notes for course ee1.1 circuit analysis 2004-05 topic 7 ...cas.ee.ic.ac.uk/people/dhaigh/ee1.1...

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 THE FREQUENCY RESPONSE FUNCTION

1.1 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 outputsignals

In the frequency domain, circuits are characterised by their frequency response function

The laboratory instruments which measure frequency content of signals and systems are thenetwork analyser, the gain-phase test set and the spectrum analyzer

We begin with a definition of frequency response function and explore its use through exampleslater

1.2 Definitions

Consider a general linear circuit having no internal independent sources, with a sinusoidal input (orexcitation) 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( ) = YX=Ym∠θ + φXm∠θ

=YmXm

∠φ

We have included the frequency ω as an argument for the system function because, in the case ofcircuits containing inductors and/or capacitors, the ratio of the output phasor to the input phasor isfrequency dependent

Topic 7 – Frequency response and filtering

2

It is common practice to write H(jω) for H(ω) because ω can arise only from the impedance of aninductor (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 foreach 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 aparticular frequency

We adopt the same idea for the system function H(ω)

We define the amplitude response function:

A ω( ) = H ω( ) = YmXm

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( )Xmand 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 thesystem 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 generalfrequency ω to our system under test

The AC steady-state response is measured at a frequency ω1, then the input signal is changed to anew 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 thephase shift φ(ω) at each frequency

We can then plot the frequency response, that is, the gain and the phase versus frequency, asfollows:


3

Gain Phase Shift

Laboratory instruments can automatically vary the frequency, determine the gain and phase shift ateach 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 sothat 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:

Y =

1jωC

R + 1jωC

×1∠0 = 11+ jωCR

×1∠0 = 1

1+ ωCR( )2∠− tan−1 ωCR( )

Since 1∠00 is the input phasor X , we may write:

Y =

11+ jωCR

1∠0 = 11+ jωCR

X

From the definition of H(ω), we have:

H ω( ) = YX=

11+ jωCR

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:


4

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 lowfrequencies; 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 wemake 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 butalso has a frequency response function which can effectively filter signals; we consider an example

2 THE IDEA OF FILTERING

2.1 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 thatwe 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 andblock 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 itswaveform

Let us use the simple RC lowpass filter circuit for which we derived the amplitude responsefunction and plotted it:


5

A ω( ) = 1

1+ ωCR( )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 << 1RC

<< 200π rad/s

Thus, let's see what happens if we adjust it so that

�

1RC

= 20π rad/s

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 nosefrequency:

A ω( ) = 1

1+ ωCR( )2=

1

1+ ω20π

⎛⎝⎜

⎞⎠⎟2

A 2π( ) = 1

1+ 2π20π

⎛⎝⎜

⎞⎠⎟2=

1

1+ 110

⎛⎝⎜

⎞⎠⎟2= 0.995

A 200π( ) = 1

1+ 200π20π

⎛⎝⎜

⎞⎠⎟2=

1

1+ 10( )2= 0.0995

We can use superposition to compute the circuit output signal resulting from our noisy signal as theinput signal:

y t( ) = A 2π( )cos 2πt + φ 2π( )⎡⎣ ⎤⎦ + A 200π( )cos 200πt + φ 200π( )⎡⎣ ⎤⎦

= 0.995cos 2πt − 5.7( ) + 0.0995cos 200πt − 84.3( )where we have calculated the phase response at the two frequencies, but the phase is irrelevant inthis case

The signal component has been almost unchanged (amplitude reduced by about one-half of onepercent), but the noise waveform has been attenuated by a factor of 1/20


6

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 thejob 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 havebeen interchanged

We can use the voltage divider rule to obtain the frequency response function and the amplitude andphase responses:

H ω( ) = YX=

R

R + 1jωC

=1

1− j 1ωCR

=1

1+ 1ωCR

⎛⎝⎜

⎞⎠⎟

2∠ tan−1 1

ωCR⎛⎝⎜

⎞⎠⎟

A ω( ) = 1

1+ 1ωCR

⎛⎝⎜

⎞⎠⎟

2 φ ω( ) = ∠ tan−1 1

ωCR⎛⎝⎜

⎞⎠⎟

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 low-frequency 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


