Problem 1 Consider the following transfer function H(s) = 2(s10)(s100) (s1)(s1000) (a) Draw the asymptotic magnitude Bode plot for H(s). Solution: The transfer function is not in standard form to sketch the asymptotic Bode plot. With a bit of algebraic manipulation we get H(s) = 2(1s/10)(1s/100) (1s/1)(1s/1000) Using the techniques from the course, the asymptotic Bode plot is sketched and is shown in Figure 1. 6dB 6dB 20dB/dec 20dB/dec 14dB 1 10 100 1000 rad/s rad/s rad/s rad/s Figure 1: Asymptotic Bode plot for the transfer function H(s). (b) Synthesize this transfer function using opamps and first-order RC circuits (discussed in class and also shown in Table 3.4 of the handout). Use as few opamps, resistors and capacitors as possible. For this step, choose all capacitors to equal 1F. Solution: There are a number of ways we can proceed here; however, if we want to directly use the circuits and expressions in Table 3.4 of the handout, it will be easier to work with the original form of the transfer function given as H(s) = 2(s10)(s100) (s1)(s1000) To synthesize this transfer function, we can use the first and third circuit shown in Table 3.4. To see this, first partition the above transfer function into three parts. H(s) = 2 (s10) (s1) (s100) (s1000) = 2 (sz 1) (sp 1 ) }{{} H 1(s) (sz 2 ) = 2 H 1 (s)h 2 (s) (sp 2 ) }{{} H 2(s) Now, the desired transfer function is just a cascade of the individual transfer functions. Since we are using passive circuits for H 1 (s) and H 2 (s), we will have to use an opamp-based buffer (as shown in class) between these two sections to avoid loading problems. Let s take each section and see how we can synthesize it. The first transfer function H 1 (s) has the form H 1 (s) = (s10) (s1) = (sz 1) (sp 1 ) where z 1 = 10rad/s and where p 1 = 1rad/s. Since z 1 > p 1, we will utilize the third circuit from Table 3.4 as shown in Figure 2 noting that there is an extra constant term (denoted as T hi in the handout) that equals Dr. Vahe Caliskan 1 of 7 Posted: February 19, 2013
R 1 = 1/p 1 1/z 1 C 2 = 1 T(s) = T hi (sz 1) (sp 1) R 2 = 1/z 1 z 1 > p 1, T hi = R 2 /(R 1 R 2 ) = p 1 /z 1 Figure 2: The lag lowpass circuit from Table 3.4 of the text. R 2 /(R 1 R 2 ). In terms of poles and zeros, this term is just p 1 /z 1. What this means for our design is that we will have to add another section to our cascade to cancel out this extra term. Solving for the component values, we get R 1 = 1/p 1 1/z 1 = 1/11/10 = 0.9Ω R 2 = 1/z 1 = 0.1Ω C 1 = 1F T hi = p 1 /z 1 = 0.1 Using these component values with the circuit above, the synthesized transfer function is 0.1(s 10)/(s 1) which is 0.1H 1 (s); therefore, we will need an additional circuit with a gain 10 to realize H 1 (s). Now let s look at the second transfer function H 2 (s) which has the form H 2 (s) = (s100) (s1000) = (sz 2) (sp 2 ) where z 2 = 100rad/s and where p 2 = 1000rad/s. Since p 2 > z 2, we will utilize the first circuit from Table 3.4 shown in Figure 3. Unlike the third circuit, there is no additional term in front. R 3 = 1/z 2 C 3 = 1 R 4 = 1/(p 2 z 2 ) T(s) = (sz2) (sp 2) p 2 > z 2 Solving for the component values, we get Figure 3: The lead highpass circuit from Table 3.4 of the text. R 3 = 1/z 2 = 1/100= 0.01Ω R 4 = 1/(p 2 z 2 ) = 1/900= 0.00111Ω C 3 = 1F Using these component values with the circuit above, the synthesized transfer function is (s 100)/(s 1000) which is exactly H 2 (s). No additional gain circuits are needed for this section. If we cascade the two sections we have designed so far, we will get the overall transfer function given by 0.1 (s10) (s1) (s100) (s1000) Dr. Vahe Caliskan 2 of 7 Posted: February 19, 2013
Therefore, we also need to include a circuit with a constant gain of 20 to make the overall product equal to the desired transfer function. This can be done with a noninverting amplifier circuit as shown in Figure 4. The gain of the noninverting amplifier is 1R f /R i ; therefore, by choosing the ratio R f /R i = 19 will result in a gain of 20. Since this is the unscaled design, we choose R f = 19Ω and R i = 1Ω. Now the overall R 1 = 0.9Ω C 2 = 1F R 2 = 0.1Ω R 3 = 0.01Ω C 3 = 1F R 4 = 0.00111Ω R f = 19Ω R i = 1Ω Figure 4: The unscaled circuit that synthesizes H(s). transfer function of the three stages is just the product given by ( 0.1 (s10) ) ( ) (s100) (1) (s1) (s1000) (20) which synthesizes the desired H(s). The solution shown in Figure 4 is just fine but we can do a little better. Instead of using the noninverting amplifier at the end we can use it between the two passive sections in place of the buffer. This