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.
analytically calculate value of convolution at certain point
1
$\begingroup$
analysis
integration
convolution
1 Answers
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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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0If 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