Introduction
In this part we’ll see how we can design different type of controllers to make a system behave as wanted.
PD-controller
G ( s ) = K s ( 1 + s T ) , F ( s ) = K p + K d s . G(s) = \dfrac{K}{s(1 + sT)}, \quad F(s) = K_p + K_d s. G ( s ) = s ( 1 + s T ) K , F ( s ) = K p + K d s . The loop transfer function is:
L ( s ) = F ( s ) G ( s ) = K ( K p + K d s ) s ( 1 + s T ) = K T ( K p + K d s ) s 2 + 1 T s . L(s) = F(s)G(s) = \dfrac{K(K_p + K_d s)}{s(1 + sT)} = \dfrac{\frac{K}{T} (K_p + K_d s)}{s^2 + \frac{1}{T} s}. L ( s ) = F ( s ) G ( s ) = s ( 1 + s T ) K ( K p + K d s ) = s 2 + T 1 s T K ( K p + K d s ) . G r y ( s ) = L ( s ) 1 + L ( s ) = K T ( K p + K d s ) s 2 + 1 T s 1 + K T ( K p + K d s ) s 2 + 1 T s = K T ( K p + K d s ) s 2 + 1 T s + K T ( K p + K d s ) = K p K T + K d K T s s 2 + 1 + K d K T s + K p K T . \begin{align*}
G_{ry}(s) = \dfrac{L(s)}{1 + L(s)} & = \dfrac{\dfrac{\frac{K}{T} (K_p + K_d s)}{s^2 + \frac{1}{T} s}}{1 + \dfrac{\frac{K}{T} (K_p + K_d s)}{s^2 + \frac{1}{T} s}} \newline
& = \dfrac{\frac{K}{T}(K_p + K_d s)}{s^2 + \frac{1}{T} s + \frac{K}{T}(K_p + K_d s)} \newline
& = \dfrac{K_p \frac{K}{T} + K_d \frac{K}{T} s}{s^2 + \frac{1 + K_d K}{T} s + K_p \frac{K}{T}}.
\end{align*} G r y ( s ) = 1 + L ( s ) L ( s ) = 1 + s 2 + T 1 s T K ( K p + K d s ) s 2 + T 1 s T K ( K p + K d s ) = s 2 + T 1 s + T K ( K p + K d s ) T K ( K p + K d s ) = s 2 + T 1 + K d K s + K p T K K p T K + K d T K s . Which we can view as a second-order system where:
ω n = K p K T , ζ = 1 + K d K 2 T ω n = 1 + K d K 2 T K p K T . \omega_n = \sqrt{K_p \frac{K}{T}}, \quad \zeta = \dfrac{1 + K_d K}{2 T \omega_n} = \dfrac{1 + K_d K}{2T\sqrt{K_p \frac{K}{T}}}. ω n = K p T K , ζ = 2 T ω n 1 + K d K = 2 T K p T K 1 + K d K . If K d K_d K d ζ \zeta ζ
If K d K_d K d ζ \zeta ζ
Example 1 (I-controller) In this case we have a simple I-controller.
u ( t ) = K i ∫ 0 t e ( τ ) d τ ⟺ F ( s ) = K i s . u(t) = K_i \int_0^t e(\tau) \, d\tau \iff F(s) = \dfrac{K_i}{s}. u ( t ) = K i ∫ 0 t e ( τ ) d τ ⟺ F ( s ) = s K i . Given that our process is:
G ( s ) = b s + a . G(s) = \dfrac{b}{s + a}. G ( s ) = s + a b . Our loop transfer function is:
L ( s ) = F ( s ) G ( s ) = K i b s ( s + a ) . L(s) = F(s)G(s) = \dfrac{K_i b}{s(s + a)}. L ( s ) = F ( s ) G ( s ) = s ( s + a ) K i b . What should K i K_i K i 45 ∘ 45^\circ 4 5 ∘ ϕ m = 45 ∘ \phi_m = 45^\circ ϕ m = 4 5 ∘
L ( j ω ) = K i b j ω ( j ω + a ) . L(j\omega) = \dfrac{K_i b}{j\omega(j\omega + a)}. L ( j ω ) = j ω ( j ω + a ) K i b . We defined the phase margin as:
ϕ m = arg ( L ( j ω c ) ) + 180 ∘ . \phi_m = \arg(L(j\omega_c)) + 180^\circ. ϕ m = arg ( L ( j ω c )) + 18 0 ∘ . 45 ∘ = arg ( L ( j ω c ) ) + 180 ∘ arg ( L ( j ω c ) ) = − 135 ∘ − 90 ∘ − arctan ( ω c a ) = − 135 ∘ arctan ( ω c a ) = 45 ∘ ω c a = 1 ω c = a . \begin{align*}
45^\circ & = \arg(L(j\omega_c)) + 180^\circ \newline
\arg(L(j\omega_c)) & = -135^\circ \newline
-90^\circ - \arctan\left(\dfrac{\omega_c}{a}\right) & = -135^\circ \newline
\arctan\left(\dfrac{\omega_c}{a}\right) & = 45^\circ \newline
\dfrac{\omega_c}{a} & = 1 \newline
\omega_c & = a.
