# BASIC FILTERS ```19
BASICFILTERS
19-1
19-2
19-3
19-4
19-5
Low-PassFilters
High-Pass
Filters
Band-Pass
Filters
Band-StopFilters
Technology
Theoryinto Practice
Workbench(EWB)and
Electronics
PSpiceTutorialsat
http://www.prenhall.com/floyd
r INTRODUCTION
The concept of filters was introduced in Chapters 16,
ll , and 18 to illustrate applications of RC RZ, and
RIC circuits. This chapter is essentially an extension
of the earlier material and provides additional coverage of the important topic of filters.
Passivefilters are discussedin this chapter.Passive filters use various combinations of resistors,
capacitors,and inductors. In a later course, you will
study active filters which use passive components
combined with amplifiers. You have already seenhow
basic RC. RL. and RLC circuits can be used as filters.
Now, you will learn that passivefilters can be placed
in four general categoriesaccording to their response
characteristics:
low-pass,high-pass,band-pass,and
band-stop.Within each category,there are several
common types that will be examined.
In the TECH TIP assignmentin Section 19-5,
you will plot the frequency responsesof filters based
on oscilloscope measurementsand identify the types
of fllters.
r TECHnology
Theorv
Into
Practice
r CHAPTER
OBfECTTVES
O Analyze the operation of RC and RZ low-pass filters
tr Analyze the operation of RC and Rt high-pass
filters
O Analyze the operation of band-passfilters
tr Analyze the operation of band-stopfilters
762 I
BASICFILTERS
FITTERS
19-1 r LOW.PASS
A law-passfilter allows signals with lower frequencies to pass from input to output
while rej ecting higher frequencies.
After completing this section, you should be able to
I Analyze the operation of RC and RL low-pass filters
. Express the voltage and power ratios of a filter in decibels
. Determine the critical frequency of a low-pass filter
. Explain the difference between actual and ideal low-pass responsecurves
. Define roll-off
. Generatea Bode plot for a low-pass filter
. Discuss phase shift in a low-pass filter
Figure 19-1 shows a block diagram and a generalresponsecurve for a low-pass filter. The
range of low frequenciespassedby a low-pass filter within a specified limit is called the
passband of the fllter. The point consideredto be the upper end of the passbandis at the
critical frequency,/f, as illustrated in Figure 19-1(b). The critical frequency is the frequency at which the fllter's output voltage is 10.77oof the maximum. The filter's critical
frequency is also called the cutoff frequency, breakfrequency, or -3 dB frequency because
the output voltage is down 3 dB from its maximum at this frequency.The term dB (decibel) is a commonly used one that you should understandbecausethe decibel unit is used
in filter measurements.
vor,
Vn
0.70'1u&quot;
(a)
(b)
F I G U R E1 9 - 1
Low-passfi.lter block diagram and generalresponsecurve.
Decibels
The basis for the decibel unit stemsfrom the logarithmic responseof the human ear to the
intensity of sound. The decibel is a logarithmic measurementof the ratio of one power to
another or one voltage to another,which can be used to expressthe input-to-output relationship of a filter. The following equation expressesa power ratio in decibels:
dB= 1o
V i n /
&quot;r(,?)
(19-1)
From the properties of logarithms, the following decibel formula for a voltage ratio is
derived. This formula is correct only when both voltases are measured in the same
imoedance.
dB=2o
&quot;'(+)
LOW-PASS
FILTERS.
EXAMPTE 19-1
763
At a ceftain frequency, the output voltage of a fllter is 5 V and the input is 10 V.
Expressthe voltageratio in decibels.
so l u ti o n 'u 'o tltv^'
Z \= zo loef
' \ *l'
v/
r ol=
/
Related Problem
z0togr o.s)- 6.02
= dB
Express the ratio VourlVin= 0.85 in decibels.
RC Low-Pass
Filter
A basic RC low-pass filter is shown in Figure 19-2. Notice that the output voltage is taken
acrossthe caoacitor.
F I C U R E1 9 - 2
I
0.005pF
When the input is dc (0 Hz), the output voltage equals the input voltage becauseX6
is infinitely large. As the input frequency is increased,X6.decreasesand, as a result, Vo,,
gradually decreasesuntil a frequency is reached where X6 = R. This is the critical frequency,f,, of the filter.
Xc= R
I
_ 2nf&quot;C
D
&quot;
1
r,=
,n*-
(1e-3)
At any frequency,the output voltage magnitude is
,,
v o u-t -/-{a:)r,,,,
\
/nr*x?,
by application of the voltage-divider formula. Since X6 = R atf,, the output voltage at the
critical frequency can be expressedas
=
v&quot;'=(
v# R''' l#)'t =(Fhy' =(+)vn=0r07v'
These calculations show that the output is l0.7%aof the input when Xc= R. The frequency
at which this occurs is, by deflnition, the critical frequency.
The ratio of output voltage to input voltage at the critical frequency can be
expressedin decibelsas follows:
Vo'' = 0'707V;'
I v in = 0.101
/ v \
= -3 dB
20 log{-j!l = 20 logt0.707r
\ vi' /
764 I
BASICFILTERS
EXAMPLE 19-2
for thelow-passRC filter in Figure19-2.
Determinethecriticalfrequency
1
| =
= 318kHz
&quot;
f'=
2nRC trloo orroJosrrF)
Solution
The output voltage is 3 dB below V;, atthis frequency (Vourhasa maximum value of Vr,).
Related Problem A certain low-pass RC filter has R = 1.0 kO and C = 0.022 pF.
Determine its critical frequencv.
&quot;Roll-Off&quot; of the ResponseCurve
The blue line in Figure 19-3 shows an actual responsecurve for a low-pass filter. The
maximum output is defined to be 0 dB as a reference. Zero decibels correspondsto
Vo,, = Vin, because20 ).og(Vo,,/V;,)= 20 log 1 = 0 dB. The output drops from 0 dB to
-3 dB at the critical frequency and then continues to decreaseat a flxed rate. This pattern
of decreaseis called the roll-off of the frequency response.The red line shows an ideal
output responsethat is consideredto be &quot;flat&quot; out tojf. The output then decreasesat the
fixed rate.
ro't
\
F I C U R E1 9 - 3
Actual and ideal responsecurvesfor a low-passfihen
As you have seen, the output voltage of a low-pass filter decreasesby 3 dB when
the frequency is increasedto the critical value f,. As the frequency continues to increase
abovef,, the output voltage continuesto decrease.In fact, for each tenfold increasein frequency abovef,, there is a20 dB reduction in the output, as shown in the following steps.
Let's take a frequency that is ten times the critical frequency (f = ljf&quot;).Since R =
X, at f,, then R = l)Xc at l)f&quot; becauseof the inverse relationship of X6 andf.
The attenuation of the RC circuit is the ratio Vo6/V;, and is developed as follows:
Vout
-
Xc
v,,- ffi**r-
-
XC
-
XC
^,FoxjiE,
XC
=-L=-+=-1
f rooxT-T,-,Vxiloo.D y xct}t v 101
The dB attenuationis
= -20 dB
20 bs(Y:-\=
&quot;\ V,&quot;J 2olos(o.l)
10
=g.1
LOW-PASS
FILTERS.
765
A tenfold changein frequency is called a decade. So, for the RC circuit, the output
voltage is reduced by 20 dB for each decadeincrease in frequency.A similar result can
be derived for the high-pass circuit. The roll-off is a constant -20 dB/decadefor a basic
RC or RL filter. Figure 19-4 shows the ideal frequency responseplot on a semilog scale,
where each interval on the horizontal axis representsa tenfold increasein frequency.This
responsecurve is called a Bode plot.
*10
-20
-30
tt
-9
-40
tael
FtcuRE't9-4
RCfilter (Bodeplot).
Frequency
roll-offfor a low-pass
19-3
EXAMPTE
Make a Bode plot for the filter in Figure 19-5 for three decadesof frequency. Use
semilog graph paper.
F I G U R E1 9 - 5
I
Solution
0.005prF
The critical frequency for this low-pass filter is
|
= 31.8kHz
t,' -- 2 n R C 2n(r.0
pF)
ka)(0.005
The idealized Bode plot is shown with the red line on the semilog graph in
Figure 19-6. The approximate actual response curve is shown with the blue line.
Notice flrst that the horizontal scale is logarithmic and the vertical scale is linear.
The frequency is on the logarithmic scale, and the filter output in decibels is on the
vertical.
766 r
BASICFILTERS
O
A
U
O
O\{@€,
qO\{-\O,
S
5
u
O\{@\O
0dB
-3 dB
N
-10 dB
'out
\
vrin
\
-20 dB
\
\
\
\
-30 dB
\
\
-40 dB
10kHz
f&quot;
100kHz
t000 kHz
FICURE19-6
Bode plot for Figure 19-5. The red.line representsthe ideal responsecurtte and the blue line
repfesentsthe actual response.
The output is flat belowf, (31.8 kHz). As the frequency is increased abovef&quot;,
output drops at a -20 dB/decade rate. Thus, for the ideal curve, every time the
quency is increasedby ten, the output is reducedby 20 dB. A slight variation from
occurs in actual practice. The output is actually at -3 dB rather than 0 dB at the
cal frequency.
Related Problem
What happens to the critical frequency and roll-off rate if C i
reducedto 0.001rrF in Fisure 19-5?
LOW-PASS
FILTERS.
767
Rl Low-PassFilter
A basic Rl low-passfilter is shown in Figure 19-7. Notice that the output voltage is taken
acrossthe resistor.
F I C U R E1 9 _ 7
RL low-passfilter.
Vi,
When the input is dc (0 Hz), the output voltage ideally equals the input voltage
because X7 is a short (if R1y is neglected). As the input frequency is increased, X1
increasesand, as a result, Vou,grad:uallydecreasesuntil the critical frequency is reached.
At this point, X. = R and the frequency is
2nf,L = R
&quot;
, -
r'-
R
2nL
|
(1e-4)
}IT,\L/R)
Just as in the RC low-pass filter, Vour= 0.70JVi, and, thus, the output voltage is -3 dB
below the input voltage at the critical frequency.
19_4
EXAMPLE
Make a Bode plot for the filter in Figure 19-8 for three decadesof frequency. Use
semilog graph paper.
F I G U R E1 9 - 8
Solution
The critical frequency for this low-pass filter is
t'=
|
I
2n(L/R) 2n(4.7 rtll2.2
=?4.-skHz
kf2)
The idealized Bode plot is shown with the red line on the semilog graph in Figure 19-9. The approximate actual responsecurve is shown with the blue line. Notice
first that the horizontal scale is logarithmic and the vertical scale is linear. The frequency is on the logarithmic scale, and the filter output in decibels is on the vertical.
768 I
BASICFILTERS
0dB
-3 dB
-10 dB
vor,
u,
-20 dB
-30 dB
-40 dB
10kHz
1000kHz
10,000
kHz
F I C U R E1 9 - 9
Bodeplot for Figure 19-8. The red line is the i.deal response cufle and the blue line is
the actual responsecurve.
