Color Doppler shear wave elastography using commercial ultrasound machine with compensated transducer scanning delay

Abstract

Purpose

Tissue elasticity can be measured and mapped using color Doppler elastography. In a previous study, a binary pattern of shear waves was observed using a color flow imaging (CFI) system with matched pulse Doppler packet size as well as shear wave frequency and displacement condition. In the present study, we demonstrate the possibility of mapping shear wave velocity and resolving phantom elasticity using any commercial ultrasound machine without fulfilling that condition.

Methods

We derive a relation between Doppler autocorrelator integration time and the estimated flow velocity. The underlying principles behind the shear wave shadows captured by a typical modern ultrasound machine are investigated. The ultrasound machine measurement preset is calibrated to remove the effect of transducer array scanning delay in modifying the appearing wavenumber and thus correct the measurement error.

Results

The method was used to successfully measure the elasticity of a biological tissue-mimicking phantom and distinguish a stiff phantom from a soft phantom.

Conclusion

Using this method, the elasticity of a biological tissue-mimicking phantom can be recovered with less strict constraint. As a result, it provides more flexibility to be implemented in common ultrasound machines. This method may be practically used to help identify tissue stiffness-related disease.

Appendix

Suppose a color Doppler imaging system is used to estimate the flow velocity of a propagating sine wave in a medium with packet size N = 4. For $-\uppi /2<\Delta {\phi }_{i}<\uppi /2$, the estimated velocity is calculated by

$$\widehat{v}=\frac{4c}{4\pi {f}_{0}{T}_{\mathrm{a}}}{\mathrm{tan}}^{-1}\left(\frac{{I}_{-2}{Q}_{-1}-{I}_{-1}{Q}_{-2}+{I}_{-1}{Q}_{0}-{I}_{0}{Q}_{-1}+{I}_{0}{Q}_{1}-{I}_{1}{Q}_{0}+{I}_{1}{Q}_{2}-{I}_{2}{Q}_{1}}{{I}_{-1}{I}_{-2}+{Q}_{-1}{Q}_{-2}+{I}_{0}{I}_{-1}+{Q}_{0}{Q}_{-1}+{I}_{1}{I}_{0}+{Q}_{1}{Q}_{0}+{I}_{2}{I}_{1}+{Q}_{2}{Q}_{1}}\right).$$

By trigonometric decomposition, we would come up with

$$\widehat{v}=\frac{4c}{4\pi {f}_{0}{T}_{\mathrm{a}}}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{sin}\left(B-D\right)\mathrm{cos}\left(C-A\right)+\mathrm{sin}\left(D\right)\mathrm{cos}\left(C-Z\right)}{\mathrm{cos}\left(B-D\right)\mathrm{cos}\left(C-A\right)+\mathrm{cos}\left(D\right)\mathrm{cos}\left(C-Z\right))}\right),$$

where

$A={\phi }_{0}+{\frac{4\pi {f}_{0}}{c}\xi }_{m}\mathrm{sin}\left({\omega }_{\mathrm{b}}t+{\phi }_{\mathrm{b}}\right)\mathrm{cos}\left({\omega }_{\mathrm{b}}\frac{{T}_{\mathrm{a}}}{2}\right)$, $B=\frac{4\pi {f}_{0}}{c}{\xi }_{m}\mathrm{cos}\left({\omega }_{\mathrm{b}}t+{\phi }_{\mathrm{b}}\right)\mathrm{sin}\left({\omega }_{\mathrm{b}}\frac{{T}_{\mathrm{a}}}{2}\right)$, $C={\phi }_{0}+{\frac{4\pi {f}_{0}}{c}\xi }_{m}\mathrm{sin}\left({\omega }_{\mathrm{b}}t+{\phi }_{\mathrm{b}}\right)\mathrm{cos}\left({\omega }_{\mathrm{b}}\frac{{T}_{\mathrm{a}}}{4}\right)$, $D=\frac{4\pi {f}_{0}}{c}{\xi }_{m}\mathrm{cos}\left({\omega }_{\mathrm{b}}t+{\phi }_{\mathrm{b}}\right)\mathrm{sin}\left({\omega }_{\mathrm{b}}\frac{{T}_{\mathrm{a}}}{4}\right)$ and $Z={\phi }_{0}+\frac{4\pi {f}_{0}}{c}{\xi }_{m}\mathrm{sin}\left({\omega }_{\mathrm{b}}t+{\phi }_{\mathrm{b}}\right)$.

