Understanding the Electric Field of a Plane Wave: A Complex Vector Approach

AI Thread Summary
The discussion focuses on calculating the electric field of a plane wave using complex vector representation. The formula E(s) = E(0) * exp(-jks) is examined, where E(s) is the electric field at a distance s, and E(0) is the electric field at the reference point. The challenge arises in understanding how to multiply a complex number with a 3D vector, leading to the conclusion that the electric field can be represented as a complex vector with real and imaginary components corresponding to the electric and magnetic fields, respectively. The instantaneous electric field is derived as the real part of the product of the complex electromagnetic field and the exponential term. The discussion concludes that this approach resolves the multiplication issue since both EM(0) and exp(-jks) are complex numbers.
broli86
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Hi I've working on a computer program where I need to calculate some
electric fields. I am referring to a thesis which provides these
formulas. For eg:

The electric field for a ray(plane wave actually) at a distance s from
the reference point:

E(s) = E(0) * exp(-jks)

where E(s) = electric field at a distance s from the reference point.
E(0) = electric field at reference point.
s = distance travelled.
k = wave number = 2 * PI / wavelength
exp(-jks) as per my thesis, represents the phase variation of the
electric field along the ray.

now, electric field is always a vector(3d in my case). so the user
will input a magnitude for the electric field at the reference point
and i can calculate the reference electric field vector[ E(0) ] easily
by multiplying it with the unit direction vector the wave.

exp(-jks) expands as

cos(ks) - j sin(ks)
^ ^ ^
if E(0) is a vector of the form a x + b y + c z,
^ ^ ^
Then E(s) = (a x + b y + c z) * ( cos(ks) - j sin(ks) )

I'm really confounded at this expression because how does one multiply
a complex number and a vector ? And even if it is possible then what
about E(s) ? If E(0) was defined to be a 3d vector then E(s) must also
be a 3d vector. What does it actually mean ? If there are some EE/
physics experts who can probably figure it out then it would be really
great.
 
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Well really E(s) = Re{Eo * exp(-jks)} (ie the real part)... so in short you're supposed to ignore the sin term.
 
nicksauce said:
Well really E(s) = Re{Eo * exp(-jks)} (ie the real part)... so in short you're supposed to ignore the sin term.

I think that equation was somewhat wrong. I think the EM field can be represented as a complex vector:

A complex vector as in a 3d vector whose components are complex
numbers :
(a1 + i * b1) xhat + (a2 + i * b2) yhat + (a3 + i * b3) zhat

where xhat, yhat, zhat are unit vectors along x, y, z respectively.

one can rearrange it to get real part as the electric field vector and
imaginary part as the magnetic field vector, both at right angles to
each other :
real + i * imag

electric field: (a1 * xhat + a2 * yhat + a3 * zhat)

magnetic field: (b1 * xhat + b2 * yhat + b3 * zhat)


instantaneous electric field: Re( EM(0) * exp( -jks ) )
instantaneous magnetic field: Imag( EM(0) * exp(-jks) )

where k (wave number ) = 2 * pi / wavelength
s = distance traveled in the direction of the plane wave
EM(0) is the EM field at reference point.

Here it seems the problem won't arise while multiplying EM(0) and exp(-jks) as both are complex numbers.

Is this correct ??
 
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