Maker Pro
Maker Pro

Beam width

A

Adam

Jan 1, 1970
0
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
 
I

Ian Stirling

Jan 1, 1970
0
Adam said:
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!

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'
 
T

Tim Wescott

Jan 1, 1970
0
Adam said:
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
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
 
T

Tim Wescott

Jan 1, 1970
0
Ian said:
1.22* wavelength / diameter of antenna gives the emitting angle (in
radians)
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
 
I

Ian Stirling

Jan 1, 1970
0
Tim Wescott said:
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.

Yes, it'll never be that good.
 
D

default

Jan 1, 1970
0
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
http://en.wikipedia.org/wiki/Phased_array
 
D

default

Jan 1, 1970
0

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.
 
A

Adam

Jan 1, 1970
0
default äæÔÊå ÇÓÊ:
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.

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!?
 
A

Adam

Jan 1, 1970
0
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....!?
 
R

Rene Tschaggelar

Jan 1, 1970
0
Adam said:
default äæÔÊå ÇÓÊ:



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!?

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

Rene
 
P

Paul Hovnanian P.E.

Jan 1, 1970
0
Adam said:
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....!?

http://en.wikipedia.org/wiki/Angular_resolution#Explanation
 
T

Tim Williams

Jan 1, 1970
0
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
 
A

Adam

Jan 1, 1970
0
Tim Williams äæÔÊå ÇÓÊ:
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

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?
 
J

joseph2k

Jan 1, 1970
0
Ian said:
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'

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