A linear system’s response to an input applied earlier may persist and decay. To find the present output, add the surviving contributions from all earlier input times. For causal functions, define convolution by (f*g)(t)=∫ from zero to t of f(τ)g(t−τ)dτ. The variable τ labels an earlier time; t−τ is the age of that contribution.
Begin with f=g=1 for nonnegative time. The integral is the length of the interval, so (1*1)(t)=t. Convolution is not pointwise multiplication: the pointwise product would remain one.
Predict
If both factors equal e^(−t), predict whether their convolution is another pure exponential or includes a time factor. The product inside the integral is e^(−τ)e^(−(t−τ)), so the dependence on τ cancels. Decide what integrating this constant-in-τ expression contributes.
Experiment
The result is (e^(−t)*e^(−t))(t)=te^(−t). Trace the operation before applying a transform theorem.
Operation
Expression
Earlier-time factor
e^(−τ)
Age-dependent factor
e^(−(t−τ))
Product inside integral
e^(−t)
Integral over 0≤τ≤t
te^(−t)
The Laplace transform of each exponential is 1/(s+1). Their product is 1/(s+1)², which is the transform of te^(−t). This checks the convolution theorem in a concrete case. Also compute 1*t at t=2: the integral of 2−τ from zero to two equals two, not the pointwise product value alone as a general rule.
Understand
Under conditions permitting absolute convergence and exchange of integration, L[f*g]=F(s)G(s). The proof changes variables from the triangular region 0≤τ≤t to independent nonnegative variables τ and v=t−τ. The exponential weight splits as e^(−sτ)e^(−sv), and the double integral factors. The assumptions justify changing integration order; the geometric rearrangement alone is not enough for arbitrary divergent expressions.
For y′+ay=f(t), multiply by e^(at) and integrate from zero to t. Dividing back gives y(t)=y₀e^(−at)+∫ from zero to t of e^(−a(t−τ))f(τ)dτ. The first term carries the initial state. The second is a convolution with the causal response kernel e^(−at). If a is positive, older contributions are attenuated more strongly.
For the oscillator y″+ω²y=f(t) with zero initial position and velocity and ω positive, the kernel is h(t)=sin(ωt)/ω. Then y=∫h(t−τ)f(τ)dτ. Differentiate using the endpoint term: h(0)=0 makes the first derivative’s boundary term vanish. On the second derivative, h′(0)=1 supplies f(t). Since h″=−ω²h, the remaining integral supplies −ω²y. This verifies the equation directly for suitably continuous input.
Apply
Use the first-order kernel to compare two equal-area inputs delivered at different times. For positive a, the more recent input contributes more to the current state. Explain why equal total input does not necessarily mean equal output at a specified time.
State the initial conditions when using an impulse-response convolution. A zero-state convolution alone does not include arbitrary initial motion. Complete the practice by identifying the kernel, the integration limits, and the endpoint terms in a derivative check. Numerical quadrature may approximate the integral, but its error is separate from the model assumptions.
Continue
The final lesson returns to nonlinear systems. Linear response and superposition are powerful, but they do not automatically describe global nonlinear behavior or determine stability when the linearized rates vanish.
Apply · explain · revise
Practice and reasoning
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For f(t)=1 and g(t)=1 on t≥0, find (f*g)(3).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For f(t)=2 and g(t)=t on t≥0, find (f*g)(3).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For causal functions, (f*g)(t) is:
Compute the convolution of e^(−t) with itself for t≥0.
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Derive y(t)=y₀e^(−at)+∫₀ᵗe^(−a(t−τ))f(τ)dτ for y′+ay=f(t), y(0)=y₀.
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For y″+ω²y=f(t), ω>0 and zero initial data, state the convolution solution and verify why the kernel is sin(ωt)/ω.
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Reflect and continue
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