Be able to design filter circuits starting with a prototype circuit and use scaling to achieve desired frequency response characteristics and component values.. Be able to use the des
Trang 1556 Introduction to Frequency Selective Circuits
DESIGN
PROBLEM
PSPICE
MULTISIM
14.36 Use a 500 nF capacitor to design a bandreject filter,
as shown in Fig P14.36 The filter has a center
fre-quency of 4 kHz and a quality factor of 5
a) Specify the numerical values of R and L
b) Calculate the upper and lower corner, or cutoff,
frequencies in kilohertz
c) Calculate the filter bandwidth in kilohertz
Figure P14.36
+
•
500 nF
If
R
L
( i
+
v <>
•
14.37 A s s u m e the bandreject filter in P r o b l e m 14.36 is
PSPICE loaded with a 1 k f t resistor
MULTISIM
a) What is the quality factor of the loaded circuit?
b) What is the bandwidth (in kilohertz) of the
loaded circuit?
c) What is the upper cutoff frequency in kilohertz?
d) What is the lower cutoff frequency in kilohertz?
14.38 Design a series RLC bandreject filter using only
three components from Appendix H that comes
closest to meeting the filter specifications in
Problem 14.36
a) Draw your filter, labeling all component values
and the input and output voltages
b) Calculate the percent error in this new filter's
center frequency and quality factor when
com-pared to the values specified in Problem 14.36
14.39 Design an RLC bandreject filter (see Fig 14.28[a])
with a quality of 2.5 and a center frequency of
25 krad/s, using a 200 nF capacitor
a) Draw your circuit, labeling the component
val-ues and output voltage
b) For the filter in part (a), calculate the bandwidth
and the values of the two cutoff frequencies
14.40 The input to the RLC bandreject filter designed in
Problem 14.39 is 250cos&tf mV Find the voltage
drop across the series combination of the inductor
and capacitor when (a) co = co ( ;, (b) co = to €i ;
(c) co = <t) C 2, (d) co = 0.2&V, (e) co = 5co a
14.41 The input to the RLC bandreject filter designed in
Problem 14.39 is 250coswf mV Find the voltage drop
across the resistor when (a) co = to a ; (b) co = co c] ;
(c) co — co c2 ; (d) o> = 02co a \ (e) co = 5co n
14.42 The purpose of this problem is to investigate how a
resistive load connected across the output terminals
of the bandreject filter shown in Fig 14.28(a) affects the behavior of the filter The loaded filter circuit is shown in Fig PI4.42
a) Find the voltage transfer function VjVi
b) What is the expression for the center frequency? c) What is the expression for the bandwidth?
d) What is the expression for the quality factor?
e) Evaluate H{jco a )
f) Evaluate //(/0)
g) Evaluate / / ( / c o ) h) What are the expressions for the corner
fre-quencies co c i and co c2 1
Figure P14.42
R
• VvV-+
R t
14.43 The parameters in the circuit in Fig PI4.42
PSPICE are R = 30 ft, L = 1/JLH C = 4 pF, and
MULTISIM R h = m a
a) Find a»(„ /3 (in kilohertz), and Q
b) Find //(/0) and //(/oo)
c) Find f c2 and /c1
d) Show that if R L is expressed in ohms the Q of
the circuit is
Q = y [1 + (30/KOJ
e) Plot Q versus R h for 10 ft < R L < 300 O
PSPICE
MULTISIM
14.44 The load in the bandreject filter circuit shown in
Fig PI4.42 is 500 ft The center frequency of the fil-ter is 25 krad/s, and the capacitor is 25 nF At very low and very high frequencies, the amplitude of the sinusoidal output voltage should be at least 90% of the amplitude of the sinusoidal input voltage
a) Specify the numerical values of R and L
b) What is the quality factor of the circuit?
Trang 2Problems 557 Sections 14.1-14.5
14.45 Given the following voltage transfer function:
Ys
1010
10*
H{s)
s 2 + 50,000* + 10
a) At what frequencies (in radians per second) is
the magnitude of the transfer function equal to
unity?
b) At what frequency is the magnitude of the
trans-fer function maximum?
c) What is the maximum value of the transfer
func-tion magnitude?
14.46 Design a series RLC bandpass filter (see Fig 14.27)
PERSPECTIVE *-or detecting the low-frequency tone generated by
DEBST pushing a telephone button as shown in Fig 14.32
PROBLEM
a) Calculate the values of L and C that place the
cutoff frequencies at the edges of the DTMF
low-frequency band Note that the resistance in
standard telephone circuits is always R = 600 ft
b) What is the output amplitude of this circuit at each of the low-band frequencies, relative to the peak amplitude of the bandpass filter?
c) What is the output amplitude of this circuit at the lowest of the high-band frequencies?
