First_Order_Systems

# Thiscanbeaccomplishedbydividing topandbottomby 1 1

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Unformatted text preview: nput voltage divided by the sum of the two generalized impedances: • The output voltage, , is the product of the current and the resistance, : · · • The system parameters can be determined by placing this equation in the standard form for a first‐order system. This can be accomplished by dividing top and bottom by : 1 · 1 • Therefore the steady‐state gain, 1. This can also be determined by noting that in the absence of initial conditions the inductor can be replaced by a short circuit at ∞ . • The time constant, A slightly more complicated example Find the Transfer Function, : • Combine and the capacitor, , into a single equivalent impedance, Note that the two impedances are in parallel. 1 · 1 • Calculate the current flowing around the closed loop: . • The output voltage, : , is equal to the product of the current, · · 1 1 · 1 • The transfer function of the circuit is therefore given by: , and Find the Time Constant, , of the circuit: • The standard form for a first‐order transfer function is: 1 • To express the denominator of the circuit transfer function in standard : form, divide top and bottom by 1 • The time constant, , is therefore given by: Find the Steady‐State Gain, , of the circuit: • The steady‐state gain, , is equal to the ratio of the steady‐state value of the output to the steady‐state value of the input (specifically for a step input): ∞ ∞ • Recall that a step input of amplitude is given by: • The steady‐state value of the input is given by: ∞ • The steady‐state value of the output can be found by applying the final value theorem to : ∞ lim · · · • The steady‐state gain, , is therefore equal to: ∞ ∞ ·...
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## This document was uploaded on 03/06/2014 for the course ENG 5952 at Memorial University.

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