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		<title>k-Wave User Forum &#187; Topic: Power law absorption coefficient units</title>
		<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units</link>
		<description>Support for the k-Wave MATLAB toolbox</description>
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		<pubDate>Tue, 12 May 2026 23:07:30 +0000</pubDate>
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		<item>
			<title>alisiddiqi87 on "Power law absorption coefficient units"</title>
			<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-7702</link>
			<pubDate>Tue, 21 Jul 2020 10:24:56 +0000</pubDate>
			<dc:creator>alisiddiqi87</dc:creator>
			<guid isPermaLink="false">7702@http://www.k-wave.org/forum/</guid>
			<description>&#60;p&#62;Hi Bradley,&#60;/p&#62;
&#60;p&#62;I think NikiKeyz is right.&#60;/p&#62;
&#60;p&#62;&#38;gt;&#38;gt;To convert from dB/km to dB/cm you need to multiply by 1e-4.&#60;br /&#62;
It should be 1e-5.&#60;/p&#62;
&#60;p&#62;&#38;gt;&#38;gt;Then to convert from dB/m/kHz^y to dB/m/MHz^y requires multiplying the prefactor by a further 0.001^y&#60;br /&#62;
I think it should be 'dividing' the prefactor by a further 0.001^y&#60;/p&#62;
&#60;p&#62;Thanks and regards,&#60;br /&#62;
Ali
&#60;/p&#62;</description>
		</item>
		<item>
			<title>Bradley Treeby on "Power law absorption coefficient units"</title>
			<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-6064</link>
			<pubDate>Fri, 21 Jul 2017 13:47:13 +0000</pubDate>
			<dc:creator>Bradley Treeby</dc:creator>
			<guid isPermaLink="false">6064@http://www.k-wave.org/forum/</guid>
			<description>&#60;p&#62;The conversion should be:&#60;/p&#62;
&#60;p&#62;(0.068e2) Np / m / MHz^-b = (0.068e2 * 8.686 / 100) dB / cm / MHz^-b = (0.59) dB / cm / MHz^-b&#60;/p&#62;
&#60;p&#62;You will need to assume a value of &#60;code&#62;alpha_power&#60;/code&#62; if you want to calculate &#60;code&#62;alpha_coeff&#60;/code&#62; from the absorption at a single frequency.
&#60;/p&#62;</description>
		</item>
		<item>
			<title>v3p on "Power law absorption coefficient units"</title>
			<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-6048</link>
			<pubDate>Fri, 07 Jul 2017 17:18:14 +0000</pubDate>
			<dc:creator>v3p</dc:creator>
			<guid isPermaLink="false">6048@http://www.k-wave.org/forum/</guid>
			<description>&#60;p&#62;Thank you Bradley!&#60;br /&#62;
It was my mistake, as I understand now only alpha power should be a scalar.&#60;br /&#62;
Might I have one more assist from you. In new simulation I tried to make heterogeneous medium where one layer is skull bone and second brain. i took data from this site &#60;a href=&#34;http://www.kayelaby.npl.co.uk/general_physics/2_4/2_4_6.html&#34; rel=&#34;nofollow&#34;&#62;http://www.kayelaby.npl.co.uk/general_physics/2_4/2_4_6.html&#60;/a&#62;&#60;/p&#62;
&#60;p&#62;When I convert units I received a huge number for alpha coefficient, for example:&#60;br /&#62;
Brain: 0.068*10^2*8.686(db)*10^(6*1.3)(Hz)/100(cm)=~37*10^6...&#60;/p&#62;
&#60;p&#62;And second is possible to calculate alpha coefficient for skull when I know only alpha at certain frequency 9@3Mhz
&#60;/p&#62;</description>
		</item>
		<item>
			<title>Bradley Treeby on "Power law absorption coefficient units"</title>
			<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-6036</link>
			<pubDate>Sun, 02 Jul 2017 21:48:18 +0000</pubDate>
			<dc:creator>Bradley Treeby</dc:creator>
			<guid isPermaLink="false">6036@http://www.k-wave.org/forum/</guid>
			<description>&#60;p&#62;Hi v3p,&#60;/p&#62;
&#60;p&#62;&#60;code&#62;medium.alpha_coeff&#60;/code&#62; can be defined as a matrix (see the heterogeneous medium examples included in the MATLAB help).&#60;/p&#62;
&#60;p&#62;To make your conversion:&#60;br /&#62;
- multiply by 8.686 to convert from Nepers to dB&#60;br /&#62;
- divide by 100 to convert from per m to per cm&#60;br /&#62;
- multiply by 1e12 to convert from per Hz^2 to per MHz^2&#60;/p&#62;
&#60;p&#62;This gives:&#60;/p&#62;
&#60;p&#62;&#60;code&#62;medium.alpha_coeff = 0.26;&#60;/code&#62;&#60;/p&#62;
&#60;p&#62;Remember to set &#60;code&#62;medium.alpha_power&#60;/code&#62; to 2.&#60;/p&#62;
&#60;p&#62;Brad
&#60;/p&#62;</description>
		</item>
		<item>
			<title>v3p on "Power law absorption coefficient units"</title>
			<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-6034</link>
			<pubDate>Sat, 01 Jul 2017 12:44:08 +0000</pubDate>
			<dc:creator>v3p</dc:creator>
			<guid isPermaLink="false">6034@http://www.k-wave.org/forum/</guid>
			<description>&#60;p&#62;In addition how can I change layer alpha coeff. if it should be a scalar? :/&#60;br /&#62;
I am trying to simulate layered medium with different alpha coedd.
