Electric Circuits 7th Edition - Solution Manual

Circuit Variables Assessment Problems AP 1.1 To solve this problem we use a product of ratios to change units from dollars/year to dollars/millisecond We begin by expressing $10 billion in scientific notation: $100 billion = $100 × 109 Now we determine the number of milliseconds in one year, again using a product of ratios: year hour sec year day · · · = · 31.5576 × 109 ms 365.25 days 24 hours 60 mins 60 secs 1000 ms Now we can convert from dollars/year to dollars/millisecond, again with a product of ratios: year 100 $100 × 109 · = = $3.17/ms year 31.5576 × 109 ms 31.5576 AP 1.2 First, we recognize that ns = 10−9 s The question then asks how far a signal will travel in 10−9 s if it is traveling at 80% of the speed of light Remember that the speed of light c = × 108 m/s Therefore, 80% of c is (0.8)(3 × 108 ) = 2.4 × 108 m/s Now, we use a product of ratios to convert from meters/second to inches/nanosecond: 100 cm in (2.4 × 108 )(100) 9.45 in 2.4 × 108 m 1s · · = = · 1s 10 ns 1m 2.54 cm (109 )(2.54) ns Thus, a signal traveling at 80% of the speed of light will travel 9.45 in a nanosecond 1–1 1–2 AP 1.3 CHAPTER Circuit Variables Remember from Eq (1.2), current is the time rate of change of charge, or i = dq In dt this problem, we are given the current and asked to find the total charge To this, we must integrate Eq (1.2) to find an expression for charge in terms of current: t q(t) = i(x) dx We are given the expression for current, i, which can be substituted into the above expression To find the total charge, we let t → ∞ in the integral Thus we have ∞ qtotal = = AP 1.4 20e−5000x dx = 20 −5000x e −5000 ∞ = 20 (e∞ − e0 ) −5000 20 20 (0 − 1) = = 0.004 C = 4000 µC −5000 5000 Recall from Eq (1.2) that current is the time rate of change of charge, or i = dq In dt this problem we are given an expression for the charge, and asked to find the maximum current First we will find an expression for the current using Eq (1.2): i= = dq d 1 t = + e−αt − dt dt α α α d d t −αt d −αt e − − e dt α2 dt α dt α2 −αt t − α e−αt − −α e−αt e α α α = 0− = − 1 −αt e +t+ α α = te−αt Now that we have an expression for the current, we can find the maximum value of the current by setting the first derivative of the current to zero and solving for t: d di = (te−αt ) = e−αt + t(−α)eαt = (1 − αt)e−αt = dt dt Since e−αt never equals for a finite value of t, the expression equals only when (1 − αt) = Thus, t = 1/α will cause the current to be maximum For this value of t, the current is i= −α/α = e−1 e α α Remember in the problem statement, α = 0.03679 Using this value for α, i= e−1 ∼ = 10 A 0.03679 Problems AP 1.5 1–3 Start by drawing a picture of the circuit described in the problem statement: Also sketch the four figures from Fig 1.6: [a] Now we have to match the voltage and current shown in the first figure with the polarities shown in Fig 1.6 Remember that 4A of current entering Terminal is the same as 4A of current leaving Terminal We get (a) v = −20 V, (c) v = 20 V, i = −4 A; i = −4 A; (b) v = −20 V, (d) v = 20 V, i = 4A i = 4A [b] Using the reference system in Fig 1.6(a) and the passive sign convention, p = vi = (−20)(−4) = 80 W Since the power is greater than 0, the box is absorbing power [c] From the calculation in part (b), the box is absorbing 80 W AP 1.6 Applying the passive sign convention to the power equation using the voltage and current polarities shown in Fig 1.5, p = vi From Eq (1.3), we know that power is the time rate of change of energy, or p = dw If we know the power, we can find the dt energy by integrating Eq (1.3) To begin, find the expression for power: p = vi = (10,000e−5000t )(20e−5000t ) = 200,000e−10,000t = × 105 e−10,000t W Now find the expression for energy by integrating Eq (1.3): w(t) = t p(x) dx 1–4 CHAPTER Circuit Variables Substitute the expression for power, p, above Note that to find the total energy, we let t → ∞ in the integral Thus we have w= = AP 1.7 ∞ −10,000x × 10 e × 105 −10,000x e dx = −10,000 ∞ × 105 × 10 × 10 (e−∞ − e0 ) = (0 − 1) = = 20 J −10,000 −10,000 10,000 5 At the Oregon end of the line the current is leaving the upper terminal, and thus entering the lower terminal where the polarity marking of the voltage is negative Thus, using the passive sign convention, p = −vi Substituting the values of voltage and current given in the figure, p = −(800 × 103 )(1.8 × 103 ) = −1440 × 106 = −1440 MW Thus, because the power associated with the Oregon end of the line is negative, power is being generated at the Oregon end of the line and transmitted by the line to be delivered to the California end of the line Chapter Problems P 1.1 To begin, we calculate the number of pixels that make up the display: npixels = (1280)(1024) = 1,310,720 pixels Each pixel requires 24 bits of information Since bits comprise a byte, each pixel requires bytes of information We can calculate the number of bytes of information required for the display by multiplying the number of pixels in the display by bytes per pixel: nbytes = 1,310,720 pixels bytes · = 3,932,160 bytes/display display pixel Finally, we use the fact that there are 106 bytes per MB: 3,932,160 bytes MB = 3.93 MB/display · display 10 bytes Problems P 1.2 c = × 108 m/s so c = 1.5 × 108 m/s 1.5 × 108 m × 106 m = 1s xs P 1.3 1–5 so x= × 106 = 33.3 ms 1.5 × 108 We can set up a ratio to determine how long it takes the bamboo to grow 10 µm First, recall that mm = 103 µm Let’s also express the rate of growth of bamboo using the units mm/s instead of mm/day Use a product of ratios to perform this conversion: day hour 250 10 250 mm · · · = = mm/s day 24 hours 60 60 sec (24)(60)(60) 3456 Use a ratio to determine the time it takes for the bamboo to grow 10 µm: 10 × 10−6 m 10/3456 × 10−3 m = 1s xs P 1.4 so x= 10 × 10−6 = 3.456 s 10/3456 × 10−3 Volume = area × thickness 106 = (10 × 106 )(thickness) ⇒ thickness = P 1.5 106 = 0.10 mm 10 × 106 300 × 109 dollars 100 pennies year day hr 1.5 mm 1m · · · · · · year dollar 365.25 days 24 hr 3600 s penny 1000 mm = 1426 m/s P 1.6 Our approach is as follows: To determine the area of a bit on a track, we need to know the height and width of the space needed to store the bit The height of the space used to store the bit can be determined from the width of each track on the disk The width of the space used to store the bit can be determined by calculating the number of bits per track, calculating the circumference of the inner track, and dividing the number of bits per track by the circumference of the track The calculations are shown below Width of track = in 25,400µm = 329.87µm/track 77 tracks in Bits on a track = 1.4 MB bits side = 72,727.273 bits/track sides byte 77 tracks Circumference of inner track = 2π(1/2 )(25,400µm/in) = 79,796.453µm Width of bit on inner track = 79,796.453µm = 1.0972µm/bit 72,727.273 bits Area of bit on inner track = (1.0972)(329.87) = 361.934µm2 1–6 P 1.7 CHAPTER Circuit Variables C/m3 = 1.6022 × 10−19 × 1029 = 1.6022 × 1010 C/m3 C/m = (1.6022 × 1010 )(5.4 × 10−4 ) = 8.652 × 106 C/m Therefore, (8.652 × 106 ) m C =i × ave vel m s Thus, average velocity = P 1.8 1400 × 10−6 = 161.81 µm/s 8.652 20 × 10−6 C/s = 1.25 × 1014 elec/s 1.6022 × 10−19 C/elec 5280 ft 12 in 2.54 cm 104 µm · · · = 5.32 × 1012 µm [b] m = 3303 mi · mi ft in cm n = 23.5 Therefore, m [a] n = The number of electrons/second is approximately 23.5 times the number of micrometers between Sydney and San Francisco P 1.9 First we use Eq (1.2) to relate current and charge: i= dq = 20 cos 5000t dt Therefore, dq = 20 cos 5000t dt To find the charge, we can integrate both sides of the last equation Note that we substitute x for q on the left side of the integral, and y for t on the right side of the integral: q(t) q(0) dx = 20 t cos 5000y dy We solve the integral and make the substitutions for the limits of the integral, remembering that sin = 0: sin 5000y q(t) − q(0) = 20 5000 t = 20 20 20 sin 5000t − sin 5000(0) = sin 5000t 5000 5000 5000 But q(0) = by hypothesis, i.e., the current passes through its maximum value at t = 0, so q(t) = × 10−3 sin 5000t C = sin 5000t mC P 1.10 w = qV = (1.6022 × 10−19 )(9) = 14.42 × 10−19 = 1.442 aJ Problems P 1.11 p = (6)(100 × 10−3 ) = 0.6 W; w(t) = P 1.12 t p dt w(10,800) = 10,800 3600 s = 10,800 s hr 0.6 dt = 0.6(10,800) = 6480 J Assume we are standing at box A looking toward box B Then, using the passive sign convention p = vi, since the current i is flowing into the + terminal of the voltage v Now we just substitute the values for v and i into the equation for power Remember that if the power is positive, B is absorbing power, so the power must be flowing from A to B If the power is negative, B is generating power