Step functions, impulses, and convolution
Real systems are switched on and off, and hit by sudden shocks. Laplace transforms handle these inputs elegantly.
The unit step function
The Heaviside step \(u(t - a)\) is 0 for \(t < a\) and 1 for \(t \ge a\): a switch turned on at time \(a\). Its transform is \(\frac{e^{-as}}{s}\). Piecewise forcing, like a voltage applied from \(t = 2\) to \(t = 5\), is written \(V[u(t - 2) - u(t - 5)]\).
The second shifting theorem
Delaying a function multiplies its transform by an exponential: \(\mathcal{L}\{f(t - a)u(t - a)\} = e^{-as}F(s)\). Conversely, an \(e^{-as}\) factor in \(Y(s)\) means the corresponding time function starts at \(t = a\).
\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s)Worked example: a switched input
Solve \(y' + y = u(t - 1)\), \(y(0) = 0\). Transforming: \((s + 1)Y = \frac{e^{-s}}{s}\), so \(Y = e^{-s}\left(\frac{1}{s} - \frac{1}{s + 1}\right)\). Inverting with the shift: \(y = u(t - 1)\left(1 - e^{-(t - 1)}\right)\). Nothing happens until \(t = 1\), then \(y\) rises toward 1.
The Dirac delta
The delta function \(\delta(t - a)\) models an instantaneous impulse, like a hammer blow, with zero width and unit total area. Its transform is \(e^{-as}\). For a mass-spring struck at \(t = 0\), \(y'' + \omega^2y = \delta(t)\) gives \(Y = \frac{1}{s^2 + \omega^2}\) and \(y = \frac{1}{\omega}\sin\omega t\).
\mathcal{L}\{\delta(t-a)\} = e^{-as}Impulse response
The response of a system at rest to \(\delta(t)\) is its impulse response, whose transform is exactly the transfer function \(H(s)\). Hitting a system with an impulse and recording the response, as engineers do with bridges and audio rooms, reveals \(H\) completely.
Convolution
The convolution \((f * g)(t) = \int_0^t f(\tau)g(t - \tau)\,d\tau\) satisfies \(\mathcal{L}\{f * g\} = F(s)G(s)\). So the response to any input \(g\) is the convolution of the impulse response with \(g\). Convolution also describes blurring in images and echoes in audio.
\mathcal{L}\{f * g\} = F(s)\,G(s)Periodic inputs
For a periodic forcing with period \(T\), \(\mathcal{L}\{f\} = \frac{1}{1 - e^{-sT}}\int_0^T e^{-st}f(t)\,dt\). This handles square waves and sawtooth signals in circuits.
Common mistakes
Forgetting to shift the function along with the step: \(u(t - a)f(t)\) is not \(u(t - a)f(t - a)\). Treating the delta function as an ordinary function. Dropping the \(e^{-as}\) factor when inverting.
Key takeaways
Step functions model switches, with \(\mathcal{L}\{u(t - a)\} = e^{-as}/s\), and delays multiply transforms by \(e^{-as}\). The Dirac delta models impulses, with transform \(e^{-as}\). The impulse response's transform is the transfer function, and convolution turns into multiplication of transforms.
Flashcard
Unit step function
Check your understanding
The Laplace transform of the unit impulse delta(t) is: