Transfer Function Lecture Notes

EE/ME 574: Intermediate Controls

Review-By-Example of some of the BasicConcepts and Methods of Control System

Analysis and Design

Contents

1 Differential Eqns, Transfer Functions & Modeling 4

1.1 Example 1, Golden Nugget Airlines . . . . . . . . . . . . . . . . 4

1.2 Block diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Laplace Transforms and Transfer Functions . . . . . . . . . . . . 6

1.3.1 Laplace transform . . . . . . . . . . . . . . . . . . . . . 6

1.3.2 Some basic Laplace transform Pairs . . . . . . . . . . . . 7

1.3.3 Transfer Functions are Rational Polynomials . . . . . . . 10

1.3.4 A transfer function has poles and zeros . . . . . . . . . . 11

1.3.5 Properties of transfer functions . . . . . . . . . . . . . . . 13

1.3.6 Forms for a transfer function . . . . . . . . . . . . . . . . 14

1.3.7 Proper and strictly proper transfer functions: . . . . . . . 15

1.3.8 System type . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.3.9 Phasors and the transfer function as complex gain . . . . . 16

1.3.10 The DC gain of a transfer function . . . . . . . . . . . . . 21

1.3.11 The impulse response of a system . . . . . . . . . . . . . 22

2 Closing the loop, feedback control 23

2.1 Analyzing a closed-loop system . . . . . . . . . . . . . . . . . . 27

2.2 Common controller structures: . . . . . . . . . . . . . . . . . . . 28

2.3 Analyzing other loops . . . . . . . . . . . . . . . . . . . . . . . 29Part 1: Controls Review-By-Example (Revised: Jan 16, 2013) Page 1


3 Calculating steady-state errors (Franklin et al. sec. 4.2 (6th Ed.)) 323.1 System type and Bode standard form for the loop gain . . . . . . . 33

3.1.1 Determining steady-state error using Bode standard form . 34

3.2 Table of steady-state errors . . . . . . . . . . . . . . . . . . . . . 35

3.3 Steady-state error example . . . . . . . . . . . . . . . . . . . . . 36

3.4 Summary for steady-state error . . . . . . . . . . . . . . . . . . 36

4 Characteristics of the Step Response 37

4.1 Rise Time, tr . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4.2 Peak Time, tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4.3 Settling Time, ts . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4.4 Percent Overshoot, PO . . . . . . . . . . . . . . . . . . . . . . . 38

5 Working with the pole-zero constellation39

5.1 Basics of pole-zero maps, 1st order, p1 = . . . . . . . . . . . 395.1.1 Real Pole location and response characteristics . . . . . . 40

5.2 Basis of pole-zero maps, 2nd order, p1,2 = j . . . . . . . 415.2.1 Notation for a complex pole pair . . . . . . . . . . . . . . 42

5.2.2 Complex pole location and response characteristics . . . 43

5.2.3 The damping factor: . . . . . . . . . . . . . . . . . . . 46

5.3 Higher order systems: dominant mode . . . . . . . . . . . . . . . 47

5.4 Which mode is the dominant mode ? . . . . . . . . . . . . . . . . 48

5.5 Summary: Pole location and response characteristics . . . . . . . 50

Part 1: Controls Review-By-Example (Revised: Jan 16, 2013) Page 2


6 Design 51

6.1 Design methods . . . . . . . . . . . . . . . . . . . . . . . . . . 52

6.2 Root Locus Design . . . . . . . . . . . . . . . . . . . . . . . . . 52

7 Summary 53

8 Glossary of Acronyms 54


EE/ME 574: Intermediate Controls Section 1.1.0

1 Differential Eqns, Transfer Functions & Modeling1.1 Example 1, Golden Nugget Airlines

Dynamic systems are governed by differential equations(or difference equations, if they are discrete-time)

Example (adapted from Franklin et al. 4thed., problem 5.41, figure 5.79)

GNA

Me(t)

(t), (t)

Mp(t)

Figure 1: Golden Nugget Airlines Aircraft. Me (t) is the elevator-moment (thecontrol input), Mp (t) is the moment due to passenger movements (adisturbance input), and (t) is the aircraft pitch angle, (t) = (t) .

For example, the aircraft dynamics give:

1d2 (t)

dt2 +4d (t)

dt +5(t) = 1d Mt (t)

dt +3Mt (t) (1)

Mt (t) = Me (t)+Mp (t) ; (t) =d (t)

dt (2)

1d Me (t)

dt +10Me (t) = 7v(t) (3)

Where Mt (t) is the total applied moment.Eqn (1) has to do with the velocity of the aircraft response to Mt (t).Eqn (2) expresses that the pitch-rate is the derivative of the pitch angle.And Eqn (3) describes the response of the elevator to an input commandfrom the auto-pilot.

Main Fact: The Differential Equations come from the physics of the system.



1.2 Block diagram

A block diagram is a graphical representation of modeling equations andtheir interconnection.

Eqns (1)-(3) can be laid out graphically, as in figure 2.

Elevator servo

7s + 10

AircraftDynamics

s2 + 4 s + 5s + 3Me(t) Mt(t)

Mp(t)Integrator

1s

(t) (t)v(t)+

+

Figure 2: Block diagram of the aircraft dynamics and signals.

Signal Symbol Signal Name / Description Units

v(t) Voltage applied to the elevator drive [volts]Me (t) Moment resulting from the elevator surface [Nm]Mp (t) Moment resulting from movement of the passengers [Nm](t) Pitch Rate (angular velocity) [rad/sec](t) Aircraft pitch angle [radians]

Table 1: List of signals for the aircraft block diagram.

When analyzing a problem from basic principals, we would also have a listof parameters.

Parameter Symbol Signal Name / Description Value Units

b0 Motion parameter of the elevator [N-m/volt]... ... ...

Table 2: List of parameters for the aircraft block diagram.



1.3 Laplace Transforms and Transfer Functions

To introduce the Transfer Function (TF), we need to review the Laplacetransform.

1.3.1 Laplace transform

The Laplace transform (LT) mapsa signal (a function of time) to afunction of the Laplace variable s .

The Inverse Laplace transformmaps F (s) to f (t).

f(t) F(s)

Space of all functionsof time

Space of all functions of s

L{f(t)}

L-1{F(s)}

"Time Domain" "Frequency Domain"

Why pay attention to Laplacetransforms ?

Answers:

1. Differential equations in timedomain correspond to algebraicequations in s domain.

2. The LT makes possible thetransfer function.

3. Time domain: find y(t) for oneu(t).

4. Frequency domain: find Guy (s)for all U (s).

u(t)

Space of all functionsof time

Space of all functions of s

Solve Diff Eq.

y(t)

U(s)Solve AlgebraicEqn.

Y(s)

Figure 3: Solve differential equation in time domain or solve an algebraicequation in s domain.



1.3.2 Some basic Laplace transform Pairs

Time Domain Signal Laplace transform

Unit Impulse: f (t) = (t) F (s) = 1Scaled Impulse: f (t) = b(t) F (s) = bUnit step: f (t) = 1+ (t) = 1.0, t 0 F (s) = 1

s

Unit ramp: f (t) = t, t 0 F (s) = 1s2

Higher Powers of t: f (t) = tn, t 0 F (s) = n !sn+1

Decaying Exponential: f (t) = be t F (s) = bs+Sinusoid: f (t) = sin(wt) F (s) =

s2+2

Sinusoidal oscillation: f (t) = Bc cos( t)+Bs sin( t) F (s) = Bcs+Bss2+0s+2Oscillatory Exp. Decay: f (t) = e t (Bc cos( t)+Bs sin( t)) F (s) = Bcs+(Bc +Bs)s2+2s+2n

n =

2 +2

Table 3: Basic Laplace-transform pairs.

t=0

Area = 1.01/

t=0

1+(t)t=0

(t) , impulse function 1+ (t), unit step Unit ramp

t=0 t=0t=0

e-t

Decaying expon. A Sinusoid Oscil. decay exp.



Two theorems of the Laplace transform permit us to build transfer functions:

1. The Laplace transform obeys superposition and scaling

Given: z(t) = x(t)+ y(t) then Z (s) = X (s)+Y (s)

Given: z(t) = 2.0 x(t) then Z (s) = 2.0 X (s)

2. The Laplace transform of the derivative of a signal x(t) is s times the LTof x(t):

Given: X (s) = L {x(t)} , then L{

d x(t)dt

}= sX (s)

Putting these rules together lets us find thetransfer function of a system from its governingdifferential equation.

Systemu(t) y(t)Gp(s)

Consider a system that takes in the signal u(t) and gives y(t), governed bythe Diff Eq:

a2d2 y(t)

dt2 +a1d y(t)

dt +a0 y(t) = b1d u(t)

dt +b0 u(t) (4)

(Notice the standard form: output (unknown) on the left, input on the right.) Whatever signals y(t) and u(t) are, they have Laplace transforms. Eqn (4)

gives:

L

{a2

d2 y(t)dt2 +a1

d y(t)dt +a0 y(t)

}= L

{b1

d u(t)dt +b0 u(t)

}(5)

a2 s2 Y (s)+a1 sY (s)+a0 Y (s) = b1 sU (s)+b0U (s) (6)(

a2 s2 +a1 s +a0

)Y (s) = (b1 s +b0) U (s) (7)



Eqns (5)-(7) tell us something about the ratio of the LT of the output to theLT of the input:(

a2 s2 +a1 s +a0

)Y (s) = (b1 s +b0) U (s) (7, repeated)

Y (s)U (s)

=(b1 s +b0)

(a2 s2 +a1 s +a0)=

Output LTInput LT

= Gp (s) (8)

A Transfer Function (TF) is a ratio of the input and output LTs

Given Gp (s) =(b1 s +b0)

(a2 s2 +a1 s +a0), Y (s) = Gp (s) U (s) (9)

Where Gp (s) is the TF of the plant.

