lecture 8: frequency response analysis and stability 1

Lecture 8:

Frequency Response Analysis and Stability

1

Objectives

• Test system stability in the frequency domain using the Bode stability method.

• Explain the meaning of gain and phase margins.

2

Frequency Response: determines the response of systems variables to a sine input.

• Perfect sine disturbances occur frequently in plants

• We can learn useful generalizations about control performance and robustness.

Why do we study frequency response?

No!

Yes!

3

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.

4

0 1 2 3 4 5 6-0.4

-0.2

0

0.2

0.4

time

Y,

ou

tlet

fro

m s

yste

m

0 1 2 3 4 5 6-1

-0.5

0

0.5

1

time

U,

inle

t to

sys

tem

input

outputB

A

P

P’

Amplitude ratio = |Y(t)| max / |U(t)| max

Phase angle = phase difference between input and output


5

Amplitude ratio = |Y(t)| max / |U(t)| max

Phase angle = phase difference between input and output

))(Re(

))(Im(tan)(angle Phase

))(Im())(Re()(Ratio Amp.

jG

jGjG

jGjGjGAR

1

22

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.


6

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()

7

TC

v1

v2

We influence stability when we implement control. How do we achieve the influence we want?

STABILITY

0 20 40 60 80

100

120-40

-20

0

20

0 20 40 60 80 100 120-0.2

0

0.2

0.4

0.6

0.8

or

No!

Yes!

8

First, let’s define stability: A system is stable if all bounded inputs to the system result in bounded outputs.

Feed

Vaporproduct

LiquidproductProcess

fluidSteam

F1

F2 F3

T1 T2

T3

T5

T4

T6 P1

L1

A1

L. Key

ProcessSampleInputs

SampleOutputs

0 0.5 1 1.5-1

-0.5

0

0.5

1

0 0.5 1 1.5-1

-0.5

0

0.5

1

bou

nd

edu

nb

oun

ded

bou

nd

edu

nb

oun

ded

9

STABILITY

Set point response

)()()()(

)()()(

)(

)(

sGsGsGsG

sGsGsG

sSP

sCV

Scvp

cvp

1

The denominator determines the stability of the closed-loop

feedback system!

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.

10

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

11

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.

12

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?


13

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.


14

Let’s put the results together. GOL(s) includes all elements in the closed loop.

At the critical frequency: GOL(jpc) = -180

The amplitude ratio: |GOL(jpc) | < 1 for stability

|GOL(jpc) | > 1 for instability

Bode Stability

15

The gain margin (GM) is defined as:

GM =

Hence, if the gain margin is less than 1 (negative in dB), the system is unstable.

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.

16

)(180PM gcjG

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!

17

Example

Find the gain and phase margin for the following process under proportional control. Is the system stable?

Find the delay margin as well.

18

.)1(

2)(

2se

ssG

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.

19

dB63.2

735.01

2)(

2

pcjG

rad/sec.31.1

.tan2)( 1

pc

pcpcpcjG

Answer, continued

• To find the phase margin, we need the gain crossover frequency.

• At this frequency,

20

rad/sec.1

11

2)(

2

gc

gcjG

.3.147

rad57.22

1

)1(tan21

)(tan2)(1

1

gcgcgcjG

Answer, continued

• Therefore, the phase margin (PM) is found as

• And the delay margin (DM) is obtained as

21

.rad57.0

7.32

3.147180PM

sec.57.0

PMDM

gc

s=tf(‘s’);

G=2/(s+1)^2*exp(-s);

margin(G)

We can check our result using the command “margin” in Matlab.

22

• Let us now use Simulink to understand the interpretation of gain margin and phase margins.

• For this purpose, we build the following model:

23

Closed-loop step response (original process with proportional controller gain = 1)

The system is stable.

24

Closed-loop step response with proportional controller of gain 1.4.

The system is

unstable.

Remember that the gain margin is 1.36.25

Closed-loop step response with proportional controller of gain 1 and additional delay of 0.6 sec.

Remember that the phase margin tells us that additional delay must not exceed 0.57.

The system is

unstable.

26

lecture 8: frequency response analysis and stability 1

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