Sonic Estimation of Elasticity via Resonance: A New Method of Assessing Hemostasis

Abstract

Uncontrolled bleeding threatens patients undergoing major surgery and in care for traumatic injury. This paper describes a novel method of diagnosing coagulation dysfunction by repeatedly measuring the shear modulus of a blood sample as it clots in vitro. Each measurement applies a high-energy ultrasound pulse to induce a shear wave within a rigid walled chamber, and then uses low energy ultrasound pulses to measure displacements associated with the resonance of that shear wave. Measured displacements are correlated with predictions from finite difference time domain models, with the best fit corresponding to the modulus estimate. In our current implementation each measurement requires 62.4 ms. Experimental data was analyzed using a fixed-viscosity algorithm and a free-viscosity algorithm. In experiments utilizing human blood induced to clot by exposure to kaolin, the free-viscosity algorithm quantified the shear modulus of formed clots with a worst-case precision of 2.5%. Precision was improved to 1.8% by utilizing the fixed-viscosity algorithm. Repeated measurements showed a smooth evolution from liquid blood to a firm clot with a shear modulus between 1.4 and 3.3 kPa. These results show the promise of this technique for rapid, point of care assessment of coagulation.

Appendices

Appendix 1: Finite Difference Formulation

Our formulation of the FDTD model is built upon a velocity – stress formulation of the shear wave equation.19,32 We expand upon traditional formulations by including a term to account for viscous loss.

$$\rho \frac{\partial }{\partial t}\vec{v}\left( {\vec{x},t} \right) = \nabla \cdot \sigma \left( {\vec{x},t} \right) + \vec{F}\left( {\vec{x},t} \right)$$

(5)

$$\frac{\partial }{\partial t}\sigma \left( {\vec{x},t} \right) = \left( {\mu + \eta \frac{\partial }{\partial t}} \right)\left[ {\left( {\nabla \vec{v}\left( {\vec{x},t} \right)} \right) + \left( {\nabla \vec{v}\left( {\vec{x},t} \right)} \right)^{T} } \right].$$

(6)

Our analysis assumes an axisymmetric test chamber, which suggests the use of cylindrical coordinates. Representing Eq. (5) in cylindrical coordinates and expanding the vector velocity into its constituent components yields:

$$\frac{{\partial \sigma_{rr} }}{\partial r} + \frac{1}{r}\frac{{\partial \sigma_{r\phi } }}{\partial \phi } + \frac{{\partial \sigma_{rz} }}{\partial z} + \frac{1}{r}\left( {\sigma_{rr} - \sigma_{\phi \phi } } \right) + F_{r} = \rho \frac{{\partial v_{r} }}{\partial t}$$

(7)

$$\frac{{\partial \sigma_{r\phi } }}{\partial r} + \frac{1}{r}\frac{{\partial \sigma_{\phi \phi } }}{\partial \phi } + \frac{{\partial \sigma_{\phi z} }}{\partial z} + \frac{2}{r}\sigma_{r\phi } + F_{\phi } = \rho \frac{{\partial v_{\phi } }}{\partial t}$$

(8)

$$\frac{{\partial \sigma_{{r{\text{z}}}} }}{\partial r} + \frac{1}{r}\frac{{\partial \sigma_{{\phi {\text{z}}}} }}{\partial \phi } + \frac{{\partial \sigma_{zz} }}{\partial z} + \frac{1}{r}\sigma_{rz} + F_{z} = \rho \frac{{\partial v_{z} }}{\partial t}.$$

(9)

We further simplify by recognizing that the only body force, the applied ultrasound radiation force, acts only in the $z$ direction. We set $F_{\phi } = F_{r} = 0$. Since our test chamber and the applied radiation force are entirely axisymmetric, we set all dependencies upon $\phi$ equal to zero. Applying these simplifications to Eq. (7) yields:

$$\frac{{\partial \sigma_{rz} }}{\partial z} = \rho \frac{{\partial^{2} u_{r} }}{{\partial t^{2} }}$$

(10)

$$0 = 0$$

(11)

$$\frac{{\partial \sigma_{{r{\text{z}}}} }}{\partial r} + \frac{1}{r}\sigma_{rz} + F_{z} = \rho \frac{{\partial^{2} u_{z} }}{{\partial t^{2} }}.$$