so the power must be flowing from B to A [a] p = (20)(15) = 300 W [b] p = (100)(−5) = −500 W [c] p = (−50)(4) = −200 W [d] p = (−25)(−16) = 400 W P 1.13 hr · 1–7 300 W from A to B 500 W from B to A 200 W from B to A 400 W from A to B [a] p = vi = (−20)(5) = −100 W Power is being delivered by the box [b] Leaving [c] Gaining P 1.14 [a] p = vi = (−20)(−5) = 100 W, so power is being absorbed by the box [b] Entering [c] Losing P 1.15 [a] In Car A, the current i is in the direction of the voltage drop across the 12 V battery(the current i flows into the + terminal of the battery of Car A) Therefore using the passive sign convention, p = vi = (−40)(12) = −480 W Since the power is negative, the battery in Car A is generating power, so Car B must have the ”dead” battery 1–8 CHAPTER Circuit Variables [b] w(t) = t w(90) = 1.5 = 1.5 · p dx; 90 60 s = 90 s 480 dx w = 480(90 − 0) = 480(90) = 43,200 J = 43.2 kJ P 1.16 p = vi; w= t p dx Since the energy is the area under the power vs time plot, let us plot p vs t Note that in constructing the plot above, we used the fact that 60 hr = 216,000 s = 216 ks p(0) = (6)(15 × 10−3 ) = 90 × 10−3 W p(216 ks) = (4)(15 × 10−3 ) = 60 × 10−3 W w = (60 × 10−3 )(216 × 103 ) + (90 × 10−3 − 60 × 10−3 )(216 × 103 ) = 16.2 kJ P 1.17 [a] To find the power at 625 µs, we substitute this value of time into both the equations for v(t) and i(t) and multiply the resulting numbers to get p(625 µs): v(625 µs) = 50e−1600(625×10 −6 ) − 50e−400(625×10 i(625 µs) = × 10−3 e−1600(625×10 −6 ) −6 ) = 18.394 − 38.94 = −20.546 V − × 10−3 e−400(625×10 −6 ) = 0.0018394 − 0.003894 = −0.0020546 A p(625 µs) = (−20.546)(−0.0020546) = 42.2 mW [b] To find the energy at 625 µs, we need to integrate the equation for p(t) from to 625 µs To start, we need an expression for p(t): p(t) = v(t)i(t) = (50)(5 × 10−3 )(e−1600t − e−400t )(e−1600t − e−400t ) Problems 1–9 = (e−3200t − 2e−2000t + e−800t ) Now we integrate this expression for p(t) to get an expression for w(t) Note we substitute x for t on the right side of the integral w(t) = t (e−3200x − 2e−2000x + e−800x )dx e−3200x e−2000x e−800x + − = −3200 1000 800 t 1 1 e−3200t e−2000t e−800t + − − + − = −3200 1000 800 −3200 1000 800 = e−3200t e−2000t e−800t + − + 5.625 × 10−4 −3200 1000 800 Finally, substitute t = 625 µs into the equation for w(t): w(625 µs) = [−4.2292 × 10−5 + 2.865 × 10−4 − 7.5816 × 10−4 + 5.625 × 10−4 ] = 12.137 µJ [c] To find the total energy, we let t → ∞ in the above equation for w(t) Note that this will cause all expressions of the form e−nt to go to zero, leaving only the constant term 5.625 × 10−4 Thus, wtotal = [5.625 × 10−4 ] = 140.625 µJ P 1.18 [a] v(20 ms) = 100e−1 sin = 5.19 V i(20 ms) = 20e−1 sin = 1.04 A p(20 ms) = vi = 5.39 W [b] p vi = 2000e−100t sin.2 150t 1 − cos 300t = 2000e−100t 2 = 1000e−100t − 1000e−100t cos 300t w = = = w ∞ 1000e−100t dt − e−100t 1000 −100 ∞ ∞ 1000e−100t cos 300t