Transfer Functions, Poles and Response

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The idea

For a linear time-invariant differential equation, the transfer function is $G(s)=Y(s)/U(s)$ under zero initial conditions. Its inverse transform is the impulse response. A nonzero initial state contributes an additional term and is not included in this ratio.

Visual guide

VISUAL GUIDEPole location determines whether a mode decays

Stable mode e⁻²ᵗ
Unstable mode eᵗ

A pole at −2 produces an impulse-response shape e⁻²ᵗ, while a pole at +1 produces eᵗ. The latter grows and can make a bounded-input response unbounded. These curves show time-domain shapes, not the complex s-plane.

Method and assumptions

Transform the input-output equation with initial values set to zero. Identify poles and zeros of the reduced rational expression. For a proper rational causal transfer function, poles strictly in the left half-plane imply a decaying impulse response and bounded-input bounded-output stability. Hidden canceled modes require a separate internal-state analysis.

Worked example: a low-pass system

For $y'+2y=2u$ , $G=2/(s+2)$ . The impulse response is $2e^{-2t}$ and the unit-step response is $1-e^{-2t}$ . DC gain is $G(0)=1$ , and the time constant is $1/2$ .

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

A transfer function H(s)=Y(s)/U(s) normally describes which response?

The response with zero initial conditions

Every response regardless of initial conditions

Only the homogeneous response

Hint 1 · Find a starting point

Initial-state terms enter transformed equations separately.

Hint 2 · Take the next step

Set those terms to zero to isolate input-to-output behavior.

Show the reasoning

Answer: The response with zero initial conditions

H describes the zero-state input response; a nonzero initial state adds its own response.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: an unstable pole

For $y'-y=u$ , $G=1/(s-1)$ . The impulse response $e^t$ grows. A unit step gives $y=e^t-1$ , so a bounded input can produce an unbounded output. The positive pole reflects that instability.

Interpreting the result

Frequency response uses $G(i\omega)$ after stable transients have decayed. A pole-zero cancellation can hide an unstable internal mode, so transfer behavior alone should not be treated as a complete state-stability proof.

Practice

What is the DC gain of $3/(s+6)$ ?
What is its time constant?
Why must initial conditions be zero when forming $Y/U$ ?

Show worked solutions

$3/6=1/2$ .
$1/6$ in the time unit used.
Otherwise $Y$ includes free response unrelated to the input, so the ratio is not a system-only transfer function.

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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