(12)

We follow a similar strategy to expand Eq. (6), to yield:

$$\frac{{\partial \sigma_{{r{\text{z}}}} }}{\partial t} = \left( {\mu + \eta \frac{\partial }{\partial t}} \right)\left( {\frac{{\partial v_{r} }}{\partial z} + \frac{{\partial v_{z} }}{\partial r}} \right).$$

(13)

Collecting Eqs. (10)–(12) and Eq. (13) yields:

$$\rho \frac{{\partial v_{r} }}{\partial t} = \frac{{\partial \sigma_{rz} }}{\partial z}$$

(14)

$$\rho \frac{{\partial v_{z} }}{\partial t} = \frac{{\partial \sigma_{{r{\text{z}}}} }}{\partial r} + \frac{1}{r}\sigma_{rz} + F_{z}$$

(15)

$$\frac{{\partial \sigma_{{r{\text{z}}}} }}{\partial t} = \left( {\mu + \eta \frac{\partial }{\partial t}} \right)\left( {\rho \frac{{\partial v_{r} }}{\partial t} + \rho \frac{{\partial v_{z} }}{\partial t}} \right).$$

(16)

Equations (14)–(16) form a system of partial differential equations that predict how radiation force will induce shear waves and how those induced shear waves will interact. This system of equations is amenable to a finite difference solution using a staggered grid approach, similar to the Yee method.35 The finite difference representation of Eqs. (14)–(16) are:

$$v_{{r_{j,k}^{i + 1} }} = v_{{r_{j,k}^{i} }} + \frac{\Delta t}{\rho \, \Delta z}\left( {\sigma_{{rz_{{j,k + \frac{1}{2}}}^{i} }} - \sigma_{{rz_{{j,k - \frac{1}{2}}}^{i} }} } \right)$$

(17)

$$v_{{z_{{j + \frac{1}{2},k + \frac{1}{2}}}^{i + 1} }} = v_{{z_{{j + \frac{1}{2},k + \frac{1}{2}}}^{i} }} + \frac{\Delta t}{\rho }\left( {\frac{1}{\Delta r}\left( {\sigma_{{rz_{{j + 1,k + \frac{1}{2}}}^{i} }} - \sigma_{{rz_{{j,k + \frac{1}{2}}}^{i} }} } \right) + \frac{1}{{2\left( {j + \frac{1}{2}} \right)\Delta r}}\left( {\sigma_{{rz_{{j,k + \frac{1}{2}}}^{i} }} - \sigma_{{rz_{{j + 1,k + \frac{1}{2}}}^{i} }} } \right) + F_{{z_{{j + + \frac{1}{2},k + \frac{1}{2}}}^{i} }} } \right)$$

(18)

$$\sigma_{{rz_{{j,k + \frac{1}{2}}}^{i + 1} }} = \sigma_{{rz_{{j,k + \frac{1}{2}}}^{i} }} + \frac{\mu \, \Delta t}{\Delta z}\left( {v_{{r_{j,k + 1}^{i} }} - v_{{r_{j,k}^{i} }} } \right) + \frac{\mu \, \Delta t}{\Delta z}\left( {v_{{r_{{j + \frac{1}{2},k + + \frac{1}{2}}}^{i} }} - v_{{r_{{j - \frac{1}{2},k + \frac{1}{2}}}^{i} }} } \right) + \frac{\eta }{\Delta z}\left( {v_{{r_{j,k + 1}^{i + 1} }} - v_{{r_{j,k + 1}^{i} }} - v_{{r_{j,k}^{i + 1} }} + v_{{r_{j,k}^{i} }} } \right) + \frac{\eta }{\Delta z}\left( {v_{{z_{{j + \frac{1}{2},k + \frac{1}{2}}}^{i + 1} }} - v_{{z_{{j + \frac{1}{2},k + \frac{1}{2}}}^{i} }} - v_{{z_{{j - \frac{1}{2},k + \frac{1}{2}}}^{i + 1} }} + v_{{z_{{j - \frac{1}{2},k + \frac{1}{2}}}^{i} }} } \right).$$

(19)

This formulation was implemented in MATLAB (Natick, MA) using the staggered grid approach shown in Fig. 12 with a default sampling interval of 100 $\mu$m. Each computational cycle consisted of two steps. In the first, the velocity components were computed via Eqs. (17) and (18). In the next step shear was computed using Eq. (19). time–dependent velocities were integrated numerically using cumulative trapezoidal integration ( cumtrapz in MATLAB) to determine the time–varying displacement.

