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### Course: Differential equations>Unit 3

Lesson 2: Properties of the Laplace transform

# Laplace transform of the dirac delta function

Figuring out the Laplace Transform of the Dirac Delta Function. Created by Sal Khan.

## Want to join the conversation?

• At the end of the video ( - ) it is stated that:

L{d(t)} = 1
L{d(t-c)} = e^(-cs)

Both d(t) and d(t-c) represent a line of the same height, one line is on 0 and the other is on c. Since both d(t) and d(t-c) are lines of the same height, howcome their area (and therefore their result to the laplace-transform) are different?

Thanks a lot for your help and another great thanks to Khan. Very helpful indeed.
• if c=0, e^(-cs)=1, so those two equations you wrote are the same
Its not that the delta functions have different areas, just that the laplace transforms look different
• At , Sal finds the Laplace transform of the delta function where c is 0. He concludes that the result is 1, but surely it should be 1/2, as only half of the spike is within the range from 0 to infinity.
• Through the magic of calculus, the entire area of 1 is exactly at zero. No part of it is "hanging off" either side of zero.
• Can someone explain me intuitively the difference between phase and time shifting in Laplace , I don't mean the formula , rather meaning behind it ?
• at why is f(c)=1? shouldn't it be just f(0) since c=0? have we defined what f(x) is?
• integrate f(t-c) from infinity to zero, which includes the value of ZERO. If you recall the last video (Dirac delta function), it did mention the value of τ could be infinite small to make infinite high in y-axis. In this case, the integration range is mainly dependent on -τ and τ. Assumed τ infinitely close to 0, then the range can be assumed as zero.
Back to your question. when c=0, 0 is still in the integration domain and therefore the calculated result is still the unit.
• What is the inverse laplace of "s"
• The inverse Laplace transform of F(s)=s is actually 𝛿'(t), the derivative of the Dirac delta function.
In general the inverse Laplace transform of F(s)=s^n is 𝛿^(n), the nth derivative of the Dirac delta function.
This can be verified by examining the Laplace transform of the Dirac delta function (i.e. the 0th derivative of the Dirac delta function) which we know to be 1 =s^0.
• So the inverse Laplace of 1 is the Dirac delta function?
• Why do the limits for the dirac delta function not matter? Sal said doesn't matter 0 to infinity or -infinity to infinity. I don't understand.
• it's because of the definition of the dirac delta function. the function, which is actually a type of general function called a distribution, has the properties that the function has a value of zero everywhere except at zero, where it's "infinite", and any definite integral of this function which includes zero in its bounds will have a value of 1, and zero otherwise.
(1 vote)
• Seems like you are assuming c>0
(1 vote)
• yes, he writes it at