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EE40 Lec 12Transfer Function Bode Plots FiltersTransfer Function, Bode Plots, Filters
Prof. Nathan Cheung
10/08/2009
Reading: Hambley Chapter 6.1-6.5
Slide 1EE40 Fall 2009 Prof. Cheung
OUTLINE
– Fourier Analysis Concept– Transfer Function – dB scale– First Order Low-Pass and High-Pass Filters– Bode PlotsBode PlotsLog magnitude vs log frequency plotLinear Phase vs log frequency plotLinear Phase vs log frequency plot– Asymptotic Frequency Behavior
Slide 2EE40 Fall 2009 Prof. Cheung
Fourier Synthesis of Waveforms
Any waveform can be constructed by adding sinusoids that have the proper amplitudes, f i d h
Example: Fourier Components of A Square Wave
frequencies, and phase.
p p q
Slide 3EE40 Fall 2009 Prof. Cheung
Transfer Function
• Transfer function is a function of frequency– A complex quantity– Both magnitude and phase are function of
frequency
Two Port filter networkVin Vout
( )VV ( )( )
( )
outout in
in
VfV
H f
θ θ
θ
= = ∠ −
∠
out
in
VHV
H(f)
Slide 4EE40 Fall 2009 Prof. Cheung
( )H f θ= ∠H(f)
Example: Response of a Two-Frequency Signal
H(f1) ≠ H(f2)i lin general
Slide 5EE40 Fall 2009 Prof. Cheung
Cascade of Two-Port Networks
( ) ( ) ( )fHfHfH 21 ×=
Slide 6EE40 Fall 2009 Prof. Cheung
Origin of Decibel (dB) unitThe bel (abbreviated B) is named after Alexander Graham Bell, who did much pioneering work with sound and the way o r ears respond to so nd O r earsour ears respond to sound. Our ears respond to sounds ranging from an intensity less than 10 -16 W/cm2 to intensities larger than 10 -4 W/cm2intensities larger than 10 4 W/cm(where we begin to experience pain). This is a range of more than 1x1012 times from the softest to the loudest sounds. Logarithm provide a convenient way to represent these values, because they compress this scale into a range of 12, rather than a range of a billion. A bel is defined as the logarithm of a power ratio. It gives us a way to compare power l l i h h h d i h
Slide 7EE40 Fall 2009 Prof. Cheung
levels with each other and with some reference power.
The Decibel (dB) Unit
• A bel (symbol B) is a unit of measure of ratios of power levels
B = log (P /P ) where P and P are– B = log10(P1/P2) where P1 and P2 are power levels.
– one bel corresponds to a ratio of 10:1 10dB• The bel is too large for everyday
use, so the decibel (dB), equal to 0.1B, is more commonly used sc
ale
, y1dB = 10 log10(P1/P2)
Log 1
0s
1B• dB are used to measure
– Electric power, Gain or loss of amplifiers, transmission loss of optical
1B
Slide 8EE40 Fall 2009 Prof. Cheung
amplifiers, transmission loss of optical fibers…..
Decibel Exercise
• Exercise: – Express a power of 50 mW in decibels relative to 1 watt.
10 l (50 10 3) 13 dB10 log10 (50 x 10-3) = - 13 dB
• Exercise:• Exercise: – Express a power of 100 mW in decibels relative to 2 mW.
10 log10 (50) = 17 dB. g10 ( )
• dBm to express absolute values of power relative to a milliwatt. – dBm = 10 log10 (power in milliwatts / 1 milliwatt)
100 mW = 20 dBm
Slide 9EE40 Fall 2009 Prof. Cheung
– 100 mW = 20 dBm– 10 mW = 10 dBm
Decibel (dB) as voltage or current ratios
22 IVP ∝∝1
2
1 Vlog20Vlog10 =
ale)20dB
22 Vlog20
Vlog10 =
g 10
sca
1
2
1 Ilog20Ilog10 =
or I
( lo
1B22 I
log20I
log10
V
o1B
Slide 10EE40 Fall 2009 Prof. Cheung
Note: 1 Bel of voltage yields 2 Bels of Power
Some useful dB values to memorize
|H(f)| |H(f)|dB
100 4010 201 01 0
0.1 -200 01 400.01 -40
2 6√2 3√2 3
1/√2 -35 (=10/2) 14 (=20-6)
Slide 11EE40 Fall 2009 Prof. Cheung
( ) ( )
Bode PlotsA Bode plot is a straight line approximation of H(ω)
• Plot of transfer function magnitude vs. frequency
i i th 20 l f th it d f th t f– y-axis is the 20•log of the magnitude of the transfer function in dB and x-axis is ω
– x -axis is ω or f in log scalex axis is ω or f in log scale
dB
log(f) or log(ω)
Slide 12EE40 Fall 2009 Prof. Cheung
Bode Plots• Plot of transfer function phase vs. frequency
– y-axis is the phase of transfer function in degrees (li l )(linear scale)
– x -axis is ω or f in log scale
degrees
log(f) or log(ω)
Slide 13EE40 Fall 2009 Prof. Cheung
Decade , Octave and Log Scale
Log10 scale
How about the Log ω axis ?
