Why use the Laplace transform?
It turns differential equations into algebra while preserving the story of transients and stability.
First, picture it
Solving a motor differential equation anew for every input is tedious. In the s-domain, differentiation becomes multiplication by s (plus initial-condition terms), so connected blocks can be multiplied or divided like ordinary expressions.
What the model says
The Laplace transform weighs a time signal by exp(−st), with s=σ+jω. The real part σ describes exponential growth or decay; the imaginary part ω describes oscillation. Transfer functions assume zero initial conditions when relating input to output. A nonzero starting state needs separate terms. The Fourier/frequency response is the s=jω slice when it exists; it is not a different plant.
ℒ{dx/dt}=sX(s)-x(0); G(s)=Y(s)/U(s) | zero initial stateThe x(0) term is why a transfer function alone does not describe every initial condition.
A pole's real part sets growth or decay; its imaginary part sets oscillation frequency. The boundary itself needs separate analysis.
Make it concrete
For τ dy/dt+y=u, zero initial state gives G(s)=1/(τs+1). Its pole at −1/τ predicts decay on time scale τ.
The s-domain is a calculation lens, not a claim that physical time disappears. Always return to a time response or frequency response for interpretation.
Check your understanding+
What does a pole with positive real part suggest about a free response?
It contains an exponentially growing mode, so a continuous-time linear system is unstable.