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16.3. Note that only the magnitude of thefrequency-response function is shown.

Back

It is understood, however, that the phase

distortion of the input signal also should be

small, within the pass band (the allowed frequency

range). Practical filters are less than ideal. Their

frequency-response functions do not exhibit sharp

cutoffs as in Figure 16.5 and, furthermore, some

phase distortion will be unavoidable.

A special type of band-pass filter that is

widely used in acquisition and monitoring of

vibration signals (e.g., in vibration testing) is the

tracking filter. This is simply a band-pass filter

with a narrow pass band that is frequency

tunable. The center frequency (the mid-value) of

the pass band is variable, usually by coupling it

to the frequency of a carrier signal. In this

manner, signals whose frequency varies with

some basic variable in the system (e.g., rotor

speed, frequency of a harmonic excitation signal,

frequency of a sweep oscillator) can be accurately

tracked in the presence of noise. The

inputs to a tracking filter are the signal that is being tracked and the variable tracking frequency

(carrier input). A typical tracking filter that can simultaneously track two signals is schematically

shown in Figure 16.6.

Filtering can be achieved using digital filters as well as analog filters. Before digital signal

processing became efficient and economical, analog filters were exclusively used for signal filtering

and they are still widely used. In an analog filter, the signal is passed through an analog circuit.

(a)

Magnitude

fc = Cutoff Frequency

fc Frequency f

(b)

G( f )

fc f

(c)

fc1 fc2 f

(d)

f f c1 fc2

FIGURE 16.5 Ideal filter characteristics: (a) low-pass

filter; (b) high-pass filter; (c) band-pass filter; (d) bandreject

(notch) filter.

Tracking

Filter

Input Channel1 Output Channel 1

Input Channel 2 Output Channel 2

Carrier Input

(Tracking Frequency)

FIGURE 16.6 Schematic representation of a twochannel

tracking filter.

Signal Conditioning and Modification 16-15

The dynamics of the circuit will be such that the desired signal components will be passed through

and the unwanted signal components will be rejected. Earlier versions of analog filters employed

discrete circuit elements such as discrete transistors, capacitors, resistors, and even discrete

inductors. Since inductors have several shortcomings, including susceptibility to electromagnetic

noise, unknown resistance effects, and large size. These days, they are rarely used in filter circuits.

Furthermore, owing to well-known advantages of IC devices, analog filters in the form of

monolithic IC chips are today extensively used in modem applications and are preferred over

discrete-element filters. Digital filters that employ digital signal processing to achieve filtering are

also widely used nowadays.

16.3.1 Passive Filters and Active Filters

Passive analog filters employ analog circuits containing only passive elements, such as resistors and

capacitors (and sometimes inductors). An external power supply is not needed in a passive filter.

Active analog filters employ active elements and components, such as transistors and operational

amplifiers in addition to passive elements. Since external power is needed for the operation of the

active elements and components, an active filter is characterized by the need of an external power

supply. Active filters are widely available in a monolithic IC form and are usually preferred over

passive filters.

Advantages of active filters include the following:

1. Loading effects are negligible because active filters can provide a very high input impedance and

very low output impedance.

2. They can be used with low-level signals because signal amplification and filtering can be provided

by the same active circuit.

3. They are widely available in a low cost and compact IC form.

4. They can be easily integrated with digital devices.

5. They are less susceptible to noise from electromagnetic interference than passive filters.

Commonly mentioned disadvantages of active filters are the following:

1. They need an external power supply.

2. They are susceptible to “saturation”-type nonlinearity at high signal levels.

3. They can introduce many types of internal noise and unmodeled signal errors (offset, bias

signals, etc.).

Note that advantages and disadvantages of passive filters can be directly inferred from the disadvantages

and advantages of active filters as given above.

16.3.1.1 Number of Poles

Analog filters are dynamic systems and they can be represented by transfer functions, assuming

linear dynamics. The number of poles of a filter is the number of poles in the associated transfer

function. This is also equal to the order of the characteristic polynomial of the filter transfer

function (i.e., order of the filter). Note that poles (or eigenvalues) are the roots of the characteristic

equation.

