Chapter 5 Frequency-Domain Analysis NUAA-Control System Engineering.

Chapter 5 Frequency-Domain Analysis

NUAA-Control System Engineering

Control System Engineering-2008

Content in Chapter 55-1 Frequency Response (or Frequency

Characteristics)

5-2 Nyquist plot and Nyquist stability criterion

5-3 Bode plot and Bode stability criterion


5-1 Frequency Response


A Perspective on the Frequency-Response Design Method

The design of feedback control systems in industry is probably accomplished using frequency-response methods more than any other.

Advantages of frequency-response design:

-It provides good designs in the face of uncertainty in the plant model

-Experimental information can be used for design purposes.Raw measurements of the output amplitude and phase of a plant undergoing a sinusoidal input excitation are sufficient to design a suitable feedback control.

-No intermediate processing of the data (such as finding poles and zeros) is required to arrive at the system model.


Frequency response

The frequency response of a system is defined as the steady-state response of the system to a sinusoidal input signal.

0( ) sinr t R t 0( ) sin( )y t tY

For a LTI system, when the input to it is a sinusoid signal, the resulting output , as well as signals throughout the system, is sinusoidal in the steady-state;

G(s)

H(s)

The output differs from the input waveform only in amplitude and phase.


The closed-loop transfer function of the LTI system:

( ) ( )( )

( ) 1 ( ) ( )

Y s G sM s

R s G s H s

For frequency-domain analysis, we replace s by jω: ( ) ( )

( )( ) 1 ( ) ( )

Y j G jM j

R j G j H j

The frequency-domain transfer function M(jω) may be expressed in terms of its magnitude and phase:

( ) ( ) ( )M j M j M j magnitude phase


The magnitude of M(jω) is

( )( )

1 ( ) ( )

( )

1 ( ) ( )

G jM j

G j H j

G j

G j H j

The phase of M(jω) is

( ) ( )

( ) 1 ( ) ( )MM j j

G j G j H j

( )M j

A

0 c

0

( )M j Gain-phase characteristics of an ideal low-pass filter

( )0

c

c

AM j

Gain characteristic

Phase characteristic


Example. Frequency response of a Capacitor

Consider the capacitor described by the equationdv

i Cdt

where v is the input and i is the output. Determine the sinusoidal steady-state response of the capacitor.Solution.The transfer function of the capacitor is

( )( )

( )

I sM s Cs

V s

So ( )M j Cj

Computing the magnitude and phase, we find that

( )M j Cj C

( ) 90MM j


( )M j Cj C

( ) 90MM j

( ) ( ) ( )I j M j V j

Gain characteristic:

Phase characteristic:

For a unit-amplitude sinusoidal input v, the output i will be a sinusoid with magnitude Cω, and the phase of the output will lead the input by 90°.

Note that for this example the magnitude is proportional to the input frequency while the phase is independent of frequency.

Output:


Resonant peak rM

Resonant frequency r

Bandwidth BW

0

( )M j

BWr

0.707

rM

( )M j

0

Cutoff rate

Typical gain-phase characteristic of a control system

Frequency-Domain Specifications

( )0

r

d M j

d


Frequency response of a prototype second-order system

Closed-loop transfer function:2

2 2

( )( )

( ) 2n

n n

Y sM s

R s s s

Its frequency-domain transfer function:2

2 2

( )( )

( ) ( ) 2 ( )n

n n

Y jM j

R j j j

Define nu

2

1( )

1 2M ju

j u u


The magnitude of M(ju) is

2 2 2 1/2

1( )

[(1 ) (2 ) ]M ju

u u

The phase of M(ju) is

12

2( ) ( ) tan

1M

uM j j

u

The resonant frequency of M(ju) is

( )0

d M ju

du 21 2ru

With , we haver r nu 21 2r n

Since frequency is a real quantity, it requires 21 2 0

So 0.707

2

1

2 1rM

Resonant peak


According to the definition of Bandwidth

2 2 2 1/2

1 1( ) 0.707

[(1 ) (2 ) ] 2M ju

u u

2 2 4 2(1 2 ) 4 4 2u

With , we havenu

2 4 2 1/2[(1 2 ) 4 4 2]nBW


Resonant peak

2

1

2 1rM

Resonant frequency

21 2r n

For a prototype second-order system ( )0.707

Bandwidth 2 4 2 1/2[(1 2 ) 4 4 2]nBW

depends on only.

