Imaging of Blood Flow with US

2. Imaging of Blood Flow with US • Ultrasound images of flow, whether color flow or spectral Doppler, are essentially obtained from measurements of movement. • In ultrasound scanners, a series of pulses is transmitted to detect movement of blood. Echoes from stationary tissue are the same from pulse to pulse. • Echoes from moving scatterers exhibit slight differences in the time for the signal to be returned to the receiver • These differences can be measured as a direct time difference or, more usually, in terms of a phase shift from which the ‘Doppler frequency’ is obtained

4. Speckle Measurement • One way to do this is track speckle patterns over time using a cross-correlation technique.

5. Doppler Measurment • Another way to do this is take advantage of the Doppler effect

6. As can be seen from Figures there has to be motion in the direction of the beam; if the flow is perpendicular to the beam, there is no relative motion from pulse to pulse. The size of the Doppler signal is dependent on: • (1) Blood velocity: as velocity increases, so does the Doppler frequency; • (2) Ultrasound frequency: higher ultrasound frequencies give increased Doppler frequency. As in B-mode, lower ultrasound frequencies have better penetration. • (3) The choice of frequency is a compromise between better sensitivity to flow or better penetration; • (4 The angle of insonation: the Doppler frequency increases as the Doppler ultrasound beam becomes more aligned to the flow direction (the angle q between the beam and the direction of flow becomes smaller). This is of the utmost importance in the use of Doppler ultrasound. The implications are illustrated schematically in the following fig.

7. Effect of the Doppler angle in the sonogram. (A) higher-frequency Doppler signal is obtained if the beam is aligned more to the direction of flow. In the diagram, beam (A) is more ali)gned than (B) and produces higher-frequency Doppler signals. The beam/flow angle at (C) is almost 90° and there is a very poor Doppler signal. The flow at (D) is away from the beam and there is a negative signal.

8. All types of Doppler ultrasound equipment employ filters to cut out the high amplitude, low-frequency Doppler signals resulting from tissue movement, for instance due to vessel wall motion. Filter frequency can usually be altered by the user, for example, to exclude frequencies below 50, 100 or 200 Hz. This filter frequency limits the minimum flow velocities that can be measured.

9. Colour Doppler Imaging

10. B-mode image is optimized • Direction of the doppler beam is adjusted • Area of interest is set • Then the coloured image is acquired

14. Frequency Shift Formula Stationary source, reflector is moving: • During an interval of duration D, the number of cycles received will be f0Dplus the number of cycles that reflector came across due to its movement. The reflector traveled a distance vD, and thus came across vD/λ additional cycles: # of cycles in D = f0D + vD/λ

15. and thus the effective frequency and frequency shift seen by the reflector is: • For positive v (moving towards the transmitter) the frequency shifts upwards, for negative v downwards.

16. Stationary reflector, source is moving: • Here the emitted wavefronts are compressed due to the movement of the source:

17. The displacement of the source in 1 cycle is v/f0 and thus the new wavelength is short by that amount: • the effective frequency and frequency shift are:

18. Transmitterand Receiver Moving • In US, the reflector is both a listener and a source and so we have both shifts occuring. The net frequency shift is: • If the transducer array is not alligned to movement direction the net shift can be found by multiplying the result with cosθ , where θ is the angle between the movement direction and the normal of transducer plane.

19. Comments: • Observe that the size of the frequency shift is proportional to f0. • For physiological blood velocities in the body: • v = 50 cm/s = 0.5 m/s in major vessels. c = 1500 m/s =>relative shift = 2v/c = 6.7x10-4 and for f0 = 5 MHz, fshift = 3.3 kHz • All Doppler measurements only account for velocities towards/away from the transducer.

20. Continious Wave Doppler US • In CW imaging, the US signal is transmitted continuously and thus pulsing cannot be used to get depth resolution. • This also means that the receiver must be a different transducer than the transmitter. Typically, one uses the intersection of two beams to create an area of localization:

21. The received signal will have components related to static reflectors at f0 and perhaps a full spectrum of other signals related to components at different velocities:

22. In order to distinguish between f0 and shifted components at feff, the received signal needs to be observed for a long enough interval: Tobs. The frequency resolution will be 1/Tobs and the velocity resolution will be: • Δv*Tobs : The distance that travelled by the min. velocity during Tobs. That distance must be at least λ/2. Thus:

23. For example, we want velocity resolution of ∆v = 5 cm/s = 5x10-5 mm/μs and for f0 = 5 MHz, Tobs must be at least 3 ms. This is much longer than the round trip delay for a pulse (which is typically on the order of 100-200 μs). • Comments: • -  CW Doppler has excellent velocity resolution and is capable of measuring an entire velocity spectrum from the region of localization • -  CW Doppler has very poor range resolution (no pulsing) • -  It requires separate transmit and receive transducers • -  It is very slow for scanning a large space

24. Pulsed Wave Doppler US • In pulsed Doppler imaging, a train of pulses is sent along a single beam repeatedly. • Only one pulse is propagating through the body at any point in time so that maximum rate for pulsing is limited by the roundtrip time for the pulses. • if rmax = 15 cm, then the pulse repetition time is TPR = 2rmax/c = 200 μs. Thus the pulse repetition frequency is fPR = 5 kHz.

25. In pulsed Doppler, the pulses are “sampling” the velocity shifts and thus, any frequency shifts larger than fPR/2 will be aliased (e.g. to negative frequencies or equivalently, negative velocities). • Therefore the maximum reasonable frequency shift (thus the frequency resolution) is fPR/2.

26. Another major consideration for pulsed Doppler relates to velocity resolution. Velocities are mapped to frequency shifts and in order to detect small frequency shifts, we need to observe in time for a long period of time: • Δvdes*Tobs : The distance that travelled by the min. velocity during Tobs. That distance must be at least λ/2. Thus:

27. Another constraint on total observation time, Tobs: must also be long enough so that the fastest moving scattererinteracts with the pulse for that length of time. • Since Dopper detects velocities towards and away from the detector only, we need scatterers to remain within the depth resolution element long enough to establish the desired velocity resolution. • If TL is the pulse length (in time), the resolution element will be Δz = c/2 * TL. The maximum speed scatterer must travel that distance in duration shorter than Tobs.

28. Yields, • Equating two expressions for Tobs, we get: • that is, the number of cycles for the length of the pulse must be larger than the number of velocity steps that one wishes to resolve. L L

29. Therefore, pulsed Dopper US uses much longer pulses than imaging US and thus, has poorer spatial (depth) resolution. Likewise, the frequency f0 is also limited by the desired depth resolution,and this, ofcourse, affects lateral resolution. Another way to think about this is to recognize that one has to have a narrow spectrum pulse in order to detect small shifts in frequency and long pulses have a narrow spectrum.

30. Remarks about PW US Doppler • Depth of region of interest, rmax determines fPR and thus the frequency shift (velocity) at which aliasing occurs. (Depth-velocity resolution relation problem) • Maximum velocity determines frequency f0 • f0and TL, determine the maximum frequency shift, • e.g. for max depth 9.6 cm and maximum velocity=> fPR= 8 kHz (c = 1540 m/s) • maximum shift will be 4 kHz. • for vmax = 50 cm/s, this sets f0,max = 6.1 MHz: Remember the fshift=2*f0*v/c => 4KHz < f0*(0.5 m/s) / (1540 m/s) =>Max f0 = 6.1 MHz • for a velocity resolution of 2 cm/s, we then need a pulse length T = 4.1 ms this corresponds to a depth resolution of ∆z = 6.25 cm. (Last equation)