Homework Assignment Question State and then explain in 2 3 sentences, the advantage of switched capacitor filters compared to continuous-time active filters. (3 points) Continuous time filters use resistors and capacitors in conjunction with amplifiers. Switchedcapacitor (SC) filters use small capacitors that are rapidly switched between nodes, to simulate large resistances. These SC resistors are then combined with small capacitors and amplifier to form active filters. The switch frequency determines the value of the resistors and pass-band and cutoff frequencies. Thus, it is very easy to make SC filter electrically-tunable. Because largevalue resistors are simulated with small capacitors, the IC real estate requires for SC filter are is small; this is advantageous in ICs. For an engineer the advantage of SC filters include: wide selection of ICs, ease of use, and low cost. Question 2 (Filters) Determine maximum allowable pass-band variation (in db) for an antialiasing filter that precedes a 6-bit A/D converter, assuming one want to use the full resolution of the A/D converter. (6 points) Assume system where V max = V. Every bit in a 6-bit system represents /2 6 V. The filter variation should be less than this, otherwise one will not utilize the full resolution of the A/D/. Thus, the filter variation should be less than 20 log( 2 6 ) = 32 0 6 db
Question 3 Determine the transfer function H(s) for the filter below. Then determine the magnitude of the frequency response, namely H(jω). (8 points) Denote the resistors with R and the capacitors with C. Further, let R ( sc) Z f = R = R + ( sc) + src The transfer function the first stage is H (s) = Z f R = R R + src = + src The stages are identical so the overall transfer function is H(s) = (H ) 3 = ( + src) 3 One way to determine H(jω) is to set s jω in the expression and then determine from the resulting 3 rd order function. A much simpler approach is to determine H (jω) 3 as follows Then H (jω) = + jωrc = + (ωrc) 2 H(jω) = H (jω) 3 = ( + (ωrc) 2 ) 3 2 2
Question 4 A second order filter has poles at s = 2 ± j 3 2. The transmission is zero at jω = ±2 rad/s and is unity at dc (ω = 0). Determine the transfer function. (0 points) The poles give the denominator: H(s) = K N(s) D(s) D(s) = (s p )(s + p 2 ) = s + 2 j 3 2 s + 2 + j 3 2 = s 2 + s + The zeros give the numerator: Thus N(s) = (s z )(s z 2 ) = (s + j2)(s j2) = s 2 + 4 At s = 0, H(s) =, so that K = 4. Thus H(s) = K s2 + 4 s 2 + s + H(s) = s 2 + 4 4 s 2 + s + 3
Question 5 (Filters) Sketch the transmission specifications for a low-pass filter having a passband defined by f khz and a stopband defined by f 2 khz. A max = 0.5 db, A min = 50 db. The maximum transmission is 0 db. Be sure to label every axis, indicate A max, A min, ω c, ω s, hatch the area that meets the specifications, and so on. (5 points) Any filter with transfer function that falls within the hatched are in the figure below, meets the specifications. ω c = 2π(000) rad s ω s = 2π(2000) rad s A max = 0.5 db, A min = 50 db Question 6 Find the order n for a low-pass Butterworth filter with f c = khz and f s = 2 khz. The maximum passband ripple is db and the stopband gain is 40 db. Remember, the Butterworth response is H(jω) = + ε 2 (ω ω c ) 2n A(jω) = 20 log 0 H(jω) = 20 log 0 + ε 2 (ω ω c ) 2n The maximum ripple (A max ) of db occurs at ω = ω c, so that A max = db = 20 log 0 + ε 2 ε = 0.509 Further, at the stopband frequency ω s the attenuation is A min, so that A min = 40 db = 20log 0 + ε 2 (2 ) 2n Solving for n yields n = 7.6. Thus use n = 8 for the order of the filter. (5 points) 4
Question 7 Consider the prototype low-pass active filter below. It has a cutoff frequency of rad/s. Scale the values such that the Ω resistors are 0K, and the cutoff frequency is khz. Draw the scaled filter. (8 points) To keep the gain the same, the -Ω and 0.68-Ω resistors are changes to 0K and 6.8K respectively. Changing the cutoff frequency from rad/s to khz means the time-constants are 2π 0 3 = 6.282 0 3 times shorter. The -Ω resistors will become 0K, so that that -F capacitors will become (6.282 0 3 )(0 0 3 ) = 0.059 µf. The resulting circuit is below. 5
Question 8 (Sensitivity) A sensitivity analysis of the filter below shows that S Q R = 0.5. An engineer designs a filter using this topology such that Q = 00. He calculates R = 5.7K, and then uses the closest % value of 5.8K. Ignoring the other components, what is the resulting filter Q? (6 points) Thus, Thus S Q R = R Q δq Q = 0.5 Q R δr R δq Q = 0.5 δr 5.8 5.7 = 0.5 = 3.9 0 3 R 5.7 δq = ( 3.9 0 3 )Q = ( 3.9 0 3 )(00) = 0.39 Thus the actual Q is 00 0.39 = 99.68 6