In our discussion, we will show simplified versions of filters, typically consisting of a single filter stage.

The performance of such a basic filter can be improved at the expense of circuit complexity (and an

increased pole count). Only simple discrete-element circuits are shown for passive filters. Simple

operational-amplifier circuits are given for active filters. Even here, much more complex devices are

commercially available, but our purpose is to illustrate underlying principles rather than to provide

descriptions and data sheets for commercial filters.

16-16 Vibration and Shock Handbook

16.3.2 Low-Pass Filters

The purpose of a low-pass filter is to allow through all signal components below a certain (cutoff)

frequency and block all signal components above that cutoff. Analog low-pass filters are widely used as

antialiasing filters in digital signal processing. An error known as aliasing will enter the digitally processed

results of a signal if the original signal has frequency components above half the sampling frequency. (Half

the sampling frequency is called the Nyquist frequency.) Hence, aliasing distortion can be eliminated if,

prior to sampling and digital processing, the signal is filtered using a low-pass filter with its cutoff set at

Nyquist frequency. This is one of numerous applications of analog low-pass filters. Another typical

application would be to eliminate high-frequency noise in a measured vibration response.

A single-pole, passive low-pass filter circuit is shown in Figure 16.7(a). An active filter corresponding

to the same low-pass filter is shown in Figure 16.7(b). It can be shown that the two circuits have identical

transfer functions. Hence, it might seem that the opamp in Figure 16.7(b) is redundant. This is not true,

however. If two passive filter stages, each similar to Figure 16.7(a), are connected together, the overall

transfer function is not equal to the product of the transfer functions of the individual stages. The reason

for this apparent ambiguity is the circuit loading that arises due to the fact that the input impedance of

the second stage is not sufficiently larger than the output impedance of the first stage. However, if two

active filter stages, similar to those in Figure 16.7(b), are connected together, such loading errors will be

negligible because the opamp with feedback (i.e., a voltage follower) introduces a very high input

impedance and very low output impedance, while maintaining the voltage gain at unity.

To obtain the filter equation for the scenario depicted in Figure 16.7(a), note that, since the output is

open circuit (zero load current), the current through capacitor C is equal to the current through resistor

R: Hence,

dvo

dt ¼

vi 2 vo

dvo

dt þ vo ¼ vi ð16:17Þ

where the filter time constant is

t ¼ RC ð16:18Þ

From Equation 16.17, it follows that the filter transfer function is

vi ¼ GðsÞ ¼

ðts þ 1Þ ð16:19Þ

From this transfer function, it is clear that an analog low-pass filter is essentially a lag circuit (i.e., it

provides a phase lag).

It can be shown that the active filter stage in Figure 16.7(b) has the same input/output equation.

First, since current through an opamp lead is almost zero, we have from the previous analysis of the

passive circuit stage

vi ¼

ðts þ 1Þ ðiÞ

in which vA is the voltage at the node point A. Now, since the opamp with feedback resistor is in fact a

voltage follower, we have

vA ¼ 1 ðiiÞ

Next, by combining Equation i and Equation ii, we obtain Equation 16.19, as required. As mentioned

earlier, a main advantage of the active filter version is that the resulting loading error is negligible.

Signal Conditioning and Modification 16-17

The frequency-response function corresponding to Equation 16.19 is obtained by setting s ¼ jv; thus

GðjvÞ ¼

ðtjv þ 1Þ ð16:20Þ

This gives the response of the filter when a sinusoidal signal of frequency, v; is applied. The

magnitude lGðjvÞl of the frequency-transfer function gives the signal amplification and phase angle

/GðjvÞ gives the phase lead of the output signal with respect to the input. The magnitude curve

(Bode magnitude curve) is shown in Figure 16.7(c). Note from Equation 16.20 that, for small

frequencies (i.e., v p 1=t), the magnitude is approximately unity. Hence, 1=t can be considered the

FIGURE 16.7 A single-pole low-pass filter: (a) a passive filter stage; (b) an active filter stage; (c) the frequencyresponse

characteristic; (d) a two-pole, low-pass Butterworth filter.

