Electric Circuits, 9th Edition P59 potx

556 Introduction to Frequency Selective Circuits DESIGN PROBLEM PSPICE MULTISIM 14.36 Use a 500 nF capacitor to design a bandreject filter, as shown in Fig. P14.36. The filter has a center fre- quency of 4 kHz and a quality factor of 5. a) Specify the numerical values of R and L. b) Calculate the upper and lower corner, or cutoff, frequencies in kilohertz. c) Calculate the filter bandwidth in kilohertz. Figure P14.36 + • 500 nF If R L ( i + v <> • 14.37 Assume the bandreject filter in Problem 14.36 is PSPICE loaded with a 1 kft resistor. MULTISIM a) What is the quality factor of the loaded circuit? b) What is the bandwidth (in kilohertz) of the loaded circuit? c) What is the upper cutoff frequency in kilohertz? d) What is the lower cutoff frequency in kilohertz? 14.38 Design a series RLC bandreject filter using only three components from Appendix H that comes closest to meeting the filter specifications in Problem 14.36. a) Draw your filter, labeling all component values and the input and output voltages. b) Calculate the percent error in this new filter's center frequency and quality factor when com- pared to the values specified in Problem 14.36. 14.39 Design an RLC bandreject filter (see Fig. 14.28[a]) with a quality of 2.5 and a center frequency of 25 krad/s, using a 200 nF capacitor. a) Draw your circuit, labeling the component val- ues and output voltage. b) For the filter in part (a), calculate the bandwidth and the values of the two cutoff frequencies. 14.40 The input to the RLC bandreject filter designed in Problem 14.39 is 250cos&tf mV. Find the voltage drop across the series combination of the inductor and capacitor when (a) co = co ( ;, (b) co = to €i ; (c) co = <t) C 2, (d) co = 0.2&V, (e) co = 5co a . 14.41 The input to the RLC bandreject filter designed in Problem 14.39 is 250coswf mV. Find the voltage drop across the resistor when (a) co = to a ; (b) co = co c] ; (c) co — co c2 ; (d) o> = 02co a \ (e) co = 5co n . 14.42 The purpose of this problem is to investigate how a resistive load connected across the output terminals of the bandreject filter shown in Fig. 14.28(a) affects the behavior of the filter. The loaded filter circuit is shown in Fig. PI4.42. a) Find the voltage transfer function VjVi. b) What is the expression for the center frequency? c) What is the expression for the bandwidth? d) What is the expression for the quality factor? e) Evaluate H{jco a ). f) Evaluate //(/0). g) Evaluate //(/co). h) What are the expressions for the corner fre- quencies co c i and co c2 1 Figure P14.42 R • VvV- + R t 14.43 The parameters in the circuit in Fig. PI4.42 PSPICE are R = 30 ft, L = 1/JLH. C = 4 pF, and MULTISIM Rh=ma a) Find a» ( „ /3 (in kilohertz), and Q. b) Find //(/0) and //(/oo). c) Find f c2 and / c1 . d) Show that if R L is expressed in ohms the Q of the circuit is Q = y [1 + (30/KOJ. e) Plot Q versus R h for 10 ft < R L < 300 O. PSPICE MULTISIM 14.44 The load in the bandreject filter circuit shown in Fig. PI4.42 is 500 ft. The center frequency of the fil- ter is 25 krad/s, and the capacitor is 25 nF. At very low and very high frequencies, the amplitude of the sinusoidal output voltage should be at least 90% of the amplitude of the sinusoidal input voltage. a) Specify the numerical values of R and L. b) What is the quality factor of the circuit? Problems 557 Sections 14.1-14.5 14.45 Given the following voltage transfer function: Ys. 10 10 10* H{s) s 2 + 50,000* + 10 a) At what frequencies (in radians per second) is the magnitude of the transfer function equal to unity? b) At what frequency is the magnitude of the trans- fer function maximum? c) What is the maximum value of the transfer func- tion magnitude? 14.46 Design a series RLC bandpass filter (see Fig. 14.27) PERSPECTIVE *- or detecting the low-frequency tone generated by DEBST pushing a telephone button as shown in Fig. 14.32. PROBLEM a) Calculate the values of L and C that place the cutoff frequencies at the edges of the DTMF low-frequency band. Note that the resistance in standard telephone circuits is always R = 600 ft. b) What is the output amplitude of this circuit at each of the low-band frequencies, relative to the peak amplitude of the bandpass filter? c) What is the output amplitude of this circuit at the lowest of the high-band frequencies? 14.47 Design a DTMF high-band bandpass filter similar r!£E!S?.« to tne low-band filter design in Problem 14.46. Be PERSPECTIVE » DESIGN" sure to include the fourth high-frequency tone, 1633 Hz, in your design. What is the response ampli- tude of your filter to the highest of the low- frequency DTMF tones? 