The output is flat below/f (74.5 kIIz). As the frequency is increasedabove/f, the
output drops at a -20 dB/decade rate. Thus, for the ideal curve, every time the frequency is increasedby ten, the output is reducedby 20 dB. A slight variation from this
occurs in actual practice. The output is actually at -3 dB rather than 0 dB at the critical frequency.
Related Problem What happens to the critical frequency and roll-off rate if I is
reducedto I mH in Figure 19-8?
PhaseShift in a Low-PassFilter
The RC low-pass filter acts as a lag network. Recall from Chapter 16 that the phaseshift
from input to output is expressedas
*,un '(*)
e = _-eo&quot;
\ R
/
At the critical frequency, Xc = R and, therefore, 6 = -45'. As the input frequencyis
reduced, @ decreasesand approaches0o when the frequency approacheszero. Figure
19-10, illustratesthis phasecharacteristic.
FILTERS.
HICH-PASS
769
F I G U R E1 9 - 1 0
Phasecharacteristic of a low-passfilter.
The Rt low-passfilter alsoactsas a lag network.Recallfrom Chapter17 that the
as
phaseshift is expressed
,lX,\
0 = _tan |r_J
As in the RC filter, the phaseshift from input to outputis -45&quot; at the critical frequency
belowf,.
for frequencies
anddecreases
sEcTtoN19-1
REVIEW
1. In a certain low-pass filter, f, = 2.5 WIz What is its passband?
2. In a certainlow-passfilter.R = 100 Q andX6= 2 Cl at a frequency./. Determine
Y ou,at fl when V;, = 5/0&quot; V rms.
3. Vout= 400mV, and Vn= 1.2V. Expressthe ratio VouylVi,in dB'
FILTERS
19_2I HIGH.PASS
A high-passfilter allows signals with higher frequencies to pass ftom input to output
while rej ecting lower fre quencies.
After completing this section, you should be able to
I Analyze the operation of RC and RL high-pass filters
. Determine the critical frequency of a high-pass filter
. Explain the difference between actual and ideal responsecurves
. Generatea Bode plot for a high-pass filter
. Discuss phase shift in a high-pass filter
Figure 19-1 I shows a block diagram and a general responsecurve for a high-pass filter.
The frequency consideredto be the lower end of the passbandis called the critical frequency.Just as in the low-pass filter, it is the frequency at which the output isl0.7%o of
the maximum, as indicated in the f,gure.
770 I
BASICFILTERS
Vo't
Vin
0.701un
G;;itn;t
frequencies
+Passesthese
frequencies
F I G U R E1 9 - 1 1
High-passfiher blnck diagram and responsecune.
RC High-Pass
Filter
A basic RC high-passfilter is shown in Figure I9-I2. Notice that the output voltageis
taken acrossthe resistor.
F I G U R E1 9 - 1 2
RC hish-pass &quot; filter.
L
r/
vi,,
&quot;-lF-1--
V ou,
I
=R
-L
:
When the input frequency is at its critical value, Xc = R and the output voltageis
O.707vi,,just as in the caseof the low-pass filter. As the input frequency increasesabove
f,, Xs decreasesand, as a result, the output voltage increasesand approachesa value equal
to V;,. The expressionfor the critical frequency of the high-pass filter is the sameas for
the low-pass filter.
, t&quot;-
|
2nRC
Belowf,, the output voltage decreases(rolls ofl at a rate of -20 dB/decade.Figure
19-13 showsan actualand an ideal responsecurve for a high-passfilter.
Ideal
0.07fc
0-3
o.tf,
dB
*20 dB
-40 dB
v^,,.
-f
(dB)
F I G U R E1 9 - 1 3
Actual and i.dealresponsecumesfor a high-passfilter
19_5
EXAMPLE
Make a Bode plot for the fllter in Figure 19-14 for three decadesof frequency. Use
semilog graph paper.
F I G U R E1 9 - 1 4
vtr4
0.047pF
Solution
The critical frequency for this high-pass filter is
t &quot;' -
l
2nRC
2n(3300)(0.047pF)
= 10.3kHz = 10kHz
The idealized Bode plot is shown with the red line on the semilog graph in Figure 19-15. The approximate actual responsecurve is shown with the blue line. Notice
first that the horizontal scale is logarithmic and the vertical scale is linear. The frequency is on the logarithmic scale, and the filter output in decibels is on the vertical.
a
A
U
O,{@€
a !
@€.
a
5
u
6
{@\O
Vou,
u,
100Hz
1000Hz
F I G U R E1 9 - 1 5
Bode plot for Figure 19-14. The reil line is the i.deal response cume and the blue line is the
actual response curae.
The output is flat beyond/| (approximately 10 kHz). As the frequency is reduced
below /|, the output drops at a -20 dB/decade rate. Thus, for the ideal curve, every
time the frequency is reduced by ten, the output is reduced by 20 dB. A slight variation from this occurs in actual practice. The output is actually at -3 dB rather than
0 dB at the critical frequency.
Related Problem If the frequency for the high-pass filter is decreasedto 10 Hz,
what is the output to input ratio in decibels?
772 T BASICFILTERS
RLHigh-Pass
Filter
A basic Rl high-pass filter is shown in Figure 19-16. Notice that the output is taken
acrossthe inductor.
FIGURE
19-16
RL high-passfilter.
When the input frequency is at its critical value, X, = R, and the output voltageis
0.707V1. As the frequency increases abovef,, X; increasesand, as a result, the output
voltage increasesuntil it equals V,,. The expressionfor the critical frequency of the highpass fllter is the same as for the low-pass fllter.
