SC · LTI and convolution

The Step Response

The output a system produces when a unit step goes in is its step response. See for yourself that it is the running integral of the impulse response, and that in a stable system it settles at a finite final value.

Does s settle to its final value

The upper figure is the impulse response h(τ) = e^(−0.8τ) sin 3τ, and the green area from 0 up to the slide time t is the step response s(t) at that time. The lower figure is that s(t) curve. As t grows, s climbs past a peak, comes back down and settles at the dashed final value of about 0.311.

impulse response h(τ)

step response s(t) = running areafinal value s(∞)

time t0.30

Current and final value

s(0.30) = 0.105  s(∞) = 0.311

It is still early: the first crest of h is piling up and s rises fast. Starting from 0, s already passes the final value at t ≈ 0.6 but shows no sign of stopping yet; this is the transient stretch in which the output is finding its place.

Not settled

A step makes the impulse response accumulate

The unit step u(t) was impulses piled up from the left up to t (A2). Because the system is LTI, it answers an integrated input with an integrated output, so the step response is the impulse response piled up the same way: s(t) = ∫₀ᵗ h(τ) dτ. The green area on the screen draws exactly this value at every time.

Observes(t) = ∫ h(τ) dτ

The step response is the impulse response accumulated.

Chooseh(t) = ?

Differentiate the step response for the impulse response.

Fill ins(∞) = ?

The final value is the total area of h.

On your owns = h ?

The step response is the convolution of h with the step.

s and h are an integral-derivative pair

Piling up is integration, so conversely, differentiating the step response gives back the impulse response: h(t) = ds/dt. In practice an exact impulse is hard to make but a step is easy, so the impulse response is often obtained by measuring the step response and differentiating it. Just as the impulse and the step were a pair in A2, the two responses are a pair too.

The final value is the DC gain

In a stable system h dies away to 0, so the running area stops as well and s settles at a finite final value. That value, s(∞) = ∫₀^∞ h(τ) dτ, is the size of the output for a constant input (DC), that is, the DC gain. In this figure ∫₀^∞ e^(−0.8τ) sin 3τ dτ = 3/(0.8² + 3²) ≈ 0.311, and the peak of 0.446 is an overshoot about 43% above it. Rise time, overshoot and settling time are all practical measures read from this step response, and control design returns to them in detail.

Back to the first screen

As the positive crest of h was added, s rose quickly and passed the final value; then the negative trough was added and it swung, and once h had died away to 0 it settled. The green area up to the slide time was exactly s(t) at every moment. The step response is the impulse response integrated once, differentiating it brings back h, and the value where it settles is the DC gain. In this way a single impulse response completes the time-domain picture of an LTI system.

The step response s(t) is the output an LTI system produces when the unit step u(t) goes in, and it is the running integral of the impulse response: s(t) = ∫₀ᵗ h(τ) dτ = (h ∗ u)(t), and conversely h(t) = ds/dt. In a stable system s settles at a finite final value s(∞) = ∫₀^∞ h(τ) dτ, the DC gain. Rise time, overshoot and settling time are all practical measures read from s.

On to the next chapter

So far the LTI system has been drawn completely in the time domain. With a single impulse response and convolution you can find the output for any input, but convolution takes a lot of work. The next chapter, Fourier, opens a new view in which signals are sums of sinusoids. Feed a sinusoid into an LTI system and a sinusoid of the same frequency comes out with only its size and phase changed, so the laborious convolution of the time domain turns into simple multiplication in the frequency domain. The starting point is the next unit, the Fourier series.