Download presentation

Presentation is loading. Please wait.

Published byRoxana Merrow Modified over 2 years ago

1
Fundamentals of Electric Circuits Chapter 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2
Overview This chapter will introduce the idea of the transfer function: a means of describing the relationship between the input and output of a circuit. Bode plots and their utility in describing the frequency response of a circuit will also be introduced. The concept of resonance as applied to LRC circuits will be covered as well Finally, frequency filters will be discussed. 2

3
Frequency Response Frequency response is the variation in a circuit’s behavior with change in signal frequency. This is significant for applications involving filters. Filters play critical roles in blocking or passing specific frequencies or ranges of frequencies. Without them, it would be impossible to have multiple channels of data in radio communications. 3

4
Transfer Function One useful way to analyze the frequency response of a circuit is the concept of the transfer function H(ω). It is the frequency dependent ratio of a forced function Y(ω) to the forcing function X(ω). 4

5
5 Transfer Function Four possible transfer functions:

6
6 Example 1 For the RC circuit shown below, obtain the transfer function Vo/Vs and its frequency response. Let v s = V m cosωt.

7
7 Solution: The transfer function is, The magnitude is The phase is Low Pass Filter

8
8 Example 2 Obtain the transfer function Vo/Vs of the RL circuit shown below, assuming v s = V m cosωt. Sketch its frequency response.

9
9 14.2 Transfer Function (6) Solution: The transfer function is, The magnitude is The phase is High Pass Filter

10
Zeros and Poles To obtain H(ω), we first convert to frequency domain equivalent components in the circuit. H(ω) can be expressed as the ratio of numerator N(ω) and denominator D(ω) polynomials. Zeros are where the transfer function goes to zero. Poles are where it goes to infinity. They can be related to the roots of N(ω) and D(ω) 10

11
Decibel Scale We will soon discuss Bode plots. These plots are based on logarithmic scales. The transfer function can be seen as an expression of gain. Gain expressed in log form is typically expressed in bels, or more commonly decibels (1/10 of a bel) 11

12
Bode Plots One problem with the transfer function is that it needs to cover a large range in frequency. Plotting the frequency response on a semilog plot (where the x axis is plotted in log form) makes the task easier. These plots are referred to as Bode plots. Bode plots either show magnitude (in decibels) or phase (in degrees) as a function of frequency. 12

13
Standard Form The transfer function may be written in terms of factors with real and imaginary parts. For example: This standard form may include the following seven factors in various combinations: –A gain K –A pole (jω) -1 or a zero (jω) –A simple pole 1/(1+jω/p 1 ) or a simple zero (1+jω/z 1 ) –A quadratic pole 1/[1+j2 2 ω/ ω n + (jω/ ω n ) 2 ] or zero 1/[1+j2 1 ω/ ω n + (jω/ ω k ) 2 ] 13

14
Bode Plots In a bode plot, each of these factors is plotted separately and then added graphically. Gain, K: the magnitude is 20log 10 K and the phase is 0°. Both are constant with frequency. Pole/zero at the origin: For the zero (jω), the slope in magnitude is 20 dB/decade and the phase is 90°. For the pole (jω) -1 the slope in magnitude is -20 dB/decade and the phase is -90° 14

15
Bode Plots Simple pole/zero: For the simple zero, the magnitude is 20log 10 |1+jω/z 1 | and the phase is tan -1 ω/z 1. Where: This can be approximated as a flat line and sloped line that intersect at ω=z 1. This is called the corner or break frequency 15

16
Bode Plots The phase can be plotted as a series straight lines From ω=0 to ω≤z 1 /10, we let =0 At ω=z 1 we let =45° For ω≥10z 1, we let = 90° The pole is similar, except the corner frequency is at ω=p 1, the magnitude has a negative slope 16