-
l
t'- 2n1ua1
PhaseShiftin a High-Pass
Filter
Both the RC and the RI high-pass filters act as lead networks. Recall from Chapters16
and l7 that the phase shift from input to output for the RC lead network is
-t(X.\
o=tan
(n/
and for the RL lead network is
d = 90&quot;-,&quot;&quot;- /4\
\ n /
At the critical frequency,Xr= R and, therefore,6 = 45o.As the frequencyis increased,
f decreasestoward 0o, as shown in Figure 19-17.
F I G U R E1 9 _ 1 7
Phasecharacteristic of a high-passfilten
BAND-PASS
FILTERS.
EXAMPLE 19-6
773
(a) In Figure 19-18, find the value of C so that X6 is ten times less than R at an input
frequencyof l0 kHz.
(b) If a 5 V sine wave with a dc level of 10 V is applied, what are the output voltage
magnitude and the phase shift?
F I C U R E1 9 - 1 8
c
,*&quot;___)l_1_r*,
l*'&quot;
t_
Solution
(a) The valueof C is determinedasfollows:
f2)= 68 O
Xc = 0.1R= 0.1(680
C=
|
2nlXs
2n(10kHzX68A)
=0.234uF
The neareststandardvalueof C is 0.22 uF.
(b) The magnitudeof the sinewaveoutputis determinedasfollows:
\-R
680f)
I
\.,
/
V o ,=, ,l
l\',,,=l--+15V=4.98V
t Y R z +X t t
\ V t o 8 o Q ) ' + i 6 8Q ) ' r
The phaseshift is
-,1X.\
,
' / 6 8O \
@=ran
o
\ /=,un \680o/=t.,
=
At 7 19 kHz, which is a decadeabovethe critical frequency,the sinusoidaloutput is almostequalto theinputin magnitude,andthephaseshift is very small.The
l0 V dc level hasbeenfilteredout anddoesnot appearat the output.
RelatedProblem Repeatparts(a) and (b) of the exampleif R is changedto 220 Q.
stcTtoN19-2
REVIEW
1. The input voltage of a high-pass filter is 1 V. What is Vo,, at the critical frequency?
2. In a certain high-pass filter, V;, = 1020&quot; V R = 1.0 kO, and Xy = 15 kfl. Determine V,,,.
FILTERS
19_3r BAND.PASS
A band-passfiIter allows a certein band offreqaencies to pass snd attenuates or
rejects allfrequencies below und ubove the passband.
After completing this section, you should be able to
I Analyze the operation ofband-pass filters
. Show how a band-passfilter is implemented with low-pass and high-pass fllters
. Define bandwidth
774 T
BASICFILTERS
. Explain the series-resonantband-passfilter
. Explain the parallel-resonantband-passfilter
. Calculate the bandwidth and output voltage of a band-passfilter
Figure 19-19 showsa typical band-passresponsecurve.
vor,
v
0;70'7v^N
(- 3 dB)
F I G U R E1 9 - 1 9
Typicalband-passresponsecurve,
Low-Pass/H
igh-PassFilter
A combination of a low-pass and a high-pass filter can be used to form a band-pass fil.
ter, as illustrated in Figure 19-20. The loading effect of the secondfilter on the first must
be taken into account.
F I G U R E1 9 - 2 0
Low-passand high-passfilters usedto form a band-pass
filter.
If the critical frequency,fat1, of the low-pass filter is higher than the critical frequency,,fanl, of the high-pass filter, the responsesoverlap. Thus, all frequencies except
those betweenf,q1 andf,11,are eliminated, as shown in Figure 19-21.
F I G U R E1 9 - 2 1
Overlappingresponsecumes of a high-pass/low-pass
filter.
BAND.PASS
FILTERS.
775
The bandwidth of a band-pass filter is the range of frequencies for which
the current, and therefore the output voltage, is equal to or greater than
70.7Vo of its value at the resonant frequency.
As vou know. bandwidth is often abbreviatedBW
EXAMPLE19-7
filter withf, = 2 kllz anda low-passfilter withf, = 2.5 kHz areusedto
A high-pass
constructa band-pass
passband?
= 2.5 kllz - 2 kHz = 500 Hz
BW = f&lt;o - [email protected])
RelatedProhlem If fdn = 9 kHz andthe bandwidthis I .5 kHz, what is f,q,y?
Solution
Filter
SeriesResonantBand-Pass
A type of series resonant band-passfilter is shown in Figure 19-22. As you learned in
Chapter 18, a series resonant circuit has minimum impedance and maximum current at
the resonantfrequency,/|. Thus, most of the input voltage is dropped across the resistor
at the resonant frequency. Therefore, the output acrossR has a band-passcharacteristic
with a maximum output at the frequency of resonance.The resonant frequency is called
the center frequency,/6. The bandwidth is determinedby the quality factor, Q, of the circuit and the resonantfrequency,as was discussedin Chapter 18. Recall rhat Q = ytlp.
r, *---lu[f!
'*l'i
v*,
l--1__=4
c**
Series
resonant
+ R
t
l_
:
n
&quot;
t
t
t
, f
fo
F I G U R E1 9 - 2 2
Seriesresonantband-pass
filter.
A higher value of Q results in a smaller bandwidth. A lower value of Q causesa
larger bandwidth. A formula for the bandwidth of a resonantcircuit in terms of B is stated
in the following equation:
u*=t
(1e-5)
776 T
BASICFILTERS
EXAMPLE 19-B
Determine the output voltage magnitude at the center frequency ffi) and the bandwidth
for the filter in Figure 19-23.
