ProbSolv_Chapter14

ProbSolv_Chapter14 - CHAPTER 14 — FREQUENCY RESPONSE List...

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Unformatted text preview: CHAPTER 14 — FREQUENCY RESPONSE List of topics for this chapter : Transfer Function Bode Plots Series Resonance Parallel Resonance Passive Filters Active Filters Scaling TRANSFER FUNCTION Problem 14.1 ' Given the circuit in Figure 14.1 and i(t) = Icos(cot) amps, find the transfer function H(0)) = V() / I and sketch the frequency response. 10 Q 20 mH i(t) 0 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((o), we need to obtain the frequency-domain equivalent of the circuit by replacing resistors, inductors, and capacitors with their impedances R, jcoL, and l/jcoC respectively. Then, use any circuit technique to obtain H(03). The frequency response of the circuit can be obtained by plotting 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 V1 10 Q V0 I ‘l E jw(0.02) Q It seems obvious that nodal analysis will yield two equations with three unknowns (1, V1, and V0). These equations can be manipulated to eliminate V1 in order to find H(w). > ATTEMPT a problem solution. Using nodal analysis, AVl—O Vl—V Atnodel: -1+ 10 + 10 =0 Vo — V1 V0 — 0 + 7— : 0 10 JCO(0.02) At node 2 : Using the equation for node 1, find V1 in terms of V0 and I . 101=2V1—V0 . _Vo+101 V —l’- 51 I“ 2 ‘2+ Simplify the equation for node 2. jco(Vo — V1) + 500Vo = 0 (jw+500)Vo—jcoV1=0 Now, substitute the equation for V1 into the simplified equation for node 2. ‘ (jco+ 500)V0 -— jco(0.5V0 + 51) = 0' I ‘ jco/2+500 '1000+jm Therefore, jml 0 H =———— (w) 1000+jco > 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, i.e., V = I Z. . 244 Label the necessary variables for this technique. 10 Q lo I 0 jco(0.02) Q I _,__£__I__1_9_I ° " 10+10+ jco(0.02) “ 20+ jw(0.02) V =ZI_WI_ jwlo ° ‘ 20+ jco(0.02) _1000+jco XL_ jcolO I "1000+jw Our check for accuracy was successful. > Has the problem been solved SATISFACTORILY? If so, present the solution; if not, then return to “ALTERNATIVE solutions” and continue through the process again. . This problem has been solved satisfactorily. jcolO H =————— (0’) 1000+ jco The freg uency response is 10' (1000, 7.071) Magnitude (unitless) 103 10 Frequency (radians/second) 245 Problem 14.2 Given the circuit in Figure 14.2 and Vin (t) = Vin cos(0)t) volts, find the . transfer function H(0)) = Vout/ Vin and sketch the frequency response. 10kg luF Figure 14.2 Transform the circuit to the frequency domain. 10 k!) 410%) Q Clearly, Vout = jw(0.01)I d I Vin an = . 106 104 J +jo)(0.01) (0 Thus, '106 Vin =[104 - J m + jco(0.01))1 So, _ jw(0.01)1 _ JO) V _ ' 6 J a ( 1‘”) m 1 4 _ ° . 6 ___~___ [0 m +_]0)(001)I 10 +JCO 0) Therefore, H(O)) = —Jmi0T 106 + {03—- —-—) 0) 246 The freg uency response is O. 1 .0 (O .0 on .0 ‘1 .0 a 0.5 Magnitude (unitless) 10“ 105 10° 10 10 Frequency (radians/second) Problem 14.3 Given the circuit in Figure 14.3 and i(t) = Iin cos((ot) amps, find the transfer function H(c0),= Vout / I in and sketch the frequency response. Figure 14.3 The fre uenc res onse is 10000 9000 8000 7000 6000 5000 4000 Magnitude (unitless) 3000 2000 1000 0 . 1o3 10‘ 105 Frequency (radians/second) 247 for the transfer function 104 Hm) : to 106 1 ' —~—- +Jiloz 0) ) WW BODE PLOTS Problem 14.4 Sketch the Bode plots, both magnitude and phase, given the following transfer function in the s-domain. 