- #1

EvLer

- 458

- 0

Hi everyone,

I have this problem and just need someone to check or correct:

[tex] f(t) = -4e^{-3|t|}(u(t + 3) - u(t-1))[/tex]

where u(t) is step function: u(t) = 1 for t >= 1 and 0 elsewhere;

so, i guess I need to break abs value into 2 cases and have 2 different equations? anyway, here's what I have if someone would be so kind and check my work (at first I applied linearity property and distributed [tex]-4e^{-3|t|}[/tex]:

1. for t > 0:

[tex] L[f(t)] = \frac{-4}{s+3} + \frac{4e^{-(s+3)}}{s+3} [/tex]

2. for t < 0:

[tex] L[f(t)] = \frac{-4}{s-3} + \frac{4e^{-(s-3)}}{s-3} [/tex]

thanks in advance!

ps: i guess one thing I should explain is that by definition of unilateral laplace transform, even though first part of signal starts at -3 we do not consider it, what we are doing is one-sided Laplace transform, so I started integrating from 0

edit: to (hopefully) increase chances that someone looks at this here's the Lapl. trnsf. that are relevant:

[tex] L[u(t-k)] = \frac{e^{-sk}}{s}[/tex]

[tex]L[e^{at}f(t)] = F(s-a)[/tex]

but you probably know this anyway :shy:

I have this problem and just need someone to check or correct:

[tex] f(t) = -4e^{-3|t|}(u(t + 3) - u(t-1))[/tex]

where u(t) is step function: u(t) = 1 for t >= 1 and 0 elsewhere;

so, i guess I need to break abs value into 2 cases and have 2 different equations? anyway, here's what I have if someone would be so kind and check my work (at first I applied linearity property and distributed [tex]-4e^{-3|t|}[/tex]:

1. for t > 0:

[tex] L[f(t)] = \frac{-4}{s+3} + \frac{4e^{-(s+3)}}{s+3} [/tex]

2. for t < 0:

[tex] L[f(t)] = \frac{-4}{s-3} + \frac{4e^{-(s-3)}}{s-3} [/tex]

thanks in advance!

ps: i guess one thing I should explain is that by definition of unilateral laplace transform, even though first part of signal starts at -3 we do not consider it, what we are doing is one-sided Laplace transform, so I started integrating from 0

^{-}the first part of the expression.edit: to (hopefully) increase chances that someone looks at this here's the Lapl. trnsf. that are relevant:

[tex] L[u(t-k)] = \frac{e^{-sk}}{s}[/tex]

[tex]L[e^{at}f(t)] = F(s-a)[/tex]

but you probably know this anyway :shy:

Last edited: