Suppose we have a theoretical LTI causal system with zero initial conditions and we apply a sinusoidal input at t=0. If the system is initially at rest, will the output be equal to H(ω)cos(ωt+Arg(H(w))) for every t>0 or t has to tend to infinity to get that output? Given that the output is the sum of the ZS and the ZI response, we should expect it to happen immediately, however I am not sure that is indeed correct. In fact I think that the eigenfunction property for LTI systems holds for infinite time after the sinusoidal input has been applied to the system. Could someone clarify that one for me? Thanks!
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\$\begingroup\$ Nothing happens immediately. \$\endgroup\$– Andy akaJan 17, 2022 at 13:54
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\$\begingroup\$ In practical systems, yes. In a model of system, what impedes us from reaching the steady state immediately, if the transient response is zero? \$\endgroup\$– dimenJan 17, 2022 at 13:56
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\$\begingroup\$ If the transient response is zero then, whatever caused that to be zero is what impedes you. \$\endgroup\$– Andy akaJan 17, 2022 at 13:58
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The system output will be the superposition/linear addition of two solutions, the steady state one which occurs from t=0 to infinity, and the transient one which also starts at t=0, but dies out to infinity.
In order to see only the steady state output, the transient solution has to be zero.
If we take our system as a parallel RL, then zero initial conditions means zero current in the L. If we apply a cosine voltage, then the initial transient is zero, and the system settles straight into its steady state. If we apply a sine voltage, then the initial transient dies out with an L/R time constant. This is why 'zero voltage switching' is the worst possible way to apply AC power to a transformer.
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\$\begingroup\$ Another possible system with zero transient is the sum of a lowpass, 1/(s+1), with a highpass, s/(s+1), which sum to 1. \$\endgroup\$ Jan 17, 2022 at 17:14