Frequency and amplitude spectra of cosine wave
john baez
A*cos (2*pi*f0*t)
is
A/2 * (Dirac(f-f0) + Dirac(f+f0))
That is two impulses, one at a positive and one at a negative frequency.
This frequency spectrum is the same as the amplitude spectrum since it is
real (no imaginary part).
I have repeatedly run into the assertion that the amplitude spectrum is
A * (Dirac(f-f0))
This doesn't seem right. First of all, what happened to the negative
frequencies? I thought that the negative half of an amplitude spectrum
was never plotted because it was redundant--just identical to the positive
side. I didn't think you multiplied the positive half by 2.
My confusion is compounded by this idea. Suppose you are finding the
amplitude response of a system--say a lens. You present a cosine grating (i.e.
a cosine + mean luminance). You would measure the amplitude and phase of
the output (image) grating. If the grating is low frequency, it will pass
through perfectly. It seems to me that you will get an amplitude of A, not
A/2 (forget about the DC component).
Can anyone help clear this up for me?
Thanks!!
Bill Simpson
Carlos Antunes
>The Fourier transform of a cosine wave
>A*cos (2*pi*f0*t)
>is
>A/2 * (Dirac(f-f0) + Dirac(f+f0))
>That is two impulses, one at a positive and one at a negative frequency.
>This frequency spectrum is the same as the amplitude spectrum since it is
>real (no imaginary part).
>I have repeatedly run into the assertion that the amplitude spectrum is
>A * (Dirac(f-f0))
Maybe the wave is being represented as A*exp(i*2*pi*f0*t). It is very common
and very useful to do this, as it simplifies calculations. Then, you will
only have to take the real part of the result, to get the right answer.
Regards,
Carlos Antunes.
--
Carlos Antunes @ SoftSousa, Lda Tel: +351-1-3975303
cm...@softsousa.pt Fax: +351-1-3975889
[Moderator's note: Antunes, or at least his software, appeared to be
under the impression that I, rather than Bill Simpson, posted the original
query. Anyway, he may have gotten to the crux of Simpson's puzzlement.]
te...@asl.dl.nec.com
In article <22JAN94....@uwpg02.uwinnipeg.ca> Bill Simpson writes:
> The Fourier transform of a cosine wave
>
> A*cos (2*pi*f0*t)
> is
> A/2 * (Dirac(f-f0) + Dirac(f+f0))
>
> That is two impulses, one at a positive and one at a negative frequency.
> This frequency spectrum is the same as the amplitude spectrum since it is
> real (no imaginary part).
>
> I have repeatedly run into the assertion that the amplitude spectrum is
>
> A * (Dirac(f-f0))
>
> This doesn't seem right. First of all, what happened to the negative
> frequencies? I thought that the negative half of an amplitude spectrum
> was never plotted because it was redundant--just identical to the positive
> side. I didn't think you multiplied the positive half by 2.
May I suggest a (longish!) graphical interpretation of your question?
COMPOSITE X/ARGAND SPACE
First, picture a space with one ordinary spatial dimension (x) and the
Argand (complex) plane orthogonal to it (here labeled by y and i). For
reasons of simplicity in drawing trig functions in this space, I've rotated
the Argand plane by pi/2 so that the real dimension of the Argand plane
(y) will be vertical. (The added commas on +i indicate the axis that is
"nearer" to the reader.):
(1) y
| -i
| /
| /
|/
-x ---+--------------------------------------------------------- x
,/|
,/ |
,/ |
+i |
-y
COSINES IN X/ARGAND SPACE
Now in this space a cosine function is represented like this:
(2) y
| -i
| /
,'"`. ,'"`. ,'"`. ,'"`.
/ |/ \ / \ / \ / \
-x /---+---\-------/-------\-------/-------\-------/-------\---- x
,/| \ / \ / \ / \
,/ | `._,' `._,' `._,' `._
,/ |
i |
-y
That is, it is purely a real function y(x), and thus really does not need
the third (imaginary) dimension (i) to represent it. It is like a flat
drawing residing in a three-dimensional space, with no "volume" to it.
REPRESENTING FREQUENCY DELTA FUNCTIONS IN X/ARGAND SPACE
However... What if you want to do a Fourier transform on the all-real (2)
cosine function? What is the proper representation of a Dirac delta
functions that result from such a transformation?
It is _not_ the waveform (2). Although Dirac delta functions are _real_
when represented in frequency space (where as you have noted they are just
simple point locations along the frequency axis), they are _not_ so simple
when translated back to real space. Specifically, Dirac delta functions
in frequency space always appear as infinitely long _helices_ when shown
in the space introduced in (2):
(3) y
| -i
| /
,'"`. ,'"`. ,'"`. ,'"`.
|/ @ , @ , @ , @
-x ----+---@---------------@---------------@---------------@---- x
,/| @ ' @ ' @ ' @
,/ | `._,' `._,' `._,' `._
,/ |
i |
-y
You have to use your imagination there a bit, but all I've really drawn is
a stretched-out Slinky centered around the x axis. You will note that this
function is quite distinct from the "flat" cosine function (2) -- this one
has a definite extent and even a "volume" (enclosed 3D region) in space (1).
