Transcript System Responses (7-3 to 7-5)

System Responses
Dr. Holbert
March 24, 2008
Lect16
EEE 202
1
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
Lect16
EEE 202
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Second-Order ODE
• Recall: the second-order ODE has a form of
d 2 x(t )
dx(t )
2
 20
 0 x(t )  f (t )
2
dt
dt
• For zero-initial conditions, the transfer function
would be


X( s) s 2  20 s  02  F( s)
X( s )
1
H (s) 
 2
F( s) s  20 s  02
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EEE 202
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Second-Order ODE
• The denominator of the transfer function is
known as the characteristic equation
• To find the poles, we solve :
s  2 0 s    0
2
2
0
which has two roots: s1 and s2
 20  (20 ) 2  402
s1 , s2 
 0  0  2  1
2
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Damping Ratio () and
Natural 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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EEE 202
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Characteristic Equation Roots
The roots of the characteristic equation
determine whether the complementary
(natural) solution wiggles
2
s1  0  0   1
s2  0  0  2  1
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EEE 202
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1. Real and Unequal Roots
• If  > 1, s1 and s2 are real and not equal
xc (t )  K1e
    2 1  t
0
0


 K 2e
    2 1  t
0
0


• This solution is overdamped
Lect16
EEE 202
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1
0.8
0.8
0.6
i(t)
i(t)
Overdamped
0.6
0.4
0.2
0
0.0E+00
0.4
0.2
0
5.0E-06
1.0E-05
-0.2
0.0E+00
5.0E-06
1.0E-05
Time
Time
Both of these graphs have a response of the form
i(t) = K1 exp(–t/τ1) + K2 exp(–t/τ2)
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2. Complex Roots
• If  < 1, s1 and s2 are complex
• Define the following constants:
  0 (Damping Coefficient)
d  0 1   2 (Damped Natural Frequency)
s1 , s2    j d
xc (t )  e  t  A1 cos d t  A2 sin d t 
• This solution is underdamped
Lect16
EEE 202
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Underdamped
1
0.8
0.6
i(t)
0.4
0.2
0
-1.00E-05
-0.2
1.00E-05
3.00E-05
-0.4
-0.6
-0.8
-1
Time
A curve having a response of the form
i(t) = e–t/τ [K1 cos(ωt) + K2 sin(ωt)]
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3. Real and Equal Roots
• If  = 1, then s1 and s2 are real and equal
xc (t )  K1e
 0t
 K2 t e
 0t
• This solution is critically damped
Lect16
EEE 202
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IF Amplifier Example
i(t)
10W
vs(t)
+
–
769pF
159mH
This is one possible implementation of the filter
portion of an intermediate frequency (IF) amplifier
Lect16
EEE 202
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IF Amplifier Example (cont’d.)
• The ODE describing the loop current is
2
d i (t ) R di (t ) 1
1 dvs (t )


i (t ) 
2
dt
L dt
LC
L dt
2
d i (t )
di (t )
2
 20
 0 i (t )  f (t )
2
dt
dt
• For this example, what are ζ and ω0?
Lect16
EEE 202
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IF Amplifier Example (cont’d.)
1
1
 

 0  2.86 106 rad/sec
LC (159 μH)(769 pF)
2
0
R
10 W
20  
   0.011
L 159 μH
• Note that 0 = 2pf = 2p 455,000 Hz)
• Is this system overdamped, underdamped, or
critically damped?
• What will the current look like?
Lect16
EEE 202
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IF Amplifier Example (cont’d.)
i(t)
• The shape of the current depends on the
initial capacitor voltage and inductor
current
1
0.8
0.6
0.4
0.2
0
-0.2
-1.00E-05
-0.4
-0.6
-0.8
-1
1.00E-05
3.00E-05
Time
Lect16
EEE 202
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Slightly Different Example
i(t)
1kW
vs(t)
+
–
769pF
159mH
• Increase the resistor to 1kW
• Exercise: what are  and 0?
Lect16
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Different Example (cont’d.)
• The natural (resonance) frequency does
not change: 0 = 2p455,000 Hz)
• But the damping ratio becomes  = 2.2
• Is this system overdamped, underdamped,
or critically damped?
• What will the current look like?
Lect16
EEE 202
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Different Example (cont’d.)
• The shape of the current depends on the
initial capacitor voltage and inductor
current 1
i(t)
0.8
0.6
0.4
0.2
0
0.0E+00
5.0E-06
1.0E-05
Time
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Damping Summary
Damping
Poles (s1, s2)
Ratio
ζ>1
Real and unequal
ζ=1
Real and equal
Damping
Overdamped
Critically damped
0 < ζ < 1 Complex conjugate
Underdamped
pair set
ζ=0
Purely imaginary pair Undamped
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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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EEE 202
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Transient and Steady-State
Responses
• Consider a time-domain response from an earlier
example this semester
1.2
1
5 5  2 t 10 3t
f (t )   e  e
6 2
3
0.8
0.6
0.4
0.2
Steady
State
Response
Lect16
Transient
Response
0
0
1
2
Transient
Response
EEE 202
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4
5
Steady-State
Response
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Class Examples
• Drill Problems P7-6, P7-7, P7-8
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