On Jun 30, 10:11*am, Cecil Moore wrote:
On Jun 30, 7:24*am, Keith Dysart wrote:

Or are you saying that a step function does not have a Fourier
transform?

Your example has the step function disappearing during steady-state.
DC steady-state has no Fourier transform since omega equals zero. A
Fourier analysis requires a recurring function of the sum of sinusoids
involving integer multiples of omega*time. Your single step does not
recur so your conclusions are invalid.

Check the a0 coefficient in the Fourier transform. This represents
the DC component of the signal. It is perfectly valid for it to be
the only non-zero coefficient.

Without this, how would you deal with a signal such as
V(t) = 10 + 2 cos(3t)


Alternatively, one can use the standard trick for dealing with
non-repetitive waveforms: choose an arbitrary period. 24 hours
would probably be suitable for these examples and transform from
there. Still, you will have zero frequency component to deal
with, but there will be some at higher frequencies (if you
choose your function to make it so).

....Keith