(1+aTs)/(1+Ts) as lead or lag compensator

$$G_c(s)=\frac{1+aTs}{1+Ts}$$

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.

 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.)

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.