Notes for course EE1.1 Circuit Analysis 2004-05
TOPIC 7 – FREQUENCY RESPONSE AND FILTERING
Objectives

The definition of frequency response function, amplitude response and phase response

Frequency response function and filtering

Frequency response function and Fourier analysis

The series and parallel tuned circuit

2nd-order passive filters

Bode plots
1
1.1
THE FREQUENCY RESPONSE FUNCTION
General
In the time domain, circuits may be characterised by their transient response
The oscilloscope enables us to observe the transient behaviour of a system's input and output
signals
In the frequency domain, circuits are characterised by their frequency response function
The laboratory instruments which measure frequency content of signals and systems are the
network analyser, the gain-phase test set and the spectrum analyzer
We begin with a definition of frequency response function and explore its use through examples
later
1.2
Definitions
Consider a general linear circuit having no internal independent sources, with a sinusoidal input (or
excitation) waveform x(t) and an output (or response) waveform y(t):
The phasor equivalent circuit is as follows:
The input signal is represented by its phasor, having amplitude Xm and angle θ
The steady state sinusoidal response phasor will have amplitude Ym and angle θ + φ
The system function is defined by:
H (ω ) =
Y Ym ∠θ + φ Ym
=
=
∠φ
X m ∠θ
Xm
X
We have included the frequency ω as an argument for the system function because, in the case of
circuits containing inductors and/or capacitors, the ratio of the output phasor to the input phasor is
frequency dependent
Topic 7 – Frequency response and filtering
It is common practice to write H(jω) for H(ω) because ω can arise only from the impedance of an
inductor (jωL) or from the impedance of a capacitor 1/(jωC), where it is always associated with j.
H(ω) contains all the information we need to know about the circuit, provided we know its value for
each value of ω
1.3
Amplitude and phase response functions
The frequency-domain description of a signal consists of a phasor with amplitude and phase at a
particular frequency
We adopt the same idea for the system function H(ω)
We define the amplitude response function:
A (ω ) = H (ω ) =
Ym
Xm
The phase response function is defines as:
φ (ω ) = ∠H (ω )
Notice that since H(ω) is a function of ω, both A and φ are functions of ω
We can now state that the amplitude of the circuit output signal is given by:
Ym = A (ω ) X m
and its phase by
∠Y (ω ) = φ (ω ) + θ
Thus, we multiply the input signal amplitude by the system gain and add its phase angle to the
system phase shift to obtain the output amplitude and phase angle, respectively
1.4
Frequency Response measurement: Gain and Phase
Consider the following set-up:
We show a laboratory signal generator applying a cosine waveform with unit amplitude and general
frequency ω to our system under test
The AC steady-state response is measured at a frequency ω1, then the input signal is changed to a
new frequency ω2 and the response is measured again
By measuring the steady-state output signal, we can determine the value of the gain A(ω) and the
phase shift φ(ω) at each frequency
We can then plot the frequency response, that is, the gain and the phase versus frequency, as
follows:
2
Topic 7 – Frequency response and filtering
Gain
Phase Shift
Laboratory instruments can automatically vary the frequency, determine the gain and phase shift at
each frequency; they are called gain and phase test sets or network analysers
1.5
Example
Let's look at a simple RC circuit example:
Suppose that we run a test by applying a sinusoidal source at the circuit input and adjusting it so
that its amplitude is one volt and its frequency is variable
We can use AC steady-state analysis to compute the response
The AC steady-state equivalent circuit is as follows:
where we have treated the frequency ω as a variable
Using the voltage divider rule, the response phasor is:
1
1
1
jω C
Y=
× 1∠0 =
× 1∠0 =
∠ − tan −1 (ω CR )
1
2
1
+
j
ω
CR
1 + (ω CR )
R+
jω C
Since 1∠00 is the input phasor X , we may write:
Y=
1
1
1∠0 =
X
1 + jω CR
1 + jω CR
From the definition of H(ω), we have:
H (ω ) =
Y
1
=
X 1 + jω CR
Thus:
A(ω ) =
1
1+ (ωCR)
2
φ (ω ) = −tan−1 (ωCR)
The plot of the amplitude response function A(ω) of the circuit is as follows:
3
Topic 7 – Frequency response and filtering
The amplitude response of this circuit passes low frequencies and suppresses high frequencies
Higher-frequency components in any input signal will be attenuated much more than those at low
frequencies; a filter with such a response is called a lowpass filter
Furthermore, if we make the time constant RC larger, we reduce the high-frequency gain, and if we
make the time constant smaller, we increase the high-frequency gain
Thus a circuit of resistors and one or more reactive elements not only has a transient response but
also has a frequency response function which can effectively filter signals; we consider an example
2
2.1
THE IDEA OF FILTERING
Illustrative example
We explore the idea of filtering of signals in order to remove noise or interference
Suppose a sinusoidal signal is transmitted over a noisy channel that adds an interfering signal that
we wish to remove by a filter circuit
Let’s assume the "noisy" signal is described by:
x ( t ) = cos(2πt ) + 0.5cos(200πt )
The term cos(2πt ) is the desired signal and the term cos(200πt ) represents the additive noise
A plot of one period of the signal with noise is shown:
:
If we could come up with a "filter" that would pass the sinusoid at the frequency of 2π rad/s and
block the one at 200 π rad/s, we would have achieved our objective
We will assume that the noisy signal is available as a voltage source having the prescribed x(t) as its
waveform
Let us use the simple RC lowpass filter circuit for which we derived the amplitude response
function and plotted it:
4
Topic 7 – Frequency response and filtering
1
A (ω ) =
1 + (ω CR )
2
The noise in our signal is a sinusoid with a frequency 100 times that of the signal
Consider making the RC time constant such that:
2π rad/s <<
1
<< 200π rad/s
RC
Thus, let's see what happens if we adjust it so that
1
= 20π rad/s
RC
This is 10 × higher than our signal frequency and 10 × lower than the noise frequency
We can calculate the amplitude response of the circuit at the signal frequency and at the nose
frequency:
1
A (ω ) =
1 + (ω CR )
A ( 2π ) =
2
1
⎛ 2π ⎞
1+ ⎜
⎝ 20π ⎟⎠
A ( 200π ) =
1
=
2
⎛ ω ⎞
1+ ⎜
⎝ 20π ⎟⎠
1
=
⎛ 1⎞
1+ ⎜ ⎟
⎝ 10 ⎠
1
⎛ 200π ⎞
1+ ⎜
⎝ 20π ⎟⎠
2
2
=
2
= 0.995
1
1 + (10 )
2
= 0.0995
We can use superposition to compute the circuit output signal resulting from our noisy signal as the
input signal:
y ( t ) = A ( 2π ) cos ⎡⎣ 2π t + φ ( 2π ) ⎤⎦ + A ( 200π ) cos ⎡⎣ 200π t + φ ( 200π ) ⎤⎦
(
)
(
= 0.995 cos 2π t − 5.7 + 0.0995 cos 200π t − 84.3
)
where we have calculated the phase response at the two frequencies, but the phase is irrelevant in
this case
The signal component has been almost unchanged (amplitude reduced by about one-half of one
percent), but the noise waveform has been attenuated by a factor of 1/20
5
Topic 7 – Frequency response and filtering
A plot of one period of the response is shown:
There is now no visible trace of the contaminating noise
We have successfully used our frequency domain approach to come up with the circuit that did the
job we wanted it to do
The circuit we used is called a 1st order lowpass filter
The corresponding 1st order highpass filter shown:
The circuit is the same as the one we have worked with, except that the capacitor and resistor have
been interchanged
We can use the voltage divider rule to obtain the frequency response function and the amplitude and
phase responses:
H (ω ) =
A (ω ) =
Y
R
1
1
⎛ 1 ⎞
=
=
=
∠ tan −1 ⎜
1
⎝ ω CR ⎟⎠
2
X R+ 1
1− j
⎛ 1 ⎞
1+ ⎜
jω C
ω CR
⎝ ω CR ⎟⎠
1
⎛ 1 ⎞
1+ ⎜
⎝ ω CR ⎟⎠
2
⎛ 1 ⎞
φ (ω ) = ∠ tan −1 ⎜
⎝ ω CR ⎟⎠
We have:
A (ω ) ω →0 = 0
A (ω ) ω →∞ = 1
Hence, for this circuit, a high-frequency signal, considered to be the signal, is passed, and a lowfrequency signal, considered to be the noise, is blocked
The plot of the amplitude response is as for the lowpass filter but with the frequency scale inverted
2.2
Bandpass Filters and Tuning
Another useful application of frequency response is "tuning" a radio receiver
Radio stations are limited by law to have a certain bandwidth
6
Topic 7 – Frequency response and filtering
The frequency plot shows the permissible frequency limits for two hypothetical radio stations,
KOKA and KOLA
The continuous curve represents the amplitude spectrum of the very weak signal that is picked up
by the antenna of our radio receiver
The signal strength of the one we desire (say KOKA) is weaker than the other (KOLA)
If we do not select just one station and reject the other, we will hear a mixture of both stations in
our speaker
Our solution is to pass the composite signal through a bandpass filter
Ideally, the filter should have an amplitude response having the rectangular shape shown in the
figure
2.3
Other Types of Filter
Consider the spectrum of the audio signal in a power amplifier in an audio sound system:
It is unfortunately true that an amplifier with a very high gain also picks up unwanted disturbances,
and one of the most common interfering signals comes from the AC power cables
This interference is in the form of a sinusoidal signal whose frequency is 50 Hz, or 100π rad/s,
which causes a hum in the speaker
Assuming that the audio signal components in a frequency range close to the interfering frequency
of 100π rad/s are not very important to the intelligibility of the waveform, we can pass the
composite waveform through a band-reject (or bandstop) filter having the frequency response
shown above
This eliminates only the interfering waveform and passes our desired audio signal relatively
unaffected, providing the "notch" is very narrow
Thus there are a number of standard filter frequency response types that can be applied in a host of
practical situations
They are called lowpass, highpass, bandpass and band-reject filters
Their ideal gain versus frequency templates are shown:
lowpass
highpass
bandpass
7
band-reject
Topic 7 – Frequency response and filtering
Actual filters will not have these "brick wall" responses; that is, they will not change abruptly from
one value to another as the frequency changes
The first-order RC lowpass filter was far from its ideal template
With more circuit elements and more sophisticated design procedures, one can approximate the
ideal filter frequency response characteristic much more closely
There are catalogues containing tables of pre-designed filters based upon the standard types and
some standard types of approximating functions, such as Butterworth, Chebyshev and elliptic
In some applications, such as audio, the phase response of a filter has very little effect on perceived
sound so it is sufficient to consider only the amplitude response
In other applications, such as television, phase characteristics are important
In general, for distortion-less filtering of a signal, the filter gain should be constant and the phase
should be linear over the frequency range of the waveform
One standard filter type we have not mentioned is the all-pass filter
The gain is constant with frequency but the phase characteristic can be used to compensate for
nonlinear phase characteristic in another circuit
Amplitude
3
3.1
Phase
QUALITY FACTOR FOR INDUCTOR AND CAPACITOR
Definition
Tuning of a radio receiver clearly requires a bandpass filter
The simplest example of a bandpass filter is the LC tuned circuit
Ideal inductors and capacitors used in such a circuit should only store energy and not dissipate any
Practical inductors and capacitors do dissipate some energy
We start by defining a factor that measures the quality of an energy storage element
Consider the AC steady-state response of the two-terminal element shown:
We define the quality factor, Q-factor or Q, by the equation:
Q = 2π
peak stored energy
w
= 2π P
energy dissipated per cycle
wD
In general, we will pick either the voltage or the current to have zero phase angle as a reference
For an ideal lossless inductor or capacitor, the energy dissipated per cycle is zero, implying that the
Q-factor is infinite
8
Topic 7 – Frequency response and filtering
Note that Q-factor is dimensionless
3.2
Q of a lossy inductor
Inductors are constructed in the form of a coil of wire having finite resistance
Thus, a practical model of an inductor consists of an ideal inductor in series with a small resistance
r s:
We assume that the current i(t) is sinusoidal and of the form:
i( t ) = Im cos(ωt + φ )
The energy stored in an inductor is given by:
1
w ( t ) = Li 2 ( t )
2
In AC steady-state terms, the stored energy is:
1
w ( t ) = LIm2 cos 2 (ωt + φ )
2
The peak value of stored energy is:
1
w P = LIm2
2
The only element that dissipates (or absorbs) energy is the resistor which shares the same current as
the inductor
Energy is the integral of power; the energy dissipated over one full period is:
wD =
1
∫ 0 P (t)dt = T T ∫ 0 P (t)dt = T P (t)
T
T
where < > denotes average value
2
2
wD = T rs i 2 ( t ) = T rs I m
cos2 (ω t + φ ) = Trs I m
cos2 (ω t + φ )
We have the general result:
cos2 ( x ) =
1
1
1+ cos(2x )) =
(
2
2
Hence
1
w D = Trs Im2
2
Therefore, the Q of the lossy inductor is:
1 2
LIm
w
2πL ωL
QL = 2π P = 2π 2
=
=
1
wD
2
Tr
rs
s
Trs Im
2
We have used the fact that T = 1 f and f = ω 2π
9
Topic 7 – Frequency response and filtering
Example 1
A lossy inductor has a series winding resistance of 10 Ω and a nominal value of 10 mH
Find the quality factor at a frequency of 100 krad/s
We then have:
QL =
3.3
ωL 10 5 ×10−2
=
= 100
rs
10
Q of a Lossy Capacitor
A capacitor is constructed in the form of parallel metal plates (perhaps rolled up or folded in the
final construction phase) separated by some sort of dielectric
Thus the dielectric’s finite resistance can be approximated by the equivalent circuit shown:
For a sinusoidal terminal voltage:
v ( t ) = Vm cos(ωt + φ )
The energy stored on the ideal capacitor as a function of time is:
1
1
w ( t ) = Cv 2 ( t ) = CVm2 cos2 (ωt + φ )
2
2
The peak energy stored is, therefore,
1
w P CVm2
2
The energy dissipated in one period in the resistor is:
1 2
TVm2
2
wD = T P(t) = T
Vm cos (ωt + φ ) =
rp
2rp
Hence the capacitor Q-factor is given by:
1
CVm2 2πCr
wP
p
QC = 2π
= 2π 2 2 =
= ωCrp
wD
T
TVm
2rp
Example 2
A lossy capacitor has a parallel dielectric resistance of 10 MΩ and a nominal value of 10 nF
Find the quality factor at a frequency of 100 krad/s
QC = ωCrp = 10 5 ×10 7 ×10−8 = 10 4
Capacitors typically have much higher Q values than do inductors
10
Topic 7 – Frequency response and filtering
3.4
Aide-memoire for inductor and capacitor Q-factor
We note that the inductor and capacitor Q-factors may be written in the following form:
QL =
ω L XL
=
rs
rs
QC = ω Crp =
rp
1 (ω C )
=
rp
XC
where XL and XC are the inductor and capacitor reactance, respectively
We note also that the expressions are dimensionless ratios of impedances involving a wanted (X)
and an unwanted (r) quantity
Note also that the conditions that make the elements ideal, rs → 0, rp → ∞, both make Q → ∞
These considerations allow us to predict Q expressions for other combinations
For example, inductor L with parallel resistance rp must have Q → ∞ for rp → ∞; so the Q
expression must be:
QL =
rp
XL
=
rp
ωL
Capacitor C in series with resistance rs must have Q → ∞ for rs → 0; so the Q expression must be:
QC =
3.5
XC 1 ω C
1
=
=
rs
rs
ω Crs
Series-to-Parallel Transformation for a Lossy Inductor
Consider the lossy inductor:
It is possible to find a parallel circuit which is equivalent to it at a specified single frequency:
Let's compute the admittance of the series sub-circuit:
Y ( jω ) =
1
r − jω L
= s
=
rs + jω L rs2 + (ω L )2
1− j
ωL
rs
⎡ ⎛ ωL⎞2 ⎤
rs ⎢1 + ⎜
⎟ ⎥
⎢⎣ ⎝ rs ⎠ ⎥⎦
=
1 1 − jQL
rs 1 + QL2
For QL >>1, we can write:
Y ( jω ) ≅=
1 1 − jQL
1
1
1
1
1
1
=
−
j
=
+
=
+
rs QL2
QL rs QL2 rs j ω L r QL2 rs jω L
QL2 rs
rs s
At a single frequency ω, this corresponds to the following equivalent sub-circuit:
11
Topic 7 – Frequency response and filtering
where
rp' = QL2 rs
L'= L
Note that this equivalent circuit depends on frequency because QL is a function of frequency
The series and parallel circuits can be equivalent only at one frequency
3.6
The Narrow Band Approximation
The equivalence that we have just derived is strictly valid only at a single frequency because rp'
depends on QL which depends on frequency
If we are only interested in small percentage frequency changes relative to the centre frequency, we
can assume that QL is a constant whose value is that assumed at the centre frequency
This is called the narrow-band approximation
Example 3
Find the parallel equivalent circuit for the lossy inductor of Example 1 with rs = 10 Ω, L = 10 mH at
100 krad/s.
