Exercise s = 1. cos 60 ± j sin 60 = 0.5 ± j 3/2. = s 2 + s + 1. (s + 1)(s 2 + s + 1) T(jω) = (1 + ω2 )(1 ω 2 ) 2 + ω 2 (1 + ω 2 )


 Elfrieda Cecily Morris
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1 Exercise 7 Ex: 7. A 0 log T [db] T A Ex: 7. A max 0 log.05 0 log db [ ] A min 0 log 40 db 0.0 Ex: 7.3 s + js j Ts k s j s + 3 j s + 4 k s + s s + 4 k s + s + T0 k 4 k 4 s + 4 Ts 4 s + s + Ex: 7.4 s s + 4 Ts a 3 s j0.8s + 0. j0.8 s j.s + 0. j. s s + 4 a 3 s + 0.s s + 0.s +.45 From Fig. we see that the real pole is at s and thus gives rise to a factor s + in the denominator. The pair of complex conjugate poles are at cos 60 ± j sin ± j 3/ The corresponding quadratic in the denominator will be 3 3 s j s j s + s + Since the filter is of the allpole type, the transfer function will be Ts k s + s + s + Since the dc gain is unity, k and Ts Tjω s + s + s + + ω ω + ω + ω ω + ω + ω ω4 ω + ω + ω 4 Ex: 7.5 jv ω4 ω + ω 6 + ω + ω 4 + ω 6.E.D. Tjω / atω ω 3dB, thus 30 + ω3db 6 0 s ω 3dB rad/s 30 At ω 3 rad/s, we have T A 0 log T 0 log/ 730 Figure 0 log db
2 Exercise 7 Ex: 7.6 ɛ 0 A max Ex: 7.8 T Tjω + ɛ ω ω p N N 5 v p v Aω s 0 log Tjω s [ ] N 0 log + ɛ ωs ω p 0 log [ N ] 30 N 0: LHS 9.35 db N : LHS 3.87 db Use N and obtain A min 3.87 db For A min to be exactly 30 db, we need 0 log [ + ɛ.5 ] 30 ɛ A max 0 log 0.54 db Ex: 7.7 The real pole is at s The complex conjugate poles are at s cos 60 ± j sin ± j Tjω for ω<ω p. + ɛ cos [N cos ωωp ] Peaks are obtained when cos [ N cos ˆω ω p ] 0 cos [ 5 cos ˆω ω p ] 0 ˆω 5 cos k + π, k 0,, ω p [ ] k + π ˆω ω p cos, k 0,, 0 π ˆω ω p cos 0.95ω p 0 3 ˆω ω p cos 0 π 0.59ω p 5 ˆω 3 ω p cos 0 π 0 Valleys are obtained when cos [ Ncos ˇω ω p ] e w p /3 jw 5 cos ˇω ω p kπ, k 0,, kπ ˇω ω p cos 5, k 0,, 60 s 30 ˇω ω p cos 0 ω p ˇω ω p cos π 5 0.8ω p ˇω 3 ω p cos π 5 0.3ω p Ts 3 3 s + s j s j s + s + s + DC gain Ex: 7.9 ɛ 0 Amax Aω s 0 log [ + ɛ cosh N cosh ω ] s ω p 0 log [ cosh 7 cosh ] 64.9 db
3 Exercise 7 3 For A max db, ɛ A ω s 0 log [ cosh 7 cosh ] 68. db This is an increase of 3.3 db Ex: 7.0 ɛ a For the Chebyshev filter: A ω s 0 log [ cosh N cosh.5 ] 50 db N 7.4 choose N 8 Excess attenuation 0 log [ cosh 8 cosh.5 ] dB Ex: 7. efer to Fig C 03 rad/s For arbitrarily selected to be 0 k, C 0. μf The two resistors labeled can also be selected to be a convenient value, say 0 k each. Ex: 7.3 Ts ω 0 s + s + For maximally flat response, /, thus ω 0 Ts s + s + ω0 Tjω ω0 ω + j ω Tjω ω 0 ω 0 ω 0 ω + ω ω 0 b For a Butterworth filter ɛ [ ] N Aω s 0 log + ɛ ωs ω p 0 log N 50 N 5.9 choose N 6 Excess attenuation [ 0 log ] db Ex: 7. C 0 4, 0 k C C 0.0 μf H.F. gain 0 00 k Vo ω 0 ω ω 4 ω + At ω, 4 Tj + which is 3 db below the value at ω 0 0 db..e.d. Ex: 7.4 Ts sk ω0 s + s + where K is the centerfrequency gain. For 0 5 rad/s and 3dB bandwidth 0 3 rad/s, we have 3dB BW Also, for a centerfrequency gain of 0, we have K 0 Ts 0 4 s s s + 0 0
4 Exercise 7 4 s + ω0 Ex: 7.5 a Ts a ω0 s + s + ω0 Tjω a a / + a ω 0 ω ω 0 ω + ω ω 0 ω ω 0 ω 0 ω v v 0 v Figure efer to Fig. and note that at any value of T there are two frequencies, ω and ω, with this gain value. From Eq. we obtain ω ω0 ω ω ω ω ω0 ω ω ω ω ω ω ω 0 ω ω 0 ω ω ω ω ω + ω ω + ω ω 0 ω ω ω 0.E.D. Now if ω and ω differ by BW a, v b For A 3 db we have BW a 0 3/0 BW a.e.d. Ex: 7.6 From Fig. 7.6e we have ωn ω max ωn rad/s a ωn T max ω max ω0 ω0 ω max + ωmax where DC gain a ω n ω0 a T max HF transmission a ω ω BW a and if the attenuation over this band of frequencies is to be greater than A db, then by using Eq. we have 0 log 0 [ + ω 0 ω ω 0 ω ω ω 0 ω 0 A/0 ] A ω ω ω ω 0 A/0 ω ω 0 A/0 BW a 0 A/0 Ex: 7.7 Maximally flat π Arbitrarily selecting k, weget C C 5 pf L π C BW a 0 A/0
5 Exercise 7 5 Also L L 0 3 π0 5.5 mh The circuit consists of 3 sections in cascade: a Firstorder section: Ex: 7.8 L C From Exercise 7.5 above, 3dB bandwidth / π0 π60/ 6 C 6 π60 C 0 4 C.6 μf L L 0 4 π H Ex: 7.9 f 0 0 khz, f 3dB 500 Hz f 04 f 3dB Using the data at the top of Table 7., we get C 4 C 6. nf 3 5 C 3.6 k π / C 65 k π Now using the data in Table 7. for the bandpass case, we obtain K centerfrequency gain 0 eferring to Fig. 7.c, we have + r /r 0 Selecting r 0 k, we obtain r 90 k. Ex: 7.0 From Eq. 7.5, we have ω p π0 4 and 5 ω p T s s ω p s + s0.4684ω p ω p s + s0.789ω p ω p Ts 0.895ω p s ω p where the numerator coefficient is set so that the dc gain. Let 0 k dc gain / 0 k 0.895ω p 0.895ω p C C 5.5 nf π b Secondorder section with transfer function: 0.493ω P Ts s + s0.4684ω p ω P where the numerator coefficient is selected to yield a dc gain of unity. 6 C 6 Select k C.43 nf π C 4 C 6 C.43 nf 0.493ωp k ω p C c Secondorder section with transfer function: ω p Ts s + s0.789ω p ω p C 3 C 4 5
