**Kevin Schmidt
Department of Physics and Astronomy
Arizona State University
Tempe, AZ
**

There are two main cases of interest. The first is when the time dependent quantities oscillate for all time with some set of possibly incommensurate frequencies, and the second is when the time dependent quantities are nonzero only for a range of times. Of course all physical systems belong to the latter case if we wait long enough, however, for systems that are oscillating for a long time, it is simpler to view this as a process that has continued and will continue forever.

A(t) ReAe^{ - it}
| (1) |

Often product of two such quantities is needed. For example if the current
and voltage are given by complex *V* and *I* , the power is given by

V(t)I(t)
| = |
[ReIe^{ - it} | |

= |
[Ie^{ - it} + I^{ *}e^{ - it} | ||

= |
ReVI^{ *} + ReVIe^{ - i2t} .
| (2) |

() | = |
dtf (t)e^{ it}
| |

f (t)
| = |
()e^{ - it} .
| (3) |

Physically measurable functions of time are real. For *f* (*t*) real, taking the
complex conjugate gives

() = dtf (t)e^{ - it} = f (- ) .
| (4) |

() | = |
dtf (t)e^{ it}
| |

f (t)
| = |
()e^{ - it} + (- )e^{ it}
| |

= |
Re ()e^{ - it}
| (5) |

The time integral of the product of two functions

f (t)
| = |
Re ()
| |

g(t)
| = |
Re ()
| (6) |

dtf (t)g(t) = dt[(')e^{ - i't} + (')e^{ i't} | (7) |

dte^{ ixt} = 2(x)
| (8) |

dtf (t)g(t)
| = | ||

= |
Re ()()
| (9) |

= Re ()() .
| (10) |

a(t) = e^{ - int} .
| (11) |

Multiplying by
*e*^{ imt} and integrating over a period using
the orthogonality of the complex exponentials gives the Fourier series
coefficients,

= dta(t)e^{ imt} .
| (12) |

= dta(t)e^{ - imt} =
| (13) |

a(t) = + 2 Re e^{ - int} .
| (14) |

It is important to notice the factor of 2 in front of the sum.
Notice particularly that if you have a single Fourier coefficient, you
need to multiply it by a factor of 2 to get the correct amplitude for
harmonic time dependence. That this must be true is easy to show if
you take
*A*(*t*) = *ReAe*^{ - it}. You will get *A*_{1} = *A*/2 ,
and the factor of two gives the correct coefficient.

The time average of two such quantities over the period is

dta(t)b(t)
| = |
dt[ + 2 Re e^{ - int} | |

= |
+ 2 Re a_{n}b_{n}^{*} .
| (15) |

P_{n} = 2 Re a_{n}b^{ *}_{n} .
| (16) |

Let's look at setting the time dependence to zero for times |*t*| > *T*/2 .
The Fourier transform of the cutoff motion
() is then

() = dtf (t)G(t)e^{ - it}
| (17) |

() | = |
dte^{ it}e^{ - i''t}('')e^{ - i't}(')
| |

= | (')( - ') | (18) |

() = sin . | (19) |

For the case where *f* (*t*) is localized in time, taking the limit
of large *T* makes
() oscillate rapidly for
|| 1/*T*
therefore in the integral above,
the slowly varying *f* (') can be replaced by
and pulled out of the integration. The remaining integral
is the inverse transform which gives *G*(*t* = 0) = 1 . So we see that
under the integral we have
*G*() = 2() .
The time integral of the product *f* (*t*)*g*(*t*) is exactly as before.

For the case where *f* (*t*) has components that oscillate forever
at some set of frequencies, the Fourier transform will contain
delta functions. For example if
*f* (*t*) = *ReAe*^{ - it}

() = | (20) |

d( - ) = dx = 2T .
| (21) |

Applying this to the single frequency case above,

dtf^{ 2}_{c}(t) = |A|^{2}
| (22) |