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 ) G(s) = \dfrac{K}{s(1 + sT)} G ( s ) = s ( 1 + s T ) K F ( s ) = K p + K d s F(s) = K_p + K_d s 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 ) s + K p K T 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}} = \dfrac{\frac{K}{T}(K_p + K_d s)}{s^2 + \frac{1}{T} s + \frac{K}{T}(K_p + K_d s)} = \dfrac{K_p \frac{K}{T} + K_d \frac{K}{T} s}{s^2 + (1 + K_d K) s + K_p \frac{K}{T}} 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 + ( 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 \omega_n = \sqrt{K_p \frac{K}{T}} \newline ω n = K p T K ζ = 1 + K d K 2 ω n = 1 + K d K 2 K p K T \zeta = \dfrac{1 + K_d K}{2 \omega_n} = \dfrac{1 + K_d K}{2\sqrt{K_p \frac{K}{T}}} ζ = 2 ω n 1 + K d K = 2 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 = a r g ( L ( j ω c ) ) + 180 ∘ \phi_m = arg(L(j\omega_c)) + 180^\circ ϕ m = a r g ( L ( j ω c )) + 18 0 ∘ 45 ∘ = a r g ( L ( j ω c ) ) + 180 ∘ 45^\circ = arg(L(j\omega_c)) + 180^\circ 4 5 ∘ = a r g ( L ( j ω c )) + 18 0 ∘ a r g ( L ( j ω c ) ) = − 135 ∘ arg(L(j\omega_c)) = -135^\circ a r g ( L ( j ω c )) = − 13 5 ∘ − 90 ∘ − a r c t a n ( ω c a ) = − 135 ∘ -90^\circ - arctan(\dfrac{\omega_c}{a}) = -135^\circ − 9 0 ∘ − a r c t an ( a ω c ) = − 13 5 ∘ a r c t a n ( ω c a ) = 45 ∘ arctan(\dfrac{\omega_c}{a}) = 45^\circ a r c t an ( a ω c ) = 4 5 ∘ ω c a = 1 \dfrac{\omega_c}{a} = 1 a ω c = 1 ω c = a \omega_c = a ω c = 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 |L(j\omega_c)| = 1 ∣ L ( j ω c ) ∣ = 1 K i b ω c ω c 2 + a 2 = 1 \dfrac{K_i b}{\omega_c \sqrt{\omega_c^2 + a^2}} = 1 ω c ω c 2 + a 2 K i b = 1 K i b a a 2 + a 2 = 1 \dfrac{K_i b}{a \sqrt{a^2 + a^2}} = 1 a a 2 + a 2 K i b = 1 K i b a 2 a 2 = 1 \dfrac{K_i b}{a \sqrt{2a^2}} = 1 a 2 a 2 K i b = 1 K i b 2 a 2 = 1 \dfrac{K_i b}{\sqrt{2}\ a^2} = 1 2 a 2 K i b = 1 K i b = 2 a 2 K_i b = \sqrt{2}\ a^2 K i b = 2 a 2 K i = 2 a 2 b K_i = \dfrac{\sqrt{2}\ a^2}{b} K i = 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 \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 ω n = 2 a 2 = 2 4 1 a ζ = 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 a r g ( G ( j ω π ) = − 180 ∘ arg(G(j\omega_{\pi}) = -180^\circ a r g ( 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(1 + \dfrac{1}{T_i s}) = K_p(\dfrac{1 + T_i s}{T_i s}) 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:
a r g ( G ( j ω π ) ) = − 180 ∘ arg(G(j\omega_{\pi})) = -180^\circ a r g ( 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:
a r g ( G ( j ω ) ) = − 90 ∘ − 2 a r c t a n ( ω 8 ) arg(G(j\omega)) = -90^\circ - 2 arctan(\dfrac{\omega}{8}) a r g ( G ( j ω )) = − 9 0 ∘ − 2 a r c t an ( 8 ω ) − 90 ∘ − 2 a r c t a n ( ω π 8 ) = − 180 ∘ -90^\circ - 2 arctan(\dfrac{\omega_{\pi}}{8}) = -180^\circ − 9 0 ∘ − 2 a r c t an ( 8 ω π ) = − 18 0 ∘ 2 a r c t a n ( ω π 8 ) = 90 ∘ 2 arctan(\dfrac{\omega_{\pi}}{8}) = 90^\circ 2 a r c t an ( 8 ω π ) = 9 0 ∘ a r c t a n ( ω π 8 ) = 45 ∘ arctan(\dfrac{\omega_{\pi}}{8}) = 45^\circ a r c t an ( 8 ω π ) = 4 5 ∘ ω π 8 = 1 \dfrac{\omega_{\pi}}{8} = 1 8 ω π = 1 ω π = 8 \omega_{\pi} = 8 ω π = 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 ∘ ⟺ a r g ( L ( j ω c ) ) + 180 ∘ = 50 ∘ \phi_m = 50^\circ \iff arg(L(j\omega_c)) + 180^\circ = 50^\circ ϕ m = 5 0 ∘ ⟺ a r g ( 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 ) a r g ( L ( j ω ) ) = a r g ( F ( j ω ) ) + a r g ( G ( j ω ) ) arg(L(j\omega)) = arg(F(j\omega)) + arg(G(j\omega)) a r g ( L ( j ω )) = a r g ( F ( j ω )) + a r g ( G ( j ω ))
a r g ( G ( j ω ) ) = − 90 ∘ − 2 a r c t a n ( 2.4 8 ) = − 123.4 ∘ arg(G(j\omega)) = -90^\circ - 2 arctan(\dfrac{2.4}{8}) = -123.4^\circ a r g ( G ( j ω )) = − 9 0 ∘ − 2 a r c t an ( 8 2.4 ) = − 123. 4 ∘ F ( j ω ) = K p ( 1 + T i j ω T i j ω ) F(j\omega) = K_p(\dfrac{1 + T_i j\omega}{T_i j\omega}) F ( j ω ) = K p ( T i j ω 1 + T i j ω ) a r g ( F ( j ω ) ) = a r c t a n ( T i ω ) − 90 ∘ arg(F(j\omega)) = arctan(T_i \omega) - 90^\circ a r g ( F ( j ω )) = a r c t an ( T i ω ) − 9 0 ∘ a r g ( L ( j ω c ) ) + 180 ∘ = 50 ∘ arg(L(j\omega_c)) + 180^\circ = 50^\circ a r g ( L ( j ω c )) + 18 0 ∘ = 5 0 ∘ a r c t a n ( T i ω c ) − 90 ∘ − 123.4 ∘ + 180 ∘ = 50 ∘ arctan(T_i \omega_c) - 90^\circ -123.4^\circ + 180^\circ = 50^\circ a r c t an ( T i ω c ) − 9 0 ∘ − 123. 4 ∘ + 18 0 ∘ = 5 0 ∘ a r c t a n ( T i ω c ) = 83.4 ∘ arctan(T_i \omega_c) = 83.4^\circ a r c t an ( T i ω c ) = 83. 4 ∘ 2.4 T i = t a n ( 83.4 ∘ ) 2.4 T_i = tan(83.4^\circ) 2.4 T i = t an ( 83. 4 ∘ ) T i = t a n ( 83.4 ∘ ) 2.4 = 3.6 T_i = \dfrac{tan(83.4^\circ)}{2.4} = 3.6 T i = 2.4 t an ( 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 |L(j\omega_c)| = 1 ∣ L ( j ω c ) ∣ = 1 ∣ F ( j ω ) G ( j ω ) ∣ = 1 |F(j\omega)G(j\omega)| = 1 ∣ F ( j ω ) G ( j ω ) ∣ = 1 ∣ F ( j ω ) ∣ ∣ G ( j ω ) ∣ = 1 |F(j\omega)| |G(j\omega)| = 1 ∣ F ( j ω ) ∣∣ G ( j ω ) ∣ = 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 \cdot \dfrac{\sqrt{1^2 + (\omega_c \cdot T_i)^2}}{\omega_c \cdot T_i} \cdot \dfrac{1}{\omega_c \sqrt{\omega_c^2 + 8^2}^2} = 1 K p ⋅ ω c ⋅ T i 1 2 + ( ω c ⋅ T i ) 2 ⋅ ω c ω c 2 + 8 2 2 1 = 1 K p = … ≈ 166.31 K_p = \ldots \approx 166.31 K p = … ≈ 166.31