Question

A Doppler flow meter uses ultrasound waves to measure blood-flow speeds. Suppose the device emits sound at 3.5 MHz, and the speed of sound in human tissue is about 1540 m/s. What is the expected beat frequency observed at the flow meter if blood is flowing in large leg arteries at 3.0 cm/s directly away from the flow meter?

Answer 1

To solve this problem we will use the concept of the Doppler effect applied to the speed of blood, the speed of sound in the blood and the original frequency. This relationship will also be extrapolated to the frequency given by the detector and measured the change in frequencies through the beat frequency. So:

$f_{blood} = f (1-\frac{v_{blood}}{v_{snd}})$

Where

$f_{blood}$ = Frequency of the blood flow

f = Frequency of the original signal

$v_{blood}$ = Speed of the blood flow

$v_{snd}$ = Speed of sound in blood

$f''_{detector} = \frac{f_{blood}}{(1+\frac{v_{blood}}{v_{snd}})}$

$f''_{detector} = f (\frac{(1-\frac{v_{blood}}{v_{snd}})}{(1+\frac{v_{blood}}{v_{snd}})})$

$f''_{detector} = f \frac{(v_{snd}-v_{blood})}{(v_{snd}+v_{blood})}$

Now calculating the beat frequency is

$\Delta f = f-f''_{detector}$

Replacing this latest value we have that,

$\Delta f = f-f \frac{(v_{snd}-v_{blood})}{(v_{snd}+v_{blood})}$

$\Delta f = f \frac{2v_{blood}}{v_{snd}+v_{blood}}$

Replacing we have,

$\Delta f = (3.5*10^6)(\frac{2*(3*10^{-2})}{1.54*10^3+3*10^{-2}})$

$\Delta f = 136.36Hz$

Therefore the beat frequency is 136.36Hz

Answer 2

Using the beat frequency relation, the expected beat frequency observed at the flow meter would be 136.36 Hz

Given the Parameters :

• Frequency of device, F = 3.5 MHz = 3.5 × 10³ Hz
• Speed of sound in tissue = Vs = 1540 m/s
• Velocity of blood = Vb = 3 cm/s = (0.01 × 3)m/s = 0.03 m/s = 3 × 10¯² m/s
• Beat frequency = Δf

The expected beat frequency observed can be calculated uisng the relation :

• $\delta F = F \frac{2V_{b}}{V_{s} + V_{b}}$

Substituting the values into the formula :

$\delta F = 3.5 \times 10^{3} \frac{2 \times 0.03}{(1540 + 0.03}$

$\delta F = 3.5 \times 10^{3} \frac{0.06}{(1540.03}$

$\delta F = 136.36 Hz$

Therefore, the expected beat frequency observed at the flow meter will be 136.36 Hz

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