Exercise 7 Ex: 7. A 0 log T [db] T 0.99 0.9 0.8 0.7 0.5 0. 0 A 0 0. 3 6 0 Ex: 7. A max 0 log.05 0 log 0.95 0.9 db [ ] A min 0 log 40 db 0.0 Ex: 7.3 s + js j Ts k s + 3 + j s + 3 j s + 4 k s + s + 4 + 3 4 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 + 0. + j0.8s + 0. j0.8 s + 0. + j.s + 0. j. s s + 4 a 3 s + 0.s + 0.65 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 60 0.5 ± j 3/ The corresponding quadratic in the denominator will be 3 3 s + 0.5 + j s + 0.5 j s + s + Since the filter is of the all-pole 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 + 3 6 730 A 0 log T 0 log/ 730 Figure 0 log730 8.6 db
Exercise 7 Ex: 7.6 ɛ 0 A max 0 0 0 0.5088 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 [ + 0.5088.5 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 0.3654 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 60 3 0.5 ± j + 0.3654 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 + 0.5 + j s + 0.5 j s + s + s + DC gain Ex: 7.9 ɛ 0 Amax 0 0 0.5 0 0.3493 Aω s 0 log [ + ɛ cosh N cosh ω ] s ω p 0 log [ + 0.3493 cosh 7 cosh ] 64.9 db
Exercise 7 3 For A max db, ɛ 0 0. 0.5088 A ω s 0 log [ + 0.5088 cosh 7 cosh ] 68. db This is an increase of 3.3 db Ex: 7.0 ɛ 0 0 0.5088 a For the Chebyshev filter: A ω s 0 log [ + 0.5088 cosh N cosh.5 ] 50 db N 7.4 choose N 8 Excess attenuation 0 log [ + 0.5088 cosh 8 cosh.5 ] 50 55 50 5dB Ex: 7. efer to Fig. 7.4. C 03 rad/s For arbitrarily selected to be 0 k, C 0. μf 0 3 4 0 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 0.5088 [ ] N Aω s 0 log + ɛ ωs ω p 0 log + 0.5088.5 N 50 N 5.9 choose N 6 Excess attenuation [ 0 log + 0.5088.5 3] 50 0.5 db Ex: 7. C 0 4, 0 k C C 0.0 μf H.F. gain 0 00 k Vo ω 0 ω 4 0 + 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 center-frequency gain. For 0 5 rad/s and 3-dB bandwidth 0 3 rad/s, we have 3-dB BW 0 3 05 00 Also, for a center-frequency gain of 0, we have K 0 Ts 0 4 s s + 0 3 s + 0 0
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 +. 00. + 00 0.986 rad/s a ωn T max ω max ω0 ω0 ω max + ωmax where DC gain a ω n ω0 a.44 0.694 T max 0.694.44 0.97 0.97 + 00 0.97 HF transmission a 0.694 3.7 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 00 0 3 Arbitrarily selecting k, weget C C 5 pf L π0 5 0 3 C BW a 0 A/0
Exercise 7 5 Also L L 0 3 π0 5.5 mh The circuit consists of 3 sections in cascade: a First-order section: Ex: 7.8 L C From Exercise 7.5 above, 3-dB bandwidth / π0 π60/ 6 C 6 π60 C 0 4 C.6 μf L L 0 4 π60 6 4.4 H Ex: 7.9 f 0 0 khz, f 3dB 500 Hz f 04 f 3dB 500 0 Using the data at the top of Table 7., we get C 4 C 6. nf 3 5 C 3.6 k π0 4 9. 0 0 6 / C 65 k π0 4 9. 0 Now using the data in Table 7. for the bandpass case, we obtain K center-frequency 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 8.408 s + 0.895ω p s + s0.4684ω p + 0.493ω p s + s0.789ω p + 0.9883ω p Ts 0.895ω p s + 0.895ω 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 0.895 π0 4 4 0 b Second-order section with transfer function: 0.493ω P Ts s + s0.4684ω p + 0.493ω P where the numerator coefficient is selected to yield a dc gain of unity. 6 C 6 Select 3 5 0 k C.43 nf 0.493 π0 4 4 0 C 4 C 6 C.43 nf 0.493ωp.4 6 4 k 0.4684ω p C c Second-order section with transfer function: 0.9883ω p Ts s + s0.789ω p + 0.9883ω p C 3 C 4 5
