• 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