Electric Circuits, 9th Edition P55 pot

516 The Laplace Transform in Circuit Analysis Figure P13.61 h(t)' 10 40 40 / (a) *(0" 4 h(ty 40 0 10 t (b) 0 1 (c) 13.62 A rectangular voltage pulse i>,- = [«(/) - u(t - 1)] V is applied to the circuit in Fig. P13.62. Use the con- volution integral to find v n . >•', in: 13.63 Interchange the inductor and resistor in Problem 13.62 and again use the convolution inte- gral to find v Q . 13.65 a) Repeat Problem 13.64, given that the resistor in the circuit in Fig. PI3.50(a) is decreased to 10 kll. b) Does decreasing the resistor increase or decrease the memory of the circuit? c) Which circuit comes closer to transmitting a replica of the input voltage? 13.66 a) Assume the voltage impulse response of a circuit is //.(/) 0, t < 0; lOtT 4 ', t > 0. Use the convolution integral to find the output voltage if the input signal is 10*<(0 V. b) Repeat (a) if the voltage impulse response is 0, t < 0; = \ 10(1 - 20, 0 < t < 0.5 s; 0, t > 0.5 s. c) Plot the output voltage versus time for (a) and (b) for 0 < t < 1 s. 13.67 The voltage impulse response of a circuit is shown in Fig. P13.67(a). The input signal to the circuit is the rectangular voltage pulse shown in Fig. P13.67(b). a) Derive the equations for the output voltage. Note the range of time for which each equation is applicable. b) Sketch v 0 for -1 < t < 34 s. 13.64 a) Use the convolution integral to find the output voltage of the circuit in Fig. P13.50(a) if the input voltage is the rectangular pulse shown in Fig. P13.64. b) Sketch v 0 (t) versus t for the time interval 0 < t < 10 ms. Figure P13.64 »i(V) lo r(ms) Problems 517 13.68 Assume the voltage impulse response of a circuit can be modeled by the triangular waveform shown in Fig. P13.68.The voltage input signal to this circuit is the step function 10«(^) V. a) Use the convolution integral to derive the expressions for the output voltage. b) Sketch the output voltage over the interval 0 to 15 s. c) Repeat parts (a) and (b) if the area under the voltage impulse response stays the same but the width of the impulse response narrows to 4 s. d) Which output waveform is closer to replicating the input waveform: (b) or (c)? Explain. Figure P13.68 A(0 (V) 10 «(») 13.69 a) Find the impulse response of the circuit shown in Fig. P13.69(a) if v g is the input signal and v 0 is the output signal. b) Given that v q has the waveform shown in Fig. P13.69(b), use the convolution integral to find v a . c) Does v a have the same waveform as v g l Why or why not? Figure P13.69 4H v g (V) 75 0 -75 0.5 1.0 t(s) (a) (b) 13.70 a) Find the impulse response of the circuit seen in Fig. PI3.70 if v g is the input signal and v n is the output signal. b) Assume that the voltage source has the wave- form shown in Fig. P13.69(b). Use the convolu- tion integral to find v (r c) Sketch % for 0 < / < 2 s. d) Does v a have the same waveform as v„l Why or why not? Figure P13.70 13.71 The sinusoidal voltage pulse shown in Fig. P13.71(a) is applied to the circuit shown in Fig. P13.71(b). Use the convolution integral to find the value of v () at t = 75 ms. Figure P13.71 5H v, 160 n- 77/20 ir/io r(s) (b) 13.72 Use the convolution integral to find v ( , in the circuit seen in Fig. P13.72 if v t = 75u{t) V. Figure P13.72 40 0 o Vs/V— 4H J-TYYV 4 16 H 13.73 The current source in the circuit shown in Fig. P13.73(a) is generating the waveform shown in Fig. PI 3.73(b). Use the convolution integral to find v„ at t = 5 ms. 518 The Laplace Transform in Circuit Analysis Figure P13.73 i g (mA) 10 H—I—I— 1 2 3 4 -20 4 (b) 5 6 t (ms) 0.4 /itF (a) Figure P13.75 i„ {fiA) 50 0 -50 (b) 100 200 /(ms) (a) 13.74 The input voltage in the circuit seen in Fig. PI 3.74 is V; = 5[u(t) - u(t - 0.5)] V. a) Use the convolution integral to find v a . b) Sketch v a for 0 < t < 1 s. Figure P13.74 20 • /vv— 100 mH 100 mF 13.76 a) Show that if y(t) = h(() * x(t), then Y{s) H(s)X(s). b) Use the result given in (a) to find /(f) if F(s) = s(s + a)" Section 13.7 13.77 The transfer function for a linear time-invariant circuit is 13.75 a) Use the convolution integral to find v 0 in the cir- cuit in Fig. P13.75(a) if i s is the pulse shown in Fig. PI3.75(b). b) Use the convolution integral to find i 0 . c) Show that your solutions for v v and i ( , are consis- tent by calculating v a and i n at 100" ms, 100 + ms, 200" ms, and 200 + ms. H(s) V 0 4(s + 3) V g s 2 + 8s + 41 If v K = 40 cos 3/ V, what is the steady-state expres- sion for v a 'l 13.78 When an input voltage of 30u(t) V is applied to a circuit, the response