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- since 1996 -

WesDyne International (AMDATA Products)
WesDyne is an engineering and technical services firm dedicated to providing standard and custom nondestructive examinations (NDE) products and systems
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Career Discussions
Sandy
Sandy
07:22 Dec-03-2003
Resonance and crystal thickness

Hello, can anyone please help me with my question?
I read somewhere:
Resonance frequency (Fr)= (speed of sound in crystal)/(2xthickness of the crystal)

I can't really believe this because what I understand is that the crystal vibrates via voltage and then produces vibration,
or do they mean the impedance of the material?

Also I read:The so-called "fundamental resonant frequency" of an ultrasound transducer is the transmit frequency that corresponds to the half-wavelength crystal thickness.
So that:
Fr = 2 x christal thickness

So my question is:
Is the Fr of the crystal obtained when the crystal has a diameter equal to exactly half the wavelength of the ultrasound that is produced by the crystal.

Or does this depend on the acoustic impedance, or speed of sound in the crystal? Do does the Fr depend on the material?

Can someone please give me the correct information on this
or point me in the good direction?

Many thanks!




    
 
 Reply 
 
Paul A. Meyer
R & D,
GE Inspection Technologies, USA, Joined Nov 1998, 47

Paul A. Meyer

R & D,
GE Inspection Technologies,
USA,
Joined Nov 1998
47
09:21 Dec-03-2003
Re: Resonance and crystal thickness
Hello Sandy,
The fundamental mechanical resonance occurs when the thickness of the crystal is 1/2 wavelength thick.
Let
lambda = wavelength
c = acoustic velocity
f = frequency


thickness = lambda/2 (at fundamental resonance)

and

lambda = c/f

then

thickness= c/(2*f)

solving for f

f=c/(2 * thickness)

This refers to the mechanical free resonance of the crystal which is achieved by deforming it and allowing it to resonate...much like striking a tuning fork or a piano string.

If you drive the crystal with an electrical signal, the response will be at the driving frequency. The amplitude of the output will vary with driving frequency...as you approach the resonant frequency of the crystal, the output amplitude will increase.

The piezoelectric element used in a transducer has many resonances, some related to the thickness, and some related to the lateral dimensions. Most transducers are designed so that the unwanted resonances are suppressed and the fundamental dominates theoutput.

So, to answer your question, Fr depends on the thickness and and the acoustic velocity of the crystal.

I hope this answers your question.
Paul
----------- Start Original Message -----------
: Hello, can anyone please help me with my question?
: I read somewhere:
: Resonance frequency (Fr)= (speed of sound in crystal)/(2xthickness of the crystal)
: I can't really believe this because what I understand is that the crystal vibrates via voltage and then produces vibration,
: or do they mean the impedance of the material?
: Also I read:The so-called "fundamental resonant frequency" of an ultrasound transducer is the transmit frequency that corresponds to the half-wavelength crystal thickness.
: So that:
: Fr = 2 x christal thickness
: So my question is:
: Is the Fr of the crystal obtained when the crystal has a diameter equal to exactly half the wavelength of the ultrasound that is produced by the crystal.
: Or does this depend on the acoustic impedance, or speed of sound in the crystal? Do does the Fr depend on the material?
: Can someone please give me the correct information on this
: or point me in the good direction?
: Many thanks!
------------ End Original Message ------------




    
 
 Reply 
 
Sandy
Sandy
05:30 Dec-09-2003
Re: Resonance and crystal thickness
Thank you very much Paul, forgot to thank you,it answers perfectly well my question,
and still another little question(S)

It is stated that for a gaussian pulse shape the bandwidth;
df x t pulse = 1

what do they mean with gaussian?
So the bandwidth x the pulse time = 1

hz x time = 1 ?