6 Exercise 7 6 The circuit is similar to that in b above but with k C 4 C π nf Thus 6 / C 55.6 k Placing the three sections in cascade, i.e. connecting the output of the firstorder section to the input of the secondorder section in b and the output of of section b to the input of c results in the overall transfer function in Eq. 7.5 except for an inversion. Ex: 7. efer to the KHN circuit in Fig Choosing C l nf, we obtain C 5.9 k π Using Eq. 7.6 and selecting 0 k, we get f 0 k Using Eq and setting 0 k, we obtain k Highfrequency gain K.5 V/V The transfer function to the output of the first integrator is V bp sc V hp sk/ C s + s + Thus the centerfrequency gain is given by K K.5 3V/V C Ex: 7. f H 3 C F L C K F/ H s + F / L ω 0 s + s / + ω 0 given C l nf, L 0 k, then C 3.83 k π k f 0 k 0 k k H 8 ω 0 ω n H 0 L 5 DC gain K F L F k /5 5.6 k F L 3 Ex: 7.3 efer to Fig. 7.5 b C C.59 nf π d k Center frequency gain K K 0 g /K 0 00 k Ex: 7.4 efer to Fig. 7.6 and Table 7. AP entry. C 0 nf C 0 k k C C flat gain 0 0 nf / 0 k gain r 0 k 3 r gain k Ex: 7.5 From Eq we have C 0 4 s 0 4 For C C C lnf k 0 9 Thus 3 00 k
7 Exercise 7 7 From Eq we have m m k Ex: 7.6 The transfer function of the feedback network is given in Fig. 7.8a. The poles are the roots of the denominator polynomial, s +s C 3 C 3 C 4 C C 3 4 For C C 0 9 F, 3 0 5, , s + s s + s s 3 04 ± and rad/s + Ex: 7.7 efer to the circuit in Fig. 7.30a, where the transfer function is given on page 340 as sα/c 4 Ts s + s + + C C 3 C C 3 4 Now, using the component value obtained in Exercise 7.5, namely C C nf 3 00 k 4 50 k the centerfrequency gain is given by note that / 3 Tjω a α + C 4 C α 4 + α α α k 0.5 and 4 α k 0.5 Ex: 7.8 sc 3 0 C 3 sc 3 Figure Figure shows the circuit of Fig. 7.34c with partial analysis to determine the transfer function. The voltage at node X can be written as V x + sc 3 X C 4 + sc 3 Now a node equation at X takes the form V x sc 3 + sc 4 V x Substituting for V x from Eq. gives sc 3 + sc 3 +sc 4 sc 3 [s C 3 C 4 + sc ] /C 3 C 4 s + s + + C 4 C 3 C 4 / C 3 C 4 and [ C3 C 4 + ] C 4 which are identical to Eqs and 7.78, respectively..e.d. From Eq. we see that 0.E.D. Ex: 7.9 The design equations are Eqs and k
8 Exercise 7 8 and C 4 C C 3 C/4 C where / 4 0.5C C π C 5.63 nf π C nf C 3.8 nf Ex: 7.30 efer to the results in Example 7.3 a 3 / 3 +% S ωo 3 / ω O ω O % S 3 / % b 4 / 4 % S ωo 4 ω O ω O % S 4 % c Combining the results in a & b, we get ω O % ω O 0% d Using the results in c for both resistors being % high, we have ω O ω O C S ωo C C C + S ωo C C + 0% C S C C + C S C + 0 C % Ex: 7.3 f 0 f 3dB 0 MHz / DC gain Using Eq. 7.99, we obtain G m C π ma/v G m G m 0.5 ma/v G m3 G m 0.5 / ma/v G m4 G m Gain G m 0.5 ma/v Ex: 7.3 From Eqs & 7.0 C 3 C 4 T c C π pf From Eq. 7. C 5 C pf 0 From Eq. 7.3 Centrefrequency gain C 6 C 5 C 6 C pf Ex: 7.33 p L 0 π k L r o p k L 5 k Ex: 7.34 in / L π BW f 0 / 455/35 3 khz C + C in ω 0 L π nf C nf Ex: 7.35 To just meet specifications, f 0 BW n in L 45.5
9 Exercise 7 9 n in 45.5 π n in.86 k.86 n.36 C + C in n 4.47 nf ω 0L C 4.36 nf At resonance, the voltage developed across is I n in I. Vbe I/n & I c g m V be g m I/n, I c I g m/n 9. A.36 A Ex: f 0 / / Eq. 7.3 f khz C ω 0 L π pf ω O C π k
10 Chapter 7 7. Ts s +, Tjω jω + Tjω ω 0 + ω ω + [ ] Im Tjω φ ω tan e Tjω tan ω/ω O G 0 log 0 Tjω A 0 log 0 Tjω ω Tjω G A φ [V/V] [db] [db] [degrees] Ts Tjω π 04 s + π 0 4 π 04 π jω + j[ω/π 0 4 ] Tjω / ω + π 0 4 φω tan ω/π 0 4 a f khz ω π 0 3 rad/s T / V/V φ tan Peak amplitude of output sinusoid V Phase of output relative to that of input 5.7. b f 0 khz ω π 0 4 rad/s T / V/V φ tan 45 Peak amplitude of output sinusoid V Phase of output relative to that of input 45. c f 00 khz ω π 0 5 rad/s T / V/V φ tan Peak amplitude of output sinusoid 0. V Phase of output relative to that of input d f MHz ω π 0 6 rad/s T / V/V φ tan Peak amplitude of output sinusoid 0.0 V Phase of output relative to that of input Ts s + s + s + s 3 + s + s + Tjω [ j ω ω 3 + ω ] [ ω Tjω ω 3 ] + ω [ 4ω 4ω 4 + ω 6 + 4ω + 4ω 4] [ + ω 6].E.D. + ω 6 For phase angle: [ ] Im Tjω φω tan e Tjω [ ] ω ω tan 3 ω For ω 0. rad/s: Tjω / φ ω rad For ω rad/s: Tjω + 6 / / φ tan rad
11 Chapter 7 Note: G 3 db 7.6 For ω 0 rad/s: T jω / 0.00 [ ] φ tan [ ] 980 tan 99 [ ] tan rad Now consider an input of A sin ωt to Ts. The output is then given by A Tjω sin ωt + φ ω Using this result, the output to each of the following inputs will be: INPUT OUTPUT 0 sin0.t 0 sin0.t 0. 0 sint 7.07 sint sin0t 0.0 sin0t At ω 0, we have 0 log T 0dB T V/V At ω ω p, we have 0 log T A max 0. db T V/V At ω ω s, we have 0 log T A min 60 db T 0.00 V/V 7.5 efer to Fig A max 0 log db A min 0 log db Selectivity factor f s f p Figure efer to Fig.. π 04 Ts s + π 0 4 Tjω ω + j π 0 4 T / f At f f p 5 khz, we have T / A max 0 log db At f f s 0 f p 50 khz, we have T / A min 0 log 4.5 db τ s τ rad/s Ts s + where we have also used the given information on the dc transmission being unity. T + ω Since A max db, we have 0 log Tjω p db Tjω p ωp ω p rad/s
