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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.

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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 aħjb 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

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 )

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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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