13.13. Example 4: Discrete-Time Convolution

Given: f[n]={1,2,1}, g[n]={1,1} Find (f ∗g)[n].

In this example, we are given two discrete functions: f[n] = {1, 2, 1} and g[n] = {1, 1}. The goal is to find the convolution of these two functions, denoted as (f ∗ g)[n]. This involves calculating the values of (f ∗ g)[n] for different values of n, which represents the time index in discrete-time systems. The convolution operation combines these two sequences into a new sequence that describes how the input signal f interacts with the filter g over time.

Think of f[n] and g[n] as two different sequences of events, like two different musical melodies played over a set amount of time. The convolution would represent how these two melodies blend together to create a new sound, with notes overlapping according to the timing and strength of each note in the original melodies.

• n=0: f[0]g[0]=1·1=1

For n = 0, we are looking at the first index of the sequences. We compute the convolution by multiplying f[0] with g[0]. Here, f[0] = 1 and g[0] = 1, thus (f ∗ g)[0] equals 1. This is a straightforward case where both sequences have non-zero values at their starting positions.

Imagine you are mixing two paints starting with their first drops. If the first drop of paint from f is red (1) and the first drop of paint from g is blue (1), when you mix them together you simply get a small amount of the mixed color (1) right at the beginning.

• n=1: f[0]g[1]+f[1]g[0]=1·1+2·1=3

For n = 1, we compute the convolution by taking contributions from f[0] with g[1] and f[1] with g[0]. Here, f[0] = 1, g[1] = 1, f[1] = 2, and g[0] = 1. Therefore, we find (f ∗ g)[1] = (1 * 1) + (2 * 1) = 3. This shows how elements at different time indices combine during convolution.

Imagine you're adding ingredients to a recipe. For n=1, you're combining the first ingredient and the second ingredient of two different recipes: one recipe calls for 1 cup of flour (f[0]) and the other for 2 cups of sugar (f[1]). When each aspect from one recipe is combined with another (1 cup of flour with 1 cup of sugar), you end up with a total of 3 units of mixed ingredients (3).

• n=2: f[0]g[2]+f[1]g[1]+f[2]g[0]=0+2·1+1·1=3

For n = 2, the convolution is calculated by considering contributions from multiple indices. Here, g[2] is 0 (as g[n] only has values for n=0 and n=1), so its contribution is zero. Thus, we focus on the non-zero contributions from f[1] and f[2]: (f ∗ g)[2] = (0 * 1) + (2 * 1) + (1 * 1) = 3. This reflects how the convolution integrates overlapping values to create the new output.

Imagine four friends at a gathering, where you’re counting interactions based on an event timeline. For n=2, the interaction between a few friends from two different circles creates a combined score of 3 because not all groups are present at the same time (one group is missing), but those who are present can still connect and result in meaningful interactions, like friendships.

• n=3: f[1]g[2]+f[2]g[1]=0+1·1=1

For n = 3, we compute the values using f[1] and g[2], and f[2] and g[1]. Since g[2] is outside the bounds of g[n], its value is 0, so we have (f ∗ g)[3] = (0 * 1) + (1 * 1) = 1. This final calculation shows how only one term contributes to the output, illustrating how convolution captures the essence of overlapping signals even in sparse cases.

Think about a performance with last-minute guests arriving. For n=3, the combination of someone arriving at just the right time with one solo performer results only in a single collaboration (1) because only they interacted directly, and everyone else was either busy or away!

Hence, (f ∗g)[n]={1,3,3,1}

The final result of our convolution operation, (f ∗ g)[n] = {1, 3, 3, 1}, summarizes how f and g interacted across all time indices. Each number represents the strength or magnitude of the output signal at that specific time based on the blending of the two original signals. It shows how signals contribute over time, providing an important outcome in the context of signal processing.

Visualize a coaching session in sports where each player's performance (made of sequences of points scored over different time intervals) blends into the team's total score. The final score reflects contributions from all players, cumulatively layered together into a remarkable performance score (1, 3, 3, 1), demonstrating not just individual values but the strength of collaborative effort over time.

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