Exercise 7 6 The circuit is similar to that in b above but with 3 5 0 k C 4 C 6 0 4 0.9883 π0 4 0 4.6 nf 0.9883 0.789 5.56 Thus 6 / C 55.6 k Placing the three sections in cascade, i.e. connecting the output of the first-order section to the input of the second-order 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. 7.4. Choosing C l nf, we obtain C 5.9 k π0 4 9 0 Using Eq. 7.6 and selecting 0 k, we get f 0 k Using Eq. 7.63 and setting 0 k, we obtain 3 0 30 k High-frequency 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 center-frequency 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 π5 0 3 9 0 0 k f 0 k 0 k 3 0 0 90 k H 8 ω 0 ω n H 0 L 5 DC gain K F L F 3 0 6.7 k /5 5.6 k F L 3 Ex: 7.3 efer to Fig. 7.5 b C C.59 nf π0 4 4 0 d 0 0 00 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 0 4 9 0 0 5 0 50 k C C flat gain 0 0 nf / 0 k gain r 0 k 3 r gain 5 0 50 k Ex: 7.5 From Eq. 7.76 we have C 0 4 s 0 4 For C C C lnf 0 4 00 k 0 9 Thus 3 00 k
Exercise 7 7 From Eq. 7.75 we have m 4 4 4 m 00 4 50 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 + + + 0 C 3 C 3 C 4 C C 3 4 For C C 0 9 F, 3 0 5, 4 5 0 4, s + s 0 9 0 + 5 0 9 5 0 4 0 8 0 0 0 s + s 3 0 4 + 0 8 0 s 3 04 ± 9 0 8 4 0 8 0.38 0 4 and.68 0 4 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 center-frequency gain is given by note that / 3 Tjω a α + C 4 C α 4 + α α 0.5 4 α 50 00 k 0.5 and 4 α 50 00 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 3 + + ] /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. 7.77 and 7.78, respectively..e.d. From Eq. we see that 0.E.D. Ex: 7.9 The design equations are Eqs. 7.79 and 7.80. 0 k
Exercise 7 8 and C 4 C C 3 C/4 C where / 4 0.5C C π 4 0 3 C 5.63 nf π 4 0 3 0 03 C 4 5.63 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 0 + 0 + 0 0% Ex: 7.3 f 0 f 3dB 0 MHz / DC gain Using Eq. 7.99, we obtain G m C π 0 0 6 0 0.5 ma/v G m G m 0.5 ma/v G m3 G m 0.5 / 0.355 ma/v G m4 G m Gain G m 0.5 ma/v Ex: 7.3 From Eqs. 7.09 & 7.0 C 3 C 4 T c C π0 4 00 0 0 3 6.83 pf From Eq. 7. C 5 C 4 6.83 0.34 pf 0 From Eq. 7.3 Centre-frequency gain C 6 C 5 C 6 C 5 0.34 pf Ex: 7.33 p L 0 π0 6 3.8 0 6 50 3k L r o p k L 5 k Ex: 7.34 in / L 0 3 0 3 35 π 455 0 3 6 5 0 BW f 0 / 455/35 3 khz C + C in ω 0 L π 455 0 3 5 0 6 4.47 nf C 4.47 0. 4.7 nf Ex: 7.35 To just meet specifications, f 0 BW 455 0 45.5 n in L 45.5
Exercise 7 9 n in 45.5 π 455 0 3 5 0 6 650 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 40 0.65 m/n 9. A.36 A Ex: 7.36 00 f 0 / / Eq. 7.3 f 0 30.8 khz C ω 0 L π 0.7 0 6 3 0 6 73.75 pf ω O C 73.7 0 π 30.8 0 3 6.94 k
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] 0 0 0 0 0.5 0.8944 0.97 0.97 6.57 0.707 3.0 3.0 45.0 0.447 6.99 6.99 63.43 5 0.96 4. 4. 78.69 0 0.0995 0.0 0.0 84.9 00 0.00 40.0 40.0 89.43 7. Ts Tjω π 04 s + π 0 4 π 04 π 0 4 + 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 / + 0.0 0.995 V/V φ tan 0. 5.7 Peak amplitude of output sinusoid 0.995 V Phase of output relative to that of input 5.7. b f 0 khz ω π 0 4 rad/s T / 0.707 V/V φ tan 45 Peak amplitude of output sinusoid 0.707 V Phase of output relative to that of input 45. c f 00 khz ω π 0 5 rad/s T / + 00 0. V/V φ tan 0 84.3 Peak amplitude of output sinusoid 0. V Phase of output relative to that of input 84.3. d f MHz ω π 0 6 rad/s T / + 0 4 0.0 V/V φ tan 00 89.4 Peak amplitude of output sinusoid 0.0 V Phase of output relative to that of input 89.4. 7.3 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 + 0. 6 / φ ω.5 0.0 rad For ω rad/s: Tjω + 6 / / 0.707 φ tan 35.356 rad
Chapter 7 Note: G 3 db 7.6 For ω 0 rad/s: T jω + 0 6 / 0.00 [ ] φ tan 0 0 3 0 [ ] 980 tan 99 [ ] 980 80 + tan 99 58.5 4.5 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.356 0 sin0t 0.0 sin0t 4.5 7.4 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 0.977 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. 7.3. A max 0 log.05 0.4 db A min 0 log 0.0005 66 db Selectivity factor f s f p 5 4.5 Figure efer to Fig.. π 04 Ts s + π 0 4 Tjω ω + j π 0 4 T / f + 0 4 At f f p 5 khz, we have T / 5 0 3 + 0.894 0 4 A max 0 log 0.894 0.97 db At f f s 0 f p 50 khz, we have T / 50 0 3 + 0.96 0 4 A min 0 log 4.5 db 0.96 7.7 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 0.794 0.794 + ωp ω p 0.765 rad/s
