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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.

Mathematics
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Is the 'del' symbolic of using the delta function?
Absolutely yes.
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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In my textbook, the delta function is not introduced for 4 more sections.
Do you know extended derivative with delta function?
Mathematician said there is no derivative in uncontinious point but engineers said it has.
@ali110 is one of them.
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)}
Those should be f(t)...
no, there is no information in an alone point for Laplace, but I made derivative to make an information then used the Laplace.
How did you make a derivative of a function that has only a constant value?
L(f(t))=L(1)=1/s if for f(t)=t then L(f(t))=1/s^2
@ali110 is one of them.
mohan gholami in which of them?
If it never equals 't', then do I only have \[\frac{1}{s} +\frac{1}{s} +\frac{1}{s}\] ?
I believe he was saying that you were an engineer.
No, you have two points not two functions!
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
Unfortunately Laplace transform doesn't sense points unless with delta function.
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?
L(f(t))=L(3)=3L(1)=3*1/s=3/s
According to the book, the answer is 1/s. Does this mean that the laplace of t=1 and t=2 equate to 0?
can we take laplace inverse at the end?
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.
So you would have: L{f(t)} = L{1} + L{3} + L{4} -> 1/s + 0 + 0 ?
ok I solve it with integral. int(0 inf) f(t)=1/s
You know the integral change the limit points to continues function and never sense them.
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
I thought in two points 1 and 2 it has jumped so they were two alone points.
And Laplace has problem with the single points.
Ah. Okay, I will see if I can't apply this to the rest of my problems. Thank you very much.
You're welcome.
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)
But at this point I would have needed to use the integral approach since we have not reached the delta function?
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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.
But @ali110 there is no unit step and delta function! I guess Openhiem is the best refrence. Isn't it?
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
L{f(t)} = L{1} + L{3} + L{4} -> 1/s + 0 + 0=1/s
@eSpeX I GUESS
@eSpeX CHECK laplace transform linearity property (in which one to one property)
Okay.
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.
Alright, I'll have to look up limit points then because I have not seen them yet as I recall.
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
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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