# How to reflect sound waves

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1. May 7, 2015

### Dr.MIke

I have 2 types of sound reflectors on the image attached.
The angles are not proportional but you should get the idea.
The black circle is the source, the black lines are the waves and the red lines are the reflections of the waves emitted by the source.
My target is to reflect the sound waves so they end up parallel.
Wich one will work?
If booth work, Wich one do you think is better? And why?
What material reflects sound best?
Will the range of sound and amplitude rise using the reflector?
Simple questions i know, but i just want to make sure.
Thanks.

2. May 7, 2015

### rumborak

Just to make sure here, you posted the same picture twice, correct? I can't make out a difference between pic 1 and 2.

If the question is left vs right, theoretically both will work equally as well.

3. May 7, 2015

### Dr.MIke

You gotta ignore the 2nd picture. :P

4. May 7, 2015

### sophiecentaur

If you base the behaviour of the reflector on idealised 'pencil beam' rays then all parabolic reflectors will produce parallel rays if the source is at the focus. However, sound (light radio etc.) are all wave phenomena and you will always get diffraction effects. when the wave is constrained by a hole or a finite reflector. The wider the aperture of the reflector, the less will the beam be spread by diffraction so I guess the simple answer to your question is that the left hand reflector will produce the 'best' beam - i.e. with least spread at a distance.

5. May 7, 2015

### Dr.MIke

You are right, i havn't noticed that. I also havn't really given a thought about it. But that answers my question.
That answerd, Placed near the source there will also be the receiver with a distance of 2 cm from the source. Think of it as a small ultrasonic range measurement device.
When reflected by an object, is there any chance that the waves, or at least good part of it, will enter the receiver or will it go back to the source without enough spread to reach the receiver?

6. May 7, 2015

### sophiecentaur

You don't say the actual scale of the set up or the wavelength involved but 2cm sounds like very near field conditions. Try a diagram of the whole apparatus with some idea of scale. Try to make your question a bit clearer. You know the details; I don't.

7. May 7, 2015

### Dr.MIke

I hope this file can help you understand the sensor and get some information about it. (size, etc.)
Dont really want to get too much into electronics though as this ia a physics website.
By the way, if it would not be possible, i could use the source as input and output at the same time.

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8. May 7, 2015

### sophiecentaur

Are you trying to increase the range / sensitivity of this system? The unit on the spec sheet seems to be purpose built and it may be hard to improve its performance with an extra reflector but the right size of 'offset paraboloid' could increase the system gain. Using the device as both input and output would be fine as long as you got the ultrasound optics right, I think.

9. May 7, 2015

### Dr.MIke

Actually i am trying to make it more 'precise' reducing the spread to be able to use it to locate an object knowing exactly where it is. (like Infrared)
And like you said, also to improve on distance. (not sure if possible)
I just hope the reflector will not have to be too big.
And last, like you mentioned, i had to choose the offset paraboloid to make sure that no waves will be reflected directly into the surce causing wrong results.
Thanks for the help. :D

10. May 8, 2015

### sophiecentaur

You can get an idea of the directivity of a reflector with the formula
θ=λ/2d when the reflector (or aperture) is uniformly illuminated. That's only a ball park figure because your source is already directive and, being offset, there will be other factors. The spec sheet gives an idea of the directivity and you could see how close that formula agrees with the published value. Apart from the raggedy nature of any pattern, you could expect the beam width to be (sort of) inversely proportional to the aperture width so you could do a swift calculation to give you an indication of how much better you could expect. You could actually measure the pattern if there is a signal level output from the unit.

11. May 8, 2015

### tech99

I think you will find that placing the TX and RX transducers side by side is not successful. The slight offset will tilt the beam enough to lose much of the gain. It is the same problem which faced the pioneers of radar. Can you use the same transducer for TX and RX, with a switching arrangement? Alternatively, it might be possible to use separate reflectors for TX and RX. Incidentally, with the left hand offset reflector, the reflector is really a small part of a large paraboloid having the transducer at its focus, a difficult shape to make exactly. You may find that a simple cone is just as good.

12. May 8, 2015

### Dr.MIke

Yeah i making that shape won't be easy for me as im not experienced in making metal shapes or even plastic ones.
Yeah i can use the TX as RX alternativley that won't be a problem.
About the cone, is there a fixed angle it needs to have? So the waves enter the small 'hole' and leave the larger one like a beam correct?
To be honest i didn't know this was possible with a cone.

13. May 8, 2015

### Dr.MIke

Also i am not sure if there is a significant energy loss while hitting the reflector reducing so the range. I will have to do some calculation before i start doing something.
What i need is somekind of pencil beam with a good range of about 300cm.
It says on the file that range would be about 250cm without updating the device.

14. May 8, 2015

### CWatters

15. May 8, 2015

### tech99

I don't know what wavelength you are using, maybe about 8 mm. A cone with axial length of 6 wavelengths and a diameter of 4.5 wavelengths should increase the intensity by about 100 times and have a beamwidth of about 10 degrees between half power points. (Based on the microwave analogy).
A cone is simple to try. Assuming you are using 40 kHz, lambda is 8 mm.The beamwidth is not the angle of the cone, but approx 58/D degrees, where D is the diameter of the mouth in wavelengths. A suitable cone might be 10 cm diameter and 15 cm long, giving 5 degrees beamwidth. It would give a gain of about 25 times in intensity compared to your 2 cm diameter transducer used alone.

16. May 8, 2015

### tech99

Sorry I sent you two ideas by mistake. Please ignore the first design, not quite right because the transducer is already quite large diameter.

17. May 8, 2015

### Dr.MIke

That sounds like an idea if the offset paraboloid is too difficult to find or make, i will try to use a cone.
But isnt there a chance that, like Sophiecentaur said, when exiting the hole of the cone the beam might suffer from diffraction?

18. May 8, 2015

### Dr.MIke

19. May 8, 2015

### sophiecentaur

Every aperture will produce diffraction. That is what defines the beam width.

tech 99 makes a good point about the effect of the 'feeds' being in different places but it may not be too serious for a low gain reflector system. The notion of a conical wave launcher may be worth looking at (horn antennae can be effective if the beam doesn't need to be too narrow) but I think that the positioning of the two (RX and TX) feeds within the cone will still introduce a phase tilt across the mouth of the cone / horn, which produces two different main beam directions. I have a feeling that there may be a fundamental problem here that imposes limits. I would perhaps consider the advantage of doctoring the US Unit so that the two transducers are closer together.

20. May 8, 2015

### Dr.MIke

Sounds like a plan to me.
But as said, i can use one 'speaker' placed in the center of the smaller mouth of the cone to transmit and recieve at the same time (of course with a small delay between switching from transmit to receive but that should be no problem as the echo will take enough time for the device to switch).
If that shows problems, i can buy the other model CWatters showed for better results.