Linear Systems help please

[nick.sek]

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

[Psi]

I have no idea, but i see a "j" in it so it probably has imaginary numbers

[nick.sek]

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.

[IanB]

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)).

[IanB]

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.

[IanB]

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.

[Kremmen]

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.

[IanB]

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.

Isn't the magnitude e^-1 rather than unity?

[Kremmen]

I just noted that there is actually a real term inserted in the exponent. That would expand to:

exp(-1) * exp(-jt) etc

In 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.

[IanB]

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.

[Kremmen]

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.

[ejeffrey]

How would you plot this equation for two cycles?

y(t)= e^(-1-jt) + e^(-1+jt)

In 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)