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Jean Baptiste Joseph Fourier (1768 1830), French mathematician and physicist who formulated the Fourier series Photo courtesy of Deutsches Museum, Munich
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15 Amplitude of Fourier coefficient 2 1 0 _1 _2 00 Square wave 2 4 6 8 10 12 14 16 18 20 22 24 Harmonic Fourier spectrum of square wave
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Figure 63 Amplitude spectrum of square wave
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Time
Time
Fundamental frequency
Two frequency components
Time
Time
Three frequency components
Four frequency components
Time
Five frequency components
Figure 64 Evolution of a square wave from its Fourier components
Next, an expression for the load voltage, VL , may be found by connecting the load to the Th venin equivalent source circuit and by computing the result of a e simple voltage divider, as illustrated in Figure 65 and by the following equation: VL = = ZL VT ZL + Z T (65)
+ V T _
ZT + ZL VL _
Z2 ZL V (ZS + Z1 )Z2 ZS + Z1 + Z2 S ZL + ZS + Z 1 + Z 2 Z L Z2 = VS ZL (ZS + Z1 + Z2 ) + (ZS + Z1 )Z2
Figure 65 Complete equivalent circuit
6
Frequency Response and System Concepts
Thus, the frequency response of the circuit, as de ned in equation 64, is given by the expression VL Z L Z2 (j ) = HV (j ) = VS ZL (ZS + Z1 + Z2 ) + (ZS + Z1 )Z2 (66)
The expression for HV (j ) is therefore known if the impedances of the circuit elements are known Note that HV (j ) is a complex quantity (dimensionless, because it is the ratio of two voltages), and that it therefore follows that
VL (j ) is a phase-shifted and amplitude-scaled version of VS (j )
If the phasor source voltage and the frequency response of the circuit are known, the phasor load voltage can be computed as follows: VL (j ) = HV (j ) VS (j ) VL ej L = |HV |ej H VS ej S or VL ej L = |HV |VS ej ( H + S ) where VL = |HV | VS and L = H + S (610) (69) (67) (68)
Thus, the effect of inserting a linear circuit between a source and a load is best understood by considering that, at any given frequency, , the load voltage is a sinusoid at the same frequency as the source voltage, with amplitude given by VL = |HV | VS and phase equal to L = H + S , where |HV | is the magnitude of the frequency response and H its phase angle Both |HV | and H are functions of frequency
EXAMPLE 61 Computing the Frequency Response of a Circuit Using Equivalent Circuit Ideas
Problem
R1 +
+ VS _
Compute the frequency response HV (j ) for the circuit of Figure 66
C RL VL _
Solution
Known Quantities: R1 = 1 k ; C = 10 F; RL = 10 k
Find: The frequency response HV (j ) = VL (j )/VS (j ) Assumptions: None
Part I
Circuits
Analysis: To solve this problem we use an equivalent circuit approach Recognizing that RL is the load resistance, we determine the equivalent circuit representation of the circuit to the left of the load, using the techniques perfected in s 3 and 4 The Th venin e equivalent circuit is shown in Figure 67 Using the voltage divider rule and the equivalent circuit shown in the gure, we obtain the following expression
ZT VT + _ ZT = Z1 || Z2
VL =
ZL VT = ZT + Z L
ZL Z2 V S = H V VS Z1 Z2 Z1 + Z 2 + ZL Z1 + Z 2
and Z L Z2 VL (j ) = HV (j ) = VS ZL (Z1 + Z2 ) + Z1 Z2 The impedances of the circuit elements are: Z1 = 103 ; Z2 = The resulting frequency response can be calculated to be:
1 j 10 5
VT = VS
Z2 Z1 + Z2
; ZL = 104
HV (j ) = 104
104 100 j 10 5 = 110 + j 1 103 103 + + j 10 5 j 10 5
100 100 = = arctan j arctan( 110 ) 110 1102 + 2 2 + 2 e 110
Comments: The use of equivalent circuit ideas is often helpful in deriving frequency
response functions, because it naturally forces us to identify source and load quantities However, it is certainly not the only method of solution For example, nodal analysis would have yielded the same results just as easily, by recognizing that the top node voltage is equal to the load voltage, and by solving directly for VL as a function of VS , without going through the intermediate step of computing the Th venin equivalent source circuit e
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