Beam width

A

Adam 19 years ago

Hello everybody,

I have heard that in the radar and radio communication there is a law of radiation. The larger aperture of the aerial (or the more number of the aerial) The narrower beam you'll have. As I have heard it is true for NDE ultrasonic transducer which means if you use an array of transducers you'll have a narrower beam of ultrasound rather than using just one of those transducers. I am not able to understand the reason? It seems there is a paradox there! Suppose an ultrasonic transducer has 60 degrees of beam width, if you use an array of 500 of these transducers then you'll get 5 degrees of beam width or even less while you have a very much large array instead of just one transducer!!!

Is there anyone to direct me why the beam separation is much less for an array of transducers rather than just one transducer?

Many thanks for any help

Vote

I

Ian Stirling 19 years ago

1.22* wavelength / diameter of antenna gives the emitting angle (in radians)

The antenna is comprised of _all_ the emitting components. Multiple little antennas spread over an area can have a significantly narrower beam, approaching the beamwidth of an antenna covering entirely that area.

Google 'radio telescope interferometry'

Vote

T

Tim Wescott 19 years ago

No paradox. The best attainable beam width is related to the ratio of the antenna size to the wavelength. It's a fundamental limit of any optical system, or any other system that's based on waves.

Google "diffraction" or "diffraction limit". There should be some information there.

Then do the math.

Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ "Applied Control Theory for Embedded Systems" came out in April. See details at http://www.wescottdesign.com/actfes/actfes.html

Vote

T

Tim Wescott 19 years ago

Actually, that figure gives you the best emitting angle that you can expect -- you can always mess it up by using the wrong antenna design.

Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ "Applied Control Theory for Embedded Systems" came out in April. See details at http://www.wescottdesign.com/actfes/actfes.html

Vote

I

Ian Stirling 19 years ago

Yes, it'll never be that good.

Vote

D

default 19 years ago

formatting link

----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Vote

F

Fred Bloggs 19 years ago

What about this one

formatting link

Vote

D

default 19 years ago

I don't know what the original poster was suggesting. No doubt with

500 emitters all hit with the same signal you'd have plenty of interferences, but the beam width wouldn't be smaller unless the signals were phased.

Anyhow he googled in and probably won't be back.

----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Vote

A

Adam 19 years ago

default =E4=E6=D4=CA=E5 =C7=D3=CA:

t News=3D=3D----

Newsgroups

=3D----

It would be for an arry of ultrasonic transducers withought phasing them..

Lan Stirling, Yes I am aware about that formula but can not understand the reason!?

Vote

A

Adam 19 years ago

Well, perhaps it is better to ask my question with another statement, Why the opening of an ultrasonic transducer is important for directivity of the wave? Yes I know this formula: (sinx= 1.2 (wavelengh/opening

But why that is true? Suppose the wavelength of the wave is smaller than the opening of the source anyway it would be spread after propagation similar to a wavelength which is larger than the opening! How know what the mystery which I don't know is? What is the real story....!?

Vote

T

Tim Williams 19 years ago

Imagine the wave front passing the opening. No matter what angle the incident waves are travelling, the waves passing through the opening will always be angled in the direction the hole is pointing. This works for relatively small holes. For a very large hole, it might as well not exist and the incident waves will propagate on through in the same direction they started. These are two very different behaviors and the point where they start to change is when the hole is within a factor of perhaps 5 of the wavelength.

Tim

-- Deep Fryer: A very philosophical monk. Website @

formatting link

Vote

R

Rene Tschaggelar 19 years ago

net News=3D=3D----

Newsgroups

=3D----

The formula applies for all transducers in phase, of course.

Rene

--=20 Ing.Buero R.Tschaggelar -

formatting link

& commercial newsgroups -

formatting link

Vote

P

Paul Hovnanian P.E. 19 years ago

formatting link

Paul Hovnanian mailto:Paul@Hovnanian.com ------------------------------------------------------------------ Why are so many towns named after water towers?

Vote

A

Adam 19 years ago

Tim Williams =E4=E6=D4=CA=E5 =C7=D3=CA:

ey

But as i said the a wave without any angle of propagation will be spread out in the air too, so?

Does anyone know the above formula for "N" of transducers?

Vote

J

joseph2k 19 years ago

really good. The best comparable ideas were phased array radar, and synthetic aperature radar.

JosephKK Gegen dummheit kampfen die Gotter Selbst, vergebens. --Schiller

Vote

Beam width

Join the Discussion

Didn't find your answer?