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Vu(t)=v_c(t)+\frac{di(t)}{dt}L+Ri(t) (1) |
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Vu(t)=v_c(t)+\frac{di(t)}{dt}L+Ri(t) \mbox{ (1)} |
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The voltage <math>v_c(t)</math> has two components, a natural responce <math>v_n(t)</math> and a forced reponse <math>v_f(t)</math> such that: |
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v_c(t)=v_f(t)+v_n(t) (3) |
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v_c(t)=v_f(t)+v_n(t)\mbox{ (3)} |
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where <math>v_n(t)</math> is the natural response and <math>v_f(t)</math> is the forcing function.
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when <math>t>0s</math> then <math>u(t)=1</math>: |
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For <math>t>0s</math> <math>u(t)=1</math>: |
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\bigg[\frac{d^2v_n(t)}{dt^2}+\frac{R}{L}\frac{dv_n(t)}{dt}+\frac{1}{LC}v_n(t)\bigg]=0 (4) |
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\bigg[\frac{d^2v_n(t)}{dt^2}+\frac{R}{L}\frac{dv_n(t)}{dt}+\frac{1}{LC}v_n(t)\bigg]=0 \mbox{ (4)} |
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\bigg[\frac{d^2v_f(t)}{dt^2}+\frac{R}{L}\frac{dv_f(t)}{dt}+\frac{1}{LC}v_f(t)\bigg]=\frac{V}{LC} (5) |
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\bigg[\frac{d^2v_f(t)}{dt^2}+\frac{R}{L}\frac{dv_f(t)}{dt}+\frac{1}{LC}v_f(t)\bigg]=\frac{V}{LC} \mbox{ (5)} |
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The natural response and forced solution are solved separately. |
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'''Solve for <math>v_f(t):</math>''' |
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'''Solve for <math>v_f(t):</math>''' |
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v_f=V (5) |
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v_f=V \mbox{ (6)} |
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'''Solve for <math>v_n(t)</math>:''' |
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'''Solve for <math>v_n(t)</math>:''' |
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Figure 1: RCL circuit
When the switch is closed, a voltage step is applied to the RCL circuit. Take the time the switch was closed to be 0s such that the voltage before the switch was closed was 0 volts and the voltage after the switch was closed is a voltage V. This is a step function given by
where V is the magnitude of the step and
for
and zero
otherwise.
To analyse the circuit response using transient analysis, an differential equation which describes the system is formulated. The voltage around the loop is given by:
where
is the voltage across the capacitor,
is the voltage across the inductor and
the voltage across the resistor.
Substituting
into equation 1:
The voltage
has two components, a natural responce
and a forced reponse
such that:
substituting equation 3 into equation 2.
when
then
:
The natural response and forced solution are solved separately.
Solve for
Since
is a polynomial of degree 0, the solution
must be a constant such that:
Substituting into equation 5:
Solve for
:
Let:
Substituting into equation 4 gives:
Therefore
has two solutions
and
where
and
are given by:
The general solution is then given by: