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EXAMPLE 63 Frequency Response of RC Filter
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Compute the response of the RC lter of Figure 69 to sinusoidal inputs at the frequencies of 60 and 10,000 Hz
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Known Quantities: R = 1 k ; C = 047 F; vi (t) = 5 cos( t) V Find: The output voltage, vo (t), at each frequency Assumptions: None Analysis: In this problem, we know the input signal voltage and the frequency response
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of the circuit (equation 618), and we need to nd the output voltage at two different frequencies If we represent the voltages in phasor form, we can use the frequency response to calculate the desired quantities: Vo 1 (j ) = HV (j ) = 1 + j CR Vi Vo (j ) = HV (j )Vi (j ) = 1 Vi (j ) 1 + j CR
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If we recognize that the cutoff frequency of the lter is 0 = 1/RC = 2,128 rad/s, we can write the expression for the frequency response in the form of equations 620 and 621: HV (j ) = 1 j 1+ 0 |HV (j )| = 1+ 1 0
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H (j ) = arctan
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Next, we recognize that at = 120 rad/s, the ratio / 0 = 0177, and at = 20,000 , / 0 = 295 Thus we compute the output voltage at each frequency as follows: Vo ( = 2 60) = 1 Vi ( = 2 60) = 0985 5 0175 V 1 + j 0177 1 Vi ( = 2 10,000) = 00345 5 1537 V 1 + j 295
Vo ( = 2 10,000) =
6
Frequency Response and System Concepts
And nally write the time-domain response for each frequency: vo (t) = 4923 cos(2 60t 0175) V vo (t) = 0169 cos(2 10,000t 1537) V at = 2 60 rad/s at = 2 10,000 rad/s
The magnitude and phase responses of the lter are plotted in Figure 611 It should be evident from these plots that only the low-frequency components of the signal are passed by the lter This low-pass lter would pass only the bass range of the audio spectrum
Magnitude response of RC filter of Example 63 1 Amplitude ratio
0 100 101 102 103 104 105 Radian frequency, rad/s (logarithmic scale) 106
Phase response of RC filter of Example 63 0
Degrees
_100 100 101 102 103 104 105 Radian frequency, rad/s (logarithmic scale) 106
Figure 611 Response of RC lter of Example 63 Comments: Can you think of a very quick, approximate way of obtaining the answer to this problem from the magnitude and phase plots of Figure 611 Try to multiply the input voltage amplitude by the magnitude response at each frequency, and determine the phase shift at each frequency Your answer should be pretty close to the one computed analytically Focus on Computer-Aided Tools: A computer-generated solution of this problem
generated by MathCad may be found in the CD-ROM that accompanies this book
EXAMPLE 64 Frequency Response of RC Low-Pass Filter in a More Realistic Circuit
Problem
Compute the response of the RC lter in the circuit of Figure 612
Part I
Circuits
RS VS ( j ) + _
R1 + C RL VL( j ) _
Source
Filter
Load
Figure 612 RC lter inserted in a circuit
Solution
Known Quantities: RS = 50
; R1 = 200
; RL = 500
; C = 10 F
Find: The output voltage, vo (t), at each frequency Assumptions: None Analysis: The circuit shown in this problem is a more realistic representation of a
ltering problem, in that we have inserted the RC lter circuit between source and load circuits (where the source and load are simply represented in equivalent form) To determine the response of the circuit, we compute the Th venin equivalent representation e of the circuit with respect to the load, as shown in Figure 613 Let R = RS + R1 and RL 1 = Z = RL j C 1 + j CRL Then the circuit response may be computed as follows: RL Z VL 1 + j CRL (j ) = = RL VS R +Z RS + R 1 + 1 + j CRL RL RL RL + R = = RL + RS + R1 + j CRL (RS + R1 ) 1 + j CRL R The above expression can be written as follows: RL K 0667 RL + R = = H (j ) = 1 + j CRL R 1 + j CREQ 1+j 600
Comments: Note the similarity and difference between the above expression and equation 616: The numerator is different than 1, because of the voltage divider effect resulting from the source and load resistances, and the cutoff frequency is given by the expression
+ VS _
+ VL _
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