is known to be -80(H)/ 111)0, v a = (50*-*™" - 20e-™*")u{t) V. What will the steady-state response be if v g = 120 cos 6000/ V? Problems 519 13.79 The op amp in the circuit seen in Fig. P13.79 is ideal. PSPKE a) Find the transfer function VJV„. MULTISIM ' "' S b) Find v a if v g = 0.6//(0 V. c) Find the steady-state expression for v„ if v g = 2 cos 10,000/: V. Figure P13.79 13.82 The inductor L x in the circuit shown in Fig. P13.82 is carrying an initial current of p A at the instant the switch opens. Find (a) v(t); (b) / - i(/); (c) i 2 (t)', and (d) A(r), where A(f) is the total flux linkage in the circuit. Figure P13.82 y„*15kll 13.80 The operational amplifier in the circuit seen in PSPICE pig. P13.80 is ideal and is operating within its lin- MULTISIM ear region. a) Calculate the transfer function V„/V R . b) If v g = 2cos400f V, what is the steady-state expression for v (> '~! 'ifPTU R «2i: 13.83 a) Let R -» oo in the circuit shown in Fig. P13.82, and use the solutions derived in Problem 13.82 to find v(t), ii(t), and i 2 {t). b) Let R = oo in the circuit shown in Fig. P13.82 and use the Laplace transform method to find -u(f), ii(t), and i 2 {t). 13.84 There is no energy stored in the circuit in Fig. P13.84 at the time the impulsive voltage is applied. a) Find v (> (t) for t > 0. b) Does your solution make sense in terms of known circuit behavior? Explain. Figure P13.80 y r> 520kn Section 13.8 13.81 Show that after V^C C coulombs are transferred from C] to C 2 in the circuit shown in Fig. 13.47, the volt- age across each capacitor is C\V {) f(C\ + C 2 ). (Hint: Use the conservation-of-charge principle.) Figure P13.84 200 H 4 mH 55(/) mV 16 mH 13.85 The parallel combination of R 2 and C 2 in the circuit shown in Fig. P13.85 represents the input circuit to a cathode-ray oscilloscope (CRO). The parallel combination of i?j and C\ is a circuit model of a compensating lead that is used to connect the CRO to the source. There is no energy stored in C\ or C 2 at the time when the 10 V source is connected to the CRO via the compensating lead. The circuit values are Q = 4 pF, C 2 = 16 pF, R l = 1.25 Mft, and R 2 = 5 MH. a) Find v a . b) Find i 0 . c) Repeat (a) and (b) given Cj is changed to 64 pF 520 The Laplace Transform in Circuit Analysis Figure P13.85 13.89 Tliere is no energy stored in the circuit in Fig. P13.89 at the time the impulsive current is applied. a) Find v () for t > () + . b) Does your solution make sense in terms of known circuit behavior? Explain. Figure P13.89 250 nF 13.86 Show that if R\C\ = RiC 2 in the circuit shown in Fig. P13.85, v (> will be a scaled replica of the source voltage. 13.87 The switch in the circuit in Fig. PI3.87 has been closed for a long time. The switch opens at t — 0. Compute (a) «,((T); (b) /,(0 + ); (c) / 2 (<T); (d) / 2 ((T); (e) i!(r); (f) / 2 (0; and (g) v{t). Figure P13.87 t = 0 1 = uv 5A © + :8 mH ::•! 0 jr 1 v(t) k\\ 4kO |16kH 13.90 The voltage source in the circuit in Example 13.1 is changed to a unit impulse; that is, v g = 8(t). a) How much energy does the impulsive voltage source store in the capacitor? b) How much energy does it store in the inductor? c) Use the transfer function to find v a (t). d) Show that the response found in (c) is identical to the response generated by first charging the capacitor to 1000 V and then releasing the charge to the circuit, as shown in Fig. P13.90. Figure P13.90 looon k -1-* ^ r> < . i 1000 V 13.88 The switch in the circuit in Fig. P13.88 has been in position a for a long time. At t = 0, the switch moves to position b. Compute (a) ^(O - ); (b) y?(0 _ ); (c) v 3 (0-); (d) i(t); (e) ^(0 + ); (f) v 2 (0 + ){ and (g)^3(0 + ). Figure P13.88 —'VW- 20kfl 100 v (-) A 0.5 ^F; 2.0/XF; r = 0 + + 13.91 There is no energy stored in the circuit in Fig. P13.91 at the time the impulse voltage is applied. a) Find i { for t > 0 + . b) Find i 2 for t > 0 + . c) Find v a for t > 0 + . d) Do your solutions for i u / 2 , and v (} make sense in terms of known circuit behavior? Explain. i(0 1.6 IJLF: + :¾ Figure P13.91 0.5 H 1>"~ 205(/) V © IH: 1H jF. Problems 521 Sections 13.1-13.8 13.92 Assume the line-to-neutral voltage Y 0 in the 60 Hz mom circuit of Fig- 13.59 is 120 /CT V (rms). Load R CI is absorbing 1200 W; load R b is absorbing 1800 W; and load X a is absorbing 350 magnetizing VAR. The inductive reactance of the line (X{) is 1 fl. Assume V<, does not change after the switch opens. a) Calculate the initial value of i 2 (t) and i[ 0 (t). b) Find V 0 , v () (t), and v () (Q + ) using the s-domain circuit of Fig. 13.60. c) Test the steady-state component of v a using pha- sor domain analysis. d) Using a computer program of your choice, plot v 0 vs. t for 0 £ t < 20 ms. 13.93 Assume the switch in the circuit in Fig. 13.59 'ERSPECTIVE °P ens al tne instant the sinusoidal steady-state voltage v a is zero and going positive, i.e., v 0 = 120V2~sinl207rtV. a) Find v 0 {t) for t > 0. b) Using a computer program of your choice, plot v 0 (t) vs. t for 0 < t < 20 ms. c) Compare the disturbance in the voltage in part (a) with that obtained in part (c) of Problem 13.92. 