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acw260
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Homework Statement
An accelerometer is connected via a length of coaxial cable to an amplifier. The arrangement is modeled by:
(a) a Norton generator in parallel with a capacitor (CP) representing the piezoelectric crystal within the accelerometer;
(b) a lumped capacitor (CC) representing the coaxial cable; and
(c) a load (RL) representing the amplifier.
See attached for circuit diagram.
The current generated by the accelerometer (iP) is proportional to the rate of displacement of the piezoelectric crystal. Hence iP = K dx/dt. In laplace form iP(s) = KsΔx(s).
(1) Derive an expression for the laplace transfer function T(s) = ΔvL(s) / ΔiP(s).(2) Express ΔvL as a function of time (i.e. the transient response of the voltage ΔvL ) when iP is subject to a step change.(3) Using the values given in TABLE A, estimate the time taken for the voltage vL to reach its steady state value if the current iP is subject to a step change of 2 nA.
CP=1400 pF
CC=250 pF
RL=5 MΩ
TABLE A
Homework Equations
The Attempt at a Solution
Q(1).
I believe (based on internet research) that after redrawing the circuit with an equivalent single parallel capacitor, C, that the answer is
ΔvL(s) / ΔiP(s) = R / 1+sRC
However, I’m struggling with the derivation. In time-domain vL = iPR(1-e^-t/CR). But when I inverse laplace transform R / 1+sRC I get 1/C(1-e^-t/CR). I don’t understand how I can replace R with 1/C in the time-domain. Can anyone help?Q(2).
vL(t) = iPR(1-e^-t/CR).Q(3).
C = CP+CC = 1400*10^-12 + 250*10^-12 = 1.65*10^-9
Time for vL to reach steady state = 5CR = 5*5*10^6*1.65*10^-9 = 41.25*10^-3 s
However, if this is correct the magnitude of the step current is irrelevant to the steady state time for vL. Since the problem specifies a step current of 2nA I’m wary of dismissing this variable as irrelevant. Am I missing something?
Any help with the above would be greatly appreciated.