For 0, the system is unstable;

For 0< 0.707, ;

For 0.707, 1

r

r

r

M

M

M

depends on both and .

For 0< 0.707, fixed, ;

For 0.707, 0.

r n

n r

r

is directly proportional to ,

For 0 0.707, fixed, ;

n n

n

n

BW BW

BW

BW

Control System Engineering-2008Correlation between pole locations, unit-step response and the magnitude of the frequency response

2

2 22n

n ns s

( )r t ( )y t

0 1

j

0

n

1cos

0

( )M j

0dB

0.3dB

BW 2 4 2 1/2[(1 2 ) 4 4 2]nBW

21 0.4167 2.917r

n

t

0

( )y t

1.00.9

0.1 t

2/ 1max overshoot e

Control System Engineering-2008Example. The specifications on a second-order unity-feedback control system with the closed-loop transfer function 2

2 2

( )( )

( ) 2n

n n

Y sM s

R s s s

are that the maximum overshoot must not exceed 10 percent, and the rise time be less than 0.1 sec. Find the corresponding limiting values of Mr and BW analytically.Solution.Maximum overshoot:

21% 100% 10%e

Rise time:21 0.4167 2.917

0.1 (0 1)rn

t

0.6

22.917 0.4167 1 0.1 0n

2

1,2

0.4167 0.4167 4 2.917 (1 0.1 )

2 2.917n

18n


Resonant peak2

(1

2 10.707)rM

0.6 18n

For 0< 0.707, ;

For 0.707, 1r

r

M

M

1 1.04rM 0.6

Bandwidth 2 4 2 1/2[(1 2 ) 4 4 2]nBW

is directly proportional to ,

For 0 0.707, fixed, ;

n n

n

n

BW BW

BW

BW

0.6 0.707 1 1.15nBW 1.15n nBW

18n 18BW

Based on time-domain analysis, we obtain and

Frequency-domain specifications:


( )R s ( )Y s2

( 2 )n

ns s

1 zT s

( )R s ( )Y s2

( 2 )n

ns s

Closed-loop TF：

Open-loop TF：

2

2 2

( ) ( )( )

( ) 1 ( ) 2n

n n

Y s G sM s

R s G s s s

Adding a zero at 1 zs T

2

( )( 2 )

(1 )z n

n

T sG s

s s

Open-loop TF：

Closed-loop TF：

2

( )( 2 )

n

n

G ss s

2

2

2 2

1( )( )

(2 )z

z n

n

n n

ss

T s

T s

The additional zero changes

both numerator and

denominator.

Effects of adding a zero to the OL TF


As analyzing the prototype second-order system, using similar but more complicate calculation, we obtain

Bandwidth 2 4 1/2( 1 / 2 4 )nBW b b

where 2 2 3 2 4 24 4 2n n z n n zb T T

For fixed ωn and ζ, we analyze the effect of . zT


-80

-60

-40

-20

0

20

Ma

gn

itud

e (

dB

)

10-1

100

101

102

-180

-135

-90

-45

0

Ph

ase

(d

eg

)

Bode Diagram

Frequency (rad/sec)

Tz=0Tz=0.2

Tz=1Tz=5

The general effect of adding a zero the open-loop transfer function is to increase the bandwidth of the closed-loop system.