90 s +1 3 +10 ms) : ( x x ) s (s + 3)(s + 30) First, we need to modify the transfer function so that it is in a form that is easy to plot. S S S . S (90)(s+1)(s+10)_ (“Damiifl i6“) (10%“ 1'6“) s(s+3)(s+30) “ S S = S S (3)(30)(s)(—3-+1 fin) (S)(§+1 5+1) Begin with a plot showing the magnitude curve _of each term in the transfer function. H(S) = 20 log10(10) 20 20 log10(j(0 + 20 log10(1/a)) 20 loglo(j((o/10)+1) 20 log10(1/[j((2)/30)+1]) 20 logm(1/[j(c0/3)+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. 90° +*1 (i(co/10)+1) 30 100 300 (l/[j(0)/30)+1]) ‘_J (1/[j(oo/3)+1]) —90° Finally, the composite phase angle plot for the transfer function can be drawn. 249 Problem 145 Given the following Bode plot, determine the value of the transfer function, H(s), represented by the Bode plot. 0.1 1 10 100 1000 From this figure we have the following H(s), determined by looking at each break point and realizing that the slope everywhere is incremented by 20 dB per decade. Please note that the one break point between 60 = 1 and 10 is estimated to be equal to 3. l 5 +1) s ~_ 10 H(s) = i K s s s . 1 W _~ m (s + )[3 + ljtoo + lX1000 + I) where K is given by 2010g10 (K) = 40 —> K = fl Problem 14,6 . Given the following Bode plot, determine the value of the transfer function, H(s), represented by the Bode plot. ~20 dB —40 dB —60 dB 250 Decomposing the Bode plot into its basic elements we get, 20 dB 0 dB 0.01 I 0.1 20 log10(1/10) ~20 dB 1/((s/100) + 1) *__J 1/((s/0.3) + 1) l/((s/10) + 1) Therefore, + 1 ms) = i W_§_____ (—s—+lIi+l in) 0.3 10 100 SERIES RESONANCE Problem 14.7 Given the circuit in Figure 14.4, find: (a) the resonant frequency and the half-power frequencies, (b) the bandwidth and the quality factor, (0) the amplitudes of Vc (t) at coo , (01, and (02. 1 uF 10 mH 10 cos(c0t) Figure 14.4 251 (a) (b) (c) 1 The resonant frequency of a series RLC circuit is 030 = Vt;- . LC So, a) 1 1 104 d/ = —'—— = —-—-— = ra s 0 J LC 410-2 x10-6 (00 = 10 krad/s . . . . _ R The half-power frequency of a series RLC crrcult 1s 031,2 = + E + So, 4 10 +[ 10 )2+[ 1 “"3 ‘ 2x10‘2 2x10-2 10-2 x10-6 031,2 = —T- 500+ J (500)2 +108 031,2 = T 500 + 10,012 rad/s or (01 = 9.512 krad/s £02 2 10.512 krad/s The bandwidth of a series RLC circuit is B = 032 —— col . So, B =10.512—-9.512 B = l krad/ s . . . . c00 The quality factor of a serles RLC c1rcu1t 15 Q = B“. So, Q _ y: — 103 Q =19 The amplitudes of vC (t) at 030, 031 ,and (02 —j106/m s2 jco10'2 g2 1040° 10 Q 252 J R 1 (ET - ' At 0) = (00 = 10 krad/s, the inductor has a value of j104 ><10‘2 = leO ohms and the capacitor has a value of - j106 /104 = -j100'ohms. Then, w=mml but 10 ‘1: 10+j100—j100 :1 amp So, VC = -j100 volts and vC (t) = 100 cos(10,000t — 90°) volts This gives an amplitude of 100 V at 0) = 030 = 10 krad/ 5. At 0) = (1)1 = 9.512 krad/s, the inductor has a value of j9,512 x 0.01 = j95.12 ohms and the capacitor has a value of - j106/9,512 = -j105. 