Frequency Response and RLC Models

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The idea

A stable linear oscillator driven at one frequency eventually responds at that frequency after its free transient decays. Complex exponentials simplify the algebra. The physical solution is the real part; complex notation does not change the real model.

Visual guide

VISUAL GUIDEDamping changes the amplitude peak

c = 0.2
c = 1
c = 2

For unit mass, stiffness and forcing, steady amplitude is 1/√((1 − ω²)² + (cω)²). Small damping gives a sharp peak near the natural frequency; stronger damping lowers and can remove the nonzero-frequency peak.

Method and assumptions

For $mx''+cx'+kx=F_0\cos\omega t$ with $c>0$ , the complex amplitude is $X=F_0/(k-m\omega^2+ic\omega)$ . Its magnitude is $F_0/\sqrt{(k-m\omega^2)^2+(c\omega)^2}$ ; use a quadrant-aware argument for phase.

Worked example: forcing at natural frequency

With $m=k=1,c=2,F_0=1,\omega=1$ , $X=1/(2i)=-i/2$ . Thus $x_{ss}=\tfrac12\sin t$ , a quarter-cycle lag behind $\cos t$ . Damping keeps the amplitude finite.

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.

For a damped stable oscillator driven sinusoidally, what happens to the transient at long times?

It cancels the input forever.

It decays, leaving a sinusoidal steady-state response.

It always grows without bound.

Hint 1 · Find a starting point

Separate the homogeneous response from the forced response.

Hint 2 · Take the next step

Stable damping makes the homogeneous modes decay.

Show the reasoning

Answer: It decays, leaving a sinusoidal steady-state response.

After the transient fades, the forcing frequency remains, with amplitude and phase set by the system.

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

Worked example: a series RLC circuit

Using charge $q$ , the circuit equation is $Lq''+Rq'+q/C=V_0\cos\omega t$ . This matches the mechanical model with $m=L,c=R,k=1/C$ . Current is $i=q'$ , so its complex amplitude is $i\omega Q$ , with an additional phase shift and amplitude factor $\omega$ .

Interpreting the result

The forcing frequency of the largest displacement amplitude need not equal $\omega_n$ . For the standard model a nonzero peak occurs at $\omega_n\sqrt{1-2\zeta^2}$ only when $\zeta<1/\sqrt2$ . Undamped resonance needs a separate time-domain solution.

Practice

What is the displacement amplitude at $\omega=0$ ?
What is its high-frequency scaling?
Can the steady response enforce arbitrary initial data by itself?

Show worked solutions

$F_0/k$ , the static displacement.
For $m>0$ , it scales as $F_0/(m\omega^2)$ .
No. Add the homogeneous transient to satisfy the initial conditions.

Further study

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

MAKE IT YOURS

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