• Poles & Zeros• Second-Order Circuits• LCR Oscillator circuit: An example• Transient and Steady States

Lecture 17. System Response II

1

2

Pole-Zero Plot

• For a pole-zero plot place "X" for poles and "0" for zeros using real-imaginary axes

• Poles directly indicate the system transient response features

• Poles in the right half plane signify an unstable system

• Consider the following transfer function

)5.1)(4)(5(

)54)(5.3)(3()(

2

2

sss

sssssH

Re

Im

3

Second-Order Circuits

R

C

+

–

vc(t)

+ –vr(t)

L

+– vl(t)

i(t)

+–

dt

tdv

dt

tidLti

Cdt

tdiR s )()(

)(1)(

2

2

• KVL around the loop:

vr(t) + vc(t) + vl(t) = vs(t)

)()(

)(1

)( tvdt

tdiLdxxi

CtiR s

t

dt

tdv

Lti

LCdt

tdi

L

R

dt

tid s )(1)(

1)()(2

2

4

Second-Order Circuits

• For zero-initial conditions, the transfer function would be

200

2

200

2

2

1

)(

)()(

)(2)(

sss

ssH

ssss

F

X

FX

)()()(

2)( 2

002

2

tftxdt

tdx

dt

txd

• In general, a second-order circuit is described by

Characteristic Equation & Poles

• The denominator of the transfer function is known as the characteristic equation

• To find the poles, we solve :

which has two roots: s1 and s2

02 200

2 ss

12

4)2(2, 2

00

20

200

21

ss

200

2 2

1

)(

)()(

sss

ssH

F

X

6

Real and Unequal Roots: Overdamped

• If > 1, s1 and s2 are real and not equal

• The amplitude decreases exponentially over time. This solution is overdamped

tt

c eKeKtx

1

2

1

1

200

200

)(

0

0.2

0.4

0.6

0.8

1

0.0E+00 5.0E-06 1.0E-05

Time

i(t)

7

Complex Roots: Underdamped

• If < 1, s1 and s2 are complex

• Define the following constants:

• This solution is underdamped

tAtAetx

jss

ddt

c

d

d

sincos)(

,

1

21

21

20

0

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1.00E-05 1.00E-05 3.00E-05

Time

i(t)

8

Real and Equal Roots

• If = 1, then s1 and s2 are real and equal

• This solution is critically damped

ttc etKeKtx 00

21)(

9

An Example

• This is one possible implementation of the filter portion of an intermediate frequency (IF) amplifier

10769pFvs(t)

i(t)

159H

+–

)()()(

2)(

)(1)(

1)()(

2002

2

2

2

tftidt

tdi

dt

tid

dt

tdv

Lti

LCdt

tdi

L

R

dt

tid s

10

An Example (cont’d.)

• Note that 0 = 2f = 2455,000 Hz)

• Is this system overdamped, underdamped, or critically damped?• What will the current look like?

011.0μH159

102

rad/sec1086.2)pF769)(μH159(

11

0

60

20

L

R

LC

-1

-0.8

-0.6

-0.4

-0.2

00.2

0.4

0.6

0.8

1

-1.00E-05 1.00E-05 3.00E-05

Time

i(t)

11

An Example (cont’d)

• Increase the resistor to 1k• Exercise: what are and

0?

1k

769pFvs(t)

i(t)

159H

+–

• The natural (resonance) frequency does not change: 0 = 2455,000 Hz)

• But the damping ratio becomes = 2.2• Is this system overdamped, underdamped, or critically damped?• What will the current look like?

12

A SummaryDamping

RatioPoles (s1, s2) Damping

ζ > 1 Real and unequal Overdamped

ζ = 1 Real and equal Critically damped

0 < ζ < 1 Complex conjugate pair set Underdamped

ζ = 0 Purely imaginary pair Undamped

13

Transient and Steady-State Responses

• The steady-state response of a circuit is the waveform after a long time has passed, and depends on the source(s) in the circuit

13

tt eetf 32

3

10

2

5

6

5)(

SteadyState

Response

TransientResponse

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5

TransientResponse

Steady-StateResponse

14

Class Examples

• P7-6, P7-7, P7-8