How to do it in MATLAB??

[heat23]

whats the laplace transform of e^(-t) * u(t-tau)
and also e^-(t-tau) * u(t-tau)

thanks 🙂

 

[Mday]

what do you mean by u(t-tau)?

do you mean the function named u of t-tau, where tau is a constant, and t is the variable?

 

[heat23]

yes..
i guess its a shift in time or whatever

 

[heat23]

up

 

[Mday]

i'll do the first for you...

L{n} is the laplace transform of n

s is the variable of laplace

exp(t) is e^t...

--

looking at the table of laplace transforms yields this g(t)*v(t-a) ==> exp(-as)L{g(t+a)}(s), where a is a positive number.

so, for your first problem, exp(-t)*u(t-tau), g(t) is exp(-t), and a is tau.

continuing...

exp(-as) L{exp(-(t+a))} (s)

exp(-as) L{[exp(-t)][exp(-a)]} (s)

since exp(-a) is a constant, you're allowed to factor it out, of the L{}...

exp(-as-a) L{exp(-t)} (s)

continuing... after some minor rearranging... and looking in the tables once again...

exp(-a(s+1)) (1/(s+1)) (s) is your answer.

 

[Mday]

the second one is similar, but instead of g(t) = exp(-t), you have g(t) = exp(-(t-tau))... note you can FACTOR out the constant ;-)

 

[heat23]

so would the 2nd one have 3 terms in the answer?

 

[Mday]

uh depends on what you mean by term

when you do the laplace transform, you can factor out that extra constant, exp(-a) which you can integrate into that exp function...

 

[heat23]

actually for the first one i got
[[exp(-tau)]/s] * (1/(s+1)]

 

[Mday]

write out what i wrote by hand, since it's typed funny... ;-)

 

[heat23]

so exp(-tau) is a constant and the laplace of that is 1/s right?

 

[Mday]

did you use:

L{exp(at) * f(t)} (s) = F(s-a)

 

[dopcombo]

I seriously dun think people should be helping others do their hw...
how's he ever gonna learn then..

 

[Mday]

<< so exp(-tau) is a constant and the laplace of that is 1/s right? >>



uh, not quite... that's if you do a laplace transform of A constant...

by factoring out the constant i am referring to this property of the transform... L{cf(t)} = cL{f(t)}


//edit where c is a constant, and f is the function, t being the variable, L being the transform

 

[heat23]

i dont see that in my chart...
lemme look at my diff eq chart and not my crappy ee one

 

[Mday]

recall that exp(a+b) = exp(a) * exp(b)

 

[flood]

matlab makes things like this so much easier
too bad i never use it 😕

 

[heat23]

so, for your first problem, exp(-t)*u(t-tau), g(t) is exp(-t), and a is tau.

continuing...

exp(-as) L{exp(-(t+a))} (s)


are you using some kind of variable substituion?? cuz from exp(-t) u split it to exp(-as) and exp(-(t+a))

 

[heat23]

I have access to matlab if someone knows the syntax to put in, i would be grateful 🙂

 

[Mday]

what i meant was... the general lapace transformation formula for some expression in the form of

g(t) u(t-a), a>=0

is this...

exp(-as) * L{g(t+a} * (s)

with that general formula...

since YOUR expression was... exp(-t) * u(t-tau)... compared to the general formula... g(t) from the formula is your exp(-t)... since the second part is u(t-tau), with respect to the formula, a is tau...

i also used a as the character representation of tau if you're confused... if so, i apologize for it...

 

[heat23]

HEY THANKS ALOT FOR YOUR HELP!!