7

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 upby 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 inour 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 thefigure

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 frequencyof 100π rad/s are not very important to the intelligibility of the waveform, we can pass thecomposite waveform through a band-reject (or bandstop) filter having the frequency responseshown above

This eliminates only the interfering waveform and passes our desired audio signal relativelyunaffected, providing the "notch" is very narrow

Thus there are a number of standard filter frequency response types that can be applied in a host ofpractical situations

They are called lowpass, highpass, bandpass and band-reject filters

Their ideal gain versus frequency templates are shown:

lowpass highpass bandpass band-reject


8

Actual filters will not have these "brick wall" responses; that is, they will not change abruptly fromone 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 theideal filter frequency response characteristic much more closely

There are catalogues containing tables of pre-designed filters based upon the standard types andsome 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 perceivedsound 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 phaseshould 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 fornonlinear phase characteristic in another circuit

Amplitude Phase

3 QUALITY FACTOR FOR INDUCTOR AND CAPACITOR

3.1 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 energyenergy dissipated per cycle

= 2π wPwD

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 theQ-factor is infinite


9

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 resistancers:

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:

�

w t( ) = 12Li2 t( )

In AC steady-state terms, the stored energy is:

�

w t( ) = 12LIm2 cos2 ωt + φ( )

The peak value of stored energy is:

�

wP = 12LIm2

The only element that dissipates (or absorbs) energy is the resistor which shares the same current asthe inductor

Energy is the integral of power; the energy dissipated over one full period is:

�

wD = P t( )dt0T∫ = T 1

TP t( )dt0

T∫ = T P t( )

where < > denotes average value

wD = T rsi2 t( ) = T rsIm

2 cos2 ωt + φ( ) = TrsIm2 cos2 ωt + φ( )

We have the general result:

�

cos2 x( ) = 121+ cos 2x( )( ) = 1

2

Hence

�

wD = 12TrsIm

2

Therefore, the Q of the lossy inductor is:

�

QL = 2π wPwD

= 2π

12LIm2

12TrsIm

2= 2πLTrs

= ωLrs

We have used the fact that T = 1 f and f =ω 2π


10

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 = ωLrs

= 105 ×10−2

10=100

3.3 Q of a Lossy Capacitor

A capacitor is constructed in the form of parallel metal plates (perhaps rolled up or folded in thefinal 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:

�

w t( ) = 12Cv2 t( ) = 1

2CVm

2 cos2 ωt + φ( )

The peak energy stored is, therefore,

�

wP12CVm

2

The energy dissipated in one period in the resistor is:

�

wD = T P t( ) = T 1rpVm2 cos2 ωt + φ( ) = TVm

2

2rp

Hence the capacitor Q-factor is given by:

�

QC = 2π wPwD

= 2π

12CVm

2

TVm2

2rp

=2πCrpT

= ωCrp

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 =105 ×107 ×10−8 =104

Capacitors typically have much higher Q values than do inductors


11

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 =ωLrs

=XLrs

QC =ωCrp =rp

1 ωC( ) =rpXC

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 Qexpression must be:

QL =rpXL

=rpωL

Capacitor C in series with resistance rs must have Q → ∞ for rs → 0; so the Q expression must be:

QC =XCrs

=1 ωCrs

=1

ωCrs3.5 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( ) = 1rs + jωL

=rs − jωLrs2 + ωL( )2

=1− jωL

rs

rs 1+ωLrs

⎛⎝⎜

⎞⎠⎟

2⎡

⎣⎢⎢

⎤

⎦⎥⎥

=1rs1− jQL1+QL

2

For QL >>1, we can write:

Y jω( ) ≅= 1rs1− jQLQL2 =

1QL2rs

− j 1QLrs

=1

QL2rs

+1

jωLrsrs

=1

QL2rs

+1jωL

At a single frequency ω, this corresponds to the following equivalent sub-circuit:


12

where

�

rp' = QL

2rs 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, wecan 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 at100 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' = QL

2rs = ωLrs

⎡

⎣ ⎢

⎤

⎦ ⎥ 2

rs =ωL( )2rs

At ω = 100 krad/s, we have:

�

rp' =

ωL( )2

rs=

105 ×10−2( )2

10= 105 = 100 kΩ


�

rp' =

ωL( )2

rs=

99 ×103 ×10−2( )2

10= 98.01 kΩ


�

rp' =

ωL( )2

rs=

101×105 ×10−2( )2

10= 102.0 kΩ

Thus, we see that a variation of ± 1 % in the frequency results in only a ± 2% variation in theresulting parallel resistance

Hence, the use of a constant rp' of 100 kΩ over this frequency range might be acceptable


13

4 THE PARALLEL TUNED CIRCUIT

4.1 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 itsvalue on the phasor circuit diagram

The impedance of the two-terminal sub-circuit is:

Z jω( ) =jωL × 1

jωC

jωL + 1jωC

= j ωL1−ω 2LC

= jX ω( )

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 = 1LC

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


14

If we perform the series-to-parallel transformation on the lossy inductor, we can combine the tworesistors 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 mightbe present (Norton equivalent for the driving source)

The narrow band approximation is based on the assumption that the resistance R is a constant overthe frequency range of interest

Our first objective will be to find the Q of the sub-circuit at the resonant frequency

�

ω0 =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

�

ω0 =1 LC we know that the part of the sub-circuit consisting of thecapacitor 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 jω0( ) = R

Thus, the phasor terminal voltage: is:

�

V = RI = RIm∠0

In other words:

�

Vm = RImLet'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:


15

wC t( ) = 12Cv2 t( ) = 1

2CVm

2 cos2 ω0t( )

To find the energy stored by the inductor, we first determlne the inductor current in phasor form:

IL =

Vjω0L

=Vm∠0

ω0L∠90 =

Vmω0L

∠− 90

In the time domain, we have:

�

iL t( ) = Vmω0L

cos ω0t − 90( ) = Vm

ω0Lsin ω0t( )

The energy stored in the inductor is, thus:

wL t( ) = 12LiL2 t( ) = 1

2L Vm

ω0Lsin ω0t( )⎡

⎣⎢

⎤

⎦⎥2=12Vm2

ω 02Lsin2 ω0t( )

We can substitute

�

ω0 =1 LC to show that:

�

wL t( ) = 12CVm

2 sin2 ω0t( )

which we can compare with:

wC t( ) = 12CVm

2 cos2 ω0t( )

This shows that the peak energy stored in the inductor is the same as the peak energy stored in thecapacitor

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 noenergy, and vice versa

The energy is being swapped back and forth between the capacitor and the inductor, and none iscoming 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 theresistor in that period:

�

wD = 12Vm2

RT0


16

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:

Q0 = 2πwPwD

= 2π

12CVm

2

12Vm2

RT0

=ω0RC

Using LC = 1/ω02, we can write an alternative expression for Q0:

Q0 =ω0R1

Lω02

⎛

⎝⎜

⎞

⎠⎟ =

Rω0L

Thus it is shown that the Q-factor of the tuned-circuit is the same as the Q-factor of the inductor orthe 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 afunction of the general frequency ω:

We redraw the circuit in the phasor domain:

For this parallel circuit we have:

�

Z jω( ) = 11R

+ jωC + 1jωL

= 11R

+ C jω + ω02

jω

⎡

⎣ ⎢

⎤

⎦ ⎥

= 11R

+ jω0Cωω0

− ω0ω

⎡

⎣ ⎢

⎤

⎦ ⎥

= R

1+ jω0RCωω0

− ω0ω

⎡

⎣ ⎢

⎤

⎦ ⎥

= R

1+ jQ0ωω0

− ω0ω

⎡

⎣ ⎢

⎤

⎦ ⎥

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( ) = R

1+Q02 ω

ω0− ω0

ω⎡

⎣⎢

⎤

⎦⎥

2 ∠Z jω( ) = − tan−1 Q0

ωω0

−ω0ω

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

|Z(jω)| normalized to R can be plotted as follows:


