ll Support for the k-Wave MATLAB toolbox en-US Wed, 13 May 2026 01:05:15 +0000 http://bbpress.org/?v=1.0.2 q http://www.k-wave.org/forum/search.php http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-7702 Tue, 21 Jul 2020 10:24:56 +0000 alisiddiqi87 7702@http://www.k-wave.org/forum/ <p>Hi Bradley,</p> <p>I think NikiKeyz is right.</p> <p>&gt;&gt;To convert from dB/km to dB/cm you need to multiply by 1e-4.<br /> It should be 1e-5.</p> <p>&gt;&gt;Then to convert from dB/m/kHz^y to dB/m/MHz^y requires multiplying the prefactor by a further 0.001^y<br /> I think it should be 'dividing' the prefactor by a further 0.001^y</p> <p>Thanks and regards,<br /> Ali </p> http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-6064 Fri, 21 Jul 2017 13:47:13 +0000 Bradley Treeby 6064@http://www.k-wave.org/forum/ <p>The conversion should be:</p> <p>(0.068e2) Np / m / MHz^-b = (0.068e2 * 8.686 / 100) dB / cm / MHz^-b = (0.59) dB / cm / MHz^-b</p> <p>You will need to assume a value of <code>alpha_power</code> if you want to calculate <code>alpha_coeff</code> from the absorption at a single frequency. </p> http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-6048 Fri, 07 Jul 2017 17:18:14 +0000 v3p 6048@http://www.k-wave.org/forum/ <p>Thank you Bradley!<br /> It was my mistake, as I understand now only alpha power should be a scalar.<br /> 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 <a href="http://www.kayelaby.npl.co.uk/general_physics/2_4/2_4_6.html" rel="nofollow">http://www.kayelaby.npl.co.uk/general_physics/2_4/2_4_6.html</a></p> <p>When I convert units I received a huge number for alpha coefficient, for example:<br /> Brain: 0.068*10^2*8.686(db)*10^(6*1.3)(Hz)/100(cm)=~37*10^6...</p> <p>And second is possible to calculate alpha coefficient for skull when I know only alpha at certain frequency 9@3Mhz </p> http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-6036 Sun, 02 Jul 2017 21:48:18 +0000 Bradley Treeby 6036@http://www.k-wave.org/forum/ <p>Hi v3p,</p> <p><code>medium.alpha_coeff</code> can be defined as a matrix (see the heterogeneous medium examples included in the MATLAB help).</p> <p>To make your conversion:<br /> - multiply by 8.686 to convert from Nepers to dB<br /> - divide by 100 to convert from per m to per cm<br /> - multiply by 1e12 to convert from per Hz^2 to per MHz^2</p> <p>This gives:</p> <p><code>medium.alpha_coeff = 0.26;</code></p> <p>Remember to set <code>medium.alpha_power</code> to 2.</p> <p>Brad </p> http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-6034 Sat, 01 Jul 2017 12:44:08 +0000 v3p 6034@http://www.k-wave.org/forum/ <p>In addition how can I change layer alpha coeff. if it should be a scalar? :/<br /> I am trying to simulate layered medium with different alpha coedd. </p> http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-6032 Fri, 30 Jun 2017 21:19:31 +0000 v3p 6032@http://www.k-wave.org/forum/ <p>Hi,</p> <p>How can I get the right values for medium.alpha_coeff and medium.alpha_power for:</p> <p>1) Glycerol, it article absorption equal to ~ 3000 10−15 neper m−1 Hz−2?<br /> 2) Any kind of foam or porous material!?</p> <p>Thank you </p> http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-5839 Tue, 21 Feb 2017 22:44:41 +0000 NikiKeyz 5839@http://www.k-wave.org/forum/ <p>Hi!<br /> It seems to be a mistake here.<br /> dB/km/kHz^y = 1e-5 dB/cm/kHz^y = 1e-5 dB/cm/(1e-3 MHz)^y<br /> y = 1.4484<br /> (1e-3)^1.4484 = 4.5165e-5<br /> So 5.3273*x^1.4484 dB/km/kHz^y = 1.1795*x^1.4484 dB/cm/MHz^y </p> http://www.k-wave.org/forum/topic/power-law-absorption-in-skull-bone-and-water-medium#post-5730 Tue, 01 Nov 2016 22:51:50 +0000 Bradley Treeby 5730@http://www.k-wave.org/forum/ <p>Hi Michele,</p> <p>The 8.686 comes from the conversion from Nepers to dB:</p> <p>alpha_dB = 20*log10 (exp(alpha_Neper)) = alpha_Neper * 20*log10 (exp) = 8.686 * alpha_Neper </p> <p>To convert from dB/m to dB/(MHz cm), divide by 100*frequency.</p> <p>Brad. </p> http://www.k-wave.org/forum/topic/power-law-absorption-in-skull-bone-and-water-medium#post-5725 Fri, 28 Oct 2016 11:16:55 +0000 yodish 5725@http://www.k-wave.org/forum/ <p>Hello,</p> <p>just a quick question: what does the term 8.686 come from.</p> <p>Is the conversion from dB/m to dB/(MHz cm) displayed above correct?</p> <p>Thanks a lot for the answer</p> <p>Michele </p> http://www.k-wave.org/forum/topic/power-law-absorption-in-skull-bone-and-water-medium#post-5630 Wed, 10 Aug 2016 17:09:07 +0000 Bradley Treeby 5630@http://www.k-wave.org/forum/ <p>Hi Songwriter45,</p> <p>It sounds like the measured attenuation value you are referring to is actually a measurement of transmission loss, is that right? In other words, it accounts for attenuation within the skull layer in addition to attenuation due to reflections (caused by the impedance mismatch between the skull and the surrounding medium). If this is the case, then this isn't the attenuation value you want to use in k-Wave, as the attenuation due to the impedance mismatch will already be inherently included in your simulation when you define heterogeneous material properties.