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Up: PHY531-PHY532 Classical Electrodynamics

Review of Spectral Resolution

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

1 Introduction

Often harmonically varying quantities can be analysed easily. For more general time dependence, spectral resolution can be used and the results of harmonic time dependence can be applied to each frequency component.

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.

2 Harmonic Time Dependence

For a single frequency, the usual complex notation is used where a complex number A is written to mean the time dependent quantity

A(t) $\displaystyle\equiv$ ReAe - i$\scriptstyle\omega$t (1)

where $\omega$ > 0 is the frequency.

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) = $\displaystyle\left [ {\rm Re} V e^{-i \omega t} \right]\left$[ReIe - i$\scriptstyle\omega$t]   
  = $\displaystyle{\textstyle\frac{1}{4}}$$\displaystyle\left [ V e^{-i \omega t} + V^* e^{-i \omega t} \right]\left$[Ie - i$\scriptstyle\omega$t + I *e - i$\scriptstyle\omega$t]   
  = $\displaystyle{\textstyle\frac{1}{2}}$ReVI * + ReVIe - i2$\scriptstyle\omega$t . (2)
The time averaged product is the first, constant, term. The second term oscillates at frequency 2$\omega$ and averages to zero. So the time averaged power is the first term.

3 A Disturbance Localized in Time

For quantities that are arbitrary functions of time, but which typically go to zero (or at least the part we are interested in goes to zero) at t $\rightarrow$ $\pm$ $\infty$ , the quantity can be Fourier transformed,
$\displaystyle\tilde{f}$($\displaystyle\omega$) = $\displaystyle\int_{-\infty}^{\infty}$dtf (t)e i$\scriptstyle\omega$t   
f (t) = $\displaystyle\int_{-\infty}^{\infty}$$\displaystyle{\frac{d\omega}{2\pi}}$$\displaystyle\tilde{f}$($\displaystyle\omega$)e - i$\scriptstyle\omega$t . (3)
Note that where the (2$\pi$)- 1 factor is placed is arbitrary. Jackson varies from placing it as above, to placing it symmetrically with a factor of 1/$\sqrt{2\pi}$ on both integrals, to placing it on the time integral. The choice I make above is the most common in physics books, (it leads to Fourier integrals that match the physical densities of states, etc.) but be aware that you need to check carefully when reading a new book or article to make sure that you are aware of the convention used. No physical results will change with the convention. Mathematics texts generally use the symmetric form.

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

$\displaystyle\tilde{f}^{*}_{}$($\displaystyle\omega$) = $\displaystyle\int_{-\infty}^{\infty}$dtf (t)e - i$\scriptstyle\omega$t = f (- $\displaystyle\omega$) . (4)

Therefore, for real f (t) , the Fourier transform can be written in terms of just positive frequencies,
$\displaystyle\tilde{f}$($\displaystyle\omega$) = $\displaystyle\int_{-\infty}^{\infty}$dtf (t)e i$\scriptstyle\omega$t   
f (t) = $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{2\pi}}$$\displaystyle\tilde{f}$($\displaystyle\omega$)e - i$\scriptstyle\omega$t + $\displaystyle\int_{-\infty}^{0}$$\displaystyle{\frac{d\omega}{2\pi}}$$\displaystyle\tilde{f}^{*}_{}$(- $\displaystyle\omega$)e i$\scriptstyle\omega$t   
  = Re $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{\pi}}$$\displaystyle\tilde{f}$($\displaystyle\omega$)e - i$\scriptstyle\omega$t (5)

The time integral of the product of two functions

f (t) = Re $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{\pi}}$$\displaystyle\tilde{f}$($\displaystyle\omega$)   
g(t) = Re $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{\pi}}$$\displaystyle\tilde{g}$($\displaystyle\omega$) (6)
is then
$\displaystyle\int_{-\infty}^{\infty}$dtf (t)g(t) = $\displaystyle{\textstyle\frac{1}{4}}$$\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{\pi}}$$\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega'}{\pi}}$$\displaystyle\int_{-\infty}^{\infty}$dt$\displaystyle\left [ \tilde f(\omega) e^{-i \omega t}
+ \tilde f^*(\omega) e^{i \omega t} \right]\left$[$\displaystyle\tilde{g}$($\displaystyle\omega$')e - i$\scriptstyle\omega$'t + $\displaystyle\tilde{g}^{*}_{}$($\displaystyle\omega$')e i$\scriptstyle\omega$'t]      (7)
and using

