SC · LTI and convolution

Computing Convolution

To compute a convolution by hand, split time into intervals at the breakpoints where the overlap changes, and integrate the product on each interval. See piece by piece how the convolution of two rectangular pulses becomes a triangle.

When the interval changes, so does the formula

The top shows the fixed input, the flipped sliding pulse, and the green area of their product. Below is the resulting triangle. Move the slide position t and the active interval and its closed-form change. Watch how the formula switches at the breakpoints t = 0, 1, 2.

input pulseflipped pulse

output y(t) = triangle

slide position t-0.40

t < 0

y = 0

0 ≤ t ≤ 1

y = t

1 ≤ t ≤ 2

y = 2 − t

t > 2

y = 0

Active interval and formula

t < 0   y = 0  =  0.00

The two pulses do not overlap. The integration interval is empty, so y(t) = 0. This is an outer interval, t < 0 or t > 2.

No overlap

Breakpoints split the intervals

A breakpoint is the instant an edge of the flipped, slid pulse crosses an edge of the fixed pulse. Here overlap begins at t = 0, becomes complete at t = 1, and ends at t = 2. Between breakpoints the integration limits are set smoothly by max(0, t−1) and min(1, t).

Integrate on each interval

Both pulses have height 1, so their product is 1 on the overlap. The integral is then just the length of the overlap. On 0 ≤ t ≤ 1 the length is t, so y = t rises; on 1 ≤ t ≤ 2 the length is 2−t, so y = 2−t falls. The two lines meet at t = 1 to form a peak of height 1.

Observey(t) = ∫ x(τ) h(t − τ) dτ

The overlap length is what the integral measures.

Choose1 ≤ t ≤ 2: y = ?

On the second interval the length shrinks to 2−t.

Fill iny(1) = ?

Full overlap at t=1 is the peak.

On your ownwidth = 1 + 1 = ?

Output duration is the sum of input durations.

Pieces join, durations add

The interval formulas agree in value at the breakpoints, so they join without a break — convolution is always continuous. The output’s duration is the sum of the two input durations; here 1 + 1 = 2. If the two pulses had different widths, the result would be a flat-topped trapezoid instead of a triangle, but the procedure of splitting and integrating is the same.

Back to the first screen

The triangle’s three straight pieces each came from one interval’s integral. The rising piece is y = t on 0 ≤ t ≤ 1, the falling piece is y = 2−t on 1 ≤ t ≤ 2, and the zeros at both ends are the outer intervals with no overlap. The triangle’s corners sit exactly at the breakpoints t = 0, 1, 2, because at each of those t a sliding edge crossed a fixed edge. Computing a convolution is, in the end, finding these breakpoints, splitting the intervals, and integrating each one.

To compute a convolution by hand, find the breakpoints where the overlap region changes (a sliding edge crossing a fixed edge), split the time axis into intervals, and integrate the product with the right limits on each. The result is a continuous, piecewise-defined function. Rectangle ∗ rectangle = triangle, and the output’s duration is the sum of the input durations. Unequal widths give a trapezoid.

On to the next unit

Now you can handle convolution both as a picture and as a formula. The next unit, properties of LTI systems, reads a system’s properties straight from the shape of its impulse response h. If h is 0 for t < 0 the system is causal; if h integrates to a finite value it is stable. Systems wired in series or parallel merge into one h through the associative and distributive laws of convolution, and the interval-by-interval integration you just learned is the tool for finding that merged response.