Signal Processing energy conservation

  • #1

Homework Statement



Two receivers, d=5 [meters] apart, are recording an air wave signal.
The air wave travels in v=330 [m/sec] and coming from one side of the receivers.
The air wave contains all the frequencies between 10 [Hz] and 200 [Hz].

a) If we sum up the recorded signal from the two receivers, does all the energy from the original wave will be conserved ? explain.

b) The sampling rate in the receivers is Δt=0.004 [Sec], sketch the amplitude spectrum of the sum series.

Homework Equations



[itex]$\omega =2\pi f$[/itex] ,

[itex]\[\sum\limits_{n=0}^{N-1}{|}{{x}_{n}}{{|}^{2}}=\frac{1}{N}\sum\limits_{k=0}^{N-1}{|}{{X}_{k}}{{|}^{2}}\][/itex] (Parseval Theorem)

and maybe the DFT of sin/cos functions

The Attempt at a Solution



a) I know that the starting point should be the distance and the velocity.
so we get t=v/d and from that the frequency can be calculated somehow.
also the wave is a combination of known sin waves, so after DFT it transfers into a delta function.

b) Here i don't even know how start, since nothing is implied about the amplitude !
 

Answers and Replies

  • #2
Hint: Since you are summing the outputs of two sensors, you must account for the frequency-dependent phase shift between them.
 
  • #3
Thank you! but...

Hint: Since you are summing the outputs of two sensors, you must account for the frequency-dependent phase shift between them.

You are right, i calculated the phase change between the receivers, i got 66 [Hz].
but how this affects the summation ?
 
  • #4
Your Nyquist frequency is less than 200 Hz.
 
  • #5
Thank you! but...

Your Nyquist frequency is less than 200 Hz.

How did you compute the Nyquist frequency in this ?
In order to calculate the Nyquist frequency i need the sampling time interval of the receivers no ?
 
  • #6
How did you compute the Nyquist frequency in this ?
In order to calculate the Nyquist frequency i need the sampling time interval of the receivers no ?
The sampling period is given in part b of the problem as 0.004sec. The inverse of that, 250Hz is the sampling rate. The Nyquist - well, you can take it from there.
 
  • #7
Thank you

The sampling period is given in part b of the problem as 0.004sec. The inverse of that, 250Hz is the sampling rate. The Nyquist - well, you can take it from there.

Thank you,
but I'm not sure that i can use the given information in part b for calculations in part a ...

and as for part b, how can i get anything about the amplitude ?
 
  • #8
Part a is, in part, a spatial question. Draw a picture of a wave impinging on your sensor array and notice the difference in phase. A plane wave has the form [itex]A(\omega)=A_0 \exp\left[i(\omega t - \vec{k}\cdot\vec{r})\right][/itex] where the wavevector [itex]\vec{k}[/itex] points along the direction of propagation with [itex]|\vec{k}|=2\pi/\lambda[/itex]. Perform the summation, using the positions [itex]\vec{r}[/itex] of your sensors, before calculating the power.
 
  • #9
Part a is, in part, a spatial question. Draw a picture of a wave impinging on your sensor array and notice the difference in phase. A plane wave has the form [itex]A(\omega)=A_0 \exp\left[i(\omega t - \vec{k}\cdot\vec{r})\right][/itex] where the wavevector [itex]\vec{k}[/itex] points along the direction of propagation with [itex]|\vec{k}|=2\pi/\lambda[/itex]. Perform the summation, using the positions [itex]\vec{r}[/itex] of your sensors, before calculating the power.

Ok, i think i got part A:

In order to conserve all the frequencies, there must be at least 2 samples per 1 wave length. so the spatial Nyquist is then: [itex]${{k}_{\max }}=\frac{1}{2\Delta x}=0.1$[/itex]

and on the other hand:

[itex]$k=f/v\to {{k}_{\max }}=200/330\sim 0.6$ [/itex]

so we won't get all the energy, is that right ??

but I'm still struggling with part b, since nothing is said about the amplitude,
 
  • #10
No, part a has nothing whatever to do with digital sampling. Have you studied anything about wave propagation? Can you visit your TA or professor? It will be easier to get this explained face to face and with a whiteboard handy.
 
  • #11
No, part a has nothing whatever to do with digital sampling. Have you studied anything about wave propagation? Can you visit your TA or professor? It will be easier to get this explained face to face and with a whiteboard handy.

My Professor is unavailable, and there is no TA.
i know the basics of wave propagation but i just don't understand how to perform the summation.
Following your steps i get:

[itex]$u(x,t)=\sum\limits_{\omega =20\pi }^{400\pi }{[A(\omega )~{{e}^{i(\omega t)}}+B(\omega )~{{e}^{i(5k-\omega t)}}}]\,\,$[/itex]
 
  • #12
Good. Replace the summation with an integral, since the problem states that all frequencies in the band are present. Let's assume that they have equal strength, so that [itex]A(\omega)=A_0[/itex] for 20π≤ω≤400π. For line-of-sight propagation in a homogeneous medium, furthermore, A = B. Your integral then takes on a simpler form that you should be able to evaluate for power and compare to the power in the incident wave.
 

Suggested for: Signal Processing energy conservation

Replies
3
Views
911
Replies
4
Views
3K
Replies
1
Views
711
Replies
2
Views
3K
Replies
5
Views
608
Replies
4
Views
579
Replies
1
Views
541
Replies
23
Views
1K
Replies
1
Views
337
Back
Top