Important fact:

The transfer function is like the gain of the system, it is the ratio of theoutput LT to the input LT.

However, the TF depends only on the parameters of the system(coefficients a2..a0 and b0..b1 in the example), and not on the actualvalues of U (s) or Y (s) (or u(t) and y(t)).

Basic and Intermediate Control System Theory are present transferfunction-based design.

By engineering the characteristics of the TF, we engineer the system toachieve performance goals.

Controller Plantr(t) y(t)+

-

ys(t)

KcGc(s) Gp(s)

Hy(s) Sensor Dynamics

e(t) u(t)

Figure 4: Block diagram of typical closed-loop control.



1.3.3 Transfer Functions are Rational Polynomials

GNA

Me(t)

(t), (t)

Mp(t)Figure 5: Golden Nugget Airlines Aircraft. (t) [radians/second] is the

pitch-rate of the aircraft, and Mt (t) is the moment (torque) applied bythe elevator surface.

Consider the differential equation of the Aircraft pitch angle:

1d2 (t)

dt2 +4d (t)

dt +5(t) = 1d Mt (t)

dt +3Mt (t) (10)

From (10) we get the TF, take LT of both sides, and rearrange:(

s2 +4s +5)

(s) = (s +3) Mt (s) (11)(s)Mt (s)

=s +3

s2 +4s +5 Note:

We can write down the TF directly from the coefficients of the differentialequation.

We can write down the differential equation directly from the coefficientsof the TF.

Transfer functions, such as Eqn (11), are ratios of two polynomials:(s)Mt (s)

=s +3

s2 +4s +5 :numerator polymonial

denominator polynomial

We call a TF such as (11) a rational polynomial.



1.3.4 A transfer function has poles and zeros

A TF has a numerator and denominator polynomial, for example

Gp (s) =N (s)D(s)

=2s2 +8s +6

s3 + ss2 +4s +0 (12)

The roots of the numerator are called the zeros of the TF, and the roots ofthe denominator are called the poles of the TF. For example:

>> num = [2 8 6]>> den = [1 2 4 0 ]num = 2 8 6den = 1 2 4 0

>> zeros = roots(num)zeros = -3

-1>> poles = roots(den)poles = 0

-1.0000 + 1.7321i-1.0000 - 1.7321i

We can also use Matlabs system tool to find the poles and zeros

>> Gps = tf(num, den) %% Build the system objectTransfer function: 2 s^2 + 8 s + 6

-----------------

s^3 + 2 s^2 + 4 s

>> zero(Gps), ans = -3-1

>> pole(Gps), ans = 0-1.0000 + 1.7321i-1.0000 - 1.7321i



Interpreting the poles (p1, p2, ..., pn) and zeros (z1, ..., zm)

We can use the poles and zeros to write the TF in factored form:

G(s) = 2s2 +8s +6

s3 +2s2 +4s=

b2 (s z1) (s z2)(s p1) (s p2) (s p3)

=2(s +3) (s +1)

(s0) (s +11.732 j) (s +1+1.732 j) With a complex pole pair we can do two things,

1. Use a shorthandG(s) = 2(s +3) (s +1)

s (s +11.732 j)(Because poles always come in complex conjugate pairs.)

(a) Write the complex pole pair as a quadratic, rather than 1st order terms

G(s) = 2(s +3) (s +1)s (s2 +2s +4)

The zeros are values of s at which the transfer function goes to zero The poles are values of s at which the transfer function goes to infinity We can plot a pole-zero map:

>> Gps=tf([2 8 6], [1 2 4 0])>> pzmap(Gps)

4 2 02

1

0

1

2

real

imag

inar

y

PZ Map

Figure 6: Pole-Zero constellation of aircraft transfer function.



1.3.5 Properties of transfer functions

Just as a differential equation can be scaled by multiplying the left and right sidesby a constant, a TF can be scaled by multiplying the numerator and denominatorby a constant.

Monic: A TF is said to be monic if an =1. We can always scale a TF to be monic.If G1 (s) is scaled to be monic, then

G1 (s) =b0

s +a1(13)

with b0 = b0/an and a1 = a1/an .

Rational Polynomial Form: A TF is in rational polynomial form when thenumerator and denominator are each polynomials. For example

Gp (s) =2s2 +8s +6

s3 + ss2 +4s +0(14)

An example of a TF not in rational polynomial form is:

G3 (s) =2(s +3)/(s)

(s2 +2s +4)/(s +1)(15)

By clearing the fractions within the fraction, T3 (s) can be expressed in rationalpolynomial form

G3 (s) =2(s +3)/(s)

(s2 +2s +4)/(s +1)(s)

(s)

(s +1)(s +1)

=2(s +3)(s +1)(s2 +2s +4)(s)

=2s2 +8s +6s3 +2s2 +4s

Note the middle form above, which can be called factored form.



1.3.6 Forms for a transfer function

1. Rational polynomial formG(s) = bm s

m +bm1 sm1 +b1 s +b0ansn +an1 sn1 + +a1 s +a0 (16)

Example:G(s) = 3s +8

2s3 +16s2 +30s (17)

2. Factored form

G(s) = Krlg(s z1) ... (s zm)(s p1) ... (s pn) (zi and pi are zeros and poles)

=32

(s + 83

)s (s +3) (s +5) (note: zi and pi are commonly negative values)

3. Bode form

1. Bring out any poles or zeros at the origin,

2. Bring out a constant, so that the least significant coefficients of thepolynomials are 1.0

Example:

G(s) = 1sno

KBodebm sm + ...+b

1 s +1

annos

nno + ...+a1 s +1(18)

=1s1

830

38 s +1

230 s

2 + 1630 s +1

(no = 1 , KBode =

89

)

4. Root locus form

1. Bring out a factor, so that the numerator and denominator polynomialsare monic.

Example: G(s) = Krlgsm +bm1 sm1 + ...+b

1 s +b

0

sn +an1 sn1 + ...+a

1 s +a

0

(19)

=32

s + 83s3 +8s2 +15s

(Krlg =

32

)



1.3.7 Proper and strictly proper transfer functions:

m = number of zeros, n = number poles

A TF with m n is said to be proper.

When m < n the TF is said to be strictly proper

Example of a TF that is not proper:

G4 (s) =2s2 +5s +4

s +1note: m = 2,

n = 1

such a TF can always be factored by long division:

G4 (s)=2s2 +5s +4

s +1=

2s (s +1)(s +1)

+3s +4s +1

= 2s+3s +4s +1

= (2s +3)+ 1s +1

A non-proper TF such as G4 (s) has a problem: As s = j j , the gaingoes to infinity !

Since physical systems never have infinite gain at infinite frequency,physical systems must have proper transfer functions.



1.3.8 System type

System Type: A property of transfer functions that comes up often is the systemtype. The type of a system is the number of poles at the origin.

So, for example, the aircraft transfer function from elevator input to pitchrate gives a type 0 system:

Gp1 (s) =(s)Me (s)

=s +3

s2 +4s +5 poles : s =21 j , type : 0

But the TF from the elevator to the pitch angle gives a type I system:

Gp2 (s) =(s)

Me (s)=

s +3s (s2 +4s +5) poles : s = 0,21 j , type : I

If we put a PID controller in the loop, which adds a pole at the origin, thesystem will be type II.

1.3.9 Phasors and the transfer function as complex gain

To consider the complex gain, we need to represent sinusoidal signalsas phasors. A phasor is a complex number that represents a sinusoidalwaveform. A time domain signal, such as

v(t) = Av cos( t +v) (20)

is represented by the complex number

~V = Avv = a+ j b . (21)



For example, the time-domain waveform

v(t) = 3.0 cos( t +40o)

is represented by~V = 3.0400 = 2.30+1.93 j .

For a phasor to represent a sinusoidal signal, the frequency, [rad/sec] mustbe given.

Phasors in polar and rectangular coordinates

Polar Rectangular

~V = Avv ~V = a+ j b (22)

Av =

a2 +b2 , v = atan2(b, a) a = Av cos(v) , b = Av sin(v) (23)

Sinusoidal signals in polar and rectangular coordinates

v(t) = Av cos( t +v) v(t) = a cos( t)b sin( t) (24)

With a and b as given in Eqn (23).

Notice the difference of - sign between the phasor in rectangularcoordinates and the time-domain function in rectangular coordinates. Thisextra - comes from the trig identity:

cos() = cos() cos() sin() sin() (25)



In the example of figure 7,

Gp (s)u(t)=Au cos(t+u) y(t)=Ay cos(t+y)

Figure 7: An element with sinusoid in and sinusoid out.

The TF evaluated with s = j gives the complex gain of the system element

Output Phasor = Gp (s = j) Input Phasor (26)

Ayy = Gp ( j ) Auu



Example Question: If v(t) = 1.5 cos(2.0 t +0o) ,what is the response of the aircraft, (t) ?

Solution: Looking back at the transfer functions

(s)V (s)

=7

s +10s +3

s2 +4s +5

Evaluating at s = j = j 2 gives the complex gain

(s)V (s)

s=2 j=

710+2 j

3+2 j(2 j)2 +4 (2 j)+5 =

21+14 j6+82 j (27)

= 0.15122.2672 j = 0.30760.5o

The complex gain determines the output phasor:

~ = (0.30760.5o) (1.50o) = 0.46160.5 (28)

So the output, in the time domain, is:

(t) = 0.461 cos(2.0 t60.5o)



The complex gain is a function of frequency. Evaluating

Gp (s) =(s)V (s)

s= j

as a function of gives the Bode plot.

Bode plot of Gp(s)

Frequency (rad/s)

100

80

60

40

20

0

Mag

nitu

de (d

B)

Magnitude at w=2.0

101 100 101 102 103180

135

90

45

0

Phas

e (de

g)

Phase at w=2.0

Figure 8: Bode plot of Gp (s).

Summary, section 1.3.9:

1. The dynamics of the aircraft itself gave us a differential equation

2. For linear differential equations, we can get the transfer function directlyfrom the Diff Eq.

(a) And for non-linear differential equations, there is no transfer function.3. With the TF we can determine the complex gain, and find the output for

any sinusoidal input.



1.3.10 The DC gain of a transfer function

Following up on the idea that the TF gives the complex gain of a systemblock and recalling the Laplace variable

s = j (29)

the DC gain is given with s = j0, or simply

DC Gain : G(s)s=0

=B(s)A(s)

s=0(30)

In some cases T (s = j 0) may evaluate to 00 (undefined). In such cases applylHopitals rule:

DC Gain : lims0

G(s)

The example TFs give these values of DC gain

(s)V (s)

=3

s +10s +3

s2 +4s +5(s)V (s)

=3

s +10s +3

s2 +4s +51s

Evaluating the DC Gain

(s)V (s)

s=0=

310

35 =

950

(s)V (s)

s=0=

310

35

10 =

(Many systems have a DC gain of infinity, it is a common property of controlsystems.)



1.3.11 The impulse response of a system

When we have a system such as in figure 9, the Laplace transform of theoutput (a signal) is given by

Y (s) = Gp (s) U (s) (31)

A unit impulse input is a very short pulse with an area under the curve ofthe pulse of 1.0.

Since the Laplace transform of a unit-impulse is 1, then the Laplacetransform of the impulse response is the transfer function

Yimp (s) = Gp (s) Uimp (s) = Gp (s) 1 = Gp (s)

U(s)=1 Y(s)=Gp(s)System

Gp(s)u(t): impulse y(t): impulse response

t

u(t)

t

y(t)

Figure 9: The transfer function is the Laplace transform of the impulse response.

The connection between the impulse response and TF can be used todetermine the TF

Apply an impulse and measure the output, y(t). Take the LT and useGp (s) = Y (s).

The connection between the impulse response and TF helps to understandthe mathematical connection between an LT and a TF



2 Closing the loop, feedback control

Open Loop

Gp(s)u(t) y(t)

Gp(s)u(t) y(t)r(t)

Closed Loopsensor

+

-

Figure 10: A plant with TF Gp (s) in open and closed loop. Closed loop requiresa sensor.

Feedback is the basic magic of controls. A feedback controller can

Make an unstable system stable . . . . . . . . . . . . . . . . . . Helicopter autopilot

Make a stable system unstable . . . . . . . . . . . . . . . . Early fly-ball governors

Make a slow system fast . . . . . . . . . . . Motor drive, industrial automation

Make a fast system slow . . . . . . . . F16 controls, approach / landing mode

Make an inaccurate system accurate . . . . . . . . . . . . . . . . . . . . machine tools

The magic comes because closing the loop changes the TFOpen loop:

Y (s)U (s)

= Gp (s)

For the Closed loop use U (s) = R(s)Y (s), dropping the (s) arguments:

Y = Gp (RY ) = Gp RGp Y

Y (1+Gp) = Gp R

Y (s)R(s)

=Gp

1+Gp(32)

Y (s)R(s)

=Forward Path Gain

1+Loop Gain= Try (s) (33)



Example, wrapping a simple feedback loop around the aircraft dynamics

s2 + 4 s + 5s + 3Mt(t) (t)r(t)

+-

1s

Figure 11: Simple feedback of aircraft pitch angle.

(s)R(s)

=Gp (s)

1+Gp (s)=

s+3s(s2+4s+5)

1+ s+3s(s2+4s+5)

(34)

Eqn (34) is not in rational polynomial form, so

(s)R(s)

=

s+3s(s2+4s+5)

1+ s+3s(s2+4s+5)

s(s2 +4s +5

)s (s2 +4s +5) =

s +3s (s2 +4s +5) + (s +3) (35)

The closed-loop TF is still not quite in Rat Poly form, here is the final step:

(s)R(s)

=s +3

s3 +4s2 +6s +3 (36)

(s)r(t) (t)R(s)

Figure 12: Block with r (t) as input and (t) as output.



Analyzing the responseGps = tf([1 3], [1 4 5 0]) %% LTI modelsTry = tf([1 3], [1 4 6 3])figure(1), clf

[Y_open, Top] = step(Gps, 6);[Y_closed, Tcl] = step(Try, 6);plot(Top, Y_open, Tcl, Y_closed)xlabel(t [seconds]);ylabel(\Omega, pitch-rate)title(Open- and closed-loop)text(3, 1.6, Open-loop, rotation, 45)text(4, 0.8, Closed-loop)SetLabels(14)print(-deps2c, OpenClosedResponse1)

The responsecompletelychanges !

The open-loopsystem is type I

The closed-loopsystem is type 0

0 1 2 3 4 5 60

0.5

1

1.5

2

2.5

3

3.5

t [seconds]

, pi

tch

rate

Open and closedloop response, aircraft

Open

loop

Closedloop

Figure 13: Open and Closed loop response of the aircraft, the two responses havevery different characteristics.



Introduce a proportional controller gain

s2 + 4 s + 5s + 3Mt(t) (t)r(t)

+-

1s

Kce(t)

Figure 14: Feedback for aircraft pitch control, with P-type gain Kc.

Look at Kc ={

1.0, 3.0, 10,0}

Kc = 1; Try1 = tf(Kc*[1 3], [1 4 5 0]+Kc*[0 0 1 3])Kc = 3; Try2 = tf(Kc*[1 3], [1 4 5 0]+Kc*[0 0 1 3])Kc = 10; Try3 = tf(Kc*[1 3], [1 4 5 0]+Kc*[0 0 1 3])figure(1), clf

...

plot(Top, Y_open, Tcl, Y_closed1, Tcl, Y_closed2, Tcl, Y_closed3)...

System gets muchfaster as Kcincreases

System gets lessstable as Kcincreases

0 1 2 3 4 5 60

0.5

1

1.5

2

2.5

3

3.5

t [seconds]

, pi

tch

rate

Open and closedloop response, aircraft

Open

loop

Closedloop, Kc =1

Kc =3

Kc =10

Figure 15: Open and Closed loop response of the aircraft, with Kc = 1.0,Kc = 3.0, and Kc =10.0 .



2.1 Analyzing a closed-loop system

A typical, basic loop (such as velocity PI control of a motor drive) has 3components:

1. Plant (thing being controlled)

2. Controller or compensator (usually a computer, a often PLC for motordrives)

3. A sensor


-

ys(t)

Nc(s)Dc(s) KcGc(s) =

Np(s)Dp(s)Gp(s) =

Ny(s)Dy(s)Hy(s) =

Sensor Dynamics

Kce(t) u(t)

Figure 16: Basic loop with a plant, compensator and sensor.

The TF is given as

Try (s) =Y (s)R(s)

=Forward Path Gain

1+Loop Gain=

Kc Gc Gp1+Kc Gc Gp Hy

Often, for the controls engineer the plant, Gp (s), is set (e.g., the designer ofa cruise-control does not get to change the engine size).

As a controls engineer, we get to pick Gc (s) and maybe influence Hy (s)(e.g., by convincing the project manager to spring $$$ for a better sensor).



2.2 Common controller structures:

PD PI PID Lead-Lag

Proportional-Derivative Prop.-Integral Prop.-Int.-Deriv.

Gc (s) = kd s + kp Gc (s) =kps+ki

s Gc (s) =kds2+kps+ki

s Gc = Kc(s+z1) (s+z2)(s+p1) (s+p2)

Common Applications

PI: Velocity control of motor drives, temperature control (good speed andaccuracy, acceptable stability)

PD: Position control where high accuracy is not required (good speed andstability, so-so accuracy)

PID: Many many places, Astrom has estimated that 80% of controllers arePID (good, speed accuracy, stability)

Lead-Lag: Used where a pole at the origin is unacceptable, can be as goodas PID (notice, 5 parameters rather than 3)



2.3 Analyzing other loops

Input ShapingHr(s)

Controller Plant

d(t)

r(t) y(t)

Sensor Dynamics

+

-

++

+

+Vs(t)ys(t)

NcDc

KcGc(s) = NpDpGp(s) =

NyDyHy(s) =

Disturbance FilterNdDdGd(s) =

Kce(t) uc(t) up(t)

Figure 17: Basic loop with a disturbance input, d (t) , and sensor noise, Vs (t)added.

In some cases we may want to consider additional inputs and outputs.

Many systems have a disturbance signal that acts on the plant, think ofwind gusts and a helicopter autopilot.

All systems have sensor noise.

Any signal in a system can be considered an output. For example, if wewant to consider the controller effort, uc (t), arising due to the referenceinput

Tru (s) =Uc (s)R(s)

=Forward Path Gain

1+Loop Gain =Hr (s) Kc Gc (s)

1+Kc Gc (s) Gp (s) Hy (s)(37)

If we wanted to consider the error arising with a disturbance, we would have

Tde (s) =E (s)D(s)

=Gd (s) Gp (s) Hy (s) (1)1+Kc Gc (s) Gp (s) Hy (s)

(38)



2.3 Analyzing other loops (continued) As a final example, lets consider the output arising with sensor noise

Tvy (s) =Y (s)Vs (s)

=Hy (s) (1) Kc Gc (s) Gp (s)1+Kc Gc (s) Gp (s) Hy (s)

(39)

The example transfer functions, Eqns (37), (38) and (39) show someinteresting properties. The TFs are repeated here (omitting the many (s)s)

Try (s) =Hr Kc Gc Gp

1+Kc Gc Gp Hy, Tru (s) =

Hr Kc Gc1+Kc Gc Gp Hy

Ted (s) =Gd Gp Hy (1)1+Kc Gc Gp Hy

, Tvy (s) =Hy (1) Kc Gc Gp1+Kc Gc Gp Hy

The denominators are all the same

The poles are the same for any input/output signal pair

The stability and damping (both determined by the poles) are the samefor any signal pair

The numerators are different

The zeros are in general different for each input/output signal pair

Since the numerator help determine if the signal is small or large, signalsmay have very different amplitudes and phase angles



If we consider what happens as Kc , we can see what happens for veryhigh gain. For this, assume that Hr (s) = 1.0 and Gd (s) = 1.0, since thesetwo terms merely pre-filter inputs.When Kc , 1+Kc Gc Gp Hy Kc Gc Gp Hy , so

Try (s) Kc Gc GpKc Gc Gp Hy =1

Hy, Tru (s) Kc GcKc Gc Gp Hy =

1Gp Hy

Tde (s) Gd Gp Hy (1)

Kc Gc Gp Hy=GdvKc Gc

, Tvy (s) Hy (1) Kc Gc GpKc Gc Gp Hy =1

Try (s) 1/Hy (s) shows that the TF of the plant can be compensated, itdisappears from the closed-loop TF as Kc .

Try (s) 1/Hy (s) also shows that the TF of the sensor can not becompensated. If the characteristics of Hy (s) are bad (e.g., a cheap sensor)there is nothing feedback control can do about it !

Tde (s)1/Kc Gc shows that disturbances can be compensated, as Kc,errors due to disturbances go to zero ;)

Tru (s) 1/Gp Hy shows that U1 (s) does not go up with Kc, and also, if theplant has a small gain (Gp (s) is small for some s = j ) then a large controleffort will be required for a given input.

Tvy (s) 1 shows that there is no compensation for sensor noise. If thereis sensor noise, it is going to show up in the output !

Summary: Feedback control can solve problems arising with characteristics of the

plant, Gp (s), and disturbances, d (t).

Feedback control can not solve problems with the sensor, Hy (s), orsensor noise, vs (t) .



3 Calculating steady-state errors (Franklin et al.sec. 4.2 (6th Ed.)) Steady-state errors are computed from the transfer function r (t) e(t), and

the final value theorem:

Final value theorem: For a stable system with transfer functionTre (s) = E (s)/R(s),

ess = limt e(t) = lims0 sE (s) (40)

In Eqn (40) E (s) (a Laplace transform) is given by Tre (s) (a transferfunction) times R(s) (a Laplace transform):

E (s) = Tre (s) R(s) (41)

which givesess = lim

t e(t) = lims0 sTre (s) R(s) (42) To compute the final value of the error, we have to specify the input r (t).

Signals typically used to compute steady-state error are seen in table 4.

The step, ramp and acceleration input signals and the Laplace Transforms can bewritten:

r (t) =1k! t

k 1+ (t) , R(s) =1

sk+1. (43)

For k = 1,2,3 the signals and Laplace transforms are given in table 4.

k Test input signal, r (t) R(s)Step input 0 r (t) = 1+ (t) R(s) = 1s

Ramp input 1 r (t) = t 1+ (t) R(s) = 1s2

Accelerating input 2 r (t) = 12 t2 1+ (t) R(s) = 1

s3

Table 4: Typical inputs to consider for determining steady-state error.



3.1 System type and Bode standard form for the loop gain


-

ys(t)

Nc(s)Dc(s) Gc(s) =

Np(s)Dp(s)Gp(s) =

Ny(s)Dy(s)Hy(s) =

Sensor Dynamics

Figure 18: Standard feedback loop, with numerator and denominators called out.

Considering figure 18, the error transfer function is given as:

Ter (s) =1

1+Kc Gc Gp Hy=

11+Kc NcDc

NpDp

NyDy

=Dc Dp Dy

Dc Dp Dy +Kc Nc Np Ny

The poles of the loop gain are the zeros of the error transfer function ! Therefore, if the loop gain has a pole at the origin, the error transfer

function has a zero at the origin (which means its DC gain is zero) If Tre (s) has zeros at the origin, these cancel out 1s terms coming from

R(s) in the final value theorem.

The number of poles at the origin in the loop gain is a sufficiently basicproperty of feedback systems that it is given a special name: the SystemType.

The system type is given asType 0: GL (s) has no pole at the origin, no = 0

Type I: GL (s) has one pole at the origin, no = 1

Type II: GL (s) has two poles at the origin, no = 2 etc.

where no is the number of open-loop poles at the origin.Part 1: Controls Review-By-Example (Revised: Jan 16, 2013) Page 33


3.1.1 Determining steady-state error using Bode standard form

Defining the loop gain as product of gains around the loop. For the standardloop (figure 18):

GL (s) = Gc (s) Gp (s) Hy (s) (44)

Writing the loop gain in Bode standard form:

GL (s) = Gc (s) Gp (s) Hy (s) =1

snoKBode

bm sm +bm1 am1...+b1s +1an sn +an1 sn1...a1 s +1(45)

Using Eqn (45), a general solution for the steady-state error can be found:

ess = lims0

s Tre (s) R(s)

= lims0

s1

1+ 1sno KBodebm sm+bm1sm1...b1 s+1an sn+an1sn1+...a1s+1

R(s) (46)

But in the limit, all of the terms(bmsm +bm1sm1 + ..+b1s

)and

(ams

m +am1sm1 + ..+a1s)

drop out, giving

ess = lims0

s1

1+ 1sno KBodeR(s)

= lims0

ssno

sno +KBodeR(s) (47)

Eqn (47) has a particularly simple form, and is valid for step, ramp oracceleration inputs, and any system type.



3.2 Table of steady-state errors

Eqn (47) leads to a general form for the steady-state error. Recall

ess = lims0

ssno

sno +KBodeR(s) (48)

Inserting R(s) from Eqn (43), above, into the equation for steady-state errorgives

ess = lims0

ssno

sno +KBode1

sk+1(49)

where no is the system type, and k is the order of the input signal. UsingEqn (49), steady-state errors for system types 0, I and II are seen in table 5.

k r (t) R(s) Type 0 System Type I System Type II System

Step 0 1+ (t) R(s) = 1s

11+KBode 0 0

Ramp 1 t 1+ (t) R(s) = 1s2

1

KBode 0

Accel 2 12 t2 1+ (t) R(s) = 1

s3

1KBode

Table 5: Connection between system type and steady-state error.

Types 0 and I are the most common system types in practical systems. TypeII is also seen. Type III and above is quite rare.



3.3 Steady-state error example

Considering the closed loop example of figure 14, with

Gc (s) =s +3

s, Gp (s) =

1s2 +4s +5 , Hy (s) = 1 (50)

we have the loop gain in Bode standard form:

GL (s) = Gc Gp Hy =35

1s1

13s +1

15s

2 + 45s +1(51)

which shows a system type of I, and KBode = 35 . Considering table 5, wefind

ess (step) = 0 , ess (ramp) =1

KBode=

53 , ess (accel) = .

3.4 Summary for steady-state error

The standard way to compute steady-state error is through table 5. Table 5 shows that steady-state error depends on

1. The system type,2. KBode,3. The order of the input (step, ramp or acceleration).

We often use the controller to make a Type 0 system into Type I, specificallyto make the steady-state error go to zero for step inputs.

Controllers that increase the system type by one: PI PID

KBode is the product of the DC gains around the loop. By increasing thecontroller gain we increase KBode and reduce steady-state errors.



4 Characteristics of the Step Response

Weve seen that steady-state error is a characteristic of the step (ramp)response. But SSE we usually calculate from the TF.

These performance measures are defined from the step response. Generallyspeaking, we cant compute them exactly from the TF, but well seeapproximate calculations based on the dominant mode in the next section.

Rise time

Peak time

Settling time

Overshoot When we consider the step response, it is always the closed-loop step

response.

2 4 6 80

0.5

1

1.5

Time [sec]

Out

put y

(t)

Step response of complex pole pair

Overshoot2%

ess

| |T

r: 1090% Rise Time

|

Ts: 98% Settling Time 0 1 2 30

0.5

1

1.5

2

2.5

Out

put y

(t)

Step response of complex pole pair

y10

y90 y

ss(1)

yp

Mp r

ss(1)

yss

(0) rss

(0)

|t10

|t90

|tp

Time [sec]

Tr: 1090% Rise Time

Figure 19: Quantity definitions in a CL step response.

4.1 Rise Time, tr

The 10% - 90% rise time. For the step response, it is the time between crossing0.10yss and 0.90yss.

tr 1.8n

(52)Part 1: Controls Review-By-Example (Revised: Jan 16, 2013) Page 37


4.2 Peak Time, tp

Time between application of a step input and the first peak of theoutput. Undefined if the output does not have a peak, but taken to besatisfied in this case if all other speed-related requirements are met.

tp pi

(53)

4.3 Settling Time, ts

As seen in figure 19, the step response of a linear system lies within anexponentially decaying envelop. Settling time, ts, is the time requiredto assure that the response does not pass again out of a narrow bandaround yss.

Typically used: 4 time constants, e4 = 1.83%. So The 4 timeconstant settling time is the time between application of a step inputand the last time that the signal exits a band 1.83% wide around yss .

ts = maxt{abs(y(t) yss

yss) > e4} 4

d(54)

where d is the real part dominant pole location.

