lect16eee 2021 system responses dr. holbert march 24, 2008
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Lect16 EEE 202 1
System Responses
Dr. Holbert
March 24, 2008
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Lect16 EEE 202 2
Introduction
• Today, we explore in greater depth the three cases for second-order systems– Real and unequal poles– Real and equal poles– Complex conjugate pair
• This material is ripe with new terminology
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Lect16 EEE 202 3
Second-Order ODE
• Recall: the second-order ODE has a form of
• 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
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Lect16 EEE 202 4
Second-Order ODE
• 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
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Lect16 EEE 202 5
Damping Ratio () andNatural Frequency (0)
• The damping ratio is ζ• The damping ratio determines what type of
solution we will obtain:– Exponentially decreasing ( >1)– Exponentially decreasing sinusoid ( < 1)
• The undamped natural frequency is 0
– Determines how fast sinusoids wiggle– Approximately equal to resonance frequency
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Lect16 EEE 202 6
Characteristic Equation Roots
The roots of the characteristic equation determine whether the complementary (natural) solution wiggles
1
1
2002
2001
s
s
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Lect16 EEE 202 7
1. Real and Unequal Roots
• If > 1, s1 and s2 are real and not equal
• This solution is overdamped
tt
c eKeKtx
1
2
1
1
200
200
)(
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Lect16 EEE 202 8
Overdamped
0
0.2
0.4
0.6
0.8
1
0.0E+00 5.0E-06 1.0E-05
Time
i(t)
-0.2
0
0.2
0.4
0.6
0.8
0.0E+00 5.0E-06 1.0E-05
Time
i(t)
Both of these graphs have a response of the formi(t) = K1 exp(–t/τ1) + K2 exp(–t/τ2)
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Lect16 EEE 202 9
2. Complex Roots
• 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
Frequency) Natural (Damped
t)Coefficien (Damping
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Lect16 EEE 202 10
Underdamped
-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)
A curve having a response of the formi(t) = e–t/τ [K1 cos(ωt) + K2 sin(ωt)]
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Lect16 EEE 202 11
3. Real and Equal Roots
• If = 1, then s1 and s2 are real and equal
• This solution is critically damped
ttc etKeKtx 00
21)(
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Lect16 EEE 202 12
IF Amplifier Example
This is one possible implementation of the filter portion of an intermediate frequency (IF) amplifier
10
769pFvs(t)
i(t)
159H
+–
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Lect16 EEE 202 13
IF Amplifier Example (cont’d.)
• The ODE describing the loop current is
• For this example, what are ζ and ω0?
)()()(
2)(
)(1)(
1)()(
2002
2
2
2
tftidt
tdi
dt
tid
dt
tdv
Lti
LCdt
tdi
L
R
dt
tid s
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Lect16 EEE 202 14
IF Amplifier 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
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Lect16 EEE 202 15
IF Amplifier Example (cont’d.)
• The shape of the current depends on the initial capacitor voltage and inductor current
-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)
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Lect16 EEE 202 16
Slightly Different Example
• Increase the resistor to 1k• Exercise: what are and 0?
1k
769pFvs(t)
i(t)
159H
+–
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Lect16 EEE 202 17
Different Example (cont’d.)
• 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?
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Lect16 EEE 202 18
Different Example (cont’d.)
• The shape of the current depends on the initial capacitor voltage and inductor current
0
0.2
0.4
0.6
0.8
1
0.0E+00 5.0E-06 1.0E-05
Time
i(t)
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Lect16 EEE 202 19
Damping Summary
Damping Ratio
Poles (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
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Lect16 EEE 202 20
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– Constant sources give DC steady-state responses
• DC steady-state if response approaches a constant
– Sinusoidal sources give AC steady-state responses• AC steady-state if response approaches a sinusoid
• The transient response is the circuit response minus the steady-state response
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Lect16 EEE 202 21
Transient and Steady-State Responses
• Consider a time-domain response from an earlier example this semester
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
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Lect16 EEE 202 22
Class Examples
• Drill Problems P7-6, P7-7, P7-8