&#60;/p&#62;</description>
		</item>
		<item>
			<title>v3p on "Power law absorption coefficient units"</title>
			<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-6032</link>
			<pubDate>Fri, 30 Jun 2017 21:19:31 +0000</pubDate>
			<dc:creator>v3p</dc:creator>
			<guid isPermaLink="false">6032@http://www.k-wave.org/forum/</guid>
			<description>&#60;p&#62;Hi,&#60;/p&#62;
&#60;p&#62;How can I get the right values for medium.alpha_coeff and medium.alpha_power for:&#60;/p&#62;
&#60;p&#62;1) Glycerol, it article absorption equal to ~ 3000 10−15 neper m−1 Hz−2?&#60;br /&#62;
2) Any kind of foam or porous material!?&#60;/p&#62;
&#60;p&#62;Thank you
&#60;/p&#62;</description>
		</item>
		<item>
			<title>NikiKeyz on "Power law absorption coefficient units"</title>
			<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-5839</link>
			<pubDate>Tue, 21 Feb 2017 22:44:41 +0000</pubDate>
			<dc:creator>NikiKeyz</dc:creator>
			<guid isPermaLink="false">5839@http://www.k-wave.org/forum/</guid>
			<description>&#60;p&#62;Hi!&#60;br /&#62;
It seems to be a mistake here.&#60;br /&#62;
dB/km/kHz^y = 1e-5 dB/cm/kHz^y = 1e-5 dB/cm/(1e-3 MHz)^y&#60;br /&#62;
y = 1.4484&#60;br /&#62;
(1e-3)^1.4484 = 4.5165e-5&#60;br /&#62;
So 5.3273*x^1.4484 dB/km/kHz^y = 1.1795*x^1.4484 dB/cm/MHz^y
&#60;/p&#62;</description>
		</item>
		<item>
			<title>PiesangBrood2000 on "Power law absorption coefficient units"</title>
			<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-4681</link>
			<pubDate>Tue, 19 Aug 2014 12:14:27 +0000</pubDate>
			<dc:creator>PiesangBrood2000</dc:creator>
			<guid isPermaLink="false">4681@http://www.k-wave.org/forum/</guid>
			<description>&#60;p&#62;Hi Bradley,&#60;/p&#62;
&#60;p&#62;That clears things up, thanks a lot!&#60;/p&#62;
&#60;p&#62;Kind regards,&#60;br /&#62;
PiesangBrood2000
&#60;/p&#62;</description>
		</item>
		<item>
			<title>Bradley Treeby on "Power law absorption coefficient units"</title>
			<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-4678</link>
			<pubDate>Mon, 18 Aug 2014 17:21:47 +0000</pubDate>
			<dc:creator>Bradley Treeby</dc:creator>
			<guid isPermaLink="false">4678@http://www.k-wave.org/forum/</guid>
			<description>&#60;p&#62;Hi PiesangBrood2000,&#60;/p&#62;
&#60;p&#62;Yes you're correct - the units in k-Wave should be dB/cm/MHz^y. To convert from dB/km to dB/cm you need to multiply by 1e-4. The absorption in air in the audible frequency range is very low compared the absorption in tissue in the MHz range. Most of the examples in the toolbox use parameters relevant to ultrasound in tissue.&#60;/p&#62;
&#60;p&#62;Hope that helps,&#60;/p&#62;
&#60;p&#62;Brad.
