Download 17. System Responses II

Lecture 17. System Response II
• Poles & Zeros
• Second-Order Circuits
• LCR Oscillator circuit: An example
• Transient and Steady States
1
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
(s  3)( s  3.5)( s 2  4s  5)
H ( s) 
(s  5)( s 2  4)( s  1.5)
Im
Re
2
Second-Order Circuits
i(t)
• KVL around the loop:
vr(t) + vc(t) + vl(t) = vs(t)
+ vr(t) –
R
+
–
+
vc(t)
C
– vl(t) +
L
t
–
1
di (t )
R i (t )   i ( x)dx  L
 vs (t )
C 
dt
di(t ) 1
d 2i(t ) dvs (t )
R
 i(t )  L

2
dt
C
dt
dt
d 2i(t ) R di(t ) 1
1 dvs (t )


i(t ) 
2
dt
L dt
LC
L dt
3
Second-Order Circuits
• In general, a second-order circuit is described by
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
4
Characteristic Equation & Poles
X( s )
1
H ( s) 
 2
F( s ) s  20 s  02
• The denominator of the transfer function is known as the
characteristic equation
• To find the poles, we solve :
s 2  2 0 s   02  0
which has two roots: s1 and s2
 20  (20 ) 2  402
s1 , s2 
 0  0  2  1
2
Real and Unequal Roots: Overdamped
• 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


• The amplitude decreases exponentially over time. This solution is
overdamped
1
i(t)
0.8
0.6
0.4
0.2
0
0.0E+00
5.0E-06
Time
1.0E-05
6
Complex Roots: Underdamped
1
• If  < 1, s1 and s2 are complex
• Define the following constants:
 d  0 1   2
s1 , s2    j d
0.4
i(t)
  0
0.8
0.6
0.2
0
-1.00E-05
-0.2
1.00E-05
3.00E-05
-0.4
-0.6
-0.8
-1
Time
xc (t )  e  t  A1 cos d t  A2 sin d t 
• This solution is underdamped
7
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
8
An Example
i(t)
10W
vs(t)
+
–
769pF
• This is one possible
implementation of the filter
portion of an intermediate
frequency (IF) amplifier
159mH
d 2i (t ) R di(t ) 1
1 dvs (t )


i
(
t
)

dt 2
L dt
LC
L dt
d 2i (t )
di(t )
2

2



0
0 i (t )  f (t )
2
dt
dt
9
An Example (cont’d.)
1
1
 

 0  2.86 106 rad/sec
LC (159 μH)(769 pF)
R
10 W
20  
   0.011
L 159 μH
2
0
i(t)
• Note that 0 = 2pf = 2p 455,000 Hz)
• Is this system overdamped, underdamped, or critically damped?
• What will the current look like?
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
10
Time
An Example (cont’d)
i(t)
1kW
vs(t)
+
–
769pF
• Increase the resistor to 1kW
• Exercise: what are  and
0?
159mH
• 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?
11
A Summary
Damping
Ratio
Poles (s1, s2)
Damping
ζ>1
Real and unequal
Overdamped
ζ=1
Real and equal
Critically damped
Complex conjugate pair set
Underdamped
Purely imaginary pair
Undamped
0<ζ<1
ζ=0
12
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
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
Transient
Response
0
0
1
Transient
Response
2
3
4
5
Steady-State
Response
13
Class Examples
• P7-6, P7-7, P7-8
14