Problem 1 A Butterworth lowpass filter is to be designed having the loss specifications given below. The limits of the the design specifications are shown in the brick-wall characteristic shown in Figure 6.13 on page 169. Determine the following items for each of the specifications I IV below. Do not design filter circuits for this problem. (c) Determine ω 0 such that α(ω p ) = α max and α(ω s ) > α min (see Figure 6.14(a)). Also calculate α(ω s ). (d) Determine ω 0 such that α(ω s ) = α min and α(ω p ) < α max (see Figure 6.14(b)). Also calculate α(ω p ). I 0.5 30 2π1,000 2π2,330 II 0.5 20 2π 1, 000 2π 2, 000 III 1.0 35 2π 1, 000 2π 3, 500 IV 0.5 20 2π 1, 000 2π 1, 725 Solution: The results are summarized in the table below. Specs I II III IV α max (db) 0.5 0.5 1.0 0.5 α min (db) 30 20 35 20 ω p (rad/s) 2π1,000 2π1,000 2π1,000 2π1,000 ω s (rad/s) 2π2,330 2π2,000 2π3,500 2π1,725 (a) n 5.326 4.832 3.756 6.143 n ( ) 6 5 4 7 (b) ψ k ( ) ±15 0 ±22.5 0 ±45 ±36 ±67.5 ±25.71 ±75 ±72 ±51.43 ±77.14 a ± jb 0.966 ± j0.259 1 0.924 ± j0.383 1 0.707 ± j0.707 0.809 ± j0.588 0.383 ± j0.924 0.901 ± j0.434 0.259 ± j0.966 0.309 ± j0.951 0.623 ± j0.782 0.223 ± j0.975 Q k 0.518 0.5 0.541 0.5 0.707 0.618 1.307 0.555 1.932 1.618 0.802 2.247 (c) ω 0 (rad/s) 7487 7754 7439 7302 f 0 (Hz) 1192 1234 1184 1162 α(ω p ) (db) 0.5 0.5 1.0 0.5 α(ω s ) (db) 34.95 21.00 37.66 24.03 (d) ω 0 (rad/s) 8233 7937 8031 7806 f 0 (Hz) 13 1263 1278 1242 α(ω s ) (db) 30 20 35 20 α(ω p ) (db) 0.166 0.401 0.571 0.203 Dr. Vahe Caliskan 1 of 5 Posted: March 3, 2016
Problem 2 A Butterworth lowpass filter is to be designed having the loss specifications given below. Specifications 1 40 2π 1, 200 2π 3, 000 The limits of the design specifications are shown in the brick-wall characteristic shown in Figure 6.13 on page 169. The final design should only use 47kΩ resistors. In order to outline the design procedure, please provide the following items: Solution: Using the expression to determine n, we get [ ] [ α min/ 1 40/ ] 1 n = αmax/ 1 = 1/ 1 2(ω s /ω p ) 2(2π3000/2π1200) = [ 4 1 0.1 1 2(2.5) ] = 5.763 n = 6 Solution: Since n = 6, there are 6 poles (3 complex-conjugate pole pairs) which are separated by 180 /n = 180 /6 = 30. There are no poles on the negative-real axis; therefore, the first pole pair closest to the negative real axis are at ψ 1 = ±15. The rest of the pole pairs are located at ψ 2 = ±45 and ψ 3 = ±75. The location of the poles and the quality factor can be determined from angle ψ. Given the angle ψ k, the poles are located (in rectangular coordinates) at a k ± jb k = cosψ k ± jsinψ k and the corresponding quality factors are Q k = 1/(2cosψ k ). The table shown below summarizes the angles, pole locations and the quality factors. ψ k a k ±jb k Q k ±15 0.966±j0.259 0.518 ±45 0.707±j0.707 0.707 ±75 0.259±j0.966 1.932 (c) Determine ω 0 such that α(ω p ) = α max and α(ω s ) > α min. Solution: We would like the match the attenuation specifications exactly at the edge of the passband and exceed the specifications in the edge of the stopband. In this case, the 3-dB frequency can be computed by ω 0 = ω p [ α max/ 1 ] = 2π1200 1/2n [ 1/ 1 ] = 2π(1343) = 8438rad/s 1/12 (d) Design the filter and simulate the performance. Solution: This filter can be designed by using three quadratic (second-order) Sallen-Key circuit shown in Figure 6.19(b) on page 174. The unscaled circuit is designed to have a 3-dB frequency of 1 rad/s; therefore, frequency scaling factor k f = ω 0 = 8438 should be used to tune to filter to the desired frequency range. Also using a magnitude scaling factor of k m =47000 will result in 47kΩ resistors. The unscaled component values are R 1 =R 2 =1, C 1 =2Q k and C 2 =1/(2Q k ) where Q k is the quality-factor for stage k. The scaled component values are R 1 =R 2 =k m, C 1 =2Q k /(k m k f ) and C 2 =1/(2Q k k m k f ). The following table summarizes the scaled and unscaled component values for each stage of the design. The simulation of the scaled circuit and the attenuation measurements from the simulation (α(ω p ), α(ω 0 ), α(ω s )) are shown in a separate document. Dr. Vahe Caliskan 2 of 5 Posted: March 3, 2016
