Convolution with a normalised function

[TeraHammer]

Im struggling to find proof for this suspicion I have;

Given is a function f(t) and a normalised function h(t), and their convolution;

f(t) * h(t) = g(t)

Is it true that $$\int fdt = \int gdt$$ ?

 

[gerben]

That is not always true, take some asymmetrical h, for example take:

$$f(t) = t$$
$$h(t) = \frac{2}{3\sqrt{\pi}} (t-1)^{2} e^{-t^2}$$

then

$$g(t)=\int_{-\infty}^\infty f(x)h(t-x)\;dx = t +\frac{2}{3}$$

and the primitive of f does not equal the primitive of g.

 

[TeraHammer]

Thanks, well spotted, but I was regarding physical systems, i.e.

$$h(t<0)=0$$

Does it hold here?

 

[gerben]

No in many cases it does not hold, for example take:

$$f(t) = t$$
$$h(t) = \left\{\begin{array}{ll} 0 & \;\;,t<0\\ 1 & \;\;,0 \leq t \leq 1\\ 0 & \;\;,t>1\\ \end{array}\right.$$

then

$$g(t)=\int_{-\infty}^\infty f(x)h(t-x)\;dx = \int_{t-1}^t x\;dx = t -\frac{1}{2}$$

and the primitive of f does not equal the primitive of g.

 

[blue_raver22]

Yes, it is true that the integral of f(t) is equal to the integral of g(t) in this case. This can be proven by using the definition of convolution, which states that the integral of the convolution of two functions is equal to the product of their individual integrals. Since h(t) is a normalised function, its integral is equal to 1. Therefore, we can rewrite the convolution as:

f(t) * h(t) = f(t) * 1 = f(t)

This shows that the integral of f(t) is equal to the integral of g(t). Additionally, since h(t) is normalised, it represents a probability density function, meaning that the integral of g(t) represents the probability of obtaining the value t when convolving f(t) and h(t). This further supports the fact that the integral of g(t) is equal to the integral of f(t).