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Frequency and wavelength spectrums

Doug Kerr

Well-known member
The color of light is determined by its spectrum, for which the more precise technical name is power spectral density (PSD). In the usual meaning in color science this is a function of wavelength (we can think of it in its graphical presentation as a curve plotted against wavelength) that describes how the power contained by the light is distributed over different wavelengths.

In electrical engineering, the concept of power spectral density is also important. There, we most commonly speak of a function of frequency that describes how the power contained by a "signal" is distributed over different frequencies.

We can plot the PSD of something either in terms of wavelength or frequency, as best fits our context. Either will describe the same situation.

Wavelength is of course directly related to frequency, thus:

L=C/f

where L is the wavelength, f is the frequency, and C is the speed of light.

Thus we might think that the two kinds of PSD (lets speak of them as their graphic plots) differ only in the meaning of the scale of the y-axis: whether that is in terms of frequency or wavelength.

But in fact there is a far greater difference.

The word "density" in the name power spectral density refers to the concept of a density function. Lets consider a PSD of the frequency flavor. There, the ordinate (vertical value) does not tell us how much power there is at the specific frequency. In fact, assuming a "continuous" spectrum, there is zero power at any specific frequency.

If this seems startling, let me give an analogy that may help. Suppose we have an aluminum bar of circular cross section but tapering diameter. Now, imagine a plane through the rod, perpendicular to its axis at some particularly place along its length, say exactly 2 inches from some reference point. How much aluminum (in volume) is there in the rod at that place? Zero - the volume of an area on a plane is zero, so in the part of the plane that is within the rod, there is zero aluminum.

Now while we have this analogy in front of us, let's proceed. Suppose that at the point of interest the cross-sectional area of the bar is 0.001 m². Consider a "slice" of the bar of a very tiny thickness - we will call the thickness dx. Actually, this slice is a cylinder of average cross-section area 0.001 m² and length dx. Thus, its volume is 0.001 dx m³.

Said another way, the volumetric density of the rod at this point is 0.001 m³ per inch of length (assuming a very tiny length - the famous "dx" of the calculus, so we need not be concerned with the cross-sectional area changing).

Now, back to our PSDs.

We often speak of a uniform PSD (a "flat" curve of the function). For such, there is a tendency to think that the abscissa can be frequency or wavelength, whichever best fits our context, with no effect on the "shape" of the curve, given that it is flat. But far from that.

If we start with a power spectral density, in terms of frequency, that is uniform, this means that the amount of power per unit frequency is constant over the range of frequency of relevance.

Now consider plotting the same power spectral density as a function of wavelength. Now, if the amount of power per unit frequency is constant, the amount of power per unit wavelength is not, because of the nonlinear relationship between wavelength and frequency (L=C/f).

Consider a radio spectrum running from 100 Mhz (L=3.00 m) through 200 Mhz (L=1.50 m). If the power spectral density on a frequency basis is uniform with frequency, the the power in the range from 99.5-100.5 Mhz (an interval of 1.0 Mhz) is the same as that in the range from 199.5-200.5 Hhz (also an interval of 1.0 Mhz).

But the power in the range of wavelength 2.995 - 3.005 m (an interval of +0.01 m, but an interval of frequency of -0.333 Mhz) is not the same as the amount of power in the range 1.495 - 1.505 m (also an interval of -0.01 m, but -1.333 Mhz, ). It is in fact essentially 1/4 as great.

Thus the plot, on a wavelength basis, of a power spectral density that (on a frequency basis) is uniform will be a curve that declines with increasing wavelength - it is in fact an "inverse square" curve, as can be readily demonstrated with a little work with (the) calculus.

Thus, we need to be very cautious when we begin to work mathematically with a power spectral density.

In complicated theoretical work on color spaces this distinction needs to be carefully attended to.

Best regards,

Doug
 

Doug Kerr

Well-known member
Some further reading suggests that there is a fairly-widely-accepted convention for keeping separate these two kins of power spectral density functions.

