Download 21. Frequency Response

Lecture 21. Frequency Response
• Frequency Response
• Frequency Response of R, L and C
• Resonance circuit
• Examples
1
Transfer Function
• Recall that the transfer function, H(s), is
Y ( s) Output
H ( s) 

X( s )
Input
• The transfer function can be shown in a block diagram
as
X(j) ejt = X(s) est
Y(j) ejt = Y(s) est
H(j) = H(s)
• The transfer function can be separated into magnitude
and phase angle information
H(j) = |H(j)| H(j)
2
Variable-Frequency Response Analysis
• As an extension of AC analysis, we now vary the frequency and
observe the circuit behavior
• Graphical display of frequency dependent circuit behavior can be
very useful
Frequency Response of an
Audio Amplifier
3
Frequency Response of a Resistor
• Consider the frequency dependent impedance of the resistor,
inductor and capacitor circuit elements
• Resistor (R):
ZR = R 0°
Phase of ZR (°)
Magnitude of ZR ()
– So the magnitude and phase angle of the resistor impedance are
constant, such that plotting them versus frequency yields
R
Frequency
0°
Frequency
4
Frequency Response of an Inductor
• Inductor (L):
ZL = L 90°
Phase of ZL (°)
Magnitude of ZL ()
– The phase angle of the inductor impedance is a constant 90°,
but the magnitude of the inductor impedance is directly
proportional to the frequency. Plotting them vs. frequency yields
(note that the inductor appears as a short circuit at dc)
Frequency
90°
Frequency
5
Frequency Response of a Capacitor
• Capacitor (C):
ZC = 1/(C) –90°
Phase of ZC (°)
Magnitude of ZC ()
– The phase angle of the capacitor impedance is –90°, but the
magnitude of the inductor impedance is inversely proportional to
the frequency. Plotting both vs. frequency yields (note that the
capacitor acts as an open circuit at dc)
-90°
Frequency
Frequency
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Poles and Zeros
• The transfer function is a ratio of polynomials
N( s ) K ( s  z1 )( s  z2 )  ( s  zm )
H( s) 

D( s) ( s  p1 )( s  p2 )  ( s  pn )
• The roots of the numerator, N(s), are called the zeros
since they cause the transfer function H(s) to become
zero, i.e., H(zi)=0
• The roots of the denominator, D(s), are called the poles
and they cause the transfer function H(s) to become
infinity, i.e., H(pi)=
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RLC Resonance Circuits
i(t)

+ vr(t) –
R
+
–
+
C
– vl(t) +
L

I ( s) s 2  20 s  02  Vs ( s)
I ( s)
1
H ( s) 
 2
2
V
(
s
)
s

2

s


vc(t)
s
0
0
– • 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
8
RLC Resonance Circuits
1
| H ( s ) | 2
2 2
2
(  0 )  4(0 )
0 ~
BW
1
- Resonance Frequency
LC
- BW=bandwidth
Q
0
BW
9
Quality Factor (Q)
• The quality factor is a measure of the sharpness of the
resonance peak; the larger the Q value, the sharper the
peak, where BW=bandwidth
Q
0
BW
• An energy analysis of a RLC circuit provides a basic
definition of the quality factor (Q) that is used across
engineering disciplines, specifically
WS
Max Energy Stored at  0
Q  2
 2
WD
Energy Dissipated per Cycle
10
Bandwidth (BW)
• The bandwidth (BW) is the difference between the two
half-power frequencies
BW = ωHI – ωLO = 0 / Q
• Hence, a high-Q circuit has a small bandwidth
• Note that: 02 = ωLO ωHI
 LO &  HI
 1
 0 

 2Q

1
 1
2
2Q  
11
Quality Factor: RLC Circuits
• For a series RLC circuit the quality factor is
Q
0
BW
 Qseries
0 L
1
1 L



R
 0 CR R C
• For a parallel RLC circuit, the quality factor is
Q
0
BW
 Q parallel
R
C

  0 CR  R
0 L
L
12
Class Examples
• Drill Problems P9-3, P9-4, P9-5
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