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7/26/2019 Introduction to Electrical Systems
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SCHOOLOFENGINEERING
IntroductiontoElectricalandElectronic
Engineering
Part7 BodePlots,FrequencyResponse
Grant A. Ellis, PhD
1
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Frequency Response, Bode Plots,
and Resonance Our concern here is the information-bearing currents andvoltages that we call now signals
Example 1, sensors or transducers on an internalcombustion engine provide electrical signals that representtemperature, speed, throttle position and the rotational
position of the crankshaft. These signals are processed todetermine the optimum firing instant of the cylinder
Example 2, signal processing related toelectrocardiogram (plot of the electrical signals generated
by the human heart). The information extracted from thesignals is used to analyse the behaviour of a patientsheart.
Grant A. Ellis
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Frequency Response, Bode Plots,
and Resonance Signal processing is concerned with manipulating signals to extractinformation and using that information to generate other useful electrical signals Even though many signals are not sinusoidal, we can
construct any waveform by adding sinusoids (sometimes thousands of them)that have the proper amplitude, frequencies and phases.
Grant A. Ellis
The short segment of a music waveform shown in (a) is the sum of thesinusoidal components shown in (b).
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Frequency Response, Bode Plots,
and Resonance Fourier analysis is a mathematical technique for finding theamplitudes, frequencies and phases of a given waveform
The range of the frequencies of the components depends onthe type of signal under consideration (table below)
Grant A. Ellis
Frequency Ranges of Selected Signals
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Frequency Response, Bode Plots,
and Resonance Example, Fourier series of asquare wave is a combination of
sinusoidal components :
where 0 = 2/T is called thefundamental angular frequency ofthe wave
Figure shows the sum its first 5components only Approximation will get better asmore components are added
The frequency, amplitude andphases can be determined bymeasurement using a spectrumanalyzer
Grant A. Ellis
A square wave and some
of its components.
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Frequency Response, Bode Plots,
and Resonance Filters are electrical circuits that are used to retain components in a givenrange of frequencies and discard the components in another range. Example,
television antenna produces a voltage composed of signals from many transmitters.
Filters select only frequencies of a particular channel and reject others
Grant A. Ellis
When an input signal vin(t) is applied to the input port of a filter, some componentsare passed to the output port while others are not, depending on their frequencies.Thus, vout(t) contains some of the components of vin(t), but not others. Usually, theamplitudes and phases of the components are altered in passing through the filter.
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Frequency Response, Bode Plots,
and Resonance Transfer function H(f) of the two-port filter is defined to be the ratio ofthe phasor output voltage to the phasor input voltage as a function of
frequency
Because phasors are complex, the transfer function is a complex
quantity having both magnitude and phase
The magnitude of the transfer function is the ratio of the output
amplitude to the input amplitude
The phase of the transfer function is the output phase minus the input
phase
In steady-state, the output signal is sinusoidal and has the same
frequency as the input signal
in
out
V
V
fH
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Frequency Response, Bode Plots,
and ResonanceExample 6.1: Given an input signaland transfer function of the filter.Find the output.
By inspectionthe frequencyf= 0/2=1000Hz From the Figurefor his frequency:
Magnitude =3 andPhase =30
Therefore: and
or
Grant A. Ellis
The transfer function of a filter. See Examples 6.1 and 6.2.
