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

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SUMMARY

The discussion focuses on calculating the electric field of a plane wave using complex vector representation. The formula for the electric field at a distance s from a reference point is given by E(s) = E(0) * exp(-jks), where k is the wave number defined as 2 * PI / wavelength. The user seeks clarification on multiplying complex numbers with vectors and concludes that the instantaneous electric field can be represented as Re{EM(0) * exp(-jks)}. The discussion emphasizes that both EM(0) and exp(-jks) are complex numbers, allowing for valid multiplication.

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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 ??
 
Last edited:

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