# convolution operation

The convolution operation is stated below

and is equivalent to

Now, lets say we have 2 functions namely x(t) and u1(t). If we convolve the x(t) with u1(t) where u1(t) is the unit doublet function, the behavior of the output becomes

x'(t) = x(t) ※ u1(t)

we can seee that everytime we convolve an input function (which is x(t)) with the unit doublet, the output is the derivative of the input function which is x'(t).

Now, according here

The question remains, why is that the integrals above are equated to -g'(t) and not g'(-t)

Reference: http://isites.harvard.edu/fs/docs/icb.topic133182.files/5-singular_func.pdf page 4, number 2

## 1 Answer

The integrals should be equal to $-g'(-t)$, which does not change the next results when setting $t=0$.

$$\int_{-\infty}^{+\infty}x(t-\tau)u_1(\tau)\mathrm{d}\tau=(x*u_1)(t)=x'(t)=-g'(-t)$$

• I have a question. Why did the negative sign pop out of g? – WantIt Jul 4 '13 at 12:14
• @vvavepacket: this is the derivative of a composite function. Let $h:t\mapsto -t$, then we can write $x=g\circ h$, so $x'(t)=h'(t)(g'\circ h)(t)$. Since $h'(t)=-1$, we get the result above. – zuggg Jul 4 '13 at 12:24