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Frequency Response, Bode Plots,
and Resonance
Input signals with multiple frequency Components (procedurefor determining the output)
1. Determine the frequency and phasor representation for eachinput components
2. Determine the (complex) value of the transfer function for eachcomponent
3. Obtain the phasor for each output component by multiplyingthe phasor for each input component by the correspondingtransfer-function value
4. Convert the phasor for the output components into time
function of various frequencies. Add these time functions toproduce the output
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Example 6.2 Given an input signaland transfer function of the filter, find the output:
Break the input into components:
By inspection: frequencies f are 0, 1000, 2000 from Figure:
Outputs: dc output phasors:
Frequency Response, Bode Plots,
and Resonance
Grant A. Ellis
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Frequency Response, Bode Plots,and Resonance
Experimental Determination of the Transfer Function
1. Connect a sinusoidal source to the input port of the circuit2. Measure the amplitude and phase of both the input signal and the
resulting output signal
3. Divide the output phasor by the input phasor
4. Repeat 1 to 3 for various frequencies of interest
The measurement can be done with instruments such as: Voltmeters,
oscilloscopes, network analyzers, signal generators & spectrum
analyzers
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First-order Lowpass Filter
First-order lowpass filter tends to pass low-frequency componentsof input signal and reject high-frequency components. In other words, for low frequencies, the output amplitude is nearlythe same as the input and for high frequencies, the output is muchless than the inputInput current is
Output voltage is
Substitute the current
Transfer function is
If thenRC
fB21
Bffj
fH
1
1
Grant A. Ellis
A first-order lowpass filter.
Another first-order lowpass filter
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First-order Lowpass Filter
Bffj
fH 1
1
21
1
BfffH
BfffH arctan
Grant A. Ellis
Magnitude and phase of the first-order lowpass
transfer function versus frequency.
The magnitude of the complex function is
The phase is the phase of the numerator (which is zero) minus the
phase of the denominator, which is
From the plot:- For low frequencies, the magnitude is approximately unity and thephase is nearly zero (they are passed)- For high frequencies, the magnitude approaches zero (they arerejected) When f = fB themagnitude of transfer
function is 1/ . Sinceit causes the poweroutput to cut half (V2/z),fB is called half-powerfrequency
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Example 6.3 Given an input signaland the filter find the output:
Half power frequency is:
For the first input component:
-The amplitude almost retained
For the second (reduced by ); For the third (significant reduction)
Frequency Response, Bode Plots,
and Resonance
Grant A. Ellis
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The transfer function can be defined in decibels as follows:
Note from the Table, that decibels are positive formagnitudes greater than 1 and negative formagnitudes less than 1
A filter designed to eliminate components ina narrow range of frequencies, is called notchfilter (example, filter to eliminate 60 Hz ofpower line noise from the audio signals, and pass component of at otherfrequencies, called passband). Both results clearly seen inonly in decibelplot but not inlinear plot
Decibels
fHfH log20dB
Grant A. EllisTransfer-function magnitude of a notch filter
used to reduce hum in audio signals
Transfer-Function Magnitudesand their Decibel Equivalents
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When we connect the output of one two-port circuit to the inputterminals of another two-port circuit, we have a cascade connection
For cascade connection the following is trueor (multiplied) Expressing both sides in decibelsor
or
The overall transfer function of a cascade connection in decibels isthe sum of individual transfer functions in decibel (or dB).
Cascaded Two-port Network
fHfHfH 21
dB2dB1dB
fHfHfH
Grant A. Ellis
Cascade connection of two two-port circuits
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Logarithmic scale is often used for frequency when plotting the transferfunction to clearly see its variation in large range of values
On a logarithmic scale, the variable is multiplied by a given factor on toobtain equal increments of length along the axis, whereas on linear
scale, equal lengths correspond to adding a given amount to thevariable
A decade is the range of frequencies for which the ratio of the highestfrequency to the lowest one is 10 eg 2 to 20 is one decade; 50 to 5000are two decades, from 50 to 500 and from 500 to 5000Hz
f2
>f1
(range f1
to f2
)
An octave is a two-to-one change in frequency eg10 to 20 Hz is oneoctave
Logarithmic Frequency Scales
Grant A. Ellis
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A Bode plot uses a logarithmic scale for frequency to show the
magnitude of a network function in decibels versus frequency.
