Transformation of the equation

We begin by making a change of variable

${\displaystyle q(t)\ {\stackrel {\mathrm {def} }{=}}\ e^{D(t)}x(t)}$

where ${\displaystyle D(t)}$ is the time integral of the damping coefficient

${\displaystyle D(t)\ {\stackrel {\mathrm {def} }{=}}\ {\frac {1}{2}}\int _{0}^{t}\beta (\tau )\,d\tau }$.

This change of variable eliminates the damping term in the differential equation, reducing it to

${\displaystyle {\frac {d^{2}q}{dt^{2}}}+\Omega ^{2}(t)q=0}$

where the transformed frequency is defined as

${\displaystyle \Omega ^{2}(t)\ {\stackrel {\mathrm {def} }{=}}\ \omega ^{2}(t)-{\frac {1}{2}}{\frac {d\beta }{dt}}-{\frac {1}{4}}\beta ^{2}(t)}$.

In general, the variations in damping and frequency are relatively small perturbations

${\displaystyle \beta (t)=\omega _{0}{\big [}b+g(t){\big ]}}$

${\displaystyle \omega ^{2}(t)=\omega _{0}^{2}{\big [}1+h(t){\big ]}}$

where ${\displaystyle \omega _{0}}$ and ${\displaystyle b}$ are constants, namely, the time-averaged oscillator frequency and damping, respectively. The transformed frequency can then be written in a similar way as

${\displaystyle \Omega ^{2}(t)=\omega _{n}^{2}{\big [}1+f(t){\big ]}}$,

where ${\displaystyle \omega _{n}}$ is the natural frequency of the damped harmonic oscillator

${\displaystyle \omega _{n}^{2}\ {\stackrel {\mathrm {def} }{=}}\ \omega _{0}^{2}\left(1-{\frac {b^{2}}{4}}\right)}$

and

${\displaystyle f(t)\ {\stackrel {\mathrm {def} }{=}}\ {\frac {\omega _{0}^{2}}{\omega _{n}^{2}}}\left[h(t)-{\frac {1}{2\omega _{0}}}{\frac {dg}{dt}}-{\frac {b}{2}}g(t)-{\frac {1}{4}}g^{2}(t)\right]}$.

Thus, our transformed equation can be written as

${\displaystyle {\frac {d^{2}q}{dt^{2}}}+\omega _{n}^{2}{\big [}1+f(t){\big ]}q=0}$.

The independent variations ${\displaystyle g(t)}$ and ${\displaystyle h(t)}$ in the oscillator damping and resonance frequency, respectively, can be combined into a single pumping function ${\displaystyle f(t)}$. The converse conclusion is that any form of parametric excitation can be accomplished by varying either the resonance frequency or the damping, or both.

Solution of the transformed equation

Let us assume that ${\displaystyle f(t)}$ is sinusoidal with a frequency approximately twice the natural frequency of the oscillator:

${\displaystyle f(t)=f_{0}\sin(2\omega _{p}t)}$

where the pumping frequency ${\displaystyle \omega _{p}\approx \omega _{n}}$ but need not equal ${\displaystyle \omega _{n}}$ exactly. Using the method of variation of parameters, the solution ${\displaystyle q(t)}$ to our transformed equation may be written as

${\displaystyle q(t)=A(t)\cos(\omega _{p}t)+B(t)\sin(\omega _{p}t)}$

where the rapidly varying components, ${\displaystyle \cos(\omega _{p}t)}$ and ${\displaystyle \sin(\omega _{p}t)}$, have been factored out to isolate the slowly varying amplitudes ${\displaystyle A(t)}$ and ${\displaystyle B(t)}$.

We proceed by substituting this solution into the differential equation and considering that both the coefficients in front of ${\displaystyle \cos(\omega _{p}t)}$ and ${\displaystyle \sin(\omega _{p}t)}$ must be zero to satisfy the differential equation identically. We also omit the second derivatives of ${\displaystyle A(t)}$ and ${\displaystyle B(t)}$ on the grounds that ${\displaystyle A(t)}$ and ${\displaystyle B(t)}$ are slowly varying, as well as omit sinusoidal terms not near the natural frequency, ${\displaystyle \omega _{n}}$, as they do not contribute significantly to resonance. The result is the following pair of coupled differential equations:

${\displaystyle 2\omega _{p}{\frac {dA}{dt}}={\frac {1}{2}}f_{0}\omega _{n}^{2}A-\left(\omega _{p}^{2}-\omega _{n}^{2}\right)B}$

${\displaystyle 2\omega _{p}{\frac {dB}{dt}}=\left(\omega _{p}^{2}-\omega _{n}^{2}\right)A-{\frac {1}{2}}f_{0}\omega _{n}^{2}B}$.

