**Ultrasound beam attenuation**

Attenuation of a sound beam is the change in intensity, power and amplitude (Remember your decibels). The amount that they decrease are totally unrelated to speed. The further the beam travels the more attenuation occurs. This is measured in decibels and they will always be negative since both the intensity and power decreases

### Attenuation and frequency

We mainly deal with attenuation in soft tissue. In soft tissue the amount of attenuation is dependent on the frequency of the transducer and the imaging depth (the distance the wave has to travel).

Think about the transducers that we use to image shallow structures. High frequency transducers to maximize the resolution because attenuation is not an issue since we are imaging a shallow structure.If we need to image a deep structure we use a low frequency transducer since we need as much of the beam as possible to make it to the area of interest. Because of attenuation much of the beam will not penetrate. We want to lessen the amount of attenuation so we decrease the operating frequency. On large patients we use a lower frequency to be able to penetrate.

Most attenuation is attributed to absorption. This is when the energy from the beam is converted into heat. A smaller percentage is attributed to scattering and reflection.

*Attenuation:** **Greatest to smallest: Air > Bone & Lung > Soft Tissue > Water*

Attenuation in blood is equal to that in soft tissue not water. Notice the difference between the posterior enhancement of a liver cyst vs a hepatic vein.

Remember, attenuation is not related to propagation speed.

**Attenuation Coefficient**

The attenuation coefficient is the number that we use to state how readily a structure can be penetrated by the beam. If the attenuation coefficient is high then that means that there will be a high amount of attenuation (weakening) of the beam as it travels through the medium.

An easy way to determine the attenuation coefficient is to divide the main operating frequency by 2.

Attenuation Coefficient dB/cm = frequency / 2

If we have an operating frequency of 5 MHz what would our attenuation coefficient be? 5 MHz / 2 = 2.5 dB / cm

So in this case the total amount of attenuation with a beam traveling 10 cm would be what?

It would be our attenuation coefficient multiplied by the distance traveled.

2.5 dB/cm x 10 cm = 25 dB total attenuation. Remember the cm cancels out so we are left with only dB.

## Attenuation

*Attenuation*.You scored %%SCORE%% out of %%TOTAL%%.Your performance has been rated as %%RATING%%

Question 1 |

they are essentially the same thing. | |

they are directly related (when one goes up the other goes up). | |

they are unrelated. | |

they are inversely related (when one goes up the other goes down). |

Question 2 |

in waves per second | |

in micrometers | |

In decibels | |

Megahertz |

Question 3 |

the decibel setting. | |

the propagation speed. | |

the amount of attenuation. | |

the linear sequential array. |

List |

#### Franco

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These video lectures are really helpful & easy to understand. Any chance you will be posting more soon? I would like to see something on the operation of the ultrasound system settings & adjustments. There is a simulation on the SPI which is so confusing because it doesn’t adjust the image when when we adjust the settings & the purpose is to clear up the image based on the system operator adjustments. It would really be helpful if you had something dedicated to that! Thanks!