# analytically calculate value of convolution at certain point

i'm a computer science student and i'm trying to analytically find the value of the convolution between an ideal step-edge and either a gaussian function or a first order derivative of a gaussian function. In other words, given an ideal step edge with amplitude $A$ and offset $B$: $$i(t)=\left\{ \begin{array}{l l} A+B & \quad \text{if t \ge t_{0}}\\ B & \quad \text{if t \lt t_{0}}\\ \end{array} \right.$$ and the gaussian function and it's first order derivative $$g(t) = \frac{1}{\sigma \sqrt{2\pi}}e^{- \frac{(t - \mu)^2}{2 \sigma^2}}\\ g'(t) = -\frac{t-\mu}{\sigma^3 \sqrt{2\pi}}e^{- \frac{(t - \mu)^2}{2 \sigma^2}}$$ i'd like to calculate the value of both $$o(t) = i(t) \star g(t)\\ o'(t) = i(t) \star g'(t)$$ at time $t_{0}$ ( i.e. $o(t_{0})$ and $o'(t_{0}) )$. I tried to solve the convolution integral but unfortunately i'm not so matematically skilled to do it. Can you help me? Thank you in advance very much.

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We can write $i(t):=B+A\chi_{[t_0,+\infty)}$. Using the properties of convolution, we can see that \begin{align} o(t_0)&=B+A\int_{-\infty}^{+\infty}g(t)\chi_{[t_0,+\infty)}(t_0-t)dt\\ &=B+A\int_{-\infty}^0\frac 1{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)dx. \end{align} The integral can be expressed with erf-function.
thank you very much! however I was looking for something like a more explicit "closed form" formula, i.e. i'd like to calculate what's the value for example in $t_{0}$, then in $t_{0} - \sigma$ or some other points near the edge itself. – Alfatau Aug 1 '12 at 19:07
The erf-function in effect is the gaussian CDF, but it's is still in an integral form, then i'm still unable to find the value at a certain $t=t_{0}$. Moreover, with referring to the second function, i't strange to notice that $Ag(0)$ seems wrong: the correct value should be $B+A/2$ for every zero-mean gaussian function, or am I wrong? However, the problem to solve the convolution integral has not been worked out. Any other idea? Thank you all in advance for your attention. – Alfatau Aug 1 '12 at 19:07
You mean for the second? The integral of the derivative on the real line is $0$. You can get an explicit formula for $o(t)$, but it will involve erf function. Indeed, this functions has been created because we can't express with other functions a primitive of a density of a normal law (for example like we created logarithm because we can't express a primitive of $\frac 1t$ with known functions. – Davide Giraudo Aug 1 '12 at 19:07
I'm sorry: I confused first with second. But i'm still confusing: given for example the first result (i.e. $o(t) = i(t) \star g(t)$, how to get the $B+A/2$ value in $t_{0}$, supposing a zero mean gaussian function? i should solve the erf function integral, but as it involves an $e^{-t^2}$ term, i'm not able to solve it. – Alfatau Aug 1 '12 at 19:14
If it's a zero mean gaussian function, we use the fact that the function is even to get the factor $1/2$. It's more complicated in general. – Davide Giraudo Aug 1 '12 at 20:12