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
how can a perfect fourth interval be considered either consonant or dissonant?
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
$begingroup$
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
edited 1 hour ago
Qmechanic♦
108k122001245
asked 1 hour ago
Ubaid HassanUbaid Hassan
19511
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.
$endgroup$
add a comment |
$begingroup$
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
edited 1 hour ago
Qmechanic♦
108k122001245
asked 1 hour ago
Ubaid HassanUbaid Hassan
19511
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
27 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♦
23 mins ago
add a comment |
$begingroup$
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
edited 1 hour ago
Qmechanic♦
108k122001245
asked 1 hour ago
Ubaid HassanUbaid Hassan
19511
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
energy acoustics frequency wavelength medical-physics
edited 1 hour ago
Qmechanic♦
108k122001245
asked 1 hour ago
Ubaid HassanUbaid Hassan
19511
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.
edited 1 hour ago
Qmechanic♦
108k122001245
asked 1 hour ago
Ubaid HassanUbaid Hassan
19511
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.
edited 1 hour ago
Qmechanic♦
108k122001245
edited 1 hour ago
Qmechanic♦
108k122001245
edited 1 hour ago
Qmechanic♦
108k122001245
108k122001245
asked 1 hour ago
Ubaid HassanUbaid Hassan
19511
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.
asked 1 hour ago
Ubaid HassanUbaid Hassan
19511
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.
Check out our Code of Conduct.
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
27 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♦
23 mins ago
add a comment |
$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
27 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♦
23 mins ago
$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
27 mins ago
$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
27 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♦
23 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♦
23 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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
answered 1 hour ago
tomtom
6,39711627
$endgroup$
$begingroup$
Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
$endgroup$
– Ubaid Hassan
24 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♦
20 mins ago
add a comment |
$begingroup$
Higher frequency provides higher resolution.
answered 1 hour ago
akhmeteliakhmeteli
18.5k21844
$endgroup$
add a comment |
$begingroup$
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.
answered 6 mins ago
DanielTuzesDanielTuzes
1885
$endgroup$
$begingroup$
Maybe "enter link description here" is not the link description you want.
$endgroup$
– dmckee♦
4 mins ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
answered 1 hour ago
tomtom
6,39711627
$endgroup$
$begingroup$
Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
$endgroup$
– Ubaid Hassan
24 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♦
20 mins ago
add a comment |
$begingroup$
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
answered 1 hour ago
tomtom
6,39711627
$endgroup$
$begingroup$
Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
$endgroup$
– Ubaid Hassan
24 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♦
20 mins ago
add a comment |
$begingroup$
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
answered 1 hour ago
tomtom
6,39711627
$endgroup$
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
answered 1 hour ago
tomtom
6,39711627
edited 49 mins ago
answered 1 hour ago
tomtom
6,39711627
answered 1 hour ago
tomtom
6,39711627
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
24 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♦
20 mins ago
add a comment |
$begingroup$
Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
$endgroup$
– Ubaid Hassan
24 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♦
20 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
24 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
24 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♦
20 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♦
20 mins ago
add a comment |
$begingroup$
Higher frequency provides higher resolution.
answered 1 hour ago
akhmeteliakhmeteli
18.5k21844
$endgroup$
add a comment |
$begingroup$
Higher frequency provides higher resolution.
answered 1 hour ago
akhmeteliakhmeteli
18.5k21844
$endgroup$
add a comment |
$begingroup$
Higher frequency provides higher resolution.
answered 1 hour ago
akhmeteliakhmeteli
18.5k21844
$endgroup$
Higher frequency provides higher resolution.
answered 1 hour ago
akhmeteliakhmeteli
18.5k21844
answered 1 hour ago
akhmeteliakhmeteli
18.5k21844
answered 1 hour ago
akhmeteliakhmeteli
18.5k21844
answered 1 hour ago
akhmeteliakhmeteli
18.5k21844
18.5k21844
add a comment |
add a comment |
$begingroup$
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.
answered 6 mins ago
DanielTuzesDanielTuzes
1885
$endgroup$
$begingroup$
Maybe "enter link description here" is not the link description you want.
$endgroup$
– dmckee♦
4 mins ago
add a comment |
$begingroup$
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.
answered 6 mins ago
DanielTuzesDanielTuzes
1885
$endgroup$
$begingroup$
Maybe "enter link description here" is not the link description you want.
$endgroup$
– dmckee♦
4 mins ago
add a comment |
$begingroup$
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.
answered 6 mins ago
DanielTuzesDanielTuzes
1885
$endgroup$
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.
answered 6 mins ago
DanielTuzesDanielTuzes
1885
edited 3 mins ago
answered 6 mins ago
DanielTuzesDanielTuzes
1885
answered 6 mins ago
DanielTuzesDanielTuzes
1885
answered 6 mins ago
DanielTuzesDanielTuzes
1885
1885
$begingroup$
Maybe "enter link description here" is not the link description you want.
$endgroup$
– dmckee♦
4 mins ago
add a comment |
$begingroup$
Maybe "enter link description here" is not the link description you want.
$endgroup$
– dmckee♦
4 mins ago
$begingroup$
Maybe "enter link description here" is not the link description you want.
$endgroup$
– dmckee♦
4 mins ago
$begingroup$
Maybe "enter link description here" is not the link description you want.
$endgroup$
– dmckee♦
4 mins ago
add a comment |
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.
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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$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
27 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♦
23 mins ago