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)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
Chapter # 6 Frequency Analysis
1. Introduction
The transfer function H(ω) (also called the network function) is a useful analytical tool
for finding the frequency response of a circuit. In fact, the frequency response of a
circuit is the plot of the circuit’s transfer function H(ω) versus ω, with ω varying from
ω = 0 to ω →∞. A transfer function is the frequency-dependent ratio of an output to
an input. In general, a linear network can be represented by the block diagram shown
in Fig.1.
Fig. 1, Representation of a linear network.
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
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)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
Since the input and output can be either voltage or current at any place in the circuit,
there are four possible transfer functions:
Being a complex quantity, H(ω) has a magnitude H(ω) and a phase φ; that is, H(ω) =
H(ω) φ.
Example (1):
For the RC circuit in Fig. 2, obtain the transfer function Vo/Vs and its frequency
response. Let vs = Vm cos ωt.
Fig. 2
The circuit T.F. can be obtained as:
The magnitude and phase of the T.F. are given as:
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
Page 2
)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
Where,
To plot the circuit frequency response, we draw the magnitude and phase at different
values of frequencies as given in the following table:
Then the response is as shown below:
Obtaining the frequency response from the transfer function by substituting the value
of (ω) directly in the system transfer function is a tedious task. The frequency range
required in frequency response is often so wide that it is inconvenient to use a linear
scale for the frequency axis. Also, there is a more systematic way of locating the
important features of the magnitude and phase plots of the transfer function. For these
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
Page 3
)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
reasons, it has become standard practice to use a logarithmic scale for the frequency
axis and a linear scale in each of the separate plots of magnitude and phase. Such
semi-logarithmic plots of the transfer function—known as Bode plots—have become
the industry standard. Bode plots contain the same information as the non-logarithmic
plots, but they are much easier to construct.
The transfer function H(s) can be expressed as:
( )
| |
In Bode plots, the magnitude is plotted in Decibels (dB) versus frequency. The dB
quantity can be obtained as:
Moreover, the phase angle (φ) is plotted versus frequency. Both magnitude and phase
plots are made on semi-log graph paper.
2. Bode Plots
A transfer function may be written in terms of factors that have real and imaginary
parts such as:
( )
(
)
(
(
){
){
( ) }
(
) }
this is called the Bode (Standard) form of the system transfer function that may
contain seven different factors:
 Bode gain K
 Pole at origin (
)
and/or zero at origin (
)
 Real pole (
)
and/or real zero (
)
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
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)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
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 Quadratic pole {
(
(
) }
and/or quadratic zero {
) }
In constructing a Bode plot, we plot each factor separately and then combine them
graphically. The factors can be considered one at a time and then combined additively
because of the logarithms involved. For this mathematical convenience of the
logarithm, Bode plots is considered as a powerful engineering tool.
In the following subsections, we will make straight-line plots of the factors listed
above. These straight-line plots known as asymptotic (approximate) Bode plots.
2.1 Bode Gain
For the gain K, there are two cases:
K is +ve and less than one: the magnitude 20 log K is negative and the phase is 0◦;
K is +ve and greater than one: the magnitude 20 log K is positive and the phase is 0◦;
K is -ve: the magnitude remain 20 log K is negative and the phase is ±180◦;
Both of the magnitude and phase are constant with frequency. Thus the magnitude and
phase plots of the gain are shown in Fig.3.
Fig. 3, Magnitude and phase plots of Bode gain
2.2 Zero at origin
For the zero (jω) at the origin, the magnitude is 20 log10 ω and the phase is 90◦. These
are plotted in Fig. 4, where we notice that the magnitude is represented by a straight
line with slope of 20 dB/decade and intersect the 0dB line at ω=1 and extended to
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
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)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
intersect the vertical axis. But the phase is represented by straight line parallel to
horizontal axis with constant value at 90.
‫ ونمذه‬1 = ω ‫ ديسبل لكل ديكاد ويمر بخط الصفر ديسيبل عنذ‬02 ‫القيمة تمثل بخط مستقيم ميلو‬
ω ‫ درجة وتمثل بخط مستقيم مىازي لمحىر‬02 ‫أما الساوية فقيمتها ثابتة عنذ‬
Fig. 4, Magnitude and phase plots of zero at origin
In general, for multiple zeros at origin (jω)N, where N is an integer, the magnitude plot
will have a slope of (20×N) dB/decade. But the phase is (90×N) degrees.