dt e−100t −1000 [−100 cos 300t + 300 sin 300t] (100)2 + (300)2 100 = 10 − 1000 = 10 − × 10 + × 104 = J ∞ 1–10 P 1.19 CHAPTER Circuit Variables [a] s ≤ t < s: v = V; i = 20t A; p = 100t W i = 20 A; p=0W i = 20 A; p=0W i = 80 − 20t A; p = −400 + 100t W i = 80 − 20t A; p = −400 + 100t W i = −120 + 20t A; p = −600 + 100t W i = −120 + 20t A; p = −600 + 100t W i = 20 A; p=0W s < t ≤ s: v = V; s ≤ t < s: v = V; s < t ≤ s: v = −5 V; s ≤ t < s: v = −5 V; s < t ≤ s: v = V; s ≤ t < s: v = V; t > s: v = V; [b] Calculate the area under the curve from zero up to the desired time: P 1.20 w(1) = (1)(100) = 50 J w(6) = (1)(100) − 12 (1)(100) + 12 (1)(100) − 12 (1)(100) = J w(10) = (1)(100) − 12 (1)(100) + 12 (1)(100) − 12 (1)(100) + 12 (1)(100) = 50 J [a] p = vi = (100e−500t )(0.02 − 0.02e−500t ) = (2e−500t − 2e−1000t ) W dp = −1000e−500t + 2000e−1000t = dt = e500t so ln = 500t so thus 2e−1000t = e−500t p is maximum at t = 1.4 ms 17–26 CHAPTER 17 The Fourier Transform [b] In the phasor domain: Vo − 125 Vo Vo + + =0 j200 j800 120 12Vo − 1500 + 3Vo + j20Vo = Vo = Io = 1500 = 60/− 53.13◦ V 15 + j20 Vo = 75 × 10−3 /− 143.13◦ A j800 io (t) = 75 cos(40,000t − 143.13◦ ) mA P 17.30 [a] Vg s Vg s2 Vo = = 25 + (100/s) + s s + 25s + 100 H(s) = Vo s2 = ; Vg (s + 5)(s + 20) H(ω) = (jω)2 (jω + 5)(jω + 20) vg = 25ig = −450e10t u(−t) − 450e−10t u(t) V Vg = − 450 450 − −jω + 10 jω + 10 Vo (ω) = H(ω)Vg = −450(jω)2 (−jω + 10)(jω + 5)(jω + 20) + −450(jω)2 (jω + 10)(jω + 5)(jω + 20) = K2 K3 K4 K5 K6 K1 + + + + + −jω + 10 jω + jω + 20 jω + jω + 10 jω + 20 Problems K1 = 450(100) = −100 (15)(30) K2 = 450(25) = −50 (15)(15) K3 = 450(400) = 400 (30)(−15) Vo (ω) = K4 = K5 = 17–27 −450(25) = −150 (5)(15) −450(100) = 900 (−5)(10) K6 = −450(400) = −1200 (−15)(−10) −200 −800 900 −100 + + + −jω + 10 jω + jω + 20 jω + 10 vo = −100e10t u(−t) + [900e−10t − 200e−5t − 800e−20t ]u(t) V [b] vo (0− ) = −100 V [c] vo (0+ ) = 900 − 200 − 800 = −100 V [d] At t = 0− the circuit is Therefore, the solution predicts v1 (0− ) will be −350 V Now v1 (0+ ) = v1 (0− ) because the inductor will not let the current in the 25 Ω resistor change instantaneously, and the capacitor will not let the voltage across the 0.01 F capacitor change instantaneously At t = 0+ the circuit is From the circuit at t = 0+ we see that vo must be −100 V, which is consistent with the solution for vo obtained in part (c) CHAPTER 17 The Fourier Transform 17–28 P 17.31 Vo s Vo − Vg 100Vo + =0 + 25 s 100s + 125 × 104 · Vo = Io = s(100s + 125 × 104 )Vg 125(s2 + 12,000s + 25 × 106 ) sVo 100s + 125 × 104 Io s2 H(s) = = Vg 125(s2 + 12,000s + 25 × 106 ) −8 × 10−3 ω H(ω) = (25 × 106 − ω ) + j12,000ω Vg (ω) = 300π[δ(ω + 5000) + δ(ω − 5000)] Io (ω) = H(ω)Vg (ω) = io (t) = −2.4π 2π = −1.2 = 12 −2.4πω [δ(ω + 5000) + δ(ω − 5000)] (25 × 106 − ω ) + j12,000ω ω [δ(ω + 5000) + δ(ω − 5000)] jtω e dω −∞ (25 × 106 − ω ) + j12,000ω ∞ 25 × 106 ej5000t 25 × 106 e−j5000t + −j(12,000)(5000) j(12,000)(5000) e−j5000t ej5000t + −j j ◦ ◦ = 0.5[e−j(5000t+90 ) + ej(5000t+90 ) ] io (t) = cos(5000t + 90◦ ) A Problems 17–29 P 17.32 [a] From the plot of vg note that vg is −10 V for an infinitely long time before t = Therefore · vo (0− ) = −10 V There cannot be an instantaneous change in the voltage across a capacitor, so vo (0+ ) = −10 V [b] io (0− ) = A At t = 0+ the circuit is 30 − (−10) 40 = = 8A 5 [c] The s-domain circuit is io (0+ ) = Vo = Vg + (10/s) Vo = H(s) = Vg s+2 10 2Vg = s s+2 17–30 CHAPTER 17 The Fourier Transform H(ω) = jω + Vg (ω) = jω − 5[2πδ(ω)] + Vo (ω) = H(ω)Vg (ω) = 10 30 30 = − 10πδ(ω) + jω + jω jω + 10 30 − 10πδ(ω) + jω + jω jω + = 20πδ(ω) 60 20 − + jω(jω + 2) jω + (jω + 2)(jω + 5) = K0 K1 K2 K3 20πδ(ω) + + + − jω jω + jω + jω + jω + K0 = 20 = 10; Vo (ω) = K1 = 20 = −10; −2 K2 = 60 = 20; K3 = 60 = −20 −3 10 20 20πδ(ω) 10 10 20 10 + − − = + + − 10πδ(ω) jω jω + jω + jω + jω jω + jω + vo (t) = 5sgn(t) + [10e−2t − 20e−5t ]u(t) − V P 17.33 [a] Vo (Vo − Vg )s Vo + =0 + 10 4s 800 · Vo = s2 Vg s2 + 1250s + 25 × 104 Vo s2 = H(s) = Vg (s + 250)(s + 1000) H(ω) = (jω)2 (jω + 250)(jω + 1000) vg = 45e−500|t| ; Vg (ω) = · Vo (ω) = H(ω)Vg (ω) = = 45,000 (jω + 500)(−jω + 500) 45,000(jω)2 (jω + 250)(jω + 500)(jω + 1000)(−jω + 500) K2 K3 K4 K1 + + + jω + 250 jω + 500 jω + 1000 −jω + 500 Problems K1 = 45,000(−250)2 = 20 (250)(750)(750) K2 = 45,000(−500)2 = −90 (−250)(500)(1000) K3 = 45,000(−1000)2 = 80 (−750)(−500)(1500) K4 = 45,000(500)2 = 10 (750)(1000)(1500) · vo (t) = [20e−250t − 90e−500t + 80e−1000t ]u(t) + 10e500t u(−t) V [b] vo (0− ) = 10 V; Vo (0+ ) = 20 − 90 + 80 = 10 V vo (∞) = V [c] IL = 0.25sVg Vo = 4s (s + 250)(s + 1000) H(s) = 0.25s IL = Vg (s + 250)(s + 1000) H(ω) = 0.25(jω) (jω + 250)(jω + 1000) IL (ω) = 0.25(jω)(45,000) (jω + 250)(jω + 500)(jω + 1000)(−jω + 500) = K4 = K1 K2 K3 K4 + + + jω + 250 jω + 500 jω + 1000 −jω + 500 (0.25)(500)(45,000) = mA (750)(1000)(1500) iL (t) = 5e500t u(−t); · iL (0− ) = mA K1 = (0.25)(−250)(45,000) = −20 mA (250)(750)(750) K2 = (0.25)(−500)(45,000) = 45 mA (−250)(500)(1000) K3 = (0.25)(−1000)(45,000) = −20 mA (−750)(−500)(1500) · iL (0+ ) = K1 + K2 + K3 = −20 + 45 − 20 = mA Checks, i.e., iL (0+ ) = iL (0− ) = mA 17–31 17–32 CHAPTER 17 The Fourier Transform At t = 0− : vC (0− ) = 45 − 10 = 35 V At t = 0+ : vC (0+ ) = 45 − 10 = 35 V [d] We can check the correctness of our solution for t ≥ 0+ by using the Laplace transform Our circuit becomes Vo (Vo − Vg )s Vo × 10−3 −6 + =0 + + 35 × 10 + 106 s 800 4s · (s2 + 1250s + 25 × 104 )Vo = s2 Vg − (35s + 5000) vg (t) = 45e−500t u(t) V; Vg = · (s + 250)(s + 1000)Vo = 45 s + 500 45s2 − (35s + 5000)(s + 500) (s + 500) 10s2 − 22,500s − 250 × 104 · Vo = (s + 250)(s + 500)(s + 1000) = 90 80 20 − + s + 250 s + 500 s + 1000 · vo (t) = [20e−250t − 90e−500t + 80e−1000t ]u(t) V This agrees with our solution for vo (t) for t ≥ 0+ P 17.34 [a] Vg (ω) = 