F I G U R E1 9 - 2 3
Solution At /0, the impedance of the resonant circuit is equal to the winding reslstance,Ry,. By the voltage-divider formula,
*
-=\i rloo
r o?)'o
a)'u
&quot; = 9.09
v.,,-=\(n--!-\u,
v
R*)'''
The center frequency is
to-
znyTc
-
2nY (l mHX0.0022pF)
= 107kHz
At/6, the inductive reactanceis
Xr= 2nfL = 2n(107 kHz)(l mH) = 672 I
and the total resistance
is
R , o , =R + R w = 1 0 0f 2 + 1 0 O = 1 1 0Q
Therefore.the circuit Q is
x,
=e ' t t
0^ = ; : , =f672f)
fi
The bandwidthis
U*=
107kHz = l 7 . 5 k H z
f&quot;
e=-
RelatedProhlem If a I mH coil with a winding resistanceof 18 f2 replacesthe
existingcoil in Figure 19*23,how is the bandwidthaffected?
ParallelResonant
Band-Pass
Filter
A type of band-passfilter using a parallel resonant circuit is shown in Figlre 19-24.
Recall that a parallel resonantcircuit has maximum impedance at resonance.The circuit
in Figure 19-24 actsas a voltage divider. At resonance,the impedanceof the tank is much
greater than the resistance.Thus, most of the input voltage is acrossthe tank, producing
a maximum output voltage at the resonant (center) frequency.
FIGURE19-24
Parallel resonant band-passfilter.
vi,
BAND-PASS
FILTERS.
777
For frequenciesabove or below resonance,the tank impedancedrops off, and more
of the input voltage is acrossR. As a result, the output voltage acrossthe tank drops off,
creating a band-passcharacteristic.
EXAMPLE
19-9
What is the center lrequency of the filter in Figure 19-25?AssumeRw = 0 Q.
F I G U R E1 9 - 2 5
The center frequency of the filter is its resonantfrequency.
Solution
= 5.03MHz
ft= --!2nYLC 2nY(r0 pH)(100
pF)
Related Prohlem
EXAMPLE
19-10
Determine/6 in Figure 19-25 if C is changed to 1000 pF.
Determine the center frequency and bandwidth for the band-passfilter in Figure 19-26
if the inductor has a windins resistanceof 15 Q.
F I G U R E1 9 - 2 6
Solution Recall from Chapter 18 (Eq. 18-15) that the resonant(center) frequency of
a nonideal tank circuit is
&quot;
ft-6a*c/D
@
- - - - r ! l . f f -
2tYLC
2n\/(50 mHX0.01p.F
The Q of thecoil at resonance
is
, , _ I L _ 2 n f o L _ 2n(7.12kHz)(50mH) = 149
Rw
Rw
15f)
The bandwidth of the filter is
7.12 kITz
u * = 6fn= - - * - = 4 7 . 8 H 2
Note that since Q &gt; 10, the simpler formula (Eq. 18-6) could have been used to
calculate/6.
Related Problem
Knowing the value of Q, recalatlate/. using the simpler formula.
778 r
BASICFILTERS
sEcTroN19-3
REVIEW
1. For a band-passfilter, f4ny = 29.8 kIIz andf&quot;o = 30.2 kHz. What is the bandwidth?
2. A parallelresonantband-passfilter has the following values:Rw = l5 Q. L = 50 pH,
and C = 470 pF. Determinethe approximatecenterfrequency.
19_4 I BAND.STOPFITTERS
A band-stopfilter is essentially the opposite of a band-passfilter in terms of the
responses.A band-stopfilter ullows all frequencies to pass except those lying within a
certain stopband,
After completing this section, you should be able to
r Analyze the operation of band-stop filters
. Show how a band-stop filter is implemented with low-pass and high-pass filters
. Explain the series-resonantband-stop filter
. Explain the parallel-resonantband-stop filter
. Calculate the bandwidth and output voltage of a band-stop filter
Figure 19-27 showsa generalband-stopresponsecurve.
F I G U R E1 9 - 2 7
Generalband-stopresponsecurve.
Vout
V*o&quot;
0.707vnax
(-3 dB)
igh-PassFilter
Low-Pass/H
A band-stop filter can be formed from a low-pass and a high-pass filter, as shown in Figure 19-28.
F I G U R E1 9 - 2 8
Low-passand high-passfilters usedto form
a band-stopfilter.
If the low-pass critical frequency,,f,ai, is set lower than the high-pass critical frequency,f,17r,a band-stop characteristicis formed as illustrated in Figure 19-29.
BAND-STOP
FILTERSI
779
F I G U R E1 9 - 2 9
Band-stopresponsecurve.
SeriesResonantBand-StopFilter
A series resonant circuit used in a band-stop configuration is shown in Figure 19-30.
Basically, it works as follows: At the resonantfrequency,the impedanceis minimum, and
therefore the output voltage is minimum. Most of the input voltage is dropped acrossR.
At frequencies above and below resonance,the impedance increases,causing more voltage acrossthe output.
F I G U R E1 9 - 3 0
Seriesresonant band-stopfilter.
EXAMPLE
19-11
vi,
Find the output voltage magnitude at fs and the bandwidth in Figure 19-31.
FIGURE,I9-31
R
vi,
100mV
5 6 O ,p. w
2Q
L
100 mH
C
0.01rrF-
+
Solution
Since X1 = Xc &amp;t resonance,the output voltage is
2a
v*, = (-3*-\r. = (
\too
&quot; &quot; -,,
&quot; = 3.45mv
\ n + n r / &quot; - - \ s 8a 1
To determine the bandwidth, calculate the center frequency and Q of the coil.
fo=--]-=+=5.03kH2
rv
2*Ee
znr.(toO..rHxo.otrrp')
kHzXl0OmH)
n _ 4, _ 2nfL _ 2n(5.03
\t=-=R
R
58fi
3.16kO = - 5 + . 5
58f)
n w =4 = t' o= ' ,9&quot;= 92.3:H2
o
Related Prohlem
width.
s4.s
Assume Rw= 10 f) in Figure 19-31. Determine Vou,ardthe band-
7BO r
BASICFILTERS
ParallelResonantBand-StopFilter
A parallel resonantcircuit used in a band-stop configuration is shown in Figure 19-32.At
the resonant frequency, the tank impedance is maximum, and so most of the input volt
age appearsacrossit. Very little voltage is acrossR at resonance.As the tank impedance
decreasesabove and below resonance.the output voltage increases.