Then find the percentage error in rp’ varies over a frequency range of 99 krad/s to 101 krad/s
Solution
We have already computed the inductor Q to be 100 at 100 krad/s
Assuming that QL >> 1, we have the general expression for rp':
rp'
= QL2 rs
2
⎡ωL ⎤2
ωL )
(
= ⎢ ⎥ rs =
rs
⎣ rs ⎦
At ω = 100 krad/s, we have:
rp'
(ωL)
=
2
rs
10 5 ×10−2 )
(
=
10
2
= 10 5 = 100 kΩ
At ω = 99 krad/s, we have:
rp' =
(ωL)
rs
2
=
(
99 ×10 3 ×10−2
)
2
= 98.01 kΩ
10
At ω = 101 krad/s, we have:
(
2
101×10 5 ×10−2
ωL )
(
'
rp =
=
rs
10
)
2
= 102.0 kΩ
Thus, we see that a variation of ± 1 % in the frequency results in only a ± 2% variation in the
resulting parallel resistance
Hence, the use of a constant rp' of 100 kΩ over this frequency range might be acceptable
12
Topic 7 – Frequency response and filtering
4
4.1
THE PARALLEL TUNED CIRCUIT
The Lossless Tuned Circuit and Resonance
Consider the lossless LC parallel circuit:
.
We assume that the driving source is a sinusoidal current source
We wish to determine the AC steady-state response for the voltage v(t).as a function of frequency ω
The phasor form of the circuit is:
The frequency ω appears in the element impedances, so we do not have to make a special note of its
value on the phasor circuit diagram
The impedance of the two-terminal sub-circuit is:
1
ωL
jω C
Z ( jω ) =
= j
= jX (ω )
2
1
1
−
ω
LC
jω L +
jω C
jω L ×
This result means that the impedance is always purely imaginary
The reactance X(ω) (the imaginary impedance without the j multiplier) can be plotted versus ω:
The reactance goes to infinity at ω = ω0, where
ω0 =
1
LC
This phenomenon is called resonance and the frequency ω0 is called the resonant frequency
4.2
The Lossy Tuned Circuit
Now let's consider the practical situation where the inductor and the capacitor both have finite Q
13
Topic 7 – Frequency response and filtering
If we perform the series-to-parallel transformation on the lossy inductor, we can combine the two
resistors into one
We then obtain the following equivalent sub-circuit:
We note that the parallel resistor is a composite of the loss resistance rp of the capacitor, the
(transformed) parallel equivalent resistance rp’ of the inductor, and any source resistance that might
be present (Norton equivalent for the driving source)
The narrow band approximation is based on the assumption that the resistance R is a constant over
the frequency range of interest
Our first objective will be to find the Q of the sub-circuit at the resonant frequency ω 0 = 1
LC
We apply a test current source to our sub-circuit and adjust it to be a sinusoid at frequency ω0
The phasor equivalent circuit form is:
At the resonant frequency ω 0 = 1 LC we know that the part of the sub-circuit consisting of the
capacitor and the inductor presents an infinite impedance, Z'(jω0) = ∞
(this part of the circuit is the same as a lossless tuned circuit we analysed earlier)
The impedance at the two terminals of the subcircuit is
Z ( jω 0 ) = R
Thus, the phasor terminal voltage: is:
V = RI = RIm∠0
In other words:
Vm = RIm
Let's compute the peak energy stored by our sub-circuit
We know that the energy stored on the capacitor as a function of time is:
14
Topic 7 – Frequency response and filtering
wC ( t ) =
1 2
1
Cv ( t ) = CVm2 cos2 (ω 0t )
2
2
To find the energy stored by the inductor, we first determlne the inductor current in phasor form:
IL =
V
V ∠0
V
= m
= m ∠ − 90

jω 0 L ω 0 L∠90
ω0L
iL ( t ) =
Vm
V
cos ω 0 t − 90 = m sin(ω 0 t )
ω 0L
ω 0L
In the time domain, we have:
(
)
The energy stored in the inductor is, thus:
2
2
⎤
1 2
1 ⎡ Vm
1 Vm
wL ( t ) = LiL ( t ) = L ⎢
sin (ω 0t ) ⎥ =
sin 2 (ω 0t )
2
2
2 ⎣ω0 L
2ω L
⎦
0
We can substitute ω 0 = 1
LC to show that:
1
w L ( t ) = CVm2 sin 2 (ω 0 t )
2
which we can compare with:
wC ( t ) =
1
CVm2 cos2 (ω 0t )
2
This shows that the peak energy stored in the inductor is the same as the peak energy stored in the
capacitor
It also shows that the sum of the inductor and capacitor energies is constant
This means that when the capacitor is storing its maximum energy, the inductor is storing no
energy, and vice versa
The energy is being swapped back and forth between the capacitor and the inductor, and none is
coming from the source
Because the impedance Z'(jω0) = ∞ the current into the parallel LC combination is zero when ω = ωο
Hence, at ω = ωο the source current of our sub-circuit only feeds the resistor
The energy absorbed by the sub-circuit in one period is the same as the energy absorbed by the
resistor in that period:
wD =
1 Vm2
T0
2 R
15
Topic 7 – Frequency response and filtering
where T0 is the period corresponding to resonant frequency ω0
The Q of the sub-circuit at ω = ω0, which we will call Q0, is:
1
CVm2
wP
2
Q0 = 2π
= 2π
= ω 0 RC
wD
1 Vm2
T0
2 R
Using LC = 1/ω02, we can write an alternative expression for Q0:
⎛ 1 ⎞
R
Q0 = ω 0 R ⎜
=
2⎟ ω L
⎝ Lω 0 ⎠
0
Thus it is shown that the Q-factor of the tuned-circuit is the same as the Q-factor of the inductor or
the Q-factor of the capacitor where the same resistor R is used in each case
4.3
Fractional Frequency Deviation
Now let's return to our parallel tuned-circuit and compute the impedance at its terminals as a
function of the general frequency ω:
We redraw the circuit in the phasor domain:
For this parallel circuit we have:
Z ( jω ) =
=
1
1
1
=
=
2⎤
1
1
⎡
⎡ω ω ⎤
1
1
ω0
+ jωC +
+ jω 0C⎢ − 0 ⎥
+
C
j
ω
+
⎢
⎥
R
jωL R
jω ⎦ R
⎣ω 0 ω ⎦
⎣
R
R
=
⎡ ω ω0 ⎤
⎡ ω ω0 ⎤
1+ jω 0 RC⎢ − ⎥ 1+ jQ0 ⎢ − ⎥
⎣ω 0 ω ⎦
⎣ω 0 ω ⎦
This is a standard form for the tuned circuit impedance
The expression in the denominator brackets is called the fractional frequency deviation
The magnitude and phase of Z(jω) are:
Z ( jω ) =
⎡ ⎛ ω ω0 ⎞ ⎤
∠Z ( jω ) = − tan −1 ⎢Q0 ⎜
−
⎟⎥
⎣ ⎝ ω0 ω ⎠ ⎦
R
⎡
1 + Q02 ⎢
ω ω0 ⎤
−
⎥
⎣ω0 ω ⎦
2
