SC · LTI and convolution

Properties of LTI Systems

An LTI system’s properties are written in the shape of its impulse response h. Read causality from when h begins, and stability from whether the total area of its magnitude is finite. Move the decay rate and cross the boundary between stable and unstable yourself.

Does the running area hit a ceiling

The top is the causal impulse response h(t)=eᵃᵗ (t≥0). Below is the running area of |h| accumulated from the left. Lower the decay rate a until this area, instead of growing without bound, reaches a finite ceiling. That ceiling is stability.

impulse response h(t)=eᵃᵗ

running |h| areaceiling 1/|a|

decay rate a0.40

Total area and verdict

∫ |h| dt = ∞

With a > 0, h grows forever. The running area of |h| diverges with no ceiling, so ∫|h| = ∞. A tiny input can blow the output up — an unstable system.

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: h(t)=0, t<0. The h here lives only for t≥0, so it is always causal. Just check whether h’s graph is silent to the left of 0 and you read causality at once.

Stability is the total area of |h|

BIBO stability means a bounded input always yields a bounded output. The output is the convolution of input with h, so the output’s size is bounded by the input’s peak size times ∫|h(t)|dt. Hence that integral must be finite for stability. The condition is ∫|h(t)|dt < ∞ — h must be absolutely integrable.

Observeh(t) = 0, t < 0

Causal means 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 then equals 1/|a|. So the stability condition is simply a<0. This a is the pole location on the s-plane you will meet next chapter, leading to the rule that poles to the left of the imaginary axis (real part < 0) mean stability. The property you read from h’s shape gets rewritten as the pole’s position.

Back to the first screen

As you lowered the decay rate a below 0, the running |h| area changed from climbing without stop to reaching the ceiling 1/|a| and halting. 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 thus written entirely in the shape of one impulse response: when it begins and whether its area is finite.

An LTI system’s properties are read straight from its impulse response h. Causal ⟺ h(t)=0 for t<0 (no response before the impulse). BIBO-stable ⟺ ∫|h(t)|dt < ∞ (absolutely integrable), because the output is bounded by the input’s peak times this integral. The exponential response h=eᵃᵗu(t) is always causal and stable only when a<0, which leads to the next chapter’s pole location on the s-plane (real part < 0).

On to the next unit

If the impulse response tells you the whole system, what does integrating that response once give? The next unit, the step response, feeds the unit step u(t) as input and shows the output is the running integral of the impulse response. A stable system’s step response settles to a finite final value, and that value and its settling speed become the practical yardsticks for the system. The finiteness of ∫|h| you just saw is exactly what guarantees this settling.