Harmonic balance is a method used to calculate the steady-state response of nonlinear differential equations, and is mostly applied to nonlinear electrical circuits. It is a frequency domain method for calculating the steady state, as opposed to the various time-domain steady-state methods. The name "harmonic balance" is descriptive of the method, which starts with Kirchhoff's Current Law written in the frequency domain and a chosen number of harmonics. A sinusoidal signal applied to a nonlinear component in a system will generate harmonics of the fundamental frequency. Effectively the method assumes a linear combination of sinusoids can represent the solution, then balances current and voltage sinusoids to satisfy Kirchhoff's law. The method is commonly used to simulate circuits which include nonlinear elements, and is most applicable to systems with feedback in which limit cycles occur. Microwave circuits were the original application for harmonic balance methods in electrical engineering. Such circuits were well-suited because, historically, microwave circuits consist of many linear components which can be directly represented in the frequency domain, plus a few nonlinear components. System sizes were typically small. For more general circuits, the method was considered impractical for all but these very small circuits until the mid-1990s, when Krylov subspace methods were applied to the problem. The application of preconditioned Krylov subspace methods allowed much larger systems to be solved, both in the size of the circuit and in the number of harmonics. This made practical the present-day use of harmonic balance methods to analyze radio-frequency integrated circuits (RFICs).
Example
Consider the differential equation x ¨ + x 3 = 0 {\displaystyle {\ddot {x}}+x^{3}=0} . We use the ansatz solution x = A cos ( ω t ) {\displaystyle x=A\cos(\omega t)} , and plugging in, we obtain
− A ω 2 cos ( ω t ) + A 3 1 4 ( cos ( 3 ω t ) + 3 cos ( ω t ) ) = 0. {\displaystyle -A\omega ^{2}\cos(\omega t)+A^{3}{\frac {1}{4}}(\cos(3\omega t)+3\cos(\omega t))=0.}
Then by matching the cos ( ω t ) {\displaystyle \cos(\omega t)} terms, we have
ω = 3 4 A {\displaystyle \omega ={\sqrt {\frac {3}{4}}}A} , which yields approximate period T = 2 π ω ≈ 7.2552 A {\displaystyle T={\frac {2\pi }{\omega }}\approx {\frac {7.2552}{A}}} . For a more exact approximation, we use ansatz solution x = A 1 cos ( ω t ) + A 3 cos ( 3 ω t ) {\displaystyle x=A_{1}\cos(\omega t)+A_{3}\cos(3\omega t)} . Plugging these in and matching the cos ( ω t ) {\displaystyle \cos(\omega t)} , cos ( 3 ω t ) {\displaystyle \cos(3\omega t)} terms, we obtain after routine algebra:
ω = 3 4 A 1 1 + y + 2 y 2 , y = A 3 / A 1 , 51 y 3 + 27 y 2 + 21 y − 1 = 0. {\displaystyle \omega ={\sqrt {\frac {3}{4}}}A_{1}{\sqrt {1+y+2y^{2}}},\quad y=A_{3}/A_{1},\quad 51y^{3}+27y^{2}+21y-1=0.}
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