|Z(jω)| normalized to R can be plotted as follows:
16
Topic 7 – Frequency response and filtering
The peak impedance occurs at ω = ω0 the resonant frequency
The response is plotted for two different values of Q0
The larger the value of Q0, the more selective is the tuned circuit
If we were tuning in a radio station, we would need a large Q0 if there was another station very
close in frequency
4.4
Bandwidth
The passband may be defined formally as the range of frequencies for which the normalized gain is
greater than 1/√2
There are two frequencies, one above ω0 and the other below ω0, at which the response drops to this
value
We call these the upper cut-off frequency ωU and the lower cut-off frequency ωL, respectively
We define the bandwidth of the tuned circuit by:
B = ωU − ω L
Let's compute this bandwidth; since:
Z ( jω )
R
1
=
⎡
1 + Q02 ⎢
ω ω0 ⎤
−
⎥
⎣ω0 ω ⎦
at the upper and lower cut-off frequencies, we must have:
⎡
ω
Q02 ⎢
⎣ω 0
2
ω0 ⎤
− ⎥ =1
ω⎦
or
17
2
Topic 7 – Frequency response and filtering
⎡ω ω ⎤
Q0 ⎢ − 0 ⎥ = ±1
⎣ω 0 ω ⎦
Because ωU > ω0 and ωL < ω0, we see that:
⎡ω
ω ⎤
Q0 ⎢ U − 0 ⎥ = +1
⎣ ω 0 ωU ⎦
⎡ω
ω ⎤
Q0 ⎢ L − 0 ⎥ = −1
⎣ω 0 ω L ⎦
We will solve first for the upper cut-off frequency by letting
x−
1 1
=
x Q0
or
x2 −
1
x −1 = 0
Q0
The solution is
⎡ 1 ⎤2
1
x=
± ⎢
⎥ +1
2Q0
⎣ 2Q0 ⎦
We take the positive sign in order to get a positive frequency:
⎡
⎤
⎡ 1 ⎤2 ⎥
1
⎢
ωU = ω 0
+ ⎢
⎥ + 1⎥
⎢ 2Q0
⎣ 2Q0 ⎦
⎣
⎦
Exchanging ωL. for ωU and changing the sign of Q0 gives:
⎡
⎤
⎡ 1 ⎤2 ⎥
1
⎢
ωL = ω0 −
+ ⎢
⎥ + 1⎥
⎢ 2Q0
⎣ 2Q0 ⎦
⎣
⎦
The bandwidth is given by the simple expression:
B = ωU − ω L =
ω0
Q0
and we have:
Q0 =
ω0
ω0
=
B ωU − ω L
Notice that ωL and ωU are not arithmetically symmetric relative to ω0 (this would imply
ω L + ωU = 2ω 0 )
It may be shown that they are geometrically symmetric, ie ω LωU = ω 0
How might we measure ω0 and Q0 in the laboratory?
Looking back at our resonance curve, we see that we can search for the peak frequency, which is ω0
Next, we find the upper cut-off frequency by noting the frequency at which the response has
dropped to 1/√2 times the peak value (1/√2 = 0.707) and identifying it as ωU
18
Topic 7 – Frequency response and filtering
Finally, we find ωL in a similar manner and subtract them to find the bandwidth
Then we can compute Q0
4.5
Phase Shift
Although we have considered the magnitude or gain characteristic, the phase shift is important for
some applications, including TV and digital transmission
It is of interest to see how the phase shift of the parallel tuned circuit varies with frequency
We now plot φ(ω) as given by the equation we derived for two values of Q0:
⎡ ⎛ ω ω0 ⎞ ⎤
∠Z ( jω ) = − tan −1 ⎢Q0 ⎜
−
⎥
ω
ω ⎟⎠ ⎦
⎝
0
⎣
Observe that the phase shift is zero at resonance and approaches ±π/2 rad (±90o) for frequencies
below and above the resonant frequency
We can define resonance as the frequency at which the tuned-circuit impedance is purely resistive
(φ(ω) = 0)
A look at the equation for φ(ω) shows immediately that this frequency is ω0
5
5.1
THE SERIES TUNED CIRCUIT
The Lossless Series Tuned Circuit
Consider an ideal series tuned circuit:
Its behaviour is very much like that of the parallel tuned circuit, except that all the properties of the
latter which we considered on an impedance basis hold for this circuit on an admittance basis
The admittance of the series tuned circuit is:
Y ( jω ) =
1
1
1
jω
ω L
=
=
=
j
= jB(ω )
Z ( jω ) jωL + 1
L ( jω ) 2 + ω 02
ω 02 − ω 2
jωC
where
ω0 =
and B(ω) is the susceptance
19
1
LC
Topic 7 – Frequency response and filtering
A sketch of B(ω) versus ω is shown:
Notice that the susceptance approaches infinity at ω = ω0; this means that it is equivalent to a short
circuit at ω0
The admittance of the series tuned circuit behaves precisely like the impedance of the parallel tuned
circuit
5.2
The Parallel-to-Series Transformation for the Lossy Capacitor
Consider the lossy capacitor equivalent circuit:
The admittance is:
Y ( jω ) = jωC +
1
rP
Inverting to obtain the impedance:
Z ( jω ) =
1
1
rP
rP
r − jrPQC
=
=
=
= P
Y ( jω ) 1 + jω C 1 + jω CrP 1 + jQC
1 + QC2
rp
If we assume that QC >> 1 (a good approximation for a capacitor), then:
Z ( jω ) ≅
rP − jrPQC
QC2
=
rP
QC2
+
rP
r
1
= P2 +
jω CrP QC jω C
This gives the series equivalent shown:
with
rS =
rP
QC2
C'= C
Since QC depends on frequency, this equivalence is strictly valid only at the frequency for which QC
is determined
20
Topic 7 – Frequency response and filtering
5.3
The Q of the Lossy Series Tuned Circuit at Resonance
Look, once again, at the series tuned circuit, assuming that both the inductor and the capacitor are
lossy:
The series resistance R includes any driving source resistance (considered as a Thevenin equivalent)
along with the inductor series loss resistance and the transformed capacitor parallel loss resistance
Let's apply a sinusoidal test voltage source to our sub-circuit and test the Q of the entire sub-circuit
at the frequency ω0
We see right away that as jω0L + (l/jω0C) = 0, we have Z'(jω0) = l/Y'(jω0) = 0
So Z(jω0) = R
Thus, the current is in phase with the source voltage; furthermore, the voltage drop across the
inductor/capacitor series combination is zero because its series impedance is zero
Hence, we can immediately compute
1 Vm2
L 2
w
ω L
Q0 = 2π × P = 2π × 2 2R = 0
wD
R
1 Vm
T0
2 R
where T0 is the period of a sinusoid at the resonant frequency ω0
By making use of the equivalence:
ω0 =
1
LC
we can obtain an equivalent expression for Q0:
Q0 =
ω0 ⎛ 1 ⎞
1
=
⎜
⎟
2
R ⎝ Cω 0 ⎠ ω 0CR
As a check, we can see from the circuit that the infinite Q condition is R = 0
It is easy to show that the peak energies stored by the inductor and the capacitor are equal to one
another - and therefore to the peak energy stored by the sub-circuit itself.
