At this point we could solve the set of three equations directly to determine

At this point we could solve the set of three

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At this point, we could solve the set of three equations directly to determine the transfer function by eliminating variables and . However, a shorter solution can be found by noticing that the above circuit is comprised of two well-understood circuits; a non-inverting amplifier and a simple difference amplifier. Using this knowledge, we know from the difference amplifier (as connected above) that: , - 𝑉 ?? 𝑉 ?? 𝑉 ? 𝑉 ?
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63/92 We know from the non-inverting amplifier that is related to by: [ ] [ . / ] [ . / . / ] [ . / ] Combining the two: [( . / ) ] ( ) ( * ( * ( * If you wish to simplify the answer more, you could normalize by , but the above is sufficiently simplified for full marks. It is also the identical answer found when attempting the naive method.
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64/92 Question 29. The Bode magnitude and phase plots for the question are as follows: | ( )| ( ) 0.1 -100 +40 0 +0 1 -60 +40 0 -45 10 -20 +20 -34 -90 56 -5 +0 -146 -90 1000 -5 -20 -214 -45 10000 -25 -20 -270 +0 100000 -45 -270 Note that some rounding is allowed, as it would be difficult to find the exact location of some frequencies, such as . -120 -100 -80 -60 -40 -20 0 0.1 1 10 100 1000 10000 100000 Magnitude, |H(j ω )|, Decibels (dB) Frequency, ω (rad/s) -315 -270 -225 -180 -135 -90 -45 0 0.1 1 10 100 1000 10000 100000 Phase Shift (°)
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65/92 Question 30. Part(a) Note: The following response is very detailed. You can solve this purely using the two basic R-C and R-L circuits taught in class. However, we recommend you read this through, as it gives insight into more refined solutions. To begin this question, we first need to reverse the Bode plot. Tracking the changes in the ROC over the graph gives us the locations of the poles and zeros: Zero initial ROC No poles or zeros at origin ROC goes from 0 to positive ROC at At least one zero at : ( ) ( * By solving for the ROC as above, we determine that there is one zero at ROC goes from +20 to negative at At least one pole at : ( ) ( * By solving for the ROC above, we need two poles to go from +20 to -20 dB/decade Note that the pole location is approximate, resulting in the slight differences from the ideal +20/-20 dB/decade ROCs; the pole is actually located at about , yielding almost exactly +20 / -20. ROC goes from -20 to zero at Need a zero at Total transfer function: ( ) ( ) * ( )( ) ( ) + Solving for the absolute level of gain, we can pick any point far from all poles/zeros, say : | ( )| | | | ( )( ) ( ) | | |( ) ( ) ( ) ( ) * ( )( ) ( ) + Note that for the level of absolute gain, it is acceptable to assume it is close to a value of one; in fact, if you solve the above with the most correct pole location ( ), you actually get
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66/92 . If you carry through , that is fine too, so long as you correctly implement this gain in circuit. Circuit Since the two poles present are identical, the matching is trivial. Dividing into PZG blocks: Block 1 Block 2 Block 1 Note: The following table shows the results for all combinations of series/parallel R + L and R+C; unacceptable sets of values are highlighted in grey. Component Values Added Gain Stuck, cannot set , but need to do so to fix N/A Makes two poles + zero at origin N/A Makes pole + zero + additional pole at origin N/A Makes two poles only N/A Makes two zeros + pole at origin N/A N/A Makes two zeros + double pole at origin
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