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:

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

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.