CONVOLUTION: ACCUMULATION

The convolution integral is visualized as a superposition of impulse responses.

Click the Signal menu button to display options for the signal function $$f(t)$$. Click a function to select it.

Click the Weight menu button to display options for the weight function $$g(t)$$. Click a function to select it.

When a function is selected, the lower graph plane shows the graph of the signal function $$f(t)$$, and the upper plane shows the graph of the convolution $$f(t)*g(t)$$, both in gray.

Click one of the step size checkboxes to select the time step, $$\Delta u$$.

The accumulation of the convolution integral can be controlled either by clicking the [>>] button to animate, or by clicking on the time slider to pick a time value $$u$$. The yellow curve represents the convolution $$f(t)*g(t)$$ up to time $$u$$; the top cyan curve represents the convolution of the weight $$g(t)$$ with a signal which is $$f(t)$$ for $$t \lt u$$ and zero for $$t \gt u$$; it cuts off at time $$t = u$$.

To understand this visual, select $$f(t) = 1 + \cos(bt)$$ and $$g(t) = e^{-at}$$. We have a model of farm runoff into a lake. It varies seasonally, and the rate at which phosphates, say, enter the lake is given by $$f(t)$$. These chemicals are washed out of the lake, as well, at a rate proportional to the amount in the lake. Thus, of every kilogram in the lake at time $$u$$, $$g(t-u)$$ kilograms remains in the lake at time $$t$$, i.e. time $$t - u$$ later. It is assumed that at $$t = 0$$ the phosphate load is zero ("rest initial conditions").

Select stepsize $$\frac{1}{4}$$ and click on the animate key [>>]. An animation shows the weight function $$g(t)$$, in the lower screen in cyan, shifting progressively to the right, multiplied at each stage by the value of the signal for the appropriate value of $$u$$. The representation of the signal turns green as this process proceeds. These blue graphs in the lower window are multiplied by $$\frac{1}{4}$$ (the stepsize) and laid down on top of what has been contributed already, for smaller values of $$u$$.

The upper graph displays the cumulative total amount of phosphate in the lake. The yellow curve shows this accumulation: it is the convolution $$f(t)*g(t)$$. With $$u = 3$$, if you follow the yellow curve and then the top blue curve, you are tracing out the amount of phosphate in the lake resulting from runoff which proceeds at the rate $$f(t)$$ for $$t \lt u$$ but is then reduced to zero for $$t \gt u$$.

Press and hold the mouse button on either plane to see the contribution made in a given time interval of length $$\frac{1}{4}$$.

The values of the constants $$a$$ and $$b$$ in the expressions for $$f(t)$$ and $$g(t)$$ are $$a = \ln(2)$$ and $$b = 2$$.

© 2001 H. Hohn and H. Miller