Laplace Transform help!

L'prof

I have this question and I was just wondering if I'm approaching it correctly? It doesn't really look right!

Use the Second Shifting Theorem L{f(t-a)H(t-a)} = e^(-as)F(s) to find the Laplace Transform of the following function:

1. f(t) =

{cos(t - (pi/2)), t > pi/2

{0, t < pi/2

My Solution:

2. L{cos(t - (pi/2))H(t-(pi/2)} = e^(-as)F(s)

a = pi/2
f(t) = cos(t)
F(s) = s/(s^2 + 1)

L = e^(-pi/2)(s/(s^2 + 1))

The part I'm confused about is going from 1 to 2 highlighted above!

Thanks,
Jay

Michael Collins

You have it perfectly correct.

H(t) is the Heaviside Unit Step function (also denoted as U(t) in other places). So all that's doing there is making sure the function is zero for t < 0, if it wasn't, the usual Laplace Transform wouldn't work as it assumes causal signals i.e. the signals are zero before t = 0.

L'prof

Michael Collins wrote: »

You have it perfectly correct.

H(t) is the Heaviside Unit Step function (also denoted as U(t) in other places). So all that's doing there is making sure the function is zero for t < 0, if it wasn't, the usual Laplace Transform wouldn't work as it assumes causal signals i.e. the signals are zero before t = 0.

Cool, I'm just confused about going from 1 to 2 above. Could you give me some examples of functions and what I would be taking the Laplace transform of?

f(t) =

{cos(t), t > pi/2

{0, t < pi/2

would be: L{cos(t)H(t)} = e^(-as)F(s)?


f(t) =

{t^2, t > pi/2

{0, t < pi/2

would be: L{t^2H(t)} = e^(-as)F(s)?

That's where I'm getting confused!

Michael Collins

OK. So in general the Laplace Transform takes a function of one variable (usually t, for time) and converts it into a function in a different variable (usually s, for complex frequency). So here's one of the more basic examples below:

e^(-at) -> 1 / (s + a)

On the left we have the time version, f(t) and on the right we have the Laplace Transform of this, F(s). (It's assumed here that these functions are identically zero for negative time).

You can get the above transform by applying the Laplace formula:

Int(e^-(st) f(t) dt), [t = 0 -> infinity]

but in general we only do this once, and then keep a list of what the transforms of different functions are.

Another, as you had above, is

cos(at) -> s/(s^2 + a^2).

But now suppose we wanted the Laplace Transform of cos(at - pi)?

So

cos(at - pi) -> ?

We could calculate this using that integral again, or we could be a bit smarter and use the shifting theorem:

f(t-a) -> e^-(as) F(s) (leaving out the Unit Step for the moment since it confuses things)

So then

f(t) = cos(t) -> F(s) = s/(s^2 + 1)

so cos(t - pi) - > e^-(as) F(s) = e^-(pi.s) s/(s^2 + 1)

Another example would be

e^-(t-4) -> ?

we know already that

e^-(at) -> 1/(s + a)

so

e^-t -> 1/(s + 1) (since a = 1 in this example)

So now what's the transform of

e^-(t-4)

We use the above transform and the shifting theorem:

e^-(t-4) -> e^-(as) F(s) = e^(-4s) 1/(s + 1)

So have a go at finding the Laplace Transform of

sin(t-1)

Step 1: Find the transform of sin(t)
Step 2: Apply the shifting theorem

L'prof

Michael Collins wrote: »

OK. So in general the Laplace Transform takes a function of one variable (usually t, for time) and converts it into a function in a different variable (usually s, for complex frequency). So here's one of the more basic examples below:

e^(-at) -> 1 / (s + a)

On the left we have the time version, f(t) and on the right we have the Laplace Transform of this, F(s). (It's assumed here that these functions are identically zero for negative time).

You can get the above transform by applying the Laplace formula:

Int(e^-(st) f(t) dt), [t = 0 -> infinity]

but in general we only do this once, and then keep a list of what the transforms of different functions are.

Another, as you had above, is

cos(at) -> s/(s^2 + a^2).

But now suppose we wanted the Laplace Transform of cos(at - pi)?

So

cos(at - pi) -> ?

We could calculate this using that integral again, or we could be a bit smarter and use the shifting theorem:

f(t-a) -> e^-(as) F(s) (leaving out the Unit Step for the moment since it confuses things)

So then

f(t) = cos(t) -> F(s) = s/(s^2 + 1)

so cos(t - pi) - > e^-(as) F(s) = e^-(pi.s) s/(s^2 + 1)

Another example would be

e^-(t-4) -> ?

we know already that

e^-(at) -> 1/(s + a)

so

e^-t -> 1/(s + 1) (since a = 1 in this example)

So now what's the transform of

e^-(t-4)

We use the above transform and the shifting theorem:

e^-(t-4) -> e^-(as) F(s) = e^(-4s) 1/(s + 1)

So have a go at finding the Laplace Transform of

sin(t-1)

Step 1: Find the transform of sin(t)
Step 2: Apply the shifting theorem

Thanks very much, I think I'm getting the hang of it now...it's not too bad actually!

Michael Collins

Yeh, the maths of it is fairly straightforward - it's the concepts that can take a while to settle in.