13 ohms. Then, vC = -j105.13I but 10 10 I=10+j95.12—j105.13 :10—j10.01 amps So, VC = (105.132-90°)(0.7068245.03°) = 74.32—44.97° volts and vC (t) = 74.3oos(9,512t — 44.97°) volts This gives an amplitude of 74.3 V at (0 = (01 = 9.512 krad/s. At co = (02 = 10.512 krad/s, the inductor has a value of j10,512 x 0.01: j105.12 ohms and the capacitor has a value of — j106 /10,512 = -j95.13 ohms. Then, vC = -j95.13I but 10 10 I: 10+j105.12—j95.13 =10+j9.99 amps So, VC = (95.132 - 90°)(0.70752 - 44.97°) = 67.32 -134.97° volts and vC (t) = 67.3cos(10,512t—134.97°) volts This gives an amplitude of 67.3 V at 0) = (02 = 10.512 krad/s. 253 Note that the output voltage for this bandpass filter is the voltage across the resistor. It can be shown that vout (t) = 10cos(10,000t) V at a) = 030 = 10 krad/s vout (t) = 7.068cos(9,512+ 45.03) V at to = 001 = 9.512 krad/s vout (t) = 7.075 cos(10,512t —- 44.97°) V at to = a)2 = 10.512 krad/s The amplitude at the half-power frequencies is 1/ ‘5 times the maximum amplitude at the center frequency. In this case, 1 $00) = 7.071 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 (00, £01, 032, and B. 10 Q L 20 cos(0)t) 1 uF Figure 14.5 L = 1 H 030 =1krad/s w1=995radls m2=1005radls B=10rad/s PARALLEL RESONANCE Problem 14.9 Given the circuit in Figure 14.6 and I = 240° amps, find (a) 030, Q, and B, (b) col and 032, (0) power dissipated at 030 ,0)] , and 032. 254 Figure 14.6 . . . 1 (a) The resonant frequency of a parallel RLC circuit 15 (00 = E. So, 1 104 0) = m : ° duo-2x106) 030 = 1 krad/ s . . . R The quality factor of a parallel RLC c1rcu1t IS Q = 0) L . 0 So, Q _ L ’ 104 ><10'2 Q = 100 . . . 030 The bandwidth of a parallel RLC Circuit 15 B = So, I B _ a 7 100 B = 100 rad / s (b) Because this is a high Q circuit, the half-power frequencies can be written as __ B 031,2 = (00 + '2“ So, 4 _ 100 (01,2 = 10 + ‘2‘ 0)],2 = 104 -T 50 or 031 = 9.95 krad/ s 032 = 10.05 krad/s 255 (c) Find the power dissipated at 000 = 1 krad/s. So, 1 1 =§<2)2<104) 1 P = 20 kwatts 1 Since all of the current flows through the resistor at resonance, P = I i 2 R . Since col and (02 correspond to the half-power points, the power dissipated at col and 032 is 10 kwatts. Problem 14.10 Given the circuit in Figure 14.7, find the resonant frequency. 109 10 mH Figure 14.7 . 1 l l Begin by finding the parallel equivalent of the series resistor and inductor elements. The parallel equivalent is given by So, 1 1 1 ________ = _+ 10+ jun/100 Req jer 10—j03/100 1 1 m + 100+ (co/100)2 R jX 9‘! 5‘1 Thus, -jco/100 _ 1 100+(co/100)2 “ leq _ 100+(m/100)2 X“! (5/100 256 At resonance, . (0:030 and Xeq =XC X ___1_____1___E Where C _ (DC m 0310‘6 — a) BLEDLE: Thus’ (no 100 ‘ (00 or 104 +‘°—‘2’—1o6 100 ‘ cog =108 ~106 =9.9><107 (00 = 9.95 krad/s or coo E 10 krad/s «mm . PASSIVE FILTERS Problem 14.11 What type of filter is represented by the circuit in Figure 14.8? What is the cutoff frequency, or what are the corner frequencies? lOkQ 10kQ Figure 14.8 In the frequency domain, the circuit is 10k!) 257 Find the transfer function Hm) : Vent . V. in Using nodal analySiS, Vout _ Vin + Vout Vout = 0 104 —j105/03+ 104 Simplifying, , 0) Vout _ Vin + J—l—avout + Vout = 0 [2 + 'gjv — V out — in Hence, H = —— (w) 2 + jco/l 0 This transfer function looks like a typical transfer function for a lowpass filter 1 1 + ijC l 5 Vin and as to —> oo Vout = 0 , We can look at a value for 0.7071 =( 2 jive To find the cutoff frequency, find the value of 0) when IH(co)I = 0.3535 . Since the voltage starts at Vout = = 0.3535 lvin out V lH(co)l=-—-—1—O-)-2~=0.3535 4+fi (Dz 8 =4+1-(-)6 (02 = 400 a): 20 rad/s This lowpass filter has a cutoff frequency of (0C = 20 rad/ s or fC = 10/ 7t 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 Q, L = 25 mH, and C = 0.4 MP. Find the quality factor. 