# Laplace and Z Transformations Help

Discussion in 'Math' started by crazyengineer, Jun 7, 2011.

1. ### crazyengineer Thread Starter Member

Dec 29, 2010
156
2
Hello everyone

I need some help understanding the mistakes I made in the following problems

1) In the following Laplace Transformation problem
e^(-2t) [u(t)-u(t-5)]
originally I thought the answer was
[1/(s+2)]-(e^(-5s))/(s+2)
[1/(s+2)](1-e^(-5(s+2)))

2) Can anyone explain why the zero transformation of z/(z-2)^2 is n2^(n-1)u[-n-1] when |z|<1?

2. ### steveb Senior Member

Jul 3, 2008
2,433
469
For this first one, try the following ...

e^(-2t) u(t) - e^(-2t) u(t-5)

e^(-2t) u(t) - e^(-10) e^(-2t+10) u(t-5)

e^(-2t) u(t) - e^(-10) e^(-2(t-5)) u(t-5)

Now take the inverse transform using the time shifting property on the second term.

1/(s+2) - e^(-10) e^(-5s) /(s+2)

Simplify

(1/(s+2))*(1 - e^(-5s-10) )

(1/(s+2))*(1 - e^(-5(s+2)) )

Last edited: Jun 7, 2011
3. ### steveb Senior Member

Jul 3, 2008
2,433
469
I'm not sure what a zero transformation is. However, if I take the z-transform of n2^(n-1)u[-n-1] I get -z/(z-2)^2 with region of convergence |z|<2, which differs from your answer by a negative sign. So, I'm not sure if I'm understanding the question, or if I am, I may be making a mistake in the calculation.

Last edited: Jun 7, 2011
4. ### crazyengineer Thread Starter Member

Dec 29, 2010
156
2
for 1) how did you get e^(-10) e^(-2t+10) in e^(-2t) u(t) - e^(-10) e^(-2t+10) u(t-5)?

for 2) I meant z transformation. The problems came from my schuam outline book on signals and systems

5. ### steveb Senior Member

Jul 3, 2008
2,433
469
So, for the first one, e^(-2t)=e^(0) e^(-2t)=e^(-10+10) e(-2t)=e^(-10) e^(+10) e^(-2t)=e^(-10) e^(-2t+10)

So just play around with exponents via allowed rules.

For the second one, I took the z-transform and got the same answer you posted, but with a minus sign. I can post this if you want, and you can try to figure out if I made a mistake or not.