Figure 12

FDTD model staggered grid geometry.

Appendix 2: Analytical Model

This analysis is based upon the Cauchy-Navier Equation for linear elasticity. We utilize a formulation incorporating viscoelastic material properties.

$$\left( {\left( {\lambda + \mu + \left( {\eta_{p} + \eta_{s} } \right)\frac{d}{dt} \cdot } \right)\nabla \left( {\nabla \cdot \vec{u}\left( {\vec{x},t} \right)} \right)} \right) + .\left( {\left( {\mu + \eta_{s} \frac{d}{dt} \cdot } \right)\nabla^{2} \vec{u}\left( {\vec{x},t} \right)} \right) - \rho \frac{{d^{2} }}{{dt^{2} }}\vec{u}\left( {\vec{x},t} \right) = \vec{F}\left( {\vec{x},t} \right).$$

(20)

To simplify analysis we assume an infinitely long cylindrical test chamber with radiation force applied to the central axis. We apply these simplifications to the components of Eq. (20):

$$\nabla \left( {\nabla \cdot \vec{u}\left( {\vec{x},t} \right)} \right) = \nabla \left( {\frac{1}{r}\frac{d}{dr}\left( {ru_{r} } \right) + \frac{1}{r}\frac{d}{d\phi }u_{\phi } + \frac{d}{dz}u_{z} } \right)$$

(21)

$$\nabla^{2} \vec{u}\left( {\vec{x},t} \right) = \left[ {\nabla^{2} u_{r} - \frac{1}{{r^{2} }}u_{r} - \frac{2}{{r^{2} }}\frac{d}{d\phi }u_{\phi } } \right]\hat{r} + \left[ {\nabla^{2} u_{\phi } - \frac{1}{{\phi^{2} }}u_{\phi } - \frac{2}{{r^{2} }}\frac{d}{d\phi }u_{r} } \right]\hat{\phi } + \left[ {\nabla^{2} u_{z} } \right]\hat{z}.$$

(22)

Because of cylindrical symmetry and the assumed infinite length of the chamber we can set all derivatives with respect to $\phi$ and $z$ equal to zero.

$$\nabla \left( {\nabla \cdot \vec{u}\left( {\vec{x},t} \right)} \right) = \nabla \left( {\frac{1}{r}\frac{d}{dr}\left( {ru_{r} } \right)} \right)\hat{r}$$

(23)

$$\nabla^{2} \vec{u}\left( {\vec{x},t} \right) = \left[ {\nabla^{2} u_{r} - \frac{1}{{r^{2} }}u_{r} } \right]\hat{r} + 0\hat{\phi } + \left[ {\nabla^{2} u_{z} } \right]\hat{z}.$$

(24)

Expanding gradients and Laplacians as appropriate, and setting derivatives with respect to $\phi$ and $z$ equal to zero yields:

$$\nabla \left( {\nabla \cdot \vec{u}\left( {\vec{x},t} \right)} \right) = \left[ {\frac{d}{dr}\left( {\frac{1}{r}\frac{d}{dr}\left( {ru_{r} } \right)} \right)} \right]\hat{r} + 0\hat{\phi } + 0\hat{z}$$

(25)

$$\nabla^{2} \vec{u}\left( {\vec{x},t} \right) = \left[ {\frac{1}{r}\frac{d}{dr}\left( {r\frac{d}{dr}u_{r} } \right) + \frac{1}{{r^{2} }}u_{r} } \right]\hat{r} + 0\hat{\phi } + \left[ {\frac{1}{r}\frac{d}{dr}\left( {r\frac{d}{dr}u_{z} } \right)} \right]\hat{z}.$$

(26)

This result shows that there are no cross terms between the radial and axial displacements. Our system measures axial displacement, so we discard the radial displacement term after substituting Eqs. (10) and (11) into Eq. (20), and assuming that the force is applied impulsively at time zero and is a spatial impulse at the center of the measurement cylinder.