Chapter 7 3 Since A min db, we have 0 log Tjω s db Tjω s 0.5 0.5 + ω s ω s 3.85 rad/s Selectivity factor ω s 3.85 ω p 0.785 5 7.8 See Fig.. Figure shows the location of the five poles in the s plane. The real-axis pole at s 0 4 gives rise to a factor s + 0 4. The pair of conjugate poles p and p are at s 0 4 sin 8 ± j0 4 cos 8 0.309 0 4 ± j0.95 0 4 This pole pair gives rise to a factor s + 0.309 0 4 + j 0.95 0 4 s + 0.309 0 4 j0.95 0 4 s + 0.68 0 4 s + 0 8 The pair of complex conjugate poles p and p are at s 0 4 sin 54 ± j0 4 cos 54 0.809 0 4 ± j0.588 0 4 This pole pair gives rise to a factor Figure s + 0.809 0 4 + j 0.588 0 4 s + 0.809 0 4 j0.588 0 4 7.9 See Fig.. s +.68 0 4 s + 0 8 Thus the denominator polynomial of Ts is Ds s + 0 4 s + 0.68 0 4 s + 0 8 s +.68 0 4 s + 0 8 a The filter is a low-pass of the all-pole type, Ts k s + 0 4 s + 0.68 0 4 s + 0 8 s +.68 0 4 s + 0 8 Figure Since the dc gain is unity, we have 7.0 k 0 0 Ts 0 0 s + 0 4 s + 0.68 0 4 s + 0 8 s +.68 0 4 s + 0 8 b The filter is a high pass, Ts ks 5 s + 0 4 s + 0.68 0 4 s + 0 8 s +.68 0 4 s + 0 8 Since the high-frequency s gain is unity, k must be unity, thus Figure Ts s 5 s + 0 4 s + 0.68 0 4 s + 0 8 s +.68 0 4 s + 0 8
Chapter 7 4 7. Ts ks + 4 s + s + 0.5 j 0.8 s + 0.5 j 0.8 ks + 4 s + s + s + 0.89 T0 4k 0.89 k 0.89 0.5 4 Ts 0.5s + 4 s + s + s + 0.89 7.4 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 +.065 T0 4k.065 k.065 0.656 4 Ts 0.656s + 4 s + 0.5s +.065 T 0.656 ks + 4 s + 0.5 + js + 0.5 j 7.3 ss + 0 6 s + 9 0 6 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 sixth-order 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 third-order 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
Chapter 7 5 which are s ± 8 0.5 ± 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/0 0 0.05 0.3493 Using Eq. 7.5, we have Aω s 0 log[ + ɛ ω s /ω p N ] 0 log[ + 0.3493.7 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 0.45.7 4 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 + 0.45 0.5 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 Nth-order Butterworth filter, we have from Eq. 7. Tjω + ɛ ω ω p N At the 3-dB 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 +.8 4 35.7 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/0 0 0.05 0.3493 The natural modes can be determined by reference to Fig. 7.0a: /N ω p ɛ /5 0 3 0.3493.34 0 3 rad/s [ π π ] p, p sin ± jcos 0 0 0.309 ± j0.95.34 0 3 0.309 ± j0.95 [ ] 3π 3π p, p sin ± jcos 0 0.34 0 3 0.809 ± j0.588 p 3.34 0 3 7.9 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 + 6 8. db For N 4, 0 log + 8 4. db
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 0 4 0.383 ± j0.94 [ ] 3π 3π p, p sin ± jcos 8 8 π 0 4 0.94 ± j0.383 Ts ω0 4 s + 0.765 s + ω0 s +.848 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 + 3 8 38. 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/0 0 0.05 0.3439 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 0.975 rad/s 4 Figure ω cos 3π 4 0.78 rad/s