This system of linear differential equations with constant coefficients can be decoupled and solved by eigenvalue/eigenvector methods. This yields the solution

${\displaystyle {\begin{bmatrix}A(t)\\B(t)\end{bmatrix}}=c_{1}{\vec {V_{1}}}e^{\lambda _{1}t}+c_{2}{\vec {V_{2}}}e^{\lambda _{2}t}}$

where ${\displaystyle \lambda _{1}}$ and ${\displaystyle \lambda _{2}}$ are the eigenvalues of the matrix

${\displaystyle {\frac {1}{2\omega _{p}}}{\begin{bmatrix}{\frac {1}{2}}f_{0}\omega _{n}^{2}&-\left(\omega _{p}^{2}-\omega _{n}^{2}\right)\\\omega _{p}^{2}-\omega _{n}^{2}&-{\frac {1}{2}}f_{0}\omega _{n}^{2}\end{bmatrix}}}$,

${\displaystyle {\vec {V_{1}}}}$ and ${\displaystyle {\vec {V_{2}}}}$ are corresponding eigenvectors, and ${\displaystyle c_{1}}$ and ${\displaystyle c_{2}}$ are arbitrary constants. The eigenvalues are given by

${\displaystyle \lambda _{1,2}=\pm {\frac {1}{2\omega _{p}}}{\sqrt {\left({\frac {f_{0}\omega _{n}^{2}}{2}}\right)^{2}-\left(\omega _{p}^{2}-\omega _{n}^{2}\right)^{2}}}}$.

If we write the difference between ${\displaystyle \omega _{p}}$ and ${\displaystyle \omega _{n}}$ as ${\displaystyle \Delta \omega =\omega _{p}-\omega _{n}}$, and replace ${\displaystyle \omega _{p}}$ with ${\displaystyle \omega _{n}}$ everywhere where the difference is not important, we get

${\displaystyle \lambda _{1,2}=\pm {\sqrt {\left({\frac {f_{0}\omega _{n}}{4}}\right)^{2}-\Delta \omega ^{2}}}}$.

If ${\displaystyle |\Delta \omega |<{\frac {f_{0}\omega _{n}}{4}}}$, then the eigenvalues are real and exactly one is positive, which leads to exponential growth for ${\displaystyle A(t)}$ and ${\displaystyle B(t)}$. This is the condition for parametric resonance, with the growth rate given by the positive eigenvalue ${\displaystyle \lambda ={\sqrt {\left({\frac {f_{0}\omega _{n}}{4}}\right)^{2}-\Delta \omega ^{2}}}}$. Note, however, that this growth rate corresponds to the amplitude of the transformed variable ${\displaystyle q(t)}$, whereas the amplitude of the untransformed variable ${\displaystyle x(t)=q(t)e^{-D(t)}}$ can either grow or decay depending on whether ${\displaystyle \lambda t-D(t)}$ is increasing or decreasing.

Intuitive derivation of parametric excitation

The above derivation may seem like a mathematical sleight-of-hand, so it may be helpful to give an intuitive derivation. The ${\displaystyle q}$ equation may be written in the form

${\displaystyle {\frac {d^{2}q}{dt^{2}}}+\omega _{n}^{2}q=-\omega _{n}^{2}f(t)q}$

which represents a simple harmonic oscillator (or, alternatively, a bandpass filter) being driven by a signal ${\displaystyle -\omega _{n}^{2}f(t)q}$ that is proportional to its response ${\displaystyle q(t)}$.

Assume that ${\displaystyle q(t)=A\cos(\omega _{p}t)}$ already has an oscillation at frequency ${\displaystyle \omega _{p}}$ and that the pumping ${\displaystyle f(t)=f_{0}\sin(2\omega _{p}t)}$ has double the frequency and a small amplitude ${\displaystyle f_{0}\ll 1}$. Applying a trigonometric identity for products of sinusoids, their product ${\displaystyle q(t)f(t)}$ produces two driving signals, one at frequency ${\displaystyle \omega _{p}}$ and the other at frequency ${\displaystyle 3\omega _{p}}$.

${\displaystyle f(t)q(t)={\frac {f_{0}}{2}}A{\big [}\sin(\omega _{p}t)+\sin(3\omega _{p}t){\big ]}}$

Being off-resonance, the ${\displaystyle 3\omega _{p}}$ signal is attenuated and can be neglected initially. By contrast, the ${\displaystyle \omega _{p}}$ signal is on resonance, serves to amplify ${\displaystyle q(t)}$, and is proportional to the amplitude ${\displaystyle A}$. Hence, the amplitude of ${\displaystyle q(t)}$ grows exponentially unless it is initially zero.