13.94 The purpose of this problem is to show that the •ERSPEcnvE li ne -t°- neutra l voltage in the circuit in Fig. 13.59 can go directly into steady state if the load R h is disconnected from the circuit at precisely the right time. Let v 0 = V m cos( 12077/ - 0°) V, where V m = 120 V2. Assume v g does not change after R b is disconnected. a) Find the value of 6 (in degrees) so that v 0 goes directly into steady-state operation when the load R f) is disconnected. b) For the value of 6 found in part (a), find %(t) for t > 0. c) Using a computer program of your choice, plot on a single graph, for -10 ms ^ t ^ 10 ms, v a (t) before and after load R b is disconnected. I CHAPTER CONTENTS 14.1 Some Preliminaries p. 524 14.2 Low-Pass Filters p. 526 14.3 High-Pass Filters p. 532 14.4 Bandpass Filters p. 536 14.5 Bandreject Filters p. 545 Know the RL and RC circuit configurations that act as low-pass filters and be able to design RL and RC circuit component values to meet a specified cutoff frequency. Know the RL and RC circuit configurations that act as high-pass filters and be able to design RL and RC circuit component values to meet a specified cutoff frequency. Know the RLC circuit configurations that act as bandpass filters, understand the definition of and relationship among the center frequency, cutoff frequencies, bandwidth, and quality factor of a bandpass filter, and be able to design RLC circuit component values to meet design specifications. Know the RLC circuit configurations that act as bandreject filters, understand the definition of and relationship among the center frequency, cutoff frequencies, bandwidth, and quality factor of a bandreject filter, and be able to design RLC circuit component values to meet design specifications. 522 Introduction to Frequency Selective Circuits Up to this point in our analysis of circuits with sinusoidal sources, the source frequency was held constant. In this chapter, we analyze the effect of varying source frequency on circuit volt- ages and currents. The result of this analysis is the frequency response of a circuit. We've seen in previous chapters that a circuit's response depends on the types of elements in the circuit, the way the ele- ments are connected, and the impedance of the elements. Although varying the frequency of a sinusoidal source does not change the element types or their connections, it does alter the impedance of capacitors and inductors, because the impedance of these elements is a function of frequency. As we will see, the careful choice of circuit elements, their values, and their con- nections to other elements enables us to construct circuits that pass to the output only those input signals that reside in a desired range of frequencies. Such circuits are called frequency-selective circuits. Many devices that communicate via electric signals, such as telephones, radios, televisions, and satellites, employ frequency-selective circuits. Frequency-selective circuits are also called filters because of their ability to filter out certain input signals on the basis of fre- quency. Figure 14.1 on page 524 represents this ability in a sim- plistic way. To be more accurate, we should note that no practical frequency-selective circuit can perfectly or completely filter out selected frequencies. Rather, filters attenuate—that is, weaken or lessen the effect of—any input signals with frequencies outside frequencies outside a particular frequency band. Your home stereo system may have a graphic equalizer, which is an excellent example of a collection of filter circuits. Each band in the graphic equalizer is a filter that amplifies sounds (audible frequencies) in the frequency range of the band and attenuates frequencies out- side of that band. Thus the graphic equalizer enables you to change the sound volume in each frequency band. Practical Perspective Pushbutton Telephone Circuits In this chapter, we examine circuits in which the source fre- quency varies. The behavior