12 Chapter 7 3 Since A min db, we have 0 log Tjω s db Tjω s ω s ω s 3.85 rad/s Selectivity factor ω s 3.85 ω p See Fig.. Figure shows the location of the five poles in the s plane. The realaxis pole at s 0 4 gives rise to a factor s The pair of conjugate poles p and p are at s 0 4 sin 8 ± j0 4 cos ± j This pole pair gives rise to a factor s j s j s s The pair of complex conjugate poles p and p are at s 0 4 sin 54 ± j0 4 cos ± j This pole pair gives rise to a factor Figure s j s j See Fig.. s s Thus the denominator polynomial of Ts is Ds s s s s s a The filter is a lowpass of the allpole type, Ts k s s s s s Figure Since the dc gain is unity, we have 7.0 k 0 0 Ts 0 0 s s s s s b The filter is a high pass, Ts ks 5 s s s s s Since the highfrequency s gain is unity, k must be unity, thus Figure Ts s 5 s s s s s + 0 8
13 Chapter Ts ks + 4 s + s j 0.8 s j 0.8 ks + 4 s + s + s T0 4k 0.89 k Ts 0.5s + 4 s + s + s I 6 efer to Fig.. I I 5 I F F V 4 H Figure I sc s s I 3 I 7. Ts ks + 4 s + 0.5s T0 4k.065 k Ts 0.656s + 4 s + 0.5s T ks + 4 s js j 7.3 ss s Ts s 6 + b 5 s 5 + b 4 s 4 + b 3 s + b s + b 0 Note that we started with the numerator factors which represent the given transmission zeros. We used the fact that there is one transmission zero at s to write the denominator sixthorder polynomial. N 6 A sketch of the magnitude response, T, is given in Fig.. T 0 3 v, krad/s Figure I 3 I + I s + V 4 + sli 3 + s s + s + s + I 5 s V 4 ss + s + I 6 I 3 + I 5 s + + ss + s + s 3 + s + s + V 4 + I 6 s + s + + s 3 + s + s + s 3 + s + 3s + s s s 3 + s + 3s + All the transmission zeros are at s. To find the poles, we have to factor the thirdorder denominator polynomial. Toward this end, we find by inspection that one of the zeros of the denominator polynomial is at s. the polynomial will have a factor s + and can be written as s 3 + s + 3s + s + s + as + b where by equating corresponding terms on both sides we find that b a + a s 3 + s + 3s + s + s + s + and the poles are at s and at the roots of s + s + 0
14 Chapter 7 5 which are s ± ± j 7/ 0.5 ± j.33 ω s 7.5 A max 0.5 db, A min 0 db,.7 ω p Using Eq. 7.4, we obtain ɛ 0 Amax/ Using Eq. 7.5, we have Aω s 0 log[ + ɛ ω s /ω p N ] 0 log[ N ] For N 5, Aω s 4.08 db For N 6, Aω s 8.58 db For N 7, Aω s 3.5 db to meet the A s 0 db specification, we use N 7 in which case the actual minimum stopband attenuation realized is A min 3.5 db If A min is to be exactly 0 db, we can use Eq. 7.5 to obtain the new value of ɛ as follows: 0 0 log[ + ɛ.7 4 ] 00 ɛ Now, using Eq. 7.3 we can determine the value to which A max can be reduced as A max 0 log + ɛ A max 0 log db 7.6 Using Eq. 7.5, we have Aω s 0 log[ + ɛ ω s /ω p N ] For large Aω s, we can neglect the unity term in this expression to obtain Aω s 0 log [ɛ ω s /ω p N ] Substituting Aω s A min we obtain 0 log ɛ + 0N logω s /ω p A min N A min 0 log ɛ 0 logω s /ω p.e.d. 7.7 For an Nthorder Butterworth filter, we have from Eq. 7. Tjω + ɛ ω ω p N At the 3dB frequency ω 3dB we have: N ɛ ω3db /N ω 3dB ω ω p ɛ p. from Eq. 7.5 the attenuation at ω.8ω 3dB is: [ ] N A 0 log + ɛ.8ω3db 0 log ω p [ + ɛ.8 ɛ /N ] N 0 log +.8 N For the case N 7, A 0 log db 7.8 A max 0.5 db, N 5, ω p 0 3 rad/s Using Eq. 7.4, we obtain ɛ 0 Amax/ The natural modes can be determined by reference to Fig. 7.0a: /N ω p ɛ / rad/s [ π π ] p, p sin ± jcos ± j ± j0.95 [ ] 3π 3π p, p sin ± jcos ± j0.588 p f p 0 khz, A max 3 db, f s 0 khz, A min 0 db Using Eq. 7.4, we obtain ɛ 0 Amax/0 0 3/0 Using Eq. 7.5, we have Aω s 0 log[ + ɛ ω s /ω p N ] A min 0 log[ + ɛ ω s /ω p N ] 0 0 log + N For N 3, 0 log db For N 4, 0 log db
15 Chapter 7 6 N 4 The poles can be determined by reference to Fig. 7.0a: a See Fig. a. b See Fig. b. 7. /4 ω p ω p π 0 4 rad/s [ π π ] p, p sin ± jcos 8 8 π ± j0.94 [ ] 3π 3π p, p sin ± jcos 8 8 π ± j0.383 Ts ω0 4 s s + ω0 s s + ω0 where π 0 4 and where we have assumed the dc gain to be unity. Using Eq. 7., we obtain Tjω ω + ω p 8 At f 30 khz 3 f p, the attenuation is A 0 log T 0 log db 7.0 Figure From Fig. we see that at the stopband edge, ω s, the Chebyshev filter provides AdB greater attentuation than the Butterworth filter of the same order and having the same A max. 7. N 7, ω p rad/s, A max 0.5 db Using Eq. 7., we obtain ɛ 0 Amax/ From Eq. 7.8, we have Tjω + ɛ cos [N cos ω/ω p ] + ɛ cos 7 cos ω Tjω will be equal to unity at the values of ω that make cos7 cos ω 0 Since the cosine function is 0 for angles that are odd multiples of π/, the solutions to Eq. are 7 cos ω k k + π where k 0,,,... k + π ω k cos 4 We now can compute the values of ω k as ω cos π rad/s 4 Figure ω cos 3π rad/s
16 Chapter 7 7 ω 3 cos 5π rad/s 4 ω 4 cos 7π 4 0 rad/s Next, we determine the passband frequencies at which maximum deviation from 0 db occurs. From Eq. we see that these are the values of ω that make cos 7 cos ω 3 Since the magnitude of the cosine function is unity for angles that are multiples of π, the solutions to Eq. 3 are given by 7 cos ω m mπ or ω m cos mπ 7 where m 0,,,... We now can compute the values of ω m as cos0 rad/s ω cos π 7 ω cos π 7 ω 3 cos 3π 7 T, db rad/s 0.63 rad/s 0.3 rad/s This point also is indicated on the sketch in Fig.. Finally, we note that since this is a 7thorder allpole filter, for s, T will be proportional to /s 7 ; that is, T will be proportional to /ω 7, thus the asymptotic response will be db/octave. 