Frequency Resp. methodGiven:

• G(jω) as a function of ω is called the freq. resp.• For each ω, G(jω) = x(ω) + jy(ω) is a point in

the complex plane• As ω varies from 0 to ∞, the plot of G(jω) is

called the Nyquist plot

• Can rewrite in Polar Form:

• |G(jω)| as a function of ω is called the amplitude resp.

• as a function of ω is called the phase resp.

• The two plots:

With log scale-ω, are Bode plot

)()G(j)( jGjejG

)( jG

~)(

~)(log20 10

jG

jG

To obtain freq. Resp from G(s):

• Select • Evaluate G(jω) at those to get

• Plot Imag(G) vs Real(G): Nyquist• or plot

With log scale ω• Matlab command to explore: nyquist,

bode

N 21

NGGGG 21

vsGangle

vsGabs

)(

))((log20 10

To obtain freq. resp. experimentally:

• Select

• Given input to system as:

• Adjust A1 so that the output is not saturated or distorted.

• Measure amp B1 and phase φ1 ofoutput:

N 21)sin()( 11 tAtu

)sin()( 111 tBty

• Then is the freq. resp. of the system at freq ω

• Repeat for all ωK

• Either plot

or plot

1

1

1

11

1

jeA

Bor

andA

B

BodeA

B

KK

KK

K

~

~

)( Kj

K

K eA

Bimag )( Kj

K

K eA

Breal

NyquistA

B

A

BK

K

KK

K

K cos~sin

Product of T.F.

sGsGsG 21

jGjGjG 21

jGjGjG 21

21 G of Bode mag.

2

G of Bode mag.

1

G of Bode mag.

log20log20log20

jGjGjG

plots. Bode mag. add

21 G ofplot phase Bode

2

G ofplot phase Bode

1

product theG, ofplot Bode phase

jGjGjG

product theofplot phase the

plot. phase individual

product theofplot Bode :Overall

plot. Bode individual

System type, steady state tracking, & Bode plot

11

11)()(

21

sTsTs

sTsTKsGsCsG

Nba

p

sintegrator # w.r.t. typeSystem NsR

11

11

21

jTjTj

jTjTKjG N

ba

C(s) Gp(s)R(s) Y(s)

As ω → 0

Therefore: gain plot slope = –20N dB/dec.

phase plot value = –90N deg

Nj

KjG

N

KjG

log20log20log20 NKjG

90Nj

KjG N

If Bode gain plot is flat at low freq, system is “type zero”

Confirmed by phase plot flat and 0°at low freq

Then: Kv = 0, Ka = 0

Kp = Bode gain as ω→0 = DC gain

(convert dB to values)

Example

typeSystem

vKpK

aK

Steady state tracking error

Suppose the closed-loop system is stable:If the input signal is a step, ess would be =

If the input signal is a ramp, ess would be =

If the input signal is a unit acceleration, ess would be =

N = 1, type = 1Bode mag. plot has –20 dB/dec slope at

low freq. (ω→0)(straight line with slope = –20 as

ω→0)

Bode phase plot becomes flatat –90° when ω→0

Kp = DC gain → ∞Kv = K = value of asymptotic straight line

at ω = 1 =s0dB =asymptotic straight line’s 0 dB

crossing frequencyKa = 0

Example

freq. low

20slope

Asymptotic straight line

s0dB ~14

The matching phase plot at lowfreq. must be → –90°type = 1

Kp = ∞ ← position error const.Kv = value of low freq. straight line

at ω = 1= 23 dB≈ 14 ← velocity error

const.Ka = 0 ← acc. error const.

Steady state tracking error

Suppose the closed-loop system is stable:If the input signal is a step, ess would be =

If the input signal is a ramp, ess would be =

If the input signal is a unit acceleration, ess would be =

N = 2, type = 2Bode gain plot has –40 dB/dec

slope at low freq.Bode phase plot becomes flat

at –180° at low freq.

Kp = DC gain → ∞Kv = ∞ also

Ka = value of straight line at ω = 1

= s0dB^2

Example

Ka

s0dB=Sqrt(Ka)

How should the phase plot look like?

2 :freq lowAt

j

KjG

2

:linestraight

KjG

16,142

KK

4at dB 0 isit

1||

vK

pK

16 aK

Steady state tracking error

Suppose the closed-loop system is stable:If the input signal is a step, ess would be =

If the input signal is a ramp, ess would be =

If the input signal is a unit acceleration, ess would be =

System type, steady state tracking, & Nyquist plot

11

11

21

jTjTj

jTjTKjG N

ba

C(s) Gp(s)

Nj

KjG

As ω → 0

Type 0 system, N=0

Kp=lims0 G(s) =G(0)=K

0+

Kp

G(j)

Type 1 system, N=1Kv=lims0 sG(s) cannot be determined easily from Nyquist plot

0+

infinity

G(j) -j∞

Type 2 system, N=2

Ka=lims0 s2G(s) cannot be determined easily from Nyquist plot

0+

infinity

G(j) -∞

Margins on Bode plots

In most cases, stability of this closed-loop

can be determined from the Bode plot of G:– Phase margin > 0– Gain margin > 0

G(s)

freq.over -crossgain :gc dB0or 1 ,at jGgc

margin phase :PM

gcjG 180

freq.over -cross phase :pc

180pcjG

dB log20margingain : pcjGGM

in value 1 pcjG

If never cross 0 dB line (always below 0 dB line), then PM = ∞.

If never cross –180° line (always above –180°), then GM = ∞.

If cross –180° several times, then there are several GM’s.

If cross 0 dB several times, then there are several PM’s.

jG

jG

jG

jG

52

1100

ss

ssGExample:

Bode plot on next page. 11

110

51

21

ss

s

100near line dB 0 cross .1 jG100 gc

_______PM

______at 180 cross .2 pcωjG _______GM

254

252

sss

sGExample:

Bode plot on next page.

______near line dB 0 cross .1 jG

______ gc

______about is at gcωjG

_______PM

1

1

2542

251

sss

1. Where does cross the –180° lineAnswer: __________

at ωpc, how much is

2. Closed-loop stability: __________

_______jG

________ pc

jG

________GM

1

120

2

40 :Example

21

ssss

sG

1. crosses 0 dB at __________

at this freq,

2. Does cross –180° line? ________

3. Closed-loop stability: __________

_______ jG

________ gc jG

________GM

________PM

jG

Margins on Nyquist plot

Suppose:• Draw Nyquist plot

G(jω) & unit circle• They intersect at point A• Nyquist plot cross neg.

real axis at –k

in value1kGM

indicated angle :Then PM

-100 -50 0 50-200

-150

-100

-50

0

50

100

150

200 Nyquist Diagram

Real Axis

Imag

inary

Axis

-2 -1.5 -1 -0.5 0-2

-1.5

-1

-0.5

0

0.5

1

1.5

2 Nyquist Diagram

Real Axis

Imagin

ary Ax

is

-4 -2 0 2 4-10

-5

0

5

10 Nyquist Diagram

Real Axis

Imagin

ary Ax

is

-2 -1.5 -1 -0.5 0-2

-1.5

-1

-0.5

0

0.5

1

1.5

2 Nyquist Diagram

Real Axis

Imagin

ary Ax

is