F I G U R E1 9 - 3 2
Parallel resonantband-stopfilten
EXAMPLE19_12 Find the center frequency of the filter in Figure 19-33. Sketch the output response
curve showing the minimum and maximum voltages.
F I C U R E1 9 _ 3 3
Solution
C
The center frequency is
, ^- = @ - @ = 5 . 7 9 M H 2
r0
znr,(spplrso pg
z*Frc
At the center (resonant)frequency,
Xr= 2nfoL= 2n(5;79MHz)(5 pH) = 182 fl
- 182-Q=ll.g
o
. =x,
R*
8Cl
Z,= Rv,{Qz+ 1) = 8 Q(22.82+ l) = 4.17 pg
(purely resistive)
Next, use the voltage-divider formula to find the minimum output voltage magnitude.
v o , , \ .-i/, R \ , ,' ' - / 5 6 9 . 3 ) l o v = r . r 8 v
&quot; \ R+ z , ) &quot; \ + . l l k o / ' &quot;
At zero frequency, the impedance of the tank is R1ybecauseXg = * and Xt =
0 Q. Therefore, the maximum output voltage below resonanceis
v o u r '\ &quot;m--' l,
R-\,,
-- / t 9 9 * ) l o V = 9 . 8 6 v
*
\ R R * ) ' ' ' \ s o so / ' &quot; '
As the frequency increasesmuch higher than ftt, X6 approaches0 f), and Vou,
approachesVi&quot;(70 V). Figure 19-34 showsthe responsecurve.
TECHNOLOGY
THEORYINTO PRACTICE.
FIGURE19-34
781
vo*(y)
10
9.86
/ (MHz)
Related Problem What is the minimum output voltage if R = 1.0 kfl in
19-33?
sEcTtoN19-4
REVIEW
1. How does a band-stop filter differ from a band-passfilter?
2. Name threebasic ways to constructa band-stopfilter.
19-5 r TECHnologyTheory Into Practice
In this TECH TIP,you will plot thefrequencyresponses
of two typesof fi.ltersbased
on u seriesof oscilloscope
me&amp;surements
and i.ilentifythe typeof filter in euchcase.
The filters are contained in sealedmodules as shown in Figure 19-35. You are concerned
only with determining the filter resDonsecharacteristics and not the types of internal
comDonents.
F I G U R E1 9 - 3 5
Fiher modules.
FilterMeasurement
and Analysis
I
Refer to Figure 19-36. Based on the seriesof four oscilloscope measurements,create
a Bode plot for the filter under test, specify applicable frequencies, and identify the
type of filter.
r Refer to Figure 19-31 . Based on the seriesof six oscilloscope measurements,create a
Bode plot for the filter under test, specify applicable frequencies,and identify the type
of filter.
BASICFILTERS
2 V peak-to-peaksignal
from function generator
F I G U R E1 9 - 3 6
T E C H N O L O CT
YH E O R YI N T O P R A C T I C E.
Filter
module 2
IN
GND
OUT
2 V peak-to-peaksignal
fiom function generator
F l c u R E1 9 - 3 7
SECTION19-5
REVIEW
l. Explain how the waveforms in Figure 19*36 indicate the type of filter.
2. Explain how the wavefonns in Figure 19-37 indicate the type of filter.
783
784 I
BASICFILTERS
r SUMMARY
I In an RC low-pass filter, the output voltage is taken acrossthe capacitor and the output lagsthe
input.
r In an RI low-pass filter, the output voltage is taken across the resistor and the output lags the
input.
r In an RC high-pass f,lteq the output is taken acrossthe resistor and the output leads the input.
r In an Rl high-pass filter, the output is taken acrossthe inductor and the output leads the input.
r The roll-off rate of a basic RC or RZ filter is -20 dB per decade.
r A band-passfilter passesfrequenciesbetween the lower and upper critical frequenciesand rejects
all others.
r A band-stop filter rejects frequenciesbetween its lower and upper critical frequencies and passes
all others.
r The bandwidth of a resonant filter is determined by the quality factor (Q) of the circuit andthe
resonant frequency.
I Critical frequencies are also called -3 dB frequencies.
r The output voltage is 70.'/Voof its maximum at the critical frequencies.
r GLOSSARY
These terms are also in the end-of-book glossary.
Attenuation A reduction of the output signal compared to the input signal, resulting in a ratio
with a value of less than I for the output voltage to the input voltage of a circuit.
Band-pass filter A filter that passesa range of frequencieslying between two critical frequencies
and rejects frequencies above and below that range.
Band-stop filter A filter that rejects a range of frequencieslying between two critical frequencies
and passesfrequencies above and below that range.
Bode plot The graph of a filter's frequency responseshowing the change in the output voltageto
input voltage ratio expressedin dB as a function of frequency for a constant input voltage.
Center frequency ffi) The resonant frequency of a band-passor band-stop filter.
Criticaf frequency The frequency at which a filter's output voltage is 70.7Voof the maximum.
Decade A tenfold change in frequency or other parameter.