Chapter 7 7 ω 3 cos 5π 0.434 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 0 0.5 0.90 rad/s 0.63 rad/s 0.3 rad/s 0.434 0.78 0.975 0.3 0.63 0.900 This point also is indicated on the sketch in Fig.. Finally, we note that since this is a 7th-order all-pole 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 7 6 4 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 0 0. 0.5088 Next we use Eq. 7.5 to determine the attenuation at ω s ω p,as Aω s 0 log 0 + 0.5088 30.3 db b We next consider the Chebyshev filter. Its ɛ is the same as that for the Butterworth filter, that is, ɛ 0.5088 The attenuation at ω s ω p can be determined using Eq. 7. as Aω s 0 log[ + 0.5088 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.. 64.9 0 v, rad/s Figure Figure shows a sketch of T for this 7th-order 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[ + 0.3493 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 /0 0.5088 f p 3.4 khz, f s 4 khz f s.76 f p A min 35 db
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[ + 0.5088 cosh N cosh.76] 35 We attempt various values for N as follows: N Aω s 8 8.8 db 9 33.9 db 0 39.0 db Use N 0. Excess attenuation 39 35 4dB 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 0.433 0.5088 and cosh N sinh ɛ π p /ω p 0.433 sin 0.00 0.04 + j0.9978 3π p /ω p 0.433 sin 0 0.065 + j0.900 5π p 3 /ω p 0.433 sin 0 0.03 + j0.743 7π p 4 /ω p 0.433 sin 0 0.77 + j0.4586 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 0.433 sin 0 0.45 + j0.580 +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 3.4 0 3 rad/s we obtain for the five pole pairs: p, p : s + s 0.0448ω p + 0.996ω p p, p : s + s 0.30ω p + 0.844ω p p 3, p 3 : s + s 0.06ω p + 0.505ω p p 4, p 4 : s + s 0.554ω p + 0.66ω p p 5, p 5 : s + s 0.830ω p + 0.0450ω 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 0 0.996 0.844 0.505 0.66 0.0450 B 4.3 0 3 p For DC gain 0.89 we select + ɛ + 0.5088 B 4.3 0 3 p 0.89 3.84 0 3 π 3.4 0 3 0 7.60 0 40 7.5 efer to Fig. 7.3 row a. Input resistance to obtain an input resistance of k, we select k
Chapter 7 9 DC gain 0 0 0 k C π 5 0 3 C 65 pf π 5 0 3 0 03 7.6 efer to Fig. 7.3 row b. H.F. Input resistance 0 k C π 00 C π 00 0 0 3 6.63 nf High-frequency gain 0 k 7.7 efer to the op-amp-c 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 op-amp-c circuit of Fig. 7.3c. Low-frequency input resistance 0 k DC gain 0 k f Z ω Z π πc 00 πc 0 0 3 C 0.6 μf π 06 f P f 0 πc 0 0 3 πc 0 0 3 C.6 nf π 08 Figure shows a sketch of the magnitude of the transfer function. Note that the magnitude of the high-frequency 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 first-order low-pass filter and a first-order high-pass 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
Chapter 7 0 This figure belongs to Problem 7.9. 4 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 low-frequency gain of the low-pass 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 high-pass circuit. The upper 3-dB frequency of the bandpass filter is the 3-dB frequency of the low-pass circuit, that is, 50 0 3 πc C π 50 0 3 00 0 3 3.8 pf Next, we consider the high-pass circuit. The high-frequency f 50 Hz gain of this circuit is 4 / 3. To obtain a gain of 4 V/V, we select 4 00 k 3 00 5 k 4 The lower 3-dB frequency of the bandpass filter 50 Hz is the 3-dB frequency of the high-pass circuit, thus 50 πc 3 C π 50 5 0 3 0.7 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 +
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 5 0 3 rad/s, we can then determine the required value of. Finally, for C 0 nf, the required value for can be found from 08 0 0 9 The results obtained are as follows: φ 30 60 90 0 50 ω/ 0.68 0.577.73 3.73 krad/s 8.66 8.66 5.89.34 k 5.36.55 0 34.60 74.63 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.. 7.3 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.
