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be found in the CD-ROM that accompanies this book
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The importance and usefulness of the frequency response concept lies in its ability to summarize the response of a circuit in a single function of frequency, H (j ), which can predict the load voltage or current at any frequency, given the input Note that the frequency response of a circuit can be de ned in four different ways: VL (j ) VS (j ) IL (j ) IS (j )
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HV (j ) =
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HI (j ) =
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VL (j ) HZ (j ) = IS (j )
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IL (j ) HY (j ) = VS (j )
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If HV (j ) and HI (j ) are known, one can directly derive the other two expressions: VL (j ) IL (j ) HZ (j ) = (612) = ZL (j ) = ZL (j )HI (j ) IS (j ) IS (j )
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Frequency Response and System Concepts
HY (j ) =
1 VL (j ) 1 IL (j ) = = HV (j ) VS (j ) ZL (j ) VS (j ) ZL (j )
(613)
With these de nitions in hand, it is now possible to introduce one of the central concepts of electrical circuit analysis: lters The concept of ltering an electrical signal will be discussed in the next section
EXAMPLE 62 Computing the Frequency Response of a Circuit
Problem
L IL( j ) + VL( j ) _
Compute the frequency response HZ (j ) for the circuit of Figure 68
IS ( j )
Solution
Known Quantities: R1 = 1 k ; L = 2 mH; RL = 4 k
Find: The frequency response HZ (j ) = VL (j )/IS (j ) Assumptions: None Analysis: To determine expressions for the load voltage, we recognize that the load
current can be obtained simply by using a current divider between the two branches connected to the current source, and that the load voltage is simply the product of the load current times RL Using the current divider rule, we obtain the following expression for IL : 1 1 RL + j L I = I IL = RL L S 1 1 S 1+ +j + R1 R1 RL + j L R1 and I L RL VL (j ) = HZ (j ) = = IS IS RL RL L 1+ +j R1 R1
Substituting numerical values, we obtain: HZ (j ) = 4 103 08 103 = 3 1 + j 04 10 6 2 10 1+4+j 3 10
Comments: You should verify that the untis of the expression for HZ (j ) are indeed ohms, as they should be from the de nition of HZ Focus on Computer-Aided Tools: A computer-generated solution of this problem may
be found in the CD-ROM that accompanies this book
FILTERS
There are many practical, everyday applications that involve lters of one kind or another Just to mention two, ltration systems are used to eliminate impurities
Part I
Circuits
from drinking water, and sunglasses are used to lter out eye-damaging ultraviolet radiation and to reduce the intensity of sunlight reaching the eyes An analogous concept applies to electrical circuits: it is possible to attenuate (ie, reduce in amplitude) or altogether eliminate signals of unwanted frequencies, such as those that may be caused by electrical noise or other forms of interference This section will be devoted to the analysis of electrical lters Low-Pass Filters Figure 69 depicts a simple RC lter and denotes its input and output voltages by Vi and Vo The frequency response for the lter may be obtained by considering the function Vo (j ) (614) H (j ) = Vi and noting that the output voltage may be expressed as a function of the input voltage by means of a voltage divider, as follows: 1/j C 1 = Vi (j ) Vo (j ) = Vi (j ) R + 1/j C 1 + j RC Thus, the frequency response of the RC lter is 1 Vo (j ) = Vi 1 + j CR (616) (615)
RC low-pass filter The circuit preserves lower frequencies while attenuating the frequencies above the cutoff frequency, v0 = 1/RC The voltages Vi and Vo are the filter input and output voltages, respectively + Vi _ R C + Vo _
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