7.3 a Consider first the Butterworth filter. For the given A max db we find the corresponding value of ɛ using Eq. 7.4 as ɛ Next we use Eq. 7.5 to determine the attenuation at ω s ω p,as Aω s 0 log db b We next consider the Chebyshev filter. Its ɛ is the same as that for the Butterworth filter, that is, ɛ The attenuation at ω s ω p can be determined using Eq. 7. as Aω s 0 log[ cosh 6 cosh ] 56.7 db which is nearly twice as much as the attenuation provided by the Butterworth filter. A sketch of the transmission of both filter types is shown in Fig v, rad/s Figure Figure shows a sketch of T for this 7thorder Chebyshev filter, with the passband maxima and minima identified. Note that the frequencies of the maxima and minima do not depend on the value of A max. To determine the attenuation at the stopband frequency ω rad/s, we utilize the expression in Eq. 7., thus A 0 log[ cosh 7 cosh ] 64.9 db Note that we have expanded the vertical scale to show the details of the passband. 7.4 A max db ɛ 0 / f p 3.4 khz, f s 4 khz f s.76 f p A min 35 db
17 Chapter 7 8 a To obtain the required order N, weuse Eq. 7., Aω s 0 log[ + ɛ cosh N cosh ω s /ω p ] 0 log[ cosh N cosh.76] 35 We attempt various values for N as follows: N Aω s db db db Use N 0. Excess attenuation dB b The poles can be determined using Eq. 7.3, namely k π p k /ω p sin sinh N N sinh ɛ k π +j cos cosh N N sinh ɛ k,,..., N First we determine sinh N sinh ɛ sinh 0 sinh and cosh N sinh ɛ π p /ω p sin j π p /ω p sin j π p 3 /ω p sin j π p 4 /ω p sin j π + j.00 cos 0 +j.00 cos +j.00 cos +j.00 cos 3π 0 5π 0 7π 0 9π p 5 /ω p sin j j.00 cos 9π 0 p 6 p 5, p 7 p 4, p 8 p 3, p 9 p, p 0 p Each pair of complex conjugate poles, p k, p k ω p k ± j k gives rise to a quadratic factor in the denominator of Ts given by s + sω p k + ω p k + k where ω p π rad/s we obtain for the five pole pairs: p, p : s + s ω p ω p p, p : s + s 0.30ω p ω p p 3, p 3 : s + s 0.06ω p ω p p 4, p 4 : s + s 0.554ω p ω p p 5, p 5 : s + s 0.830ω p ω p The transfer function Ts can now be written as B Ts Product of five quadratic terms The value of B determines the required dc gain, specifically DC gain B ωp B p For DC gain 0.89 we select + ɛ B p π efer to Fig. 7.3 row a. Input resistance to obtain an input resistance of k, we select k
18 Chapter 7 9 DC gain k C π C 65 pf π efer to Fig. 7.3 row b. H.F. Input resistance 0 k C π 00 C π nf Highfrequency gain 0 k 7.7 efer to the opampc circuit in Fig. 7.3 row c. Assuming an ideal op amp, we have Ts Z s Z s Since both Z and Z have a parallel structure, it is far more convenient to work in terms of Y s and Y s, thus Ts Y s Y s + sc + sc C C ω Z C ω P C s + C s + C DC gain T0 HF gain T C C 7.8 We use the opampc circuit of Fig. 7.3c. Lowfrequency input resistance 0 k DC gain 0 k f Z ω Z π πc 00 πc C 0.6 μf π 06 f P f 0 πc πc C.6 nf π 08 Figure shows a sketch of the magnitude of the transfer function. Note that the magnitude of the highfrequency gain is C /C 00 or 40 db. Figure 7.9 Figure on the next page shows the bandpass filter realized as the cascade of a firstorder lowpass filter and a firstorder highpass filter. The component values are determined as follows: in To make in as large as possible while satisfying the constraint that no resistance is larger than 00 k, we select 00 k
19 Chapter 7 0 This figure belongs to Problem C 3 V C V 0 db 0 db/decade V db 0 db/decade 0 db/ decade 0 db/decade db 50 khz f 50 Hz f 50 Hz 50 khz f Figure The lowfrequency gain of the lowpass circuit is /. With 00 k, the maximum gain obtained is unity and is achieved by selecting 00 k This implies that the required gain of db or 4 V/V must be all realized in the highpass circuit. The upper 3dB frequency of the bandpass filter is the 3dB frequency of the lowpass circuit, that is, πc C π pf Next, we consider the highpass circuit. The highfrequency f 50 Hz gain of this circuit is 4 / 3. To obtain a gain of 4 V/V, we select 4 00 k k 4 The lower 3dB frequency of the bandpass filter 50 Hz is the 3dB frequency of the highpass circuit, thus 50 πc 3 C π μf 7.30 I V V efer to Fig.. /sc V + + /sc sc + V V + sc + I V sc sc + V I C I Figure sc + sc sc + sc +
20 Chapter 7 sc sc + or Ts where C Tjω jω jω + jω + jω jω/ + jω/ Tjω s /C s + /C s s + + ω/ + ω/ φω tan ω ω tan [ ] ω φ for a given phase shift φ, we can use Eq. to determine ω/. For ω rad/s, we can then determine the required value of. Finally, for C 0 nf, the required value for can be found from The results obtained are as follows: φ ω/ krad/s k function Ts s/ s, we analyze the circuit as follows: V + + sc s s + C s V V + s + C I V /C s + C V I s s + C /C s + C s s + C Ts s s s /C s + /C s s + where /C Tjω + jω + jω φω φ N ω φ D ω where φ N is the phase angle of the numerator and φ D is the phase angle of the denominator. A graphical construction showing φ N and φ D and their difference φ is depicted in Fig I C V V I v 0 Im jv f f N f D 0 v 0 e Figure Figure shows the circuit of Fig. 7.4 with and C interchanged. To determine the transfer Observe that the difference, φ, is a positive angle whose value ranges from 80 at ω 0 to 0 at ω. The two end points should also be obvious from T jω which is atω 0 and + atω 0.
