Dirac Delta Function: Scaling and Shifting

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The discussion focuses on the scaling and shifting of the Dirac delta function in discrete time. When scaling, the correct representation is 1/a * diracdelta[n/a], which maintains the integral property of the delta function. Specifically, scaling the argument by a factor of 2 results in diracdelta[2n] being equal to 1/2 * diracdelta[n], ensuring the area under the curve remains 1. Therefore, scaling does not simply increase or decrease the height but adjusts the function's width accordingly. Understanding these properties is crucial for correctly applying the Dirac delta function in various mathematical contexts.
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Just have a question about the dirac delta function. I understand how you would write it if you want to shift it but how would you scale it assuming we are using discrete time. Would you write 2*diracdelta[n] or diracdelta[2n]. Also, would that increase it or reduce it by 2 meaning that instead of it being 1 at n=0, it would be 2 instead or would it be 1/2. Does it work the same way as scaling other functions in other words? Thank you!
 
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Remember the defining property of the Dirac delta:

\int_{-\infty}^{\infty} \delta(x)dx = 1

Thus

\int_{-\infty}^{\infty} \delta(ax)dx = \frac{1}{a} \int_{-\infty}^{\infty} \delta(y)dy = \frac{1}{a}

So we can think of multipling the argument by a as being the same thing as dividing the function by a:

\delta(ax) = \frac{1}{a}\delta(x)
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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