5.4
Using fractional frequency deviation
Now compute the impedance of the lossy series tuned circuit:
21
Topic 7 – Frequency response and filtering
Z ( jω ) = R + jωL +
⎡ ω L ⎛ jω ω ⎞⎤
1
= R⎢1+ 0 ⎜
+ 0 ⎟⎥
jωC
R ⎝ ω 0 jω ⎠⎦
⎣
So the admittance is:
1
R
Y ( jω ) =
⎡
⎛ ω ω 0 ⎞⎤
− ⎟⎥
⎢1+ jQ0 ⎜
⎝ ω 0 ω ⎠⎦
⎣
If we compare this with the Z(jω) for the parallel tuned circuit developed in the preceding section,
we will see that they are identical in form
The magnitude and phase are:
Y ( jω ) =
1
R
⎛ ω ω0 ⎞
1 + Q02 ⎜
−
⎝ ω 0 ω ⎟⎠
⎡ ⎛ ω ω0 ⎞ ⎤
∠Y ( jω ) = − tan −1 ⎢Q0 ⎜
−
⎥
ω
ω ⎟⎠ ⎦
⎝
0
⎣
2
|Y(jω)| normalized to l/R and ∠ Y(jω) can be sketched:
The admittance of the series tuned-circuit has the same form as the impedance of the parallel tunedcircuit
5.5
Effect of resistor on resonant frequency
A two-terminal circuit is said to be resonant at any nonzero frequency for which the AC steadystate impedance is purely real
This does not always occur where an inductor and a capacitor produce equal and opposite
reactances, as shown in the next example
Example 6
Find the resonant frequency of the following circuit:
22
Topic 7 – Frequency response and filtering
Solution
The AC steady-state impedance is:
1
R (1 − jω CR )
R
jω C
Z ( jω ) = jω L +
= jω L +
= jω L +
2
1
1 + jω CR
1 + (ω CR )
R+
jω C
R×
=
R
1 + (ω CR )
2
⎛
⎞
CR 2
+ jω ⎜ L −
2⎟
1 + (ω CR ) ⎠
⎝
Setting the imaginary part to zero gives ω = 0 and
CR 2
L
1 ⎛
L ⎞
ω 02 =
⎜1−
⎟
LC ⎝ CR 2 ⎠
1+ (ω 0CR) =
2
ω0 =
1
L
1−
LC
CR 2
If R → ∞, we see that the resonant frequency approaches that of an ideal series tuned circuit
The second radical represents the change in resonant frequency due to finite R
6
6.1
2ND-ORDER PASSIVE FILTERS
General
The tuned circuits which we have investigated have impedance and asmittance functions which
exhibit frequency selectivity
We now look into filters; a filter may be defined as a 2-port circuit whose frequency response
function H (ω ) = Vo Vi exhibits frequency selectivity
We begin by considering some examples
6.2
Filtering Examples
The following example investigates a second-order bandpass filter:
Example 4
Find the voltage transfer function of the circuit shown and plot its gain and phase characteristics
23
Topic 7 – Frequency response and filtering
Solution
Notice that there is a series tuned circuit between the voltage source (the filter input) and the output
terminal
Calling this impedance Z(jω), we have:
H (ω ) =
Vo
RL
=
Vs Z ( jω ) + RL
Now we already know the functional form of Y(jω) for the lossy series tuned circuit:
Y ( jω ) =
1
Rs
⎛ ω ω0 ⎞
1 + jQ0 ⎜
−
⎝ ω 0 ω ⎟⎠
Hence, we obtain
H (ω ) =
RL
1
+ RL
Y ( jω )
=
RL
⎛ ω ω0 ⎞
RL + Rs + jRsQ0 ⎜
−
⎝ ω 0 ω ⎟⎠
RL
=
H0
RL + Rs
=
⎛ ω ω0 ⎞
⎛ ω ω0 ⎞
Rs
1+ j
Q0 ⎜
−
1 + jQ0' ⎜
−
⎟
RL + Rs
⎝ ω0 ω ⎠
⎝ ω 0 ω ⎟⎠
where
H0 =
RL
Rs + RL
Q0' =
Rs
Q0
Rs + RL
Thus, our voltage transfer function has the same form as the admittance of the series tuned circuit or
the impedance of the parallel tuned circuit
There are only two differences: Qo has been decreased to Q0’ by the additional resistance and Ho is
a voltage gain, not an admittance or impedance
The gain and phase variations with frequency are shown:
Note that they are identical in form with those for the parallel and series tuned circuits
The next example discusses a second-order bandstop filter
Example 5
Find the voltage gain transfer function of the circuit below and plot its gain and phase versus ω
24
Topic 7 – Frequency response and filtering
Solution
We start by recognizing that a major component of this circuit is a parallel tuned circuit, whose
impedance we have already determined
Thus, we can write the voltage gain transfer function as:
⎡
⎛ ω ω0 ⎞ ⎤
RL ⎢1 + jQ0 ⎜
−
⎥
⎝ ω 0 ω ⎟⎠ ⎦
Vo
RL
RL
⎣
H (ω ) =
=
=
=
RP
Vs Z ( jω ) + RL
+ RL R + R ⎡1 + jQ ⎛ ω − ω 0 ⎞ ⎤
⎥
P
L⎢
0⎜
⎛ ω ω0 ⎞
⎝ ω 0 ω ⎟⎠ ⎦
⎣
1 + jQ0 ⎜
−
⎝ ω 0 ω ⎟⎠
⎛ ω ω0 ⎞
⎛ ω ω0 ⎞
1 + jQ0 ⎜
−
1
+
jQ
0
⎜⎝ ω − ω ⎟⎠
⎝ ω 0 ω ⎟⎠
RL
0
=
= H0
RP + RL
⎛ ω ω0 ⎞
⎛ ω ω0 ⎞
RL
1+ j
Q0 ⎜
−
1 + jQ0' ⎜
−
⎟
RP + RL
⎝ ω0 ω ⎠
⎝ ω 0 ω ⎟⎠
where
H0 =
RL
RL + RP
Q0' =
RL
Q0
RL + RP
Let's investigate this function by converting it to Euler form by taking the magnitude and angle:
2
ω0 ⎞
− ⎟
⎝ω0 ω ⎠
2
⎛
ω0 ⎞
'2 ω
1+ Q0 ⎜
− ⎟
⎝ω0 ω ⎠
⎛ω
A(ω ) = H (ω ) = H 0
1+ Q02 ⎜
and
⎡ ⎛ ω ω ⎞⎤
⎡ ⎛ ω ω ⎞⎤
φ (ω ) = ∠H (ω ) = tan−1⎢Q0 ⎜
− 0 ⎟⎥ − tan−1⎢Q0' ⎜
− 0 ⎟⎥
ω
ω
ω
ω ⎠⎦
⎝
⎠
⎝
⎣
⎦
⎣
0
0
We show a plot of these functions:
25
Topic 7 – Frequency response and filtering
We have used ω0 = 1, H0 = 0.1 and Q0 = 20 for Q0; this makes Q0’ = 2
6.3
Type of frequency response function by inspection
For the above example, we can predict the type of response it has before we carry out detailed
analysis
Consider the behaviour of the inductor and capacitor at the extreme frequencies ω = 0 and ω → ∞
Z ω →0
Z ω →∞
Element Type
Impedance
Inductor
Z L = jω L
ZL = 0
ZL = ∞
(short-circuit) (open-circuit)
Capacitor
ZC =
1
jω C
ZC = ∞
ZC = 0
(open-circuit) (short-circuit)
Hence the inductor behaves like a short-circuit at zero frequency and an open-circuit as frequency
tends to infinity
The capacitor behaves like an open-circuit at zero frequency and a short-circuit as frequency tends
to infinity
Note that the equivalents for zero frequency are the same as the DC steady state equivalents we
used in transient analysis (setting frequency to zero implies that the analysis is a DC analysis)
Note further that at resonance, the parallel tuned-circuit is equivalent to an open-circuit and the
series tuned-circuit is equivalent to a short-circuit
We can now apply this to our band-stop circuit example:
At zero frequency the inductor is a short-circuit and at infinite frequency the capacitor is a shortcircuit; therefore vo = vs at these extreme frequencies; this implies that the amplitude response is
unity and the phase is zero
At resonance, the parallel tuned-circuit will become an open-circuit and the circuit simplifies to a
potential divider consisting of Rp and RL
The gain depends on the resistor values and is H0 = 0.1 in this case
The phase at resonance must be zero
Notice that a good notch filter (this one is not terribly good!) would have a large value of Rp relative
to the value of RL and a high Q0
Thus, the notch would be narrower and deeper
This by-inspection approach can be applied to most passive circuits in order to determine the type
of response (low-pass, high-pass, band-pass, band-stop, all-pass)
26
Topic 7 – Frequency response and filtering
6.4
Canonical Forms for 2nd-Order Filters
We now present a set of standard, or canonical, forms for the three most basic standard 2nd-order
filter types: lowpass, bandpass, highpass and bandstop:
H LP (ω ) =
H BP (ω ) =
H HP (ω ) =
1
2
⎛ jω ⎞
1 jω
⎜⎝ ω ⎟⎠ + Q ω + 1
0
0 0
1 jω
Q0 ω 0
2
⎛ jω ⎞
1 jω
⎜⎝ ω ⎟⎠ + Q ω + 1
0
0 0
⎛ jω ⎞
⎜⎝ ω ⎟⎠
0
2
2
⎛ jω ⎞
1 jω
⎜⎝ ω ⎟⎠ + Q ω + 1
0
0 0
2
H BS (ω ) =
⎛ jω ⎞
⎜⎝ ω ⎟⎠ + 1
0
2
⎛ jω ⎞
1 jω
⎜⎝ ω ⎟⎠ + Q ω + 1
0
0 0
s may be used as a place-holder for jω/ω0
These functions all have the same denominator
If we wish, we can relate these expressions to those for tuned circuit impedances and/or admittances