\end{align*} 4 5 ∘ arg ( L ( j ω c )) − 9 0 ∘ − arctan ( a ω c ) arctan ( a ω c ) a ω c ω c = arg ( L ( j ω c )) + 18 0 ∘ = − 13 5 ∘ = − 13 5 ∘ = 4 5 ∘ = 1 = a . Using the fact that ∣ L ( j ω c ) ∣ = 1 |L(j\omega_c)| = 1 ∣ L ( j ω c ) ∣ = 1
∣ L ( j ω c ) ∣ = 1 K i b ω c ω c 2 + a 2 = 1 K i b a a 2 + a 2 = 1 K i b a 2 a 2 = 1 K i b 2 a 2 = 1 K i b = 2 a 2 K i = 2 a 2 b . \begin{align*}
|L(j\omega_c)| & = 1 \newline
\dfrac{K_i b}{\omega_c \sqrt{\omega_c^2 + a^2}} & = 1 \newline
\dfrac{K_i b}{a \sqrt{a^2 + a^2}} & = 1 \newline
\dfrac{K_i b}{a \sqrt{2a^2}} & = 1 \newline
\dfrac{K_i b}{\sqrt{2} a^2} & = 1 \newline
K_i b & = \sqrt{2} a^2 \newline
K_i & = \dfrac{\sqrt{2} a^2}{b}.
\end{align*} ∣ L ( j ω c ) ∣ ω c ω c 2 + a 2 K i b a a 2 + a 2 K i b a 2 a 2 K i b 2 a 2 K i b K i b K i = 1 = 1 = 1 = 1 = 1 = 2 a 2 = b 2 a 2 . Let’s put this into L ( s ) L(s) L ( s )
L ( s ) = 2 a 2 s ( s + a ) . L(s) = \dfrac{\sqrt{2} a^2}{s(s + a)}. L ( s ) = s ( s + a ) 2 a 2 . Let’s now find G r y ( s ) G_{ry}(s) G r y ( s )
G r y ( s ) = L ( s ) 1 + L ( s ) = … = 2 a 2 s ( s + a ) + 2 a 2 = 2 a 2 s 2 + a s + 2 a 2 . G_{ry}(s) = \dfrac{L(s)}{1 + L(s)} = \ldots = \dfrac{\sqrt{2} a^2}{s(s + a) + \sqrt{2} a^2} = \dfrac{\sqrt{2} a^2}{s^2 + as + \sqrt{2} a^2}. G r y ( s ) = 1 + L ( s ) L ( s ) = … = s ( s + a ) + 2 a 2 2 a 2 = s 2 + a s + 2 a 2 2 a 2 . Second-order system!
ω n = 2 a 2 = 2 1 4 a , ζ = a 2 ω n = a 2 ( 2 1 4 a ) = 1 2 ⋅ 2 1 4 = 0.42. \begin{align*}
\omega_n & = \sqrt{\sqrt{2} a^2} = 2^{\frac{1}{4}} a, \newline
\zeta & = \dfrac{a}{2 \omega_n} = \dfrac{a}{2(2^{\frac{1}{4}} a)} = \dfrac{1}{2 \cdot 2^{\frac{1}{4}}} = 0.42.