14.47 Design a DTMF high-band bandpass filter similar
r!£E!S?.« t o t n e low-band filter design in Problem 14.46 Be
PERSPECTIVE »
DESIGN" sure to include the fourth high-frequency tone,
1633 Hz, in your design What is the response ampli-tude of your filter to the highest of the low-frequency DTMF tones?
14.48 The 20 Hz signal that rings a telephone's bell has to
P RSPECTIVE n a v e a v e ry la rge amplitude to produce a loud
DESIGN"" enough bell signal How much larger can the ring-ing signal amplitude be, relative to the low-bank DTMF signal, so that the response of the filter in Problem 14.46 is no more than half as large as the largest of the DTMF tones?
Trang 315.1 First-Order Low-Pass and High-Pass
Filters p 560
15.2 Scaling p 564
15.3 Op Amp Bandpass and Bandreject
Filters p 566
15.4 Higher Order Op Amp Filters p 573
15.5 Narrowband Bandpass and Bandreject
Filters p 586
/ C H A P T E R O B J E C T I V E S
Know the op amp circuits that behave as
first-order low-pass and high-pass filters and be able
to calculate component values for these circuits
to meet specifications of cutoff frequency and
passband gain
Be able to design filter circuits starting with a
prototype circuit and use scaling to achieve
desired frequency response characteristics and
component values
Understand how to use cascaded first- and
second-order Butterworth filters to implement
low-pass, high-pass, bandpass, and bandreject
filters of any order
Be able to use the design equations to calculate
component values for prototype narrowband,
bandpass, and bandreject filters to meet desired
filter specifications
558
Active Filter Circuits
Up to this point, we have considered only passive filter circuits,
that is, filter circuits consisting of resistors, inductors, and capaci-tors There are areas of application, however, where active cir-cuits, those that employ op amps, have certain advantages over passive filters For instance, active circuits can produce bandpass and bandreject filters without using inductors This is desirable because inductors are usually large, heavy, costly, and they may introduce electromagnetic field effects that compromise the desired frequency response characteristics
Examine the transfer functions of all the filter circuits from Chapter 14 and you will notice that the maximum magnitude does not exceed 1 Even though passive resonant filters can achieve voltage and current amplification at the resonant fre-quency, passive filters in general are incapable of amplification, because the output magnitude does not exceed the input magni-tude This is not a surprising observation, as many of the transfer functions in Chapter 14 were derived using voltage or current division Active filters provide a control over amplification not available in passive filter circuits
Finally, recall that both the cutoff frequency and the pass-band magnitude of passive filters were altered with the addition
of a resistive load at the output of the filter This is not the case with active filters, due to the properties of op amps Thus, we use active circuits to implement filter designs when gain, load varia-tion, and physical size are important parameters in the design specifications
In this chapter, we examine a few of the many filter circuits that employ op amps As you will see, these op amp circuits over-come the disadvantages of passive filter circuits Also, we will show how the basic op amp filter circuits can be combined to achieve specific frequency responses and to attain a more nearly ideal filter response Note that throughout this chapter we assume that every op amp behaves as an ideal op amp
Trang 4A H tff* ••'"M
Practical Perspective
Bass Volume Control
In this chapter, we continue to examine circuits that are
fre-quency selective As described in Chapter 14, this means that
the behavior of the circuit depends on the frequency of its
sinusoidal input Most of the circuits presented here fall into
one of the four categories identified in Chapter 14—low-pass
filters, high-pass filters, bandpass filters, and bandreject
fil-ters But whereas the circuits in Chapter 14 were constructed
using sources, resistors, capacitors, and inductors, the
cir-cuits in this chapter employ op amps We shall soon see what
advantages are conferred to a filter circuit constructed using
op amps
Audio electronic systems such as radios, tape players, and
CD players often provide separate volume controls labeled
''treble" and "bass." These controls permit the user to select
the volume of high frequency audio signals ("treble") inde-pendent of the volume of low frequency audio signals ("bass") The ability to independently adjust the amount of amplification (boost) or attenuation (cut) in these two fre-quency bands allows a listener to customize the sound with more precision than would be provided with a single volume control Hence the boost and cut control circuit is also referred to as a tone control circuit
