Problem 9.36 Design an active lowpass filter with a gain of 4, a corner frequency of1khz,andagainroll-offrateof 60 db/decade.

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1 Problem 9.36 Design an active lowpass filter with a gain of 4, a corner frequency of1khz,andagainroll-offrateof 60 db/decade. Solution: The roll-off rate of 60 db requires a three-stage LP filter, similar in design to that in Fig To secure positive gain, we need an additional fourth stage. Arbitrarily, we choose all resistors of the first three stages to be 10-kΩ resistors, and we realize the overall gain through the last stage. C f C f C f V s 40 kω V o Figure P9.36(a) The transfer function is given by H(ω) V ( o 1 4 V s 1 jω/ω c1 ) 3, with ω c1 1 R f C f. The problem states that the corner frequency of the overall filter, which we will call ω c3,shouldbe ω c3 2π f c 2π 10 3 rad/s. According to Exercise 9.14, Hence, ω c1 ω c and since R f, ω c3 0.51ω c1. 2π C f 1 R f ω c krad/s, 8.12 nf. The spectral response of the magnitude of H(ω) is shown in Fig. P9.36(b).

2 M (db) ω Figure P9.36(b)

3 Problem 9.37 Design an active highpass filter with a gain of 10, a corner frequency of 2 khz, and a gain roll-off rate of 40 db/decade. Solution: To secure a roll-off rate of 40 db/decade we need to use two stages of the circuit in Fig R f1 R f2 R C s s V s R s C s V o The two stages have the same input impedances (R s and C s ). We choose R f1 R s, R f2 100 kω. Consequently, The overall response is: with G 1 R f 1 R s 1, G 2 R f 2 R s 10. H(ω) V ( o jω/ωhp G 1 G 2 V s 1 jω/ω HP 2 jω/ωhp 10, 1 jω/ω HP ω HP 1 R s C s. The problem statement specifies a corner frequency f c 2kHzwithacorresponding angular frequency ω c given by ω c 2π f c 4π 10 3 rad/s. By definition, ω c is the angular frequency at which the magnitude of H(ω) is equal to of its maximum value. Thus, at ω ω c, jωc /ω 2 HP H(ω c ) 10 1 jω c /ω HP 7.07, which leads to with x ω c /ω HP. Solution of the above equation gives x 2 1 x x ) 2

4 Hence and ω HP 1 R s C s 1.55ω c π rad/s, C s 1 1 R s ω HP F 5.1 nf. AplotofM [db] is shown in Fig. P9.37(b) M (db) ω Figure P9.37(b)

5 Problem 9.38 The element values in the circuit of the second-order bandpass filter shown in Fig. P9.38 are: R f1 100 kω, R s1, R f2 100 kω, R s2, C f F, C s F. Generate a spectral plot for the magnitude of H(ω)V o /V s.determinethefrequencylocationsofthemaximum value of M [db] and its half-power points. C f1 Cf1 R f1 Rf1 R s1 R f2 Rs1 C Rf2 s2 Rs2 R C s2 s2 V s V o Figure P9.38: Circuit for Problem Solution: The overall transfer function is given by with H(ω) V out V s G 2 LPG 2 HP G LP R f 1 R s1 10, G HP R f 2 R s2 10, ( 1 1 jω/ω LP ) 2 ( jω/ωhp 1 jω/ω HP ω LP 1 1 R f1 C f krad/s, ω HP 1 1 R s2 C s krad/s Figure P9.38(a) displays the calculated plot of M [db]. ) 2, (1)

6 M (db) ω From the plot we determine that: Figure P9.38(b) ω krad/s ω 1 ( 3dB)94 krad/s, ω 2 ( 3dB)336 krad/s. (M(ω 0 ) db),

7 2. (a) The given system H( s ), has: 7 Zeros: f z 10 [ Hz], f 10 [ ] 1 z Hz Poles: f 10 [ Hz], f 10 [ Hz] p1 p2 The slopes are 20 db dec H(s) [db] *log H(s) H f [Hz] The actual values are calculated from the following formula H s ω ω ω ω 20 log 1 log 1 log 1 log π 10 4π 10 4π 10 4π f f f f 20 log 1 log 1 log 1 log Values can be determined by plugging the frequency into the above formula, e.g. (marked on actual plot below): H ( s j2π 1MHz) 20 log 1 log 1 log 1 log [ db] [ db] [ db] [ db] [ db]

8 Magnitude [db] vs. frequency [Hz]) Actual plot (b) H ( s j2π 10Hz) 3.01[ db] ( 2π 10 ) 39.91[ ] H s j KHz db ( 2π 10 ) 3.01[ ] H s j MHz db 3. (a) Notice that the given input impedance is real, therefore it either consists of resistors only, or resistors in addition to a combination of capacitors and inductors which cancel each other at the center frequency. 20KΩ R' Km R Km 20 1KΩ The center frequency is shifted, therefore 100KHz ω' K f ω K f 20 5KHz

9 The net scaling factors for the components of the circuit are as following R' Km R 20 R 1 1 C' C C Km K f 400 Km L' L L K f In order for the quality factor to remain the same we require ω0' ω0 2π 100KHz 2π 5KHz Q' Q B ' B B ' 500Hz B ' 10KHz ω ' 95 KHz, ω ' 105KHz c1 c2 These values may be applied to different circuits, e.g. the example shown on page 441 in the book or others band-pass filters. (b) Op-amp should be able to operate on high frequencies. Op-amp should be as ideal as possible (High input resistance, low output resistance, high gain). 4. The general form of a transfer function with two zeros and two poles H s ( z )( ) 1 z2 ( s sp )( s sp ) K s s s s 1 2 In the given problem we have the following Zeros: sz 0, s 3 1 z 2 Poles: s 1 4 j, s 1 4j Therefore H s p1 p2 ( ) K s 0 s 3 K s s 3 s 1 4j s 1 4j s 1 4j s 1 4j Applying the given assumption K s s lim H( s) 1 K 1 s s s H s s ( s 3) ( s ( 1 4j) )( s ( 1 4j) ) ( )