Question

Laplace Transform: Evaluate L{f(t)}
\[\begin {align*} f(t) &= 1 ,\ t \ge 0, \quad t \neq 1, \ t \neq 2 \\
&= 3,\ t = 1\\
&= 4,\ t = 2\end {align*} \]
Would appreciate someone explaining how to set this up and evaluate.

Answer 1

Is the 'del' symbolic of using the delta function?

Answer 2

Absolutely yes.

Answer 3

Could you please explain the logic you approach this with and how you handled the intervals where the function equaled a constant?

Answer 4

In my textbook, the delta function is not introduced for 4 more sections.

Answer 5

Do you know extended derivative with delta function?

Answer 6

Mathematician said there is no derivative in uncontinious point but engineers said it has.

Answer 7

@ali110 is one of them.

Answer 8

So far what I know of piecewise functions is that L{f(x)} = L{f_1(x)} + L{f_2(x)} + L{f_3(x)}

Answer 9

Those should be f(t)...

Answer 10

no, there is no information in an alone point for Laplace, but I made derivative to make an information then used the Laplace.

Answer 11

How did you make a derivative of a function that has only a constant value?

Answer 12

L(f(t))=L(1)=1/s
if for f(t)=t then
L(f(t))=1/s^2

Answer 13

@ali110 is one of them.

Answer 14

mohan gholami in which of them?

Answer 15

If it never equals 't', then do I only have
\[\frac{1}{s} +\frac{1}{s} +\frac{1}{s}\] ?

Answer 16

I believe he was saying that you were an engineer.

Answer 17

No, you have two points not two functions!

Answer 18

oh i am an electrical engineering student of 5th semester who got 71 marks out of 100 in laplace transform in his 4th semster:))) @eSpeX

Answer 19

Unfortunately Laplace transform doesn't sense points unless with delta function.

Answer 20

Laplace does not make sense to me on how to handle them, and with respect to this piecewise I do not see how it will be done if we have not been shown the delta function. Is it something (or similar) to the heavyside step function?

Answer 21

L(f(t))=L(3)=3L(1)=3*1/s=3/s

Answer 22

According to the book, the answer is 1/s. Does this mean that the laplace of t=1 and t=2 equate to 0?

Answer 23

can we take laplace inverse at the end?

Answer 24

No the answer is just 1/s because Laplace transform can not sense limit points, and it just follow the infinity points which defined with delta function.

Answer 25

So you would have: L{f(t)} = L{1} + L{3} + L{4} -> 1/s + 0 + 0 ?

Answer 26

ok I solve it with integral.
int(0 inf) f(t)=1/s

Answer 27

You know the integral change the limit points to continues function and never sense them.

Answer 28

1
f'(t)=3del(t-1)-3del(t-1)+4del(t-2)-4del(t-2) and f(0)=1

L(f'(t))=3e^-s-3e^-s+4e^-2s-4e^-2s=sL(f(t))-1
L(f(t))=1/s(0+0+1)=1/s

Answer 29

I thought in two points 1 and 2 it has jumped so they were two alone points.

Answer 30

And Laplace has problem with the single points.

Answer 31

Ah. Okay, I will see if I can't apply this to the rest of my problems. Thank you very much.

Answer 32

You're welcome.

Answer 33

if the points in 1 and 2 was jumped so the solution was :

1
f'(t)=3del(t-1)+4del(t-2) and f(0)=1

L(f'(t))=3e^-s+4e^-2s=sL(f(t))-1
L(f(t))=1/s(3e^-s+4e^-2s+1)

Answer 34

But at this point I would have needed to use the integral approach since we have not reached the delta function?

Answer 35

It appears that all of those examples have a range that the integral is evaluated over. So none of the laplace methods evaluate a point.

Answer 36

But @ali110 there is no unit step and delta function!
I guess Openhiem is the best refrence. Isn't it?

Answer 37

check page 11 and every problem will solve as writer show that for t not equal to 1 And 2 as in above question F=0
and Agha! i love alan V openheim as i take all his video lectures about signals and systems but in our engineerig college we study Indian professor Ghosh sumarjit
check his book on Signal and system and about fourier series more intersesting then oppenheim

Answer 38

@mahmit2012

Answer 39

L{f(t)} = L{1} + L{3} + L{4} -> 1/s + 0 + 0=1/s

Answer 40

@eSpeX I GUESS

Answer 41

@eSpeX CHECK laplace transform linearity property (in which one to one property)

Answer 42

Okay.

Answer 43

Ok. And I should mention that don't use L(3)=0 because it is not true. You can write L*(3)=0 and define L* means Laplace for limit points.

Answer 44

Alright, I'll have to look up limit points then because I have not seen them yet as I recall.

Answer 45

You have to use the unit step function for 2 and 3
3u(t-1)
4u(t-2)
for the first one I'd break it up
u(t-1) - u(t-1minus) + u(t-1 plus) -u(t-2 minus) ect
With these there's direct transformations

Answer 46

In EE the slope of the step function is an indication of bandwidth, if there was infinite bandwidth it would be a unit step