1. Frequency Response Analysis and Stability
Lecture 8:
Frequency Response Analysis and Stability

2. Objectives
Test system stability in the frequency domain using the Bode stability method.
Explain the meaning of gain and phase margins.

3. Why do we study frequency response?
Frequency Response: determines the response of systems variables to a sine input.
Why do we study frequency response?
Perfect sine disturbances occur frequently in plants
We can learn useful generalizations about control performance and robustness.
No!
Yes!

4. Frequency Response : Sine in  sine out
How do we calculate the frequency response?
The frequency response can be calculated from the transfer function by setting s = j, with = frequency and j = complex variable.

5. Frequency Response : Sine in  sine out
Amplitude ratio = |Y(t)| max / |U(t)| max
Phase angle = phase difference between input and output P 1 2 3 4 5 6
-0.4
-0.2
0.2
0.4
time
Y, outlet from system -1
-0.5
0.5
U, inlet to system B
output P’ A
input

6. Frequency Response : Sine in  sine out
Amplitude ratio = |Y(t)| max / |U(t)| max
Phase angle = phase difference between input and output
These calculations are tedious by hand but easily performed in standard programming languages.
In most programming languages, the absolute value gives the magnitude of a complex number.

7. Frequency response of mixing tank.
Time-domain
behavior.
Bode Plot - Shows
frequency response for
a range of frequencies
Log (AR) vs log()
Phase angle vs log()

8. STABILITY 20 40 60 80
100
120
-40
-20 TC v1 v2
No! or 20 40 60 80
100
120
-0.2
0.2
0.4
0.6
0.8
Yes!
We influence stability when we implement control. How do we achieve the influence we want?

9. Sample Inputs Sample Outputs Process bounded bounded unbounded
First, let’s define stability: A system is stable if all bounded inputs to the system result in bounded outputs.
Sample
Inputs
Sample
Outputs
Process
0.5 1
1.5 -1
-0.5
0.5 1
1.5 -1
-0.5
bounded
bounded
unbounded
unbounded

10. STABILITY Set point response The denominator determines
the stability of
the closed-loop
feedback system!
Set point response
Bode Stability Method
Calculating the roots is easy with standard software. However, if the equation has a dead time, the term e-s appears. Therefore, we need another method.
The method we will use next is the Bode Stability Method.

11. Bode Stability: To understand, let’s do a thought experiment
GP(s)
Gv(s)
GC(s)
GS(s)
CV(s)
CVm(s)
SP(s)
E(s)
MV(s) + -
Loop open

12. Bode Stability: To understand, let’s do a thought experiment
GP(s)
Gv(s)
GC(s)
GS(s)
CV(s)
CVm(s)
SP(s)
E(s)
MV(s) + -
Loop closed
No forcing!!
GP(s)
Gv(s)
GC(s)
GS(s)
CV(s)
CVm(s)
SP(s)
E(s)
MV(s) + -
Loop closed
No forcing!!
Under what conditions is the system stable (unstable)?
Hint: think about the sine wave as it travels around the loop once.

13. Bode Stability: To understand, let’s do a thought experiment
GP(s)
Gv(s)
GC(s)
GS(s)
CV(s)
CVm(s)
SP(s)
E(s)
MV(s) + -
Loop closed
If the sine is larger in amplitude after one cycle; then it will increase each “time around” the loop. The system will be unstable.
Now: at what frequency does the sine most reinforce itself?

14. Bode Stability: To understand, let’s do a thought experiment
GP(s)
Gv(s)
GC(s)
GS(s)
CV(s)
CVm(s)
SP(s)
E(s)
MV(s) + -
Loop closed
When the sine has a lag of 180° due to element dynamics, the feedback will reinforce the oscillation (remember the - sign).
This is the critical or phase crossover frequency, pc.

15. Bode Stability
Let’s put the results together. GOL(s) includes all elements in the closed loop.
At the critical frequency:  GOL(jpc) = -180
The amplitude ratio: |GOL(jpc) | < 1 for stability
|GOL(jpc) | > 1 for instability
The gain margin (GM) is defined as:
GM = 1 | 𝐺 𝑂𝐿 (jpc)|
Hence, if the gain margin is less than 1 (negative in dB), the system is unstable.

16. Phase Margin
Another relevant term is the phase margin. To calculate it, we need to find the gain crossover frequency (gc) which is the frequency at which the open-loop gain crosses unity.
The phase margin (PM) is the distance between the open loop phase and -180 at frequency gc
If the phase margin is negative, the system is unstable.

17. Interpretation of Gain and Phase Margins
The gain margin tells us the maximum proportional gain we are allowed to use without affecting system stability.
The phase margin tells us the maximum additional phase (and also dead time) that can be added to the loop without affecting system stability.
The best way to understand the gain and phase margins is to try a numerical example!

18. Example
Find the gain and phase margin for the following process under proportional control. Is the system stable? Find the delay margin as well.

19. Answer, continued
To find the gain margin, we need the phase crossover frequency.
At this frequency,
(system stable)
Therefore, the gain margin = 1/0.735 = 1.36 = 2.63 dB.

20. Answer, continued
To find the phase margin, we need the gain crossover frequency.
At this frequency,

21. Answer, continued Therefore, the phase margin (PM) is found as
And the delay margin (DM) is obtained as

22. We can check our result using the command “margin” in Matlab.
s=tf(‘s’); G=2/(s+1)^2*exp(-s); margin(G)

23. Let us now use Simulink to understand the interpretation of gain margin and phase margins.
For this purpose, we build the following model:

24. Closed-loop step response (original process with proportional controller gain = 1)
The system is stable.

25. Closed-loop step response with proportional controller of gain 1.4.
The system is unstable.
Remember that the gain margin is 1.36.

26. Closed-loop step response with proportional controller of gain 1 and additional delay of 0.6 sec.
The system is unstable.
Remember that the phase margin tells us that additional delay must not exceed 0.57.