Chapter 7 7.3 V V 3 C V C Figure shows a circuit composed of the cascade connection of two all-pass circuits of the type shown in Fig. 7.4. 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 60 0 6 3.3 k π 60 0 6 The value of can be selected arbitrarily, say 0 k 7.33 efer to Fig. 7.6a. Ts ω 0 s + s + 0 8 s + 5000 s + 0 8 ω max 0 4 9354 rad/s 4 Since the dc gain is unity, we have a o /ω 0 T max 4 6.066 7.34 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 second-order Butterworth filter that just meets the given passband specifications. Figure b shows the second-order 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 second-order Butterworth, we have / 0.707 and ω p rad/s. The dc gain is unity. Ts s + s + Tjω + ω N Tj + 4 A min Aω s 0 log 0 7.3 db b Chebyshev filter: For a second-order 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 ± j0.7769 for this pair of complex-conjugate poles, we have 0.38 0.6436 and ω 0 0.38 + 0.7769 0.707
Chapter 7 3 This figure belongs to Problem 7.34. Figure K Ts s + 0.6436 s + 0.707 From Fig. b we see that the dc gain is / + ɛ 0.707 K 0.707 0.707 0.5 and 0.5 Ts s + 0.6436 s + 0.707 At s j, we have Tj 0.44 0.5 0.707 4 + 0.6436 A min A 0 log 0.44 7.0 db 7.35 efer to Fig. 7.6b. For a maximally flat response, we have / and ω 3dB rad/s If the high-frequency 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 0.707 ± j0.707 7.36 Poles are at 0.5 ± j 3/. 3 0.5 + rad/s 0.5 0.5 High-frequency gain a. Ts s s + s + 7.37 f 0 0 khz π 0 4 rad/s BW f 0 500 Hz 0 Center-frequency gain 0 ω0 0 s Ts ω0 s + s + ω0 π 0 4 s s + sπ 0 3 + π 0 4
Chapter 7 4 Poles are at ± j 4 π 03 ± jπ 0 4 4 0 π 03 [ ± j40 0.9997].57 0 3 ± j39.988 Zeros are at s 0 and s. 7.38 a A second-order bandpass filter with a center-frequency 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 0,000 9000 rad/s The factor can now be found from 3-dB BW 0,000 800 9000 9000 900 4.74 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 9000 7,000 rad/s 3000 Using Eq., we obtain A min Aω s 0 log [ + ] ω s ω s ] 0 log [ + 4.74 9000 3000 3000 9000. db Figure shows a sketch of T. This figure belongs to Problem 7.38, part b. T, db 0 3. 3,000 f s 8,00 9,000 0,000 f p f 0 f p 7,000 f s f, Hz Figure
Chapter 7 5 7.39 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 0/0 0 00 the filter transfer function with unity dc gain is s + ω0 Ts s + s + ω0 s + π 60 s + s π 60 + π 60 7.40 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 5 0 5 C 0 0 3 C 5nF LC.E.D. L /ω 0 C L 0 mh 0 0 5 0 9 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 0.995 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.
Chapter 7 6 b Z,C Z,C + sc + sc Y,C + sc + sc + C + C sc s s + C + C C 0 0 + 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 + 0 0 L L + L There is a transmission zero at 0 dc due to L. 7.46 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 0 6 0 9 707 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 low-pass 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 high-pass filter with / LC, C, and a high-frequency gain of unity, thus C L
Chapter 7 7 This figure belongs to Problem 7.49. 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 center-frequency 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 + ω0 7.50 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]. 0.83 V/V. 7.5 L C 4 3 5 / Selecting 3 5 0 k, we have L C 4 C 4 0 8 C 4 L 0 8 a For L 5 H, we have C 4 5 0 8 F 0.5 μf b For L.5 H, we have C 4 0.05 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
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 4 3 5 /. Other conversion functions are possible and the circuit is known as a Generalized Impedance Converter or GIC. 7.53 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.