14.48 The 20 Hz signal that rings a telephone's bell has to P RSPECTIVE nave a ver y l ar g e amplitude to produce a loud DESIGN"" enough bell signal. How much larger can the ring- ing signal amplitude be, relative to the low-bank DTMF signal, so that the response of the filter in Problem 14.46 is no more than half as large as the largest of the DTMF tones? . . i- ' 15.1 First-Order Low-Pass and High-Pass Filters p. 560 15.2 Scaling p. 564 15.3 Op Amp Bandpass and Bandreject Filters p. 566 15.4 Higher Order Op Amp Filters p. 573 15.5 Narrowband Bandpass and Bandreject Filters p. 586 /CHAPTER OBJECTIVES Know the op amp circuits that behave as first- order low-pass and high-pass filters and be able to calculate component values for these circuits to meet specifications of cutoff frequency and passband gain. Be able to design filter circuits starting with a prototype circuit and use scaling to achieve desired frequency response characteristics and component values. Understand how to use cascaded first- and second-order Butterworth filters to implement low-pass, high-pass, bandpass, and bandreject filters of any order. Be able to use the design equations to calculate component values for prototype narrowband, bandpass, and bandreject filters to meet desired filter specifications. 558 Active Filter Circuits Up to this point, we have considered only passive filter circuits, that is, filter circuits consisting of resistors, inductors, and capaci- tors. There are areas of application, however, where active cir- cuits, those that employ op amps, have certain advantages over passive filters. For instance, active circuits can produce bandpass and bandreject filters without using inductors. This is desirable because inductors are usually large, heavy, costly, and they may introduce electromagnetic field effects that compromise the desired frequency response characteristics. Examine the transfer functions of all the filter circuits from Chapter 14 and you will notice that the maximum magnitude does not exceed 1. Even though passive resonant filters can achieve voltage and current amplification at the resonant fre- quency, passive filters in general are incapable of amplification, because the output magnitude does not exceed the input magni- tude. This is not a surprising observation, as many of the transfer functions in Chapter 14 were derived using voltage or current division. Active filters provide a control over amplification not available in passive filter circuits. Finally, recall that both the cutoff frequency and the pass- band magnitude of passive filters were altered with the addition of a resistive load at the output of the filter. This is not the case with active filters, due to the properties of op amps. Thus, we use active circuits to implement filter designs when gain, load varia- tion, and physical size are important parameters in the design specifications. In this chapter, we examine a few of the many filter circuits that employ op amps. As you will see, these op amp circuits over- come the disadvantages of passive filter circuits. Also, we will show how the basic op amp filter circuits can be combined to achieve specific frequency responses and to attain a more nearly ideal filter response. Note that throughout this chapter we assume that every op amp behaves as an ideal op amp. AH tff* ••'"M Practical Perspective Bass Volume Control In this chapter, we continue to examine circuits that are fre- quency selective. As described in Chapter 14, this means that the behavior of the circuit depends on the frequency of its sinusoidal input. Most of the circuits presented here fall into one of the four categories identified in Chapter 14—low-pass filters, high-pass filters, bandpass filters, and bandreject fil- ters. But whereas the circuits in Chapter 14 were constructed using sources, resistors, capacitors, and inductors, the cir- cuits in this chapter employ op amps. We shall soon see what advantages are conferred to a filter circuit constructed using op amps. Audio electronic systems such as radios, tape players, and CD players often provide separate volume controls labeled ''treble" and "bass." These controls permit the user to select the volume of high frequency audio signals ("treble") inde- pendent of the volume of low frequency audio signals ("bass"). The ability to independently adjust the amount of amplification (boost) or attenuation (cut) in these two fre- quency bands allows a listener to customize the sound with more precision than would be provided with a single volume control. Hence the boost and cut control circuit is also referred to as a tone control circuit. The Practical Perspective example at the end of this chapter presents a circuit that implements bass volume con- trol using a single op amp together with resistors and capaci- tors. An adjustable resistor supplies the necessary control over the amplification in the bass frequency range. Treble 559 560 Active Filter Circuits Figure 15.1 A A first-order low-pass filter. Figure 15.2 A A general op amp circuit. 