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Doppler Effect of ultrasound

  1. Aug 2, 2008 #1
    1. The problem statement, all variables and given/known data
    The calibration for a piece of ultrasound equipment states that the velocity of blood flow in the aorta is normally about 0.28 m/s, and that the detector emits a frequency of 4.20 MHz.

    a) If the ultrasound waves were directed along the blood flow and reflected from the red blood cells, what frequency would be received back at the detector? Assume that the waves travel with a speed of 1.5x10^3 m/s.

    b) In reality, the signal you receive is the beat frequency, the difference between the sent and received signals. This is what you use for comparisons. What is the beat frequency for this situation?

    c) If the cross sectional area of the aorta is partially blocked by plaque build-up on the walls, such that it is 1/2 its normal cross sectional area, what are the frequency and beat frequency received? (All other factors, apart from the decrease in cross sectional area, are considered to be constant).

    2. Relevant equations
    ƒo = ((v – vo) / (v-vs)) *ƒs
    ƒbeat = Δƒ
    speed of sound through blood = 1570 m/s

    3. The attempt at a solution

    a) ƒo = ((v – vo) / (v-vs)) *ƒs
    ƒo = ((1570 m/s - 1.5x10^3) / (1570 m/s - 0.28 m/s)) * 4.20 MHz
    ƒo = 0.18731 MHz

    b) ƒbeat = Δƒ
    ƒbeat = 4.20 MHz - 0.18731 MHz
    ƒbeat = 4.0126

    c) ƒo = ((v – vo) / (v-2*vs)) *ƒs
    ƒo = ((1570 m/s - 1.5x10^3) / (1570 m/s - (2*0.28))) * 4.20 MHz
    ƒo = 0.18732 MHz

    ƒbeat = Δƒ
    ƒbeat = 4.20 MHz - 0.18732 MHz
    ƒbeat = 4.0127 MHz

    I'm not sure if this is correct. Any help would be appreciated.
  2. jcsd
  3. Aug 2, 2008 #2
    Your Dopler formula is fishy. In your case, the detector doesn't move, so it simplifies to

    [tex]f_1 = \left(1-\frac{v}{v_s}\right)f_0[/tex]

    which is the frequency of hitting the blood cells and then

    [tex]f_2 = \left(1-\frac{v}{v_s}\right)f_1 \approx \left(1-\frac{2v}{v_s}\right)f_0[/tex]

    which is the frequency reflected from the moving source. You can make these approximations because [tex]v\ll v_s[/tex].

    You're given the speed of sound (waves) 1.5x10^3m/s, so don't invent another figure of 1570m/s.
  4. Aug 2, 2008 #3
    The reflector doesn't move, true, but the waves do move. This is how my professor explained it to me. Considering he made up the question I'm inclined to believe him. The equation given is an equation straight out of my book.

    So who am I to believe?
  5. Aug 2, 2008 #4
    Consult your book what the symbols in the equation stand for.
  6. Aug 2, 2008 #5
    "If both source and observer are moving, we combine the two Doppler shifts to obtain

    ƒo = ((1-vo/v) / (1-vs/v)) *ƒs = (v-vo / v-vs)*ƒs

    where vo is the velocity of the observer, vs is the velocity of the source, and ƒo is the observed frequency."

    This is what I thought I did. The symbol v is unaccounted for and I assumed to be the speed of sound in blood, which is also given in my book maybe 10 pages previous to the above quote.
  7. Aug 2, 2008 #6
    The source (of the sound waves) is your apparatus (both generator and reflector and observer). And what is it's speed?

    Also note that the speed of sound in blood, which is v, is the same thing as the speed of waves given in your problem.
  8. Aug 2, 2008 #7
    Okay, I suppose, from your description, the source would then be the red blood cells, which travel at 0.28 m/s. The speed of sound in blood, v, is then the speed of waves which is 1.5x10^3 m/s.

    Using the equation you supplied (since this is a terrible and confusing book anyway...)

    ƒ = (1 - (1.5x10^3 m/s / 0.28 m/s)) * 4.20 MHz = -22495.8 MHz

    So if this is true why is the answer negative? Unless I'm still confused and switched v and vs... if that's true then the final answer is 4.19 MHz
  9. Aug 2, 2008 #8
    No, the source is stationary. Your equation is OK if you know what letter means what.
  10. Aug 2, 2008 #9
    The general doppler-shif expression is [tex]f' = \frac{v+v_{0}}{v-v_{s}}f[/tex]

    Here in your problem the source is stationary. When the source is stationary you have two cases;

    * Observer moving toward the source, you use: [tex]f' = \frac{v+v_{0}}{v}f[/tex]

    * Observer moving away from the source, you use: [tex]f' = \frac{v-v_{0}}{v}f[/tex]
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