1

0.2n


( )R s ( )Y s2

( 2 )n

ns s

1

1 pT s

( )R s ( )Y s2

( 2 )n

ns s

Closed-loop TF：

2

2 2

( ) ( )( )

( ) 1 ( ) 2n

n n

Y s G sM s

R s G s s s

2

( )( 2 (1 ))

n

n p

G ss T ss

Open-loop TF：

Open-loop TF：2

( )( 2 )

n

n

G ss s

Closed-loop TF：2

3 2 2( )

(1 2 ) 2n

p n p n n

sT s T s s

Effects of adding a pole to the OL TF

Adding a pole at 1 ps T


-150

-100

-50

0

50

Ma

gn

itud

e (

dB

)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

-45

0

Ph

ase

(d

eg

)Bode Diagram

Frequency (rad/sec)

Tp=0Tp=0.5

Tp=1Tp=5

The effect of adding a pole the open-loop transfer function is to make the closed-loop system less stable, while decreasing the bandwidth.

1

0.707n


5-2 Nyquist Plot and Nyquist Criterion


Nyquist Criterion

What is Nyquist criterion used for?

G(s)

H(s)

( )R s ( )Y s

Nyquist criterion is a semigraphical method that determines the stability of a closed-loop system;

Nyquist criterion allows us to determine the stability of a closed-loop system from the frequency-response of the loop function G(jw)H(j(w)


Review about stability

Closed-loop TF: ( )

( )1 ( ) ( )

G sM s

G s H s

Characteristic equation (CE): ( ) 1 ( ) ( ) 0s G s H s

Stability conditions:

Open-loop stability: poles of the loop TF G(s)H(s) are all in the left-half s-plane.

Closed-loop stability: poles of the closed-loop TF or roots of the CE are all in the left-half s-plane.


Definition of Encircled and Enclosed

Encircled: A point or region in a complex function plane is said to be encircled by a closed path if it is found inside the path.Enclosed: A point or region in a complex function plane is said to be encircled by a closed path if it is encircled in the countclockwise(CCW) direction.

A

B

Point A is encircled in the closed path;

Point A is also enclosed in the closed path;


Number of Encirclements and Enclosures

B

AD

C

Point A is encircled once;Point B is encircled twice.

Point C is enclosed once;Point D is enclosed twice.


Mapping from the complex s-plane to the Δ(s) -planeExercise 1: Consider a function Δ(s) =s-1, please map a circle with a radius 1 centered at 1 from s-plane to the Δ(s)-plane .

1 2

3 4

2; 1

0; 1

s s j

s s j

1 2

3 4

( ) 1; ( )

( ) 1; ( )

s s j

s s j

Δ( s)-planeImj

Re[ ( )]s1( )s

2( )s

3( )s

4( )s

01

s-planej

1s

2s

3s

4s

2

1

10

Mapping


Principle the Argument Let be a single-valued function that has a finite number of poles in the s-plane. Suppose that an arbitrary closed path is chosen in the s-plane so that the path does not go through any one of the poles or zeros of ;The corresponding locus mapped in the -plane will encircle the origin as many times as the difference between the number of zeros and poles (P) of that are encircled by the s-plane locus .

N Z P In equation form:

( )s

s

( )s ( )s

s( )s

N - number of encirclements of the origin by the -plane locus( )s

Z - number of zeros of encircled by the s-plane locus( )s

P - number of poles of encircled by the s-plane locus( )s


Nyquist Path A curve composed of the imaginary axis and an arc of infinite radius such that the curve completely encloses the right half of the s-plane .

Nyquist path is in the CCW direction

s

j

s-plane

0

R

s

Note Nyquist path does not pass through any poles or zeros of Δ(s); if Δ(s) has any pole or zero in the right-half plane, it will be encircled by .s

Since in mathematics, CCW is traditionally defined to be the positive sense.