If the shear wave frequency is much smaller than PRF, we may approximate

$$\begin{aligned}\underset{\frac{{f}_{\mathrm{b}}}{\mathit{PRF}}\to 0}{\mathrm{lim}}\frac{\mathrm{cos}C-A}{\mathrm{cos}C-Z}&=\underset{\frac{{f}_{\mathrm{b}}}{\mathit{PRF}}\to 0}{\mathrm{lim}}\frac{\mathrm{cos}\left(\frac{4\pi {f}_{0}}{c}{\xi }_{m}\mathrm{sin}\left({\omega }_{\mathrm{b}}t+{\phi }_{\mathrm{b}}\right)(\mathrm{cos}\left(4\pi \frac{{f}_{\mathrm{b}}}{PRF}\right)-\mathrm{cos}\left(2\pi \frac{{f}_{\mathrm{b}}}{PRF}\right)\right)}{\mathrm{cos}\left(\frac{4\pi {f}_{0}}{c}{\xi }_{m}\mathrm{sin}\left({\omega }_{\mathrm{b}}t+{\phi }_{\mathrm{b}}\right)(\mathrm{cos}\left(2\pi \frac{{f}_{\mathrm{b}}}{PRF}\right)-1)\right)}\\ & =\frac{\mathrm{cos}\left(\frac{4\pi {f}_{0}}{c}{\xi }_{m}\mathrm{sin}\left({\omega }_{\mathrm{b}}t+{\phi }_{\mathrm{b}}\right)(\mathrm{cos}\left(0\right)-\mathrm{cos}\left(0\right)\right)}{\mathrm{cos}\left(\frac{4\pi {f}_{0}}{c}{\xi }_{m}\mathrm{sin}\left({\omega }_{\mathrm{b}}t+{\phi }_{\mathrm{b}}\right)(\mathrm{cos}\left(0\right)-1)\right)}=1.\end{aligned}$$

Under this condition, we approximate velocity by

$$\begin{aligned}\widehat{v}&\cong \frac{4c}{4\pi {f}_{0}{T}_{a}}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{sin}B-D+\mathrm{sin}D}{\mathrm{cos}B-D+\mathrm{cos}D}\right)\\& \cong \frac{4c}{4\pi {f}_{0}{T}_{a}}{\mathrm{tan}}^{-1}\mathrm{tan}\left(\frac{B-D+D}{2}\right)\\& \cong \frac{2c}{4\pi {f}_{0}{T}_{a}}B.\end{aligned}$$

(A1)

On the other hand,

$$\begin{aligned}\underset{\frac{{f}_{\mathrm{b}}}{\mathit{PRF}}\to 0}{\mathrm{lim}}\frac{B}{D}& =\underset{\frac{{f}_{\mathrm{b}}}{\mathit{PRF}}\to 0}{\mathrm{lim}}\frac{\frac{4\pi {f}_{0}}{c}{\xi }_{p}\mathrm{cos}\left({\omega }_{\mathrm{b}}t+{\phi }_{\mathrm{b}}\right)\mathrm{sin}\left(4\pi \frac{{f}_{\mathrm{b}}}{PRF}\right)}{\frac{4\pi {f}_{0}}{c}{\xi }_{p}\mathrm{cos}\left({\omega }_{\mathrm{b}}t+{\phi }_{\mathrm{b}}\right)\mathrm{sin}\left(2\pi \frac{{f}_{\mathrm{b}}}{PRF}\right)}\\ &=\underset{\frac{{f}_{\mathrm{b}}}{\mathit{PRF}}\to 0}{\mathrm{lim}}\frac{2\mathrm{cos}\left(4\pi \frac{{f}_{\mathrm{b}}}{PRF}\right)}{\mathrm{cos}\left(2\pi \frac{{f}_{\mathrm{b}}}{PRF}\right)}=2.\end{aligned}$$