\end{align*} ω n ζ = 2 a 2 = 2 4 1 a , = 2 ω n a = 2 ( 2 4 1 a ) a = 2 ⋅ 2 4 1 1 = 0.42. Example 2 (PI-controller) Let’s now do a PI-controller
Given the system:
G ( s ) = 1 s ( s + 8 ) 2 . G(s) = \dfrac{1}{s(s + 8)^2}. G ( s ) = s ( s + 8 ) 2 1 . Design a PI-controller so that the phase margin is 50 ∘ 50^\circ 5 0 ∘ ω c = 0.3 ω π \omega_c = 0.3 \omega_{\pi} ω c = 0.3 ω π
NB : This notation, ω c = C ⋅ ω π \omega_c = C \cdot \omega_{\pi} ω c = C ⋅ ω π ∣ G ( j ω c ) ∣ = 1 |G(j\omega_c)| = 1 ∣ G ( j ω c ) ∣ = 1 arg ( G ( j ω π ) ) = − 180 ∘ \arg(G(j\omega_{\pi})) = -180^\circ arg ( G ( j ω π )) = − 18 0 ∘ ω c G \omega_{cG} ω c G
A general PI-controller is written as:
F ( s ) = K p + K i s = K p ( 1 + 1 T i s ) = K p ( 1 + T i s T i s ) . F(s) = K_p + \dfrac{K_i}{s} = K_p\left(1 + \dfrac{1}{T_i s}\right) = K_p\left(\dfrac{1 + T_i s}{T_i s}\right). F ( s ) = K p + s K i = K p ( 1 + T i s 1 ) = K p ( T i s 1 + T i s ) . We know that:
arg ( G ( j ω π ) ) = − 180 ∘ , \arg(G(j\omega_{\pi})) = -180^\circ, arg ( G ( j ω π )) = − 18 0 ∘ , G ( j ω ) = 1 j ω ( j ω + 8 ) 2 . G(j\omega) = \dfrac{1}{j\omega(j\omega + 8)^2}. G ( j ω ) = j ω ( j ω + 8 ) 2 1 . Which means:
arg ( G ( j ω ) ) = − 90 ∘ − 2 arctan ( ω 8 ) . \arg(G(j\omega)) = -90^\circ - 2 \arctan\left(\dfrac{\omega}{8}\right). arg ( G ( j ω )) = − 9 0 ∘ − 2 arctan ( 8 ω ) . − 90 ∘ − 2 arctan ( ω π 8 ) = − 180 ∘ 2 arctan ( ω π 8 ) = 90 ∘ arctan ( ω π 8 ) = 45 ∘ ω π 8 = 1 ω π = 8. \begin{align*}
-90^\circ - 2 \arctan\left(\dfrac{\omega_{\pi}}{8}\right) & = -180^\circ \newline
2 \arctan\left(\dfrac{\omega_{\pi}}{8}\right) & = 90^\circ \newline
\arctan\left(\dfrac{\omega_{\pi}}{8}\right) & = 45^\circ \newline
\dfrac{\omega_{\pi}}{8} & = 1 \newline
\omega_{\pi} & = 8.