Nyquist Criterion and Nyquist Diagram j

s-plane

0

R

s

( ) 1 ( ) ( )s G s H s

Δ( s)-plane

1

Nyquist Path

Nyquist Diagram:Plot the loop function to determine the closed-loop stability

G( s)H(s)-plane

01Critical point:(-1+j0)


Nyquist Criterion and G(s)H(s) Plot j

s-plane

0

R

s

( ) ( )G s H s

G( s)H(s)-plane

01

Nyquist Path G(s)H(s) Plot

The Nyquist Path is shown in the left figure. This path is mapped through the loop tranfer function G(s)H(S) to the G(s)H(s) plot in the right figure. The Nyquist Creterion follows:

N Z P


Nyquist Criterion and Nyquist Plot j

s-plane

0

R

s

( ) ( )G s H s

G( s)H(s)-plane

01

Nyquist Path Nyquist Plot

The condition of closed-loop stability according to the Nyquist Creterion is:

N P

N - number of encirclements of (-1,j0) by the G(s)H(s) plot

Z - number of zeros of that are inside the right-half plane( )s

P - number of poles of that are inside the right-half plane( )s


1 1

1

( ) ( )( ) 1 ( ) ( )

( )

n m

i ij in

ij

s p K s zs G s H s

s p

1

1

( )( ) ( )

( )

m

iin

ij

K s zG s H s

s p

has the same poles as , so P can be obtained by counting the number of poles of in the right-half plane.

( )s ( ) ( )G s H s( ) ( )G s H s


-2 -1 0 1 2 3 4 5-4

-3

-2

-1

0

1

2

3

4Nyquist Diagram

Real Axis

Ima

gin

ary

Axi

s

An example Consider the system with the loop function

3

5( ) ( )

( 1)G s H s

s

Matlab program for Nyquist plot (G(s)H(s) plot)

>>num=5;>>den=[1 3 3 1];>>nyquist(num,den);

Question 1: is the closed-loop system stable?

Question 2: what if

3

5( ) ( ) ?

( 1)

KG s H s

s

N=0, P=0, N=-P, stable


Root Locus

Real Axis

Ima

gin

ary

Axi

s

-4 -3 -2 -1 0 1 2-3

-2

-1

0

1

2

3

*3 3

5 1( ) ( )

( 1) ( 1)

KG s H s K

s s

1. With root locus technique:

For K* varies from 0 to ∞, we draw the RL

>>num=1;>>den=[1 3 3 1];>>rlocus(num,den);

When K*=8 (K=1.6), the RL cross the jw-axis, the closed-loop system is marginally stable.

For K*>8 (K>1.6), the closed-loop system has two roots in the RHP and is unstable.

*K

*K

*K

* 0K

* 8K

Control System Engineering-20082. With Nyquist plot and Nyquist criterion:

-2 -1 0 1 2 3 4 5-4

-3

-2

-1

0

1

2

3

4Nyquist Diagram

Real Axis

Ima

gin

ary

Axi

s

>>K=1;>>num=5*K;>>den=[1 3 3 1];>>nyquist(num,den);

Nyquist plot does not encircle (-1,j0), so N=0

K=1

3

5( ) ( )

( 1)

KG s H s

s

No pole of G(s)H(s) in RHP, so P=0;

Thus N=-PThe closed-loop system is stable


-2 -1 0 1 2 3 4 5 6 7 8-6

-4

-2

0

2

4

6Nyquist Diagram

Real Axis

Ima

gin

ary

Axi

s

2. With Nyquist plot and Nyquist criterion:

>>K=1.6;>>num=5*K;>>den=[1 3 3 1];>>nyquist(num,den);

The Nyquist plot just go through (-1,j0)

K=1.6

3

5( ) ( )

( 1)

KG s H s

s


The closed-loop system is marginally stable


-5 0 5 10 15 20-15

-10

-5

0

5

10

15Nyquist Diagram

Real Axis

Ima

gin

ary

Axi

s

2. With Nyquist plot and Nyquist criterion:

>>K=4;>>num=5*K;>>den=[1 3 3 1];>>nyquist(num,den);

Nyquist plot encircles (-1,j0) twice, so N=2

K=4

3

5( ) ( )

( 1)

KG s H s

s


Thus Z=N+P=2The closed-loop system has two poles in RHP and is unstable


Nyquist Criterion for Systems with Minimum-Phase Transfer Functions

What is called a minimum-phase transfer function?A minimum-phase transfer function does not have poles or zeros in the right-half s-plane or on the jw-axis, except at s=0.