(a)

Input

Output

(b)

+ +

− −

−

Input

Output

(c)

Magnitude

(Log)

0 dB

−3 dB

Slope = −20 dB/decade

wc, wb Frequency (Log) w

Input

Outp

−

B +

(d)

16-18 Vibration and Shock Handbook

cutoff frequency vc :

vc ¼

t ð16:21Þ

Example 16.3

Show that the cutoff frequency given by Equation 16.21 is also the half-power bandwidth for the low-pass

filter. Show that, for frequencies much larger than this, the filter transfer function on the Bode magnitude

plane (i.e., log magnitude vs. log frequency) can be approximated by a straight line with slope 2 20 dB/

decade. This slope is known as the roll-off rate.

Solution

The frequency corresponding to half power (or 1/2 magnitude) is given by

lt jv þ 1l ¼

1ffiffi

2 p

t 2v 2 þ 1 ¼

t 2v2 þ 1 ¼ 2

t 2v2 ¼ 1

Hence, the half-power bandwidth is

v b ¼

t ð16:22Þ

This is identical to the cutoff frequency given by Equation 16.11.

Now, for v q 1=t (i.e., tv q 1) Equation 16.20 can be approximated by

GðjvÞ ¼

tjv

This has the magnitude

lGðjvÞl ¼

In the log scale

log10lGðjvÞl ¼ 2log10 v 2 log10 t

It follows that the log10 (magnitude) vs. log10 (frequency) curve is a straight line with slope 2 1. In other

words, when frequency increases by a factor of ten (i.e., a decade), the log10 magnitude decreases by

unity (i.e., by 20 dB). Hence, the roll-off rate is 2 20 dB/decade. These observations are shown in

Figure 16.7(c). Note that an amplitude change by a factor of

ffiffi

2 p (or power by a factor of 2) corresponds to

3 dB. Hence, when the DC (zero-frequency) magnitude is unity (0 dB), the half power magnitude

is 2 3 dB.

Cutoff frequency and the roll-off rate are the two main design specifications for a low-pass filter.

Ideally, we would like a low-pass filter magnitude curve to be flat until the required pass-band limit

(cutoff frequency) and then roll off very rapidly. The low-pass filter shown in Figure 16.7 only

Signal Conditioning and Modification 16-19

approximately meets these requirements. In particular, the roll-off rate is not as large as is desirable. We

would like a roll-off rate of at least 2 40 dB/decade and, preferably, 2 60 dB/decade in practical filters.

This can be realized by using a higher order filter (i.e., a filter having many poles). The low-pass

Butterworth filter is a widely used filter of this type.

16.3.2.1 Low-Pass Butterworth Filter

A low-pass Butterworth filter having two poles can provide a roll-off rate of 2 40 dB/decade, and one

having three poles can provide a roll-off rate of 2 60 dB/decade. Furthermore, the steeper the slope of the

roll-off, the flatter is the filter magnitude curve within the pass band.

A two-pole, low-pass Butterworth filter is shown in Figure 16.7(d). We could construct a two-pole

filter simply by connecting two single-pole stages of the type shown in Figure 16.7(b). Then, we would

require two opamps, whereas the circuit shown in Figure 16.7(d) achieves the same objective by using

only one opamp (i.e., at a lower cost).

Example 16.4

Show that the opamp circuit in Figure 16.7(d) is a low-pass filter having two poles. What is the transfer

function of the filter? Estimate the cutoff frequency under suitable conditions. Show that the roll-off rate

is 2 40 dB/decade.