15,1 First-Order Low-Pass and High-Pass Filters Consider the circuit in Fig. 15.1. Qualitatively, when the frequency of the source is varied, only the impedance of the capacitor is affected. At very low frequencies, the capacitor acts like an open circuit, and the op amp cir- cuit acts like an amplifier with a gain of -R 2 /R}. At very high frequencies, the capacitor acts like a short circuit, thereby connecting the output of the op amp circuit to ground. The op amp circuit in Fig. 15.1 thus functions as a low-pass filter with a passband gain of —R^Ri- To confirm this qualitative assessment, we can compute the transfer function H(s) = V 0 (s)/Vi(s). Note that the circuit in Fig. 15.1 has the gen- eral form of the circuit shown in Fig. 15.2, where the impedance in the input path (Z ( ) is the resistor R h and the impedance in the feedback path (Zf) is the parallel combination of the resistor R 2 and the capacitor C. The circuit in Fig. 15.2 is analogous to the inverting amplifier circuit from Chapter 5, so its transfer function is —Zf/Z- r Therefore, the transfer function for the circuit in Fig. 15.1 is H{s) Zi -Rl\ ( R ,sCj (x) c -K —. S + (li r (15.1) where and <-i- Oi r = R 7 C (15.2) (15.3) Note that Eq. 15.1 has the same form as the general equation for low-pass fil- ters given in Chapter 14, with an important exception: The gain in the pass- band, K, is set by the ratio Rj/R\ • The op amp low-pass filter thus permits the passband gain and the cutoff frequency to be specified independently. A Note About Frequency Response Plots Frequency response plots, introduced in Chapter 14, provide valuable insight into the way a filter circuit functions. Thus we make extensive use of frequency response plots in this chapter, too. The frequency response plots in Chapter 14 comprised two separate plots —a plot of the transfer function magnitude versus frequency, and a plot of the transfer function phase angle, in degrees, versus frequency. When we use both plots, they are normally stacked on top of one another so that they can share the same frequency axis. In this chapter, we use a special type of frequency response plots called Bode plots. Bode plots are discussed in detail in Appendix E, which includes detailed information about how to construct these plots by hand. You will probably use a computer to construct Bode plots, so here we 15.1 First-Order Low-Pass and High-Pass Filters 561 summarize the special features of these plots. Bode plots differ from the frequency response plots in Chapter 14 in two important ways. First, instead of using a linear axis for the frequency values, a Bode plot uses a logarithmic axis. This permits us to plot a wider range of fre- quencies of interest. Normally we plot three or four decades of frequen- cies, say from 10 2 rad/s to 10 6 rad/s, or 1 kHz to 1 MHz, choosing the frequency range where the transfer function characteristics are changing. If we plot both the magnitude and phase angle plots, they again share the frequency axis. Second, instead of plotting the absolute magnitude of the transfer function versus frequency, the Bode magnitude is plotted in decibels (dB) versus the log of the frequency. The decibel is discussed in Appendix D. Briefly, if the magnitude of the transfer function is \H(jco) |, its value in dB is given by /l dB = 201og 1() |//(y W )|. It is important to remember that while \H(jco)\ is an unsigned quantity, /4dB is a signed quantity. When A dR = 0, the transfer function magnitude is 1, since 20 log 10 (l) = 0. When A dB < 0, the transfer function magni- tude is between 0 and 1, and when A dB > 0, the transfer function magni- tude is greater than 1. Finally, note that 201og 10 |l/V5| = -3dB. Recall that we define the cutoff frequency of filters by determining the frequency at which the maximum magnitude of the transfer function has been reduced by 1/V2. If we translate this definition to magnitude in dB, we define the cutoff frequency of a filter by determining the frequency at which the maximum magnitude of the transfer function in dB has been reduced by 3 dB. For example, if the magnitude of a low-pass filter in its passband is 26 dB, the magnitude used to find the cutoff frequency is 26 - 3 = 23 dB. Example 15.1 illustrates the design of a first-order low pass filter to meet desired specifications of passband gain and cutoff frequency, and also illustrates a Bode magnitude plot of the filter's transfer function. Example 15.1 Designing a Low-Pass Op Amp Filter Using the circuit shown in Fig. 15.1, calculate values for C and R 2 that, together with R x = 1 fi, produce a low-pass filter having a gain of 1 in the passband and a cutoff frequency of 1 rad/s. Construct the transfer function for this filter and use it to sketch a Bode magnitude