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ZS = RS +
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Figure 1226 Active band-pass lter
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(1259)
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Next, we compute the closed-loop frequency response of the op-amp, as follows: ABP (j ) = ZF j CS RF = ZS (1 + j CF RF )(1 + j CS RS ) (1260)
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The form of the op-amp response we just obtained should not appear as a surprise It is very similar (although not identical) to the product of the low-pass and high-pass responses of equations 1250 and 1256 In particular, the denominator of ABP (j ) is exactly the product of the denominators of ALP (j ) and AHP (j ) It is particularly enlightening to rewrite ALP (j ) in a slightly different form, after making the observation that each RC product corresponds to some critical frequency: 1 = 1 RF CS LP = 1 R F CF HP = 1 RS C S (1261)
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It is easy to verify that for the case where HP > LP (1262)
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the response of the op-amp lter may be represented as shown in Figure 1227 in both linear and dB plots (in the gure, 1 = 1, HP = 1,000, LP = 10) The dB plot is very revealing, for it shows that, in effect, the band-pass response is the graphical superposition of the low-pass and high-pass responses shown earlier The two 3-dB (or cutoff) frequencies are the same as in ALP (j ), 1/RF CF ; and in AHP (j ), 1/RS CS The third frequency, 1 = 1/RF CS , represents the point where the response of the lter crosses the 0-dB axis (rising slope) Since 0 dB corresponds to a gain of 1, this frequency is called the unity gain frequency The ideas developed thus far can be employed to construct more complex functions of frequency In fact, most active lters one encounters in practical applications are based on circuits involving more than one or two energy-storage elements By constructing suitable functions for ZF and ZS , it is possible to realize lters with greater frequency selectivity (ie, sharpness of cutoff), as well as atter band-pass or band-rejection functions (that is, lters that either allow or reject signals in a limited band of frequencies) A few simple applications
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Amplitude response of band-pass active filter 10
Amplitude ratio
dB amplitude response of band-pass active filter 20 10
8 6 4 2 0 10 1 100 101 102 103 104 Radian frequency (logarithmic scale) 105
0 10 20 10 1 100 102 103 104 101 Radian frequency (logarithmic scale) 105
Figure 1227 Normalized amplitude response of active band-pass lter
are investigated in the homework problems One remark that should be made in passing, though, pertains to the exclusive use of capacitors in the circuits analyzed thus far One of the advantages of op-amp lters is that it is not necessary to use both capacitors and inductors to obtain a band-pass response Suitable connections of capacitors can accomplish that task in an op-amp This seemingly minor fact is of great importance in practice, because inductors are expensive to mass-produce to close tolerances and exact speci cations and are often bulkier than capacitors with equivalent energy-storage capabilities On the other hand, capacitors are easy to manufacture in a wide variety of tolerances and values, and in relatively compact packages, including in integrated circuit form Example 125 illustrates how it is possible to construct active lters with greater frequency selectivity by adding energy-storage elements to the design
EXAMPLE 125 Second-Order Low-Pass Filter
Problem
Determine the closed-loop voltage gain as a function of frequency for the op-amp circuit of Figure 1228
Solution
Known Quantities: Feedback and source impedances Find:
R1 VS
L +
+ Vout
A(j ) =
Vout (j ) VS (j )
Schematics, Diagrams, Circuits, and Given Data: R2 C = L/R1 = 0 Assumptions: Assume ideal op-amp Analysis: The expression for the gain of the lter of Figure 1228 can be determined by using equation 1245:
A(j ) =
ZF (j ) Vout (j ) = VS (j ) ZS (j )
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