Expressed in Fourier space, the multiplication ${\displaystyle f(t)q(t)}$ is a convolution of their Fourier transforms ${\displaystyle {\tilde {F}}(\omega )}$ and ${\displaystyle {\tilde {Q}}(\omega )}$. The positive feedback arises because the ${\displaystyle +2\omega _{p}}$ component of ${\displaystyle f(t)}$ converts the ${\displaystyle -\omega _{p}}$ component of ${\displaystyle q(t)}$ into a driving signal at ${\displaystyle +\omega _{p}}$, and vice versa (reverse the signs). This explains why the pumping frequency must be near ${\displaystyle 2\omega _{n}}$, twice the natural frequency of the oscillator. Pumping at a grossly different frequency would not couple (i.e., provide mutual positive feedback) between the ${\displaystyle -\omega _{p}}$ and ${\displaystyle +\omega _{p}}$ components of ${\displaystyle q(t)}$.

Introduction

A parametric amplifier is implemented as a mixer. The mixer's gain shows up in the output as amplifier gain. The input weak signal is mixed with a strong local oscillator signal, and the resultant strong output is used in the ensuing receiver stages.

Parametric amplifiers also operate by changing a parameter of the amplifier. Intuitively, this can be understood as follows, for a variable capacitor-based amplifier. Charge ${\displaystyle Q}$ in a capacitor obeys: ${\displaystyle Q=C\times V}$,
therefore the voltage across is ${\displaystyle V={\frac {Q}{C}}}$.

Knowing the above, if a capacitor is charged until its voltage equals the sampled voltage of an incoming weak signal, and if the capacitor's capacitance is then reduced (say, by manually moving the plates further apart), then the voltage across the capacitor will increase. In this way, the voltage of the weak signal is amplified.

If the capacitor is a varicap diode, then "moving the plates" can be done simply by applying time-varying DC voltage to the varicap diode. This driving voltage usually comes from another oscillator—sometimes called a "pump".

The resulting output signal contains frequencies that are the sum and difference of the input signal (f1) and the pump signal (f2): (f1 + f2) and (f1 − f2).

A practical parametric oscillator needs the following connections: one for the "common" or "ground", one to feed the pump, one to retrieve the output, and maybe a fourth one for biasing. A parametric amplifier needs a fifth port to input the signal being amplified. Since a varactor diode has only two connections, it can only be a part of an LC network with four eigenvectors with nodes at the connections. This can be implemented as a transimpedance amplifier, a traveling-wave amplifier or by means of a circulator.

Mathematical equation

The parametric oscillator equation can be extended by adding an external driving force ${\displaystyle E(t)}$:

${\displaystyle {\frac {d^{2}x}{dt^{2}}}+\beta (t){\frac {dx}{dt}}+\omega ^{2}(t)x=E(t)}$.

We assume that the damping ${\displaystyle D}$ is sufficiently strong that, in the absence of the driving force ${\displaystyle E}$, the amplitude of the parametric oscillations does not diverge, i.e., that ${\displaystyle \alpha t<D}$. In this situation, the parametric pumping acts to lower the effective damping in the system. For illustration, let the damping be constant ${\displaystyle \beta (t)=\omega _{0}b}$ and assume that the external driving force is at the mean resonance frequency ${\displaystyle \omega _{0}}$, i.e., ${\displaystyle E(t)=E_{0}\sin \omega _{0}t}$. The equation becomes

${\displaystyle {\frac {d^{2}x}{dt^{2}}}+b\omega _{0}{\frac {dx}{dt}}+\omega _{0}^{2}\left[1+h_{0}\sin 2\omega _{0}t\right]x=E_{0}\sin \omega _{0}t}$

whose solution is approximately

${\displaystyle x(t)={\frac {2E_{0}}{\omega _{0}^{2}\left(2b-h_{0}\right)}}\cos \omega _{0}t}$.

As ${\displaystyle h_{0}}$ approaches the threshold ${\displaystyle 2b}$, the amplitude diverges. When ${\displaystyle h_{0}\geq 2b}$, the system enters parametric resonance and the amplitude begins to grow exponentially, even in the absence of a driving force ${\displaystyle E(t)}$.

Advantages

1. It is highly sensitive
2. low noise level amplifier for ultra high frequency and microwave radio signal

Other relevant mathematical results

If the parameters of any second-order linear differential equation are varied periodically, Floquet analysis shows that the solutions must vary either sinusoidally or exponentially.

The ${\displaystyle q}$ equation above with periodically varying ${\displaystyle f(t)}$ is an example of a Hill equation. If ${\displaystyle f(t)}$ is a simple sinusoid, the equation is called a Mathieu equation.