As another application, we wish to design an electro-optic modulator. This
takes an input signal at a frequency 
and multiplies with another
signal at a frequency 
to produce an output at either the sum or
difference frequencies; let us consider the difference frequency 
. For example, let the two electric fields be given
by:
The polarization density P is proportional to 
, and in particular the component at
frequency 
is:
where 
.
This implies the induced dipoles, corresponding to 
, have a sinusoidal
pattern in space with wavelength 
. They also oscillate
at the temporal frequency , and hence tend to induce a wave 
at the frequency .
Suppose the material has an index of refraction that depends on frequency, 
. Then 
also depends on
frequency as:
The spatial pattern of the induced dipoles has wavelength:
while the propagating field it produces has wavelength:
In order for power at the difference frequency to emerge, we must satisfy the phase match condition: these two wavelengths must coincide so that the two effects support each other in a constructive manner. This requires:
Achieving this phase match condition is not easy in practice, and requires a combination of careful temperature control and precise crystal orientation (recall all these effects are direction dependent, although we have suppressed this fact to simplify the formulas given above).
Shockingly enough, we can predict the total powers present at these three
frequencies, using the Manley-Rowe relations. These relations apply to
(nonlinear) systems in which energy at several discrete frequencies are
present. Specifically, given two frequencies 
and 
, frequencies of
the form:
for integer m,n are present. For example, 
are permitted, as
are harmonics such as 
and so forth. Let 
denote the total
power input to the system at frequency 
; if there is a net
generation of power at this frequency, then is negative. The basic
assumption is that the system is lossless:
The Manley-Rowe relations are:
For the case where only the frequencies , and 
are
present, we can combine equations (4.14) and (4.15) to obtain:
For example, if we wish modulate a 100MHz signal with a 1GHz signal to produce a 1.1GHz output, and we wish 1mW of output power, then we require 0.0909mW at 100MHz and 0.9090mW at 1GHz. The 1GHz signal is sometimes called the pump since it provides most of the power needed in the modulation process. At the demodulator, we want to combine the 1.1GHz signal with a 1GHz signal to restore a 100MHz signal. Once again, if we input 1mW at 1.1GHz, we produce 0.0909mW at 100MHz and 0.9090mW at 1GHz; in particular, this implies the 1GHz source at the receiver must actually absorb power, and cannot contribute power to the process. Also note that whatever power gain we obtained in modulation (0.0909mW to 1mW) has been lost at the demodulator (1mW to 0.0909mW). If this were not the case, we could amplify signals simply by repeatedly modulating and demodulating them!