The Practical Perspective example at the end of this chapter presents a circuit that implements bass volume con-trol using a single op amp together with resistors and capaci-tors An adjustable resistor supplies the necessary control over the amplification in the bass frequency range
Treble
559
Trang 5560 Active Filter Circuits
Figure 15.1 A A first-order low-pass filter
Figure 15.2 A A general op amp circuit
15,1 First-Order Low-Pass
and High-Pass Filters
Consider the circuit in Fig 15.1 Qualitatively, when the frequency of the source is varied, only the impedance of the capacitor is affected At very low frequencies, the capacitor acts like an open circuit, and the op amp
cir-cuit acts like an amplifier with a gain of -R 2 /R} At very high frequencies,
the capacitor acts like a short circuit, thereby connecting the output of the
op amp circuit to ground The op amp circuit in Fig 15.1 thus functions as
a low-pass filter with a passband gain of
—R^Ri-To confirm this qualitative assessment, we can compute the transfer
function H(s) = V 0 (s)/Vi(s) Note that the circuit in Fig 15.1 has the
gen-eral form of the circuit shown in Fig 15.2, where the impedance in the input path (Z() is the resistor R h and the impedance in the feedback path
(Zf) is the parallel combination of the resistor R 2 and the capacitor C The circuit in Fig 15.2 is analogous to the inverting amplifier circuit
from Chapter 5, so its transfer function is —Zf/Z- r Therefore, the transfer function for the circuit in Fig 15.1 is
H{s)
Zi
-Rl\ (
R ,sCj
(x)c
-K — S + (lir (15.1)
where
<-i-Oi r =
R 7 C
(15.2)
(15.3)
Note that Eq 15.1 has the same form as the general equation for low-pass fil-ters given in Chapter 14, with an important exception: The gain in the
pass-band, K, is set by the ratio Rj/R\ • The op amp low-pass filter thus permits the
passband gain and the cutoff frequency to be specified independently
A Note About Frequency Response Plots
Frequency response plots, introduced in Chapter 14, provide valuable insight into the way a filter circuit functions Thus we make extensive use of frequency response plots in this chapter, too The frequency response plots in Chapter 14 comprised two separate plots —a plot of the transfer function magnitude versus frequency, and a plot of the transfer function phase angle, in degrees, versus frequency When we use both plots, they are normally stacked on top of one another so that they can share the same frequency axis
In this chapter, we use a special type of frequency response plots
called Bode plots Bode plots are discussed in detail in Appendix E, which
includes detailed information about how to construct these plots by hand You will probably use a computer to construct Bode plots, so here we
Trang 6summarize the special features of these plots Bode plots differ from the
frequency response plots in Chapter 14 in two important ways
First, instead of using a linear axis for the frequency values, a Bode
plot uses a logarithmic axis This permits us to plot a wider range of
fre-quencies of interest Normally we plot three or four decades of
frequen-cies, say from 102 rad/s to 106 rad/s, or 1 kHz to 1 MHz, choosing the
frequency range where the transfer function characteristics are changing
If we plot both the magnitude and phase angle plots, they again share the
frequency axis
Second, instead of plotting the absolute magnitude of the transfer
function versus frequency, the Bode magnitude is plotted in decibels (dB)
versus the log of the frequency The decibel is discussed in Appendix D
Briefly, if the magnitude of the transfer function is \H(jco) |, its value in dB
is given by
/ ld B = 201og1()|//(yW)|
It is important to remember that while \H(jco)\ is an unsigned quantity,
/4dB is a signed quantity When A dR = 0, the transfer function magnitude
is 1, since 20 log10(l) = 0 When A dB < 0, the transfer function
magni-tude is between 0 and 1, and when A dB > 0, the transfer function
magni-tude is greater than 1 Finally, note that
2 0 1 o g1 0| l / V 5 | = - 3 d B Recall that we define the cutoff frequency of filters by determining
the frequency at which the maximum magnitude of the transfer function
has been reduced by 1/V2 If we translate this definition to magnitude in
dB, we define the cutoff frequency of a filter by determining the frequency
at which the maximum magnitude of the transfer function in dB has been
reduced by 3 dB For example, if the magnitude of a low-pass filter in its
passband is 26 dB, the magnitude used to find the cutoff frequency is
26 - 3 = 23 dB
