Baixe sadiku 5ª edição c...eletricos e extras - chapter 14 frequency response e outras Notas de estudo em PDF para Engenharia Elétrica, somente na Docsity! CHAPTER 14 - FREQUENCY RESPONSE List of topics for this chapter : Transfer Function Bode Plots Series Resonance Parallel Resonance Passive Filters Active Filters Scaling +—- >> 4 TRANSFER FUNCTION Problem 14.1 ' Given the circuit in Figure 14.1 and i(t) = Icos(wt) amps, find the transfer function H(0) = V,/ 1 and sketch the frequency response. no MA 0 + it) D) E 100 vt) 8 20mH Figure 14.1 » Carefully DEFINE the problem. Each component is labeled completely. The problem is clear. > PRESENT everything you know about the problem. To obtain the transfer function, H(), we need to obtain the frequency-domain equivalent of the circuit by replacing resistors, inductors, and capacitors with their impedances R, joL, and 1/jwC respectively. Then, use any circuit technique to obtain H(c). The frequency response of the circuit can be obtained by ploiting the magnitude and phase of the transfer function as the frequency varies. > Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success. After transforming the circuit from the time domain to the frequency domain, we can use nodal analysis, mesh analysis, or basic circuit analysis to find the transfer function. Let's examine the frequency-domain equivalent circuit in order to make the best choice. 243 Transforming the circuit to the frequency domain yields v; 109 v. AM I O g 10 s jo(0.02) O Tt seems obvious that nodal analysis will yield two equations with three unknowns (1, Vi, and V,). These equations can be manipulated to eliminate V, in order to find H(w). ATTEMPT a problem solution. Using nodal analysis, MD MV Atnode 1: -1+ io +40 —º Atnade? WoW N,-0 tmode 2: IO * jo(0.02) Using the equation for node 1, find V, in terms of V, and I. 101=2V,-V, e V,+101 V, = 7 +51 Simplify the equation for node 2. jo(V, — V)+500V, = 0 (o +500)V,-joV, =0 Now, substitute the equation for V, into the simplified equation for node 2. (Jo + 500) V, — jo(0.5V, +5D=0 (jo/2+500)V, = jo51 V jus joio o 1 j0/2+500 1000+jo Therefore, joio H(g)= 007 (9) = T3oo + jo EVALUATE the solution and check for accuracy. A check of our solution can be done using basic circuit analysis. Find the output voltage as the current through the inductor multiplied by the impedance of the inductor, ie., V=I Z. o 244 The frequency response is e 1 r r 09 08 Magnitude (unitless) oo eco so no E ada q o o 1et 10º 1º 10 10º Frequency (radians/second) 10º 3, Problem 14,3 Given the circuit in Figure 14.3 and i(t) = 1, cos(wt) amps, find the transfer function H().= Vyy/ Im and sketch the frequency response. + e 1 s 10k0 10 mH a) LuF Figure 14.3 The frequency response is 10000 9000 8000 q 7000 1 6000 5000 4000 3000 2000 1000 q o 0 10º o! 10º Frequency (radians/second) 247 for the transfer function 10º H(o) = nu Nes o] Jão” BODE PLOTS Problem 14.4 Sketch the Bode plots, both magnitude and phase, given the following transfer function in the s-domain. 90Xs + ts +10 no = CNEHDE+O) s(s+3)Xs +30) First, we need to modify the transfer function so that it is in a form that is easy to plot. 8 s (90Xs+ Ds +10) cono(ê+ (5.1) coisa To ) H(g)= s(s+3(s+30) s “oco + ds = + ) NE 3 + ks + ) Begin with a plot showing the magnitude curve of each term in the transfer function. 