36 36 72jω − = − jω + jω (4 − jω)(4 + jω) Problems Vo (s) = (16/s) Vg (s) 10 + s + (16/s) H(s) = 16 16 Vo (s) = = Vg (s) s + 10s + 16 (s + 2)(s + 8) H(ω) = 16 (jω + 2)(jω + 8) Vo (ω) = H(ω) · Vg (ω) = = 17–33 1152jω (4 − jω)(4 + jω)(2 + jω)(8 + jω) K2 K3 K4 K1 + + + − jω + jω + jω + jω K1 = 1152(4) =8 (8)(6)(12) K2 = 1152(−4) = 72 (8)(−2)(4) K3 = 1152(−2) = −32 (6)(2)(6) K4 = 1152(−8) = −32 (12)(−4)(−6) · Vo (jω) = 72 32 32 + − − − jω + jω + jω + jω · vo (t) = 8e4t u(−t) + [72e−4t − 32e−2t − 32e−8t ]u(t)V [b] vo (0− ) = V [c] vo (0+ ) = 72 − 32 − 32 = V The voltages at 0− and 0+ must be the same since the voltage cannot change instantaneously across a capacitor P 17.35 Vo (s) = 30 40 600(s + 10) 10 + − = s s + 20 s + 30 s(s + 20)(s + 30) Vo (s) = H(s) · 15 s · H(s) = 40(s + 10) (s + 20)(s + 30) · H(ω) = 40(jω + 10) (jω + 20)(jω + 30) 17–34 CHAPTER 17 The Fourier Transform · Vo (ω) = vo (ω) = 40(jω + 10) 1200(jω + 10) 30 · = jω (jω + 20)(jω + 30) jω(jω + 20)(jω + 30) 60 80 20 + − jω jω + 20 jω + 30 vo (t) = 10sgn(t) + [60e−20t − 80e−30t ]u(t) V P 17.36 [a] f (t) = [b] W = [c] W = [d] π π 2π −∞ eω ejtω dω + ∞ (1/π) dt = 2 (1 + t ) π ∞ −2ω 0 ∞ ∞ e−2ω dω = 1e π −2 e−ω ejtω dω = ∞ 1/π + t2 dt J = 2 (1 + t ) 2π = J 2π 0.9 , − e−2ω1 = 0.9, e−2ω dω = 2π ω1 = (1/2) ln 10 ∼ = 1.15 rad/s ω1 e2ω1 = 10 P 17.37 Io = Ig R RCsIg = R + (1/sC) RCs + H(s) = s Io = Ig s + (1/RC) RC = (100 × 103 )(1.25 × 10−6 ) = 125 × 10−3 ; H(s) = s ; s+8 Ig (ω) = 30 × 10−6 jω + H(ω) = Io (ω) = H(ω)Ig (ω) = jω jω + 30 × 10−6 jω (jω + 2)(jω + 8) 1 = =8 RC 0.125 Problems 17–35 ω(30 × 10−6 ) √ |Io (ω)| = √ ( ω + 4)( ω + 64) |Io (ω)|2 = K1 K2 900 × 10−12 ω = + 2 (ω + 4)(ω + 64) ω + ω + 64 K1 = (900 × 10−12 )(−4) = −60 × 10−12 (60) K2 = (900 × 10−12 )(−64) = 960 × 10−12 (−60) |Io (ω)|2 = W1Ω = = = π 960 × 10−12 60 × 10−12 − ω + 64 ω2 + ∞ |Io (ω)|2 dω = 120 × 10−12 ω tan−1 π 960 × 10−12 π ∞ − ∞ 60 × 10−12 dω − ω + 64 π ω 30 × 10−12 tan−1 π %= 120 π 30 π × 10−12 = (60 − 15) × 10−12 = 45 pJ · − · π π Between and rad/s W1Ω = ∞ 120 30 tan−1 − tan−1 × 10−12 = 7.14 pJ π π 7.14 (100) = 15.86% 45 P 17.38 [a] Vg (ω) = 60 (jω + 1)(−jω + 1) 0.4 (jω + 0.5) H(s) = Vo 0.4 ; = Vg s + 0.5 Vo (ω) = 24 (jω + 1)(jω + 0.5)(−jω + 1) Vo (ω) = 32 −24 + + jω + jω + 0.5 −jω + H(ω) = vo (t) = [−24e−t + 32e−t/2 ]u(t) + 8et u(−t) V ∞ dω +4 ω2 17–36 CHAPTER 17 The Fourier Transform [b] |Vg (ω)| = [c] |Vo (ω)| = [d] Wi = [e] Wo = 60 + 1) (ω 24 √ (ω + 1) ω + 0.25 ∞ 0 −∞ = 32 + −2t 900e e−2t dt = 1800 −2 64e2t dt + ∞ −2t [576e ∞ ∞ = 900 J (−24e−t + 32e−t/2 )2 dt − 1536e−3t/2 + 1024e−t ] dt = 32 + 288 − 1024 + 1024 = 320 J Problems [f] |Vg (ω)| = 60 , +1 ω2 |Vg2 (ω)| = 3600 (ω + 1)2 Wg = 3600 π = 3600 π ω + tan−1 ω 2 ω +1 = 1800 + tan−1 = 863.53 J π = Wo = + 1)2 (ω + 0.25) 768 1024 1024 − − ω + 0.25 (ω + 1)2 (ω + 1) 1024 · · tan−1 2ω π −768 P 17.39 Io = ω + tan−1 ω ω +1 2048 384 1024 tan−1 − + tan−1 − tan−1 π π π 319.2 × 100 = 99.75% 320 sIg 0.5sIg = 0.5s + 25 s + 50 H(s) = s Io = Ig s + 50 H(ω) = jω jω + 50 I(ω) = 2 = 319.2 J %= 576 (ω −1024 tan−1 ω = 863.53 × 100 = 95.95% 900 · % = [g] |Vo (ω)|2 = dω (ω + 1)2 12 jω + 10 Io (ω) = H(ω)I(ω) = 12(jω) (jω + 10)(jω + 50) 17–37 17–38 CHAPTER 17 The Fourier Transform 12ω |Io (ω)| = (ω + 100)(ω + 2500) |Io (ω)|2 = = 144ω (ω + 100)(ω + 2500) ω2 Wo (total) = = 150 −6 + + 100 ω + 2500 π 150dω − + 2500 π ∞ ω2 ω tan−1 π 50 ∞ − ∞ ω2 6dω + 100 ω 0.6 tan−1 π 10 ∞ = 1.5 − 0.3 = 1.2 J Wo (0–100 rad/s) = 0.6 tan−1 (2) − tan−1 (10) π π = 1.06 − 0.28 = 0.78 J Therefore, the percent between and 100 rad/s is 0.78 (100) = 64.69% 1.2 P 17.40 [a] |Vi (ω)|2 = × 104 ; ω2 |Vi (100)|2 = × 104 = 4; 1002 |Vi (200)|2 = × 104 =1 2002 Problems [b] Vo = RCVi Vi R = R + (1/sC) RCs + H(s) = s Vo ; = Vi s + (1/RC) H(ω) = jω jω + 100 |Vo (ω)| = × 104 , ω + 104 |Vo (100)|2 = [c] W1Ω = π = 100 ≤ ω ≤ 200 rad/s; × 104 = 2; 104 + 104 200 100 |Vo (200)|2 = × 104 × 104 dω = − ω2 π ω |Vo (ω)|2 = 0, × 104 = 0.8 × 104 200 100 × 104 200 ∼ − = = 63.66 J π 100 200 π = π = 106 10−3 1000 = = = 100 RC (0.5)(20) 10 |ω| 200 200 √ ·√ = |ω| ω + 104 ω + 104 |Vo (ω)|2 = [d] W1Ω 17–39 200 100 × 104 ω × 104 · tan−1 dω = ω + 10 π 100 400 [tan−1 − tan−1 1] ∼ = 40.97 J π 200 100 elsewhere 17–40 CHAPTER 17 The Fourier Transform P 17.41 [a] Vi (ω) = H(s) = A ; a + jω |Vi (ω)| = √ s ; s+α H(ω) = WIN = ∞ = A2 π A2 ; 2a when α = a we have ω dω A2 = π (ω + a2 )2 a dω − a + ω2 a2 dω (a2 + ω )2 a π −1 ∞ A2 π ω2 A2 dω = (a2 + ω )2 4a WOUT (a) = 0.5 − = 0.1817 or WOUT (total) π Therefore ω α2 + ω ω A2 (a2 + ω )(α2 + ω ) a A2 4aπ WOUT (total) = |H(ω)| = √ (a2 + ω )(α2 + ω ) A2 e−2at dt = WOUT (a) = jω ; α + jω ωA Therefore |Vo (ω)| = Therefore |Vo (ω)|2 = A + ω2 a2 18.17% [b] When α = a we have WOUT (α) = = π ω A2 dω (a2 + ω )(α2 + ω ) α A2 π where K1 = α a2 K1 K2 + dω +ω α + ω2 a a2 − α2 and K2 = −α2 a2 − α2 Therefore WOUT (α) = α A2 απ a tan−1 − 2 π(a − α ) a WOUT (total) = Therefore π A2 π A2 − α a = π(a2 − α2 ) 2 2(a + α) α απ WOUT (α) · a tan−1 − = WOUT (total) π(a − α) a √ For α = a 3, this ratio is 0.2723, √ or 27.23% of the output energy lies in the frequency band between and a √ [c] For α = a/ 3, the ratio is 0.1057, √ or 10.57% of the output energy lies in the frequency band between and a/ ... 0.3t2 + 0.005t3 W = − 0.6t + 0.015t2 = 0 < t < 40 s Problems 1–13 Use a calculator to find the two solutions to this quadratic equation: t1 = 8.453 s; t2 = 31.547 s Now we must find which of these... power delivered to the circuit and the power absorbed by the circuit is 96,975 − 60,975 = 36,000 One-half of this difference is 18,000W, so it is likely that pc or pf is in error Either the voltage... Elements The results are shown in the figure below: [a] To find i1 , write a KVL equation around the left-hand loop, summing voltages in a clockwise direction starting below the 5V source: −5 V + 54,000i1