The Bode plots can be closely approximated by straight-line
segments that are relatively easy to draw and then make quick
estimate of transfer functions
Given a transfer function magnitude
Convert to decibel
Substitute and modify
since log(1)=0
Since then
Bode Plots
2
dB 1log10
Bf
ffH
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Plot of
If f>fB (high frequency range)
then
and the amplitude is approximated by the straight slope line on the right-hand side of the graph These two straight-line asymptotes intersect at the half-power frequency
fB, called the corneror break frequency The slope of the high-frequency asymptote is -20 dB per decade The asymptotes are in error at the corner by only
(3 dB frequency).
Bode Plots (amplitude)
2
dB 1log10BfffH
Grant A. Ellis
Magnitude Bode plot for first-order lowpass filter.
20 dB / decade= 6 dB / octave
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The phase of transfer function
By observation, the phase
approaches zero at very low
frequencies, equals -45 at the
break frequency, andapproaches -90 at high
Frequencies
The curve can be approximated by the following straight-linesegments:
Bode Plots (phase)
Grant A. Ellis
Phase Bode plot for thefirst-order lowpass filter.
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The analysis of the circuit for thefirst-order highpass filter similar tothat of the lowpass filter
if
then
The magnitude is
The phase is
Amplitude goes to 0 fordc signal
First- Order Highpass Filter
RCfB
2
1
B
B
ffj
ffjfH
1in
out
V
V
Grant A. Ellis
Magnitude and phase for the first-order highpass
transfer function.
First-order highpass filter.
RCf2j1
RCf2j
V
VfH
in
out
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For Bode Plot, the magnitude is converted to decibels (dB) and alogarithmic frequency scale is used
If the second term in amplitude expression is almost zero (f >fB the resulting magnitude is almost 0 (horizontal asymptote
Intersection is at fB
Bode Plots First-Order Highpass Filter
Grant A. EllisBode plots for the first-order highpass filter.
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Example 6.4: Highpass filter with magnitude -30 dB at frequency
f = 60 Hz. Define the break frequency of this filter.
Low-frequency asymptote slopes at the rate of 20db / decade
Then the number of decades from -30 dB to 0 dB is:
According to the definition of the decade:
Solving the equation, we obtain:
Highpass filter (example)
Grant A. Ellis
31.6 * 60 = 1896
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Series Resonance
LCf
2
10
Grant A. Ellis
The RLC circuit (resonant circuit) have better performance (inpassing desired signals and rejecting undesired signals that are
relatively close in frequency) than first-order filters
In a resonant circuit, when a sinusoidal source of a proper
frequency is applied, voltages much larger than the source voltage
can appear in the circuit
The impedance seen by the source
in this circuit is
The resonant frequency f0 is defined to
be the one at which the impedance is purely
resistive (i.e. the total reactance is zero), that means
or
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The quality factor QS is defined to be ratio of the reactance of theinductance at the resonant frequency to the resistance:
Excluding L in this equation (L is from the resonant frequencyequation), we obtain:
Note:
Substituting L and C from the equations above into theimpedance formula we obtain:
Series Resonance
R
LfQs
02
CRfQs
02
1
f
f
f
fjQRfZ ss
0
0
1
Grant A. Ellis
CRf2
1
R
Lf2Q
0
0s
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Plots of the normalized magnitude and the phase of the impedanceversus normalized frequency f / f0 shows that the impedancemagnitude is minimum at the resonant frequency (f = f0) (equal to
what?) As the quality factor Q increases, the minimum becomes sharper.
For f < f0 the impedance has a capacitive nature and for f > f0it has an inductive nature.
Series Resonance
f
f
f
fjQRfZ ss
0
0
1
Grant A. Ellis
Plots of normalized magnitude and phase for the impedance of the
series resonant circuit versus frequency.