R 1 =1Ω R 2 =1Ω C 1=2Q C 2 = 1 2Q + Stage Q-factor Unscaled Scaled (k f =8438, k m =47000) k Q k R 1 R 2 C 1 =2Q k C 2 =1/(2Q k ) R 1 R 2 C 1 C 2 1 0.518 1 1 1.036F 0.965F 47kΩ 47kΩ 2.6nF 2.435nF 2 0.707 1 1 1.414F 0.707F 47kΩ 47kΩ 3.566nF 1.783nF 3 1.932 1 1 3.864F 0.259F 47kΩ 47kΩ 9.742nF 652.582pF Dr. Vahe Caliskan 3 of 5 Posted: March 3, 2016
Problem 3 A Butterworth lowpass filter is to be designed having the loss specifications given below. Specifications 0.5 25 2π 2, 200 2π 4, 250 The limits of the design specifications are shown in the brick-wall characteristic shown in Figure 6.13 on page 169.The final design should only use 33nF capacitors. In order to outline the design procedure, please provide the following items: Solution: Using the expression to determine n, we get [ ] [ α min/ 1 25/ ] 1 n = αmax/ 1 = 0.5/ 1 2(ω s /ω p ) 2(2π4250/2π2200) = [ 2.5 1 0.05 1 2 (1.932) ] = 5.966 n = 6 Solution: Since n = 6, there are 6 poles (3 complex-conjugate pole pairs) which are separated by 180 /n = 180 /6 = 30. There are no poles on the negative-real axis; therefore, the first pole pair closest to the negative real axis are at ψ 1 = ±15. The rest of the pole pairs are located at ψ 2 = ±45 and ψ 3 = ±75. The location of the poles and the quality factor can be determined from angle ψ. Given the angle ψ k, the poles are located (in rectangular coordinates) at a k ± jb k = cosψ k ± jsinψ k and the corresponding quality factors are Q k = 1/(2cosψ k ). The table shown below summarizes the angles, pole locations and the quality factors. ψ k a k ±jb k Q k ±15 0.966±j0.259 0.518 ±45 0.707±j0.707 0.707 ±75 0.259±j0.966 1.932 (c) Determine ω 0 such that α(ω s ) = α min and α(ω p ) < α max. Solution: We would like the match the attenuation specifications exactly at the edge of the stopband and fall below the specifications in the edge of the passband. In this case, the 3-dB frequency can be computed by ω 0 = ω s [ α min/ 1 ] = 2π4250 1/2n [ 25/ 1 ] = 2π(2631) = 16533rad/s 1/12 (d) Design the filter and simulate the performance. Solution: This filter can be designed by using three quadratic (second-order) Sallen-Key circuit shown in Figure 6.19(a) on page 174. Unlike the circuit shown in Figure 6.19(b), this circuit has equal capacitor values but does have a dc gain of K = 3 1/Q. Since we want to overall dc gain to be unity, we can make the dc gain of each stage equal to unity. This can be accomplished by replacing the first 1Ω resistor by a voltage divider as shown in the following page. Resistors R a and R b provide an attenuation by a factor of K while keeping the equivalent resistance equal to 1Ω. The unscaled circuit is designed to have a 3-dB frequency of 1 rad/s; therefore, frequency scaling factor k f =ω 0 =16533 should be used to tune to filter to the desired frequency range. Since we want the final capacitor values to be 33nF, we should choose a magnitude scaling factor k m =1/(k f (33nF)) = 1833. The following table summarizes the scaled and unscaled component values for each stage of the design. The simulation of the scaled circuit and the attenuation measurements from the simulation (α(ω p ), α(ω 0 ), α(ω s )) are shown in a separate document. Dr. Vahe Caliskan 4 of 5 Posted: March 3, 2016
R 1 =1Ω R 2 =1Ω C 1 =1F + R a =K C 2 =1F R y =2 1/Q R b = K K 1 R x =1Ω Stage Q-factor Unscaled k Q k K k =3 1/Q k R a =K k R b =K k /(K k 1) R 2 C 1 C 2 R x R y =2 1/Q k 1 0.518 1.068 1.068 15.674 1 1 1 1 0.068 2 0.707 1.586 1.586 2.707 1 1 1 1 0.586 3 1.932 2.482 2.482 1.675 1 1 1 1 1.482 Stage Scaled (k f =16533, k m =1833) k R a R b R 2 C 1 C 2 R x R y 1 1.958kΩ 28.728kΩ 1.833kΩ 33nF 33nF 1.833kΩ 124.908Ω 2 2.907kΩ 4.962kΩ 1.833kΩ 33nF 33nF 1.833kΩ 1.074kΩ 3 4.550kΩ 3.069kΩ 1.833kΩ 33nF 33nF 1.833kΩ 2.717kΩ Dr. Vahe Caliskan 5 of 5 Posted: March 3, 2016