Shifted by Log (2 ) or 0 798Shifted by Log10(2π) or 0.798
Note: Log x•y = Log x + Log y !!ω =2πf (rad/sec)
Slide 14EE40 Fall 2009 Prof. Cheung
g y g g y
Frequency Response
• The shape of the frequency response of the complex ratio of phasors VOUT/VIN is a convenient means of classifying a circuit behavior and identifying keyclassifying a circuit behavior and identifying key parameters.
OUT
VV
GainBreak point OUT
VV
G i
Break point
Low Pass
INV Gain INV
High Pass
Gain
Frequency
Low Pass
Frequency
High Pass
Slide 15EE40 Fall 2009 Prof. Cheung
*These are log ratio vs log frequency plots
A Quick Quiz: High Pass or Low Pass?
Slide 16EE40 Fall 2009 Prof. Cheung
Example: Low Pass CircuitR2
++A = 100
R1 = 100,000 Ω
−
+AVT
R1+
VTVOUT
CVIN R2 = 1000 Ω
C = 10 uF
OUT
VV)(HFunctionTransfer =ω
C = 10 uF
INV
cOUT AZ=
V/100C/1i
]/[j1A)(H
Bωω+=ω
AA)jwC/1(AV
cRIN ZZ +V s/rad100CR/1with 2B ==ω
Slide 17EE40 Fall 2009 Prof. Cheung
)]CR/1/(j1[A
)CRj1(A
)Cj/1R)jwC/1(A
VV
222IN
OUT
ω+=
ω+=
ω+=
First Order Bode Magnitude Plot
• The break frequency ωΒ describes the frequency where the trends on the Bode plot are changed
12/12
B ])/(1[1log20Alog20)(Hlog20ωω+
+=ω
dB dB
40 dB
0 dBωB ωB
0 dB
-20dB/dec
Slide 18EE40 Fall 2009 Prof. Cheung
Bode Magnitude Plot – First-order Low-Pass Filter1
dB2/12
B ])/(1[1log20Alog20)(Hlog20ωω+
+=ω
10010040 ωB = 100A
2100
|1|100
=+ j
0
20Actual value =
0
-2010 100 10001 Radian
Frequency
logω10000
Slide 19EE40 Fall 2009 Prof. Cheung
Slope = -20dB/dec or -6dB/octave for ω>> ωB
Bode Phase Plot – First-order Low-pass Filter
180)/(tan
]/[j1A)(H B
1
B
ωω=ωω+
∠=ω∠ −
90
Phas
e s/rad100CR/1with 2B ==ω
-90
010 100 10001 Radian
Frequency (log scale)
-180
4545001000100 ==∠
=∠ PhasePhase
Actual value at ωB is -45o
Slide 20EE40 Fall 2009 Prof. Cheung
45450452
|1|
−=−=∠
=+
Phasej
Phase
Bode Plot: Actual versus Straight Line Approximation
First Order Low-Pass Filter
Magnitude Phase
Slide 21EE40 Fall 2009 Prof. Cheung
Bode Plot of a High-Pass Filter
( )( )
cout
/j1/j
RLjLj
VV)(H ωω
=ω
==ω LR/=ω( )cs /j1RLjV)(
ωω++ω LRc /ωPole: will cause decrease by
20db/dec after ω
Slide 22EE40 Fall 2009 Prof. Cheung
20db/dec after ωc=> so magnitude is flat after ωc
Bode Magnitude Plot of a High-Pass Filter
( ) ( ) |/j1
1|log20|/j|log20|)(H|log20c
c ωω++ωω=ω
Slide 23EE40 Fall 2009 Prof. Cheung
Bode Phase Plot of a High-Pass Filter
( ) 1/j
)(H
∠+ωω∠=
ω∠
( ) ( )
( )o
cc
190
/j1/j
∠+=
ωω+∠+ωω∠=
