Solution Found!

Confirm that the energy in the TEmn mode travels at the

Chapter 9, Problem 30P

(choose chapter or problem)

Get Unlimited Answers
QUESTION:

Problem 30P

Confirm that the energy in the TEmn mode travels at the group velocity. [Hint: Find the time averaged Poynting vector and the energy density  (use Prob. 9.12 if you wish). Integrate over the cross section of the wave guide to get the energy per unit time and per unit length carried by the wave, and take their ratio.]

Reference prob 9.12

In the complex notation there is a clever device for finding the time average of a product. Suppose f (r, t) = A cos (k · r – ω + δa) and g(r, t) = B cos (k · r − ωt + δb). Show that  where the star denotes complex conjugation. [Note that this only works if the two waves have the same k and ω, but they need not have the same amplitude or phase.] For example,

Questions & Answers

QUESTION:

Problem 30P

Confirm that the energy in the TEmn mode travels at the group velocity. [Hint: Find the time averaged Poynting vector and the energy density  (use Prob. 9.12 if you wish). Integrate over the cross section of the wave guide to get the energy per unit time and per unit length carried by the wave, and take their ratio.]

Reference prob 9.12

In the complex notation there is a clever device for finding the time average of a product. Suppose f (r, t) = A cos (k · r – ω + δa) and g(r, t) = B cos (k · r − ωt + δb). Show that  where the star denotes complex conjugation. [Note that this only works if the two waves have the same k and ω, but they need not have the same amplitude or phase.] For example,

ANSWER:

Solution 30P

Step 1 of 5:

In the question, our aim is to show that the energy in the 54CodeCogsEqn.gif mode travels at the group velocity.

We have to start with the time averaged Poynting vector A1CodeCogsEqn.gif and the energy density a2CodeCogsEqn.gif and then integrate them over the cross section of the waveguide to get the energy per unit time and energy per unit length and take their ratio to prove that it will be same as the group velocity.

Add to cart


Study Tools You Might Need

Not The Solution You Need? Search for Your Answer Here:

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back