$\displaystyle\int_{-\infty}^{\infty}$dte ixt = 2$\displaystyle\pi$$\displaystyle\delta$(x) (8)

and that $\omega$ and $\omega$' are positive, this becomes
$\displaystyle\int_{-\infty}^{\infty}$dtf (t)g(t) = $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{2\pi}}$$\displaystyle\left [
\tilde f(\omega) \tilde g^*(\omega)
+\tilde f(\omega) \tilde g^*(\omega) \right]$   
  = Re $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{\pi}}$$\displaystyle\tilde{f}$($\displaystyle\omega$)$\displaystyle\tilde{g}^{*}_{}$($\displaystyle\omega$) (9)
If the integral of f (t)g(t) is called P , the differential P per frequency range d$\omega$ is

$\displaystyle{\frac{dP}{d\omega}}$ = $\displaystyle{\textstyle\frac{1}{\pi}}$Re $\displaystyle\tilde{f}$($\displaystyle\omega$)$\displaystyle\tilde{g}^{*}_{}$($\displaystyle\omega$) . (10)

4 Periodic Time Dependence

Periodic motion is a common situation. For example electrons executing a circular orbit in a synchrotron move periodically. As noted above, in a real synchroton, the electrons are injected at some time, and eventually collisions or other mechanisms cause them to stop orbiting. Therefore we can analyze this system exactly as in the previous section if we cut of the motion. Let's defer that analysis and instead note that any function that is periodic with period 2$\pi$/$\omega_{0}^{}$ can be written in terms of the complete set of functions on this interval e - i$\scriptstyle\omega_{0}$nt for integer n , - $\infty$ < n < $\infty$ ,

a(t) = $\displaystyle\sum_{n=-\infty}^{\infty}$$\displaystyle\tilde{a}_{n}^{}$e - i$\scriptstyle\omega_{0}$nt . (11)

You can apply Sturm-Liouville to show this.

Multiplying by e i$\scriptstyle\omega_{0}$mt and integrating over a period using the orthogonality of the complex exponentials gives the Fourier series coefficients,

$\displaystyle\tilde{a}_{m}^{}$ = $\displaystyle{\frac{\omega_0}{2\pi}}$$\displaystyle\int_{0}^{2\pi/\omega_0}$dta(t)e i$\scriptstyle\omega_{0}$mt . (12)

Similarly to the previous case, for real a(t) ,

$\displaystyle\tilde{a}_{-m}^{}$ = $\displaystyle{\frac{\omega_0}{2\pi}}$$\displaystyle\int_{0}^{2\pi/\omega_0}$dta(t)e - i$\scriptstyle\omega_{0}$mt = $\displaystyle\tilde{a}_{m}^{*}$ (13)

so that again only positive frequencies are needed

a(t) = $\displaystyle\tilde{a}_{0}^{}$ + 2 Re $\displaystyle\sum_{n=1}^{\infty}$$\displaystyle\tilde{a}_{n}^{}$e - i$\scriptstyle\omega_{0}$nt . (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 - i$\scriptstyle\omega_{0}$t. You will get A1 = A/2 , and the factor of two gives the correct coefficient.

The time average of two such quantities over the period is

$\displaystyle{\frac{\omega_0}{2\pi}}$$\displaystyle\int_{0}^{2\pi/\omega_0}$dta(t)b(t) = $\displaystyle{\frac{\omega_0}{2\pi}}$$\displaystyle\int_{0}^{2\pi/\omega_0}$dt$\displaystyle\left [
\tilde a_0
+ 2 ~ {\rm Re} \sum_{n=1}^\infty \tilde a_n e^{-i \omega_0 n t} \right]\left$[$\displaystyle\tilde{b}_{0}^{}$ + 2 Re $\displaystyle\sum_{m=1}^{\infty}$$\displaystyle\tilde{b}_{m}^{}$e - i$\scriptstyle\omega_{0}$nt]   
  = $\displaystyle\tilde{a}_{0}^{}$$\displaystyle\tilde{b}_{0}^{}$ + 2 Re $\displaystyle\sum_{n=1}^{\infty}$anbn* . (15)
As a check, notice that for a single frequency this goes to the harmonic result. If the time average of a(t)b(t) is called P , the contribution Pn from frequency n$\omega_{0}^{}$ to P is

Pn = 2 Re  anb *n . (16)

5 Relating the various forms

A particle orbiting in a precessing elliptic orbit that never closes is an example where the frequencies are incommensurate and neither of the above methods will work directly. However, either taking a period very long so that the frequencies are nearly commensurate with that period or cutting off the motion so that it is confined to a large but finite interval will allow us to use either the Fourier series or integral. Taking the appropriate limit then gives a general result.