1

10( 1)( )

( 10)

sG s

s

2

10( 1)( )

( 10)

sG s

s

Consider the transfer functions

Both transfer functions have the same magnitude for all frequencies

1 2( ) ( )G j G j

But the phases of the two transfer functions are drastically different.


0

5

10

15

20

Ma

gn

itud

e (

dB

)

10-2

10-1

100

101

102

103

0

45

90

135

180

Ph

ase

(d

eg

)

Bode Diagram

Frequency (rad/sec)

1 2( ) ( )G j G j

1( )G j

2 ( )G j

A minimum-phase system (all zeros in the LHP) with a given magnitude curve will produce the smallest change in the associated phase, as shown in G1.


Consider the loop transfer function:

( ) ( ) ( )L s G s H s

If L(s) is minimum-phase, that is, L(s) does not have any poles or zeros in the right-half plane or on the jw-axis, except at s=0

Then P=0, where P is the number of poles of Δ(s)=1+G(s)H(s), which has the same poles as L(s).Thus, the Nyquist criterion (N=-P) for a system with L(s) being minimum-phase is simplified to

0N


For a closed-loop system with loop transfer function L(s) that is of minimum-phase type, the system is closed-loop stable , if the Nyquist plot (L(s) plot) that corresponds to the Nyquist path does not enclose (-1,j0) point. If the (-1,j0) is enclosed by the Nyquist plot, the system is unstable.

The Nyquist stability can be checked by plotting the segment of L(jw) from w= ∞ to 0.

0N

Nyquist criterion for systems with minimum-phase loop transfer function

Control System Engineering-2008Example Consider a single-loop feedback system with the loop transfer function

( ) ( ) ( )( 2)( 10)

KL s G s H s

s s s

Analyze the stability of the closed-loop system.Solution.Since L(s) is minimum-phase, we can analyze the closed-loop stability by investigating whether the Nyquist plot enclose the critical point (-1,j0) for L(jw)/K first.( ) 1

( 2)( 10)

L j

K j j j

w=∞:

( )0 270

L j

K

w=0+

:

( 0)90

L j

K

0

Imj

Real

0

Im[ ( ) ] 0L j K


1Im[ ( ) ] Im[ ] 0

( 2)( 10)L j K

j j j

20 /rad s

The frequency is positive, so 20 /rad s

1( 20) 0.004167

20( 20 2)( 20 10)L j K

j j j

1. 240 ( 20) 1K L j

the Nyquist plot does not enclose (-1,jw);2. 240 ( 20) 1K L j

the Nyquist plot goes through (-1,jw);

3. 240 ( 20) 1K L j

the Nyquist plot encloses (-1,jw).

stable

marginally stable

unstable


-30 -25 -20 -15 -10 -5 0 5 10-20

-15

-10

-5

0

5

10

15

20Root Locus

Real Axis

Ima

gin

ary

Axi

s1

( ) ( ) ( )( 2)( 10)

L s G s H s Ks s s

>>z=[]>>p=[0, -2, -10];>>k=1>>sys=zpk(z,p,k);>>rlocus(sys);

240K

By root locus technique

Control System Engineering-2008 Relative Stability

Gain Margin and Phase Margin

0

Imj

Real

0

1

For a stable system, relative stability describes how stable the system is.In time-domain, the relative stability is measured by maximum overshoot and damping ratio.

In frequency-domain, the relative stability is measured by resonance peak and how close the Nyquist plot of L(jw) is to the (-1,j0) point.

The relative stability of the blue curve is higher than the green curve.