4.4 Percent Overshoot, PO

Difference between peak response and yss, the steady-state response,expressed as a fraction of yss:

PO = 100 max(y(t)) yssyss

, epi12 (55)

where [] is the damping factor. If y(t) never goes above yss, PO = 0.



5 Working with the pole-zero constellation

5.1 Basics of pole-zero maps, 1st order, p1 = You have seen that:

Each real pole is associated with a mode of the system response.

A real pole gives the terms of Eqn (57), as seen in figure 20.

Y (s) =C1

s +=

3s +2

y(t) = C1 e t = 3e2t (56)

3 2 1 0 12

1

0

1

2PoleZero Map

Real Axis [sec1]

Imag

inar

y Ax

is [se

c1 ]

Splane

Faster

0 1 2 30

1

2

3

Time (secs)

Ampl

itude e

t = et/ = e2 t

h(t) =

|

C1 / e

Impulse response

Figure 20: First order pole and impulse response.

A real pole has these characteristics: y(t) et/ where [sec] = 1/ is the time constant. Further to the left indicates a faster response (smaller ). The pole-zero constellation does not show either the KDC or the Krlg of

the TF or the amplitude of the impulse response.



5.1.1 Real Pole location and response characteristics

Example 1: Shifting poles to the left accelerates transient decay

Two example first-order responses are shown.X Case: = 4 [sec1] = 1/4 [sec]

Case: = 16 [sec1] = 1/16 [sec]

20 15 10 5 0 55

0

5

PZ map showing two systems

Real Axis [sec1]

Imag

inar

y Ax

is [se

c1 ]

Faster Decay

Figure 21: A change in pole location changes the decay rate and damping.

0 0.5 10

0.5

1

1.5

Time (secs)Am

plitu

de

= 0.25 [sec]

X System step response

0 0.5 10

0.5

1

1.5

Time (secs)

Ampl

itude

= 0.06 [sec]

System step response

Figure 22: A change in changes the time constant.



5.2 Basis of pole-zero maps, 2nd order, p1,2 = j And we have seen for second order:

Each complex pole pair gives a mode of the system response.

A complex pole pair gives the terms of Eqn (57), as seen in figure 23. Using the Laplace transform pair

F (s) =Bcs +(Bc+Bs )

s2 + ss +2n f (t) = e t (Bc cos( t)+Bs sin( t))

one finds

Y (s) =b1 s +b0

s2 +2s +(2 +2)=

2s +14s2 +3s +18.25 (57)

y(t) = Ae t cos( t +) = 3.4e1.5tcos(4 t53.97o) (58)

with si = j , n =

2 +2 , : damping factor, = n

8 6 4 2 0 25

0

5PoleZero Map

Real Axis [sec1]

Imag

inar

y Ax

is [se

c1 ] Splane

0 2 42

1

0

1

2

3

Time (secs)

Ampl

itude

et = et/ = e1.5t

3.40 cos(4t 54.0o)

Impulse response

Figure 23: Second order pole and impulse response.



5.2.1 Notation for a complex pole pair

A complex pole pair can be expressed in polar or rectangular coordinates:

3 2 1 0 1 26

4

2

0

2

4

6PZ Map, p1, p1

* =1.5 j 4

Real Axis [sec1]

Imag

inar

y Ax

is [se

c1 ]

j

n

1

Splane

p1

p1*

Figure 24: Complex pole pair with n and defined.

Term Description Given by Units or Decay Rate p1 =+ j [sec1]

or d Damped Nat. Freq. p1 =+ j [rad/sec]n Natural Freq. 2n =

(2 +2

)[rad/sec]

Pole Angle = atan2(, ) [deg] or Damping Factor = /n Dimless, []

Table 6: Factors derived from the location of a complex pole.(Note: Franklin et al. often use , d and .)

Rectangular Polarp1 =+ j p1 = n (90+)

= n = sin() =

12 n = /n

H (s) =

(s +)2 +2H (s) =

n

12s2 +2n s +2n

Table 7: The terms of table 6 relate to rectangular or polar form for the poles.



5.2.2 Complex pole location and response characteristics

Example 1: Shifting poles to the left accelerates transient decay

n PO

X Case: 1 [sec1] 4 [rad/sec] 4.12 [rad/sec] 0.25 [Dimless] 44%

Case: 4 [sec1] 4 [rad/sec] 5.66 [rad/sec] 0.71 [Dimless] 4%

8 6 4 2 0 2 46

4

2

0

2

4

6PZ map showing two systems

Real Axis [sec1]

Imag

inar

y Ax

is [se

c1 ]

Faster Decay

8 6 4 2 0 2 46

4

2

0

2

4

6

7 5 3 1

0.9

0.7 0.5 0.3

0.9

0.7 0.5 0.3

PZ map with sgrid (indicates and n)

Real Axis [sec1]

Imag

inar

y Ax

is [se

c1 ]

Figure 25: A change in changes the decay rate and damping.

0 2 40

0.5

1

1.5

Time (secs)

Ampl

itude

= 0.2446 % Overshoot


0 2 40

0.5

1

1.5

Time (secs)

Ampl

itude

= 0.714 % Overshoot


Figure 26: Step response: a change in , is unchanged.



Example 2: Shifting poles out vertically increases oscillation frequency

n PO



30 20 10 0 10

20

10

0

10

20


Real Axis [sec1]

Imag

inar

y Ax

is [se

c1 ]

Faster Oscillation

30 20 10 0 10

20

10

0

10

20

28 20 12 4

0.9

0.7 0.5 0.3

0.9

0.7 0.5 0.3


Real Axis [sec1]

Imag

inar

y Ax

is [se

c1 ]

Figure 27: A change in changes the oscillation frequency and damping.

0 1 20

0.5

1

1.5

Time (secs)Am

plitu

de

= 0.714 % Overshoot


0 1 20

0.5

1

1.5

Time (secs)

Ampl

itude

= 0.2446 % Overshoot


Figure 28: Step response: a change in .



Example 3: Shifting poles out radially rescales time

n PO



30 20 10 0 1020

10

0

10

20


Real Axis [sec1]

Imag

inar

y Ax

is [se

c1 ]

30 20 10 0 1020

10

0

10

20

28 20 12 4

0.9

0.7 0.50.3

0.9

0.7 0.50.3


Real Axis [sec1]

Imag

inar

y Ax

is [se

c1 ]

Figure 29: A radial change in pole location changes the decay rate and oscillationfrequency, but not the damping.

0 1 20

0.5

1

1.5

Time (secs)

Ampl

itude

= 0.714 % Overshoot

= 0.25 [sec]


0 0.2 0.40

0.5

1

1.5

Time (secs)

Ampl

itude

= 0.714 % Overshoot

= 0.06 [sec]


Figure 30: Maintaining , time is rescaled.

Note: The S-plane has units of [sec1].



5.2.3 The damping factor:

Plugging = /n back into the 2nd order form gives:

H (s) =b0

s2 +a1 s +a0=

b0s2 +2n s +2n

(59)

giving: = a1

2n=

a12a0 (60)

is defined by Eqn (60) for either real poles: ( 1.0), ora complex pole pair: (0.0 < < 1.0).

As illustrated in figure 28, above, on the range 0.0 < < 1.0, the dampingfactor relates to percent overshoot. For a system with two poles and nozeros, the percent overshoot is given by Eqn (61) and plotted in figure 31

P.O. = 100epi/

12 (61)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

100

Damping factor [.]

Perc

ent O

vers

hoot

Figure 31: Percent overshoot versus damping fact. Exact for a system with twocomplex poles and no zeros, and approximate for other systems.



5.3 Higher order systems: dominant mode

The response of a higher order system is the superposition of the responsesof the modes:

1st order modes; 2nd order modes.

Typically (not quite always) one mode will dominate. In the example offigure 32, the slower second-order mode dominates.

15 10 5 010

5

0

5

10PoleZero Map

Real Axis [sec1]

Imag

inar

y Ax

is [se

c1 ]

Dom. Mode

0 2 40

0.5

1

1.5

Time (secs)

Ampl

itude

Step response

Step response Dom. mode

0 2 40

0.5

1

1.5

Time (secs)

Ampl

itude

Fast secondorder mode

0 2 40

0.5

1

1.5

Time (secs)

Ampl

itude

Firstorder mode

Figure 32: Example high-order system showing dominant mode in step response.

Recall: the goal of studying the S-plane is to tie controller specifications topole locations.

The dominant mode dominates the response.

The root-locus design method operates by placing the dominant poles inthe desired region of the S-plane(Other poles can be anywhere in stable region).



Dominant mode (continued) As the figure 32 Step response plot shows, the dominant mode is only an

approximation to the true step response.

For high-order systems (n 3), design control by:

1. Use dominant mode concept and approximation for initial design; then

2. Fine tune controller based on actual (high order) response Iterative process of design and analysis.

5.4 Which mode is the dominant mode ?

The dominant mode is:

A slow mode, with a

Large residue.

On the S-plane this means that the dominant mode is (almost always):

The slowest mode (other modes die out faster).

Exceptionally, one must also consider the size of the residue, the dominantmode should be:

Near to other poles; for example, s = 0 pole from step input(Graphical method: being near other poles gives a large residue).

Far from zeros(Graphical method: being far from zeros gives a large residue).



5.3 Dominant mode (continued) A second example is presented in figure 33.

Which mode is dominant in figure 33 ? StudentExercise

15 10 5 0 510

5

0

5

10PoleZero Map

Real Axis [sec1]

Imag

inar

y Ax

is [se

c1 ]

0 5 100

0.5

1

Time (secs)

Ampl

itude

Step response

Step responseDominant mode

0 5 100

0.5

1

Time (secs)

Ampl

itude

Slow secondorder mode

0 5 100

0.5

1

Time (secs)

Ampl

itude

Fast secondorder mode

Figure 33: Second example showing dominant mode in step response.