NEGATIVE AND POSITIVE FREQUENCY DELTAS IN X/ARGAND SPACE
Each Dirac function also has a definite "handedness" when represented in
space (1). That is, its spiral can be either left-handed or right-handed:
(4a) Right-handed Delta function helix:
y
| -i
| /
,'"`. ,'"`. ,'"`. ,'"`.
|/ @ , @ , @ , @
-x ----+---@---------------@---------------@---------------@---- x
,/| @ ' @ ' @ ' @
,/ | `._,' `._,' `._,' `._
,/ |
i |
-y
(4b) Left-handed Delta function helix:
y
| -i
| /
,'"`. ,'"`. ,'"`. ,'"`.
|/ . @ . @ . @ .
-x ----+-----------@---------------@---------------@------------ x
,/| ` @ ` @ ` @ `
,/ | `._,' `._,' `._,' `._
,/ |
i |
-y
And how does this this "handedness" translate into frequency space?
Not too surprisingly, it translates into a _sign change_ for the helix,
without changing its frequency. This just makes good sense, really, as
(4a) and (4b) clearly _do_ have the same frequency -- they just take the
contrary approaches in how they curl around the x axis to achieve that
frequency. And for that matter, (4a) and (4b) also have the exactly the
same frequency as the "flat" cosine function (2).
ADDING NEGATIVE AND POSITIVE DELTA FUNCTIONS
Now here's an interesting question: What exactly would happen if one were
to _add_ the amplitudes of (4a) and (4b) along the x axis? What kind of
function would be the result of such a linear combination of the two?
Intuitively, one might be tempted to guess that adding adding negative and
positive helices of equal magnitude might result in a "null" or zero
function. Wrong! What actually happens is much more interesting: The
two helices _cancel out each others imaginary dimensions_, while at the
same time adding up _constructively_ in the real dimension y to give a
doubling of amplitude in y:
(5a) Overlaid left-and-right handed helices:
y
| -i
| /
'^` . , '^` . , '^` . , '^` .
@|@ `. ,' @ @ `. ,' @ @ `. ,' @ @ `.
-x ----+-@-----------@---@-----------@---@-----------@---@------ x
,/| @ ` ' @ @ ` ' @ @ ` ' @ @ `
,/ | ` .x, ' ` .x, ' ` .x, ' ` .x
,/ |
i |
-y
(5a) Resulting point-by-point (linear) sum of amplitudes:
y
| -i (All y amplitudes times 2)
| /
,'"`. ,'"`. ,'"`. ,'"`.
/ |/ \ / \ / \ / \
-x /---+---\-------/-------\-------/-------\-------/-------\---- x
,/| \ / \ / \ / \
,/ | `._,' `._,' `._,' `._
,/ |
i |
-y
Shucky darns! It's our old friend, the no-imagination cosine function,
only scaled by a factor of two from the addition of its _two_ Dirac delta
(Fourier transform) components! From this we can quickly gather that the
correct Fourier components are two "in phase" spirals (e^ixk) functions
with the same frequency as the cosine function, but opposite signs and
each with half the amplitude as the cosine.
You can, if you wish, "pretend" that the cosine function is simply the
_projection_ of a same-amplitude helix onto the xy plane -- that is, by
simply "throwing away" the imaginary part, instead of canceling it out
by linear summing with the f* (inverted sign) version of the function.
For many types of waves (e.g., sound) this is entirely satisfactory, since
for such waves the imaginary part is nothing more than a mathematical
convenience without any physical implications.
However, for _quantum_ waves, the imaginary part of the wave is very much
a participant in determining the final outcome of an experiment. It has
a significant impact on how the waves interfere to produce detectable
particle distributions. Thus for quantum waves, it is not really valid
to take the simple projection of an e^ixk helix and expect your amplitudes
to stay unchanged after the Fourier transform. The simplest interpretation
is simply that the resulting Fourier components -- the left and right handed
helices -- should have half the amplitude of the original cosine.
IMPLICATIONS OF REAL-ONLY FUNCTIONS IN QUANTUM MECHANICS
There is an interesting general principle here, by the way. By adding
helical (e^ixk) wave functions together in various ways, you can in some
cases _eliminate_ the imaginary component of the wave function to give a
nice, real-only function such as in the cosine example above. Physically,
such combinations represent especially interesting wave functions that have
_real_ physical distributions of where the particle can be found. Atomic
electron wave functions and other types of "bound" wave functions are good
examples of such "real world" probability distributions that are ultimately
the result of adding together Dirac components with opposite sign (helicity).
Intuitively, you can get a little bit of feel for what's going on by noting
that in the classic "particle in a box" quantum mechanical problem, the
particle is required to bounce _back and forth_ between the walls. Each
"bounce" changes the sign of its momentum, and thus the handedness of the
helix that represents its momentum in space (1). When the scales are small
enough, time/energy uncertainty then permits the two _oppositely signed_
helices representing particle motion to "merge" and become a single "real
only" cosine wave function. Because it is real, the distribution of the
particle ceases to be continuous in real space, and distinct nodes at which
the particle cannot be found are formed where the cosine function crosses
the x axis.
References: The above is largely home-grown, but hardly radical -- it's
just a mapping of standard QM functions into x/Argand space. I note that
Penrose uses a nearly identical helix representation Dirac delta functions
in his popular book "The Emperor's New Mind," if you would like to see a
bit more explanation of it. (I've skipped over the closely related issue
of e^ixk helices in momentum space, which Penroses briefly addresses there.)
.....
My apologies for such a long explanation of a simple question, but I do find
the graphical representations of such question most interesting, and I think
that in at least some cases they can provide a better "feel" for what is
going on in the more precise mathematical representations. A wee bit of
visual intuition can be helpful at times, I suspect.
Cheers,
Terry Bollinger