• Power spectral density as a function of frequency is called power spectral density (PSD).

• Power spectral density as a function of wavelength is called spectral power distribution (SPD).

They are of course both density functions, and they of course both show distribution of power over the range of the spectrum.

So how can we go wrong!

Now how do we remember which is which?

Best regards,

Doug
 
Now how do we remember which is which?

Hi Doug,

Well, by observing the abscissa (horizontal axis)? We photographers usually talk about Spectral wavelengths, not the physical frequency (both have their uses, e.g. the dual behavior of photons as waves in polarization or as energy for the determination of threshold sensitivity).

So for photographers the wavelength, thus SPD, is usually the more probable when we speak of spectrally uniform reflection of a "White balance" target. We do recognize that the energy of shorter wavelengths (e.g. X-rays) is greater than of longer wavelengths (e.g. Radio waves). Yet, in the limited sensitivity range of common silicon based photovoltaic sensors, 1 photon (regardless of wavelength) seems to generate 1 electon Volt of charge, which may be an approximation, but it's commonly used in combination with the Poisson probability distribution to predict photon (shot) noise characteristics.

Thanks for pointing out the potential pitfalls.

Cheers,
Bart
 

Doug Kerr

Well-known member
Hi, Bart,

Well, by observing the abscissa (horizontal axis)?
No. I meant how do we decide which of the two names do we apply to each case. There is nothing about the two names that makes one uniquely suited for one situation.

There are some suggestions that "density" does not properly apply to the function of wavelength, but I have not seen any arguments for that notion. In both cases, the ordinate tells us the amount of power per unit span of the abscissa (again, in the limit, where the span approaches zero) - the classical concept of a density function.

We photographers usually talk about Spectral wavelengths, not the physical frequency (both have their uses, e.g. the dual behavior of photons as waves in polarization or as energy for the determination of threshold sensitivity).
Indeed.

So for photographers the wavelength, thus SPD, is usually the more probable when we speak of spectrally uniform reflection of a "White balance" target.

Actually, for reflectance, the two functions are the same. Only the scale of the abscissa differs.

The reason is that when we speak of "spectral reflectance", the concept of density isn't involved. The reflectance at any wavelength (or frequency) is that ratio of the reflected power to the incident power (in any small span of wavelength, or frequency).

There is a 1/pi in there when SI units are used, but for convenience I will ignore that for the moment .​

So if we consider the reflected power within a tiny span of wavelength, centered about some wavelength, and the incident power within that same tiny span, and they are the same, then the reflectance at that wavelength is 1.

Now, if we consider the reflected power in some tiny span of frequency, centered about the frequency that corresponds to that same wavelength, and the incident power in that same tiny span of frequency, and they are the same, again the reflectance at that frequency is 1. And if one of those is true, so is the other.

If this is true (either in the wavelength model or the frequency model) across the entire band of wavelength/frequency of interest, then the surface is a spectrally-uniform ("neutral" - some would say "white") reflector.

The difference between this and the matter of spectral density is that here we never divide the amount of power within some small span of wavelength [or frequency] by that span of wavelength [or frequency]. We just use it as is.

So a "spectrally uniform" reflecting surface will have a flat plot of reflectance vs. wavelength, and also a flat plot of reflectance vs. frequency.

Interesting.

Thanks for your insights.

Best regards,

Doug
 

Doug Kerr

Well-known member
There are several complications in the matter of naming the various density functions of interest here.

One is the word "spectrum", which has a spectrum of different meanings (to use it with one of its own many meanings).

The second is the fact that in a density function, there are two "bases" of interest.

One is the "denominator" of the density; that is, a density is the amount of some "stuff" per unit of some variable. Thus, a type of density has the unit kg/m³. The "base" of that density is volume (in m³).

The full formal name of that ratio, by the way, is "volumetric density of mass" (or "mass volumetric density").

The second is of what variable is that ratio considered a function - it varies depending on what?

