 
5919 views  Career Discussions   Sandy
 Sandy
 07:22 Dec032003 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 socalled "fundamental resonant frequency" of an ultrasound transducer is the transmit frequency that corresponds to the halfwavelength 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!
 
   Paul A. Meyer R & D, GE Inspection Technologies, USA, Joined Nov 1998, ^{47}   09:21 Dec032003 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 socalled "fundamental resonant frequency" of an ultrasound transducer is the transmit frequency that corresponds to the halfwavelength 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 
 
   Sandy
 Sandy
 05:30 Dec092003 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 socalled "fundamental resonant frequency" of an ultrasound transducer is the transmit frequency that corresponds to the halfwavelength 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 
 
   Paul A, Meyer R & D, GE Inspection Technologies, USA, Joined Nov 1998, ^{47}   01:01 Dec102003 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 socalled "fundamental resonant frequency" of an ultrasound transducer is the transmit frequency that corresponds to the halfwavelength 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 
 
  Aerospace Systems  Automated Ultrasonic InspectionUSL are specialists in the design and manufacture of turnkey
ultrasonic inspection systems for aer ... ospace applications. From
monolithic composites to complex honeycomb structures. This video
shows just a few examples of what is possible, find out more at:
www.ultrasonicsciences.co.uk > HDCR 35 NDT Computed Radiography SystemPortable highresolution CR scanner for all radiography applications  weld testing, profile images ... and aerospace. No matter what type of radiographic testing you are performing, the unique TreFoc Technology of the HDCR 35 NDT imaging plate scanner always guarantees the highest image quality. > 

We use technical and analytics cookies to ensure that we will give you the best experience of our website  More Info
s