Solution

To obtain the filter equation, we write the current balance equations. Specifically, the sum of the currents

through R1 and C1 passes through R2: The same current passes through C2 because current through the

opamp lead must be zero. Hence,

vi 2 vA

R1 þ C1

dt ðvo 2 vAÞ ¼

vA 2 vB

R2 ¼ C2

dvB

dt ðiÞ

Also, since the opamp with a feedback resistor Rf is a voltage follower (with unity gain), we have

vB ¼ vo ðiiÞ

From Equation i and Equation ii, we obtain

vi 2 vA

R1 þ C1

dvo

2 C1

dvA

dt ¼ C2

dvo

dt ðiiiÞ

vA 2 vo

R2 ¼ C2

dvo

dt ðivÞ

Now, defining the constants

t1 ¼ R1C1 ð16:23Þ

t2 ¼ R2C2 ð16:24Þ

t3 ¼ R1C2 ð16:25Þ

and introducing the Laplace variable, s; we can eliminate vA by substituting Equation iv into Equation iii;

thus

vi ¼

½t1t2s2 þ ðt2 þ t3Þs þ 1􀀉 ¼

½s2 þ 2zv2

n þ v2

n􀀉 ð16:26Þ

This second-order transfer function becomes oscillatory if ðt2 þ t3Þ2 , 4t1t2: Ideally, we would like to

have a zero resonant frequency, which corresponds to a damping ratio value z ¼ 1=

ffiffi

2 p : Since the

undamped natural frequency is

vn ¼

ffiffiffiffiffiffi

t1t2 p ð16:27Þ

16-20 Vibration and Shock Handbook

the damping ratio is

z ¼

t2ffiffiþffiffitffiffi3ffi

4t1t2 p ð16:28Þ

and the resonant frequency is

vr ¼

ffiffiffiffiffiffiffiffiffiffi

1 2 2z 2

vn ð16:29Þ

we have, under ideal conditions (i.e., for vr ¼ 0),

ðt2 þ t3Þ2 ¼ 2t1t2 ð16:30Þ

The frequency-response function of the filter is (see Equation 16.26)

GðjvÞ ¼

½v2

n 2 v2 þ 2jzvnv􀀉 ð16:31Þ

Now, for v p vn; the filter frequency response is flat with a unity gain. For v q vn; the filter frequency

response can be approximated by

GðjvÞ ¼ 2

In a log (magnitude) vs. log (frequency) scale, this function is a straight line with slope equals to 2 2.

Hence, when the frequency increases by a factor of ten (i.e., one decade), the log10 (magnitude) drops by

2 units (i.e., 40 dB). In other words, the roll-off rate is 2 40 dB/decade. Also, vn can be taken as the filter

cutoff frequency. Hence,

vc ¼

ffiffiffiffiffiffi

t1t2 p ð16:32Þ

It can be easily verified that, when z ¼ 1=

ffiffi

2 p ; the frequency is identical to the half-power bandwidth (i.e.,

the frequency at which the transfer function magnitude becomes 1=

ffiffi

2 p ).

Note that, if two single-pole stages (of the type shown in Figure 16.7(b)) are cascaded, the resulting

two-pole filter has an overdamped (nonoscillatory) transfer function, and it is not possible to achieve

z ¼ 1=

ffiffi

2 p ; as in the present case. Also, note that a three-pole, low-pass Butterworth filter can be

obtained by cascading the two-pole unit shown in Figure 16.7(d) with a single-pole unit as shown in

Figure 16.7(b). Higher order low-pass Butterworth filters can be obtained in a similar manner by

cascading an appropriate selection of basic units.

16.3.3 High-Pass Filters

Ideally, a high-pass filter allows through it all signal components above a certain (cutoff) frequency and

blocks off all signal components below that frequency. A single-pole, high-pass filter is shown in Figure 16.8.

As for the low-pass filter that was discussed earlier, the passive filter stage (Figure 16.8(a)) and the active

filter stage (Figure 16.8(b)) have identical transfer functions. The active filter is desirable, however, because

of its many advantages, including negligible loading error due to the high input impedance and low output

impedance of the opamp voltage follower that is present in this circuit.

