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ELG4135: Tutorial on Chapter 12 Q1 (12.5) A low-pass filter is specified to have Amax = 1 dB and Amin = 10 dB. It is found that its specifications can be just met with a single-time-constant RC circuit having a time constant of 1 s and a dc transmission of unity. What must ω p and ω s of this filter be? What is the selectivity factor? Solution: The frequency response of the low-pass filter can be plotted as T ( jω ) dB 0 -1 -10 ω ωp ωs An R-C low-pass filter can be plotted as R Vi Vo C The voltage transfer function T ( jω ) is then 1 V 1 jωC = T ( jω ) = o = 1 Vi 1 + jωRC R+ jωC Given τ = RC = 1 s, 1 j ωC T ( jω ) = 1 1 + jω T ( jω ) is then The magnitude of 1 | T ( jω ) |= 1+ ω 2 At the pass-band edge: −1 −1 | T ( jω p ) |= 10 20 . (Note: 10 20 is the linear scale of -1 dB loss.) ie. 1 1+ ωp = 10 2 −1 20 => ω p = 0.5088 rad/s At the stop-band edge: | T ( jω s ) |= 10 ie. 1 1 + ωs 2 −10 20 . (Note: 10 = 10 −10 20 is the linear scale of -10 dB loss.) −10 20 => ω s = 3 rad/s =>Selectivity= ωs ωp =5.9 # Q2 (12.11) Analyze the RLC network of Fig. P11.11 to determine its transfer function T ( s ) = Vo ( s ) and hence its Vi ( s ) poles and zeros. (1) 1 Ω 2 H (2) (3) Vi Vo 1 F 1 F 1 Ω Solution: The easiest way to solve the circuit is to use nodal analysis at nodes (1), (2) and (3). At node (3), ΣI = 0 Vo Vo Vo − V1 + + =0, 1 1/ s 2s where V1 is the voltage at node (2). -> V1 = Vo (2 s + 2 s + 1) 2 (a) At node (2), ΣI = 0 V1 − Vi V1 V1 − Vo + + =0, 1 1/ s 2s -> V1 (2s 2 + 2s + 1) = Vo + 2sVi Substitute (a) to (b) Vo (2s 2 + 2s + 1) 2 = Vo + 2sVi => T ( s ) = Vo ( s) 0.5 = 3 Vi ( s ) s + 2s 2 + 2s + 1 =>There is no zero. =>Poles are given by. s 3 + 2s 2 + 2s + 1 = 0 ( s + 1)( s 2 + s + 1) = 0 1 3 s = −1 , s = − ± j 2 2 # (b) Q3 (12.12) Determine the order N of the Butterworth filter for which Amaz = 1 dB, Amin ≥ 20 dB, and the selectivity ratio ωs = 1 .3 . ωp What is the actual value of minimum stopband attenuation realized? If Amin is to be exactly 20 dB, to what value can Amax be reduced? Solution: From equation (12.15) A(ω s ) = 10 log[1 + ε 2 ( ωs 2 N ) ] = Amin ωp A ωs 2N 1 + ε ( ) = 10 10 ωp min 2 Amin ω 10 10 − 1 ( s )2N = ωp ε2 Amin ω 10 10 − 1 log[( s ) 2 N ] = log[ ] ωp ε2 N= 10 log[ Amin 10 −1 ε ωs 2 log[( ) ] ωp 2 ] (a) From equation (11.14), as Amaz = 1 dB, ε = 10 Amax 10 1 10 − 1 = 10 − 1 = 0.5088 Substitute Amin ≥ 20 dB, ε , and the selectivity ratio ωs = 1 .3 ωp into (a) N ≈ 11.3 => Choose N=12 The actual value of stopband attenuation can be calculated using N=12 as A(ω s ) = 10 log[1 + ε 2 ( ωs 2N ) ] = 27.35 dB ωp If Amin is to be exactly 20 dB, from equation (12.15) ε2 = 10 Amin 10 −1 ω ( s )2N ωp = 0.1824 , with Amin = 20 dB and N=12 From equation (11.13), Amax = 10 log(1 + ε 2 ) = 0.73 dB # Q4 (12.14) 3 Find the natural modes of a Butterworth filter with a 1-dB bandwidth of 10 rad/s and N=5. Solution: From equation (11.14), as Amaz = 1 dB, Amax 10 1 10 ε = 10 − 1 = 10 − 1 = 0.5088 3 And ω p = 10 rad/s, N=5, the solutions can be found graphically. (The method is shown in Fig. 11.10) The radius of the circle is P1 = ω 0 e π π π π ± j( + ) 2 2N P2 = ω0 e P3 = ω0 e = −269.96 ± j830.84 π ± j( + + ) 2 2N N ± j (π ) = −706.75 ± j 513.49 = −873.59 π π N π N 1 ω 0 = ω p ( )1 / N = 873.59 ε 2N Q5 (12.22) By cascading a first-order op amp-RC low-pass circuit with a first-order op amp-RC high-pass circuit one can design a wideband bandpass filter. Provide such a design for the case the midband gain is 12 dB and the 3-dB bandwidth extends from 100 Hz to 10 kHz. Select appropriate component values under the constraint that no resistors higher than 100 kohms are to be used, and the input resistance is to be as high as possible. Solution: C1 R4 R1 R2 Vo R3 12 / 20 Gain= 10 C2 = 3.98 ≈ 4 Want Ri=R1 large, so R1=100 kOhm Total gain=Alp*Ahp=4 Alp=-R2/R1 => R2=-Alp*R1 and R2<100 kOhm, so Alp=-1, R2=100 kOhm From Fig. 11.14, R2 C1 = C1 = 1 R2ω0,lp = 1 ω 0,lp , 1 = 0.159 nF 2π (10 ×10 3 )100 ×10 3 Ahp=-4, R4=100 kOhm, R3=25 kOhm From Fig. 11.14, R3C 2 = C1 = 1 R2ω0,lp = 1 ω 0,hp , 1 = 63.7 nF 2π × (100) × 25 ×10 3