17

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 veryclose in frequency

4.4 Bandwidth

The passband may be defined formally as the range of frequencies for which the normalized gain isgreater than 1/√2

There are two frequencies, one above ω0 and the other below ω0, at which the response drops to thisvalue

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 −ωL

Let's compute this bandwidth; since:

Z jω( )R

=1

1+Q02 ω

ω0− ω0

ω⎡

⎣⎢

⎤

⎦⎥2

at the upper and lower cut-off frequencies, we must have:

�

Q02 ωω0

− ω0ω

⎡

⎣ ⎢

⎤

⎦ ⎥ 2

=1

or


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�

Q0ωω0

− ω0ω

⎡

⎣ ⎢

⎤

⎦ ⎥ = ±1

Because ωU > ω0 and ωL < ω0, we see that:

�

Q0ωUω0

− ω0ωU

⎡

⎣ ⎢

⎤

⎦ ⎥ = +1 Q0

ωLω0

− ω0ωL

⎡

⎣ ⎢

⎤

⎦ ⎥ = −1

We will solve first for the upper cut-off frequency by letting

�

x − 1x

= 1Q0

or

�

x2 − 1Q0

x −1= 0

The solution is

�

x = 12Q0

± 12Q0

⎡

⎣ ⎢

⎤

⎦ ⎥ 2

+1

We take the positive sign in order to get a positive frequency:

�

ωU = ω012Q0

+ 12Q0

⎡

⎣ ⎢

⎤

⎦ ⎥ 2

+1⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

Exchanging ωL. for ωU and changing the sign of Q0 gives:

�

ωL = ω0 −12Q0

+ 12Q0

⎡

⎣ ⎢

⎤

⎦ ⎥ 2

+1⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

The bandwidth is given by the simple expression:

�

B = ωU −ωL = ω0Q0

and we have:

Q0 =ω0B

=ω0

ωU −ωL

Notice that ωL and ωU are not arithmetically symmetric relative to ω0 (this would implyωL +ωU = 2ω0 )

It may be shown that they are geometrically symmetric, ie

�

ωLωU = ω0How 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 hasdropped to 1/√2 times the peak value (1/√2 = 0.707) and identifying it as ωU


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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 forsome 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:

∠Z jω( ) = − tan−1 Q0ωω0

−ω0ω

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

Observe that the phase shift is zero at resonance and approaches ±π/2 rad (±90o) for frequenciesbelow 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 THE SERIES TUNED CIRCUIT

5.1 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 thelatter 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( ) = 1Z jω( )

= 1

jωL + 1jωC

= 1L

jωjω( )2 + ω0

2 = j ω Lω02 −ω2

= jB ω( )

where

�

ω0 = 1LC

and B(ω) is the susceptance


20

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 shortcircuit at ω0

The admittance of the series tuned circuit behaves precisely like the impedance of the parallel tunedcircuit

5.2 The Parallel-to-Series Transformation for the Lossy Capacitor

Consider the lossy capacitor equivalent circuit:

The admittance is:

�

Y jω( ) = jωC + 1rP

Inverting to obtain the impedance:

Z jω( ) = 1Y jω( ) =

11rp

+ jωC=

rP1+ jωCrP

=rP

1+ jQC=rP − jrPQC1+QC

2

If we assume that QC >> 1 (a good approximation for a capacitor), then:

Z jω( ) ≅ rP − jrPQCQC2 =

rPQC2 +

rPjωCrP

=rPQC2 +

1jωC

This gives the series equivalent shown:

with

�

rS = rPQC

2 C'= C

Since QC depends on frequency, this equivalence is strictly valid only at the frequency for which QC

is determined


21

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 arelossy:

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-circuitat 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 theinductor/capacitor series combination is zero because its series impedance is zero