plot of the filter's frequency response. Solution Equation 15.2 gives the passband gain in terms of Ri and R 2 , so it allows us to calculate the required value of R 2 ' R 2 = KR X = (1)0) = in. Equation 15.3 then permits us to calculate C to meet the specified cutoff frequency: C 1 R 2 co c 1 (1)(1) = IF. The transfer function for the low-pass filter is given by Eq. 15.1: H(s) K- S + ft). s + 1 562 Active Filter Circuits The Bode plot of |//(/OJ)| is shown in Fig. 15.3. This is the so-called prototype low-pass op amp filter, because it uses a resistor value of 1 Q. and a capaci- tor value of 1 F, and it provides a cutoff frequency of 1 rad/s. As we shall see in the next section, proto- type filters provide a useful starting point for the design of filters by using more realistic component values to achieve a desired frequency response. 10 I s 1¾ -10 15 -20 \ ^ - \ \ 0.1 0.5 1.0 w (rad/s) 5.0 10 Figure 15.3 • The Bode magnitude plot of the low-pass filter from Example 15.1. You may have recognized the circuit in Fig. 15.1 as the integrating amplifier circuit introduced in Chapter 7. They are indeed the same cir- cuit, so integration in the time domain corresponds to low-pass filtering in the frequency domain. This relationship between integration and low-pass filtering is further confirmed by the operational Laplace transform for integration derived in Chapter 12. The circuit in Fig. 15.4 is a first-order high-pass filter. This circuit also has the general form of the circuit in Fig. 15.2, only now the impedance in the input path is the series combination of /? t and C, and the impedance in the feedback path is the resistor R 2 . The transfer function for the circuit in Fig 15.4 is thus H(s) -Z Figure 15.4 • A first-order high-pass filter. = -K- s -+- u) c where and R 2 K = i- (15.4) (15.5) a> ( . = R { C (15.6) 15.1 First-Order Low-Pass and High-Pass Filters 563 Again, the form of the transfer function given in Eq. 15.4 is the same as that given in Eq. 14.20, the equation for passive high-pass filters. And again, the active filter permits the design of a passband gain greater than 1. Example 15.2 considers the design of an active high-pass filter which must meet frequency response specifications from a Bode plot. Example 15.2 Designing a High-Pass Op Amp Filter Figure 15.5 shows the Bode magnitude plot of a high-pass filter. Using the active high-pass filter cir- cuit in Fig. 15.4, calculate values of Ri and R 2 that produce the desired magnitude response. Use a 0.1 /xF capacitor. If a 10 kO load resistor is added to this filter, how will the magnitude response change? Solution Begin by writing a transfer function that has the magnitude plot shown in Fig. 15.5. To do this, note that the gain in the passband is 20 dB; therefore, AT = 10. Also note that the 3 dB point is 500 rad/s. Equation 15.4 is the transfer function for a high- pass filter, so the transfer function that has the mag- nitude response shown in Fig. 15.5 is given by //(5) = -10^ s + 500 We can compute the values of R { and R 2 needed to yield this transfer function by equating the transfer function with Eq. 15.4: H(s) = -10s HRT/RI)S s + 500 s + (1/RiQ Equating the numerators and denominators and then simplifying, we get two equations: 10 = 500 1 RiC Using the specified value of C (0.1 /xF), we find R t = 20 kft, R 2 = 200 kil. The circuit is shown in Fig. 15.6. Because we have made the assumption that the op amp in this high-pass filter circuit is ideal, the addition of any load resistor, regardless of its resistance, has no effect on the behavior of the op amp. Thus, the magnitude response of a high-pass filter with a load resistor is the same as that of a high-pass filter with no load resistor, which is depicted in Fig. 15.5. 30 20 10 tt 0 <_ -10 -20 -30 -40 1 5 10 50 100 5001000 500010,000 o) (rad/s) Figure 15.5 • The Bode magnitude plot of the high-pass filter for Example 15.2. / / / \/ i < / / A / 1 200 kil 20 left °*if F ^ 1(— Figure 15.6 • The high-pass filter for Example 15.2. 564 Active Filter Circuits •ASSESSMENT PROBLEMS Objective 1—Know the op amp circuits that behave as first order low-pass and high-pass filters and be able to calculate their component values 15.1 Compute the values for R 2 and C that yield a high-pass filter with a passband gain of 1 and a cutoff frequency of 1 rad/s if R { is 1 ft. (Note: This is the prototype high-pass filter.) Answer: R 2 = 1 ft, C = 1 F. NOTE: Also try Chapter Problems 15.6 and 15.10. 