Let's look at setting the time dependence to zero for times |t| > T/2 . The Fourier transform of the cutoff motion $\tilde{f}_{c}^{}$($\omega$) is then

$\displaystyle\tilde{f}_{c}^{}$($\displaystyle\omega$) = $\displaystyle\int_{-\infty}^{\infty}$dtf (t)G(t)e - i$\scriptstyle\omega$t (17)

where G(t) = 1 for |t| < T/2 and zero otherwise. Writing both f (t) and G(t) in terms of their Fourier transforms and performing the time integral, this becomes the convolution,
$\displaystyle\tilde{f}_{c}^{}$($\displaystyle\omega$) = $\displaystyle\int_{-\infty}^{\infty}$dte i$\scriptstyle\omega$t$\displaystyle\int_{-\infty}^{\infty}$$\displaystyle{\frac{d\omega''}{2\pi}}$e - i$\scriptstyle\omega$''t$\displaystyle\tilde{G}$($\displaystyle\omega$'')$\displaystyle\int_{-\infty}^{\infty}$$\displaystyle{\frac{d\omega'}{2\pi}}$e - i$\scriptstyle\omega$'t$\displaystyle\tilde{f}$($\displaystyle\omega$')   
  = $\displaystyle\int_{-\infty}^{\infty}$$\displaystyle{\frac{d\omega'}{2\pi}}$$\displaystyle\tilde{f}$($\displaystyle\omega$')$\displaystyle\tilde{G}$($\displaystyle\omega$ - $\displaystyle\omega$') (18)

$\displaystyle\tilde{G}$($\displaystyle\omega$) = $\displaystyle{\textstyle\frac{2}{\omega}}$sin $\displaystyle\left ( \frac{\omega T}{2} \right)\,$. (19)

For the case where f (t) is localized in time, taking the limit of large T makes $\tilde{G}$($\omega$) oscillate rapidly for |$\omega$| $\gg$ 1/T therefore in the integral above, the slowly varying f ($\omega$') can be replaced by $\omega$ 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 $\lim_{T \rightarrow \infty}^{}$G($\omega$) = 2$\pi$$\delta$($\omega$) . 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 - i$\scriptstyle\omega_{0}$t

$\displaystyle\tilde{f}$($\displaystyle\omega$) = $\displaystyle{\textstyle\frac{1}{2}}$$\displaystyle\left [ A \delta(\omega-\omega_0)
+A^* \delta(\omega+\omega_0) \right]$ (20)

where the second negative frequency delta function can be dropped if the transforms are written in terms of positive frequencies only. In this case, the result of calculating the time integral of f 2(t) will be infinite (i.e. the integral of the square of a delta function). This is of course correct. What we want, however, in this case is the time average not the time integral. Therefore, for these delta function contributions we do the time integral first before taking T to infinity. Delta function contributions in $\tilde{f}$($\omega$) like $\delta$($\omega$ - $\omega_{0}^{}$) get replaced in $\tilde{f}_{c}^{}$($\omega$) by $\tilde{G}$($\omega$ - $\omega_{0}^{}$) . Time integrals which contain a square of $\tilde{G}$($\omega$ - $\omega_{0}^{}$) are of the same form as any other well behaved function of $\omega$ ,

$\displaystyle\lim_{T\rightarrow \infty}^{}$$\displaystyle{\textstyle\frac{1}{\pi}}$$\displaystyle\int_{0}^{\infty}$d$\displaystyle\omega$$\displaystyle\tilde{G}^{2}_{}$($\displaystyle\omega$ - $\displaystyle\omega_{0}^{}$) = $\displaystyle\lim_{T\rightarrow \infty}^{}$$\displaystyle{\frac{2 T}{\pi}}$$\displaystyle\int_{-\infty}^{\infty}$dx$\displaystyle{\frac{\sin^2(x)}{x^2}}$ = 2T . (21)

Applying this to the single frequency case above,

$\displaystyle\int_{-\infty}^{\infty}$dtf 2c(t) = $\displaystyle{\frac{T}{2}}$|A|2 (22)

and the time average is as before. In general, this shows that delta function squared contributions in the frequency integral can be replaced by 2T times the coefficient; the remainder of the time integration can be done as in the case where the motion has a finite duration.

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