Linear Systems help please - Page 1
Forum section: electronics / beginners

author: nick.sek
author_rank: Regular Contributor
timestamp: May 09, 2012, 04:05:26 am
content: Hey EEVblog family, need your help to get into the swing of things.I've been out of school for four months, for a co-op and my math is rusty, but how would you solve this?How would you plot this equation for two cycles?y(t)= e^(-1-jt) + e^(-1+jt)any input would be helpful, thanks nick

author: Psi
author_rank: Super Contributor
timestamp: May 09, 2012, 04:10:43 am
content: I have no idea, but i see a "j" in it so it probably has imaginary numbers

author: nick.sek
author_rank: Regular Contributor
timestamp: May 09, 2012, 04:13:12 am
content: yes it is imaginary numbers... I know the really fun stuff... the book store at my school is only getting 8 copies and they don't come to Sept. so I ordered from amazon - but I still got to wait a week... it driving me nuts not learning.... lol.

author: IanB
author_rank: Super Contributor
timestamp: May 09, 2012, 04:16:28 am
content: No idea what you do with that in an electrical engineering sense, but in a mathematical sense y is a complex function of time with a real part and an imaginary part. To plot it with time on the horizontal axis you would have to plot the real and imaginary parts separately.Use mathematical identities as follows:e^(a+ib)= (e^a)(e^ib)= (e^a)(cos b + i sin b)Expand your two terms in this fashion and collect real and imaginary parts together. Now you can plot Re(y(t)) and Im(y(t)).

author: IanB
author_rank: Super Contributor
timestamp: May 09, 2012, 04:19:52 am
content: Note you could also plot it in the complex plane with t as a parametric variable, in which case I think you would perhaps get something looking like a spiral.

author: IanB
author_rank: Super Contributor
timestamp: May 09, 2012, 04:22:03 am
content: In polar coordinates you would have a magnitude r and a (phase) angle theta. You could plot each of these against time and the result might have more meaning in an electrical sense.

author: Kremmen
author_rank: Super Contributor
timestamp: May 09, 2012, 04:52:11 am
content: Those are counter-rotating unit vectors in the polar notation (in the Re,Im plane).The vector sum y oscillates between 1,j0 and -1,j0.

author: IanB
author_rank: Super Contributor
timestamp: May 09, 2012, 04:57:56 am
content: Quote from: Kremmen on May 09, 2012, 04:52:11 amIsn't the magnitude e^-1 rather than unity?

author: Kremmen
author_rank: Super Contributor
timestamp: May 09, 2012, 05:13:01 am
content: I just noted that there is actually a real term inserted in the exponent. That would expand to:exp(-1) * exp(-jt) etcIn the first term there is no imaginary part so it would act as a multiplication constant altering the vector length. So they would be shorter that unit vectors but the rotation staus the same.

author: IanB
author_rank: Super Contributor
timestamp: May 09, 2012, 05:26:05 am
content: OK, let's work the whole thing out.We have:y(t) = exp(-1-jt) + exp(-1+jt)This is:exp(-1) exp(-jt) + exp(-1) exp(jt)= exp(-1) [ exp(-jt) + exp(jt) ]= exp(-1) [ cos(-t) + j sin(-t) + cos(t) + j sin(t) ]= exp(-1) [ cos(t) - j sin(t) + cos(t) + j sin(t) ]= exp(-1)[ 2 cos(t) ]= 2 exp(-1) cos(t)Hence y(t) is a real function of t looking like a (co)sine wave of constant magnitude 2e^-1 and period 2 pi.

author: Kremmen
author_rank: Super Contributor
timestamp: May 10, 2012, 06:47:13 am
content: Yep, that opens it nicely.Just to avoid confusion it is maybe worth noting that during the tractation the imaginary terms canceled out. So what remains is a purely real component that oscillates along the Re axis.It is of course the vector sum of the 2 original exponentials. Those describe vectors that rotate in oppsite directions always maintaining identical real parts and imaginary parts with same magnitude but opposite sign. The vector sum of the imaginary parts always cancels. This would be the way an electrician sees it, i guess.So, a cosine wave to be sure but strictly one-dimensional real one.

author: ejeffrey
author_rank: Super Contributor
timestamp: May 10, 2012, 03:11:42 pm
content: Quote from: nick.sek on May 09, 2012, 04:05:26 amIn addition to what everyone else says, double check that the equation is correct. What you have written is somewhat odd notation since the e^-1 is just a multiplicative factor. Nothing wrong with it if that is correct, but you should make sure the equation isn't the following:y(t) = e^((-1-j)t) + e^((-1+j)t)This would be an exponentially decaying cosine wave:y(t) = 2*e^-t*cos(t)