Hence, we can immediately compute

�

Q0 = 2π × wPwD

= 2π ×

12LVm

2

R212Vm2

RT0

= ω0LR

where T0 is the period of a sinusoid at the resonant frequency ω0

By making use of the equivalence:

ω0 =1LC

we can obtain an equivalent expression for Q0:

Q0 =ω0R

1Cω0

2⎛

⎝⎜

⎞

⎠⎟ =

1ω0CR

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 oneanother - 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:


22

�

Z jω( ) = R + jωL + 1jωC

= R 1+ ω0LR

jωω0

+ ω0jω

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥

So the admittance is:

�

Y jω( ) =

1R

1+ jQ0ωω0

− ω0ω

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥

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( ) =1R

1+Q02 ω

ω0− ω0

ω⎛⎝⎜

⎞⎠⎟

2 ∠Y jω( ) = − tan−1 Q0

ωω0

−ω0ω

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

|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 tuned-circuit

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 steady-state impedance is purely real

This does not always occur where an inductor and a capacitor produce equal and oppositereactances, as shown in the next example

Example 6

Find the resonant frequency of the following circuit:


23

Solution

The AC steady-state impedance is:

Z jω( ) = jωL +R × 1

jωC

R + 1jωC

= jωL +R

1+ jωCR= jωL +

R 1− jωCR( )1+ ωCR( )2

=R

1+ ωCR( )2+ jω L −

CR2

1+ ωCR( )2⎛

⎝⎜

⎞

⎠⎟

Setting the imaginary part to zero gives ω = 0 and

�

1+ ω0CR( )2 = CR2

L

ω02 = 1

LC1− L

CR2⎛ ⎝ ⎜

⎞ ⎠ ⎟

ω0 = 1LC

1− LCR2

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 2ND-ORDER PASSIVE FILTERS

6.1 General

The tuned circuits which we have investigated have impedance and asmittance functions whichexhibit frequency selectivity

We now look into filters; a filter may be defined as a 2-port circuit whose frequency responsefunction 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


24

Solution

Notice that there is a series tuned circuit between the voltage source (the filter input) and the outputterminal

Calling this impedance Z(jω), we have:

H ω( ) = VoVs

=RL

Z jω( ) + RLNow we already know the functional form of Y(jω) for the lossy series tuned circuit:

Y jω( ) =1Rs

1+ jQ0ωω0

− ω0ω

⎛⎝⎜

⎞⎠⎟

Hence, we obtain

H ω( ) = RL1Y jω( ) + RL

=RL

RL + Rs + jRsQ0ωω0

− ω0ω

⎛⎝⎜

⎞⎠⎟

=

RLRL + Rs

1+ j RsRL + Rs

Q0ωω0

− ω0ω

⎛⎝⎜

⎞⎠⎟

=H0

1+ jQ0' ω

ω0− ω0

ω⎛⎝⎜

⎞⎠⎟

where

�

H0 = RLRs + RL

Q0' = Rs

Rs + RLQ0

Thus, our voltage transfer function has the same form as the admittance of the series tuned circuit orthe impedance of the parallel tuned circuit

There are only two differences: Qo has been decreased to Q0’ by the additional resistance and Ho isa 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


25

Solution

We start by recognizing that a major component of this circuit is a parallel tuned circuit, whoseimpedance we have already determined

Thus, we can write the voltage gain transfer function as:

H ω( ) = VoVs

=RL

Z jω( ) + RL=

RLRP

1+ jQ0ωω0

− ω0ω

⎛⎝⎜

⎞⎠⎟

+ RL=

RL 1+ jQ0ωω0

− ω0ω

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

RP + RL 1+ jQ0ωω0

− ω0ω

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

=RL

RP + RL

1+ jQ0ωω0

− ω0ω

⎛⎝⎜

⎞⎠⎟

1+ j RLRP + RL

Q0ωω0

− ω0ω

⎛⎝⎜

⎞⎠⎟

= H0

1+ jQ0ωω0

− ω0ω

⎛⎝⎜

⎞⎠⎟

1+ jQ0' ω

ω0− ω0

ω⎛⎝⎜

⎞⎠⎟

where

H0 =RL

RL + RP Q0

' =RL

RL + RPQ0

Let's investigate this function by converting it to Euler form by taking the magnitude and angle:

�

A ω( ) = H ω( ) = H01+ Q0

2 ωω0

− ω0ω

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

1+ Q0'2 ω

ω0− ω0ω

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

and

�

φ ω( ) = ∠H ω( ) = tan−1 Q0ωω0

− ω0ω

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥ − tan−1 Q0

' ωω0

− ω0ω

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥

We show a plot of these functions:


26

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 detailedanalysis

Consider the behaviour of the inductor and capacitor at the extreme frequencies ω = 0 and ω → ∞

Element Type Impedance Z ω→0 Z ω→∞

Inductor ZL = jωL ZL = 0(short-circuit)

ZL = ∞(open-circuit)

Capacitor ZC =1jωC

ZC = ∞(open-circuit)

ZC = 0(short-circuit)

Hence the inductor behaves like a short-circuit at zero frequency and an open-circuit as frequencytends to infinity

The capacitor behaves like an open-circuit at zero frequency and a short-circuit as frequency tendsto infinity

Note that the equivalents for zero frequency are the same as the DC steady state equivalents weused 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 theseries 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 short-circuit; therefore vo = vs at these extreme frequencies; this implies that the amplitude response isunity and the phase is zero

At resonance, the parallel tuned-circuit will become an open-circuit and the circuit simplifies to apotential 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 relativeto 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 typeof response (low-pass, high-pass, band-pass, band-stop, all-pass)


27

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-orderfilter types: lowpass, bandpass, highpass and bandstop:

HLP ω( ) = 1

jωω0

⎛⎝⎜

⎞⎠⎟

2+ 1Q0

jωω0

+1

HBP ω( ) =1Q0

jωω0

jωω0

⎛⎝⎜

⎞⎠⎟

2+ 1Q0

jωω0

+1

HHP ω( ) =jωω0

⎛⎝⎜

⎞⎠⎟

2

jωω0

⎛⎝⎜

⎞⎠⎟

2+ 1Q0

jωω0

+1

HBS ω( ) =jωω0

⎛⎝⎜

⎞⎠⎟

2+1

jωω0

⎛⎝⎜

⎞⎠⎟

2+ 1Q0

jωω0

+1

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 admittancesby restructuring the denominator polynomial:

D ω( ) = jωω0

⎛⎝⎜

⎞⎠⎟

2+1Q0

jωω0

+1 = jωω0

1Q0

1+ jQ0ωω0

−ω0ω

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

The second factor is one that we easily recognize to be the denominator of our tuned circuitadmittance or impedance

The above forms clearly show the following limiting values:

ω → 0 ω → ω 0 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 littlehigher than Q0; it may be shown that:


28

�

HLP ω( ) max = Q0

1− 12Q0

2

at ω p = ω0 1− 12Q0

2

We can plot amplitude and phase response for the lowpass filter for a range of Q values

Amplitude response |H(jω)| Phase 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:

�

ω p = ω0

1− 12Q0

2

The plots of gain and phase are the same as for the lowpass filter but with the frequency scaleinverted about ω0

7 MORE GENERAL APPROACH TO 2ND ORDER PASSIVE FILTER DESIGN

7.1 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 ofpossible compete circuits; the short-circuit shows a possible positions for a voltage source

The impedance looking into the short-circuit is:

Z = Z1 + Z2 + Z3We have dropped the (jω) or (ω) because the use of upper case letters shows that we are working inthe frequency domain

We now form new 2-port circuits by introducing a source into the above circuit in such a way thatde-activation of the source gives the original circuit:


29

Circuit H jω( )

VoVi

=Z3

Z1 + Z2 + Z3

VoVi

=Z2 + Z3

Z1 + Z2 + Z3

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 de-activated circuit; therefore the denominator of the circuit depends on the de-activated circuit; itgoverns 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 possiblecompete 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 +Y3We now form new circuits by introducing sources into the above circuit in such a way that de-activation of the new circuits gives the original circuit:

Circuit H jω( )

VoVi

=Y1

Y1 +Y2 +Y3

VoVi

=Y1 +Y2

Y1 +Y2 +Y3

We have used voltage division to determine the frequency response functions for these two circuits


30

Notice that in both cases the denominator of the expression is the same as the admittance of the de-activated circuit; again the denominator of the circuit depends on the de-activated circuit; it governsthe natural response of the circuit independently of sources

We are now ready to consider replacing general impedances and admittances by R, L and Celements

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:

Circuit H(jω) Type

H jω( ) = 1jω( )2 LC + jωCR +1 Lowpass

H jω( ) = jω( )2 LCjω( )2 LC + jωCR +1

Highpass

H jω( ) = jωCRjω( )2 LC + jωCR +1 Bandpass

The frequency response functions can be confirmed by use of the voltage division rule

The second architecture generates only one useful circuit:

Circuit H(jω) Type

H jω( ) = jω( )2 LC +1jω( )2 LC + jωCR +1

Bandstop

Notice that all of these four circuits have the same denominator expression; this is expected becausethey 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:

Circuit H(jω) Type

H jω( ) = 1jω( )2 LC + jωL R +1

Lowpass

H jω( ) = jω( )2 LCjω( )2 LC + jωL R +1

Highpass


31

H jω( ) = jωL Rjω( )2 LC + jωL R +1 Bandpass

The fourth architecture generates only one useful circuit:

Circuit H(jω) Type

H jω( ) = jω( )2 LC +1jω( )2 LC + jωL R +1

Bandstop

The four circuits based on parallel admittances have the same denominator expression as expected

The general 2nd order denominator expression is:

D ω( ) = jωω0

⎛⎝⎜

⎞⎠⎟

2+1Q0

jωω0

+1

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 =1ω02

The second design equation for the first set of circuits is:

CR =1

ω0Q0=

LCQ0

R =1Q0

LC

The second design equation for the first set of circuits is:

LR=

1ω0Q0

=LCQ0

R = Q0LC

The 2nd order allpass response can not be produced by such simple circuits; we either need to use apassive circuit with a lattice structure or we can realise it with an active circuit using an operationalamplifier or transistors

8 BODE PLOTS

8.1 Introduction

The Bode plot method is a way of rapidly sketching the gain and phase response of a circuit from itstransfer function


32

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:

VxVi

=1

1+ 1+ 105

jω⎛

⎝⎜

⎞

⎠⎟

=jω

j2ω +105=12

jωjω + 5 ×104

Let the voltage of the voltage source, 100Vx , be denoted Vy ; then we have:

VyVx

= 100

and

VoVy

=Z2

Z1 + Z2=

1Z1Y2 +1

=1

2 12+ jω107⎛

⎝⎜⎞⎠⎟ +1

=12

107

jω +107

Putting the three gain terms together, we have:

H ω( ) = VoVi

=VoVy

VyVx

VxV1

=12

107

jω +107×100 × 1

2jω

jω + 5 ×104=

25 ×107 jω( )jω + 5 ×104( ) jω +107( )

We will be using this result later, so we will write it using symbols:


33

�

H ω( ) =K jω( )

jω + p1( ) jω + p2( )8.2 Form of the System Function

By generalizing upon the preceding example, we see that the system function of any responsevariable of a circuit constructed from our standard supply of elements, namely R, L, C, anddependent sources, has the form

�

H ω( ) = VoVi

= Kjω + z1( ) jω + z2( ) ... jω + zm( )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 agraph 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 frequencyrange 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 ofapproximately 1

As we wish to investigate this effect more closely, we adjust the scale of our plot so that a gain ofone 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-whereA(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 = 20logA ω( )The logarithm is to the base 10, not to the base e


34

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 thelogarithm of a power ratio:

