CHAPTER 6
Frequency Response, Bode
Plots, and Resonance
1. State the fundamental concepts of Fourier
analysis.
2. Determine the output of a filter for a given
input consisting of sinusoidal components
using the filter’s transfer function.
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3. Use circuit analysis to determine the
transfer functions of simple circuits.
4. Draw first-order lowpass or highpass filter
circuits and sketch their transfer functions.
5. Understand decibels, logarithmic frequency
scales, and Bode plots.
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6. Draw the Bode plots for transfer functions of
first-order filters.
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Frequency Response, Bode Plots, and Resonance
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Fourier Analysis
All real-world signals are sums of sinusoidal
components having various frequencies,
amplitudes, and phases.
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Chapter 6
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Frequency Response, Bode Plots, and Resonance
Filters
Filters process the sinusoid components of an input
signal differently depending of the frequency of
each component. Often, the goal of the filter is to
retain the components in certain frequency ranges
and to reject components in other ranges.
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Frequency Response, Bode Plots, and Resonance
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Transfer Functions
The transfer function H(f ) of the two-port filter
is defined to be the ratio of the phasor output
voltage to the phasor input voltage as a function
of frequency:
Vout
H(f )=
Vin
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Vout
H(f )=
Vin
The magnitude of the transfer function shows
how the amplitude of each frequency component
is affected by the filter. Similarly, the phase of
the transfer function shows how the phase of each
frequency component is affected by the filter.
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Frequency Response, Bode Plots, and Resonance
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Determining the output of a filter
for an input with multiple
components:
1. Determine the frequency and phasor
representation for each input component.
2. Determine the (complex) value of the transfer
function for each component.
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3. Obtain the phasor for each output
component by multiplying the phasor for each
input component by the corresponding
transfer-function value.
4. Convert the phasors for the output
components into time functions of various
frequencies. Add these time functions to
produce the output.
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Frequency Response, Bode Plots, and Resonance
Linear circuits behave as if they:
1. Separate the input signal into components
having various frequencies.
2. Alter the amplitude and phase of each
component depending on its frequency.
3. Add the altered components to produce
the output signal.
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Chapter 6
Frequency Response, Bode Plots, and Resonance
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Chapter 6
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Chapter 6
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FIRST-ORDER LOWPASS
FILTERS
1
fB =
2πRC
H(f ) =
1
1 + ( f fB )
2
 f 
1
∠H ( f ) = − arctan  
H(f )=
1 + j( f f B )
 fB 
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H(f ) =
1
1 + ( f fB )
2
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Chapter 6
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DECIBELS, THE CASCADE
CONNECTION, AND LOGARITHMIC
FREQUENCY SCALES
H ( f ) dB = 20 log H ( f )
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Cascaded Two-Port Networks
H ( f ) = H1 ( f )× H 2 ( f )
H ( f ) dB = H 1 ( f ) dB + H 2 ( f ) dB
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Logarithmic Frequency Scales
On a logarithmic scale, the variable is
multiplied by a given factor for equal
increments of length along the axis.
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A decade is a range of frequencies for which
the ratio of the highest frequency to the
lowest is 10.
 f2 
number of decades = log 
 f1 
An octave is a two-to-one change in frequency.
 f 2   log( f 2 f1 ) 
number of octaves = log 2   = 

 f1   log(2 ) 
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BODE PLOTS
A Bode plot shows the magnitude of a network
function in decibels versus frequency using a
logarithmic scale for frequency.
1
H(f )=
1 + j( f f B )
H ( f ) dB
ELECTRICAL
  f 2 
= −10 log 1 +   
  f B  
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1. A horizontal line at zero for f < fB /10.
2. A sloping line from zero phase at fB /10 to –90° at 10fB.
3. A horizontal line at –90° for f > 10fB.
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FIRST-ORDER HIGHPASS
FILTERS
Vout
j( f f B )
=
H(f )=
Vin 1 + j ( f f B )
1
fB =
2πRC
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SERIES RESONANCE
Resonance is a phenomenon that can be
observed in mechanical systems and
electrical circuits.
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f0 =
1
2π LC
2πf 0 L
Qs =
R
1
Qs =
2πf 0CR

 f
f 0 
Z s ( f ) = R 1 + jQs  − 
f 
 f0

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Series Resonant Circuit as a
Bandpass Filter
VR
1
=
Vs 1 + jQs ( f f 0 − f 0 f )
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B = fH − fL
B
f H ≅ f0 +
2
f0
B=
Qs
B
f L ≅ f0 −
2
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PARALLEL RESONANCE
1
Zp =
(1 R ) + j 2πfC − j (1 2πfL)
f0 =
1
2π LC
R
Qp =
2πf 0 L
Q p = 2πf 0CR
R
Zp =
1 + jQ p ( f f 0 − f 0 f )
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Ideal Filters
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Second-Order Lowpass Filter
− jQ s ( f 0 f )
Vout
H(f )=
=
Vin 1 + jQ s ( f f 0 − f 0 f )
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DIGITAL SIGNAL
PROCESSING
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Conversion of Signals from
Analog to Digital Form
If a signal contains no components with
frequencies higher than fH, the signal can be
exactly reconstructed from its samples, provided
that the sampling rate fs is selected to be more
than twice fH.
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Digital Lowpass Filter
y (n ) = ay (n − 1) + (1 − a )x (n )
τ T
a=
1+τ T
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