21 Chapter V V 3 C V C Figure shows a circuit composed of the cascade connection of two allpass circuits of the type shown in Fig We require that each circuit provide 0 phase shift at ω π 60 rad/s. From the data in Fig. 7.4a, φω tan ωc 0 tan ωc ωc tan 60 3 π k π The value of can be selected arbitrarily, say 0 k 7.33 efer to Fig. 7.6a. Ts ω 0 s + s s s ω max rad/s 4 Since the dc gain is unity, we have a o /ω 0 T max There are many possibilities, but only two are optimal. The first is the Butterworth filter which exhibits maximum flatness of T at ω 0. The second is the Chebyshev for which T exhibits equiripple response in the passband. Figure See Figure on the next page. Figure a shows the secondorder Butterworth filter that just meets the given passband specifications. Figure b shows the secondorder Chebyshev filter that just meets the given passband specifications. Note that no stopband specifications were given otherwise the problem would be overspecified; we are simply asked to calculate the attenuation at ω s ω p rad/s. For both filters, since A max 3 db, we have ɛ a Butterworth filter: For a secondorder Butterworth, we have / and ω p rad/s. The dc gain is unity. Ts s + s + Tjω + ω N Tj + 4 A min Aω s 0 log db b Chebyshev filter: For a secondorder Chebyshev filter, the poles are given by Eq. 7.3, which for N, ω p, and ɛ yields π p, p sin sinh 4 sinh π ±j cos cosh 4 sinh 0.38 ± j for this pair of complexconjugate poles, we have and ω
22 Chapter 7 3 This figure belongs to Problem Figure K Ts s s From Fig. b we see that the dc gain is / + ɛ K and 0.5 Ts s s At s j, we have Tj A min A 0 log db 7.35 efer to Fig. 7.6b. For a maximally flat response, we have / and ω 3dB rad/s If the highfrequency gain is unity, then we have a and s Ts s + s + The two zeros are at s 0. The poles are complex conjugate and given by s ± j 4 ± j 4 ± j ± j Poles are at 0.5 ± j 3/ rad/s Highfrequency gain a. Ts s s + s f 0 0 khz π 0 4 rad/s BW f Hz 0 Centerfrequency gain 0 ω0 0 s Ts ω0 s + s + ω0 π 0 4 s s + sπ π 0 4
23 Chapter 7 4 Poles are at ± j 4 π 03 ± jπ π 03 [ ± j ] ± j Zeros are at s 0 and s a A secondorder bandpass filter with a centerfrequency gain of unity arbitrary has the transfer function s / Ts s + s / + ω0 Tjω / T ω / ω 0 ω + ω / + ω ω ω 0 between ω and ω can be determined by considering the second term in the denominator of Eq., as follows: ω 0 ω ω ω ω Cross multiplying and collecting terms results in ω ω ω 0.E.D. b Since the two edges of the passband must be geometrically symmetric around, we have ω P ω P 800 0, rad/s The factor can now be found from 3dB BW 0, The geometric symmetry of T enables us to find ω s from ω s ω s ω 0 v v 0 v v Figure From Fig. we see that at each T below the peak value there are two frequencies ω < and ω > with the same T. The relationship ω s ,000 rad/s 3000 Using Eq., we obtain A min Aω s 0 log [ + ] ω s ω s ] 0 log [ db Figure shows a sketch of T. This figure belongs to Problem 7.38, part b. T, db ,000 f s 8,00 9,000 0,000 f p f 0 f p 7,000 f s f, Hz Figure
24 Chapter efer to the problem statement of Exercise 7.5. For our case, we have π 60 rad/s BW a π 6 rad/s A 0 db the required is BW a 0 A/0 π 60 π 6 0 0/ the filter transfer function with unity dc gain is s + ω0 Ts s + s + ω0 s + π 60 s + s π 60 + π Since near the poles, T exhibits a peak, and near the zeros it exhibits a dip, the results shown in Fig. are obtained. + sly C + sl sc + Ts s s s LC + s L + /LC s + s C + LC [which is Eq. 7.34] LC C C [which is Eq. 7.35] 7.43 C C C 5nF LC.E.D. L /ω 0 C L 0 mh Figure 7.4 An increase in p increases the magnitude of the peak. On the other hand, an increase in z increases the magnitude of the dip. the results shown in Fig. are obtained. p > z 7.44 a If L became.0l, we obtain.0lc.005 LC LC decreases by 0.5%. b If C increases by %, similar analysis as in a results in decreasing by 0.5%. c is independent of the value of no change. p < z Figure 7.4 efer to Fig. 7.7c. Using the voltage divider rule, we obtain Z C Z C + sl v 7.45 a As ω reaches 0 dc, we get C /sc + C + C sc sc C C + C As ω reaches, C C + C No transmission zero are introduced.
25 Chapter 7 6 b Z,C Z,C + sc + sc Y,C + sc + sc + C + C sc s s + C + C C C C C C + C There is a transmission zero at s 0, due to C. c sl sl + L L L + L 0 L L + L L L + L No transmission zeros. d sl sl + L L L + L There is a transmission zero at 0 dc due to L efer to Fig. 7.8c. Using the voltage divider rule, we obtain Z L Z L + sc + sc Y L + sc + sl + sc + s LC s s + s C + LC 7.47 efer to the circuit in Fig. 7.8b. C ω 0 LC 0 L 0 9 L mh 7.48 efer to the circuit in Fig. 7.8d. Since, at the center frequency, the LC circuit acts as an open circuit, the circuit reduces to. We can change the gain to 0.5, by splitting into two equal parts, each equal to. The voltage divider, will provide at, a gain of 0.5. The factor will be determined by the parallel equivalent of and. Since the latter is equivalent to, will remain unchanged, as required. The resulting circuit is shown in Fig.. Figure 7.49 Using superposition, we find the resulting Ts in each of the three cases and then sum the results. Thus see Figure on the next page: a When x is lifted off ground and connected to V x, the circuit becomes as shown in Fig. a, and the transfer function becomes that of a lowpass filter with / LC, C, and dc gain of unity, thus ω0 V x s + s ω 0 + b With y lifted off ground and connected to V y, the resulting circuit [shown in Fig. b], is that of a highpass filter with / LC, C, and a highfrequency gain of unity, thus C L
26 Chapter 7 7 This figure belongs to Problem L C V x V y C L L C V z 3 a b c Figure V y s s + s + c With z lifted off ground and connected to V z, the resulting circuit shown in Fig. c is that of a bandpass filter with / LC, C, and a centerfrequency gain of unity, thus 3 V c s + s s ω0 + ω0 Now summing,, and 3 gives ω0 s V y + s V z + ω0 V x ω0 s + s + ω efer to Fig. 7.8g. ω n / LC / LC + C ω n + C C. + C C 3 C 0. C At frequencies, T approaches unity because L approaches a short circuit and C and C approach open circuits. At frequencies, T approaches C C + C + C C [see Fig. 7.8b] V/V. 7.5 L C / Selecting k, we have L C 4 C C 4 L 0 8 a For L 5 H, we have C F 0.5 μf b For L.5 H, we have C μf 5 nf c For L 0.5 H, we have C 4.5 nf 7.5 a Figure on the next page shows the analysis. It is based on assuming ideal op amps that exhibit virtual short circuits between their input terminals, and draw zero currents into their input terminals. Observe that: The circuit is fed at port with a voltage V. The virtual short circuit between the input terminals of each of A and A cause the voltage at port to be V V With port terminated in an impedance Z 5, the current that flows out of port is I V Z 5 3 Following the analysis indicated, we find the input current into port as I Z Z 4 Z Z 3 Z 5
27 Chapter 7 8 This figure belongs to Problem 7.5. Z I V Z A 4 Z Z Z 3 Z 5 V V Z 4 Z 3 Z 4 I V V /Z 5 Z Z Z 3 Z 4 0 V 0 V Z Z 4 Z Z 3 Z 5 V Z 4 /Z 3 Z 5 V V V /Z 5 V /Z 5 V Z 4 Z 3 Z 5 V V Z 4 Z 5 Z 5 A V Z Z in Z 3 Z Z 4 Z 5 I Figure This current could have been written as Z Z 4 I V I Z Z 3 4 Equations and describe the operation of the circuit: It propagates the voltage applied to port, V, and to port, V V ; and whatever current is drawn out of port is multiplied by the function Z Z 4 /Z Z 3 and appears at port. 5 The result of the two actions in 4 is that the input impedance looking into port becomes V Z V I I Z Z 4 /Z Z 3 Z Z 3 V Z Z 4 I or Z Z 3 Z Z 5 Z Z 4 b If port is terminated in an impedance Z 6, the input impedance looking into port can be obtained by invoking the symmetry of the circuit, thus Z Z 4 Z Z 6 Z Z 3 the circuit behaves as an impedance transformer with the transformation ratio from Z Z 3 port to port being and from port to Z Z 4 Z Z 4 port being. Z Z 3 Since Z, Z, Z 3, and Z 4 can be arbitrary functions of s, the transformation ratio can be an arbitrary function of s. Thus the circuit can be used as an impedance converter. For instance, the particular selection of impedances in Fig. 7.0a results in the transformation ratio from port to port being sc 4 3 /. As a result, the circuit in Fig. 7.a converts a resistance 5 into an inductance L C /. Other conversion functions are possible and the circuit is known as a Generalized Impedance Converter or GIC Figure on the next page shows the suggested circuit together with the analysis. The input impedance looking into port is Z s V I For s jω we have s C 4 C 6 3 Z jω ω C 4 C 6 which is a negative resistance whose magnitude depends on frequency ω; thus it is called a Frequency Dependent Negative esistance or FDN.