Gain Margin (GM)(for minimum-phase loop transfer functions)

0

Imj

Real

0

p

Phase crossover

Phase crossover frequency ωp

( ) 180pL j

For a closed-loop system with L(jw) as its loop transfer function, it gain margin is defined as

10

10

1gain margin (GM) = 20log

( )

20log ( ) dB

p

p

L j

L j

( )pL j

L(jw)-plane


10(stableWhen ( ) 1 , log () GM 0) 0L j L j

(closer to the or( ) GMigin) (more stable)L j

(closer to -1) (les( s stab e) G l )ML j

10(unstablWhen ( ) 1 , loe) GM<) 0 0g (L j L j

10(marginally sWhen ( ) 1 , logtable) GM) 0 0( =L j L j

Gain margin represents the amount of gain in decibels (dB) that can be added to the loop before the closed-loop system becomes unstable.

10

10

1gain margin (GM) = 20log

( )

20log ( ) dB

p

p

L j

L j


Phase Margin (PM)(for minimum-phase loop transfer functions)

Gain margin alone is inadequate to indicate relative stability when system parameters other the loop gain are subject to variation.

( ) 1gL j 0

Imj

Real

AB

1

With the same gain margin, system represented by plot A is more stable than plot B.

Gain crossover frequency ωg

( ) 1gL j

Phase margin:

phase margin (PM) = ( ) 180gL j

( )gL j

PM

Control System Engineering-2008Example Consider the transfer function10

( )( 1)

G ss s

Draw its Nyquist plot when w varies from 0 to ∞.

Solution. Substituting s=jw into G(s) yields: 10( )

( 1)G j

j j

The magnitude and phase of G(jw) at w=0 and w=∞ are computed as follows.

0 0 0

0 0 0

10 10lim ( ) lim lim

( 1)

10 10lim ( ) lim lim 90

( 1)

G jj j

G jj j j

2

10lim ( ) lim 0

( 1)

10 10lim ( ) lim lim 180

( 1)

G jj j

G jj j

Thus the properties of the Nyquist plot of G(jw) at w=0 and w=∞ are ascertained.

Next we determine the intersection…

Control System Engineering-2008If the Nyquist plot of G(jw) intersects with the real axis, we have

Im[ ( )] 0G j 2

4 2 4 2

10 10 10( )

( 1)G j j

j j

4 2

100

This means that the G(jw) plot intersects only with the real axis of the G(jw)-plane at the origin.

Similarly, intersection of G(jw) with the imaginary axis:

which corresponds to the origin of the G(jw)-plane.

The conclusion is that the Nyquist plot of G(jw) does not intersect any one of the axes at any finite nonzero frequency.

Re[ ( )] 0G j

At w=0, Re[ ( )] 10G j At w=∞, Re[ ( )] 0G j

Control System Engineering-2008Example Consider a system with a loop transfer function as

2500( )

( 5)( 50)L s

s s s

Determine its gain margin and phase margin.

Solution.Phase crossover frequency ωp:

Im[ ( )] 0 15.88 rad/secpL j

10GM = 20log ( ) 14.80 dBpL j

Gain margin:

Gain crossover frequency ωg:

( ) 1 6.22 rad/secg gL j

Phase margin:

PM = ( ) 180 31.72gL j

0

Imj

0

1

( ) 0.182pL j 15.88p

6.22g 31.72

0.182


Advantages of Nyquist plot:

-By Nyquist plot of the loop transfer function, the closed-loop stability can be easily determined with reference to the critical point (-1,j0).

-It can analyze systems with either minimum phase or nonminimum phase loop transfer function.

Disadvantages of Nyquist plot:

-By Nyquist plot only, it is not convenient to carry out controller design.


5-3 Bode Plot


Bode PlotThe Bode plot of the function G(jw) is composed of two plots: -- the amplitude of G(jw) in decibels (dB) versus log10w or w -- the phase of G(jw) in degrees as a function of log10w or w.

