Why use the Laplace transform?
It turns differential equations into algebra while preserving the story of transients and stability.
Turn a first-order differential equation into a transfer function and read its time scale.
Convolution computes an output by adding contributions over time. Laplace transforms turn that calculation into multiplication for zero-state LTI models.
Review the previous lesson: Build a response from tiny impulsesFirst, 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.
Build the idea step by step
A differential equation describes change through time. The Laplace transform rewrites it in terms of a variable called s. The useful rule is that a time derivative becomes multiplication by s, with a correction for the starting value. This turns many linear differential equations into algebra. We solve that algebra, then interpret the result back in time. The variable s can be complex: its real part describes growth or decay and its imaginary part describes oscillation. You do not need to picture a new physical world.
Read the equations
A derivative becomes multiplication by the complex variable s, plus a term for the starting value.
The input-to-output transfer function sets every initial state to zero; x collects those states.
What the symbols mean
- complex frequency variable, in inverse seconds
- Laplace transform of output
- Laplace transform of input
- zero-state transfer function
- all state values at the start
- output rate and initial output
A transfer function describes the input-to-output relation when every initial state is zero. In this first-order example that means the starting output is zero. A charged capacitor or already moving cart needs an additional initial-state term.
A pole's real part sets growth or decay; its imaginary part sets oscillation frequency. The boundary itself needs separate analysis.
A two-second first-order plant
Use a plant with a 2 second time constant and zero initial output. Input and output have the same normalized units.
Transform and solve for output divided by input.
The pole is at minus 0.5 inverse seconds. A unit step reaches about 63% after 2 seconds.
More detail and model assumptions
From a changing signal to an algebraic equation
Start with τẏ+y=u and y(0)=0. Taking the Laplace transform gives (τs+1)Y(s)=U(s), so G(s)=Y/U=1/(τs+1). You can now connect this block to a controller without solving the differential equation again for every input. The transform has reorganized the same dynamics; it has not removed them.
The initial value matters: with y(0)=y₀, the equation is (τs+1)Y=U+τy₀. G=Y/U describes the zero-initial-state input–output part, while τy₀ describes a separate free response. Never divide the entire expression by U and pretend an initial condition is another input transfer function.
Why the complex variable helps
A term e^(st), with s=σ+jω, combines a growth or decay rate σ with an oscillation rate ω. A pole at s=−1/τ therefore predicts decay; a pole with positive real part predicts growth. This is why a pole diagram says something about a time trace before you draw that trace.
Later, substituting s=jω asks how a stable system responds to a sustained sinusoid at frequency ω. It is a particular view of the same model, not a mysterious new physical domain. A frequency response need not exist for an unstable system in the usual steady-sinusoid sense.
Connect the example to the lab
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.
Make a prediction before moving a control.
- PREDICTDoes one G(s) describe two different starting speeds?
- CHANGE ONE THINGFor τẏ+y=u, write the transformed equation once with y(0)=0 and once with y(0)=y₀.
- OBSERVEIdentify the term that changed while G(s) stayed the same.
- EXPLAINExplain why the transfer function requires a stated initial-state assumption.
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+
Can the transfer function alone describe a cart that was already moving when the input began?
No. A transfer function describes the zero-initial-state input response. Add the response due to initial position and velocity as well.Why the next lesson follows
A transfer function's poles and zeros show how a model's modes appear at its chosen output.