We might be interested in the (volume) density of the air at ground level (that being in terms of kg/m³) as a function of time of day.

We can express the density itself in the terms of (the) differential calculus, as:

dm/dv

where m is the quantity mass and v is the quantity volume. The use of the differential notation reminds us that, strictly, we are interested in the ratio of the amount of mass to the corresponding volume in the limit as the amount of volume approaches zero.

Now, if we are interested in this density as a function of time of day, we can use this notation to remind us of that:

dm/dv (t)​

where t is the time of day. Thus means, "the ratio dm/dv as a function of t".

Now, when we have the customary power spectral density of the "frequency" flavor, what we really have is:

dP/df (f)​

Now, frequency, f, plays two quite distinct roles:

• The "denominator" of the ratio (the "per" quantity; the f of df).

• The value of which this ratio is a function - the f of (f).

And that is common for density functions, but not ubiquitous (as we saw a little earlier).

For example, we could be interested in the amount of power per hertz of frequency span, but we want to know the value of that ratio for different values of wavelength. In other words, we are interested in:

dP/df (L)​

where L is wavelength.

However, the usual power spectral density is in fact:

The frequency density of power, as a function of frequency.
And the function with which we are commonly interested in colorimetry is in fact:

The wavelength density of power, as a function of wavelength.

Best regards,

Doug
 

Doug Kerr

Well-known member
In this figure we have plotted the power spectral density (wavelength outlook) (called, by convention in this case, the spectral power distribution, SPD) of a certain kind of light. We cover the range of wavelength from 400-700 nm, this being a handy gross representation of the range of visible wavelengths.

The total power in the light is uniformly distributed over that range of wavelength.

psd_11.gif


This is essentially the spectrum of CIE standard illuminant "E".

The ordinate (vertical coordinate) is power per unit wavelength, and of course since the abscissa (horizontal axis) is wavelength, the ordinate is a function of wavelength.

This is reflected by my notation for the ordinate:

Spectrum_ordinate_01.gif


P' refers to a power density. The subscript lambda (the symbol, for wavelength) means that the density is per unit of wavelength. The lambda in parentheses means that we look at this value as a function of wavelength.

Because the "denominator" of the ordinate is in fact of the same dimensionality and has the same unit as the abscissa, and since both quantities are plotted on linear scales, then the geometric area under the curve over the range of wavelengths considered is equal to the total power of the signal (0.3 W, in this case).

Now we consider the very same light on the basis of its power spectral density (frequency outlook) (called, by convention in this case, the power spectral density, PSD).

psd_12.gif


The range of frequencies for the curve - about 375-750 Thz - is (with just a tiny bit of rounding) consistent with the range of wavelengths on the previous figure. The little square and round markers help us realize which end of this curve corresponds to which end of the previous one.

Now the ordinate (vertical coordinate) is power per unit frequency, and of course since the abscissa (horizontal axis) is frequency, the ordinate is a function of frequency.

Because the "denominator" of the ordinate is in fact of the same dimensionality and has the same unit as the abscissa, and since both quantities are plotted on linear scales, then the geometric area under the curve over the pertinent range of wavelengths is equal to the total power of the signal (0.3 W, in this case). (It certainly had better come out that way, since it is the same light!)

The reason the curve is "flat" in one case and "decreasing, concave upward" in the other is not because the horizontal axis is wavelength in one case and frequency in the other. It is because the "base" of the density (the "per" quantity) is wavelength in one case and frequency in the other.

We can see this distinction in this figure, a presentation we do not often see:

psd_14.gif


Here the ordinate is power per unit wavelength, but we plot it as a function of frequency (perfectly legitimate).

Since this is the same light as before, the power per unit wavelength is the same as before: constant with wavelength, and thus constant with frequency (constant is constant). And we see that from the plot.

In this case, since the "base" of the density is not the same as the abscissa, the area under the curve does not relate to the total power (which is however, as noted, the same as before).

Best regards,

Doug
 
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