Plo;ing H(s): Bode Plots
Frequency Response:
Transfer Functions &
Bode Plots II
§  Manipulate transfer function
§  Put H(s) into standard form
§  Identify terms to plot & identify zeros
and poles
§  Examples
EGR 220, Chapter 14
April 19, 2016
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Find H(s): Voltage Gain
Notice presence of zeros and poles
Example: Amplitude & Phase
H (s) =
200 jω
( jω + 2)( jω +10)
H dB (s) = ?
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ϕ H (s) = ?
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Special ω: Zeros & Poles
Example: Amplitude Plot
H ( s) =
H ( s) =
200 jω
( jω + 2)( jω + 10)
N ( s)
200 jω
200s
=
=
D( s) ( jω + 2)( jω + 10) ( s + 2)(s + 10)
•  What happens to H(s) when (s = jω)
H dB ( s) = ?
o s = 0
o s = -2
o s = -10
•  In terms of algebra,
o Zero: H(s) = 0 when __________________?
o Pole: H(s) = ∞ when __________________?
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Example: Phase Plot
200 jω
( jω + 2)( jω + 10)
=?
H ( s) =
φH ( s )
2
±1
jω
jω
jω
N(s) Ks (1+ z1 ) (1+ z2 )(1+ zn )
H(s) =
=
D(s)
(1+ jω p1 ) (1+ jω p2 )(1+ jω pm )
Possible Terms in H(s)
Amplitude terms:
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H (s) = 20 log10 K ± 20 log10 jω + 20 log10 1+ ( ωz1 ) …+ 20 log10 1+ ( ωzn )
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− 20 log10 1+ ( ωp1 ) …− 20 log10 1+ ( pωm )
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4.  A quadratic pole or zero
Phase terms:

− ∠tan −1
! 1+ s $
z&
#
# 1 &
"
%
( )…+ ∠tan ( )
( )…− ∠tan ( )
∠H (s) = 0 + ∠ ± 90 + ∠tan
ω
p1
−1 ω
z1
−1
−1
2
1.  A gain K, where K is constant
2.  A zero (s), or pole (1/s), at the origin
!
$
3.  A zero or pole of the form: ! 1+ s $
z & or # 1 &
#
ω
zn
ω
pm
2
1
&
%
# 1+ s &
p%
"
!
$
1
&
or ##
2&
" (1+ s p) %
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Example: Bode Plot for H(s)
#
"
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1. A constant term “K”
1.  Put H(s) into standard (simplified) form
2.  Identify the components (constant gain,
zeros and poles...)
3.  Plot the amplitude and phase
H(s) =
€
5( s + 2)
s( s + 10)
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2a. A zero at the origin:
2b. A pole at the origin:
1 1
H (s) = =
s jω
H (s) = s = jω
H (s) dB = 20 log10 (?)
H (s) dB = 20 log10 (?)
∠H (s) = ?
∠H (s) = ?
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3a. Real axis zero:
(
1+ jω
H (s) =
a
)=(
1+ s
a
3a. Real axis zero:
)
1
1
H (s) dB = 20 log10 (?)
1+ jω ) (1+ s )
(
a
a
H (s) =
=
Magnitude:
ω << a
1
1
H (s) dB = 20 log10 (?)
∠H (s) = ?
Angle: 0
H ( jω ) dB ≈ 20 log10 10 = 0
∠H (s) = ?
Here a = 1
∠H ( jω ) =
ω << a
ω
a
ω >> a
H ( jω ) dB ≈ 20 log10
ω=a
H ( jω ) dB = 20 log10 2 = 3
ω >> a
∠H ( jω ) =
ω=a
H ( jω ) dB ≈ 20 log10 10 0 = 0
ω
a
H ( jω ) dB = 20 log10 2 = 3
H ( jω ) dB ≈ 20 log10
∠H ( jω ) =
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3b. Real axis pole:
b
b
b
b
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4. Quadratic zeros and poles
•  The effect of the squared term?
o log(x)n = n log(x)
o With dB: 20 log(x)±2 = ± 40 log(x)
•  The slope of the magnitude plot is ± 40dB/
decade
o Phase
•  At ω = ωn à φ = ±90º (‘zero’ is +; ‘pole’ is – )
•  As ω→∞ à φ = ± 180º
•  The effect of two zeros or poles is 2x the effect
of one
o See pages 621 – 624
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Example: Bode Plot for H(s)
* Action of Zeros and Poles *
1.  Put H(s) into standard (simplified) form
2.  Identify the components (constant gain,
zeros and poles...)
3.  Plot the amplitude and phase
•  Zero
o Action: Increase amplitude plot; begin to
pass frequencies
o … At the corner frequency
•  Pole
H(s) =
o Action: Decrease amplitude plot; begin
to block frequencies
o … At the corner frequency
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€
5( s + 2)
s( s + 10)
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Example: Bode Plot for H(s)
1.  Put H(s) into standard form
2.  Identify the components
3.  Plot the amplitude and phase
H(s) =
5( s + 2)
s( s + 10)
€
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Example 2: Bode Plot
1. Put H(s) into standard form
2. Identify the components
3. Plot
G( s) =
( s + 2) 2
s( s + 5)(s + 10)
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Summary
Deduce the Transfer Function
•  Possible elements of transfer function, to
be plotted as Bode plot
•  Heuristics for Bode plot
•  Action of zeros and poles
•  Examples
o Deduce transfer function from Bode plot
o Create Bode plot from transfer function
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