2.3 Pole at origin
For the pole (jω)-1 at the origin, the magnitude is -20 log10 ω and the phase is -90◦.
These are plotted in Fig. 5, where we notice that the magnitude is represented by a
straight line with slope of -20 dB/decade and intersect the 0dB line at ω=1 and
extended to intersect the vertical axis. But the phase is represented by straight line
parallel to horizontal axis with constant value at -90.
‫ ونمذه‬1 = ω ‫ ديسبل لكل ديكيذ ويمر بخط الصفر ديسيبل عنذ‬-02 ‫القيمة تمثل بخط مستقيم ميلو‬
ω ‫ درجة وتمثل بخط مستقيم مىازي لمحىر‬-02 ‫أما الساوية فقيمتها ثابتة عنذ‬
In general, for multiple poles at origin (jω)-N, where N is an integer, the magnitude
plot will have a slope of - (20×N) dB/decade. But the phase is - (90×N) degrees.
Fig. 5, Magnitude and phase plots of pole at origin
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
Page 6
)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
2.4 Real Zero
The magnitude of a real zero (
is obtained from
|
) is obtained from
|, and the phase
( ). We notice that:
|
- For small values of ω, the magnitude is
- For large values of ω, the magnitude is
|
|
|
| |
From the above two points, we can approximate the magnitude of real zero by two
straight lines ( at ω  0 : a straight line with zero slope with zero magnitude) and (at
ω  ∞ : a straight line with slope 20 dB/decade). At the frequency ω = z1 where the
two asymptotic lines meet is called the corner frequency. Thus the approximate
magnitude plot is shown in Fig. 6. The actual plot for real zero is also shown in that
figure. Notice that the approximate plot is close to the actual plot except at the corner
frequency, where ω = z1 and the deviation is
|
|
√
.
‫ والنمذه‬z1 = ω ‫ ديسبل لكل ديكيذ ويمر بخط الصفر ديسيبل عنذ‬+02 ‫القيمة تمثل بخط مستقيم ميلو‬
‫ ونصل بينهما بخط مستقيم‬02 = ‫) الساوية‬10Z1( ‫ ثم ديكيذ بعذ‬،‫) الساوية = صفر‬z1/10( ‫ ديكيذ قبل‬:‫الساوية‬
‫ درجة لكل ديكيذ‬54 ‫ليكىن ميل الخط‬
Fig. 6, Magnitude and phase plots of real zero
The phase angle of real zero that given as
( ) is represented as a straight-line
approximation, φ = 0 for ω ≤ z1/10, φ = 45◦for ω = z1, and φ = 90◦ for ω ≥ 10z1 as
shown in Fig. 6. The straight line has a slope of 45 per decade.
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
Page 7
)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
2.5 Real Pole
The magnitude of a real pole (
)
is obtained from
|
|, and the
( ). We notice that:
phase is obtained from
|
- For small values of ω, the magnitude is
|
|
- For large values of ω, the magnitude is
|
| |
From the above two points, we can approximate the magnitude of real pole by two
straight lines ( at ω  0 : a straight line is with zero slope and zero magnitude) and (at
ω  ∞ : the straight line is with slope -20 dB/decade). At the frequency ω = p1 where
the two asymptotic lines meet is called the corner frequency. Thus the approximate
magnitude plot is shown in Fig. 7. The actual plot for real pole is also shown in that
figure. Notice that the approximate plot is close to the actual plot except at ω = p1, the
deviation is
|
|
√
.
‫ والنمذه‬p1 = ω ‫ ديسبل لكل ديكيذ ويمر بخط الصفر ديسيبل عنذ‬-02 ‫القيمة تمثل بخط مستقيم ميلو‬
The phase angle of real pole that given as
( ) is represented as a straight-line
approximation, φ = 0 for ω ≤ p1/10, φ = -45◦for ω = p1, and φ = -90◦ for ω ≥ 10p1 as
shown in Fig. 7. The straight line has a slope of -45 per decade.
‫ ونصل بينهما بخط مستقيم‬02- = ‫) الساوية‬10p1( ‫ ثم ديكيذ بعذ‬،‫) الساوية = صفر‬p1/10( ‫ ديكيذ قبل‬:‫الساوية‬
‫ درجة لكل ديكيذ‬-54 ‫ليكىن ميل الخط‬
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
Page 8
)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
Fig. 7, Magnitude and phase plots of real pole
2.6 Quadratic Zero
The magnitude of a quadratic zero {
(
|
|
(
) |
(
|(
) |
) } is obtained as
) |
|(
)|
Thus, the amplitude plot consists of two straight asymptotic lines: one with zero slope
for ω < ωn and the other one with slope −40 dB/decade for ω > ωn, with ωn as the
corner frequency. Figure 8 shows the approximate and actual amplitude plots. Note
that the actual plot depends on the damping ratio ξ2 as well as the corner frequency ωn.
The significant peaking in the neighborhood of the corner frequency should be added
to the straight-line approximation if a high level of accuracy is desired. However, we
will use the straight-line approximation for the sake of simplicity.
The phase plot is a straight line with a slope of 90◦ per decade starting at ωn/10 and
ending at 10ωn, as shown in Fig. 8. We see again that the difference between the
actual plot and the straight-line plot is due to the damping factor.
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
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)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
Fig. 8, Magnitude and phase plots of quadratic zero
For the quadratic pole {
(
) } the plots shown in Fig. 6 are inverted
because the magnitude plot has a slope of -40 dB/decade while the phase plot has a
slope of -90◦ per decade.
3. Stability Analysis Using Bode Plots
The gain crossover frequency g is defined as the frequency at which the total
magnitude equals 0 dB. Therefore, its value can be determined from the intersection of
the total magnitude line with the 0 dB line as shown in Fig. 9. On the other hand, the
phase crossover frequency p is defined as the frequency at which the total phase
equals -180. Therefore, its value can be determined from the intersection of the total
phase line with the -180 line as shown in Fig. 9.
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
Page 10
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‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
Fig. 9, Gain and Phase Margins
The system Gain Margin GM in dB can be determined by calculating the total
magnitude at  = p. Also, he system Phase Margin PM in degrees can be determined
by calculating the total phase at  = g.
4. Plotting Bode Plots Using Matlab
To specify the frequency range for Bode plots, use the command:
>> logspace(d1,d2)
This generate 50 points logarithmically equally spaced between 10d1 and 10d2.
For example, if we need Bode plot starts at 0.1 rad/sec and finish at 100 rad/sec, enter
the command:
>> logspace(-1,2)
If we need to change the number of points between d1 and d2 rather than 50, use the
command:
>> logspace(d1,d2,n)
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
Page 11
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where n is the number of points to be generated.
For example, to generate 100 points between 1 rad/sec and 1000 rad/sec, use:
>> W=logspace(0,3,100)
To draw the Bode plot, we use the command
>> bode(num,den,W)
Suppose we need to draw the Bode plot for the control system:
(
)
( )
(
)(
)
So we write the following Matlab code
The magnitude and phase plots are as shown below in Fig. 10, and it is clear that this
is an unstable system.
Fig. 10, Magnitude and phase plots
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
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Example:
Construct the Bode blot for the transfer function:
( )
(
(
)(
)
)(
)
Then calculate the gain and phase crossover frequency.
5. Series Resonance
The concept of resonance applies in several areas of science and engineering.
Resonance occurs in any system that has a complex conjugate pair of poles; it is the
cause of oscillations of stored energy from one form to another. It is the phenomenon
that allows frequency discrimination in communications networks. Resonance occurs
in any circuit that has at least one inductor and one capacitor.
Consider the series RLC circuit shown in Fig. 11.
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
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‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
From the above equations for inductive reactance, if the Frequency is increased the
overall inductive reactance would also increase. As the frequency approaches infinity
the inductors reactance would also increase towards infinity with the circuit element
acting like an open circuit.
The same is also true for the capacitive reactance formula above but in reverse. If the
Frequency is increased the overall capacitive reactance would decrease. As the
frequency approaches infinity the capacitors reactance would reduce to zero causing
the circuit element to act like a perfect conductor of 0Ω’s (short circuit).
