8092 Part b Remember that transient response for an LTI systemcircuit with an

8092 part b remember that transient response for an

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80/92 Part (b) Remember that transient response for an LTI system/circuit with an applied unit-step input can be roughly predicted by the pole locations. In this case, we have the following component values and transfer function denominator: ( ) ( ) ( ) Given a pair of complex-conjugate poles as the only ones in the denominator, we expect the response to look roughly like the red line below: This is called an under-damped response, as covered in the notes. Note that we do not have increasing oscillation, as all poles have negative real parts (notice that when solved).
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81/92 Question 41. Since the graph range is already given to us, the graph for this function is relatively simple: Calculate starting point using leftmost point, : | ( )| | ( ) . / | No poles or zeros at origin, so magnitude graph continues at ROC = 0 until , when it changes to Graph continues at until , where it changes to ( ) . This ROC continues indefinitely. -80.0 -60.0 -40.0 -20.0 0.0 20.0 0.1 1 10 100 1000 Magnitude Plot
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82/92 For the phase plot, it starts at ; there are no poles or zeros at the origin, and the pure gain term is positive. At , ROC changes from 0 to +45 (zero starts) At , ROC changes from +45 to ( ) (4 poles start) At , ROC changes from to (zero ends) At , ROC changes from to (4 poles end) -270.0 -225.0 -180.0 -135.0 -90.0 -45.0 0.0 45.0 0.1 1 10 100 1000 Phase Plot
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83/92 Question 42. Find the transfer function of the following circuit. We will assume that the lower connection for and is at potential, in which case we have an inverting amplifier. We can use either ( ) , or derive from equations: Equation 1 . / Solving: ( )( ) ( ) ( )( ) This equals if you use the direct method. Either is acceptable on a quiz.
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84/92 Question 43. Find the transfer function of the following circuit. To begin, we assume a virtual ground for the op-amp (i.e., ). We will need three equations to solve this circuit. We label the second unknown node (junction of , , and ) as the unknown . Equation 1 (Non-Inverting Input) Equation 2 (Node ) Equation 3 (Inverting Input) Solving: ( ) ( )
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85/92 Question 44. Find the transfer function ( ) of the following circuit. Hint: Substitute component values immediately for this question to avoid wasting time on solving. To solve this circuit, we must write three equations; one for each op-amp input pin ( ) and one for the unknown node, (junction of , , , and ). Equation 1 (Inverting Input) Equation 2 (Non-Inverting Input) || || Equation 3 (Node ) Solving (this one is quite long, much longer than a quiz problem, especially if solved without substitution). With substitution: ( )
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86/92 Question 45. Find the transfer function, ( ) ( ) ( ) , of the following circuit. To solve, we can first assume that all op-amps have virtual ground, so in this case (all non-inverting inputs tied to ground). Note that we have a set of standard op-amp configurations which can make our solution much simpler; each op-amp creates an inverting amplifier configuration.
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