EE221 Circuits II. Chapter 14 Frequency Response
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1 EE22 Circuits II Chapter 4 Frequency Response
2 Frequency Response Chapter 4 4. Introduction 4.2 Transfer Function 4.3 Bode Plots 4.4 Series Resonance 4.5 Parallel Resonance 4.6 Passive Filters 4.7 Active filters 2
3 4. Introduction What is Frequency Response of a Circuit? It is the variation in a circuit s behavior with change in signal frequency and may also be considered as the variation of the gain and phase with frequency. 3
4 4.2 Transfer Function The transfer function H(ω) of a circuit is the frequency-dependent ratio of a phasor output Y(ω) (voltage or current ) to a phasor input X(ω) (voltage or current). H( ω) Y( ω) = = H( ω) X( ω) φ 4
5 4.2 Transfer Function Four possible transfer functions: H( ω ) = Voltage gain = V o( ω) V ( ω) i H( ω ) = Transfer Impedance = V o( ω) I ( ω) i H( ω) Y( ω) = = H( ω) φ X( ω) H( ω ) = Current gain = Io( ω) I ( ω) i H( ω ) = Transfer Admittance = Io( ω) V ( ω) i 5
6 4.2 Transfer Function Example For the RC circuit shown below, obtain the transfer function Vo/Vs and its frequency response. Let v s = V m cosωt. 6
7 4.2 Transfer Function Solution: The transfer function is H( ω) = V V o s jωc = = R + / jω C + jω RC The magnitude is H( ω) = 2, + ( ω / ω o ) The phase is ω φ = tan ω o ω o =/RC 7 Low Pass Filter
8 4.2 Transfer Function Example 2 Obtain the transfer function Vo/Vs of the RL circuit shown below, assuming v s = V m cosωt. Sketch its frequency response. 8
9 4.2 Transfer Function Solution: The transfer function is H( ω) = V V o s jω L = = R + jω L R + jω L, H( ω ) = The magnitude is ω 2 + ( o ) ω High Pass Filter The phase is φ = 90 tan ω o = R/L ω ω o 9
10 4.4 Bode Plots Bode Plots are semilog plots of the magnitude (in db) and phase (in deg.) of the transfer function versus frequency. H H = db He jφ = 20 log 0 H 0
11 Bode Plot of Gain K
12 Bode Plot of a zero (jω) 2
13 Bode plot of a zero 3
14 Bode Plot of a quadratic pole 4
15 5
16 Example 6
17 Example 7
18 Example 2 8
19 4.4 Series Resonance Resonance is a condition in an RLC circuit in which the capacitive and inductive reactance are equal in magnitude, thereby resulting in purely resistive impedance. Resonance frequency: Z = R + j ( ω L ω C ) ωo = rad/s or LC fo = Hz 2π LC 9
20 4.4 Series Resonance The features of series resonance: Z= R+ j( ωl ) ωc The impedance is purely resistive, Z = R; The supply voltage Vs and the current I are in phase, so cos θ = ; The magnitude of the transfer function H(ω) = Z(ω) is minimum; The inductor voltage and capacitor voltage can be much more than the source voltage. 20
21 4.4 Series Resonance Bandwidth B The frequency response of the resonance circuit current is Z = R + j ( ω L ) I = I = R 2 V m + ( ω L / ω C) 2 ω C The average power absorbed by the RLC circuit is P( ω) = 2 I 2 R The highest power dissipated occurs at resonance: P( ω o ) = 2 V R 2 m 2
22 4 4 Series Resonance Half-power frequencies ω and ω 2 are frequencies at which the dissipated power is half the maximum value: P( ω ) = P( ω ) = 2 (V / 2) R 2 m 2 = 2 Vm 4R The half-power frequencies can be obtained by setting Z equal to 2 R. R 2 + ω = + 2L R ( ) 2L LC ω R 2 2 = + ω o = ω ω2 + 2L R ( ) 2L LC Bandwidth B B = ω ω 2 22
23 4.4 Series Resonance Quality factor, Q = Peak energy stored in the Energy dissipated by the in one period at resonance circuit circuit = ω o L R = ω CR o The relationship between the B, Q and ωo: R B ω= 2 o QL o ω The quality factor is the ratio of its resonant frequency to its bandwidth. If the bandwidth is narrow, the quality factor of the resonant circuit must be high. If the band of frequencies is wide, the quality factor must be low. 23
24 4.4 Series Resonance Example 3 A series-connected circuit has R = 4 Ω and L = 25 mh. a. Calculate the value of C that will produce a quality factor of 50. b. Find ω and ω 2, and B. c. Determine the average power dissipated at ω = ω o, ω, ω 2. Take V m = 00V. 24
25 4.5 Parallel Resonance It occurs when imaginary part of Y is zero Y = R + j ( ω C ) ω L Resonance frequency: ω o = rad/s or fo = LC 2π LC Hz 25
26 4.5 Parallel Resonance Summary of series and parallel resonance circuits: characteristic Series circuit Parallel circuit ω o LC LC Q ωol R or ω RC o R ω L o or ω RC o B ω o Q ω o Q ω, ω 2 Q 0, ω, ω 2 ω o + ( ) ± 2Q B ω o ± 2 ω 2Q 2 o 2 ωo ω o + ( ) ± 2Q 2Q B ω o ± 2 26
27 4.5 Resonance Example 4 Calculate the resonant frequency of the circuit in the figure shown below. Answer: ω = 9 = rad/s 27
28 4.6 Passive Filters A filter is a circuit that is designed to pass signals with desired frequencies and reject or attenuate others. Passive filter consists of only passive element R, L and C. There are four types of filters. Low Pass High Pass Band Pass Band Stop 28
29 Low Pass Filter (Passive) 29
30 High Pass Filter (Passive) 30
31 Band Pass Filter (Passive) See Equation 4.33 for corner Frequencies ω and ω 2 3
32 Band Reject Filter (Passive) 32
33 Low Pass Filter (Active) 33
34 High Pass Filter (Active) 34
35 Band Pass Filter (Active) 35
36 Band Reject Filter (Active) 36
37 Magnitude and Frequency Scaling Example: 4 th order low-pass filter Corner Frequency: rad/sec Resistance: Ω Corner Frequency: 00π krad/sec Resistance: 0 kω 37
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