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16th April 2020, 01:31 #1
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Timeinvariant problem
Hey guys so im stuck with determining if the following system is timeinvariant.The system looks as following
y(t) = Re{sin(t)x(t)}+ Im{jcos(t)x∗(t)}
I did all of the steps with the sin(tt0) and jcos(tt0) and also the y2 = x1 (tt0). But i cant seem to be able to finish the analysis. Could anyone help?

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16th April 2020, 02:28 #2
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Re: Timeinvariant problem
By inspection, the y(t) is full of nontrivial func(t)
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.

16th April 2020, 03:45 #3
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Re: Timeinvariant problem
Perhaps you should decompose x(t) and its complex conjugate x*(t) into their arbitrary Real and Imaginary parts a±jb and make the necessary algebraic manipulation to see what comes out.

Part of the world that you live in, You are the part that you're giving ( Renaissance )

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16th April 2020, 20:30 #4
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Re: Timeinvariant problem
Hmmm I didn't think of that. I will try it and see if it helped, thanks for the help

16th April 2020, 20:59 #5
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Re: Timeinvariant problem
I presume you have a complex output, that is (as said by andre_teprom)
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

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16th April 2020, 22:00 #6
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Re: Timeinvariant problem
Ohhhh i see i was susposed to consider the entire system within the formula of a+jb.Okay so now a and b should be x1 right? and in the part y= a(t)*sin(t) that indicates that the number a(t) is conjugated? Not multiplication. So now if i would to insert numbers for t and t0 i should be getting diffrent end results?
Thanks for your help u helped me a lot.

17th April 2020, 02:24 #7
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Re: Timeinvariant problem
substituting in you system definition we have:
y(t) = a(t)*sin(t) + a(t)*cos(t)
a(t) = 1 / ( sin(t) + cos(t) )
Part of the world that you live in, You are the part that you're giving ( Renaissance )

17th April 2020, 02:59 #8
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Re: Timeinvariant problem
Actually, and this is probably on me, im susposed to determine which one of these is it. It doenst say PROVE that its a timevariant or timeinvariant function, but it just say check which one it is.
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