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Lecture 6: Frequency Response Analysis and Stability. Objectives. To study system stability in the frequency domain using the Bode stability method. To understand the concepts of gain and phase margins. Frequency Response: determines the response of systems variables to a sine input.

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Lecture 6: Frequency Response Analysis and Stability

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Lecture 6 frequency response analysis and stability

Lecture 6:

Frequency Response Analysis and Stability


Lecture 6 frequency response analysis and stability

Objectives

  • To study system stability in the frequency domain using the Bode stability method.

  • To understand the concepts of gain and phase margins.


Lecture 6 frequency response analysis and stability

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 use sine to characterize time-varying inputs, specially disturbances

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

No!

Yes!

Yes!


Lecture 6 frequency response analysis and stability

Frequency Response : Sine in  sine out

  • How do we calculate the frequency response?

  • We could use dynamic simulation

  • - Lots of cases at every frequency

  • - Can be done for non-linear systems

  • For linear models, we can use the transfer function

  • - Remember that the frequency response can be calculated by setting s = j , with = frequency and j = complex variable.


Lecture 6 frequency response analysis and stability

0.4

0.2

Y, outlet from system

0

-0.2

-0.4

0

1

2

3

4

5

6

time

1

0.5

X, inlet to system

0

-0.5

-1

0

1

2

3

4

5

6

time

Frequency Response : Sine in  sine out

Amplitude ratio = |Y’(t)| max / |X’(t)| max

Phase angle = phase difference between input and output

P

B

output

P’

A

input


Lecture 6 frequency response analysis and stability

Frequency Response : Sine in  sine out

Amplitude ratio = |Y’(t)| max / |X’(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.


Lecture 6 frequency response analysis and stability

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()


Lecture 6 frequency response analysis and stability

0.8

0.6

0.4

0.2

20

0

v1

0

-0.2

0

20

40

60

80

100

120

-20

TC

100

-40

0

20

40

60

80

120

v2

STABILITY

No!

or

Yes!

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


Lecture 6 frequency response analysis and stability

1

1

0.5

0.5

0

0

-0.5

-0.5

-1

-1

1.5

1.5

0

0

0.5

0.5

1

1

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

bounded

bounded

unbounded

unbounded


Lecture 6 frequency response analysis and stability

STABILITY

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.


Lecture 6 frequency response analysis and stability

SP(s)

E(s)

CV(s)

MV(s)

+

+

GC(s)

Gv(s)

GP(s)

+

-

CVm(s)

Loop open

GS(s)

Bode Stability: To understand, let’s do a thought experiment


Lecture 6 frequency response analysis and stability

No forcing!!

No forcing!!

SP(s)

SP(s)

E(s)

E(s)

CV(s)

CV(s)

MV(s)

MV(s)

+

+

+

+

GC(s)

GC(s)

Gv(s)

Gv(s)

GP(s)

GP(s)

+

+

-

-

CVm(s)

CVm(s)

Loop closed

Loop closed

GS(s)

GS(s)

Bode Stability: To understand, let’s do a thought experiment

Under what conditions is the system stable (unstable)?

Hint: think about the sine wave as it travels around the loop once.


Lecture 6 frequency response analysis and stability

SP(s)

E(s)

CV(s)

MV(s)

+

+

GC(s)

Gv(s)

GP(s)

+

-

CVm(s)

Loop closed

GS(s)

Bode Stability: To understand, let’s do a thought experiment

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?


Lecture 6 frequency response analysis and stability

SP(s)

E(s)

CV(s)

MV(s)

+

+

GC(s)

Gv(s)

GP(s)

+

-

CVm(s)

Loop closed

GS(s)

Bode Stability: To understand, let’s do a thought experiment

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.


Lecture 6 frequency response analysis and stability

Bode Stability

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

At the critical frequency:  GOL(pcj) = -180

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

|GOL(pcj) | > 1 for instability

The gain margin (GM) is define as:

GM =

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


Phase margin

Phase Margin

Another relevant term is the phase margin. To calculate it, we need to find the gain crossover frequency(gc) whichis 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

PM =

If the phase margin is negative, the system is unstable.


Interpretation of gain and phase margins

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 these two points is to try a numerical example!


Example

Example

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

.

Find the delay margin as well.


Answer

Answer

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

    .

  • This gives rad/sec.

  • At this frequency,

    =

    (system stable)

    dB.

  • Therefore, the gain margin = 1/0.735 = 1.36 = 2.63 dB.


Answer1

Answer

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

    .

  • This gives rad/sec.

  • At this frequency,

     = --2tan-1()

    = -1-2tan-1(1)

    = - 1 -

    = -2.57 rad

    = -147.3.


Answer2

Answer

  • Therefore, the phase margin (PM)

    = + 

    = 0.57 rad

    = 32.7.

  • The delay margin, DM

    =

    =

    = 0.57 sec.


Lecture 6 frequency response analysis and stability

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

s=tf(‘s’);

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

margin(G)


Lecture 6 frequency response analysis and stability

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

  • For this purpose, we build the following model:


Closed loop step response original process with proportional controller gain 1

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

The system is stable.


Closed loop step response with proportional controller of gain 1 4

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

The system is unstable.

Remember that the gain margin is 1.36.


Closed loop step response with proportional controller of gain 1 and additional delay of 0 6 sec

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.


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