Phase Change in a transmission line

[David J]

Homework Statement

(a) A transmission line has a length, $l$, of 0.4λ. Determine the phase change, $\beta l$, that occurs down the line.

Homework Equations

$\beta=\frac{\omega}{f\lambda}$ or $\beta=\frac{2\pi}{\lambda}$

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 don't 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 canceled 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

 

[jim hardy]

What is f

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.

My question is, is my understanding of this correct ?

I think you do understand it, just you can't believe it's so easy, ω = 2πƒ


good luck in your studies.

 

[David J]

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