1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Phase Change in a transmission line

  1. Jan 29, 2017 #1

    David J

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    (a) A transmission line has a length, ##l##, of 0.4λ. Determine the phase change, ##\beta l##, that occurs down the line.

    2. Relevant equations
    ##\beta=\frac{\omega}{f\lambda}## or ##\beta=\frac{2\pi}{\lambda}##

    3. The attempt at a solution
    This question was posted a couple of years ago. I wanted to ask questions in that original thread but the last post requested people not to use old threads but start new ones.
    Thus this post.
    My notes give me the equation
    ##\beta=\frac{\omega}{f\lambda}##
    The original solution that was posted back in 2015 is shown below and used this equation.

    β = ω / fλ
    ∴ βℓ = ω 0.4λ / fλ
    = ω 0.4 / f
    = 0.4x2π f / f
    = 0.8π (=2.513 to 3d.p.)

    ##\beta=\frac{\omega}{f\lambda}## I understand this
    The question is asking me to determine the phase change which is ##\beta l## or ##\beta## multiplied by ##l##

    so ##\beta l =\frac{\omega(0.4\lambda)}{f\lambda}## What exactly has happened here?

    I can see that the ##0.4\lambda## has been introduced to the equation but I dont understand why it has been inserted at that point

    I can see below that the ##\lambda## have been removed as they cancel each other out
    ##=\frac{\omega(0.4)}{f}##

    This next part I am struggling to understand. Suddenly there is a ##2\pi f## introduced
    so now we have ##=\frac{0.4(2\pi f)}{f}## Does ##\omega## in some way ##=2\pi##??

    In the next step the ##f## appear to have cancelled each other out and we are left with
    ##=0.8\pi=2.513## and I believe this answer is correct in radians

    However

    During the initial post it was recommended to use the equation ##\beta=\frac{2\pi}{\lambda}##

    The statement was What was with the ω and f? β = 2π/λ so βl = (2π/λ)(0.4λ) = 2.513 radians.

    So from this I am guessing that ##\omega## does in fact ##=2\pi##

    ##\beta=\frac{2\pi}{\lambda}##

    I need to multiply ##\beta## by ##0.4l## so ##\beta l=\frac{2\pi}{\lambda}(0.4\lambda)##

    The ##\lambda## cancel out so we are left with ##\beta l=(2\pi)(0.4)##

    so ##\beta l=2.513##

    The answer is the same as the first equation.

    I have drawn this post out a bit by trying to explain my understanding of all of this. I understand the second method better than the first because in the first we had to deal with ##f## for which I did not have a value.

    My question is, is my understanding of this correct ???
    Does ##\omega=2\pi##
    What is ##f##

    Thanks
     
  2. jcsd
  3. Jan 29, 2017 #2

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2016 Award

    I'd say it's a safe bet the author uses ƒ for frequency in cycles per second
    and ω for frequency in radians per second
    of course there being 2π radians in a cycle, ω = 2πƒ

    β then is radians per unit length ? 2π radians per wavelength?
    We often use wavelengths instead of inches to describe transmission lines and antennas. After you've used Smith charts a couple times it'll be intuitive.
    I think you do understand it, just you can't believe it's so easy, ω = 2πƒ


    good luck in your studies.
     
  4. Jan 30, 2017 #3

    David J

    User Avatar
    Gold Member

    Yes I think i understand it a bit better now and it is easier than I thought
    When you have this equation, βl=(2π)(0.4), the result relies purely on the ##\lambda## (wavelength) value because ##2\pi## is obviously fixed. It makes sense now.

    Thanks for the advice with this
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Phase Change in a transmission line
  1. Phase change? (Replies: 2)

  2. Transmission Line (Replies: 5)

  3. Transmission line (Replies: 3)

Loading...