k-Wave

1. Songwriter45
   

   Member
   Joined: Sep '15
   Posts: 7

   Hello.

   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".

   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".

   Could you tell me if this method is right?

   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.

   Thank you in advance.

   Posted 8 years ago #

2. Bradley Treeby
   

   Developer
   Joined: Mar '10
   Posts: 1,643

   Hi Songwriter45,

   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 in this paper).

   Is your reference value of -12 dB for the same frequency?

   Brad.

   Posted 8 years ago #

3. Songwriter45
   

   Member
   Joined: Sep '15
   Posts: 7

   Hi Brad.

   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.
   The only data I know from this paper is that skull attenuation measured at 500 kHz is 60 Np/m.

   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.

   Posted 8 years ago #

4. Bradley Treeby
   

   Developer
   Joined: Mar '10
   Posts: 1,643

   Hi Songwriter45,

   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.

   Hope that helps,

   Brad.

   Posted 8 years ago #

5. yodish
   

   Member
   Joined: Oct '16
   Posts: 1

   Hello,

   just a quick question: what does the term 8.686 come from.

   Is the conversion from dB/m to dB/(MHz cm) displayed above correct?

   Thanks a lot for the answer

   Michele

   Posted 8 years ago #

6. Bradley Treeby
   

   Developer
   Joined: Mar '10
   Posts: 1,643

   Hi Michele,

   The 8.686 comes from the conversion from Nepers to dB:

   alpha_dB = 20*log10 (exp(alpha_Neper)) = alpha_Neper * 20*log10 (exp) = 8.686 * alpha_Neper

   To convert from dB/m to dB/(MHz cm), divide by 100*frequency.

   Brad.

   Posted 8 years ago #

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