Without loss of generality, the following transfer function is used to illustrate the construction of the Bode Plot

1 22 2

1

(1 )(1 )( )

(1 )(1 2 / / )jn n

K T s T sG s

s s s s

where K, T1, T2, τ1, ζ, ωn are real constants. It is assumed that the second-order polynomial in the denominator has complex conjugate zeros.


The magnitude of G(jw) in dB is obtained by multiplying the logarithm (base 10) of |G(jw)| by 20; we have

1 22 2

1

(1 )(1 )( )

( )(1 )(1 2 / / )n n

K jT jTG j

j j j

Substituting s=jw into G(s) yields

10

10 10 1 10 2

2 210 10 1 10

( ) 20log ( )

20log 20log 1 20log 1

20log 20log 1 20log 1 2 / /

dB

n n

G j G j

K jT jT

j j j

The phase of G(jw) is

1 2

2 21

( ) (1 ) (1 )

(1 ) (1 2 / / )n n

G j K jT jT j

j j


In general, the function G(jw) may be of higher order andhave many more factored terms. However, the above two equations indicate that additional terms in G(jw) would simply produce more similar terms in the magnitude and phase expressions, so the basic method of construction of the Bode plot would be the same. In general, G(jw) can contain just four simple types of factors:

1. Constant factor: K2. Poles or zeros at the origin of order p: (jw)±p

3. Poles or zeros at s =-1/T of order q: (1+jwT )±q

4. Complex poles and zeros of order r: (1 + j2ζω/ωn-ω2/ω2

n)


1. Real constant K

2020log

constantdBK K

0 0

180 0

KK

K


2. Poles or zeros at the origin,( ) pj

Magnitude of ( ) in dB:pj

20 20( ) 20log ( ) 20 log dBp p

dBj j p

For a given p, it is a straight line with the slope:

1010

20 log 20 dB/decadelog

dp p

d

At =1, ( ) 0.p

dBj

Thus a unit change in corresponds to a change of ±20 dB in the magnitude.

10log

So these lines pass through the 0dB axis at ω =1.


Phase of ( ) :

90

pj

p

Magnitude of ( ) :pj

2020 log dBp


3. (a) Simple zero 1+jwT Consider the function( ) 1G j j T

where T is a positive real constant.

The magnitude of G(jw) in dB is 2 2

10 10( ) 20log ( ) 20log 1dB

G j G j T

At very low frequencies, 1T

10( ) 20log 1 0 dBdB

G j

( is neglected when compared with 1.) 2 2T

At very high frequencies, 1T 2 2

10 10( ) 20log 20log dBdB

G j T T

This represents a straight line with a slope of 20dB

The two lines intersect at:1/T

(corner frequency)

Control System Engineering-2008The steps of making of sketch of 1 dBj T

Step 1: Locate the corner frequency w=1/T on the frequency axis;Step 2: Draw the 20dB/decade line and the horizontal line at 0 dB with the two lines intersecting at w=1/T.Step 3: Sketch a smooth curve by locating the 3-dB point at the corner frequency and the 1-dB points at 1 octave above and below the corner frequency.


The phase of G(jw)=1+jwT is1( ) tanG j T

At very low frequencies, ( ) 0G j

At very high frequencies, ( ) 90G j

Control System Engineering-20083. (b) Simple pole, 1/(1+jwT)

Consider the function 1( )

1G j

j T

The magnitude of G(jw) in dB is 2 2

10 10( ) 20log ( ) 20log 1dB

G j G j T

At very low frequencies, 1T

10( ) 20log 1 0 dBdB

G j

At very high frequencies, 1T 2 2

10 10( ) 20log 20log dBdB

G j T T

This represents a straight line with a slope of -20dB

The two lines intersect at:1/T

(corner frequency)

The phase of G(jw): 1( ) tanG j T

For w varies from 0 to ∞, varies from 0°to -90°. ( )G j



4. Complex poles and zeros

Consider the second-order transfer function2

2 2 2 2

1( )

2 1 (2 ) (1 )n

n n n n

G ss s s s

We are interested only in the case when ζ ≤ 1, since otherwise G(s) would have two unequal real poles, and the Bode plot can be obtained by considering G(s) as the product of two transfer functions with simple poles.

