Why use ultrasound for medical imaging? The 2019 Stack Overflow Developer Survey Results Are...

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Why use ultrasound for medical imaging?

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Why use ultrasound for medical imaging?



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is there an upper frequency limit to ultrasound?How is it possible for an Ultrasound device to correctly interpret a negative density change in tissue?How specifically does an MRI machine build an image from received radio wavesIs it possible to send modulated ultrasound wave from underwater to air?Why the Doppler Ultrasound beam needs to be looking directly down at a pipeIntensity of an ultrasound beam?Dynamic range of ultrasound machine expressed in dBUltrasound wave/beam generationDoppler effect of sound waves in bloodAveraging speed of ultrasound between two differnt boundaires












1












$begingroup$


What advantage does ultrasound have over sound between 20-20000Hz that it is used in medical imaging over sound in that frequency range?










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Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    your edits are quite helpful, but can I ask why you do edits so frequently?
    $endgroup$
    – Ubaid Hassan
    32 mins ago










  • $begingroup$
    Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
    $endgroup$
    – dmckee
    29 mins ago
















1












$begingroup$


What advantage does ultrasound have over sound between 20-20000Hz that it is used in medical imaging over sound in that frequency range?










share|cite|improve this question









New contributor




Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    your edits are quite helpful, but can I ask why you do edits so frequently?
    $endgroup$
    – Ubaid Hassan
    32 mins ago










  • $begingroup$
    Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
    $endgroup$
    – dmckee
    29 mins ago














1












1








1





$begingroup$


What advantage does ultrasound have over sound between 20-20000Hz that it is used in medical imaging over sound in that frequency range?










share|cite|improve this question









New contributor




Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




What advantage does ultrasound have over sound between 20-20000Hz that it is used in medical imaging over sound in that frequency range?







energy acoustics frequency wavelength medical-physics






share|cite|improve this question









New contributor




Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Qmechanic

108k122001245




108k122001245






New contributor




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asked 1 hour ago









Ubaid HassanUbaid Hassan

19511




19511




New contributor




Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    your edits are quite helpful, but can I ask why you do edits so frequently?
    $endgroup$
    – Ubaid Hassan
    32 mins ago










  • $begingroup$
    Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
    $endgroup$
    – dmckee
    29 mins ago


















  • $begingroup$
    your edits are quite helpful, but can I ask why you do edits so frequently?
    $endgroup$
    – Ubaid Hassan
    32 mins ago










  • $begingroup$
    Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
    $endgroup$
    – dmckee
    29 mins ago
















$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
32 mins ago




$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
32 mins ago












$begingroup$
Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
$endgroup$
– dmckee
29 mins ago




$begingroup$
Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
$endgroup$
– dmckee
29 mins ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

I think the simple answer here is resolution.



Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.



If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by



$$lambda = {c over f} $$



so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....



The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate



$$lambda = 0.001 {rm m} = 1 {rm mm}$$



At 20000 Hz $lambda = 75$ mm






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
    $endgroup$
    – Ubaid Hassan
    30 mins ago






  • 1




    $begingroup$
    The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
    $endgroup$
    – dmckee
    26 mins ago



















1












$begingroup$

Higher frequency provides higher resolution.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 points.



    The waves are coming from behind the specimen passing through the points or the waves are reflected on the points. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 point, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.



    On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
    $$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
    where $lambda$ corresponds to the wavelength and $d$ is the size of point.



    Two points are called resolvable if their picture on the detector are distinguishable, for example, if their Airy-disk are non-overlapping. That means, it will limit how close the distinguishable points can be. With a given pixel and size properties of the detector you can only modify $lambda$, and as it can be seen, lowering $lambda$ makes it possible to resolve even closer objects.



    And as it is known,
    $$lambda = c/f,$$
    where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Maybe "enter link description here" is not the link description you want.
      $endgroup$
      – dmckee
      10 mins ago












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    I think the simple answer here is resolution.



    Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.



    If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by



    $$lambda = {c over f} $$



    so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....



    The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate



    $$lambda = 0.001 {rm m} = 1 {rm mm}$$



    At 20000 Hz $lambda = 75$ mm






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
      $endgroup$
      – Ubaid Hassan
      30 mins ago






    • 1




      $begingroup$
      The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
      $endgroup$
      – dmckee
      26 mins ago
















    5












    $begingroup$

    I think the simple answer here is resolution.



    Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.



    If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by



    $$lambda = {c over f} $$



    so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....



    The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate



    $$lambda = 0.001 {rm m} = 1 {rm mm}$$



    At 20000 Hz $lambda = 75$ mm






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
      $endgroup$
      – Ubaid Hassan
      30 mins ago






    • 1




      $begingroup$
      The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
      $endgroup$
      – dmckee
      26 mins ago














    5












    5








    5





    $begingroup$

    I think the simple answer here is resolution.



    Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.



    If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by



    $$lambda = {c over f} $$



    so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....



    The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate



    $$lambda = 0.001 {rm m} = 1 {rm mm}$$



    At 20000 Hz $lambda = 75$ mm






    share|cite|improve this answer











    $endgroup$



    I think the simple answer here is resolution.



    Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.



    If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by



    $$lambda = {c over f} $$



    so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....



    The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate



    $$lambda = 0.001 {rm m} = 1 {rm mm}$$



    At 20000 Hz $lambda = 75$ mm







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 55 mins ago

























    answered 1 hour ago









    tomtom

    6,39711627




    6,39711627












    • $begingroup$
      Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
      $endgroup$
      – Ubaid Hassan
      30 mins ago






    • 1




      $begingroup$
      The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
      $endgroup$
      – dmckee
      26 mins ago


















    • $begingroup$
      Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
      $endgroup$
      – Ubaid Hassan
      30 mins ago






    • 1




      $begingroup$
      The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
      $endgroup$
      – dmckee
      26 mins ago
















    $begingroup$
    Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
    $endgroup$
    – Ubaid Hassan
    30 mins ago




    $begingroup$
    Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
    $endgroup$
    – Ubaid Hassan
    30 mins ago




    1




    1




    $begingroup$
    The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
    $endgroup$
    – dmckee
    26 mins ago




    $begingroup$
    The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
    $endgroup$
    – dmckee
    26 mins ago











    1












    $begingroup$

    Higher frequency provides higher resolution.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Higher frequency provides higher resolution.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Higher frequency provides higher resolution.






        share|cite|improve this answer









        $endgroup$



        Higher frequency provides higher resolution.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        akhmeteliakhmeteli

        18.5k21844




        18.5k21844























            1












            $begingroup$

            Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 points.



            The waves are coming from behind the specimen passing through the points or the waves are reflected on the points. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 point, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.



            On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
            $$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
            where $lambda$ corresponds to the wavelength and $d$ is the size of point.



            Two points are called resolvable if their picture on the detector are distinguishable, for example, if their Airy-disk are non-overlapping. That means, it will limit how close the distinguishable points can be. With a given pixel and size properties of the detector you can only modify $lambda$, and as it can be seen, lowering $lambda$ makes it possible to resolve even closer objects.



            And as it is known,
            $$lambda = c/f,$$
            where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Maybe "enter link description here" is not the link description you want.
              $endgroup$
              – dmckee
              10 mins ago
















            1












            $begingroup$

            Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 points.



            The waves are coming from behind the specimen passing through the points or the waves are reflected on the points. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 point, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.



            On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
            $$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
            where $lambda$ corresponds to the wavelength and $d$ is the size of point.



            Two points are called resolvable if their picture on the detector are distinguishable, for example, if their Airy-disk are non-overlapping. That means, it will limit how close the distinguishable points can be. With a given pixel and size properties of the detector you can only modify $lambda$, and as it can be seen, lowering $lambda$ makes it possible to resolve even closer objects.



            And as it is known,
            $$lambda = c/f,$$
            where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Maybe "enter link description here" is not the link description you want.
              $endgroup$
              – dmckee
              10 mins ago














            1












            1








            1





            $begingroup$

            Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 points.



            The waves are coming from behind the specimen passing through the points or the waves are reflected on the points. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 point, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.



            On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
            $$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
            where $lambda$ corresponds to the wavelength and $d$ is the size of point.



            Two points are called resolvable if their picture on the detector are distinguishable, for example, if their Airy-disk are non-overlapping. That means, it will limit how close the distinguishable points can be. With a given pixel and size properties of the detector you can only modify $lambda$, and as it can be seen, lowering $lambda$ makes it possible to resolve even closer objects.



            And as it is known,
            $$lambda = c/f,$$
            where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.






            share|cite|improve this answer











            $endgroup$



            Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 points.



            The waves are coming from behind the specimen passing through the points or the waves are reflected on the points. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 point, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.



            On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
            $$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
            where $lambda$ corresponds to the wavelength and $d$ is the size of point.



            Two points are called resolvable if their picture on the detector are distinguishable, for example, if their Airy-disk are non-overlapping. That means, it will limit how close the distinguishable points can be. With a given pixel and size properties of the detector you can only modify $lambda$, and as it can be seen, lowering $lambda$ makes it possible to resolve even closer objects.



            And as it is known,
            $$lambda = c/f,$$
            where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 9 mins ago

























            answered 11 mins ago









            DanielTuzesDanielTuzes

            1885




            1885












            • $begingroup$
              Maybe "enter link description here" is not the link description you want.
              $endgroup$
              – dmckee
              10 mins ago


















            • $begingroup$
              Maybe "enter link description here" is not the link description you want.
              $endgroup$
              – dmckee
              10 mins ago
















            $begingroup$
            Maybe "enter link description here" is not the link description you want.
            $endgroup$
            – dmckee
            10 mins ago




            $begingroup$
            Maybe "enter link description here" is not the link description you want.
            $endgroup$
            – dmckee
            10 mins ago










            Ubaid Hassan is a new contributor. Be nice, and check out our Code of Conduct.










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            Ubaid Hassan is a new contributor. Be nice, and check out our Code of Conduct.












            Ubaid Hassan is a new contributor. Be nice, and check out our Code of Conduct.
















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