\frac{1}{aT}$** and a **pole at $s=-\frac{1}{T}$**. Whether it is lead or lag depends only on $a$. #### Phase of the network $$ \angle G_c(j\omega)=\tan^{-1}(a\omega T)-\tan^{-1}(\omega T) $$ #### Case 1: $a>1$ — lead compensator - $a\omega T>\omega T$, so the phase is **positive** at all frequencies. - Zero ($-1/aT$) is **nearer the origin** than the pole ($-1/T$). - Maximum lead at $\omega_m=\frac{1}{T\sqrt a}$: $\sin\phi_m=\frac{a-1}{a+1}$; high-frequency gain $=a$ ($20\log a$ dB). - Use: improve phase margin, damping and speed (transient response). #### Case 2: $a<1$ — lag compensator - $a\omega T<\omega T$, so the phase is **negative**. - Pole ($-1/T$) is **nearer the origin** than the zero ($-1/aT$). - High-frequency gain $=a<1$: attenuation of $20\log(1/a)$ dB; DC gain 1. - Use: improve steady-state accuracy (with the loop gain raised by $1/a$), or lower the crossover to gain PM. ```text Lead (a>1): -x------o-----+ sigma -1/T -1/aT 0 Lag (a<1): -o------x-----+ sigma -1/aT -1/T 0 ``` ($a=1$ gives $G_c=1$, no compensation.) | | $a>1$ | $a<1$ | |---|---|---| | Type | lead | lag | | Phase | positive | negative | | Gain at high $\omega$ | $a$ (boost) | $a$ (cut) | **Example:** $G_c=\frac{1+0.5s}{1+0.1s}$ ($a=5$, $T=0.1$): lead, maximum lead $\sin^{-1}\frac{4}{6}=41.8^\circ$ at $\omega_m=\frac{1}{0.1\sqrt5}=4.47$ rad/s. $G_c=\frac{1+2s}{1+20s}$ ($a=0.1$, $T=20$): lag. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For Gc = (1 + aTs)/(1 + Ts), where are the zero and pole? A. Zero at -1/(aT), pole at -1/T B. Zero at -1/T, pole at -1/(aT) C. Zero at -aT, pole at -T D. Zero at the origin, pole at -1/T Answer: A. Numerator gives zero -1/(aT); denominator gives pole -1/T. (Key point: Zero and pole) 2. For which a is this network a lead compensator? A. a < 1 B. a > 1, because the phase atan(aTw) - atan(Tw) is positive and the zero is nearer the origin C. a = 1 D. Any a, if T is large Answer: B. With a > 1 the zero is nearer the origin, giving positive phase at all frequencies. (Key point: Lead condition) 3. For a > 1, what are the frequency and value of maximum lead? A. wm = T sqrt a; sin(phi_m) = (a+1)/(a-1) B. wm = 1/(aT); sin(phi_m) = 1/a C. wm = 1/(T sqrt a); sin(phi_m) = (a-1)/(a+1) D. wm = 1/T; phi_m = 90 deg Answer: C. wm is the geometric mean of the zero and pole frequencies. (Key point: Maximum lead) 4. For a < 1, what does the network do? A. Acts as a lead with gain a B. Passes all frequencies unchanged C. Boosts high frequencies by 20log(a) dB D. Acts as a lag compensator with high-frequency gain a (attenuation 20log(1/a) dB) and DC gain 1 Answer: D. With a < 1 the pole is nearer the origin and phase is negative. 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