#### arhzz

##### Newbie level 3

*y(t) = Re{sin(t)x(t)}+ Im{jcos(t)x∗(t)}*

I did all of the steps with the sin(t-t0) and jcos(t-t0) and also the y2 = x1 (t-t0). But i cant seem to be able to finish the analysis. Could anyone help?

- Thread starter arhzz
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I did all of the steps with the sin(t-t0) and jcos(t-t0) and also the y2 = x1 (t-t0). But i cant seem to be able to finish the analysis. Could anyone help?

expressions so how can it be time invariant?

I think the equation probably has some typos but:

sin(t) is not time invariant

x(t) might or might not be.

jcos(t) is not time invariant

x (as a constant) is undefined

(t) is obviously not time invariant.

Only if x(t) and x both equal zero, could the larger

equation be time invariant.

x(t) = a(t) + jb(t)

substituting in you system definition we have:

y(t) = a(t)*sin(t) + a(t)*cos(t)

to check if it's time invariant we have to calculate first the output when the intput [that is x(t)] is time shifted by "to". Let's call it yo(t):

yo(t) = a(t+to)*sin(t) + a(t+to)*cos(t)

now we have to calculate the output when the system is time shifted by the same amount "to"

y(t+to) = a(t+to)*sin(t+to) + a(t+to)*cos(t+to)

since yo(t) <> y(t+to) the system in NOT time invariant

Thanks for your help u helped me a lot.

Seems like there is only particular case at which the above function could be time invariant would be if a(t) could elliminate ( sin(t) + cos(t) ) :substituting in you system definition we have:

y(t) = a(t)*sin(t) + a(t)*cos(t)

Which is not valid at some values of t, since it has discontinuity, so perhaps we could assume that y(t) berhaves more like a time-variant function.a(t) = 1 / ( sin(t) + cos(t) )