5.5 Summary: Pole location and response characteristics

S-plane

Stable Region (LHP)

Unstable Region (RHP)

Marginally Stable Region(the j axis)

X

X

p1

p1*

Faster Decay

FasterOscillation

Better Damping

Figure 34: Decay rate, oscillation freq. and damping depend upon pole location.

Figure 35: S-plane and responses.



6 Design

In some sense, Design is the opposite of Analysis

PerformanceSpecifications

Completed Controller Design

Analysis

DesignFigure 36: Design = Analysis1

In Analysis, we use mathematical methods to determine performancespecifications from a completed controller design (all structure andparameters specified).

In Design, we use whatever method works !

Mathematical

Gut feeling

Trial and error

Calling a colleague with experience

to determine a controller structure and parameters to meet performancegoals.



6.1 Design methods

The major design methods are:

Root locus

Speed, Stability: determine by determining pole locations

Accuracy: increase the system type, check the SSE

Frequency response ()

Speed: Bandwidth and Cross-over frequency directly from bode plot

Stability: Phase margin, Gain margin directly from bode plot

Accuracy: tracking accy., disturbance rejection directly from bode plot State Space methods ()

Design using state-space design methods, check speed, stability andaccuracy from the step response

6.2 Root Locus Design Devised by Walter R. Evans, 1948 (1920 -

1999)W.R. Evans, Control system synthesis byroot locus method, Trans. AIEE, vol. 69,pp. 6669, 1950.

(Amer. Institute of Elec. Engs becameIEEE in 1963)

Evans was teaching a course in controls,and a student (now anonymous) asked aquestion about an approximation.



7 Summary

Dynamic systems are governed by differential equations Every input or output signal of the system has a unique Laplace transform For linear systems,

The ratio of Laplace transforms, however, does not depend on the signals.

The ratio depends only on properties of the system. We call it the transferfunction. For example:

Guy (s) =b1 s +b0

s2 +a1 s +a0(62)

Transfer function gives us many results

The pole locations tells us the stability and damping ratio

We can get approximate values for rise time, settling time and otherperformance measures

Control system analysis is the process of determining performance from asystem model and controller design

We have various tools for evaluating performance, including

* Pole locations * Steady-state error * Step response * Bode plot

Control systemDesign =Analysis1

it is the process of determining a controller design given a system modeland performance goals.

The root locus method is one method for controller design.



8 Glossary of Acronyms

LT: Laplace Transform

TF: Transfer Function

FPG: Forward path gain

LG: Loop Gain, (also Gl (s))

RHP: Right-half of the S plane (unstable region)

LHP: Left-half of the S plane (stable region)

SSE: Steady-state error

PO: Percent Overshoot

LF: Low frequency (as in LF gain). Frequencies below the cross-overfrequency.

HF: High frequency (as in HF gain). Frequencies above the cross-overfrequency.

CL: Closed loop

OL: Open loop

P, PD, PI, PID: Proportional-Integral-Derivative control, basic and very commoncontroller structures.



Frequency-Response Based Analysis

Contents

1 Introduction 3

1.1 Review of complex numbers and arithmetic . . . . . . . . . . . . 5

1.1.1 Complex Conjugate: . . . . . . . . . . . . . . . . . . . . 61.1.2 Complex Arithmetic: . . . . . . . . . . . . . . . . . . . 7

2 Frequency Response and the Bode plot 10

2.1 Sketching a Bode plot by hand . . . . . . . . . . . . . . . . . . 12

2.1.1 Why make a Bode plot by hand . . . . . . . . . . . . . . 12

2.1.2 Sketching individual elements . . . . . . . . . . . . . . . 13

2.1.3 Example 1, One pole no zeros . . . . . . . . . . . . . . . 14

2.1.4 Details of the break point . . . . . . . . . . . . . . . . . 15

2.1.5 Example 2, Two poles one zero . . . . . . . . . . . . . . 16

2.1.6 Example 3, A pole at the origin . . . . . . . . . . . . . . 17

2.2 The Bode plot of the loop gain . . . . . . . . . . . . . . . . . . . 19

2.3 Bode standard form for the loop gain . . . . . . . . . . . . . . . . 23

2.3.1 Plotting terms of the form KBode ( j )no A . . . . . . . . 252.3.2 Contrasting KBode with Krlg . . . . . . . . . . . . . . . . 26

2.3.3 Plotting terms of the form ( j +1) (a real pole or zero) 272.3.4 Plotting a complex pole or zero pair . . . . . . . . . . . . 31

2.3.5 Summary of Bode plot plotting rules . . . . . . . . . . . 35

2.4 Examples: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

Part 2: Frequency Response Analysis (Revised: Mar 26, 2013) Page 1


2.4.1 Example 1: Sketch the bode plot for K G(s) . . . . . . . 37

2.4.2 Example 2: Sketch the bode plot for a system with acomplex pole pair . . . . . . . . . . . . . . . . . . . . . . 40

2.5 Steady state errors . . . . . . . . . . . . . . . . . . . . . . . . . 43

2.5.1 Determining KBode and SSE using the Matlab bode()command . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.6 Summary of Frequency Response and the Bode plot . . . . . . . . 49

3 Stability Considerations, Phase and Gain Margins 50

3.1 Marginal Stability . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.2 Definitions for phase and gain margin . . . . . . . . . . . . . . . 51

3.3 Examples, stability, marginal stability and instability . . . . . . . 52

3.3.1 A challenge with applying gain and phase margin . . . . . 56

3.3.2 Example with the pattern of figure 38(a) . . . . . . . . . . 583.4 Interpreting Phase and Gain Margin . . . . . . . . . . . . . . . . 60

4 Summary: Analysis by Frequency Response 61



1 Introduction

For a linear system a sinusoidal input produces a sinusoidal output,(after all transients have died out)

Figure 1 shows the response of the system

G(s) = 1s +10

1s+10

u(t) y(t)G(s)=

for the inputu(t) = sin(10 t)

Once the startup transient dies out, an input with = 10 gives and outputwith = 10.

0 1 2 3 4 5 6 7 8 9 100.1

0.05

0

0.05

0.1

0.15

0.2

Syst

em O

utpu

t

Time [seconds]

Zerostate Response of G(s) to sin(10 t)

Figure 1: A sinusoidal input to a linear system results in a sinusoidal output.



We can determine the steady-state output sinusoid from the input and thetransfer function

Representing the input as a phasor

Waveform : Phasor A Complex Number

u(t) = Au cos( t +u) ~U = Auu

Looking at the system

Output PhasorInput Phasor

=~Y~U

= Complex Gain = G(s)s= j

=1

j +10

For example, with u(t) = sin(10 t) and G(s) = 1/(s +10), then y(t) isdetermined from:

~Y~U

= G(s)s= j 10

=1

j 10+10 =1

10

2 45o

The input as a phasor is give by

u(t) = sin(10 t) = cos(10 t90o) ~U = 1.090o (with = 10[rad/sec])

The output phasor is given by:

~Y = (Input Phasor) (Complex Gain) = (1.090o)(

110

2 45o

)

~Y =1

10

2135o y(t) = 1

10

2cos(10 t135o)

(This is the waveform in figure 1)



1.1 Review of complex numbers and arithmetic(See Franklin et al., Appendix B)

Certain polynomials have no solutions which are real numbers,

x2 +1 = 0 or x2 =1

Italian mathematicians of the 1500s began considering polynomials anddiscussing solutions involving

1 In 1637 Ren Descartes coins the term imaginary number to suggest that

work on the subject was an illusion. It turns out that imaginary numbers solve a wide range of problems in signal

processing, medical image reconstruction, impedance analysis, etc.

Any complex number is made up of a real and imaginary part

A = + j real(A) = imag(A) =

j (or i for mathematicians) is the complex number

j2 =1 (1)

Higher powers of j are given from:j3 = j2 j = 1 j

j4 = j2 j2 =1 1 = 1j5 = j2 j2 j = j

etc.



Complex numbers correspond to a point on the plane

a+b j

Complex Plain

Real

Imag

A <

V

V*

Figure 2: A complex number = A point on the complex plain.

A complex number (point on the complex plain) can be represented in polaror rectangular coordinates

Polar RectangularV = Avv V = a+ j b (2)

Av =

a2 +b2 a = A cos(v) (3)v = atan2(b, a) b = A sin(v) (4)

What the complex plane reveals is that the real and imaginary parts areorthogonal.

1.1.1 Complex Conjugate:

The complex conjugate of a number V is written V.With V = Avv = a+ j b , then V = Avv = a j bV+V gives a real number, V+V = 2a+ j 0VV also gives a real number,(a+ j b)(a j b) = (a2 +b2)+ j (abab) = (a2 +b2)+ j0Part 2: Frequency Response Analysis (Revised: Mar 26, 2013) Page 6


1.1.2 Complex Arithmetic:

The operations of complex arithmetic are given as:

Polar Rectangular

V = A , W = B V = a+ j b , W = c+ j d

Addition : (none) V+W = (a+ c)+ j (b+d) (5)

Subtraction : (none) VW = (a c)+ j (bd) (6)

Multiplication : VW = AB+ VW = (a+ j b) (c+ j d) (7)

= (acbd)+ j (ad +bc)

Division : VW = A/BVW =

(ac+bd)+ j (bcad)c2 +d2 (8)

Where the equation for division in rectangular coordinates comes fromwriting:

VW =

VWWW

and multiplying out the terms.