will use one less opamp than the design of Figure 4 but will use the same noninverting amplifier section as before. The optimal design is shown in Figure 5. In this form, the noninverting section is both a buffer and and amplifier. The buffering is due to the fact the first section sees an infinite input impedance looking into the positive terminal of the opamp and the second section sees zero impedance looking into the output of the opamp. R 1 = 0.9Ω R 3 = 0.01Ω C 2 = 1F R 2 = 0.1Ω R f = 19Ω C 3 = 1F R 4 = 0.00111Ω R i = 1Ω Figure 5: The optimal, unscaled circuit that synthesizes H(s). Now the overall transfer function of the three stages is just the product given by ( 0.1 (s10) ) ( ) (s100) (20) (s1) (s1000) which synthesizes the desired H(s). Dr. Vahe Caliskan 3 of 7 Posted: February 19, 2013
(c) Now use magnitude scaling (with an appropriate value for k m ) to make all capacitors equal to 1µF which is a more practical value. Solution: Now we need to scale the components so that all capacitors are 1µF. We can do this by using magnitude scaling factor k m. With magnitude scaling, the new and old values of the resistors and capacitors are related by R new = k m R old C new = 1 k m C old Since all of the old capacitors are 1F and the new capacitors need to be 1µF, the magnitude scaling factor k m is equal to C old /C new = 10 6. The final magnitude-scaled circuits are shown in Figures 6 and 7. R 1 = 900kΩ C 2 = 1µF R 2 = 100kΩ R 3 = 10kΩ C 3 = 1µF R 4 = 1.11kΩ R f = 19kΩ R i = 1kΩ Figure 6: The magnitude scaled circuit that synthesizes H(s). R 1 = 900kΩ R 3 = 10kΩ C 2 = 1µF R 2 = 100kΩ R f = 19kΩ C 3 = 1µF R 4 = 1.11kΩ R i = 1kΩ Figure 7: The optimal, magnitude scaled circuit that synthesizes H(s). Note that since R f and R i are just providing a gain of 20 and are not setting any pole or zero locations, they can be magnitude scaled independently. Thus, any practical resistor combination that gives R f /R i = 19 is acceptable. Dr. Vahe Caliskan 4 of 7 Posted: February 19, 2013
(d) Use a circuit simulator of your choice to verify that the transfer function of the circuit matches what you derived in (a). Solution: Either circuit (Figure 6 or 7) that was designed in part (c) can be simulated to verify the validity of the design procedure. Each circuit will produce the same transfer function; however, the design of Figure 7 usesonelessopamptoimplement thedesiredtransferfunctionsowewill useittoverifyourdesignprocedure. Figure 8 shows the LTspice schematic which implements the cascade design of Figure 7. The circuit schematic is setup to perform an ac sweep from 0.01 rad/s to 100 krad/s. Figure 9 shows the results of the small-signal ac simulation which is a magnitude Bode plot for the transfer function. Figure 9 plots 20log H(jω) (db) vs. ω(rad/s) and has the features that we predicted by our asymptotic Bode plot from part (a). You will notice that the Bode plot of Figure 9 deviates from our straight-line approximations by 3dB at the corner frequencies. Furthermore, the bottom part of the curve never gets to 14dB. These deviations are expected and are just the inherent shortcomings of the asymptotic Bode plot. Figure 8: LTspice schematic of the filter. Dr. Vahe Caliskan 5 of 7 Posted: February 19, 2013
Figure 9: Simulation results from LTspice showing 20 log H(jω) (db) vs. ω(rad/s). The asymptotic Bode plot derived in part (a) is shown in blue. (e) Use frequency scaling on the circuit of part (c) to move the frequency response up by three decades. Using a simulator verify that the frequency response is indeed three decades higher than that of part (c). Solution: If we frequency scale up by a factor of three decades, we need k f = 1000. Frequency scaling will only change the values of the capacitors leaving the resistor values unchanged. Starting from the old values in part (c), the new component values are given by R new = R old C new = 1 C old = 1 1µF = 1nF k f 1000 So to move up the response by three decades, we need to make all capacitors equal 1nF. Figure 10 shows the LTspice schematic which implements the frequency cascade design. The circuit schematic is setup to perform an ac sweep from 10 rad/s to 100 Mrad/s. Figure 11 shows the results of the small-signal ac simulation which is a magnitude Bode plot for the transfer function. Dr. Vahe Caliskan 6 of 7 Posted: February 19, 2013
Figure 10: LTspice schematic of the frequency scaled filter. Figure 11: Simulation results from LTspice showing 20 log H(jω) (db) vs. ω(rad/s). The asymptotic Bode plot is shown in blue. Dr. Vahe Caliskan 7 of 7 Posted: February 19, 2013