Sinusoidal Steady State and Frequency Response

Definition Let $t$ be time and let primes denote derivatives with respect to $t$. Consider

with real constants $m > 0$, $c \ge 0$, $k > 0$, a force amplitude $F_0 > 0$ and a driving angular frequency $\omega \ge 0$, and let $p(r) = mr^2 + cr + k$ be the characteristic polynomial. When $p(i\omega) \ne 0$, the complex gain at $\omega$ is

The condition $p(i\omega) \ne 0$ holds for every real $\omega$ when $c > 0$, and for every $\omega \ne \sqrt{k/m}$ when $c = 0$. The reason it works is that $(e^{i\omega t})' = i\omega e^{i\omega t}$ and $(e^{i\omega t})'' = -\omega^2 e^{i\omega t}$, so a trial $Ze^{i\omega t}$ with a complex constant $Z$ turns the left side into $p(i\omega)Ze^{i\omega t}$. Matching $F_0e^{i\omega t}$ gives $Z = H(i\omega)F_0$. Since the coefficients are real, the real part of a complex solution solves the equation with the real part of the force, and $F_0\cos\omega t = \operatorname{Re}(F_0e^{i\omega t})$. The imaginary unit is only a tool for the calculation. The force and the displacement are real.

Definition For the drive $F_0\cos\omega t$ with $p(i\omega) \ne 0$, the sinusoidal steady state is the particular solution $x_s(t) = \operatorname{Re}(H(i\omega)F_0e^{i\omega t})$. Writing the gain in polar form as $H(i\omega) = \lvert H(i\omega) \rvert e^{-i\phi}$, it is

a sinusoid at the driving frequency. It is the only solution of the form $a\cos\omega t + b\sin\omega t$. When $c > 0$, every solution approaches it as $t \to \infty$, because the difference of two solutions solves the homogeneous equation and decays. That is why it is called the steady state. Without damping the same formula still gives a periodic particular solution, but other solutions do not approach it.

Definition The frequency response of the oscillator is the complex gain viewed as a function of the driving angular frequency, $\omega \mapsto H(i\omega)$ for $\omega \ge 0$. It is usually shown as two real curves against $\omega$, the amplitude ratio $\lvert H(i\omega) \rvert$ and the phase lag $\phi(\omega)$. Because the equation is linear, a drive that is a finite sum of sinusoids at different frequencies has as its steady state the sum of the separate steady states, each scaled and delayed according to its own frequency. With $m$, $c$ and $k$ positive, the amplitude ratio starts at $1/k$ when $\omega = 0$, behaves like $1/(m\omega^2)$ for large $\omega$, and has a peak at a positive frequency only when $c^2 < 2mk$.

Definition The amplitude ratio at the angular frequency $\omega$ is the modulus of the complex gain,

where $\lvert p(i\omega) \rvert^2 = (k - m\omega^2)^2 + c^2\omega^2$. It is the amplitude of the steady displacement divided by the amplitude of the force, in meters per newton. Geometrically, $\lvert p(i\omega) \rvert$ is the length of the vector from $0$ to the point $(k - m\omega^2, c\omega)$ of the complex plane, and the ratio is its reciprocal.

Definition Write $\operatorname{atan2}(y, x)$ for the angle in $(-\pi, \pi]$ of the point $(x, y) \ne (0, 0)$. The phase lag at $\omega$ is the angle of the point $(k - m\omega^2, c\omega)$ that represents $p(i\omega)$,

so that $p(i\omega) = \lvert p(i\omega) \rvert e^{i\phi}$ and $H(i\omega) = \lvert H(i\omega) \rvert e^{-i\phi}$. For $c > 0$ and $\omega > 0$ the second coordinate $c\omega$ is positive, so $0 < \phi < \pi$, and the steady state $A\cos(\omega t - \phi)$ reaches each peak $\phi/\omega$ seconds after the most recent peak of the force. The lag is less than $\pi/2$ when $\omega < \sqrt{k/m}$, equal to $\pi/2$ when $\omega = \sqrt{k/m}$, and greater than $\pi/2$ beyond that. The one-argument formula $\arctan(c\omega/(k - m\omega^2))$ gives the same angle only when $k - m\omega^2 > 0$. Above the natural frequency it returns a negative angle in the wrong quadrant, and at the natural frequency it is undefined.

Definition Let $c > 0$ and consider the drive $F_0\cos\omega t$ with initial values $x(0)$ and $x'(0)$. The solution splits as

where $x_s$ is the sinusoidal steady state and the transient $x_{\text{tr}}$ is the solution of the homogeneous equation $mx'' + cx' + kx = 0$ with $x_{\text{tr}}(0) = x(0) - x_s(0)$ and $x_{\text{tr}}'(0) = x'(0) - x_s'(0)$. Because $c > 0$, both characteristic roots have negative real part, so the transient decays to zero and $x - x_s \to 0$ as $t \to \infty$. The steady state depends only on the drive, and the transient carries the initial data.