Eulers Formula:

Recall the transcendental functions:

ex = 1+ x+x2

2!+

x3

3! +x4

4!+

cos x = 1 x2

2!+

x4

4!+

sin x = x x3

3! +x5

5! +

Eulers formula gives meaning to the complex exponential.

e j = 1+ j + j2 22!

+j3 33! +

j4 44!

+j5 55! + (9)

Using the fact that j2 =1 , j4 = 1, etc.

e j = 1+ j 2

2! j

3

3! +44!

+j 55! +

=

(1

2

2!+

44!

+ )

+ j(

3

3! +55! +

)

e j = cos + j sin (10)

From Eqns (2) and (10) we find that

A = Ae j = A cos()+ j A sin() = a+ j b (11)



If we write the phasor~V = A (12)

then

v(t) = A cos( t +) = A real(

e j( t+))

= A real(

e j t e j)

We can imagine a complex signal (t) = e j( t+) as a vector on the complexplane. The real signal is the projection of the vector onto the real axis.

Complex Plain

Real

Imag

A ej(t+)rotationt

v(t) = e j ( t+)

= e j t e j

Figure 3: Phasor rotating on the complex plane.

Rotation by an angle (either phase shift or by t) is just multiplicationby e j .



2 Frequency Response and the Bode plot

Considering the steady state, a sinusoidal input gives a sinusoidal output The relationship between the input and output is given by the complex gain

Which is given by the transfer function evaluated at s = j

In general, the G(s = j ) a function of frequency

In polar coordinates, a complex number has a magnitude and phase

The magnitude and phase are a function of frequency

101 100 101 102 1035040302010

010

Gai

n dB

LowPass Filter

G(s)=3/(s+3)

20dB/decade

101 100 101 102 10390

60

30

0

Frequency [radians/second]

Phas

e [de

g]

Lagging phase

101 100 101 102 10310

01020304050

Gai

n dB

HighPass Filter

G(s)=(s+10)/10

+20dB/decade

101 100 101 102 1030

30

60

90


Phas

e [de

g]

Leading phase

Low Pass Filter High Pass Filter

G(s) = 3s +3 G(s) =

s +1010

Figure 4: Bode plot: First-order response characteristic.

1st order low-pass:

20dB / decade roll-off90o phase at high freq.

1st order high-pass:

+20dB / decade

+90o phase at high freq.



2nd orderG(s) =

2n

s2 +2n s +2n(13)

101 100 101 10260

40

20

0

20

Gai

n dB

2nd order mode, =0.28

G(s)=10/(s2 + 1.0 s + 10)

40dB/decade

101 100 101 102180150120

906030

0


Phas

e [de

g]

101 100 101 10260

40

20

0

20

Gai

n dB

2nd order mode, =0.056

G(s)=10/(s2 + 0.2 s + 10)

40dB/decade

101 100 101 102180150120

906030

0


Phas

e [de

g]

2nd order, moderate damping 2nd order, light damping

G(s) = 10s2 +1.0s +10 G(s) =

10s2 +0.2s +10

Figure 5: Bode plot: Second-order response characteristic. The smaller thedamping, the higher the resonant peak.

2nd order response:

40 dB / decade roll-off180o phase at high frequency

Sharpness of the transition depends on the damping:

Moderate to high damping: shallow peak, smooth phase transition

Low damping: sharp peak, abrupt phase transition



2.1 Sketching a Bode plot by hand

2.1.1 Why make a Bode plot by hand

Given Gc (s) Gp (s) computers are great for making the exact Bode plot Given a structure for Gc (s) and approximate values, computers can be pretty

good at fine tuning the parameters (adaptive control) Computers are no good at all at determining a good structure for Gc (s), or

finding the correct ball park for parameters

An approximate sketch of the Frequency Response is a very powerful wayto understand the over-all characteristics of a system.

The Bode plot is a tool on which Engineering judgement and intuition canbe based.

Some key ideas of the bode plot are:

The axes:

Frequency axis is logarithmic Magnitude is plotted in decibels (dB) Phase is plotted in degrees

Decibels: a logarithmic scale

|G(s)|dB = 20 log10 (|G(s)|)

Multiplying the loop gain by a factor corresponds to adding decibels|40|dB = 32dB , |50|dB = 34dB , |40 50|dB = 32+34 dB

Multiplicative factors:2+6dB , 10= +20 dB , 20+26dB

0.56dB , 0.1=20dB , 0.0526dB



Advantage of working in dB:

Gains of blocks in series multiply,G1

u(t) y(t)G2

G1 = A1 1 , G2 = A2 2 , G1 G2 = A1 A21 +2 = A1212

But Bode plot terms in series add

Magnitude : 20 log10 (A1 A2) = 20 log10 (A1)+20 log10 (A2)

Phase : 12 = 1 +2

2.1.2 Sketching individual elements

Low frequency asymptote given aslim0

G( j ) (14)

High frequency asymptote given aslim

G( j ) (15)

Poles and zeros make contributions as a function of

A pole adds a slope of 1 and 90o of phase lag

A pole pair adds a slope of 2 and 180o of phase lag

A zero adds a slope of +1 and 90o of phase lead

A zero pair adds a slope of +2 and 180o of phase lead



2.1.3 Example 1, One pole no zeros

101 100 101 102 1035040302010

010

Gai

n dB

One pole no zeros

HighFreq Asymptote LowFreq Asymptote

20dB/decade

101 100 101 102 10390

60

30

0


Phas

e [de

g]

10/(2 j ) gives 90o

G(s) = 102s +8

Break Point: = 4 [rad/sec]

Low Freq Asymp:

G( j 2) 100 j +8

2.0 dB0o High Freq Asymp:

(|2 j |>> 8)

G( j ) 102 j +0

102

90o

Figure 6: One pole, no zero. Break points:

Freqs at which poles (or zeros) introduce transitions into the Bode plot. Also, the break-point frequency is the frequency at which the real and

imaginary parts of the pole (or zero) are equal. Examples:

Gp (s) =s +5

s +10Break point frequency of zero: = 5 [rad/sec].Break point frequency of pole: = 10 [rad/sec].

Gp (s) =3s +1

0.02 s +1Break point frequency of zero: = 1/3 [rad/sec].Break point frequency of pole: = 50 [rad/sec].



2.1.4 Details of the break point

101 100 101 102 1035040302010

010

Gai

n dB

Break Point

Gain reduce by 1/sqrt(2)

101 100 101 102 10390

60

30

0


Phas

e [de

g]

Phase shift of 45o

Example:

G(s) = 102s +8

The break point is at = 4.0 [rad/sec]

At the break point,the real and imaginaryparts are equal.

Figure 7: The break point is at = 4.0 [rad/sec].

The break point is the frequency at which the real and imaginary parts ofthe term (pole or zero) are equal

The break point marks the transition from where the real term dominatesto where the imaginary part dominates

ConsiderG(s) = 1

s +1The break point is = 1.0 [rad/sec]

G(s = 1.0 j) = 1j +1 =1245o



2.1.5 Example 2, Two poles one zero

G(s) = 40s +160s2 +21s +20 =

40(s +4)(s +1)(s +20) = 8

1s0

14 s +1

(s +1)( 1

20 s +1) (16)

101 100 101 102 103302010

0102030

Gai

n dB

One pole no zeros

101 100 101 102 10390

60

30

0


Phas

e [de

g]

Figure 8: Two poles, one zero.

The two asymptotes are very simple Low Freq Asymptote:

lims j 0

G( j ) = lims j 0

(40s +160

s2 +21s +20

)=

16020 = 18.1dB0

o

High Freq Asymptote:

lims j

G( j ) = lims j

(40s +160

s2 +21s +20

)=

40ss2

=40s

=40j =

40

90o

Three break points: = 1(down), = 4(up), = 20(2x down)Part 2: Frequency Response Analysis (Revised: Mar 26, 2013) Page 16


2.1.6 Example 3, A pole at the origin

G(s) = 40s +160s2 +20s =

40(s +4)s(s +20) = 8

1s1

14 s +1( 1

20 s +1) (17)

101 100 101 102 10320

0

20

40

Gai

n dB

One pole no zeros

Mag=40 (32dB)

Mag=8 (16 dB)

101 100 101 102 10390

60

30

0


Phas

e [de

g]

Figure 9: A Type I system element.

Asymptotes, think about the points where the asymptotes hit = 1.0.

Low Freq Asymptote:

lims j 0

G( j ) = lims j 0

8 1s1

14 s +1( 120 s +1

) = lim0

8 1j =8

90o

High Freq Asymptote (same as example 2):

lims j

G( j ) = lims j

8 1s1

14 s +1( 120 s +1

) = 401s

=40j =

40

90o



Two break points: = 4(up), = 20(down)



2.2 The Bode plot of the loop gain

The Bode plot of elements in series is given by adding the plots of theindividual elements

The loop gain is open loop. It will turn out that we can establish the stability, disturbance rejection, noise

rejection and other properties from the loop gain, without ever solving forthe closed-loop poles.

Example 5, A basic loop

r(t) y(t)+

-

ys(t)

Kc Gc(s) Gp(s)

Hy(s)

e(t) u(t)

Figure 10: A basic control loop.

With:

Kc Gc (s) = 2(s +5)

Gp (s) =1s

Hy (s) =20

s +20

The loop gain is given as:

K GL = Kc Gc Gp Hy =40 (s +5)s (s +20) (18)



101 100 101 102 10360

40

20

0

20

Gai

n dB

Gp(s)=1/s

101 100 101 102 103100

80

60

40

20

0


Phas

e [de

g]

101 100 101 102 10340

30

20

10

0

10

Gai

n dB

Hy(s)=20/(s+20)

101 100 101 102 10390

60

30

0


Phas

e [de

g]

Gp (s) =1s

Hy (s) =20

s +20

Figure 11: Components of the loop gain, the plant and sensor.