28 Chapter 7 9 This figure belongs to Problem A I V s C 4 C C 4 0 V C 6 0 sc 6 V V sc 6 V 0 s C 4 C 6 3 sc6 V V V sc 6 V A Z Figure 7.54 KV A K V Y 6 V I 0 i V 3 C 0 4 V 5 sc 4 Y 6 3 V 6 C 6 Y 6 V Y 6 V 0 Y 6 V sc 4 Y 6 3 V 0 A Figure Figure shows the circuit together with the analysis details. At the input we can write V + I i 5 V + sc 4Y V where V K and Y sc 6 KV K + sc 4 sc K /C 4 C s + s + C 6 6 C 4 C which is a lowpass function with / C 4 C / C 6 6 C DC gain K which are the same as the data given in Table 7...E.D.
29 Chapter For a fifthorder Butterworth filter with A max 3 db, and thus ɛ, and ω p 0 3 rad/s, the poles can be determined using the graphical construct of Fig. 7.0a, thus The pole pair p, p has 0 3 rad/s and π sin The pole pair p, p has 0 3 rad/s and 3π 03 sin The realaxis pole p 3 is at a s 0 3 rad/s the transfer function Ts is Ts 03 s s + s s + s The firstorder factor, T s 03 s can be realized by the circuit in Fig. 7.3a with 00 k arbitrary but convenient value. C C 0.0 μf 0 nf The secondorder transfer function T s 0 6 s + s can be realized by the circuit in Fig. 7.a with C 4 C 6 C 0 nf practical value 3 5 C 00 k / C k K The secondorder transfer function T 3 s 0 6 s + s can be realized using the circuit in Fig. 7.a with C 4 C 6 C 0 nf practical value k C k C K The complete filter circuit is obtained as the cascade connection of the three filter sections, as shown in Fig. on the next page efer to Fig. 7.e and the LPN entry in Table 7.. Select C 4 C 0 nf 3 5 C.59 k π k ω0 0 C 6 C nf ω n C 6 C nf K 7.57 f 0 khz Selecting, C 4 C 6 C.0 nf, we obtain 3 5 C π k k C r r 0 k arbitrary
30 Chapter 7 This figure belongs to Problem k 00 k 0 nf 00 k 00 k 00 k 0 nf 00 k 6.8 k 0 nf 00 k 00 k 00 k 0 nf 00 k 6.8 k 0 nf Figure C Ts K C 6 + C 6 s + /C 4 C s + s + C 6 + C 6 6 C 4 C 6 + C ω n C 4 C ω 0 C 4 C 6 + C C 6 + C 6 6 Dividing Eq. by Eq., we obtain ω0 C 6 4 ωn C 6 + C 6 Selecting C 4 C 6 + C 6 C practical value, then from Eq. 4 we obtain C 6 C and ω0 ω n C 6 C C 6 Selecting 3 5, and using Eq., we obtain ω 0 C C Using Eq. 3, we obtain 5 6 C Finally, from the expression for Ts, we see that DC gain K
31 Chapter efer to Fig. 7.f and to the HPN entry in Table 7.. Ts s + /C 4 C K s + s + + C 6 6 C 4 C ω n C 4 C ω 0 C 4 C 6 3 ω0 ωn 5 5 where ω ω n [ 5 5 / ωn ] Choosing C 4 C 5 C practical value and 3 4, weget ω 0 C 3 C / C C 6 6 C 6 6 Finally, C K Highfrequency gain 7.60 a Ts s s s + s eplacing s by s/0 5, we obtain Ts s s s s s s s + s b Firstorder section: T s s where the dc gain is made unity. This function can be realized using the circuit of Fig. 7.3a with C nf arbitrary but convenient value C k For dc gain of unity, we have 3.7 k Secondorder LPN section: T s 0.68s s + s Selecting, we obtain C 4 C 6 + C 6 C nf and 3 5 then C k C 6 C ω0 ω n nf pf C nf 38 pf k C Finally, K DC gain The complete circuit is shown in Fig. on the next page. 7.6 Bandpass with f 0 khz, 3dB bandwidth of 50 Hz, thus f 0 BW khz 50 Hz 40 efer to the circuit in Fig. 7.4a. Using C 0 nf
32 Chapter 7 3 This figure belongs to Problem 7.60, part b. Figure then C π k Select 0 k then f 0 k Select k then k K Centerfrequency gain K a efer to the circuits in Fig. 7.4 and the transfer function in Eq. 7.66, that is, Ts K F/ H s s F / B + F / L ω0 s + s + F s s H / B + H / L ω0 K H s + s + For this to be an allpass function, that is, Ts Flat gain s s / + ω 0 s + s / + ω 0 then H L and H B That is, L H B.E.D. Flat gain K F.E.D. H b 0 5 rad/s, 4 and flat gain 0, thus Selecting C nf then C 0 k Selecting 0 k then f 0k Selecting 0 k then 3 70 k Selecting L H 0 k then B H k
33 Chapter 7 4 Now, K 4.75 Flat gain 0 K F H F 0 H K k 7.63 Consider Fig. 7.4 and Eq. 7.66, that is, Ts K F/ H s s F / B + F / L ω0 s + s / + ω0 For this to be the transfer function of a notch filter, that is, s + ωn Ts G s + s + where G is the highfrequency gain, then by equating the coefficients of the corresponding numerator terms, we obtain B 3 ω n H L ω 0 H L G K F H where K thus G ω0 ω n F H 4 F G 5 H / Equations 3, 4, and 5 are the design equations for the resistors associated with the summer. Observe that the value of one of the three resistors, L, H, and F can be arbitrarily selected Using Eq with B, we obtain F s + H / L ω0 K H ω0 s + s + ω0 ω n H L ω 0 Now if H and L can have % tolerances, the worst case will be when one is at the highest possible value and the other at the lowest possible value, for instance, ω n HN.0 LN 0.99 where HN and LN are the nominal values. In this case ω n.0 ω n.0 The other case yields ω n 0.99 Thus the worstcase percentage deviation between ω n and is % efer to Fig. 7.6 and Table 7.. Using C 0 nf, then C k d 0 0 k Select r 0 k 3 If the dc gain is unity, then HF gain ω n ω HF gain C C highfrequency gain nf /ω n HF gain k 7.66 Using Eq with, we have r C s s + 3 C CC C s + s C + C ω z / CC
34 Chapter z C CC r C 3 C r From Eqs. and we see that trimming ω z and z can proceed in the following sequence: a Trim to adjust ω z. This will affect z. b Trim 3 to adjust z. This will not affect ω z s Ts s s + s a eplacing s by s/0 5, we obtain Ts s s s + s b Firstorder section: T s s which is made to have a dc gain of unity, as required. This function can be realized by the circuit in Fig. 7.3a. Selecting C nf arbitrary but convenient we have C k For a dc gain of unity, we have 3.7 k Secondorder LPN section: T s 0.68s s + s where the dc gain is unity. efer to Fig. 7.6 and Eq Selecting C nf then C k.0504 d k Select r 0 k Now, C C highfrequency gain nf 68 pf 3 /ω n HF Gain k The complete circuit is shown in Fig.. This figure belongs to Problem k 3.7 k 35.9 k 3.7 k nf 68 pf nf 9.76 k nf 0 k 0 k 9.76 k Figure