Decibel A logarithmic measurement of the ratio of one power to another or one voltage t0
another,which can be used to expressthe input-to-output relationship of a filter.
Passband The range of frequenciespassedby a filter.
Roll-off
r FORMULAS
The rate of decreaseof a filter's frequencv resDonse.
(1e-1)
dB= 1oron
&quot; \/fu\
P-l
Power ratio in decibels
(re-2)
dB=2or&quot;*
- fp)
Voltage ratio in decibels
\V* I
(1e-3)
, t'-
7
2nRC
Critical frequency
(re-4)
t &quot;' -
I
2n(LlR)
Critical frequency
(1e-s)
BW=b
o
Bandwidth
PROBLEMS.
r SEIF.TEST
785
1 . The maximum output voltage of a certain low-pass filter is 10 V. The output voltage at the critical frequency is
(a) 10 V
(b) 0 V
(c) 7.07Y
(d) 1.414v
2. A sinusoidal voltage with a peak-to-peak value of 15 V is applied to an RC low-pass filter. If
the reactanceat the input frequency is zero, the output voltage is
(a) 15 V peak-to-peak
(b) zero
(c) 10.6 V peak+o-peak
(d) 7.5 V peak-to-peak
3. The
same signal in Question 2 is applied to an RC high-pass filter. If the reactanceis zero at
the input frequency, the output voltage is
(a) 15 V peak-to-peak
(b) zero
(c) 10.6V peak-to-peak
(d) 7.5 V peak-to-peak
4. At the critical frequency, the output of a filter is down from its maximum by
(b) -3 dB
(a) 0 dB
(c) -20 dB
(d) -6 dB
If the output of a low-pass RC fllter is 12 dB below its maximum at f = 1 kJIz, then at f =
l0 kHz, the output is below its maximum by
(a) 3 dB
(b) l0 dB
(c) 20 dB
(d) 32 dB
6. In a passive filter, the ratio Vou,lVi, is called
(a) roll-off
1
(b) gain
(c) attenuation
(d) critical reduction
For each decadeincreasein frequency above the critical frequency, the output of a low-pass filter decreasesby
(a) 20 dB
(b) 3 dB
(c) 10 dB
(d) 0 dB
8. At the critical frequency, the phase shift through a high-pass filter is
(a) 90&quot;
(b) 0'
(c) 45&quot;
(d) dependenton the reactance
9. In a seriesresonantband-passfilter, a higher value of Q results in
(a) a higher resonant frequency
(b) a smaller bandwidth
(c) a higher impedance
(d) a larger bandwidth
10. At seriesresonance,
(a)Xr=Yt
(b) Xc&gt;XL
G) Xg&lt;XL
1 1 . In a certain parallel resonant band-passfilter the resonant frequency is 10 kHz. If the bandwidth is 2 kJ1z, the lower critical frequency is
(a) 5 kHz
t2.
(b) 12 kIIz
(c) 9 kHz
(d) not determinable
In a band-passfilter, the output voltage at the resonant frequency is
(a) minimum
(b) maximum
(c) 10.7Voof maximum
(d) 70.'7Voof minimum
13. ln
a band-stop filter, the output voltage at the critical frequenciesis
(a) minimum
(b) maximum
(c) 10.7Voof maximum
(d) 70.77aof minimum
14, At a suffrciently high value of Q, the resonant frequency for a parallel resonant filter is ideally
(a) much greater than the resonant frequency of a series resonant filter
(tr) much less than the resonant frequency of a series resonant filter
(c) equal to the resonant frequency of a seriesresonant filter
r PROBLEMS
More dfficult problems are indicated by an asterisk (*).
SECTION19-1 Low-Pass
Filters
1. In a certain low-pass filter, Xc = 500 O and R = 2.2 kQ. What is the output voltage (Vo,,)
when the input is l0 V rms?
2. A certain low-pass filter has a critical frequency of 3 kHz. Determine which of the following
frequenciesare passedand which are rejected:
(a) 100 Hz
(b) I kHz
(c) 2WIz
(d) 3 kHz
(e) 5 kHz
786 T
BASICFILTERS
Determine the output voltage (V,,,) of each filter in Figure 19-38 at the specified frequency
when V;&quot; = 10 V.
u&quot;
(a)f = 60Hz
I
Vou,
(.b)f = 400H2
(c)/=
u,
(d) f = zkJlz
lkHz
F I G U R E1 9 - 3 8
4. What is/f for each f,lter in Figure 19-38? Determine the output voltage atlf in each case
whenVr&quot;=5Y.
For the filter in Figure 19-39, calculate the value of C required for each of the following critical frequencies:
(a) 60 Hz
(b) 500 Hz
(c) I kHz
(d) 5 kHz
F I G U R E1 9 - 3 9
14,*----ML-----?_----------o%.
22oa
I
T
.
l_
6. Determinethe critical frequency for each switch position on the switched filter network of
Figure 19-40.
F I G U R E1 9 - 4 0
R
10ko
7. Sketch a Bode plot for each part of Problem 5.
8. For each following case,expressthe voltage ratio in dB:
(a)V*=lY,Vo,,=l\
( b ' )v i &quot; = 5 Y , V o u , = J \ /
( c ) V * = l 0 v , V o n = ' 7 . 0 7Y
(.d)Vi&quot;=25Y,Vo,,= 5y
9. The input voltage to a low-pass RC filter is 8 V rms. Find the output voltage at the following
dB levels:
(a) -1 dB
(b) -3 dB
(c) -6 dB
(d) -20 dB
10. For a basic RC low-pass filter, find the output voltage in dB relative to a 0 dB input for the
following frequencies (f&quot; = | Wlz):
(a) 10 kHz
(b) 100 kHz
(c) 1 MHz
. 787
PROBLEMS
SECTION19-2 High-PassFilters
11. In a high-pass fiItel X6 = 500 A and R = 2.2 kA. What is the output voltage (V,ur) when
Vi,=lOVrms?