20 logu(t0) q 20 20 logyo(jo + 1) + 20 logro(1/0) 20 Jogra(j(/10)+1) + + 4 + + > + 20 logyo(1/Fj(09/30)+H]) 20 logyo(L/[j(0/3)+1]) 248 Now, combine, or add, the curves to acquire the composite magnitude (dB) plot of the transfer function. Note that the dashed curve shows the approximation to the actual curve. 40 + The individual phase angle plots for cach component of the transfer function are shown below. 90º + 1 «, ase À Gi(on10)+1) Í 100 300 sed i(o30+) 1 «o -90º T Finally, the composite phase angle plot for the transfer function can be drawn. 90º + As d 249 (a) The resonant frequency of a series RLC circuit is So, 1 1 1 ie 3, == ===> = 10º radi Lc Vioêxi0s 0 o, =10krad/s R RY (1). The half-power frequency of a series RLC circuítis 0,, = FE + (E) + ( . 21" VOL) CALC So, RO ( 10 ) ( 1 ) 2 ="? *yloxi02) "oz 108 0,2 = 500 44/ (500)? +10º 0, = 500+10,012 rad's or 0, = 9,512 krad/s o, = 10.512 krad/s (b) The bandwidth of a series RLC circuítis B=0, -0,. So, B=10.512-9.512 B=lkrad/s o The quality factor of a series RLC circuit is Q = Re So, 10º Q=75 Q=10 (e) The amplitudes of vc(t) at 05, O and O, B 10% Q jo1o?Q 1020º 252 Ato=0,=10 krad/s, the inductor has a value of j10º x 10? = 100 ohms and the capacitor has a value of - jL0S /10* =-j100 ohms. Then, Ve =-1001 but pe =T0+5100=4100 7! “MP So, Ve =-j100 volts and ve(t) = 100cos(10,000t — 90º) volts This gives an amplitude of 100 V ato =, = 10 krad/s. Ato =0,=9.512 krad/s, the inductor has a value of j9,512 x 0.01 = 95.12 ohms and the capacitor has a value of - jiO* [9,512 = -;105.13 ohms. Then, Ve =-]105.131 but 10 10 [= o75.12-0105.137 10 p10,01 “MPS So, Ve = (105.132 -90º%0.706845.03º) = 74.32 - 44.97º volts and Velt) = 74.3c08(9,512t — 44.97º) volts This gives an amplitude of 74.3 V ato =w, = 9.512 krad/s. Ato=0, = 10.512 krad/s, the inductor has a value of j10,512x 0.01 = j105.12 ohms and the capacitor has a value of - jl0º /10,512 = -;95.13 ohms. Then, Ve =-]95.131 but 10 10 = Tor ]0s12-]9513 7 10+)9,99 “PPS So, Ve = (95.132 -90)%0.70752 - 44.979 = 67.3/-134.97º volts and ve(t) = 67.3c08(10,5121-134.97º) volts This gives an amplitude of 67.3 V ato =, = 10.512. krad/s. 253 Note that the output voltage for this bandpass filter is the voltage across the resistor. Tt can be shown that 0 Vou (t) = 10 cos(10,000t) V ato=0a,=10 krad's Vou(t) = 7.068 cos(9,512+ 45.03) V ato=0o =9.512 krad's Vou (t) = 7.075 cos(10,512t — 44.979) V ato=0w, = 10.512 krad/s The amplitude at the half-power frequencies is Y/ «2 times the maximum amplitude at the center frequency. In this case, 1 5 19=791 where the calculated amplitudes of 7.068 volts and 7.075 volts are quite close to the expected half-power value of 7.071 volts. Problem 14.8 Given the circuit in Figure 14.5, find the value of L so that we have a Q of 100. Also, find 05, 0,, Q,, and B. 10 L AMA ——UU0O 20 cos(wt) O 75 LuF 0 Figure 14.5 L=1H o = lkrad/s 0, =995rad/s 0, = 1005 rad/s B=l0rad/s + + PARALLEL RESONANCE Problem 14.9 Given the circuit in Figure 14.6 and = 220º amps, find (a) O, Q,andB, (b) o and 0, (c) power dissipated at 0,,0,, and O. e 254 Atresonance, where Xç=THE Gero 10º o 10º Thus, o 1007 o, or 0 10º +Tog = 10 0; =10*-10º =9.9x107 O = 9.95krad/s or 9, = 10krad/s + e PASSIVE FILTERS Problem 14.11 What type of filter is represented by the circuit in Figure 14.87 What is the cutoff frequency, or what are the corner frequencies? 