35 Chapter efer to Fig. 7.9 and Eqs and 7.76: C C C nf / 3 and 4 /4 4 C / k k k 7.69 V b V a / 3 3 V b sc C sc 3 4 C 3 V a [ sc + C C 3 + sc 3 4 V a V b sc + + C + 3 C sc 3 4 sc C C 3 sc 3 4 s + s + + C C 3 C C 3 4 s + s C 3 C 3 C 4 C C 3 4 which is identical to the expression given in Fig. 7.8a..E.D. ] 7.70 efer to Fig. 7.8a. V a a 0 C X C b V b V a / 3 sc V b V x V x / 4 4 Figure V b ts s + s C + C s + s 3 C + C s + s/τ + /τ s + s3/τ + /τ If the network is placed in the negativefeedback path of an ideal infinitegain op amp, as in Fig. 7.4, the poles will be given by the roots of the numerator polynomial, thus efer to Fig.. The voltage at node X can be written as V x V a sc V a + sc 3 V b V a 3 Writing a node equation at X gives V b V a 3 + sc V b V x V x V b sc + 3 V a 3 V x V b sc sc Substituting for V x from Eq. gives V b sc V a + sc + 4 sc 3 V b + sc 4 sc 3 V a τ and /τ /τ 0.5 the poles will be coincident at s /τ 7.7 sc 4 V x 0 V C 4 0 sc 3 sc 3 C 3 X Figure V x
36 Chapter 7 7 The circuit is shown in Fig.. The voltage at node X can be found as V x 0 sc 3 sc 3 A node equation at X can be written as sc 3 + sc 4 V x + V x 0 Substituting for V x from Eq., we obtain sc 3 + sc 4 sc 3 sc sc 3 + s C 3 C 4 + sc 3 + sc 4 sc 4 s C 3 C 4 + sc s/c 3 s + s + + C 4 C 3 C 4 For, C 4 C, and C 3 C/36, we obtain s36/c s + s C + 36 C This is a bandpass function with 6 C 6/C /C 3 and Centerfrequency gain 8 Using the design equations 7.75 and 7.76, we obtain 3 4 /4 C where π f 0 BW 0 5 C 5 π 0 4 Selecting C 0 nf we obtain k π C C 0 nf k /C 4 Centerfrequency gain 7.73 C + C V/V C C Figure The circuit is shown in Fig.. The transfer function can be obtained using the equation on page 340 with α, thus s + s + C C s/c C C 3 4 Figure The circuit is shown in Fig.. The voltage at node X is given by V x β sc β + sc β sc
37 Chapter 7 8 Writing a node equation at X gives β + V x /4 + sc V x 0 Substituting for V x from Eq. and collecting terms gives V x β s + s C [ + ] + 4 β C s + s C + 4 C We observe that, as expected, C a To obtain an allpass function, we set [ + ] C β C β + But, β + b To obtain a notch function, we set [ + ] 0 C β β + or, equivalently, 7.74 The analysis is shown in Fig.. The voltage at node X is given by V x + 3 sc + sc 3 A node equation at X provides V x 4 + sc V x sc V x 4 + sc 0 Substituting for V x from Eq. and collecting terms, we obtain s s + s C C C C 3 4 This is a highpass function with a highfrequency gain of unity. To obtain a maximally flat response with ω 3dB 0 4 rad/s and using C C C 0 nf then ω 3dB 0 4 rad/s / 3 + C C / k ω 0 C C k 7.07 k C X 4 C 3 3 V x / 4 sc V x Figure Figure
38 Chapter 7 9 Figure shows a graphical construct to determine the poles of the fifth order Butterworth filter. The pair of complex conjugate poles p and p have a frequency ω 3dB π 0 4 rad/s and a factor sin 8.68 The pair of complex conjugate poles p and p have ω 3dB π 0 4 rad/s and a factor, sin The realaxis pole p 3 is at s π 0 4 rad/s The first secondorder section can be realized using the circuit in Fig. 7.34c. The design equations are k C 4 C C 3 C/4 Here,.68, thus C C C C.68 π C 5.5 nf π C nf 49 pf C nf The second secondorder section also can be realized using the circuit in Fig. 7.34c. Here, 0 k C 4 C C 3 C 4 where Thus C C C C 0.68 π C.97 nf π C 3.9 nf C 4.97 nf The firstorder section can be realized using the circuit in Fig. 7.3a with 0 k C.59 nf π The complete circuit is shown in Fig. below efer to Fig. 7.3 and let the network n have a transfer function V a V b s s + s + which is a bandpass with a unity centerfrequency gain. The complementary network in b will have a transfer function This figure belongs to Problem 7.75, part b. Figure
39 Chapter 7 30 V a V c V a V b s s + s + s + ω 0 s + s + which is a notch function. S L L / L S C C C / S 0 For we have L C L L L C C LC C L C C/L C/L / S L L L S C C C Figure As an example, consider the LC bandpass circuit shown in Fig. a. It has the transfer function T s V s o C s + s C + LC Interchanging the input terminal with ground, we obtain the circuit shown in Fig. b. Straightforward analysis shows that this circuit has the transfer function s + T s LC s + s C + LC which is a notch function. Observe that T s T s that is, the circuits in a and b are complementary For the circuit in Fig. 7.8b we have /LC Ts s + s/c + /LC C LC L For we have L LC / L L 3/ C / L C C 0 S 7.78 a y uv S y x uv x x uv v u x x u x x u + v x x v S u x + Sv x uv + u v x x uv b y u/v S y x y x x y u/v x x u/v u xv v x u u v xv v x u u x x u v x x v S u x Sv x c y ku S y x y x x y k u x x ku u x x u S u x d y u n S y x y x x y nu x n u x u n
40 Chapter 7 3 n u x n u ns u x e y f u, u f x S y x y x x y f u u x u u x f u u [ ][ f u u f x u f u x [ ][ f u u f x u f u x S f u S f x S y u Su x x u ] x f x 7.79 The highpass filter in Fig. 7.33b is derived from the feedback loop in Fig. 7.9; thus it will exhibit the same sensitivities relative to the opamp gain as that of the circuit in Fig These have been derived in Example 7.3 and given by S A 0 S A A 7.80 Using Eqs and 7.79, we have C 3 C 3 C3 C 4 C3 C 4 C 4 + C 3 C3 C 4 S C 3 C 3 C 3 ω O Clearly, S C 3 S C 4 S S C 3 C 3 C3 C 4 + C 4 C 3 S C 3 S C C 4 C ] / / + / / + S / / / + / If S 0. Similarly, S From Table 7. we have C4 C / C 6 6 C C 4 C 4 S C 4 C 4 C 4 Similarly, S C 6 S S 3 S ωo 5 ω O S ωo Now for : S S C C 6 C 6 S + 6 S C C 4 C 4 S,, efer to the circuit in Fig. 7.35f. Vi I o g m, G m g m, But, g m, k n I D, k n I/ k n I
