# I can't understand why δ[2n] = δ[n] and δ(2t) = 1/2 * δ(t)

• Jncik
In summary, the dirac delta function is a mathematical concept used to represent an infinitely sharp impulse at a specific point. In continuous time, the integral of the dirac delta function is equal to 1. However, when dealing with a scaled version of the function, such as δ(2t), the integral is no longer equal to 1, but rather half of it. This is because the variable u is just a dummy variable and can be replaced with any other variable, making the integral equal to 2. In discrete time, the summation over the integral is equal to 1, but for a scaled version, such as δ[2n], the values for 2n may not be integers, resulting in a loss
Jncik
where δ is the dirac delta function, I'm using the Greek alphabet on my keyboard

how can I prove this?

for the continuous time, we have that

$$\int_{-\infty}^{+\infty}\delta (t) dt = 1$$

so by having

δ(2t)$$\int_{-\infty}^{+\infty}\delta (2t) dt = 1$$

I use $$2t = u <=> 2dt = du <=> dt = du/2$$

hence $$\int_{-\infty}^{+\infty}\delta (2t) dt = 1 => \frac{1}{2} \int_{-\infty}^{+\infty}\delta (u) du = 1$$

but since u is just a variable, we have that

$$\int_{-\infty}^{+\infty}\delta (u) du = 2$$

in discrete time the summation over the integral of $$(-\infty, +\infty)$$

is 1.

now, what does δ[2n] mean?

## The Attempt at a Solution

what I understand is that, when we have normal functions in discrete time, and for example we know that from -2 to 2 the values are not zero we say that

-2<=2n<=2

hence

-1<=n<=1

and the step now will be not 1, but 0.5, but in discrete time, we can see the values only when n is an integer, hence we will lose some of them

now for dirac delta we know that for n = 0 it is 1. hence we have

0<=2n<=0
0<=n<=0

hence again, for n = 0, it will be not 0...

but I'm not really sure if this logic is correct

I mean, why can't I use the same thing for the continuous time and say that δ(2t) = δ(t)?

Last edited:
Jncik said:
where δ is the dirac delta function, I'm using the Greek alphabet on my keyboard

how can I prove this?

for the continuous time, we have that

$$\int_{-\infty}^{+\infty}\delta (t) dt = 1$$

so by having

δ(2t)

$$\int_{-\infty}^{+\infty}\delta (2t) dt = 1$$

I use $$2t = u <=> 2dt = du <=> dt = du/2$$

hence

$$\int_{-\infty}^{+\infty}\delta (2t) dt = 1 => \frac{1}{2} \int_{-\infty}^{+\infty}\delta (u) du = 1$$

but since u is just a variable, we have that

$$\int_{-\infty}^{+\infty}\delta (u) du = 2$$
...

It is not true that $\displaystyle \int_{-\infty}^{+\infty}\delta (2t) dt = 1 \,.$

But it is true that $\displaystyle \int_{-\infty}^{+\infty}\delta (t) dt =\int_{-\infty}^{+\infty}\delta (u) du = 1 \,.$ The variables u & t are what we call 'Dummy' variables.

Therefore, $\displaystyle \int_{-\infty}^{+\infty}\delta (2t) dt =\frac{1}{2}\int_{-\infty}^{+\infty}\delta (u) du = \frac{1}{2} \,.$

## 1. Why is the Dirac delta function equal to zero for even values of n?

The Dirac delta function is defined as an impulse function that is zero everywhere except at the origin, where it is infinite. When n is an even number, δ[2n] represents two impulses of the same magnitude at two different points on the x-axis (2n and -2n). These two impulses cancel each other out, resulting in a value of zero for δ[2n].

## 2. How does the Dirac delta function behave when multiplied by a constant?

When multiplied by a constant, the Dirac delta function is also multiplied by that constant. This means that δ(at) = a * δ(t), where a is the constant. This can be seen in the example given, where δ(2t) = 2 * δ(t).

## 3. Why is the Dirac delta function defined as infinite at the origin?

The Dirac delta function is defined as infinite at the origin because it represents an impulse or a spike at a single point. This means that at the point of origin, the function has an infinite value, and it is zero everywhere else.

## 4. What is the significance of the Dirac delta function in mathematics and physics?

The Dirac delta function is used to model a wide range of phenomena in mathematics and physics. It is commonly used in signal processing, control theory, and quantum mechanics. Its ability to represent impulses and effectively "concentrate" a function at a single point makes it a powerful tool in these fields.

## 5. How does the Dirac delta function relate to the Kronecker delta function?

The Dirac delta function and the Kronecker delta function are closely related but serve different purposes. The Dirac delta function is a continuous function that represents an impulse at a single point, while the Kronecker delta function is a discrete function that represents a 1 at a specific value and 0 everywhere else. In other words, the Kronecker delta function is the discrete counterpart of the Dirac delta function.

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