16K0 AM + Yin 28 M pF Vout £ 10k0Q Figure 14.8 In the frequency domain, the circuit is 10 k92 ANY + M” Vim O JM O Vou E 10KQ 257 Find the transfer function V, H() = (0) V. Using nodal analysis, Vou — Na + Yam Vem =0 10 -jl0%/o 10 Simplifying, o Vo = Vi + Tg Voa + Vau =0 o +35 Va = V, (2 18) au = Vim Hence, 1 H(o) = 7——— 0) =5Tiao This transfer function looks like a typical transfer function for a lowpass filter 1 1+ joRC 1 Since the voltage starts at Voy = 3 Yu andas wo > 0 Vi = 0, we can look ata value for 0.7071 Voa (2rom Ja =0.3535 |V, To find the cutoff frequency, find the value of q when Ho) = 0.3535. wo= 20 rad/s This lowpass filter has a cutoff frequency of wc = 20 rad/s or fo = 10/n Hz. 258 Problem 14.12 [14.43] Determine the range of frequencies that will be passed by a series RLC bandpass filter with R = 10 9, L=25 mb, and C=0.4 pF. Find the quality factor. 1 1 = ESTO keradi VLC ((25x10º)0.4x 10%) ves poB. dO 4 d/; =1 70,025 0 "0 es Thus, 10 = 4— Q=5"%4 Q=25 This is a high Q circuit so we can use 9.8 0,=0,-B/2=10-0.2=9.8 krad/s or fi= Ea 1.5597kHz 10.2 0, =0,+B/2=10+0.2=10.2krdis or f=""D=1.6234k Therefore, 1.5597 kHz < f < 1.6234 kHz Problem 14.13 What type of filter is represented by the circuit in Figure 14.9? What is the cutoff frequency, or what are the corner frequencies? 10 k92 10 uF Ir IR + Vin O Voe É 10H0 Figure 14.9 This highpass filter bas a cutoff frequency of mç; = 26.55 rad/s or fç = 4.226 Hz. 259 or Re=dos md 0 2mf, ánxio? Clearly, the capacitor becomes a short circuit at high frequencies. Hence, the high frequency gain is Re cio R=25R RCA or (=2. If we let R = 10 K€2, then R, = 25 k0, and C=oonxi0or = 7958 nF. + + SCALING Problem 14.16 [14.63] | For the circuit in Figure 14.12, (a) draw the new circuit after it has been scaled by K,, = 200 and K, = 10º. (b) obtain the Thevenin equivalent impedance at terminals a-b of the scaled circuit at w=10"rad/s. o 1H as “UOUON K fo $29 ? 051, Figure 14.12 (a) R'=K,R = (2002) = 400 9 KL (2001) L =” qo 5% mH C=D"—= "001, = 0.25 pF 262 We now have a new cireuit, 20 mH a “BUOO E T 0.25 uF E 4000 f 05L (b) Insert a 1 amp source at the terminals a-b. a Vi V2 f GUI k LHC) s R Sos b = Atnode 1: Atnode 2: V, VV, VV. V 1 i 1 2 1 240. az GOT To t05L=-2 But, 1, =sCV.. So, the nodal equations become MV MV; M =Y, A 1=sCV, + L +0.5sC V, =R Solving for V,, v- sL+R 1 S?LC+0.5SCR +1 po SER mT 1 sLC+0SSCR+I Ato=10! Zoo (10º 20x 107) + 400 á Tm (J10)(20x10?)(0.25 x 10) + 0.5CjO* (0.25 x 108 X400) +1 400 + j200 . v => 055]05 = 600 — j200 Zn = 632.52-18.43º0 263 Problem 14.17 Given the circuit in Figure 14,13, find the values necessary to scale this circuit, increasing the comer frequency to 100 rad/s. Use a 1 uF capacitor. 20. AA 2F e 29 AN No. + + Vinlt) Voult) Figure 14.13 To scale the circuit in Figure 14.13 from o = 1/4 rad/sto o! = 100 rad/s using a 1 HF capacitor, the feedback resistor and the input resistor must be 10 KO. 264