Therefore

$$D\cong \frac{B}{2}.$$

In this approximation, a similar result is obtained by

$$\begin{aligned}\widehat{v}& \cong \frac{4c}{4\pi {f}_{0}{T}_{\mathrm{a}}}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{sin}B-\frac{B}{2}\mathrm{cos}C-A+\mathrm{sin}\frac{B}{2}\mathrm{cos}C-Z}{\mathrm{cos}B-\frac{B}{2}\mathrm{cos}C-A+\mathrm{cos}\frac{B}{2}\mathrm{cos}C-Z)}\right)\\ & \cong \frac{4c}{4\pi {f}_{0}{T}_{\mathrm{a}}}{\mathrm{tan}}^{-1}\left(\frac{\left(\mathrm{cos}C-A+\mathrm{cos}C-Z\right)\mathrm{sin}\frac{B}{2}}{\left(\mathrm{cos}C-A+\mathrm{cos}C-Z\right)\mathrm{cos}\frac{B}{2})}\right)\cong \frac{2c}{4\pi {f}_{0}{T}_{\mathrm{a}}}B.\end{aligned}$$

(A2)

If we reduce the packet size to N = 2 by removing the second and fourth pulses from the calculation, the velocity estimation is

$$\begin{aligned}\widehat{v}& =\frac{2c}{4\pi {f}_{0}{T}_{\mathrm{a}}}{\mathrm{tan}}^{-1}\left(\frac{{I}_{-2}{Q}_{0}-{I}_{0}{Q}_{-2}+{I}_{0}{Q}_{2}-{I}_{2}{Q}_{0}}{{I}_{0}{I}_{-2}+{Q}_{0}{Q}_{-2}+{I}_{2}{I}_{0}+{Q}_{2}{Q}_{0}}\right)\\ & =\frac{2c}{4\pi {f}_{0}{T}_{\mathrm{a}}}{\mathrm{tan}}^{-1}\left(\frac{2a\mathrm{sin}B\left(\mathrm{sin}A\mathrm{sin}Z+\mathrm{cos}A\mathrm{cos}Z\right)}{2a\mathrm{cos}B\left(\mathrm{cos}A\mathrm{cos}Z+\mathrm{sin}A\mathrm{sin}Z\right)}\right)\\ & =\frac{2c}{4\pi {f}_{0}{T}_{\mathrm{a}}}{\mathrm{tan}}^{-1}\left(\mathrm{tan}B\right)\\& =\frac{2c}{4\pi {f}_{0}{T}_{\mathrm{a}}}B.\end{aligned}$$

(A3)

If we reduce further to a single packet size, leaving only the first and last pulses in the calculation, the velocity estimation becomes

$$\begin{aligned}\widehat{v}&=\frac{c}{4\pi {f}_{0}{T}_{\mathrm{a}}}{\mathrm{tan}}^{-1}\left(\frac{{I}_{-2}{Q}_{2}-{I}_{2}{Q}_{-2}}{{I}_{2}{I}_{-2}+{Q}_{2}{Q}_{-2}}\right)\\&=\frac{c}{4\pi {f}_{0}{T}_{\mathrm{a}}}{\mathrm{tan}}^{-1}\left(\frac{{a}^{2}2\mathrm{cos}B\mathrm{sin}B}{{a}^{2}({\mathrm{cos}}^{2}B-{\mathrm{sin}}^{2}B)}\right)\\&=\frac{c}{4\pi {f}_{0}{T}_{\mathrm{a}}}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{sin}2B}{\mathrm{cos}2B}\right)\\&=\frac{2c}{4\pi {f}_{0}{T}_{\mathrm{a}}}B.\end{aligned}$$

(A4)

This derivation suggests that omitting ultrasound pulses other than the first and last signals does not significantly affect the estimation of sinusoid flow velocity.