15.2 Compute the resistor values needed for the low-pass filter circuit in Fig. 15.1 to produce the transfer function -20,000 H(s) = s + 5000 Use a 5 /JLF capacitor. Answer: R x = 10 ft, R 2 = 40 ft. Component scale factors • 15,2 Scaling In the design and analysis of both passive and active filter circuits, working with element values such as 1 ft, 1 H, and 1 F is convenient. Although these values are unrealistic for specifying practical components, they greatly simplify computations. After making computations using conven- ient values of R, L, and C, the designer can transform the convenient val- ues into realistic values using the process known as scaling. There are two types of scaling: magnitude and frequency. We scale a circuit in magnitude by multiplying the impedance at a given frequency by the scale factor k m . Thus we multiply all resistors and inductors by k m and all capacitors by l/k m . If we let unprimed variables represent the initial values of the parameters, and we let primed variables represent the scaled values of the variables, we have R' = k m R, L' = k m L, and C = C/k, (15.7) Note that k m is by definition a positive real number that can be either less than or greater than 1. In frequency scaling, we change the circuit parameters so that at the new frequency, the impedance of each element is the same as it was at the original frequency. Because resistance values are assumed to be independ- ent of frequency, resistors are unaffected by frequency scaling. If we let kf denote the frequency scale factor, both inductors and capacitors are multi- plied by \/kf. Thus for frequency scaling, R' = R, L' = L/k f , and C = C/k f . (15.8) The frequency scale factor kf is also a positive real number that can be less than or greater than unity. A circuit can be scaled simultaneously in both magnitude and frequency. The scaled values (primed) in terms of the original values (unprimed) are R = k m R, k, L' = -^ L, C = 7 k m kf C. (15.9) 15.2 Scaling 565 The Use of Scaling in the Design of Op Amp Filters To use the concept of scaling in the design of op amp filters, first select the cutoff frequency, co c , to be 1 rad/s (if you are designing low- or high- pass filters), or select the center frequency, w 0 , to be 1 rad/s (if you are designing bandpass or bandreject filters).Then select a 1 F capacitor and calculate the values of the resistors needed to give the desired passband gain and the 1 rad/s cutoff or center frequency. Finally, use scaling to compute more realistic component values that give the desired cutoff or center frequency. Example 15.3 illustrates the scaling process in general, and Example 15.4 illustrates the use of scaling in the design of a low-pass filter. Example 15.3 Scaling a Series RLC Circuit The series RLC circuit shown in Fig. 15.7 has a cen- ter frequency of Vl/LC = 1 rad/s, a bandwidth of R/L = 1 rad/s, and thus a quality factor of 1. Use scaling to compute new values of R and L that yield a circuit with the same quality factor but with a cen- ter frequency of 500 Hz. Use a 2 /xF capacitor. Figure 15.7 • The series RLC circuit for Example 15.3. Solution Begin by computing the frequency scale factor that will shift the center frequency from 1 rad/s to 500 Hz. The unprimed variables represent values before scaling, whereas the primed variables repre- sent values after scaling. o)' () 277-(500) l f I = 3141.59. Now, use Eq. 15.9 to compute the magnitude scale factor that, together with the frequency scale factor, will yield a capacitor value of 2 fxF: km ~~ J_C_ k f C 1 (3141.59)(2 X 10" (1 ) = 159.155. Use Eq. 15.9 again to compute the magnitude- and frequency-scaled values of R and L: R' V k m R 159.155 ft, L = 50.66 mH. With these component values, the center fre- quency of the series RLC circuit is VTJLC = 3141.61 rad/s or 500 Hz, and the band- width is R/L = 3141.61 rad/s or 500 Hz; thus the quality factor is still 1. Example 15.4 Scaling a Prototype Low-Pass Op Amp Filter Use the prototype low-pass op amp filter from Example 15.1, along with magnitude and frequency scaling, to compute the resistor values for a low- pass filter with a gain of 5, a cutoff frequency of 1000 Hz, and a feedback capacitor of 0.01 /xF. Construct a Bode plot of the resulting transfer func- tion's magnitude. Solution To begin, use frequency scaling to place the cutoff frequency at 1000 Hz: k f = to' c /to c = 2TT(1000)/1 = 6283.185, where the primed variable has the new value and the unprimed variable has the old value of the cut- off frequency. Then compute the magnitude scale factor that, together with kf = 6283.185, will scale the capacitor to 0.01 /xF: 1 k f C (6283.185)(10^) 15,915.5. Since resistors are scaled only by using magnitude scaling, R\ = /¾ = k m R = (15,915.5)(1) = 15,915.511. . specifications. 558 Active Filter Circuits Up to this point, we have considered only passive filter circuits, that is, filter circuits consisting of resistors, inductors, and capaci- tors. There. Component scale factors • 15,2 Scaling In the design and analysis of both passive and active filter circuits, working with element values such as 1 ft, 1 H, and 1 F is convenient. Although these