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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.

For the second one, things are easier since we can compute the integrals: \begin{align} o'(t_0)&=A\int_{-\infty}^0g'(t)dt=Ag(0). \end{align}

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    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.2012-08-01