( )c/j190
ωω+∠+=
Slide 24EE40 Fall 2009 Prof. Cheung
Poles and Zeros
• The transfer function H(ω) = M(ω) ejφ(ω)
can be written as:H(ω) = A1(ω) A2(ω) … An(ω)H(ω) A1(ω) A2(ω) … An(ω)
Zero
( ) ( )
⋅ωω+⋅=ωω+
•=ω z 1/j110)/(j110)(H
ZeroExample
( ) ( )
ωω+
⋅ωω+⋅=ωω+
•=ωp
zp /j1
/j110)/(j1
10)(H
Slide 25EE40 Fall 2009 Prof. Cheung
Pole
Poles and Zeros
• With H(ω) = A1(ω) A2(ω) … An(ω)
MagnitudeM [dB] = 20 log|A1| + 20 log|A2| + … + 20 log|An|
= A [dB] + A [dB] + + A [dB]= A1 [dB] + A2 [dB] + … + An [dB]
Phaseφ(ω) = φA1(ω) + φA2(ω) + … + φAn(ω)
Thi t B d l t ( it d d
Slide 26EE40 Fall 2009 Prof. Cheung
• This means we can generate Bode plot (magnitude and phase) by adding the poles and zeros!
Poles And Zero Exercise
• Generate the Magnitude Bode Plots of H(ω).
( )( )ω+ω+ω
j1000j10010)(H( )( )( )( )ω+ω+
=ωj10j10
)(H 4
Slide 27EE40 Fall 2009 Prof. Cheung
( )( )Poles And Zero Exercise
( )( )( )( )
( )( )4
1000/j1100/j1100010010j10j10
j1000j10010)(H
ω+ω+••ω+ω+
ω+ω+=ω
( )( )( )( )
( )( )44
1000/j1100/j11010/j110/j11010
1000/j1100/j1100010010
ω+ω+ω+ω+•
ω+ω+••=
( )( )( )( )410/j110/j1
jjω+ω+
=
Slide 28EE40 Fall 2009 Prof. Cheung
A More Challenging Bode Plot Example
( ) ( )( )50/j14000j
5/j1400H2
ω+•ω•ω+•
=ω
Zeros: Two zeros at 5 rad/s ( a double zero)
( )50/j14000j ω+•ω•
( )Poles: one pole at 0 rad/s and one at 50 rad/s;
1 |400/j4000| l t1. |400/j4000| plotRatio = 1/10 => -20dB constant for all ω
2. |1/ω| plot| | plog 0 does not show up in log scale ( -∞ !!! )However we know magnitude is decreasing at -20 dB/dec and equals -20 dB at 1 rad/s. We can now
Slide 29EE40 Fall 2009 Prof. Cheung
dB/dec and equals 20 dB at 1 rad/s. We can now draw the 1/ω plot around ω =1
A More Challenging Bode Plot Example
( ) ( )( )50/j14000j
5/j1400H2ω+•
=ω( ) ( )50/j14000j ω+•ω•
3. |(1+jω/5)2| plot0dB below 5 rad/sslope= +40 dB/dec beyond 5 rad/sslope= +40 dB/dec beyond 5 rad/s
4. |1/(1+jω/50)| plot0dB b l 50 d/0dB below 50 rad/sslope= -20 dB/dec beyond 50 rad/s
Slide 30EE40 Fall 2009 Prof. Cheung
Magnitude
3
11 42
Slide 31EE40 Fall 2009 Prof. Cheung
Phase
( ) ( ) ( )50/j115/j1
j1
4000400H 2
ω+∠+ω+∠+
ω∠+∠=ω∠
( ) ( )50/j115/j12)90(0 o
ω+∠+ω+∠+−+=
5 10 20 50050 5000
Slide 32EE40 Fall 2009 Prof. Cheung
Bode Plot: Quick Summary
In addition, we need to checkneed to check H(ω) as ω -> 0 and H(ω) as ω -( )> ∞.
Slide 33EE40 Fall 2009 Prof. Cheung
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