1 1 coo =——=————————=1o krad/s x/LC ,/(25 x10‘3)(0.4 x10'6) R 10 B = — = ‘_—‘ = 4 L 0.025 00 rad/S Thus, Q_ 91-32 ‘ B “ o. Q=_5_ This is a high Q circuit so we can use 9.8 (01 =030—B/2=10—0.2=9.8 krad/s or fl 2-2::15597kHz 10.2 (02 :030+B/2=10+0.2=10.2 krad/s or f2 =~§=L6234kflz Therefore, 1.5597 kHz < f < 1.6234 kHz Pro’blem 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? 10k§2 Vin O 10 k9 Figure 14.9 This highpass filter has a cutoff frequency of (0C = 26.55 rad/ s or fC = 4.226 Hz . 259 ACTIVE FILTERS Problem 14.14 What type of filter is represented by the circuit in Figure 14.10? What is the cutoff frequency, or what are the corner frequencies? 100 kg Figure 14.10 In the frequenCy domain, the circuit is Find the transfer function V011! H(03) = V. m 260 Using nodal analysis, . Va_Vin+Va—Vout+Va—Vout_0 105 105 -j106/co ‘ where Va = Vb = 0. Simplifying, ‘Vm _ Vout —ji_(_)Vout : 0 (1+ .39.}, _ J 1 0 out — 1n Hence, H 0) = ——-—— ( ) 1+ jm/ 10 This transfer function looks like a typical transfer function for a lowpass filter 1 1+ ijC This lowpass filter has a cutoff frequency of (DC = 10 rad/ s or fc = 1.5915 Hz. Problem 14.15 [14. 55] Design the filter in Figure 14.11 to meet the following requirements : » (a) It must attenuate a signal at 2 kHz by 3 dB compared with its value at 10 MHz. (b) It must provide a steady-state output of V0 (t) = 10 sin(27t x 108 t + 180°) volts for an input vi (t) = 4 sin(27t x108 t) volts. Figure 14.11 This is a highpass filter with . f0 = 2 kHz, (x)C = 27th =1/(RC) 261 or 1 1 RC 2 27:1; 2 471x103 Clearly, the capacitor becomes a short circuit at high frequencies. Hence, the high frequency gain is -Rf -10 ———=—— or R = 2.5R R 4 f IfweletR:10kQ,then Rf=25kQ,andC== =7.958nF. 40001: x 104 SCALING Problem 14.16 [14.63] For the circuit in Figure 14.12, (a) draw the new circuit after it has been scaled by Km = 200 and Kf = 104. (b) obtain the Thevenin equivalent impedance at terminals a—b of the scaled circuit at co = 104 rad/s. Figure 14.12 (a) R' = KmR = (200)(2) = 400 :2 K L (200)(1) I: ——"—1— = — = 20 L K 104 262 We now have a new circuit, b _ (b) Insert a 1 amp source at the terminals a—b. a V1 SL V2 HM 1A 0 0 0.51K b —: At node 1 : At node 2 : . V V —V V —V V 1: 1 1 2 1 2 = __2_ 1/(sC) + sL sL +0‘SIX R But, Ix = sCVl. So, the nodal equations become V1 —V2 V1 —V2 V2 1—sCV1 + SL sL +0.55CV1=E Solving for V], V __ sL+R ‘ ' $2LC+O.5sCR+l Z _ 1a _ ,__sIg_R_.__ Th " 1 " 52LC+0.SSCR+1 '104 20 '3 “03:10:, Zn: . 4 2 a (J )(_6x10 )fr4400 (J10 ) (20x10 )(0.25x10 )+0.5(]10 )(0.25x10'6)(400)+1 z — 9991299 — 600 '200 ‘ Th ‘ 0.5+j0.5 “ “J . zTh = 632.54-18.43°§2 263 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. 291. Figure 14.13 To scale the circuit in Figure 14.13 from 0) = 1/ 4 rad/s to 03’ = 100 rad/s using a 1 HF capacitor, the feedback resistor and the in ut resistor must be 10 k9. ‘ 264 ...
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ProbSolv_Chapter14 - CHAPTER 14 — FREQUENCY RESPONSE List...

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