Question 9 (Filters) An engineer designs an 8 th order active filter and uses a cascade implementation. The following table summarizes requirements for the four 2 nd order stages. Write down the cascade order. That is, should the order be A, B, C, D, or C, D, B, A, etc.? Explain briefly. (3 points) 2 nd Order Section Normalized f o Q A 0.997 4.20 B 0.85 4.266 C 0.265 0.753 D 0.584.956 Everything else being equal, one should order the filters so that their bandwidth becomes progressively narrower. That is, the sections should be cascaded in order of ascending Q. This avoids loss of dynamic range and filter accuracy due to possible signal clipping. Thus, use this cascade: C, D, B, A However, if there are sections with high Qs, which could amplify internal noise, one could have a different ordering. Section A has a high Q, but probably not high enough to warrant a different order. Question 0 Find the order n for a low-pass Butterworth filter with f c = khz and f s = 2 khz. The maximum passband ripple is db and the stopband gain is 40 db. Remember, the Butterworth response is H(jω) = + ε 2 (ω ω c ) 2n A(jω) = 20 log 0 H(jω) = 20 log 0 + ε 2 (ω ω c ) 2n The maximum ripple (A max ) of db occurs at ω = ω c, so that A max = db = 20 log 0 + ε 2 ε = 0.509 Further, at the stopband frequency ω s the attenuation is A min, so that A min = 40 db = 20log 0 + ε 2 (2 ) 2n Solving for n yields n = 7.6. Thus use n = 8 for the order of the filter. 7 (8 points)
Question (Switched-Capacitors) For the switched-capacitor circuit below, the parameters are C = 30 pf, C 2 = 5 pf, C F = 2 pf. The clock frequency is 00 khz. Determine the lowfrequency gain and the cutoff frequency. (6 points) The switched capacitors C and C 2 function as resistors with values R = f C C = 333.3 kω, and R 2 = f C C 2 = 2 MΩ respectively. At low frequencies C F is an open circuit and the lowfrequency gain is A V = R 2 R = C C 2 = 6 The cutoff frequency is determined by C F and the switched capacitor R 2 f 3dB = = 2πR 2 C F 2π(2 0 6 = 6.63 khz )2 0 2 8
Question 2 (Filters) After designing a narrowband bandpass filter, an engineer measured the frequency response around the resonance frequency by driving the filter with a 60 mv sinusoidal signal and measuring the output voltage. The following table summarizes the results. F (khz).85.9.95 2 2.05 2. 2.5 2.25 2.3 2.35 2.4 Vo (V) 0.43 0.49 0.55 0.6 0.67 0.7 0.77 0.68 0.59 0.54 0.5 Use this information and determine the f L, f H, f 0, Q, and H 0BP (0 points) Normalize the output voltage to V. The 3-dB frequencies f L and f H are where the output voltage drops to 0.707 V. From the table, f L is between.9 and.95 khz. Estimate f L.925 khz. Further, f H is between 2.3 and 2.35 khz. Estimate f H 2.325 khz. F (khz).85.9.95 2 2.05 2. 2.5 2.25 2.3 2.35 2.4 Vo (V) 0.56 0.64 0.7 0.79 0.87 0.92.0 0.89 0.77.70 0.5 The resonance frequency f 0 is the geometric mean of f L and f H f 0 = f L f H =.925 2.325 = 2.2 khz The bandwidth of the filter is f H f L =.925 2.325 = 0.4 khz The filter Q is Q = f 0 BW = 2.2 0.4 = 5.3 Referring to the original table, at the resonance frequency the output voltage is 0.77 V. The input voltage is 60 mv. Thus H 0BP = 700 60 = 2.8 9
Question 3 An engineer measures the 3-dB frequencies of a symmetric bandpass filter as 9.34 khz and 0.2 khz. What is the resonant frequency (f 0 ) and Q of the filter? (3 points) For a symmetric narrowband bandpass filter f 0 = f L f H = 9.34 0.2 = 9.77 khz The bandwidth of the filter is 0.2 9.34 = 0.87 khz Question 4 Find the phase margin for a voltage follower that is driving a 0.0 μf capacitive load. Assume the op-amp has an open loop gain of A o of 00 db, an f T of MHz, and an output resistance of 250 Ω. Assume the op-amp is ideal except for A o, f T and R o. (5 points) Our approach is to find an expression for the loop gain T(s) and then find the frequency where T(jω) =. For the op-amp f B = f T A o = 0 6 0 5 = 0 Hz. Thus Now T(s) = A(s)β(s) = A o + s + sr o C = 05 ω + s s + 0 20π 4 0 5 0 5 T(jω) = + ω ω 20π 2 + 4 0 5 2 T(jω) = tan ω 20π ω tan 4 0 5 Set T(jω) = and solve for ω (using a spreadsheet or calculator) to find f = 248 khz. Substitute this in the expression for T(jω) to find T(jω) = 66. Thus, the phase margin is 80 66 = 34 0