The filter equation is obtained by considering current balance in Figure 16.8(a), noting that the output

is in open circuit (zero load current). Accordingly,

dt ðv1 2 voÞ ¼

Signal Conditioning and Modification 16-21

dvo

dt þ vo ¼ t

dvi

dt ð16:33Þ

in which the filter time constant is

t ¼ RC ð16:34Þ

Introducing the Laplace variable, s; the filter transfer function is obtained as

vi ¼ GðsÞ ¼

ðts þ 1Þ ð16:35Þ

Note that this corresponds to a lead circuit (i.e., an overall phase lead is provided by this transfer

function). The frequency-response function is

GðjvÞ ¼

tjv

ðtjv þ 1Þ ð16:36Þ

Since its magnitude is zero for v p 1=t and is unity for v q 1=t; we have the cutoff frequency

vc ¼

t ð16:37Þ

Signals above this cutoff frequency are allowed undistorted by an ideal high-pass filter, and signals

below the cutoff are completely blocked off. The actual behavior of the basic high-pass filter discussed

(a)

Input Output

(b)

+ +

− −

−

Input

Output

(c)

Magnitude

(Log)

0 dB

−3 dB

Slope = –20 dB/decade

wc Frequency (Log) w

FIGURE 16.8 A single-pole high-pass filter: (a) a passive filter stage; (b) an active filter stage; (c) frequencyresponse

characteristic.

16-22 Vibration and Shock Handbook

above is not perfect, as observed from the frequency-response characteristic shown in Figure 16.8(c).

It can be easily verified that the half-power bandwidth of the basic high-pass filter is equal to the cutoff

frequency given by Equation 16.37, as in the case of the basic low-pass filter. The roll-up slope of the

single-pole high-pass filter is 20 dB/decade. Steeper slopes are desirable. Multiple-pole, high-pass

Butterworth filters can be constructed to give steeper roll-up slopes and reasonably flat pass-band

magnitude characteristics.

16.3.4 Band-Pass Filters

An ideal band-pass filter passes all signal components within a finite frequency band and blocks off

all signal components outside that band. The lower frequency limit of the pass band is called the

lower cutoff frequency ðvc1Þ; and the upper frequency limit of the band is called the upper cutoff

frequency ðvc2Þ:

The most straightforward way to form a band-pass filter is to cascade a high-pass filter of cutoff

frequency vc1 with a low-pass filter of cutoff frequency vc2: Such an arrangement is shown in Figure 16.9.

The passive circuit shown in Figure 16.9(a) is obtained by connecting the circuits shown in Figure 16.7(a)

and Figure 16.8(a). The passive circuit shown in Figure 16.9(b) is obtained by connecting a voltage

follower opamp circuit to the original passive circuit. Passive and active filters have the same transfer

function, assuming that loading problems are not present in the passive filter. Since loading errors can be

serious in practice, however, the active version is preferred.

(a)

Input

Output

(b)

+ +

− −

Input

Output

C2 −

(c)

Magnitude

(Log)

0 dB

−20 dB/decade

Frequency (Log) w

R1 A

20 dB/decade

wc1 wc2

FIGURE 16.9 Band-pass filter: (a) a basic passive filter stage; (b) a basic active filter stage; (c) frequency-response

characteristic.

Signal Conditioning and Modification 16-23

To obtain the filter equation, first consider the high-pass portion of the circuit shown in Figure 16.9(a).

Since the output is open circuit (zero current), we have from Equation 16.35:

vA ¼

t2s

ðt2s þ 1Þ ðiÞ

in which

t2 ¼ R2C2 ð16:38Þ

Next, writing the current balance at node A of the circuit, we have

vi 2 vA

R1 ¼ C1

dvA

dt þ C2

dt ðvA 2 voÞ ðiiÞ

Introducing the Laplace variable, s; we obtain

vi ¼ ðt1s þ t3s þ 1ÞvA 2 t3svo ðiiiÞ

in which

t1 ¼ R1C1 ð16:39Þ

and

t3 ¼ R1C2 ð16:40Þ

Now, on eliminating vA by substituting Equation i in Equation iii, we obtain the band-pass filter transfer

function

vi ¼ GðsÞ ¼

t2s

½t1t2s2 þ ðt1 þ t2 þ t3Þs þ 1􀀉 ð16:41Þ

We can show that the roots of the characteristic equation

t1t2s2 þ ðt1 þ t2 þ t3Þs þ 1 ¼ 0 ð16:42Þ

are real and negative. The two roots are denoted by 2vc1 and 2vc2 and they provide the two cutoff

frequencies shown in Figure 16.9(c). It can be verified that, for this basic band-pass filter, the roll-up

slope is þ20 dB/decade and the roll-down slope is 2 20 dB/decade. These slopes are not sufficient in

many applications. Furthermore, the flatness of the frequency response within the pass band of the basic

filter is not adequate either. More complex (higher order) band-pass filters with sharper cutoffs and

flatter pass bands are commercially available.

16.3.4.1 Resonance-Type Band-Pass Filters

There are many applications where a filter with a very narrow pass band is required. The tracking filter

mentioned in the beginning of the section on analog filters is one such application. A filter circuit with a

sharp resonance can serve as a narrow-band filter. Note that the cascaded RC circuit shown in Figure 16.9

does not provide an oscillatory response (the filter poles are all real) and, hence, it does not form a

resonance-type filter. A slight modification to this circuit using an additional resistor, R1; as shown in

Figure 16.10(a), will produce the desired effect.

To obtain the filter equation, note that, for the voltage follower unit

vA ¼ vo ðiÞ

16-24 Vibration and Shock Handbook

Next, since current through an opamp lead is zero, for the high-pass circuit unit (see Equation 16.35), we

have

vB ¼

t2s

ðt2s þ 1Þ ðiiÞ

in which

t2 ¼ R2C2

Finally, current balance at node B gives

vi 2 vB

R1 ¼ C1

dvB

dt þ C2

dt ðvB 2 vAÞ þ

vB 2 vo

or, by using the Laplace variable, we obtain

vi ¼ ðt1s þ t3s þ 2ÞvB 2 t3svA 2 vo ðiiiÞ

Now, by eliminating vA and vB in the equations from Equation i to Equation iii, we obtain the filter

transfer function

vi ¼ GðsÞ ¼

t2s

½t1t2s2 þ ðt1 þ t2 þ t3Þs þ 2􀀉 ð16:43Þ

(a)

(b)

Magnitude

Frequency w

Input

Output

−

C1 R2

M / 2

Δ ω

wc1wr wc2

FIGURE 16.10 A resonance-type narrow-band-pass filter: (a) an active filter stage; (b) frequency-response

characteristic.

Signal Conditioning and Modification 16-25

It can be shown that, unlike Equation 16.41, the present characteristic equation

t1t2s2 þ ðt1 þ t2 þ t3Þs þ 2 ¼ 0 ð16:44Þ

can possess complex roots.

Example 16.5

Verify that the band-pass filter shown in Figure 16.10(a) can have a frequency response with a resonant

peak as shown in Figure 16.10(b). Verify that the half-power bandwidth Dv of the filter is given by 2zvr

at low damping values. (Note: z ¼ damping ratio and vr ¼ resonant frequency.)

Solution

We may verify that the transfer function given by Equation 16.43 can have a resonant peak by showing

that the characteristic equation (Equation 16.44) can have complex roots. For example, if we use

parameter values C1 ¼ 2; C2 ¼ 1; R1 ¼ 1; and R2 ¼ 2; we have t1 ¼ 2; t2 ¼ 2; and t3 ¼ 1: The

corresponding characteristic equation is

4s2 þ 5s þ 2 ¼ 0

It has the roots

^ j

ffiffi

7 p

is obviously complex.

To obtain an expression for the half-power bandwidth of the filter, note that the filter transfer function

may be written as

GðsÞ ¼

ðs2 þ 2zvns þ v2

nÞ ð16:45Þ

in which

vn ¼ undamped natural frequency

z ¼ damping ratio

k ¼ a gain parameter

The frequency-response function is given by

GðjvÞ ¼

kjv

½v2

n 2 v2 þ 2jzvnv􀀉 ð16:46Þ

For low damping, resonant frequency vr ø vn: The corresponding peak magnitude M is obtained by

substituting v ¼ vn in Equation 16.46 and taking the transfer function magnitude; thus