by restructuring the denominator polynomial:
2
⎛ jω ⎞
⎛ ω ω0 ⎞ ⎤
1 jω
jω 1 ⎡
D (ω ) = ⎜
+
+1=
−
⎢1 + jQ0 ⎜
⎥
⎟
Q0 ω 0
ω 0 Q0 ⎣
⎝ ω0 ⎠
⎝ ω 0 ω ⎟⎠ ⎦
The second factor is one that we easily recognize to be the denominator of our tuned circuit
admittance or impedance
The above forms clearly show the following limiting values:
ω→0
ω→ω0 ω→∞
HLP
1
–jQ0
0
HBP
0
1
0
HHP
0
–jQ0
1
HBS
1
0
1
If Q0 >> 1, then the lowpass and highpass filters have substantial gain at the frequency ω0
However, for the lowpass filter, the actual maximum gain occurs slightly below ω0 and is a little
higher than Q0; it may be shown that:
27
Topic 7 – Frequency response and filtering
H LP (ω ) max =
Q0
1−
at
1
2Q02
ω p = ω 0 1−
1
2Q02
We can plot amplitude and phase response for the lowpass filter for a range of Q values
Phase response ∠H(jω)
Amplitude response |H(jω)|
It can be seen that for lower values of Q the frequency of peak gain is significantly less than ω0
Notice also that the higher the value of Q, the steeper the phase curve at ω = ω0
For the highpass filter, the maximum gain is the same but it occurs slightly above ω0:
ω0
ωp =
1−
1
2Q02
The plots of gain and phase are the same as for the lowpass filter but with the frequency scale
inverted about ω0
7
7.1
MORE GENERAL APPROACH TO 2ND ORDER PASSIVE FILTER DESIGN
General circuit architectures
Rather than looking at specific examples, in this section, we take a more general approach
Consider three general impedances connected in series:
We can regard this circuit as having been obtained by de-activating the sources in a number of
possible compete circuits; the short-circuit shows a possible positions for a voltage source
The impedance looking into the short-circuit is:
Z = Z1 + Z 2 + Z 3
We have dropped the (jω) or (ω) because the use of upper case letters shows that we are working in
the frequency domain
We now form new 2-port circuits by introducing a source into the above circuit in such a way that
de-activation of the source gives the original circuit:
28
Topic 7 – Frequency response and filtering
H ( jω )
Circuit
Vo
Z3
=
Vi Z1 + Z 2 + Z 3
Vo
Z2 + Z 3
=
Vi Z1 + Z 2 + Z 3
We have used voltage division to determine the frequency response functions for these two circuits
Notice that in both cases the denominator of the expression is the same as the impedance of the deactivated circuit; therefore the denominator of the circuit depends on the de-activated circuit; it
governs the natural response of the circuit independently of sources
Consider now three general admittances connected in parallel:
We can regard this circuit as having been obtained by de-activating a source in a number of possible
compete circuits; the open-circuit shows a possible position for a current source
The admittance of the circuit looking into the open-circuit is:
Y = Y1 + Y2 + Y3
We now form new circuits by introducing sources into the above circuit in such a way that deactivation of the new circuits gives the original circuit:
H ( jω )
Circuit
Vo
Y1
=
Vi Y1 + Y2 + Y3
Vo
Y1 + Y2
=
Vi Y1 + Y2 + Y3
We have used voltage division to determine the frequency response functions for these two circuits
29
Topic 7 – Frequency response and filtering
Notice that in both cases the denominator of the expression is the same as the admittance of the deactivated circuit; again the denominator of the circuit depends on the de-activated circuit; it governs
the natural response of the circuit independently of sources
We are now ready to consider replacing general impedances and admittances by R, L and C
elements
7.2
Generation of 2nd order RLC circuits
The first of the four architectures, based on series impedances, can generate three useful circuits,
which we show together with their frequency response functions:
H(jω)
Circuit
H ( jω ) =
H ( jω ) =
H ( jω ) =
Type
1
( jω )
2
LC + jω CR + 1
( jω )2 LC
( jω )2 LC + jω CR + 1
jω CR
( jω )2 LC + jω CR + 1
Lowpass
Highpass
Bandpass
The frequency response functions can be confirmed by use of the voltage division rule
The second architecture generates only one useful circuit:
H(jω)
Type
2
jω ) LC + 1
(
H ( jω ) =
( jω )2 LC + jω CR + 1
Bandstop
Circuit
Notice that all of these four circuits have the same denominator expression; this is expected because
they all reduce to the same form when the input voltage source is deactivated
The third architecture, based on parallel admittances, can generate three useful circuits:
H(jω)
Circuit
H ( jω ) =
H ( jω ) =
30
Type
1
( jω )2 LC + j ω L
R +1
( jω )2 LC
( jω )2 LC + j ω L
R +1
Lowpass
Highpass
Topic 7 – Frequency response and filtering
H ( jω ) =
jω L R
( jω )
2
LC + j ω L R + 1
Bandpass
The fourth architecture generates only one useful circuit:
H(jω)
Circuit
H ( jω ) =
Type
( jω )2 LC + 1
( jω )2 LC + j ω L
R +1
Bandstop
The four circuits based on parallel admittances have the same denominator expression as expected
The general 2nd order denominator expression is:
2
⎛ jω ⎞
1 jω
D (ω ) = ⎜
+
+1
⎟
Q0 ω 0
⎝ ω0 ⎠
where ω0 is the resonant frequency and Q0 is the Q-factor
Comparing this with the expressions in the above tables, the first design equation is:
LC =
1
ω 02
The second design equation for the first set of circuits is:
CR =
R=
1
LC
=
ω 0Q0
Q0
1
Q0
L
C
The second design equation for the first set of circuits is:
L
1
LC
=
=
R ω 0Q0
Q0
R = Q0
L
C
The 2nd order allpass response can not be produced by such simple circuits; we either need to use a
passive circuit with a lattice structure or we can realise it with an active circuit using an operational
amplifier or transistors
8
8.1
BODE PLOTS
Introduction
The Bode plot method is a way of rapidly sketching the gain and phase response of a circuit from its
transfer function
31
Topic 7 – Frequency response and filtering
Here, we will concentrate on the gain plot only
Let's start with an example:
Example 7
Find the transfer function H (ω ) = Vo Vi for the circuit shown:
This circuit is a model for a voltage amplifier, such as the "preamp" in a stereo system
Note that the circuit includes a voltage-controlled voltage source
Solution
As we are interested in frequency response, we use the phasor equivalent circuit:
If use a unit of impedance of kΩ. and a unit of current of mA, the unit of voltage will remain V
We can analyse the circuit in stages:
Vx
=
Vi
1
⎛ 10 5 ⎞
1 + ⎜1 +
jω ⎟⎠
⎝
=
jω
j2ω + 10 5
=
1
jω
2 jω + 5 × 10 4
Let the voltage of the voltage source, 100Vx , be denoted Vy ; then we have:
Vy
Vx
= 100
and
Vo
Z2
1
1
1 10 7
=
=
=
=
⎛1
2 jω + 10 7
Vy Z1 + Z 2 Z1Y2 + 1
7⎞
2 ⎜ + jω 10 ⎟ + 1
⎝2
⎠
Putting the three gain terms together, we have:
H (ω ) =
25 × 10 7 ( jω )
Vo Vo Vy Vx 1 10 7
1
jω
=
=
×
100
×
=
2 jω + 5 × 10 4
Vi Vy Vx V1 2 jω + 10 7
jω + 5 × 10 4 jω + 10 7
(
We will be using this result later, so we will write it using symbols:
32
)(
)
Topic 7 – Frequency response and filtering
H (ω ) =
8.2
K ( jω )
( jω + p1 )( jω + p2 )
Form of the System Function
By generalizing upon the preceding example, we see that the system function of any response