\end{align*} − 9 0 ∘ − 2 arctan ( 8 ω π ) 2 arctan ( 8 ω π ) arctan ( 8 ω π ) 8 ω π ω π = − 18 0 ∘ = 9 0 ∘ = 4 5 ∘ = 1 = 8. From our initial condition:
ω c = 0.3 ⋅ ω π = 2.4. \omega_c = 0.3 \cdot \omega_{\pi} = 2.4. ω c = 0.3 ⋅ ω π = 2.4. A phase margin of 50 ∘ 50^\circ 5 0 ∘
ϕ m = 50 ∘ ⟺ arg ( L ( j ω c ) ) + 180 ∘ = 50 ∘ . \phi_m = 50^\circ \iff \arg(L(j\omega_c)) + 180^\circ = 50^\circ. ϕ m = 5 0 ∘ ⟺ arg ( L ( j ω c )) + 18 0 ∘ = 5 0 ∘ . We actually don’t need to calculate what the loop transfer function is here, we can use the fact that L ( s ) = F ( s ) G ( s ) L(s) = F(s)G(s) L ( s ) = F ( s ) G ( s ) arg ( L ( j ω ) ) = arg ( F ( j ω ) ) + arg ( G ( j ω ) ) \arg(L(j\omega)) = \arg(F(j\omega)) + \arg(G(j\omega)) arg ( L ( j ω )) = arg ( F ( j ω )) + arg ( G ( j ω ))
arg ( G ( j ω c ) ) = − 90 ∘ − 2 arctan ( 2.4 8 ) = − 123.4 ∘ . \arg(G(j\omega_c)) = -90^\circ - 2 \arctan\left(\dfrac{2.4}{8}\right) = -123.4^\circ. arg ( G ( j ω c )) = − 9 0 ∘ − 2 arctan ( 8 2.4 ) = − 123. 4 ∘ . F ( j ω ) = K p ( 1 + T i j ω T i j ω ) , arg ( F ( j ω ) ) = arctan ( T i ω ) − 90 ∘ . F(j\omega) = K_p\left(\dfrac{1 + T_i j\omega}{T_i j\omega}\right), \quad \arg(F(j\omega)) = \arctan(T_i \omega) - 90^\circ. F ( j ω ) = K p ( T i j ω 1 + T i j ω ) , arg ( F ( j ω )) = arctan ( T i ω ) − 9 0 ∘ . arg ( L ( j ω c ) ) + 180 ∘ = 50 ∘ arctan ( T i ω c ) − 90 ∘ − 123.4 ∘ + 180 ∘ = 50 ∘ arctan ( T i ω c ) = 83.4 ∘ 2.4 T i = tan ( 83.4 ∘ ) T i = tan ( 83.4 ∘ ) 2.4 = 3.6. \begin{align*}
\arg(L(j\omega_c)) + 180^\circ & = 50^\circ \newline
\arctan(T_i \omega_c) - 90^\circ - 123.4^\circ + 180^\circ & = 50^\circ \newline
\arctan(T_i \omega_c) & = 83.4^\circ \newline
2.4 T_i & = \tan(83.4^\circ) \newline
T_i & = \dfrac{\tan(83.4^\circ)}{2.4} = 3.6.
\end{align*} arg ( L ( j ω c )) + 18 0 ∘ arctan ( T i ω c ) − 9 0 ∘ − 123. 4 ∘ + 18 0 ∘ arctan ( T i ω c ) 2.4 T i T i = 5 0 ∘ = 5 0 ∘ = 83. 4 ∘ = tan ( 83. 4 ∘ ) = 2.4 tan ( 83. 4 ∘ ) = 3.6. Now, let’s use that ∣ L ( j ω c ) ∣ = 1 |L(j\omega_c)| = 1 ∣ L ( j ω c ) ∣ = 1
∣ L ( j ω c ) ∣ = 1 ∣ F ( j ω c ) G ( j ω c ) ∣ = 1 ∣ F ( j ω c ) ∣ ∣ G ( j ω c ) ∣ = 1 K p ⋅ 1 2 + ( ω c ⋅ T i ) 2 ω c ⋅ T i ⋅ 1 ω c ( ω c 2 + 8 2 ) 2 = 1 K p = … ≈ 166.31. \begin{align*}
|L(j\omega_c)| & = 1 \newline
|F(j\omega_c)G(j\omega_c)| & = 1 \newline
|F(j\omega_c)| |G(j\omega_c)| & = 1 \newline
K_p \cdot \dfrac{\sqrt{1^2 + (\omega_c \cdot T_i)^2}}{\omega_c \cdot T_i} \cdot \dfrac{1}{\omega_c \left(\sqrt{\omega_c^2 + 8^2}\right)^2} & = 1 \newline
K_p & = \ldots \approx 166.31.
\end{align*} ∣ L ( j ω c ) ∣ ∣ F ( j ω c ) G ( j ω c ) ∣ ∣ F ( j ω c ) ∣∣ G ( j ω c ) ∣ K p ⋅ ω c ⋅ T i 1 2 + ( ω c ⋅ T i ) 2 ⋅ ω c ( ω c 2 + 8 2 ) 2 1 K p = 1 = 1 = 1 = 1 = … ≈ 166.31.