My lecturer ask me to answer this question but i can't seem to find any explanation on google after searching for quite some time.
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This question looks more like physics.... – DonAntonio May 28 '14 at 09:48
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Change the tag , Thanks – Alex May 28 '14 at 09:49
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What happens as $t$ gets larger? If you still can't answer then, then maybe look transient up in a dictionary. – May 28 '14 at 09:50
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I think about it . Thanks – Alex May 28 '14 at 09:57
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Setting C=2, R=3 and w=5, I get a sine wave for the steady state, and in engineering terms (process dynamics and control), a first order system for the transient. – Mats Granvik May 28 '14 at 10:00
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I search for more info on it as i never learn anything on process, dynamics and control. Thanks – Alex May 28 '14 at 10:09
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Assuming that your circuit is driven with a sinusoidal input, then the steady-state response will be some form of a sinusoid. However, while the system is still adjusting to the new input (think, a car driving over a pot hole), its response will have certain characteristics (bouncing up and down for a while) that do not last indefinitely. These characteristics are called the transient response, whereas the response characteristics that last indefinitely (for the duration of the steady input) are called the steady-state response.
FundThmCalculus
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I can understand a little about what your trying to say but i cant seem find a answer to the question. – Alex May 28 '14 at 10:28
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Notice that that the steady state component of the voltage will never go to zero as time progresses (lim t-> inf V_steadystate =...). Whereas the transient component will go to zero (lim t-> inf V_transient = 0). Thus, since it decays away, it must be the transient component of the voltage. – FundThmCalculus May 28 '14 at 12:06
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In steady state, it is oscillatory in this case and stays on forever, or until you switch the power off:
It is of the form $ A cos(... t) +B sin(... t) $
Transient means it is going to stay on only for a short time like the exponential decay as in this case:
$C e^{(-...t)} $
(...) is some constant.
Narasimham
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