M ¼

2zvn ð16:47Þ

At half-power frequencies, we have

lGðjvÞl ¼

Mffiffi

2 p

kv ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðv2

n 2 v2Þ2 þ 4z2v2

nv2

p ¼

ffiffi

2 p zvn

16-26 Vibration and Shock Handbook

This gives

ðv2

n 2 v2Þ2 ¼ 4z2v2

nv2 ð16:48Þ

the positive roots of which provide the pass band frequencies vc1 and vc2. Note that the roots are given by

n 2 v2 ¼ ^2zvnv

Hence, the two roots, vc1 and vc, satisfy the following two equations:

c1 þ 2zvnvc1 2 v2

n ¼ 0

c2 2 2zvnvc2 2 v2

n ¼ 0

Accordingly, by solving these two quadratic equations and selecting the appropriate sign, we obtain

vc1 ¼ 2zvn þ

ffiffiffiffiffiffiffiffiffiffiffiffiffi

n þ z2v2

ð16:49Þ

vc2 ¼ zvn þ

ffiffiffiffiffiffiffiffiffiffiffiffiffi

n þ z2v2

ð16:50Þ

The half-power bandwidth is

Dv ¼ vc2 2 vc1 ¼ 2zvn ð16:51Þ

Now, since vn ø vr; for low z we have

Dv ¼ 2zvr ð16:52Þ

A notable shortcoming of a resonance-type filter is that the frequency response within the

bandwidth (pass band) is not flat. Hence, quite nonuniform signal attenuation takes place inside the

pass band.

16.3.5 Band-Reject Filters

Band-reject filters, or notch filters, are commonly used to filter out a narrow band of noise components

from a signal. For example, 60 Hz line noise in signals can be eliminated by using a notch filter with a

notch frequency of 60 Hz.

An active circuit that could serve as a notch filter is shown in Figure 16.11(a). This is known as the

Twin T circuit because its geometric configuration resembles two T-shaped circuits connected together.

To obtain the filter equation, note that the voltage at point P is vo because of unity gain of the voltage

follower. Now, we write the current balance at nodes A and B; thus

vi 2 vB

R ¼ 2C

dvB

dt þ

vB 2 vo

dt ðvi 2 vAÞ ¼

R=2 þ C

dt ðvA 2 voÞ

Next, since the current through the positive lead of the opamp (voltage follower) is zero, we have the

current through point P as

vB 2 vo

R ¼ C

dt ðvo 2 vAÞ

These three equations are written in the Laplace form as

vi ¼ 2ðts þ 1ÞvB 2 vo ðiÞ

tsvi ¼ 2ðts þ 1ÞvA 2 tsvo ðiiÞ

vB ¼ ðts þ 1Þvo 2 tsvA ðiiiÞ

Signal Conditioning and Modification 16-27

in which

t ¼ RC ð16:53Þ

Finally, eliminating vA and vB in Equation i to Equation iii, we obtain

vi ¼ GðsÞ ¼ ðt2s2 þ 1Þ

ðt2s2 þ 4ts þ 1Þ ð16:54Þ

The frequency-response function of the filter is

GðjvÞ ¼ ð1 2 t2v2Þ

ð1 2 t2v2 þ 4jtvÞ ð16:55Þ

with s ¼ jv: Note that the magnitude of this function becomes zero at frequency

vo ¼

t ð16:56Þ

This is known as the notch frequency. The magnitude of the frequency-response function of the notch

filter is sketched in Figure 16.11(b). It is noticed that any signal component at frequency vo will be

completely eliminated by the notch filter. Sharp roll-down and roll-up are needed to allow the other

(desirable) signal components through without too much attenuation.

Whereas the previous three types of filters achieve their frequency-response characteristics through the

poles of the filter transfer function, a notch filter achieves its frequency-response characteristic through

its zeros (roots of the numerator polynomial equation). Some useful information about filters is

summarized in Box 16.2.

1 1

(a)

(b)

Magnitude

Frequency w

Input

Output

−

R/2 2C

wo = =

FIGURE 16.11 A notch filter: (a) an active Twin T filter circuit; (b) frequency-response characteristic.

16-28 Vibration and Shock Handbook

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16.3. Note that only the magnitude of thefrequency-response function is shown.

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