Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronic Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

Time-invariant problem

arhzz

Newbie level 3
Joined
Apr 16, 2020
Messages
4
Helped
Reputation
0
Reaction score
0
Trophy points
1
Activity points
28
Hey guys so im stuck with determining if the following system is time-invariant.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(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?
 

dick_freebird

Advanced Member level 5
Joined
Mar 4, 2008
Messages
6,682
Helped
1967
Reputation
3,936
Reaction score
1,788
Trophy points
1,393
Location
USA
Activity points
53,596
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.
 

andre_teprom

Super Moderator
Joined
Nov 7, 2006
Messages
9,108
Helped
1131
Reputation
2,280
Reaction score
1,101
Trophy points
1,403
Location
Brazil
Activity points
52,966
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.
 

arhzz

Newbie level 3
Joined
Apr 16, 2020
Messages
4
Helped
Reputation
0
Reaction score
0
Trophy points
1
Activity points
28
Hmmm I didn't think of that. I will try it and see if it helped, thanks for the help
 

albbg

Advanced Member level 4
Joined
Nov 7, 2009
Messages
1,196
Helped
406
Reputation
812
Reaction score
377
Trophy points
1,363
Location
Italy
Activity points
9,200
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
 

arhzz

Newbie level 3
Joined
Apr 16, 2020
Messages
4
Helped
Reputation
0
Reaction score
0
Trophy points
1
Activity points
28
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.
 

andre_teprom

Super Moderator
Joined
Nov 7, 2006
Messages
9,108
Helped
1131
Reputation
2,280
Reaction score
1,101
Trophy points
1,403
Location
Brazil
Activity points
52,966
substituting in you system definition we have:

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

a(t) = 1 / ( sin(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.
 

arhzz

Newbie level 3
Joined
Apr 16, 2020
Messages
4
Helped
Reputation
0
Reaction score
0
Trophy points
1
Activity points
28
Actually, and this is probably on me, im susposed to determine which one of these is it. It doenst say PROVE that its a time-variant or time-invariant function, but it just say check which one it is.
 

Toggle Sidebar

Welcome to EDABoard.com

Sponsor

Sponsor

Design Fast


X
Top