variable of a circuit constructed from our standard supply of elements, namely R, L, C, and
dependent sources, has the form
H (ω ) =
Vo
( jω + z1 )( jω + z2 ) ... ( jω + zm )
=K
Vi
( jω + p1 )( jω + p2 ) ... ( jω + pn )
We call the factors (jω + zl), (jω + z2) ... (jω + zm) the zero factors of H(w) and the factors (jω + p1),
(jω + p2) ... (jω + n) the pole factors of H(ω)
K is the scale factor
In the previous example, there is only one zero factor jω with z1 = 0
There are two finite pole factors: p1 = 5 × 104 and p2 = 107
The scale factor is K = 25 × 107
These zero and pole factors are all simple, or of order one, ie none are repeated
If we plot the gain frequency response of a circuit such as the example circuit, we will obtain a
graph with the general shape shown:
It is zero at DC (ω = 0) and also at infinite frequency (ω → ∞)
The response is constant over a very wide frequency range; the value of gain over this frequency
range is called the midband gain Amb
8.3
Decibels
Consider the following problem:
We investigate an experimental plot of A(ω) taken in the lab and discover that Amb = 100
We notice something strange about the plot in the low frequency range where A(ω) has a value of
approximately 1
As we wish to investigate this effect more closely, we adjust the scale of our plot so that a gain of
one corresponds to a height of, say, 1 cm
If we wish, however, to closely investigate the shape of the plot in the midband range as well-where
A(ω) is l00, we find that our plot must be at least 100 cm tall
This would be a very unwieldy plot to handle
For this reason, we perform a nonlinear transformation on our scale by setting:
AdB = 20log A(ω )
The logarithm is to the base 10, not to the base e
33
Topic 7 – Frequency response and filtering
There is a historical reason for the factor of 20
The Bel (after Alexander Graham Bell, inventor of the telephone) was first defined as the unit of the
logarithm of a power ratio:
⎛P ⎞
Power ratio in Bels = log⎜ 2 ⎟
⎝ P1 ⎠
It was then discovered that for most applications this unit was too large; thus, the decibel became
the standard
This was defined by the equation
⎛P ⎞
Power ratio in deciBels = 10log⎜ 2 ⎟
⎝ P1 ⎠
Assuming that powers are developed in resistors with sinusoidal voltage waveforms, we note that:
Pi =
1 Vm2
2 R
Thus, if both powers are measured relative to the same resistance value, then:
⎛V 2 ⎞
⎛V ⎞
Power ratio in deciBels = 10log⎜ 22 ⎟ = 20log⎜ 2 ⎟
⎝ V1 ⎠
⎝ V1 ⎠
Thus, we have the usual equation:
AdB = 20log A(ω )
There are a number of common values that occur frequently:
A(ω)
AdB
0.001
–60
0.01
–40
0.1
–20
1/2
–6
1/√2
-3
1
0
√2
+3
2
+6
10
+20
100
+40
1000
+60
Notice that a gain of 0 dB does not mean that the response is zero-it means that it has exactly the
same magnitude as the input
Observe also the special values √2, 1/√2 have dB values of ±3, and 2, 1/2 have dB values of ±6
A gain of 1/x has the same magnitude in dB as that of x but opposite sign
34
Topic 7 – Frequency response and filtering
Doubling or halving gain corresponds to adding or subtracting 6 dB
Multiplying or dividing gain by a factor of 10 corresponds to adding or subtracting 20 dB
Example 8
Find the values of gain corresponding to 26 dB, –100 dB, and 34 dB
Solution
We recall that log(xy) = log(x) + log(y); thus, adding dB values corresponds to multiplying the
actual gain values
26 dB = 20 dB + 6 dB
20 dB corresponds to a gain of 10 and 6 dB to a gain of 2
Thus, 26 dB corresponds to a gain of 20
–100 dB = 201og(x)
x = 10–100/20 = 10–5
34 dB = 40 dB – 6 dB
Gain is 100 × 0.5 = 50.
Logarithms of a few basic positive integers are worth remembering:
log(2) = 0.301 log(3) = 0.477 log(5) = 0.699 log(7) = 0.845
Other values can be easily calculated:
log(4) = log(22) = 2log(2) = 0.602
log(9) = log(32) = 2 log(3) = 0.954
20log(30) = 20log(10 × 3) = 20log(10) + 20log(3) = 20 + 20 ×0.477 = 20 + 9.54 = 29.54 dB
8.4
The Logarithmic Frequency Scale
Now look back at the typical amplifier gain curve and consider the frequencies ωU and ωL
Suppose we find in our lab experiment that ωU = 105 rad/s and ωL = 10 rad/s
Suppose we want to investigate both regions in the frequency plot carefully, so we make 10 rad/s
correspond to 1 cm of horizontal length on our plot
We find that in order to include ωU = 105 rad/s we must make the plot 104 = 10,000 cm long
For this reason, we perform a logarithmic transformation on the frequency axis:
ω '= log(ω )
Note that we do not have the factor of 20 present in this case as we do for the gain
Consider a normal ω and a log(ω) scale superimposed:
35
Topic 7 – Frequency response and filtering
We note that multiplying any given frequency value by 10 results in adding one unit of log
frequency; we call such a frequency change a decade
Dividing a given frequency by a factor of 10 results in subtracting one unit of log frequency
Similarly, multiplying or dividing by a factor of 2 results in adding or subtracting 0.3 unit of log
frequency
We call such a frequency change an octave
The name comes from music, where there are eight notes in a frequency increment of a factor of 2
Note that when using a logarithmic frequency scale, the scale may be labelled in ω or log(ω) units
as shown above
Logarithmic frequency scales are also commonly used when the frequency is in Hz
8.5
Gain and phase for the general frequency response function
A Bode plot is a plot of the gain A(ω) in dB versus log(ω) and of the phase φ(ω) versus log(ω)
We start from the general expression for the system function of a circuit:
H (ω ) =
Vo
( jω + z1 )( jω + z2 ) ... ( jω + zm )
=K
Vi
( jω + p1 )( jω + p2 ) ... ( jω + pn )
We then take the absolute value and note that the absolute value of a product is the product of the
absolute values and that the absolute value of a ratio is the ratio of the absolute values:
A(ω ) = H (ω ) = K
jω + z1 jω + z2 ... jω + zm
jω + p1 jω + p2 ... jω + pn
Finally, we take the base 10 logarithm and multiply by 20 to get the dB value:
AdB (ω ) = 20log K + 20log jω + z1 + 20log jω + z2 + ...+ 20log jω + zm
− 20log jω + p1 − 20log jω + p2 − ...− 20log jω + pn
Our interest is to develop quick approximate methods for sketching the Bode plot
Let's investigate the different types of factor one at a time
We assume that all the zero and pole constants (zk and pk, respectively) are real
8.6
The Scale Factor
We look first at the scale factor, 20log|K|
We have sketched this factor:
The solid line represents a scale factor with magnitude greater than unity (positive dB value) and
the dotted line one with a magnitude less than unity
36
Topic 7 – Frequency response and filtering
The horizontal axis is labelled with values of ω, but distances are plotted in terms of log(ω). Thus,
the origin is at ω = 1 rad/s
8.7
The General Factor
Now suppose that q finite zeros or q finite pole constants are equal
Then we have zl = z2 = ... = zq = a or p1 = p2 = ... = pq = a in the general system function
In this case, there is a zero or a pole factor of order q
We will consider factors of the form.