</p> <p>Hope that helps,</p> <p>Brad. </p> http://www.k-wave.org/forum/topic/power-law-absorption-in-skull-bone-and-water-medium#post-5610 Wed, 27 Jul 2016 14:38:18 +0000 Songwriter45 5610@http://www.k-wave.org/forum/ <p>Hi Brad.</p> <p>The attenuation value of -12 dB that I'm talking about is obtained by simulation, as shown in a public paper of University of Columbia (amplitude in the focus in presence of skull compared to the one in absence of skull) and in my simulations I get -4 dB, not -12 dB, so my value of attenuation of skull might be wrong.<br /> The only data I know from this paper is that skull attenuation measured at 500 kHz is 60 Np/m.</p> <p>Reaserchers of Columbia are using an "alpha_coeff_skull=20 dB/(MHz^1.1 cm)" but with an "alpha_poewer=1.1", not 2. If I use the same alpha_power they use, but employing my conversion method, I get ~11 dB/(MHz^1.1 cm), not 20, so I get lower attenuation of focused beam when propagating through skull. </p> http://www.k-wave.org/forum/topic/power-law-absorption-in-skull-bone-and-water-medium#post-5600 Wed, 27 Jul 2016 09:04:17 +0000 Bradley Treeby 5600@http://www.k-wave.org/forum/ <p>Hi Songwriter45,</p> <p>You're method of converting the units of attenuation looks correct. To me, the value of attenuation ~20 dB/(MHz^2 cm) is on the high side, especially if this is for the compressional wave (e.g., see Table 1 and Figure 5 <a href="http://www.homepages.ucl.ac.uk/~rmapbtr/papers/JOURN_23_2014_Treeby_JASA_PowerLawAbsorptionElastic.pdf">in this paper</a>).</p> <p>Is your reference value of -12 dB for the same frequency?</p> <p>Brad. </p> http://www.k-wave.org/forum/topic/power-law-absorption-in-skull-bone-and-water-medium#post-5589 Fri, 15 Jul 2016 23:17:51 +0000 Songwriter45 5589@http://www.k-wave.org/forum/ <p>Hello.</p> <p>I'm studying FUS transcranial propagation through a human skull submerged in water, but I'm having many problems when defining absorption properties of both mediums skull and water. I'm getting confused when obtaining "medium.alpha_coef" and "medium.alpha_power". </p> <p>Let me explain: I have an attenuation of 60 Np/m on skull measured at 500 kHz, so I define "alpha_ref = 60 Np/m", "f_ref = 0.5e6 Hz", "alpha_power = 2" and finally I obtain "alpha_coeff_skull = alpha_ref*8.686*((1e6/f_ref)^alpha_power)/100". For water I do the same: attenuation of 2.5 Np/m measured at 1 MHz, so I define "alpha_ref = 2.5 Np/m", "f_ref = 1e6 Hz", the same alpha power as in skull "alpha_power = 2" and finally I obtain "alpha_coeff_water = alpha_ref*8.686*((1e6/f_ref)^alpha_power)/100".</p> <p>Could you tell me if this method is right?</p> <p>The problem I have is that I read in many papers that skull generates a global attenuation of about -12 dB, and I get only -4 dB, so I thought attenuation parameters were wrong.</p> <p>Thank you in advance. </p> http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-4681 Tue, 19 Aug 2014 12:14:27 +0000 PiesangBrood2000 4681@http://www.k-wave.org/forum/ <p>Hi Bradley,</p> <p>That clears things up, thanks a lot!</p> <p>Kind regards,<br /> PiesangBrood2000 </p> http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-4678 Mon, 18 Aug 2014 17:21:47 +0000 Bradley Treeby 4678@http://www.k-wave.org/forum/ <p>Hi PiesangBrood2000,</p> <p>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.</p> <p>Hope that helps,</p> <p>Brad. </p> http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-4673 Thu, 14 Aug 2014 13:37:08 +0000 PiesangBrood2000 4673@http://www.k-wave.org/forum/ <p>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. </p> http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-4672 Thu, 14 Aug 2014 13:30:12 +0000 PiesangBrood2000 4672@http://www.k-wave.org/forum/ <p>Hi Ben,</p> <p>Thanks, that does help.<br /> 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?</p> <p>Kind regards,<br /> PiesangBrood2000 </p> http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-4671 Wed, 13 Aug 2014 20:50:46 +0000 bencox 4671@http://www.k-wave.org/forum/ <p>Hi PiesangBrood2000,</p> <p>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). </p> <p>So 5.3273 dB/km/kHz^y = 5.3273*0.001^(y+1) dB/m/MHz^y = 2.4x10^-7 dB/m/MHz^y.</p> <p>Hope that helps,<br /> Ben </p> http://www.k-wave.org/forum/topic/power-law-absorption-coefficient-units#post-4670 Tue, 12 Aug 2014 12:49:28 +0000 PiesangBrood2000 4670@http://www.k-wave.org/forum/ <p>Hi,</p> <p>I'm struggling to get the right values for medium.alpha_coeff and medium.alpha_power. I'm working from the "Attenuation of sound in air" table from<br /> <a href="http://www.kayelaby.npl.co.uk/general_physics/2_4/2_4_1.html" rel="nofollow">http://www.kayelaby.npl.co.uk/general_physics/2_4/2_4_1.html</a>.<br /> 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].<br /> 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)]?</p> <p>Kind regards,<br /> PiesangBrood2000 </p>