Example 15.1 illustrates the design of a first-order low pass filter to
meet desired specifications of passband gain and cutoff frequency, and
also illustrates a Bode magnitude plot of the filter's transfer function
Using the circuit shown in Fig 15.1, calculate values
for C and R 2 that, together with R x = 1 fi, produce
a low-pass filter having a gain of 1 in the passband
and a cutoff frequency of 1 rad/s Construct the
transfer function for this filter and use it to sketch a
Bode magnitude plot of the filter's frequency
response
Solution
Equation 15.2 gives the passband gain in terms of
Ri and R 2 , so it allows us to calculate the required
value of R 2 '
R 2 = KR X
= i n
Equation 15.3 then permits us to calculate C to meet the specified cutoff frequency:
R 2 co c
1 (1)(1)
= I F The transfer function for the low-pass filter is given by Eq 15.1:
K-S + ft)
s + 1
Trang 7562 Active Filter Circuits
The Bode plot of | / / ( / O J ) | is shown in Fig 15.3 This
is the so-called prototype low-pass op amp filter,
because it uses a resistor value of 1 Q and a
capaci-tor value of 1 F, and it provides a cutoff frequency
of 1 rad/s As we shall see in the next section,
proto-type filters provide a useful starting point for the
design of filters by using more realistic component
values to achieve a desired frequency response
10
I
s
1¾
-10
15
- 2 0
\
^
-\
\
w (rad/s)
5.0 10
Figure 15.3 • The Bode magnitude plot of the low-pass filter from Example 15.1
You may have recognized the circuit in Fig 15.1 as the integrating amplifier circuit introduced in Chapter 7 They are indeed the same cir-cuit, so integration in the time domain corresponds to low-pass filtering in the frequency domain This relationship between integration and low-pass filtering is further confirmed by the operational Laplace transform for integration derived in Chapter 12
The circuit in Fig 15.4 is a first-order high-pass filter This circuit also has the general form of the circuit in Fig 15.2, only now the impedance in the input path is the series combination of /?t and C, and the impedance in
the feedback path is the resistor R 2 The transfer function for the circuit
in Fig 15.4 is thus
H(s)
-Z
Figure 15.4 • A first-order high-pass filter
=
-K-s -+- u) c
where
and
R 2
i-(15.4)
(15.5)
a> ( =
Trang 815.1 First-Order Low-Pass and High-Pass Filters 563
Again, the form of the transfer function given in Eq 15.4 is the same as that
given in Eq 14.20, the equation for passive high-pass filters And again, the
active filter permits the design of a passband gain greater than 1
Example 15.2 considers the design of an active high-pass filter which
must meet frequency response specifications from a Bode plot
Figure 15.5 shows the Bode magnitude plot of a
high-pass filter Using the active high-pass filter
cir-cuit in Fig 15.4, calculate values of Ri and R 2 that
produce the desired magnitude response Use a
0.1 /xF capacitor If a 10 kO load resistor is added to
this filter, how will the magnitude response change?
Solution
Begin by writing a transfer function that has the
magnitude plot shown in Fig 15.5 To do this, note
that the gain in the passband is 20 dB; therefore,
AT = 10 Also note that the 3 dB point is 500 rad/s
Equation 15.4 is the transfer function for a
high-pass filter, so the transfer function that has the
mag-nitude response shown in Fig 15.5 is given by
/ / ( 5 ) = - 1 0 ^
s + 500
We can compute the values of R { and R 2 needed to
yield this transfer function by equating the transfer
function with Eq 15.4:
H(s) = -10s HRT/RI)S
s + 500 s + (1/RiQ
Equating the numerators and denominators and
then simplifying, we get two equations:
RiC
Using the specified value of C (0.1 /xF), we find
R t = 20 kft, R 2 = 200 kil
The circuit is shown in Fig 15.6
Because we have made the assumption that
the op amp in this high-pass filter circuit is ideal,
the addition of any load resistor, regardless of its
resistance, has no effect on the behavior of the op
amp Thus, the magnitude response of a high-pass
filter with a load resistor is the same as that of a
high-pass filter with no load resistor, which is
depicted in Fig 15.5
30
20
10
tt 0
<_ -10
- 2 0
- 3 0
-40
1 5 10 50 100 5001000 500010,000
o) (rad/s)
Figure 15.5 • The Bode magnitude plot of the high-pass filter for
Example 15.2
/ /
/
\/
i <
/
/
A / 1
200 k i l
20 left °*ifF
^ 1(—
Figure 15.6 • The high-pass filter for Example 15.2
Trang 95 6 4 Active Filter Circuits
• A S S E S S M E N T PROBLEMS
Objective 1—Know the op amp circuits that behave as first order low-pass and high-pass filters and be able to
calculate their component values
15.1 Compute the values for R 2 and C that yield a
high-pass filter with a passband gain of 1 and a
cutoff frequency of 1 rad/s if R { is 1 ft (Note:
This is the prototype high-pass filter.)