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
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But as the frequency approaches zero or DC level, the capacitors reactance would
rapidly increase up to infinity causing it to act like a very large resistance acting like
an open circuit condition. This means then that capacitive reactance is ―Inversely
proportional‖ to frequency for any given value of capacitance and this shown below:
At a higher frequency XL is high and at a low frequency XC is high. Then there must
be a frequency point were the value of XL is the same as the value of XC and there is.
If we now place the curve for inductive reactance on top of the curve for capacitive
reactance so that both curves are on the same axes, the point of intersection will give
us the series resonance frequency point, ( ƒ0 or ω0 ) as shown below.
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
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‫قشم اهلندسة الكهربائية‬
The input impedance is
We can see then that at resonance, the two reactances cancel each other out thereby
making a series LC combination act as a short circuit with the only opposition to
current flow in a series resonance circuit being the resistance, R. i.e. Z = R.
Therefore at resonance the impedance of the series circuit is at its minimum value and
equal only to the resistance, R of the circuit. The circuit impedance at resonance is
called the ―dynamic impedance‖ of the circuit and depending upon the frequency, XC
(typically at high frequencies) or XL (typically at low frequencies) will dominate
either side of resonance as shown below.
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
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Since the current flowing through a series resonance circuit is the product of voltage
divided by impedance, at resonance the impedance, Z is at its minimum value, ( =R ).
Therefore, the circuit current at this frequency will be at its maximum value of V/R as
shown below.
The current angle (φ) is changing according to the frequency. When ω = 0, the
impedance is pure capacitive and (φ) is +90. When ω→∞, the impedance is pure
inductive and (φ) is – 90. At resonance the current angle is 0 as shown below:
The circuit power factor (p.f.) can be obtained by taking the cosine of angle (φ) as
shown below:
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
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‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
At specific frequency before resonant frequency, the current waveform leads the
voltage waveform by certain angle (θ). Also, at specific frequency after resonant
frequency, the current waveform lags the voltage waveform by certain angle (θ). But
at resonant frequency, both current and voltage waveforms are in phase as shown:
If the series RLC circuit is driven by a variable frequency at a constant voltage, then
the magnitude of the current, I is proportional to the impedance, Z, therefore at
resonance the power absorbed by the circuit must be at its maximum value as P = I2Z.
If we now adjust the frequency until the average power absorbed by the resistor in the
series resonance circuit is half that of its maximum value at resonance, we produce
two frequency points called the half-power frequencies which are -3dB down from
maximum, taking 0dB as the maximum current reference.
These -3dB points give us a current value that is 70.7% of its maximum resonant value
which is defined as:
0.5( I2 R ) = (0.707 x I)2 R
Then the point corresponding to the lower frequency at half the power is called the
―lower cut-off frequency‖, labeled ƒ1 with the point corresponding to the upper
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
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‫قشم اهلندسة الكهربائية‬
frequency at half power being called the ―upper cut-off frequency‖, labeled ƒ2. The
distance between these two points, i.e. ( ƒ2 – ƒ1 ) is called the Bandwidth, (BW) and is
the range of frequencies over which at least half of the maximum power and current is
provided as shown.
The frequency response of the circuits current magnitude above, relates to the
―sharpness‖ of the resonance in a series resonance circuit. The sharpness of the peak is
measured quantitatively and is called the Quality factor, Q of the circuit. As shown in
figure below, at small resistance the BW is small and the quality factor is high. As the
circuit resistance increases, the BW is increased and the quality factor is decreased.
This mean the quality factor is inversely proportional with the bandwidth:
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
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‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
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‫قشم اهلندسة الكهربائية‬
The quality factor relates the maximum or peak energy stored in the circuit (the
reactance) to the energy dissipated (the resistance) during each cycle of.
This means:
Note that at resonance: The inductor voltage and capacitor voltage can be much more
than the source voltage.
At half-power, the circuit impedance (Z) equals √2
Solving for ω,
A resonant circuit is designed to operate at or near its resonant frequency. It is said to
be a high-Q circuit when its quality factor is equal to or greater than 10. For high-Q
circuits (Q ≥ 10), the half-power frequencies are, for all practical purposes,
symmetrical around the resonant frequency and can be approximated as:
We see that a resonant circuit is characterized by five related parameters: the two halfpower frequencies ω1 and ω2, the resonant frequency ω0, the bandwidth B, and the
quality factor Q.