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Define transfer function of the filter:
Plot of the magnitude of transfer function:
Series Resonant Circuit as a
Bandpass Filter
ffffjQ1
1fH
V
V
00ss
R
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Discussions: At lower frequencies f < f0 theimpedance is capacitive in nature, itreduces the circuit current and VRbecomes small compared to VS At higher frequencies f > f0 theimpedance is inductive in nature, itstill reduces the circuit current and
again VR becomes small compared to VS
At f=f0 the impedance becomes minimum (equal R) and VR becomesequal to VS
As a result, if a source has components ranging in frequencyabout the resonant frequency f0 the components of the source close tothat frequency will have only a small change, but other components willbe significantly reduced. So Bandpass filter passes the
components centered at the resonant frequency the
rest of components are partly rejected.
Series Resonant Circuit as aBandpass Filter
Grant A. Ellis
Transfer-function magnitude
|VR / Vs| for the series resonantbandpass-filter.
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Discussions (continued): There are two half-powerfrequencies for series resonant circuit
when the transfer function magnitudehas fallen from its maximum by aFactor 1/ =0.707, i.e. fL and fH
The bandwidth B of this filter is the
difference between the half-powerfrequencies:
It can be shown that
For QS >> 1, the following approximation is valid:
Series Resonant Circuit as a
Bandpass Filter
LH ffB
sQ
fB 0
20
BffH
20
BffL
Grant A. Ellis
The bandwidth B is equal to thedifference between the half-power
frequencies.
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Example 6.5Given a filter circuit Resonant frequency:
Quality factor and bandwidth:
Half-power frequencies: Reactances:
Voltages across at resonant frequencies:
Grant A. Ellis
Series resonant circuit.
Series Resonance
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Parallel Resonance
fLjfCjRZp
2121
1
LCf
2
10
Lf
RQ
p02
CRfQp 02
ffffjQ
RZ
pp
001
Parallel resonant circuit The impedance is:
At the resonant frequency the impedance is purely resistive (sameas the series circuit)
The quality factor is defines as (reciprocal of series)and can be written as
Then impedance can be written as:then V
out
= I ZP
Half-power frequencies
and the band width:
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Ideal Filters
Ideal filters pass components in the desired frequency rangewith no change in amplitude or phase and totally reject the
components in the undesired frequency range.
They are ideal and can be just approximated by real circuits. fLandfH are cutoff frequencies.
Grant A. Ellis
Transfer functions of ideal filters.
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Ideal Filters
The input signal vin consists of a 1-kHz sine wave plus high-frequency
noise. By passing vin through an ideal
lowpass filter with the proper cutoff
frequency, the sine wave is passedand the noise is rejected, resulting in
a clean output signal.
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Second-Order Lowpass Filter
ffffjQffjQ
fHs
s
00
0
in
out
1
V
V
LCf
2
10
R
LfQs
02
Grant A. Ellis
It is based on series resonant circuit
For QS >> 1 the highpeak is reached in thevicinity of f0 To cut the peak andmake it flat we choose
QS=1 (more preciselyQS = 0.707). The transfer functionat this quality factor is said tobe maximally flat and is calledButterworth function. The transfer function for thesecond order filter falls more rapidlythan for the first order filter (different slope) Lowpass filter circuits and their
transfer-function magnitudes versus
frequency.
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Second-Order Highpass Filter
Grant A. Ellis
Second-order highpass filter and its transfer-function magnitude
versus frequency for several values of Qs
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Filter design Example 6.7.
Grant A. Ellis
Second-Order Highpass Filter
Design a filter which passes components higher than 1 kHz and rejects components
Less than 1 kHz using a second order circuit with L = 50 mH.
F0.507
1050102
1
Lf2
1C
CL2
1f
3-620
0
22
1B
fQ 0s To provide a nearly constant transfer function up to 1 kHz
314.11
1050102Q
Lf2QCf2
1R -33
s
0
s0
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Second-Order Bandpass Filter
Grant A. Ellis
Second-order bandpass filter and its transfer-function magnitudeversus frequency for several values of Q
s
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Second-Order Band-Reject (Notch)
Filter
Grant A. Ellis
Second-order band-reject filter and its transfer-function magnitude versus
frequency for several values of Qs.