Answer: R 2 = 1 ft, C = 1 F
NOTE: Also try Chapter Problems 15.6 and 15.10
15.2 Compute the resistor values needed for the
low-pass filter circuit in Fig 15.1 to produce the transfer function
-20,000
H(s) =
s + 5000 Use a 5 /JLF capacitor
Answer: R x = 10 ft, R 2 = 40 ft
Component scale factors •
15,2 Scaling
In the design and analysis of both passive and active filter circuits, working with element values such as 1 ft, 1 H, and 1 F is convenient Although these values are unrealistic for specifying practical components, they greatly simplify computations After making computations using
conven-ient values of R, L, and C, the designer can transform the convenconven-ient
val-ues into realistic valval-ues using the process known as scaling
There are two types of scaling: magnitude and frequency We scale a
circuit in magnitude by multiplying the impedance at a given frequency by
the scale factor k m Thus we multiply all resistors and inductors by k m and
all capacitors by l/k m If we let unprimed variables represent the initial
values of the parameters, and we let primed variables represent the scaled values of the variables, we have
Note that k m is by definition a positive real number that can be either less than or greater than 1
In frequency scaling, we change the circuit parameters so that at the new frequency, the impedance of each element is the same as it was at the original frequency Because resistance values are assumed to be
independ-ent of frequency, resistors are unaffected by frequency scaling If we let kf
denote the frequency scale factor, both inductors and capacitors are
multi-plied by \/kf Thus for frequency scaling,
The frequency scale factor kf is also a positive real number that can be less
than or greater than unity
A circuit can be scaled simultaneously in both magnitude and frequency The scaled values (primed) in terms of the original values (unprimed) are
R = k m R,
k,
L' = - ^ L,
C =
7
Trang 1015.2 Scaling 565
The Use of Scaling in the Design of Op Amp Filters
To use the concept of scaling in the design of op amp filters, first select
the cutoff frequency, co c , to be 1 rad/s (if you are designing low- or
high-pass filters), or select the center frequency, w0, to be 1 rad/s (if you are
designing bandpass or bandreject filters).Then select a 1 F capacitor and
calculate the values of the resistors needed to give the desired passband
gain and the 1 rad/s cutoff or center frequency Finally, use scaling to
compute more realistic component values that give the desired cutoff or
center frequency
Example 15.3 illustrates the scaling process in general, and
Example 15.4 illustrates the use of scaling in the design of a low-pass filter
The series RLC circuit shown in Fig 15.7 has a
cen-ter frequency of V l / L C = 1 rad/s, a bandwidth of
R/L = 1 rad/s, and thus a quality factor of 1 Use
scaling to compute new values of R and L that yield
a circuit with the same quality factor but with a
cen-ter frequency of 500 Hz Use a 2 /xF capacitor
Figure 15.7 • The series RLC circuit for Example 15.3
Solution
Begin by computing the frequency scale factor that
will shift the center frequency from 1 rad/s to
500 Hz The unprimed variables represent values
before scaling, whereas the primed variables repre-sent values after scaling
o)' () 277-(500)
Now, use Eq 15.9 to compute the magnitude scale factor that, together with the frequency scale factor,
will yield a capacitor value of 2 fxF:
k f C
1 (3141.59)(2 X 10"(1) = 159.155 Use Eq 15.9 again to compute the magnitude- and
frequency-scaled values of R and L:
R'
V
k m R 159.155 ft,
L = 50.66 mH
With these component values, the center
fre-quency of the series RLC circuit is VTJLC = 3141.61 rad/s or 500 Hz, and the band-width is R/L = 3141.61 rad/s or 500 Hz; thus the
quality factor is still 1
Use the prototype low-pass op amp filter from
Example 15.1, along with magnitude and frequency
scaling, to compute the resistor values for a
low-pass filter with a gain of 5, a cutoff frequency of
1000 Hz, and a feedback capacitor of 0.01 /xF
Construct a Bode plot of the resulting transfer
func-tion's magnitude
Solution
To begin, use frequency scaling to place the cutoff
frequency at 1000 Hz:
k = to' /to = 2TT(1000)/1 = 6283.185,
where the primed variable has the new value and the unprimed variable has the old value of the cut-off frequency Then compute the magnitude scale
factor that, together with kf = 6283.185, will scale
the capacitor to 0.01 /xF:
1
k f C (6283.185)(10^) 15,915.5
Since resistors are scaled only by using magnitude scaling,
R\ = /¾ = k m R = (15,915.5)(1) = 15,915.511