SC · LTI and convolution

Properties of LTI Systems

The properties of an LTI system are written in the shape of its impulse response h. You read causality from when h begins and stability from whether the total area of |h| is finite, and by moving the decay rate you cross the boundary between stable and unstable yourself.

Does the running area hit a ceiling

The upper figure is the causal impulse response h(t) = eᵃᵗ (t ≥ 0), and the lower figure is the running area of |h| piled up from the left. Lower the decay rate a far enough below 0 and, instead of growing without end, this area stops at the ceiling 1/|a| marked by the green dashed line; the existence of that ceiling is what stability means.

impulse response h(t)=eᵃᵗ

running |h| areaceiling 1/|a|

decay rate a0.40

Total area and verdict

∫ |h| dt = ∞

Since a > 0, h grows without end. The running area of |h| diverges with no ceiling, so ∫|h| = ∞, and this is an unstable system in which even a small input sends the output running away.

Unstable

Causality is when h begins

The impulse arrives at t = 0. A causal system cannot react before that, so its impulse response must be 0 for t < 0; in symbols, h(t) = 0 for t < 0. The h on this screen is alive only for t ≥ 0, so whatever a is, it is always causal. Just check whether the graph of h stays quiet to the left of 0 and you read causality at once.

Stability is the total area of |h|

BIBO stability means that a bounded input always produces a bounded output. The output is the convolution of the input with h, so its size can never exceed the input's largest size times ∫|h(t)| dt. That is why this integral must be finite for stability: the condition is ∫|h(t)| dt < ∞, that is, h must be absolutely integrable.

Observeh(t) = 0, t < 0

If causal, h is 0 for t < 0.

Choosestable: ∫ |h(t)| dt ?

Stable means the absolute integral is finite.

Fill in∫₀^∞ eᵃᵗ dt = ?

For a<0 the integral is −1/a = 1/|a|.

On your ownstable ⟺ a ?

An exponential response is stable only for a<0.

Read it from the sign of the pole

For h = eᵃᵗu(t), the integral ∫₀^∞ eᵃᵗ dt is finite only when a < 0, and its value is 1/|a|. So the stability condition is simply a < 0, and with a = −1 the ceiling is 1, with a = −0.5 it is 2: the closer a gets to 0, the higher the ceiling. This a is the pole location on the s-plane you will meet in the next chapter, and it leads to the rule that a system is stable when its poles lie to the left of the imaginary axis (real part < 0). The property you read from the shape of h gets rewritten as the location of a pole.

Back to the first screen

When the decay rate a went below 0, the running area of |h|, which had been growing without stopping, changed into one that reaches the ceiling 1/|a| and stops. The existence of that ceiling, the fact that ∫|h| is finite, is exactly stability. Causality came for free, because h was 0 throughout t < 0. The two key properties of an LTI system are written in this way in the shape of a single impulse response: when it begins and whether its area is finite.

The properties of an LTI system are read directly from its impulse response h. Causal ⟺ h(t) = 0 for t < 0 (no reaction before the impulse). BIBO stable ⟺ ∫|h(t)| dt < ∞ (absolutely integrable), because the output can never exceed the input's largest size times this integral. The exponential response h = eᵃᵗu(t) is always causal and stable only when a < 0, which leads to the pole location on the s-plane (real part < 0) in the next chapter.

On to the next unit

If the impulse response holds everything about a system, what is that response integrated once? In the next unit, the step response, the unit step u(t) goes in as the input, and you confirm that the output is the running integral of the impulse response. The step response of a stable system settles at a finite final value, and that final value and how fast it settles become the yardsticks by which a system is judged in practice. The finiteness of ∫|h| you just saw is exactly what guarantees that settling.