�

Power ratio in Bels = log P2P1

⎛

⎝ ⎜

⎞

⎠ ⎟

It was then discovered that for most applications this unit was too large; thus, the decibel becamethe standard

This was defined by the equation

�

Power ratio in deciBels = 10log P2P1

⎛

⎝ ⎜

⎞

⎠ ⎟

Assuming that powers are developed in resistors with sinusoidal voltage waveforms, we note that:

�

Pi = 12Vm2

RThus, if both powers are measured relative to the same resistance value, then:

�

Power ratio in deciBels = 10log V22

V12

⎛

⎝ ⎜

⎞

⎠ ⎟ = 20log V2

V1

⎛

⎝ ⎜

⎞

⎠ ⎟

Thus, we have the usual equation:

�

AdB = 20logA 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 thesame 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


35

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 theactual 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/scorrespond 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:


36

We note that multiplying any given frequency value by 10 results in adding one unit of logfrequency; 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 logfrequency

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) unitsas 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( ) = VoVi

= Kjω + z1( ) jω + z2( ) ... jω + zm( )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 theabsolute values and that the absolute value of a ratio is the ratio of the absolute values:

�

A ω( ) = H ω( ) = Kjω + z1 jω + z2 ... jω + zmjω + p1 jω + p2 ... jω + pn

Finally, we take the base 10 logarithm and multiply by 20 to get the dB value:

�

AdB ω( ) = 20logK + 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) andthe dotted line one with a magnitude less than unity


37

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( )q = jω q = ωq

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 ωq( ) = 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 toobtain 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


38

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( )q = jω + a q

Finally, taking the logarithm, we have

�

F ω( ) dB = 20log jω + a( )q = 20qlog jω + a = 20qlog ω2 + a2

Plotting the log term is not so easy!

For this reason, we resort to an approximation, noting that:

�

F ω( ) dB ≅20qlog ω( ); ω >> a20qlog 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 weassume 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 curveoccurs 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 3qdB 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:


39

�

Error ω = 2 a( ) = Error ω = 12a

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = ±0.97q ≅ q dB

A decade above and below the break frequency, we have

�

Error ω = 10 a( ) = Error ω = 110

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( ) = VoVi

=105 jω( )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 toproblems in presentation

Now we simply add all four plots order to obtain the overall Bode gain plot:


40

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 foreach 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 anyconvenient 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 valuesof 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 breakfrequency or downward by an amount –20q dB/dec at each pole break frequency, where q is theorder 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 ×105 jω +10( ) jω +100( ) jω + 500( )jω( )2 jω +1000( ) jω +104( )

Solution

The first step is to mark the zeros and poles on the log(ω) axis; we use for o for zeros and × forpoles:

The bulk of the solution consists merely of steps in sketching the graph, so we merely sketch thefinal 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 ×105 10( ) 100( ) 500( )1( )2 1000( ) 104( ) = 2 ×104 ≡ 86 dB


41

The exact plot is 3 dB above our linearized one at the corner frequency at 10 rad/s and 3 dB belowat 104 rad/s

The errors at the other break frequencies cannot be determined so easily because they are notwidely 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 associatedzero 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 orderof 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 breakupward by +20q dB/dec. at each break frequency associated with a zero of order q and downwardby -20q dB/dec. at each break frequency associated with a pole of order q. Continue this until thelast 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

R × 1jωC

R + 1jωC

+ R

= 1+ jωCR2 + jωCR

= jω +100jω + 200


42

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-frequencyassymptotic gain value is 0 dB

This checks with the circuit, because the capacitor is a short circuit for high frequencies and anopen 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:

�

HLP ω( ) = 1

jωω0

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

+ 1Q0

jωω0

⎛

⎝ ⎜

⎞

⎠ ⎟ +1

Denominator can only be factorised into two real factors

�

D ω( ) = jω + α( ) jω + β( )only if Qo < 1/2

Amplitude response |HLP(jω)| Phase 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 variablerather 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

notes for course ee1.1 circuit analysis 2004-05 topic 7 ...cas.ee.ic.ac.uk/people/dhaigh/ee1.1...

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