41 Chapter 7 3 G m kn I For G m 0.5 ma/v and k n 0.5 ma/v,we have I I 0.5 ma Since G m is proportional to I, tuning G m in the range ±5% requires tuning I in the range 0.95 I nominal to.05 I nominal, which is approximately ±0% of the nominal value G m G m A/V ma/v Since the output terminal is connected back to the input, the output resistance appears in effect in parallel with the resistance /G m, thus the actual resistance realized is o k G m For the integrator in Fig. 7.36b, we have G m V sc Unitygain frequency G m πc G m π 5 0 G m 0.34 ma/v 7.86 Both o and C o will appear in parallel with C, thus G m + sc + C o o The transfer function realized will be V G m o + sc + C o The integrator time constant is τ C + C o /G m C + C o G m C The quantity + C o C represents the error factor. For the error to be less than %, we must have C o C 0.0 C 00C o the smallest value of C is 00C o. Figure The circuit is shown in Fig., for which we can write G m4 G m V G m V + G m3 V 3 G m G m4 V G m G m4 V + G m3 G m4 V 3 To obtain V V + 3V 3 we select G m G m4 G m G m4 G m3 3 G m4 Frequency of the lowfrequency pole C + C o o C o If this frequency is to be at least two decades lower than the unity gain frequency G m C, then 0.0 G m C o C G m 00 o the smallest value of G m must be 00/ o efer to the circuit in Fig. 7.36c and its transfer function in Eq. 7.9, namely G m sc + G m
42 Chapter 7 33 Pole frequency G m πc G m π 0 G m 0.5 ma/v DC gain G m G m 0 G m G m G m.5 ma/v 7.88 Figure Figure shows the circuit. A node equation at X yields sc G m + G m + sc sc G m G m + sc + sc G m sc G m + sc + C 7.89 a efer to the circuit in Fig. P7.89. The output current of the G m transconductor is G m V. V G m sc V This is the input voltage to the negative transconductor G m. Thus the output current of G m, which is equal to I, will be I G m V G mg m sc V Z in V C s I G m G m which is that of an inductance L, C L.E.D. G m G m b To obtain an LC resonator, we connected a capacitor C from node to ground and a resistance realized by the transconductor G m3, as shown in Fig. below. c A fourth transconductor G m4 is used to feed a current G m4 to node, as shown in Fig.. The resulting circuit is identical to that in Fig. 7.37b except here C C C. d With analogy to the identical circuit in Fig. 7.37b, V / will be a secondorder bandpass filter with a transfer function given by Eq. 7.93, and V / will be a secondorder lowpass filter with a transfer function given by Eq V and, V s + s G m3 C s + s G m C sg m4 /C + G mg m C G m G m4 /C + G mg m C This figure belongs to Problem 7.89, part b. G m3 G m4 C V G m G m C V Figure
43 Chapter Using Eqs and 7.96, we have G m G m C C Gm G m C G m3 C Selecting G m G m G m3 G m, we obtain G m C C 3 C 4 C Selecting C C, then from Eq. 4 we have C C and from Eq. 3 we have G m C 7.9 The resulting circuit is shown in Fig.. For V we can write V G m V G m5 sc A node equation at X can be written as G m V + sc 3 V + G m4 + G m3 V + sc V 0 Substituting for V from Eq. into Eq. and collecting terms results in the transfer function V C3 G s m4 s + G mg m5 C + C 3 C + C 3 C + C 3 C G m3 s + s + C + C 3 C + C 3 C 7.9 f 0 5 MHz, 5, Centerfrequency gain 5 G mg m The design equations are given by 7.99, 7.00, and 7.0. Thus G m C where C C C 5pF G m π ma/v G m G m ma/v G m3 G m ma/v 5 G m4 G m Gain ma/v This figure belongs to Problem 7.9. G m G m V X V G m G m C C 3 C G m3 V sc 3 V G m4 G m4 G m3 V G m5 G m5 Figure
44 Chapter eq T c C f c C For C pf, C eq 5M For C 5 pf, eq M For C 0 pf, eq 500 k Change transferred CV 0 pc For f 0 00 khz, average current is given by 7.96 From Eqs and 7.0, C 3 C 4 T c C π pf From Eq. 7., C 5 C pf 50 From Eq. 7.3, Centerfrequency gain C 6 C 5 C 6 C pf 7.97 ω 3dB 0 3 rad/s / and DC gain f c 00 khz, C C C 5pF I AV T pc From Eqs and 7.0, 0. μa For each clock cycle, the output will change by the same amount as the change in voltage across C. V /C pc 0 pf 0. V For V 0. V for each Clock cycle, the amplifier will saturate in 0 V 00 cycles 0. V slope V t 0 4 V/s 0V 00 cycles / From Eqs and 7.0, C 3 C 4 T c C π pf From Eq. 7., C 5 C pf 0 From Eq. 7.3, Centerfrequency gain C 6 C 5 C 6 C pf C 3 C 4 T c C pf From Eq. 7., C 5 C / 0.07 pf The dc gain of the lowpass circuit is DC gain C 6 C 4 For DC gain, C 6 C pf 7.98 efer to Figs. and on next page. For the BJT we have g m I C ma 40 ma/v V T 0.05 V r π β 00 g m 40 5k C π 0 pf Miller capacitance C μ + g m L pf Total capacitance C + C π + C Miller pf
45 Chapter 7 36 This figure belongs to Problem 7.98, part a. L 5 k s 0 k V s L 0.5 H V be C 00 pf ma Z in Figure Z r p C m g m L in C p LCtotal Figure 80.3 Mrad/s Total effective parallel resistance 0 k 5k 3.33 k L dB BW V be V s r π + s r π g m L V be V s 66.7V s V s 66.7 V/V 7.99 p L π k 967 khz C ω 0 L.53 nf π For a 3dB bandwidth of khz, we have which requires a parallel resistance of L π k the additional parallel resistance required, a, can be determined from a p a a 7.84 k 7.00 f 0 π LC π khz Equivalent parallel resistance 3 9k p L π
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