$$\left( {\left( {\mu + \eta_{s} \frac{d}{dt} \cdot } \right)\left( {\frac{{d^{2} }}{{dr^{2} }}u_{z} \left( {r,t} \right) + \frac{1}{r}\frac{d}{dr}u_{z} \left( {r,t} \right)} \right)} \right) - \rho \frac{{d^{2} }}{{dt^{2} }}u_{z} \left( {r,t} \right) = \delta \left( t \right)\delta \left( r \right)/r.$$

(27)

We now utilize the Hankel Transform as defined in Eq. (28)26 to solve this partial differential equation.

$$\tilde{u}\left( {\xi_{i} ,t} \right) = \mathop \smallint \limits_{0}^{{R_{c} }} ru\left( {r,t} \right)J_{0} \left( {\xi_{i} r} \right)dr,$$

(28)

where $R_{c}$ is the radius of the cylinder, $J_{0} \left( x \right)$ is 0th order Bessel function of the first kind, and $\xi_{i} = \lambda_{i} /R_{c}$ where $\varvec{\lambda}_{\varvec{i}}$ is the i^th root of the 0th order Bessel function of the first kind. Performing the Hankel Transform on Eq. (27) yields:

$$- \mu \xi_{i}^{2} \tilde{u}\left( {\xi_{i} ,t} \right) - \eta_{s} \xi_{i}^{2} \frac{d}{dt}\tilde{u}\left( {\xi_{i} ,t} \right) - \rho \frac{{d^{2} }}{{dt^{2} }}\tilde{u}\left( {\xi_{i} ,t} \right) = \delta \left( t \right).$$

(29)

The application of the Hankel Transform has reduced a partial differential equation to an ordinary differential equation. Assuming that $\hat{u}\left( {\xi_{i} ,0} \right) = 0$ and that $\frac{d}{dt}\tilde{u}\left( {\xi_{i} ,t} \right) = 0$ then:

$$\tilde{u}\left( {\xi_{i} ,t} \right) = \frac{1}{\sqrt A }e^{{ - t\eta_{s} \xi_{i}^{2} /2\rho }} \left( {e^{ - t\sqrt A /2\rho } - e^{t\sqrt A /2\rho } } \right).$$

(30)

where $A = \xi_{i}^{2} \left( {\eta_{s}^{2} \xi_{i}^{2} - 4\mu \rho } \right)$. The finite Hankel Transform can be inverted using Eq. (31) [Equation 45 on page 83 of Ref. 26].

$$f\left( x \right) = \frac{2}{{R_{c}^{2} }}\mathop \sum \limits_{i = 1}^{\infty } \tilde{f}\left( {\xi_{i} } \right)\frac{{J_{0} \left( {\xi_{i} x} \right)}}{{\left[ {J_{1} \left( {\xi_{i} R_{c} } \right)} \right]^{2} }}.$$

(31)

Applying the inversion formula of Eq. (31) to Eq. (30) yields an expression for the time–dependent displacement.

$$u\left( {r,t} \right) = \frac{2}{{R_{c}^{2} }}\mathop \sum \limits_{i = 1}^{\infty } \frac{{J_{0} \left( {\xi_{i} r} \right)}}{{\sqrt A \left[ {J_{1} \left( {\xi_{i} R_{c} } \right)} \right]^{2} }}e^{{ - t\eta_{s} \xi_{i}^{2} /2\rho }} \times \left( {e^{ - t\sqrt A /2\rho } - e^{t\sqrt A /2\rho } } \right).$$

(32)

In comparing Eq. (33) to results from a simplified FDTD model we see that analytical model consistently exhibits high frequency content that is not visible in the FDTD model. We believe that this discrepancy is a result of the finite spatial sampling interval of the FDTD model. We have found empirically that the analytical model is in good agreement with the FDTD model when we include only terms up to those for which there is at least 1.5 samples per half cycle of the Bessel function. This truncated model, shown as Eq. (33) shows excellent agreement with the simplified FDTD model.

$$\hat{u}\left( {r,t} \right) = \frac{2}{{R_{c}^{2} }}\mathop \sum \limits_{i = 1}^{{2R_{c} /3\Delta x}} \frac{{J_{0} \left( {\xi_{i} r} \right)}}{{\sqrt A \left[ {J_{1} \left( {\xi_{i} R_{c} } \right)} \right]^{2} }}e^{{ - t\eta_{s} \xi_{i}^{2} /2\rho }} \times \left( {e^{ - t\sqrt A /2\rho } - e^{t\sqrt A /2\rho } } \right),$$