Chapter 7 9 This figure belongs to Problem 7.53. A I V s C 4 C 6 3 0 3 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 6 3 5 V where V K and Y 6 6 + sc 6 KV K + sc 4 sc 6 + 3 5 6 K /C 4 C 6 3 5 s + s + C 6 6 C 4 C 6 3 5 which is a low-pass function with / C 4 C 6 3 5 / C 6 6 C 4 3 4 DC gain K which are the same as the data given in Table 7...E.D.
Chapter 7 0 7.55 For a fifth-order 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 03 0 3 0.309 0 0.309.68 The pole pair p, p has 0 3 rad/s and 3π 03 sin 0 3 0.809 0 0.809 0.68 The real-axis pole p 3 is at a s 0 3 rad/s the transfer function Ts is Ts 03 s + 0 0 6 3 0 3 s + s.68 + 06 0 6 s + s 0 3 0.68 + 06 The first-order factor, T s 03 s + 0 3 can be realized by the circuit in Fig. 7.3a with 00 k arbitrary but convenient value. C C 0.0 μf 0 nf 0 3 00 03 The second-order transfer function T s 0 6 s + s 03.68 + 06 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 0 3 0 0 9 6 / C 00 6.8 k K The second-order transfer function T 3 s 0 6 s + s 03 0.68 + 06 can be realized using the circuit in Fig. 7.a with C 4 C 6 C 0 nf practical value 3 5 00 k C 6 0.68 00 6.8 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. 7.56 efer to Fig. 7.e and the LPN entry in Table 7.. Select C 4 C 0 nf 3 5 C.59 k π 0 0 3 0 0 9 6.59 5.9 k ω0 0 C 6 C 0 6.94 nf ω n C 6 C 6.94 3.06 nf K 7.57 f 0 khz Selecting, C 4 C 6 C.0 nf, we obtain 3 5 C π 0 3 0 0 9 79.6 k 6 7.96 59. k C r r 0 k arbitrary
Chapter 7 This figure belongs to Problem 7.55. 00 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 6 7.58 Ts K C 6 + C 6 s + /C 4 C 6 3 5 s + s + C 6 + C 6 6 C 4 C 6 + C 6 3 5 ω n C 4 C 6 3 5 ω 0 C 4 C 6 + C 6 3 5 3 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
Chapter 7 7.59 efer to Fig. 7.f and to the HPN entry in Table 7.. Ts s + /C 4 C 6 3 5 K s + s + + C 6 6 C 4 C 6 3 5 5 ω n C 4 C 6 3 5 ω 0 C 4 C 6 3 ω0 ωn 5 5 where 5 5 5 ω 5 0 5 ω n [ 5 5 / 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 High-frequency gain 7.60 a Ts 0.4508s +.6996 s + 0.794s + s0.786 +.0504 eplacing s by s/0 5, we obtain Ts s 0.4508 +.6996 0 0 s s 0 + 0.794 5 0 + s 0.786 +.0504 0 05 0.4508 0 5 s +.6996 0 0 s + 0.794 0 5 s + s0.786 0 5 +.0504 0 0 b First-order section: T s 0.794 05 s + 0.794 0 5 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 0.794 0 5 0 9 3.7 k For dc gain of unity, we have 3.7 k Second-order LPN section: T s 0.68s +.6996 0 5 s + s0.786 0 5 +.0504 0 0 Selecting, we obtain C 4 C 6 + C 6 C nf and 3 5 then C.0504 05 0 9 9.76 k C 6 C ω0 ω n 9.0504 00 0 0.68 nf.6996 0 0 68 pf C 6 0.68 0.38 nf 38 pf.0504 0.786 3.679 6 3.679 9.76 35.9 k C Finally, K DC gain The complete circuit is shown in Fig. on the next page. 7.6 Bandpass with f 0 khz, 3-dB bandwidth of 50 Hz, thus f 0 BW khz 50 Hz 40 efer to the circuit in Fig. 7.4a. Using C 0 nf
Chapter 7 3 This figure belongs to Problem 7.60, part b. Figure then C π 0 3 0 0 9 7.96 k Select 0 k then f 0 k Select k then 3 80 79 k K 40.975 Center-frequency gain K.975 40 79 7.6 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 all-pass 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 0 5 0 9 Selecting 0 k then f 0k Selecting 0 k then 3 70 k Selecting L H 0 k then B H 4 0 40 k