By letting s=jw, G(s) becomes

2 2

1( )

[1 ( )] 2 ( )n n

G jj


2 2

1( )

[1 ( )] 2 ( )n n

G jj

The magnitude of G(jw) in dB is

10

2 2 2 2 210

( ) 20log ( )

20log [1 ( )] 4 ( )

dB

n n

G j G j

At very low frequencies, / 1n

10( ) 20log 1 0 dBdB

G j

At very high frequencies, / 1n

410 10( ) 20log ( ) 40log ( ) dBn ndB

G j

This equation represents a straight line with a slope of 40 dB decade in the Bode plot coordinates.

The two lines intersect at:n

(corner frequency)


The reason for this is that the amplitude and phase curves of the second-order G(jw) depend not only on the corner frequency wn, but also on the damping ratio ζ, which does not enter the asymptotic curve.

The actual magnitude curve of G(jw) in this case may differ strikingly from the asymptotic curve.


The phase of G(jw) is given by

2

1 2( ) tan 1

n n

G j

Control System Engineering-2008ExampleConsider the following transfer function

10( 10)( )

( 2)( 5)

sG s

s s s

Sketch its Bode Plot.

Solution.Letting s=jw, we have10( 10)

( )( 2)( 5)

jG j

j j j

Reformulating it into the form for Bode Plot

1

1 2

(1 ) 10(1 0.1 )( )

(1 )(1 ) (1 0.5 )(1 0.2 )

K jT jG j

j j j j j j

1 1 210, 0.1, 0.5, 0.2K T where

So G(jw) has corner frequencies at w=10,2 and 5 rad/sec.


1. Bode plot of K=10

19

19.5

20

20.5

21

Ma

gn

itud

e (

dB

)

10-1

100

101

102

103

-1

-0.5

0

0.5

1

Ph

ase

(d

eg

)Bode Diagram

Frequency (rad/sec)


-60

-40

-20

0

20

Ma

gn

itud

e (

dB

)

10-1

100

101

102

103

-91

-90.5

-90

-89.5

-89

Ph

ase

(d

eg

)

Bode Diagram

Frequency (rad/sec)

2. Bode Plot of the component with pole at origin : jw

magnitude curve: a straight line with slope of 20 dB/decade, passing through the w=1 rad/sec point on the 0-dB axis.


20

30

40

50

60

Ma

gn

itud

e (

dB

)

10-1

100

101

102

103

0

45

90

Ph

ase

(d

eg

)

Bode Diagram

Frequency (rad/sec)

Corner frequency: w=1/0.1=10 rad/sec

3. Bode plot of simple zero component 1+j0.1w


-80

-60

-40

-20

0

Ma

gn

itud

e (

dB

)

10-1

100

101

102

103

-90

-45

0

Ph

ase

(d

eg

)

Bode Diagram

Frequency (rad/sec)


4. Bode plot of simple pole componet 1/(1+j0.5w)


-60

-50

-40

-30

-20

-10

Ma

gn

itud

e (

dB

)

10-1

100

101

102

103

-90

-45

0

Ph

ase

(d

eg

)

Bode Diagram

Frequency (rad/sec)


5. Bode plot of simple pole component 1/(1+j0.2w)

Control System Engineering-2008|G(jw)|dB is obtained by adding the component curves together, point by point.Bode Plot:Gain crossover point: |G(jw)|dB cross the 0-dB axisPhase crossover point: where the phase curve cross the -180°axis.


Nyquist Plot (Polar Plot) : The gain-crossover point is where , The phase crossover point is where .

( ) 1G j ( ) 180G j

Chapter 5 Frequency-Domain Analysis NUAA-Control System Engineering.

Documents

frequency response slide

input frequency

lti system

system model

order system

frequency characteristics

magnitude phase slide

frequencyresponse methods