(33)

where $\Delta x$ is the spatial sampling interval of the FDTD model. The terms of the sum take on two different forms depending upon whether $A > 0$, in which case the system is overdamped, or $A < 0$, in which case the system is underdamped. In the underdamped case the first term of the series is a sine wave under a decaying exponential envelope. Assuming that the system has very low damping, the resonant frequency is:

$$f_{0} = {\raise0.7ex\hbox{${\sqrt A }$} \!\mathord{\left/ {\vphantom {{\sqrt A } {2\pi 2\rho }}}\right.\kern-0pt} \!\lower0.7ex\hbox{${2\pi 2\rho }$}} = {\raise0.7ex\hbox{${ - i\sqrt {\xi_{1}^{2} \left( {\eta_{s}^{2} \xi_{1}^{2} - 4\mu \rho } \right)} }$} \!\mathord{\left/ {\vphantom {{ - i\sqrt {\xi_{1}^{2} \left( {\eta_{s}^{2} \xi_{1}^{2} - 4\mu \rho } \right)} } {4\pi \rho }}}\right.\kern-0pt} \!\lower0.7ex\hbox{${4\pi \rho }$}} \approx {\raise0.7ex\hbox{${ - i\sqrt {\xi_{1}^{2} \left( { - 4\mu \rho } \right)} }$} \!\mathord{\left/ {\vphantom {{ - i\sqrt {\xi_{1}^{2} \left( { - 4\mu \rho } \right)} } {4\pi \rho }}}\right.\kern-0pt} \!\lower0.7ex\hbox{${4\pi \rho }$}} = \frac{{\xi_{1} }}{2\pi }\sqrt {\frac{\mu }{\rho }} = \frac{2.4048}{{2\pi R_{c} }}\sqrt {\frac{\mu }{\rho }} ,$$

(34)

where the factor of $2\varvec{\pi}$ is required to convert from radial to temporal frequency. The shear modulus can be estimated from the resonant frequency:

$$\mu = \rho \left( {\frac{{2\pi f_{0} R_{c} }}{2.4048}} \right)^{2} .$$

(35)

The concept of resonant frequency is only meaningful when the resonant chamber is under-damped. The system is under-damped when $A < 0$, or equivalently when:

$$\mu > \frac{{\eta_{s}^{2} \xi_{i}^{2} }}{4\rho }.$$

(36)

For our geometry assuming a blood density of 1060 kg/m³ and a shear viscosity of approximately 0.25 Pa s, the system will be under-damped for moduli greater than 19.3 Pa.

Appendix 3: Cramér–Rao Lower Bound on SEER

We begin our analysis from the formulation derived by Scharf in Ref. 22 The lower bound on the variance $C_{ii}$ of estimates of the parameter $\theta_{i}$ is:

$$C_{ii} \ge \left( {J^{ - 1} } \right)_{ii}$$

(37)

where $J$ is the Fisher information matrix. The elements of the Fisher information matrix, for a single realization of data, are defined by Equation 6.134 of Ref. 22:

$$J_{ij} = \frac{1}{2}tr\left[ {R^{ - 1} \frac{\partial R}{{\partial \theta_{i} }}R^{ - 1} \frac{\partial R}{{\partial \theta_{j} }}} \right] + \frac{{\partial m^{t} }}{{\partial \theta_{i} }}R^{ - 1} \frac{\partial m}{{\partial \theta_{j} }},$$

(38)

where $tr\left[ X \right]$ is the trace of matrix $X$, $R$ is the expected covariance matrix of the observed data, and $m$ is the expected mean of the observed data. In SEER the observed data is the vector of time–displacement estimates. We assume that the noise in the displacement estimates is generally uncorrelated so that the covariance matrix can be modeled as an identity matrix scaled by the variance of the displacement estimates.