Chapter 7 4 Now, K 4.75 Flat gain 0 K F H F 0 H K 0 0.75 57. 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 high-frequency 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. 7.64 Using Eq. 7.66 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 worst-case percentage deviation between ω n and is %. 7.65 efer to Fig. 7.6 and Table 7.. Using C 0 nf, then C 0 5 0 0 9 k d 0 0 k Select r 0 k 3 If the dc gain is unity, then HF gain ω n ω 0 0 5 HF gain.3 0 5 0.597 C C high-frequency gain 0 0.597 5.9 nf /ω n HF gain k 7.66 Using Eq. 7.68 with, we have r C s s + 3 C CC C s + s C + C ω z / CC
Chapter 7 5 3 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 7.67 0.4508s +.6996 Ts s + 0.794s + s0.786 +.0504 a eplacing s by s/0 5, we obtain Ts 0.4508 0 5 s +.6996 0 0 s + 0.794 0 5 s + s 0.786 0 5 +.0504 0 0 b First-order section: T s 0.794 05 s + 0.794 0 5 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 0.794 0 5 0 9 3.7 k For a dc gain of unity, we have 3.7 k Second-order LPN section: T s 0.68s +.6996 0 0 s + s0.786 0 5 +.0504 0 0 where the dc gain is unity. efer to Fig. 7.6 and Eq. 7.68. Selecting C nf then C.0504 05 0 9 9.76 k.0504 d 9.76 35.9 k 0.786 Select r 0 k Now, C C high-frequency gain 0.68 0.68 nf 68 pf 3 /ω n HF Gain 9.76.0504.6996 0.68 9.76 k The complete circuit is shown in Fig.. This figure belongs to Problem 7.67. 9.76 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
Chapter 7 6 7.68 efer to Fig. 7.9 and Eqs. 7.75 and 7.76: C C C nf / 3 and 4 /4 4 C / 0 5 4.4 k 0 5 0 9 3 4.4 k 4 7.07 k 7.69 V b V a / 3 3 V b sc + 3 + + C sc 3 4 C 3 V a [ 3 + 4 + sc + C C 3 + sc 3 4 V a V b sc + + C + 3 C sc 3 4 sc + + + C + 3 4 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 negative-feedback path of an ideal infinite-gain 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 3 4 4 + sc Substituting for V x from Eq. gives V b sc + 3 3 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
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 4 + + 0 sc 3 + s C 3 C 4 + sc 3 + sc 4 sc 4 s C 3 C 4 + sc 3 + + 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 Center-frequency gain 8 Using the design equations 7.75 and 7.76, we obtain 3 4 /4 C where π 0 0 3 f 0 BW 0 5 C 5 π 0 4 Selecting C 0 nf we obtain 0 5.9 k π 0 4 0 0 9 C C 0 nf 3 5.9 k 4 4 5.9 59. 00 /C 4 Center-frequency gain 7.73 C + C 3 00 50 V/V 4 3 7.7 3 4 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 4 3 + 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
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 all-pass 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 3 4 3 + sc V x 4 + sc 0 Substituting for V x from Eq. and collecting terms, we obtain s s + s + + 3 C C C C 3 4 This is a high-pass function with a high-frequency 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 / 0 4 3 0 0 9 3 0 8 0 4 4.4 k ω 0 C C 3 4 0 8 0 8 0 8 4.4 0 3 4 4 00 k 7.07 k 4.4 7.75 3 C X 4 C 3 3 V x / 4 sc V x Figure Figure
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 54 0.68 The real-axis pole p 3 is at s π 0 4 rad/s The first second-order section can be realized using the circuit in Fig. 7.34c. The design equations are 7.77 7.80. 0 k C 4 C C 3 C/4 Here,.68, thus C C 3 4.68 0.095C C.68 π 0 4.68 C 5.5 nf π 0 4 0 03 C 3 0.49 nf 49 pf C 4 5.5 nf The second second-order section also can be realized using the circuit in Fig. 7.34c. Here, 0 k C 4 C C 3 C 4 where 0.68. Thus C C 3 4 0.68 0.655C C 0.68 π 0 4 0.68 C.97 nf π 0 4 04 C 3.9 nf C 4.97 nf The first-order section can be realized using the circuit in Fig. 7.3a with 0 k C.59 nf π 0 4 04 The complete circuit is shown in Fig. below. 7.76 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 center-frequency gain. The complementary network in b will have a transfer function This figure belongs to Problem 7.75, part b. Figure
