ECE 210: Analog Signal Processing

Deriving the Impulse Response h(t)h(t) from the Step Response ys(t)y_s(t)

System Selection

System Context (Circuit Domain)

A system that acts as a finite-time moving average window. The step response ramps up and then flattens out, generating an impulse response with two impulses.

ys(t)=u(t)−u(t−2) y_s(t) = u(t) - u(t-2)
h(t)=δ(t)−δ(t−2) h(t) = \delta(t) - \delta(t-2)

Why do these relate? Because the input u(t)u(t) is the integral of δ(t)\delta(t), the step response ys(t)y_s(t) is the integral of the impulse response h(t)h(t). Therefore, h(t)=ddtys(t)h(t) = \frac{d}{dt}y_s(t).

The Core Concept

h(t)=ycont′(t)⏟Continuous Derivative+∑[ys(tk+)−ys(tk−)]⏟Jump Sizeδ(t−tk) h(t) = \underbrace{y'_{cont}(t)}_{\text{Continuous Derivative}} + \sum \underbrace{\left[y_s(t_k^+) - y_s(t_k^-)\right]}_{\text{Jump Size}} \delta(t - t_k)

1. Unit-Step Response ys(t)y_s(t)

2. Continuous Derivative ycont′(t)y'_{cont}(t)

3. True Impulse Response h(t)h(t)

4. Comparison: Missing the Jumps?

Instructor Notes & Pedagogical Design

How to use this applet:

Use the dropdown to select different LTI systems. Scrub your mouse across any plot to view instantaneous values synced across all domains. This tool visually proves that standard continuous differentiation is insufficient for finding h(t)h(t) if the step response contains jump discontinuities.

Misconceptions Addressed:

  • Missing the Impulse: Students routinely differentiate e−tu(t)e^{-t}u(t) as −e−tu(t)-e^{-t}u(t), forgetting the chain rule on u(t)u(t). Card 4 makes the missing δ(t)\delta(t) graphically obvious.
  • Physical connection: Tooltips connect abstract ys(t)y_s(t) expressions to real circuit voltages (e.g., across a capacitor vs. a resistor).
  • Time-invariance: The delayed pulse preset shows that shifting the input directly shifts the output and the resulting impulses.