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
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Example:
A series RLC circuit produced a resonant curve with half-power frequencies of 432 Hz
and 454 Hz. If Q = 20,
a) What is the resonant frequency (f0) of the circuit?
b) What is the bandwidth (B)?
c) If the inductance is hundred times the capacitance, determine the circuit resistance,
inductance and capacitance.
f1 = 432 Hz, f2 = 454 Hz, and Q = 20
B = f2 - f1 = 454 - 432 = 22 Hz
B = 2 (22) = 138.23 rad/s
##
o = B×Q = 138.23×20 = 2764.6 rad/s
##
Since L = 100 C
Also
√
√
L = 100 ×36.62×10-6 = 3.62 mH
Since B = R/L
R = B×L = 138.23×3.62×10-3 = 0.5 Ω
Example:
A series RLC circuit has a mid-band admittance of 25 ×10
−3
S, quality factor of 80
and a resonant frequency of 200 k rad/s. Calculate the values of R, L and C. Find the
bandwidth and the half-power frequencies.
Y at resonance = 1/R = 0.025 S
R = 1/0.025 = 40 Ω
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
Page 21
)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
Resonant frequency (ω0) = 200000 rad/s
Quality factor (Q) = 80
Then the bandwidth can be obtained as B = ω0 / Q = 200000/80 = 2500 rad/s ##
the half power frequencies ω1 and ω2 can be obtained from
ω1 = ω0 - B/2 = 200000 - 1250 = 198750 rad/s ###
ω1 = ω0 + B/2 = 200000 + 1250 = 201250 rad/s ###
Since B = R/L
Then L = R/B = 40/2500 = 0.016 H ###
Also Q = 1/( ω0RC)
Then C = 1/( ω0RQ) = 1/(200000*40*80) = 1.5625*10-9 F = 1.5625 nF
##
as check
√
√
Example:
In an electronic device, a series circuit is employed that has a resistance of 100 , a
capacitive reactance of 5 k, and an inductive reactance of 300  when used at 2
MHz. Find the resonant frequency and bandwidth, half power frequency, and the
quality factor of the circuit.
C = 15.9155 pF
L = 23.87 H
√
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
Page 22
)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
Since Q > 10
6. Parallel Resonance
A parallel circuit containing a resistance, R, an inductance, L and a capacitance, C will
produce a parallel resonance (also called anti-resonance) circuit when the resultant
current through the parallel combination is in phase with the supply voltage. At
resonance there will be a large circulating current between the inductor and the
capacitor due to the energy of the oscillations, and then parallel circuits produce
current resonance.
The circuit admittance is given by:
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
Page 23
)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
Resonance occurs when the imaginary part of Y is zero:
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
Page 24
)0202 ‫حتليل الدوائر الكهربائية (كهر‬
‫ كلية اهلندسة‬- ‫جامعة سلمان بن عبد العزيز‬
‫ أمحد مصطفى حشني‬.‫د‬
‫قشم اهلندسة الكهربائية‬
If we replace
We can obtain the all equations for parallel resonance.
Example:
A parallel resonance circuit has a resistance of 2 k Ω and half-power frequencies of
86 kHz and 90 kHz:
(a) Design the appropriate capacitance and inductance for that condition,
(b) Determine the resonant frequency
(c) Calculate the bandwidth and the quality factor
R = 2000 Ω
f1 = 86000 Hz  ω1 = 540.3539 krad/s
f2 = 90000 Hz  ω2 = 565.4867 krad/s
B = f2 - f1 = 90 - 86 = 4 kHz = 25.1327 krad/s
For parallel resonance,
Chapter Six: Frequency Analysis
By Dr. Ahmed Mustafa Hussein
EE2020
Page 25
‫جامعة سلمان بن عبد العزيز ‪ -‬كلية اهلندسة‬
‫حتليل الدوائر الكهربائية (كهر ‪)0202‬‬
‫قشم اهلندسة الكهربائية‬
‫د‪ .‬أمحد مصطفى حشني‬
‫√‬
‫‪Page 26‬‬
‫‪EE2020‬‬
‫‪By Dr. Ahmed Mustafa Hussein‬‬
‫√‬
‫‪Chapter Six: Frequency Analysis‬‬