17
Bode Plots Quadratic pole/zero: The magnitude of the quadratic pole 1/[1+j2 2 ω/ ω n + (jω/ ω n ) 2 ] is - 20log 10 [1+j2 2 ω/ ω n + (jω/ ω n ) 2 ] This can be approximated as: Thus the magnitude plot will be two lines, one with slope zero for ω<ω n and the other with slope -40dB/decade, with ω n as the corner frequency 17

18
Bode Plots The phase can be expressed as: This will be a straight line with slope of - 90°/decade starting at ω n /10 and ending at 10 ω n. For the quadratic zero, the plots are inverted. 18

19
Bode Plots 19

20
Bode Plots 20

21
Figure 14.14

22
Resonance The most prominent feature of the frequency response of a circuit may be the sharp peak in the amplitude characteristics. Resonance occurs in any system that has a complex conjugate pair of poles. It enables energy storage in the firm of oscillations It allows frequency discrimination. It requires at least one capacitor and inductor. 22

23
Series Resonance A series resonant circuit consists of an inductor and capacitor in series. Consider the circuit shown. Resonance occurs when the imaginary part of Z is zero. The value of ω that satisfies this is called the resonant frequency 23

24
Series Resonance At resonance: –The impedance is purely resistive –The voltage V s and the current I are in phase –The magnitude of the transfer function is minimum. –The inductor and capacitor voltages can be much more than the source. 24

25
25 Half-power frequencies ω 1 and ω 2 are frequencies at which the dissipated power is half the maximum value: The half-power frequencies can be obtained by setting Z equal to √2 R. Bandwidth B

26
Quality Factor The “sharpness” of the resonance is measured quantitatively by the quality factor, Q. It is a measure of the peak energy stored divided by the energy dissipated in one period at resonance. It is also a measure of the ratio of the resonant frequency to its bandwidth, B 26

27
27 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 1 and ω 2, and B. c. Determine the average power dissipated at ω = ω o, ω 1, ω 2. Take V m = 100V.

28
28

29
Parallel Resonance The parallel RLC circuit shown here is the dual of the series circuit shown previously. Resonance here occurs when the imaginary part of the admittance is zero. This results in the same resonant frequency as in the series circuit. 29

30
30 Resonance frequency: It occurs when imaginary part of Y is zero

31
Series Resonance The relevant equations for the parallel resonant circuit are: 31

32
Figure 14.29

33
Passive Filters A filter is a circuit that is designed to pass signals with desired frequencies and reject or attenuate others. A filter is passive if it consists only of passive elements, R, L, and C. They are very important circuits in that many technological advances would not have been possible without the development of filters. 33

34
Passive Filters There are four types of filters: –Lowpass passes only low frequencies and blocks high frequencies. –Highpass does the opposite of lowpass –Bandpass only allows a range of frequencies to pass through. –Bandstop does the opposite of bandpass 34

35
Lowpass Filter A typical lowpass filter is formed when the output of a RC circuit is taken off the capacitor. The half power frequency is: This is also referred to as the cutoff frequency. The filter is designed to pass from DC up to ω c 35

36
Highpass Filter A highpass filter is also made of a RC circuit, with the output taken off the resistor. The cutoff frequency will be the same as the lowpass filter. The difference being that the frequencies passed go from ω c to infinity. 36

37
Bandpass Filter The RLC series resonant circuit provides a bandpass filter when the output is taken off the resistor. The center frequency is: The filter will pass frequencies from ω 1 to ω 2. It can also be made by feeding the output from a lowpass to a highpass filter. 37

38
Bandstop Filter A bandstop filter can be created from a RLC circuit by taking the output from the LC series combination. The range of blocked frequencies will be the same as the range of passed frequencies for the bandpass filter. 38

39
39 Example 5 For the circuit in the figure below, obtain the transfer function Vo(ω)/Vi(ω). Identify the type of filter the circuit represents and determine the corner frequency. Take R1=100 =R2 and L =2mH. Answer:

40
40

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google