12. A high-pass filter has a critical frequency of 50 Hz. Determine which of the following frequencies are passedand which are rejected:
(a) 1 Hz
(b) 20 Hz
(c) 50 Hz
(d) 60 Hz
(e) 30 kHz
13. Determine the output voltage of each filter in Figure 19-41 at the specified frequency when
Vi,= l0Y'
rin
(a) f = 6ottz
\$) f = a00Hz
(c)/=
(d) f = zkHz
lkHz
FIGURE 19-41
14. What isf, for each filter in Figure 19-41? Determine the output voltage atf, in each case
(V- = l0 V)'
15. Sketch the Bode plot for each filter in Figure 19-41.
*16. Determinef, for each switch position in Figure 19-42.
FIGURE19-42
Voutt
R2
1.0ko
'-'-ll
Vout2
&quot;2 o.otp
v,, n)
3
0.015pF
R
R5
2.2kO
l
4
VoutZ
Voutl
SECTION19-3 Band-Pass
Filters
17. Determine the center frequencv for each filter in Fisure 19-43.
FTGURE
19-43
r
v'[email protected]
lznrLI
0.01pF
R
7sa
(a)
(b)
7BB T
BASICFILTERS
18. Assuming that the coils in Figure 19-43 have a winding resistance of 10 Q, find the bandwidth for each filter.
19. What are the upper and lower critical frequenciesfor each filter in Figure 19-43? Assumethe
20. For each fllter in Figure 1944, find the center frequency of the passband.Neglect R1y.
F I G U R E1 9 - 4 4
vi,
21. If the coils in Figure 1944 have a winding resistanceof 4 Q, what is the output voltageat
resonancewhen V;, = 120Y?
*,,) Determine the separationof center frequencies for all switch positions in Figure 19-45. Do
any of the responsesoverlap? Assume Rw = 0 C) for each coil.
F I G U R E1 9 - 4 5
Lr
50 pH
R
4.
100prH
L3
1000pF
L2
0.001prF
*23. Design a band-passfilter using a parallel resonant circuit to meet all of the following specifications: BI4z= 500 Hz; Q = 40; and [email protected]*\ = 20 mA, V61^*y= 2.5 Y.
SECTION19-4 Band-Stop
Filters
24. Determine the center frequencv for each filter in Fisure 19-46.
PROBLEMS.
789
FrcuRE19-46
vi,
r
0.0022 1lF
(a)
0.041 1tF
(b)
25. For each filter in Figure 1947 , find the center frequency of the stopband.
26. If the coils in Figtre 1941 have a winding resistanceof 8 Q, what is the output voltage
resonancewhen V;n= 50 V?
F I G U R E1 9 - 4 7
Vin
*27. Determine the values of L1 and L2 in Figure 19-48 to pass a signal with a frequency of
I2O0 kHz and stop (reject) a signal with a frequencv of 456 kHz.
F I G U R E1 9 - 4 8
EWBTroubleshooting
and Analysis
Theseproblems require your EWB compact disk.
28. Open file PROI9-28.EWB and determine if there is a fault. If so, find the fault.
29, Open file PRO19-29.EWB and determine if there is a fault. If so, find the fault.
30. Open file PRO19-30.EWB and determine if there is a fault. If so, find the fault.
31. Open file PROI9-3l.EWB and determine if there is a fault. If so, find the fault.
32. Open file PRO19-32.EWB and determine if there is a fault. If so, find the fault.
33. Open file PRO19-33.EWB and determine if there is a fault. If so, find the fault.
34. Open file PRO19-34.EWB and determine the center frequency of the circuit.
35. Open file PRO19-35.EWB and determine the bandwidth of the circuit.
790 I
BASICFILTERS
TO SECTION
REVIEWS
Section19-1
1. The passbandis 0 Hz to 2.5 kJ{z.
2. You,= 1002-88.9' mV rms
3, 20 Iog(V&quot;,,/V*) = -9.54 dB
Section19-2
l. Vou,=0.707Y
2. You,= 9.9813.81&quot;Y
Section19-3
l. BW= 30.2kIlz - 29.8kJIz= 400Hz
2. f6= I.04MHz
Section
'19-4
l. A band-stop filter rejects, rather than passes,a certain band of frequencies.
2. High-pass/low-passcombination, series resonant circuit, and parallel resonant circuit
Section19-5
l. The waveforms indicate that the output amplitude decreaseswith an increase in frequency asin
a low-pass filter.
2. The waveforms indicate that the output amplitude is maximum at 10 kHz and drops off above
and below as in a band-passfllter.
TO RELATED
PROBLEMS
FOR
EXAMPLES
t9-l -1.41 dB
l9-2 7.23kHz
l9-3 f, increases
194 f&quot; increases
to 350 kHz. Roll-off rateremains-20 dB/decade.
-60
19-s
dB
19-6 C = 0.723y.F;Vou,= 4.98V; d = 5.7&quot;
l9-7 10.5kHz
19-8 B![/increases
to 18.8kHz.
l9-9 1.59MHz
19-10 7.12kHz (no significantdifference)
l9-ll V&quot;il = 15.2mV; BW = l05Hz
t9-12 t94V
r. (c)
TO SELF.TEST e. (b)
2. (b)
10. (a)
3. (a)
11. (c)
4. (b)
12. (b)
s.(d)
13. (c)
6. (c)
1.4.(c)
/. (al
8. (c)
```