Sandy

----------- Start Original Message -----------
: Hello Sandy,
: The fundamental mechanical resonance occurs when the thickness of the crystal is 1/2 wavelength thick.
: Let
: lambda = wavelength
: c = acoustic velocity
: f = frequency
:
: thickness = lambda/2 (at fundamental resonance)
: and
: lambda = c/f
: then
: thickness= c/(2*f)
: solving for f
: f=c/(2 * thickness)
: This refers to the mechanical free resonance of the crystal which is achieved by deforming it and allowing it to resonate...much like striking a tuning fork or a piano string.
: If you drive the crystal with an electrical signal, the response will be at the driving frequency. Theamplitude of the output will vary with driving frequency...as you approach the resonant frequency of the crystal, the output amplitude will increase.
: The piezoelectric element used in a transducer has many resonances, some related to the thickness, and some related to the lateral dimensions. Most transducers are designed so that the unwanted resonances are suppressed and the fundamental dominates the output.
: So, to answer your question, Fr depends on the thickness and and the acoustic velocity of the crystal.
: I hope this answers your question.
: Paul
: : Hello, can anyone please help me with my question?
: : I read somewhere:
: : Resonance frequency (Fr)= (speed of sound in crystal)/(2xthickness of the crystal)
: : I can't really believe this because what I understand is that the crystal vibrates via voltage and then produces vibration,
: : or do they mean the impedance of the material?
: : Also I read:The so-called "fundamental resonant frequency" of an ultrasound transducer is the transmit frequency that corresponds to the half-wavelength crystal thickness.
: : So that:
: : Fr = 2 x christal thickness
: : So my question is:
: : Is the Fr of the crystal obtained when the crystal has a diameter equal to exactly half the wavelength of the ultrasound that is produced by the crystal.
: : Or does this depend on the acoustic impedance, or speed of sound in the crystal? Do does the Fr depend on the material?
: : Can someone please give me the correct information on this
: : or point me in the good direction?
: : Many thanks!
------------ End Original Message ------------




    
 
 Reply 
 
Paul A, Meyer
R & D,
GE Inspection Technologies, USA, Joined Nov 1998, 47

Paul A, Meyer

R & D,
GE Inspection Technologies,
USA,
Joined Nov 1998
47
01:01 Dec-10-2003
Re: Resonance and crystal thickness
Hi Sandy,

A continuous wave signal has only one frequency component and that signal will continue for all time. A pulse can be represented as a summation of continuous wave signals of various frequency and amplitude. The Fourier or frequency spectrum of a pulse shows the distribution of energy as a function of frequency necessary to create the prescribed pulse. The Gaussian shape you reference means that the frequency spectrum has a Gaussian, or "normal", distribution with frequency. This is the same "normal" distribution you encounter in statistics.

However, if I understand your question, it refers to a different situation. You can calculate the frequency spectrum of a pulse through the use of a mathematical process called a Fourier Transform. With the advent of digital computers an approximation algorithm was developed that calculates the transform at discrete frequencies rather that as a continuous function. For practical purposes, this is sufficient and much quicker computationally. In thatresult, the frequency increment (df) calculated is equal to 1/(length of pulse in time). Rearranging this equation yields: df x t pulse = 1 I do not believe that a Gaussian spectrum shape is necessary for this equation to be true.

Thanks,
Paul

----------- Start Original Message -----------
: Thank you very much Paul, forgot to thank you,it answers perfectly well my question,
: and still another little question(S)
: It is stated that for a gaussian pulse shape the bandwidth;
: df x t pulse = 1
: what do they mean with gaussian?
: So the bandwidth x the pulse time = 1
: hz x time = 1 ?
: Sandy
: : Hello Sandy,
: : The fundamental mechanical resonance occurs when the thickness of the crystal is 1/2 wavelength thick.
: : Let
: : lambda = wavelength
: : c = acoustic velocity
: : f = frequency
: :
: : thickness = lambda/2 (at fundamental resonance)
: : and
: : lambda = c/f
: : then
: : thickness= c/(2*f)
: : solving for f
: : f=c/(2 * thickness)
: : This refers to the mechanical free resonance of the crystal which is achieved by deforming it and allowing it to resonate...much like striking a tuning fork or a piano string.
: : If you drive the crystal with an electrical signal, the response will be at the driving frequency. The amplitude of the output will vary with driving frequency...as you approach the resonant frequency of the crystal, the output amplitude will increase.
: : The piezoelectric element used in a transducer has many resonances, some related to the thickness, and some related to the lateral dimensions. Most transducers are designed so that the unwanted resonances are suppressed and the fundamental dominates the output.
: : So, to answer your question, Fr depends on the thickness and and the acoustic velocity of the crystal.
: : I hope this answers your question.
: : Paul
: : : Hello, can anyone please help me with my question?
: : : I read somewhere:
: : : Resonance frequency (Fr)= (speed of sound in crystal)/(2xthickness of the crystal)
: : : I can't really believe this because what I understand is that the crystal vibrates via voltage and then produces vibration,
: : : or do they mean the impedance of the material?
: : : Also I read:The so-called "fundamental resonant frequency" of an ultrasound transducer is the transmit frequency that corresponds to the half-wavelength crystal thickness.
: : : So that:
: : : Fr = 2 x christal thickness
: : : So my question is:
: : : Is the Fr of the crystal obtained when the crystal has a diameter equal to exactly half the wavelength of the ultrasound that is produced by the crystal.
: : : Or does this depend on the acoustic impedance, or speed of sound in the crystal? Do does the Fr depend on the material?
: : : Can someone please give me the correct information on this
: : : or point me in the good direction?
: : : Many thanks!
------------ End Original Message ------------




    
 
 Reply 
 

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