Convolution Demo and Visualization

This page can be used as part of a tutorial on the convolution of two signals. It lets the user visualize and calculate how the convolution of two functions is determined - this is ofen refered to as graphical convoluiton.

The tool consists of three graphs.

Top graph: Two functions, h(t) (dashed red line) and f(t) (solid blue line) are plotted in the topmost graph. As you choose new functions, these graphs will be updated.

Middle graph: The middle graph shows three separate functions and has an independent variable (i.e., x-axis) of λ. This is important - in this graph λ varies and t is constant. Shown are

The variable λ does not appear in the final convolution, it is merely a dummy variable used in the convolution integral (see below).

Bottom graph: The bottom graph shows y(t), the convolution of h(t) and f(t), as well as the value of "t" specified in the middle graph (you can change the value of t by clicking and dragging within the middle or bottome graph).

\[y(t)=\int_{-\infty}^{+\infty}{h(t-\lambda)\cdot f(\lambda)\cdot d\lambda}=\int_{0^-}^t{h(t-\lambda)\cdot f(\lambda)\cdot d\lambda}\]
Click here to see why the two integrals are equal despite the different limits of integration.

The constant value of t from the middle graph is indicated by a black dot on the bottom two graphs. You can change the value of t by entering a value into the box below, or you can click and move the mouse horizontally in either of the two lower graphs.

The value of the convolution at this time is y() = .
This value is calculated numerically and may differ slightly from an exact value.

You can select two functions, h(t) and f(t) to be convolved. You can also choose to show the complete solution or just the solution until a specified time. This is useful when learning because you can try to figure out what the convolution looks like to test your understanding.

Select h(t):  
Select f(t):  
Show complete solution?  

The functions used for the current example are (note: all functions are implicitly multiplied by the unit step function, γ(t), so they are equal to 0 for t<0).

h(t)=,
and
f(t)= .

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To help your understanding, you can get an explanation of the convolution of select pairs of h(t) and f(t). Use the drop down below to select values two functions, and an explanation of the shape of the convolution will appear above.


References

© Copyright 2005 to 2019 Erik Cheever    This page may be freely used for educational purposes.

Erik Cheever       Department of Engineering         Swarthmore College