Microwave Interference and Signal Detection: Understanding Homework Question

[Jimmy87]

Homework Statement

I have attached the question to this thread as it requires looking at a diagram.

Homework Equations

None

The Attempt at a Solution

We are supposed to attempt the question and mark it but I have a few questions if people could kindly answer for me please. In the main 6 mark question I said in my answer that the path length difference from M and H would vary between 0 - 14mm (half a wavelength). If you look at the answer it says 0 - one full wavelength (28mm). I don't see how that is possible? If it is moved from plate M to 7mm away how could it vary anymore than half a wavelength. Also, the next 3 mark question requires using the path length difference of 14mm to get the answer. Unless it reflected of H again a second time but then if you go down that route it could reflect multiple times and have a path length difference greater than one wavelength.

My second point of confusion is about other stuff the answer allows. It says allow full credit for (A^2+a^2)/(A^2-a^2). How is this calculation useful to the question? I get that when M and H are on top of each other the combined amplitude is A+a and when H is 7mm it would be A-a but how does finding a ratio of their squares help answer the question?

Many thanks for any help given.

 

[haruspex]

I agree with you re the variation being only half a wavelength. In fact, path length difference generally refers to the unsigned difference, since that is what you care about for resulting amplitude, so no matter how far the sheet is displaced the path difference cannot be more than half a cycle.

Wrt the square of amplitude ratio, the detectability of the intervening sheet will depend on whether the variation in power detected is sufficiently great in relation to the average power. However, I do not see how they get that particular expression. When the amplitudes simply add, isn't the power (A+a)²? But I am no expert on this matter, so I could be wrong.

 

[Jimmy87]

I agree with you re the variation being only half a wavelength. In fact, path length difference generally refers to the unsigned difference, since that is what you care about for resulting amplitude, so no matter how far the sheet is displaced the path difference cannot be more than half a cycle.

Wrt the square of amplitude ratio, the detectability of the intervening sheet will depend on whether the variation in power detected is sufficiently great in relation to the average power. However, I do not see how they get that particular expression. When the amplitudes simply add, isn't the power (A+a)²? But I am no expert on this matter, so I could be wrong.

Thanks. I am confused when you say that you can never get a path length difference of more than half a wavelength. Surely if plate H is 14mm away the difference is 28mm - one full wavelength? What do you mean by unassigned difference?

Do you think the square ratio formula is to put a numerical answer to the difference in the signals being big enough to be detectable? So if 'a' is much less than 'A' the equation would be close to one so the bigger the difference, the closer the equation is to zero and the more detectable the difference? Is that why? I have no idea why you would square them because I thought you do this to keep the answer positive but 'A' will always be bigger than 'a'?

 

[haruspex]

What do you mean by unassigned difference?

We do not care about whole wavelengths in the difference, so a difference of x is equivalent to a difference of x+λ. And if the individual lengths are x and y then we only care about the unsigned difference |x-y|. So consider y=x+2λ/3. The difference is the same as for y+λ and x, which would be λ/3.

Do you think the square ratio formula is to put a numerical answer to the difference in the signals being big enough to be detectable?

Sure, but they used the wrong formula.
If you have a wave $A\sin(\omega t)$ the power it carries is $\frac{\int ( A\omega\sin(\omega t))^2.dt}{\int dt} = \frac 12 (A\omega)^2$.
If there are two superimposed waves of the same frequency in the same direction, amplitudes A and B, phase difference φ, we can find the power by adding the wave equations, squaring and integrating to get $\frac 12(A^2+2AB\cos(\phi)+B^2)\omega^2$.
It is not a coincidence that this is the same formula as for the squared magnitude of a vector sum: $|\vec x+\vec y|^2=|\vec x|^2+2|\vec x.\vec y|+|\vec y|^2 = |\vec x|^2+2|\vec x||\vec y|\cos(\theta)+|\vec y|^2$, where θ is the angle between the vectors.

When in phase, this gives (omitting the factor ω²) ½(A+B)², and when 180 degrees out of phase ½(A-B)². So the ratio they should have quoted is $\left(\frac{A+B}{A-B}\right)^2$.
For the signal to be detectable, this needs to be measurably different from 1. If B is very small it might not be.

Note that the ratio they quoted cannot be right since it implies that with a>A the power ratio would be negative!
Also, note that if the two frequencies are not close to being in an integer ratio then it is equivalent to a rapidly changing phase difference, and also to a fixed phase difference of 90 degrees. In all three cases we get ½(A²+B²). I would guess whoever wrote the question had that in mind.