F (ω ) = ( jω + a)
q
If the F(ω) term is a qth-order zero factor, it will appear in the numerator
If it is a pole factor, it will appear in the denominator
Consider the first case for which a = 0
Let's consider:
F (ω ) = ( jω )
q
We start by taking the absolute value, to obtain:
F (ω ) = ( jω ) = jω = ω q
q
q
We have used the fact that the absolute value of a product is the product of the absolute values
Finally, we take the logarithm and multiply by 20:
( )
F (ω ) dB = 20log ω q = 20qlog(ω )
We have plotted this factor assuming a qth-order zero factor at ω = 0
As one must subtract the factor if it represents a pole, we have merely changed the sign in order to
obtain the plot for a pole factor:
A 10-fold increase or decrease in frequency results in a 20q dB increase or decrease in the factor
37
Topic 7 – Frequency response and filtering
This is a slope of ±20q dB/decade
It is also ±6q dB/octave
Suppose, now, that a ≠ 0:
We now have:
F (ω ) = ( jω + a)
q
Next, we take the absolute value:
F (ω ) = ( jω + a) = jω + a
q
q
Finally, taking the logarithm, we have
F (ω ) dB = 20log ( jω + a) = 20qlog jω + a = 20qlog ω 2 + a 2
q
Plotting the log term is not so easy!
For this reason, we resort to an approximation, noting that:
⎧⎪20qlog(ω ); ω >> a
F (ω ) dB ≅ ⎨
⎪⎩20qlog a; ω << a
Note that we have used magnitude signs around the parameter a for it could possibly be negative
8.8
The Straight-Line Approximation
We now show the approximation involved in constructing a linearised Bode gain plot where we
assume that the factor is equal to 20qlog(ω) for ω ≥ |a| and to 20q log|a| 1 for ω ≤ |a|
The resulting plots for a zero factor and for a pole factor are as follows:
The frequency ω = |a| where the staight lines intersect is called the break frequency
Note that the extrapolation of the sloping lines always pass through the origin at (1 rad/sec, 0 dB)
8.9
The Approximation Error
It can be sown that the maximum error between the straight line approximation and the actual curve
occurs at the break frequency w = |a| and is given by:
Error(ω = a ) = ±3q dB
The plus sign will hold for a zero factor and the negative sign for a pole factor
The true plot is above the approximate one by 3q dB at a zero break frequency and is below it by 3q
dB at a pole break frequency
For a single zero or single pole for which q = 1, the maximum error is ±3 dB
An octave above and below the break frequency, we have:
38
Topic 7 – Frequency response and filtering
⎛
1 ⎞
Error(ω = 2 a ) = Error⎜ω = a ⎟ = ±0.97q ≅ q dB
⎝
2 ⎠
A decade above and below the break frequency, we have
⎛
1
Error(ω = 10 a ) = Error⎜ω =
⎝
10
⎞
a ⎟ = ±0.04q ≅ 0 dB
⎠
for moderately small orders of q
We now give an example showing how to sketch a Bode gain plot.
Example 9
Suppose that a circuit or system has the system function
H (ω ) =
Vo
( jω )
= 10 5
Vi
( jω + 10)( jω + 1000) 2
Sketch the linearized Bode gain plot.
Solution
The scale factor is K = 105, or 100 dB
It is plotted as follows:
There is a simple zero factor at ω = 0
The corresponding factor is plotted as follows:
For the simple pole factor (jω + 10) the break frequency is 10 rad/s; the plot is as follows:
Finally, we plot the Bode factor corresponding to the pole factor (jω + 1000)2 of order 2:
In this last plot, we have not preserved the same scale as for the others because this would lead to
problems in presentation
Now we simply add all four plots order to obtain the overall Bode gain plot:
39
Topic 7 – Frequency response and filtering
Notice that we have sketched in the curve, which falls below the linearised approximation
Because the break frequencies are widely separated, the error at each is almost that computed for
each in isolation, i.e., –3 dB at 10 rad/s and –6 dB at 1000 rad/s
The gain for the flat part of the curve may be fixed by adding the individual figures at any
convenient frequency, eg at 10 rad/sec
A(10) = 100 + 20 − 20 −120 = −20 dB
8.10 The Quick Method
There is a much faster way of doing this type of plot
Looking at the Bode plot produced in the above example, we could have started at very low values
of frequency with a line of positive slope dictated by the order of the zero factor at ω = 0
Then we could make the slope break upward by an amount 20q dB/dec at each zero break
frequency or downward by an amount –20q dB/dec at each pole break frequency, where q is the
order of each zero or pole
When any frequency being considered is below a given break frequency, we use the constant value;
when it is above that break frequency, we use ±20qlog(ω)
Example 10
Use the quick method to sketch the Bode gain plot for the following transfer function:
H (ω ) = 4 ×10 5
( jω + 10)( jω + 100)( jω + 500)
( jω ) 2 ( jω + 1000)( jω + 10 4 )
Solution
The first step is to mark the zeros and poles on the log(ω) axis; we use for o for zeros and × for
poles:
The bulk of the solution consists merely of steps in sketching the graph, so we merely sketch the
final shape, starting from the low end of the frequency scale:
We can evaluate the approximate magnitude at ω = 1 rad/s to fix the vertical axis scale:
H (1) ≅ 4 ×10 5
(10)(100)(500) = 2 ×10 4 ≡ 86
(1) 2 (1000)(10 4 )
40
dB
Topic 7 – Frequency response and filtering
The exact plot is 3 dB above our linearized one at the corner frequency at 10 rad/s and 3 dB below
at 104 rad/s
The errors at the other break frequencies cannot be determined so easily because they are not
widely separated from their nearest neighbours
The ‘quick method procedure’ for sketching Bode gain plots can be summarised as follows:
1. Draw a frequency axis.
2. Evaluate the approximate magnitude at one single value of frequency to fix the vertical scale
(typically at a low value of frequency, perhaps ω = 1 rad/s) and draw the vertical axis. Note that
"approximate" means to use the linear approximation for each of the factors.
3. Label all the finite break frequencies with an o for a zero and an x for a pole. If the associated
zero or pole is of order higher than one, label its order in parentheses.
4. Start the Bode plot at low values of frequency with a slope of ±20q dB/dec., where q is the order
of the zero or pole at ω = 0, if any. If there are none, q = 0 and the initial slope is zero.
5. Imagine the frequency to increase slowly. Continue drawing the plot, causing the slope to break
upward by +20q dB/dec. at each break frequency associated with a zero of order q and downward
by -20q dB/dec. at each break frequency associated with a pole of order q. Continue this until the
last break frequency has been included.
Example 12
Find the transfer function and sketch the linearised Bode gain and phase plots for the circuit shown:
Solution
We let the common resistor value be R and the capacitor value be C
The phasor domain equivalent circuit is as follows:
We use the voltage divider rule:
H (ω ) =
R
1+ jωCR jω + 100
=
=
1
2
+
j
ω
CR
jω + 200
R×
jωC
+R
1
R+
jωC
41
Topic 7 – Frequency response and filtering
The gain at low frequencies is 1/2, or –6 dB
There is one finite zero factor (jω + 100) and one finite pole factor (jω + 200)
This gives the following gain plot:
Notice that a slope of 20 dB/decade is the same as a slope of 6 dB/octave; so the high-frequency
assymptotic gain value is 0 dB
This checks with the circuit, because the capacitor is a short circuit for high frequencies and an
open circuit for low frequencies
The Bode plot method is not applicable to complex poles and zeros:
Eg 2nd order lowpass filter based on LC tuned-circuit:
H LP (ω ) =
1
2
⎛ jω ⎞
1 ⎛ jω ⎞
⎜ ⎟ +
⎜ ⎟ +1
⎝ ω 0 ⎠ Q0 ⎝ ω 0 ⎠
Denominator can only be factorised into two real factors D(ω ) = ( jω + α )( jω + β ) only if Qo < 1/2
Phase response ∠HLP(jω)
Amplitude response |HLP(jω)|
In the case of complex poles and zeros, a computer or calculator is needed to plot A(ω) and φ(ω)
9
CONCLUSIONS
In this topic, we have extended the concept of phasor by allowing the frequency ω to be a variable
rather than a constant
We defined frequency response function, amplitude response and phase response
We looked at series and parallel tuned circuits and defined Q-factor and bandwidth
This led to a discussion of filtering and we considered 2nd-order passive filters
Finally we looked at Bode plots for rapid approximate sketching of circuit amplitude responses
42