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.

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- eSpeX

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

- anonymous

Absolutely yes.

- eSpeX

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

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## More answers

- eSpeX

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

- anonymous

Do you know extended derivative with delta function?

- anonymous

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

- anonymous

@ali110 is one of them.

- eSpeX

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)}

- eSpeX

Those should be f(t)...

- anonymous

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

- eSpeX

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

- anonymous

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

- anonymous

@ali110 is one of them.

- anonymous

mohan gholami in which of them?

- eSpeX

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

- eSpeX

I believe he was saying that you were an engineer.

- anonymous

No, you have two points not two functions!

- anonymous

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

- anonymous

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

- eSpeX

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?

- anonymous

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

- eSpeX

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

- anonymous

can we take laplace inverse at the end?

- anonymous

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.

- eSpeX

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

And Laplace has problem with the single points.

- eSpeX

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

- anonymous

You're welcome.

- anonymous

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)

- eSpeX

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

- anonymous

##### 1 Attachment

- eSpeX

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.

- anonymous

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

- anonymous

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

- anonymous

- anonymous

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

- anonymous

@eSpeX I GUESS

- anonymous

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

- eSpeX

Okay.

- anonymous

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.

- eSpeX

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

- KenLJW

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

- KenLJW

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

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