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Frequency Representation	Fourier Series / Transforms	Background Material

Convolution can be thought of as a function “weighted” by another function. The convolution symbol can be represented algebraically by the following formulas for continuous and discrete functions, respectively. Another way to think about convolution in a formulistic way is "flip and shift." To convolve (the verb form of convolution is convolve, not convolute) two functions, we "flip" one of them and slide it across the other one. This will be demonstrated later in the examples.

Just as there are two types of signals, discrete and continuous, there is continuous convolution to go with the continuous signals and discrete convolution to go with discrete signals. The equation for continous convolution is:

The equation for discrete convolution is similar but we replace the integral with a summation:

Convolution abides by some multiplicative rules that we are already familiar with.

Let's try to work an example with 2 continuous signals.

Continuous Example:

Let's convolve the following two continuous functions drawn below using the "flip and shift" method.

We can flip the first function and slide it across the second function. Let's see how this looks pictorally for discrete signals.

First, we need to pick one signal and flip it.

Next, according to our convolution equation, we need to time shift that graph by t. So exactly, what does this mean? This means what originally corresponded to 0 on the graph now corresponds to 't'. We can see that in the leftmost figure, the y = t signal has been flipped across the y-axis and time shifted by an arbitrary value of 't'. We notice that when t is less than 0 (on the blue graph), there are no overlapped regions, thus, the convolution of these two functions is also zero. When t slides to the right and crosses the zero mark, we begin to see overlap in the function and we can perform the integral to solve for the function for when t is greater than 0.

To solve this, we need to draw upon our knowledge of integration-by-parts from calculus I. Let's review that quickly.

After we determine which part of our integral corresponds to u, v, du, and dv, we can integrate and get the solution for the region 0 ≤ t ≤ 1.

When the left side (t-1) of the second function has slid passed zero, we've reached the next region of integration (t > 1). Well, one may ask, what's different between this region and the previous region? The answer lies with the limits of integration. In the previous integration, the region of overlap between the two functions occurred from τ=0 to τ=t (Remember, we are integrating with respect to τ, which means our x-axis is τ not t - This is where 80% of the mistakes with convolution occur!!!). However, in this region of integration, the limits of integration are from τ=t-1 to τ=t because the overlapped region spans the entire triangle. Let's take a look at how the math looks.

Once again, we use integration-by-parts to solve this.

Now, if we put everything together and graph it, we get the solution given below.

Discrete Example:

Let's convolve the following two discrete functions drawn below using the "flip and shift" method.

In the discrete case, we can also flip the first function and slide it across the second function. Let's see how this looks pictorally for discrete signals.

Next, according to our convolution equation, we need to time shift that graph by t. So exactly, what does this mean? This means what originally corresponded to 0 on the graph now corresponds to 't'. Since we originally had data points occurring at τ = -2 and τ = -3 after we flipped the graph, those data points after flipping the graph and time-shifting by 't' will become τ = t-2 (originally τ = -2 after the flip and t = 2 before the flip) and τ = t-3 (originally τ = -3 after the flip and t = 3 before the flip).

Here, in the first graph of the series of graphs, we see that when t-2 has not "slid" past τ=1, there is no overlap in the graphs. Thus, we can say that when t-2 is less than 1 (t-2 < 1), our solution y(t < 3) = 0. In the second graph, we see the first sign of overlap between the two graphs, which occurs when τ = t-2 reaches τ = 1. To determine the value of the solution that occurs at y(t=3), we need to multiply and sum across all of the overlapped points. In this case, the graph sliding from the left has a value of 1 overlapping with the stationary graph with value of 1. Therefore, y(t=3) = 1 (because 1 times 1 is 1). We then moved into the territory where the graph sliding from the left overlaps completely with the stationary graph. This occurs in the region where t-3 ≤ 1 (the left side of the sliding graph overlaps or slides past the left side of the stationary graph) and t-2 ≤ 4 (the right side of the sliding graph overlaps but has not slid past the right side of the stationary graph). Solving the equalities, we arrive at the solution that this region is y(4 ≤ t ≤ 6). The value at y(4 ≤ t ≤ 6) becomes 2*1 + 1*1=3, thus, y(4 ≤ t ≤ 6) = 3. We see in the fourth figure that the sliding graph shifts out at t-3 = 4 or t = 7 so y(t = 7) = 2. And finally, when t-3 ≥ 5 or t ≥ 8, we arrive at the other end where the two graphs cease to overlap once again.

We can sum this solution up in equation form as:

and graphically as:

For a list of applications of convolution, refer to:

http://en.wikipedia.org/wiki/Convolution