Question

\[\huge \mathcal L \lbrace e^{-at} - e^{-bt} \rbrace\]sounds like a trick question....

Answer 1

Hint: Above expression equals \[L(e^{-at}) - L(e^{-bt}) \]

Answer 2

that's it?

Answer 3

how do i get hte limits? the s> thingies?

Answer 4

Upper bound = infinity
Lower Bound: 0. Nothing about the problem says anything about special conditions for the bounds, so assume that you use the normal ones.

Answer 5

no not that

Answer 6

the s > thingies

Answer 7

....Clarify?

Answer 8

i dont know what it's called...like \[\mathcal L \lbrace \cos kt \rbrace = \frac{s}{s^2 + k^2}; \; \text{for} \; s>0\]
that s > thingy

Answer 9

Oh....wait then its s > 0 like normal. For the most part, its always s > 0 for just one function you're trying to transform. It will be different if, let's say, you have a piecewise function and there are different bounds for each equation.

Answer 10

no..it's not s>0 :/

Answer 11

Wait, ok now I get it. First, do the laplace transform. Then find which region the transform is UNDEFINED or DOESN"T EXIST.

Answer 12

The transform simplifies to \[1/(s+a) - 1/(s+b) = (b-a)/{(s+a)(s+b)}\]

Answer 13

So its undefined when s = a or s = b. So the answer should be:
All real numbers s expect a and b.

Answer 14

my book says s>max(-a, -b)

idk what that means

Answer 15

Oops sorry I mean "except -a and -b". This basically means either -a or -b depending on which is the largest.

Answer 16

what do you call that s> thingy?

Answer 17

"All values of s greater than"

Answer 18

no i meant i was calling it "limits thingy" earlier but you didnt understand it so what is it actually called?

Answer 19

Umm I don't know. I don't think it's a formal word for this.

Answer 20

oh well thanks

Answer 21

\(\mathsf{s}\) is here frequency domain I guess...

Answer 22

*frequency only..