Representation and Analysis of Dynamical Systems
1h15
The exercises are independent.
Note: most parameters are intentionally given without physical units. The numerical values are
intentionally chosen in order to allow easy computation by hand.
Exercise 1: Bode diagram and closed loop bandwidth
We consider the system given by the following transfer function:
𝑌(𝑠)
1
=
𝑈(𝑠) (1 + 𝑠)(1 + 2𝑠)(1 + 10𝑠)
𝐹(𝑠) =
The Bode diagrams of this open-loop transfer function are given in figure 1:
20𝑙𝑜𝑔10(𝐾) ≈ 10
0.3 𝑟𝑎𝑑/𝑠
𝑃𝑀 ≈ 60°
Figure 1: Bode diagram of F
The system is controlled according to the block diagram of figure 2
𝑟
+
𝑢
𝐹(𝑠)
𝐾
−
Figure 2: control
1
𝑦
Question 1:
What is the value of the constant gain 𝐾 that will ensure a closed loop bandwidth of 0.3 𝑟𝑎𝑑. 𝑠 −1 ? (An
approximate response obtained from the Bode diagram is ok).
Read on the Bode diagram: 20𝑙𝑜𝑔10(𝐾) ≈ 10 ⇒ 𝐾 ≈ 100.5 ⇒ 𝐾 ≈ 3sult accepted: 𝐾 ∈ [2
4]
Question 2:
For this gain 𝐾 what is the static gain in closed loop?
𝐾
Static gain of the open loop is K. So static gain of the closed loop is 1+𝐾
Result accepted: static gain ∈ [0.6 0.8]
Question 3:
For this gain 𝐾 what is the phase margin?
Read on the diagram: 𝑃𝑀 ≈ 60°
Accepted 𝑃𝑀 ∈ [50 70]
Question 4:
For this gain 𝐾 three candidates for the closed loop step response are given in figure 3. Which one is the
correct one. You must justify with at least two arguments.
Correct is plot 1:
-
Static gain of plot 3 is much too small
Damping of plot 2 is much too small for a phase margin in the range [50
2
70]
Figure 3: step response
Question 5:
In order to cancel the static error we replace the pure proportional controller 𝐾 by a PI controller:
𝐾(𝑠) = 𝐾 × (1 +
𝜔𝑖
)
𝑠
In the figure 4 we represent the step response with two values of 𝜔𝑖 : 𝜔1 = 0.08 and 𝜔2 = 0.3
Give the correct correspondence between the step response and the value of 𝜔𝑖 . Explain.
𝜔1 corresponds to plot 1 and 𝜔2 corresponds to plot 2
𝜔1 is significantly less than the closed loop bandwidth. The PI will add almost no extra phase and
consequently not modify the phase margin
3
Figure 4: step response with a PI controller
4
Exercise 2: Coupled tanks
We consider a three tanks system as depicted in figure 5. Each tank (1, 2 and 3) as a level of water given
by 𝐿1 (𝑡), 𝐿2 (𝑡) and 𝐿3 (𝑡). The first tank is filled with water with a pump controlled by a control voltage
𝑈(𝑡)
𝑈(𝑡)
𝐿1 (𝑡)
𝐿2 (𝑡)
𝐿3 (𝑡)
Figure 5: Three tanks system
The level of each tank is given by the following set of differential equations where each coefficient is
strictly positive:
𝑑𝐿1 (𝑡)
= −𝛼1 √𝐿1 (𝑡) + 𝛽 𝑢(𝑡)
𝑑𝑡
𝑑𝐿2 (𝑡)
= −𝛼2 √𝐿2 (𝑡) + 𝛼1 √𝐿1 (𝑡)
𝑑𝑡
𝑑𝐿3 (𝑡)
{ 𝑑𝑡 = −𝛼3 √𝐿3 (𝑡) + 𝛼2 √𝐿2 (𝑡)
Question 1:
We consider this system as single input (control input 𝑢(𝑡)) and single output (level of tank 3, 𝑙3 (𝑡)).
What is the water level equilibrium (𝐿1 , 𝐿2 , 𝐿3 ) when the input voltage 𝑈(𝑡) = 𝑈 is constant?
5
𝛽𝑈 2
)
𝛼1
𝛼1 2
𝛽𝑈 2
𝐿2 = ( ) 𝐿1 = ( )
𝛼2
𝛼2
2
𝛽𝑈
𝐿3 = ( )
{
𝛼3
𝐿1 = (
Question 2:
Give the linearized system of differential equations near this equilibrium point. You’ll note 𝐿𝑖 (𝑡) = 𝐿𝑖 +
𝑙𝑖 (𝑡) where 𝐿𝑖 corresponds to the equilibrium level and 𝑙𝑖 (𝑡) the small variation of the level near this
equilibrium and 𝑈(𝑡) = 𝑈 + 𝑢(𝑡)
𝑑𝑙1 (𝑡)
𝛼1
=−
𝑙1 (𝑡) + 𝛽𝑢(𝑡) = −𝑎1 𝑙1 (𝑡) + 𝛽𝑢(𝑡)
𝑑𝑡
2√𝐿1
𝑑𝑙2 (𝑡)
𝛼2
𝛼1
=−
𝑙2 (𝑡) +
𝑙1 (𝑡) = −𝑎2 𝑙2 (𝑡) + 𝑎1 𝑙1 (𝑡)
𝑑𝑡
2√𝐿2
2√𝐿1
𝑑𝑙3 (𝑡)
𝛼3
𝛼2
=−
𝑙3 (𝑡) +
𝑙2 (𝑡) = −𝑎3 𝑙3 (𝑡) + 𝑎2 𝑙2 (𝑡)
2√𝐿3
2√𝐿2
{ 𝑑𝑡
Question 3:
Is it possible to control independently the level of each tank using only the control input 𝑢(𝑡)?
Check controllability matrix:
−
𝐴=
𝛼1
2√𝐿1
𝛼1
2√𝐿1
[
0
[𝐵
0
−
0
𝛼2
2√𝐿2
𝛼2
2√𝐿2
𝐴𝐵
−𝑎1
= [ 𝑎1
0
0
−
𝛼3
0
−𝑎2
𝑎2
0
𝛽
0 ] 𝐵 = [0 ]
−𝑎3
0
2√𝐿3 ]
𝛽
𝐴 𝐵] = [ 0
0
2
−𝛽𝑎1
𝛽𝑎1
0
Rank is 3 (assuming non-null coefficients)
Question 4:
Give the transfer function between 𝑢(𝑡) and 𝑙3 (𝑡)
6
𝛽𝑎12
− 𝛽𝑎1 𝑎2 ]
𝛽𝑎1 𝑎2
−𝛽𝑎12
𝐿3 (𝑠) =
𝑎2
𝑎1
𝛽
𝑈(𝑠)
(𝑠 + 𝑎3 ) (𝑠 + 𝑎2 ) (𝑠 + 𝑎1 )
Remark: transfer function is of order 3, confirms full observability and controllability
7