101 100 101 102 103100

80604020

020

Gai

n dB

Gc(s) Hy(s)=20/(s

2+20s)

101 100 101 102 103180150120

9060

Frequency [radians/second]Ph

ase

[deg]

Put together the Bodeplots of Gp (s) and Hy (s)

G(s)= 20s2 +20s =

20s(s +20)

Low Freq Asymp:G( j ) = 20/20s

= 190o

High Freq Asymp:G( j ) = 20/s2

= 202

180o

Break Point: = 20 [rad/sec]

Figure 12: Components of the loop gain, combining the plant and sensor.

Note that at high freq. this system has a roll-off of -40 dB per decade (slopeof -2), and has 180o degree of phase lag.



101 100 101 102 103100

80604020

020

Gai

n dB

Gc(s) Hy(s)=20/(s

2+20s)

101 100 101 102 103180150120

9060


Phas

e [de

g]

101 100 101 102 1030

20406080

100120

Gai

n dB

Kc G

c(s)=2s+10

101 100 101 102 1030

306090


Phas

e [de

g]Break point

Gp (s)Hy (s) =1s

20s +20 Kc Gc (s) = 2 (s +5) = 2s +10

Figure 13: Components of the loop gain, the plant and sensor.

101 100 101 102 10340

20

0

20

40

60

Gai

n dB

Loop Gain Kc G

c(s) Gp(s) Hy(s)

101 100 101 102 103100

80

60

40

20

0


Phas

e [de

g]

Some things to notice: The phase angle corre-

sponds to the slope.-20 dB per decadegives -90 [deg]

The zero comes at = 5 [rad/sec], whilethe sensor pole comes at = 20 [rad/sec]

So the +90o of Gc (s)wins out for a rangeof




What the Bode plot would look like if the sensor pole came first (had lowerfrequency) than the PD compensator zero

101 100 101 102 103100

80604020

020

Gai

n dB

101 100 101 102 103180150120

9060


Phas

e [de

g]

101 100 101 102 1030

20406080

100120

Gai

n dB

101 100 101 102 1030

306090


Phas

e [de

g]

Gp (s)Hy (s)=1s

202s +20 Kc Gc (s)= 2 (0.25s +5)= 0.5s+10

Figure 15: Components of the loop gain, faster compensator, slower sensor.

101 100 101 102 10360

40

20

0

20

40

Gai

n dB

Loop Gain Kc G

c(s) Gp(s) Hy(s)

101 100 101 102 103120

100

80

60

40

20


Phas

e [de

g]

Some things to notice: The Bode plots of the

components look quitesimilar,

The bode plot of the loopgain is quite different(especially the phase)

Conclusion:

The relative speed ofpoles and zerosmatters




2.3 Bode standard form for the loop gain

The loop gain of a linear system can be written in a standard form, calledBode form:

K GL (s) = KBode1

snoG

L (s)

= KBode1

sno

(1 s +1) (2 s +1) (a s +1) (b s +1) (s2/2n +2s/n +1)

(19)

where

K GL (s) is the loop gain

KBode is the DC gain of the loop, after poles or zeros at the orgin arepulled out using the 1/sno term

sno , no is the number of poles at the origin; no is the system type

GL (s) is the transfer function of poles and zeros not at the origin,scaled so that the DC gain of GL (s) is 1.0.

1 , 2 , ...[seconds] are time constants of the (real) zerosa , b , ...[seconds] are time constants of the (real) polesn [radians/second] and [] are the frequency and damping of acomplex pole pair

Eqn (19) is composed of three classes of terms:

1. KBode ( j )no , a constant term and poles or zeros at the origin2. ( j +1)1 , first order terms

3.[( j

n

)2+2

( j n

)+1

]1, second order terms.

We will see how each of these terms contributes to a Bode plot.



Example,

K GL (s) = Kc Gc (s) Gp (s) Hy (s) = 2 (s +5)1s

20s +20 = 2

1s

20(s +5)s +20(20)

1. Pull out any singularities, poles or zeros, at the origin

K GL (s) =1s

220 (s +5)

s +20 , so no = 1

no = 1 indicates there is a pole at the origin.

2. Scale the polynomical terms of GL (s) to have constant coefficients of1.0, putting the loop gain in Bode standard form:

K GL (s) = KBode1

sno

b(s)a(s)

=520

201s

(15 s +1

)1

20s +1= 5.0 1

s

(15s +1

)( 1

20s +1)

3. Identify the time constants

K GL (s) = KBode1

sno

(1 s +1)(a s +1)

= 5.0 1s

(15 s +1

)( 1

20 s +1) (21)

which gives 1 = 15 [seconds], a = 120 [seconds] , KBode = 5.0 and no = 1

Eqn (21) is the loop gain example (20), expressed in Bode standard form.



2.3.1 Plotting terms of the form KBode ( j )no A

This term combines the DC gain with poles or zeros at the origin. The magnitude curve is a straight line, with a slope of no20 dB / decade The phase curve is a flat line, at an angle of 90no degrees Examples of KBode ( j )no :

101 100 101 102 10340

20

0

20

40

60

Gai

n dB

101 100 101 102 103100

80

60

40

20

0


Phas

e [de

g]

101 100 101 102 10360

40

20

0

20

40G

ain

dB

101 100 101 102 1030

20

40

60

80

100


Phas

e [de

g]

KBode ( j )no = 12s1 , no = +1 KBode ( j )no = 0.05s+1, no =1

101 100 101 102 10310

0

10

20

30

40

Gai

n dB

101 100 101 102 10350

0

50


Phas

e [de

g]

KBode ( j )no = 4 , no = 0

Figure 17: First term of the loop gain

The most common cases are: no = 0 (Type 0 system)no = 1 (Type I system)



2.3.2 Contrasting KBode with Krlg

Given an example TFGL (s) =

2s +104s2 +3s +8 (22)

we can write the TF in two standard forms:

Bode Form:

K GL (s)= KBode1

sno

(1 s +1) (2 s +1) (a s +1) (b s +1) (s2/2n +2s/n +1)

(19, repeated)

For example:

GL (s) = KBode1

snoG

L (s) =

108

1s0

(15)

s +1(12)

s2 +(3

8)

s +1(23)

Root Locus Form:

Krlg(s z1) (s z2)

(s pa) (s pb) (s2 +2n s +2n)

where z1 =1/1, z2 =1/2, pa =1/a, pb =1/b, etc. For example:

GL (s) = Krlg GL (s) =

24

s +5s2 +

(34)

s +2(24)

For making bode plots, we use Bode form and determine KBode For determining K from a root-locus plot we use root-locus form and

determine Krlg.



2.3.3 Plotting terms of the form ( j +1) (a real pole or zero)

First-order terms are understood by their asymptotes and the break point

101 100 10130

20

10

0

10

Gai

n dB Break Point

101 100 10190

60

30

0


Phas

e [de

g]

Break Point

Figure 18: First-order asymptotes, a = 0.5 [seconds].

The first-order term is

G1 (s) =1

a s +1=

1j a +1

Low-frequency asymptote, | j a|> 1

G1 (s) =1

a s +1 1j a =

1a

90o (26)

The high-frequency asymptote has

A slope of 1 ,

A phase of 90o

Intercepts = 1 at 20 log10 (1/a) (shown at triangle)



Characteristics of the first-order break point

101 100 10130

20

10

0

10

Gai

n dB Break Point

101 100 10190

60

30

0


Phas

e [de

g]

Break Point

Figure 20: First-order term, a = 0.5 [seconds].

The break point is at bp a = 1, or bp = 1/a At the break point:

G1 (s) =1

j a +1 =1

j 1+1 =12

45o (27)

The slope of the phase curve at = bp gives the phase transition on therange

=[0.2bp ... 5bp

](a 25 : 1 range of frequency !)

On the magnitude curve, the break point is down (or up, for a zero)by 1/

2 or -3 dB



A zero has the same characteristics as a pole, but with the opposite sense.

100 101 102 103604020

0204060

Gai

n dB

Intercept

100 101 102 1030

30

60

90


Phas

e [de

g]

Break Point

Figure 21: First-order asymptotes, 1 = 0.02 [seconds].

The first-order term for a zero is:

G1 (s) = (s1 +1) (28)

( j 1 +1) has a low-frequency asymptote of 10o

( j 1 +1) has a high-frequency asymptote of 1+90o

The high-frequency asymptote has a slope of +1

The intercept of the high-frequency asymptote is |1|dB at = 1.0

The break point is at bp 1 = 1 []

The phase angle at the break point is +450



2.3.4 Plotting a complex pole or zero pair

Second-order terms are also understood by their asymptotes and break point

100 101 10240

20

0

20

Gai

n dB

Break Point Intercept

100 101 102180150120

906030

0


Phas

e [de

g]

180o degrees

0o degrees

Break Point

Figure 22: Second-order asymptotes, n = 4.0 [radians/second], = 0.2 [] .

The second-order term is

G(s) = 1( j n

)2+2

( j n

)+1

=1

(

n

)2+2 j

(n

)+1

Low-frequency asymptote, | j /n|> 1:

G(s) = 1

(

n

)2+2 j

(n

)+1

1(

n

)2 =(n

)2180o (30)

The high-frequency asymptote has

A slope of -2,

A phase of 180o

Intercepts = 1 at 20 log10(2n) (shown at triangle)



Characteristics of the second-order break point

10140

20

0

20

Gai

n dB

Break Point, 20 log10(1/2)

101180150120

906030

0


Phas

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