&#60;/p&#62;</description>
		</item>
		<item>
			<title>PiesangBrood2000 on "Power law absorption coefficient units"</title>
			<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-4673</link>
			<pubDate>Thu, 14 Aug 2014 13:37:08 +0000</pubDate>
			<dc:creator>PiesangBrood2000</dc:creator>
			<guid isPermaLink="false">4673@http://www.k-wave.org/forum/</guid>
			<description>&#60;p&#62;Also this value of 2.4X10^-7 for alpha_coeff seems very small, compared to the 0.75 used in the example problems. I assume the examples aren't for waves in air (sound speed of 1500 m/s) as well as being in the MHz range.
&#60;/p&#62;</description>
		</item>
		<item>
			<title>PiesangBrood2000 on "Power law absorption coefficient units"</title>
			<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-4672</link>
			<pubDate>Thu, 14 Aug 2014 13:30:12 +0000</pubDate>
			<dc:creator>PiesangBrood2000</dc:creator>
			<guid isPermaLink="false">4672@http://www.k-wave.org/forum/</guid>
			<description>&#60;p&#62;Hi Ben,&#60;/p&#62;
&#60;p&#62;Thanks, that does help.&#60;br /&#62;
In the k-Wave manual it gives the units for alpha_coeff as [dB/(MHz^y cm)] not [dB/(MHz^y m)], is this correct? If so then it would only be necessary to multiply the 2.4X10^-7 by a factor of 0.01 right?&#60;/p&#62;
&#60;p&#62;Kind regards,&#60;br /&#62;
PiesangBrood2000
&#60;/p&#62;</description>
		</item>
		<item>
			<title>bencox on "Power law absorption coefficient units"</title>
			<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-4671</link>
			<pubDate>Wed, 13 Aug 2014 20:50:46 +0000</pubDate>
			<dc:creator>bencox</dc:creator>
			<guid isPermaLink="false">4671@http://www.k-wave.org/forum/</guid>
			<description>&#60;p&#62;Hi PiesangBrood2000,&#60;/p&#62;
&#60;p&#62;To convert from dB/km to dB/m requires multiplying the prefactor by 0.001. Then to convert from dB/m/kHz^y to dB/m/MHz^y requires multiplying the prefactor by a further 0.001^y. So to convert your 5.3273 dB/km/kHz^y to dB/m/MHz^y as required by k-Wave, you need to multiply by 0.001^(y+1). &#60;/p&#62;
&#60;p&#62;So 5.3273 dB/km/kHz^y = 5.3273*0.001^(y+1) dB/m/MHz^y = 2.4x10^-7 dB/m/MHz^y.&#60;/p&#62;
&#60;p&#62;Hope that helps,&#60;br /&#62;
Ben
&#60;/p&#62;</description>
		</item>
		<item>
			<title>PiesangBrood2000 on "Power law absorption coefficient units"</title>
			<link>http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-4670</link>
			<pubDate>Tue, 12 Aug 2014 12:49:28 +0000</pubDate>
			<dc:creator>PiesangBrood2000</dc:creator>
			<guid isPermaLink="false">4670@http://www.k-wave.org/forum/</guid>
			<description>&#60;p&#62;Hi,&#60;/p&#62;
&#60;p&#62;I'm struggling to get the right values for medium.alpha_coeff and medium.alpha_power. I'm working from the &#34;Attenuation of sound in air&#34; table from&#60;br /&#62;
&#60;a href=&#34;http://www.kayelaby.npl.co.uk/general_physics/2_4/2_4_1.html&#34; rel=&#34;nofollow&#34;&#62;http://www.kayelaby.npl.co.uk/general_physics/2_4/2_4_1.html&#60;/a&#62;.&#60;br /&#62;
From their data I got an approximate fitting formula of y=5.3273*x^1.4484 (for 50% humidity) where y is the attenuation coefficient in [dB/km] and x is the frequency in [kHz].&#60;br /&#62;
I assume since medium.alpha_coeff is measured in [dB/(MHz^y cm]) I cant use 5.3273 as is. Would 0.053273 be the correct value for medium.alpha_coeff in [dB/(MHz^y cm)]?&#60;/p&#62;
&#60;p&#62;Kind regards,&#60;br /&#62;
PiesangBrood2000
&#60;/p&#62;</description>
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