of these circuits varies as the source frequency varies, because the impedance of the reac- tive components is a function of the source frequency. These frequency-dependent circuits are called filters and are used in many common electrical devices. In radios, filters are used to select one radio station's signal while rejecting the signals from others transmitting at different frequencies. In stereo systems, filters are used to adjust the relative strengths of the low- and high-frequency components of the audio signal. Filters are also used throughout telephone systems. A pushbutton telephone produces tones that you hear when you press a button. You may have wondered about these tones. How are they used to tell the telephone system which button was pushed? Why are tones used at all? Why do the tones sound musical? How does the phone system tell the dif- ference between button tones and the normal sounds of peo- ple talking or singing? The telephone system was designed to handle audio signals—those with frequencies between 300 Hz and 3 kHz. Thus, all signals from the system to the user have to be audible—including the dial tone and the busy signal. Similarly, all signals from the user to the system have to be audible, including the signal that the user has pressed a button. It is important to distinguish button signals from the normal audio signal, so a dual-tone-multiple-frequency (DTMF) design is employed. When a number button is pressed, a unique pair of sinusoidal tones with very precise frequencies is sent by the phone to the telephone system. The DTMF frequency and timing specifications make it unlikely that a human voice could pro- duce the exact tone pairs, even if the person were trying. In the central telephone facility, electric circuits monitor the audio signal, listening for the tone pairs that signal a number. In the Practical Perspective example at the end of the chapter, we will examine the design of the DTMF filters used to deter- mine which button has been pushed. 523 524 Introduction to Frequency Selective Circuits Input signal Filter Output signal Figure 14.1 • The action of a filter on an input signal results in an output signal. We begin this chapter by analyzing circuits from each of the four major categories of filters: low pass, high pass, band pass, and band reject. The transfer function of a circuit is the starting point for the frequency response analysis. Pay close attention to the similarities among the trans- fer functions of circuits that perform the same filtering function. We will employ these similarities when designing filter circuits in Chapter 15. 14.1 Some Preliminaries Vi(s) Figure 14.2 A A circuit with voltage input and output. Recall from Section 13.7 that the transfer function of a circuit provides an easy way to compute the steady-state response to a sinusoidal input. There, we considered only fixed-frequency sources. To study the frequency response of a circuit, we replace a fixed-frequency sinusoidal source with a varying- frequency sinusoidal source. The transfer function is still an immensely useful tool because the magnitude and phase of the output signal depend only on the magnitude and phase of the transfer function H{ja)). Note that the approach just outlined assumes that we can vary the fre- quency of a sinusoidal source without changing its magnitude or phase angle. Therefore, the amplitude and phase of the output will vary only if those of the transfer function vary as the frequency of the sinusoidal source is changed. To further simplify this first look at frequency-selective circuits, we will also restrict our attention to cases where both the input and output signals are sinusoidal voltages, as illustrated in Fig. 14.2. Thus, the transfer function of interest to us will be the ratio of the Laplace transform of the output volt- age to the Laplace transform of the input voltage, or H(s) — V 0 (s)/Vi(s). We should keep in mind, however, that for a particular application, a current may be either the input signal or output signal of interest. The signals passed from the input to the output fall within a band of frequencies called the passband. Input voltages outside this band have their magnitudes attenuated by the circuit and are thus effectively pre- vented from reaching the output terminals of the circuit. Frequencies not in a circuit's passband are in its stopband. Frequency-selective circuits are categorized by the location of the passband. One way of identifying the type of frequency-selective circuit is to examine a frequency response plot. A frequency response plot shows how a circuit's transfer function (both amplitude and phase) changes as the source frequency changes. A frequency response plot has two parts. One is a graph of \H(jai)\ versus frequency w. This part of the plot is called the magnitude plot. The other part is a graph of d(Jw) versus frequency w. This part is called the phase angle plot. The ideal frequency response plots for the four major categories of fil- ters are shown in Fig. 14.3. Parts (a) and (b) illustrate the ideal plots for a low-pass and a high-pass filter, respectively. Both filters have one pass- band and one stopband, which are defined by the cutoff frequency that separates them. The names low pass and high pass are derived from the magnitude plots: a low-pass filter passes signals at frequencies lower than the cutoff frequency from the input to the output, and a high-pass filter passes signals at frequencies higher than the cutoff frequency. Thus the terms low and high as used here do not refer to any absolute values of fre- quency, but rather to relative values with respect to the cutoff frequency. Note from the graphs for both these filters (as well as those for the bandpass and bandreject filters) that the phase angle plot for an ideal filter varies linearly in the passband. It is of no interest outside the passband because there the magnitude is zero. Linear phase variation is necessary to avoid phase distortion. \H(ja>)\ I em 0° 9{ju c ) - Passband 1//(/0,)1 1 Stopband d(jw c ) 0° - Stopband Passband (a) (b) d(ja>) 0° 0(/w c2 ) h)\ Stopband Pass- band Stopband 0) cl (O c2 0) \ - \ m i 0° 0(M:l) M Passband Stop band Passband <u c ] a> C 2 (o (c) (d) Figure 14.3 • Ideal frequency response plots of the four types of filter circuits, (a) An ideal low-pass filter, (b) An ideal high-pass filter, (c) An ideal bandpass filter, (d) An ideal bandreject filter. The two remaining categories of filters each have two cutoff frequen- cies. Figure 14.3(c) illustrates the ideal frequency response plot of a bandpass filter, which passes a source voltage to the output only when the source frequency is within the band defined by the two cutoff frequencies. Figure 14.3(d) shows the ideal plot of a bandreject filter, which passes a source voltage to the output only when the source frequency is outside the band defined by the two cutoff frequencies. The bandreject filter thus rejects, or stops, the source voltage from reaching the output when its fre- quency is within the band defined by the cutoff frequencies. In specifying a realizable filter using any of the circuits from this chap- ter, it is important to note that the magnitude and phase angle characteris- tics are not independent. In other words, the characteristics of a circuit that result in a particular magnitude plot will also dictate the form of the phase angle plot and vice versa. For example, once we select a desired form for the magnitude response of a circuit, the phase angle response is also determined. Alternatively, if we select a desired form for the phase angle response, the magnitude response is also determined. Although there are some frequency-selective circuits for which the magnitude and phase angle behavior can be independently specified, these circuits are not presented here. The next sections present examples of circuits from each of the four filter categories. They are a few of the many circuits that act as filters. You should focus your attention on trying to identify what properties of a cir- cuit determine its behavior as a filter. Look closely at the form of the transfer function for circuits that perform the same filtering functions. Identifying the form of a filter's transfer function will ultimately help you in designing filtering circuits for particular applications. All of the filters we will consider in this chapter are passive filters, so called because their filtering capabilities depend only on the passive . frequencies. Such circuits are called frequency-selective circuits. Many devices that communicate via electric signals, such as telephones, radios, televisions, and satellites, employ frequency-selective. frequency. These frequency-dependent circuits are called filters and are used in many common electrical devices. In radios, filters are used to select one radio station's signal while. pro- duce the exact tone pairs, even if the person were trying. In the central telephone facility, electric circuits monitor the audio signal, listening for the tone pairs that signal a number.