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. 7.77 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
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 high-pass 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 op-amp gain as that of the circuit in Fig. 7.9. These have been derived in Example 7.3 and given by S A 0 S A A 7.80 Using Eqs. 7.78 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 4 + 4 ] / / + / / + S / / / + / If S 0. Similarly, S 0 7.8 From Table 7. we have C4 C 6 3 5 / C 6 6 C 4 3 5 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 6 6 6 6 6 + S C C 6 C 6 S + 6 S C C 4 C 4 S,, 3 4 7.8 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
Chapter 7 3 G m kn I For G m 0.5 ma/v and k n 0.5 ma/v,we have 0.5 0.5I 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. 7.83 G m G m 0 3 0 3 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 00 0.99 k G m 7.84 7.85 For the integrator in Fig. 7.36b, we have G m V sc Unity-gain frequency G m πc 0 0 6 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 low-frequency 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. 7.87 efer to the circuit in Fig. 7.36c and its transfer function in Eq. 7.9, namely G m sc + G m
Chapter 7 33 Pole frequency G m πc 0 0 6 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 second-order bandpass filter with a transfer function given by Eq. 7.93, and V / will be a second-order low-pass filter with a transfer function given by Eq. 7.94. 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
Chapter 7 34 7.90 Using Eqs. 7.95 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, Center-frequency 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 5 0 6 5 0 0.785 ma/v G m G m 0.785 ma/v G m3 G m 0.785 0.57 ma/v 5 G m4 G m Gain 0.785 5 5 0.785 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
Chapter 7 35 7.93 eq T c C f c C For C pf, 00 0 3 C eq 5M 00 0 3 0 For C 5 pf, eq M 00 0 3 5 0 For C 0 pf, eq 500 k 00 0 3 0 0 7.94 Change transferred CV 0 pc For f 0 00 khz, average current is given by 7.96 From Eqs. 7.09 and 7.0, C 3 C 4 T c C π 0 4 6.83 pf 0 0 00 03 From Eq. 7., C 5 C 4 6.83 0.6 pf 50 From Eq. 7.3, Center-frequency gain C 6 C 5 C 6 C 5 0.6 pf 7.97 ω 3dB 0 3 rad/s / and DC gain f c 00 khz, C C C 5pF I AV T pc 00 03 From Eqs. 7.09 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 /00 0 3 7.95 From Eqs. 7.09 and 7.0, C 3 C 4 T c C π 0 4.5 pf From Eq. 7., 500 0 3 0 C 5 C 4.5 0.6 pf 0 From Eq. 7.3, Center-frequency gain C 6 C 5 C 6 C 5 0.6 pf C 3 C 4 T c C 0 3 0.05 pf 00 0 3 5 0 From Eq. 7., C 5 C 4 0.05 / 0.07 pf The dc gain of the low-pass circuit is DC gain C 6 C 4 For DC gain, C 6 C 4 0.05 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 0.5 + 40 5 00.5 pf Total capacitance C + C π + C Miller 00 + 0 + 00.5 30.5 pf
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 0.5 0 6 30.5 0 Total effective parallel resistance 0 k 5k 3.33 k L 3.33 0 3 83 80.3 0 6 0.5 0 6 3-dB BW V be V s r π + s 80.3 06 83 r π g m L V be 40 5 3 V s 66.7V s V s 66.7 V/V 7.99 p L π 0 6 0 0 6 50 5.7 k 967 khz C ω 0 L.53 nf π 0 6 0 0 6 For a 3-dB bandwidth of khz, we have 06 0 83.3 3 which requires a parallel resistance of L π 0 6 0 0 6 83.3 5.3 k the additional parallel resistance required, a, can be determined from a p a 5.7 5.3 a 7.84 k 7.00 f 0 π LC π 36 0 6 0 3 0 838.8 khz Equivalent parallel resistance 3 9k p L 9 0 3 π 838.8 0 3 36 0 6 47.4