$$R = {\sigma_{x}^{2} }I,$$

(39)

where $\sigma_{x}^{2}$ is the variance of the displacement estimates. Assuming that the error covariance does not depend upon the parameter (either $\theta_{i}$ or $\theta_{j}$) and that the covariance can be represented as in (39), then (38) can be simplified as:

$$J_{{ij}} = \frac{{\partial m^{t} }}{{\partial \theta _{i} }}R^{{ - 1}} \frac{{\partial m}}{{\partial \theta _{j} }} = \frac{{\partial m^{t} }}{{\partial \theta _{i} }}\left[ {{{\sigma _{x}^{2} }}I} \right]^{{ - 1}} \frac{{\partial m}}{{\partial \theta _{j} }} = \frac{1}{{\sigma _{x}^{2} }}\frac{{\partial m^{t} }}{{\partial \theta _{i} }}\frac{{\partial m}}{{\partial \theta _{j} }}.$$

(40)

For the free-viscosity estimator we estimate both the viscosity and the modulus, so the Fisher Information Matrix is:

$$J = \left[ {\begin{array}{*{20}c} {\frac{1}{{\sigma_{x}^{2} }}\frac{{\partial m^{t} }}{{\partial \theta_{\mu } }}\frac{\partial m}{{\partial \theta_{\mu } }}} & {\frac{1}{{\sigma_{x}^{2} }}\frac{{\partial m^{t} }}{{\partial \theta_{\mu } }}\frac{\partial m}{{\partial \theta_{\eta } }}} \\ {\frac{1}{{\sigma_{x}^{2} }}\frac{{\partial m^{t} }}{{\partial \theta_{\eta } }}\frac{\partial m}{{\partial \theta_{\mu } }}} & {\frac{1}{{\sigma_{x}^{2} }}\frac{{\partial m^{t} }}{{\partial \theta_{\eta } }}\frac{\partial m}{{\partial \theta_{\eta } }}} \\ \end{array} } \right] = \frac{1}{{\sigma_{x}^{2} }}\left[ {\begin{array}{*{20}c} {\frac{{\partial m^{t} }}{{\partial \theta_{\mu } }}\frac{\partial m}{{\partial \theta_{\mu } }}} & {\frac{{\partial m^{t} }}{{\partial \theta_{\mu } }}\frac{\partial m}{{\partial \theta_{\eta } }}} \\ {\frac{{\partial m^{t} }}{{\partial \theta_{\eta } }}\frac{\partial m}{{\partial \theta_{\mu } }}} & {\frac{{\partial m^{t} }}{{\partial \theta_{\eta } }}\frac{\partial m}{{\partial \theta_{\eta } }}} \\ \end{array} } \right].$$

(41)

Substituting Eqs. (41) into (37) yields the following lower bounds on estimation in the free-viscosity algorithm:

$$C_{\mu \mu } \ge \frac{{\sigma_{x}^{2} }}{{\frac{{\partial m^{t} }}{{\partial \theta_{\mu } }}\frac{\partial m}{{\partial \theta_{\mu } }}\frac{{\partial m^{t} }}{{\partial \theta_{\eta } }}\frac{\partial m}{{\partial \theta_{\eta } }} - \frac{{\partial m^{t} }}{{\partial \theta_{\mu } }}\frac{\partial m}{{\partial \theta_{\eta } }}\frac{{\partial m^{t} }}{{\partial \theta_{\eta } }}\frac{\partial m}{{\partial \theta_{\mu } }}}} \frac{{\partial m^{t} }}{{\partial \theta_{\eta } }}\frac{\partial m}{{\partial \theta_{\eta } }}$$

(42)

$$C_{\eta \eta } \ge \frac{{\sigma_{x}^{2} }}{{\frac{{\partial m^{t} }}{{\partial \theta_{\mu } }}\frac{\partial m}{{\partial \theta_{\mu } }}\frac{{\partial m^{t} }}{{\partial \theta_{\eta } }}\frac{\partial m}{{\partial \theta_{\eta } }} - \frac{{\partial m^{t} }}{{\partial \theta_{\mu } }}\frac{\partial m}{{\partial \theta_{\eta } }}\frac{{\partial m^{t} }}{{\partial \theta_{\eta } }}\frac{\partial m}{{\partial \theta_{\mu } }}}}\frac{{\partial m^{t} }}{{\partial \theta_{\mu } }}\frac{\partial m}{{\partial \theta_{\mu } }},$$

(43)

where $\mu$ is modulus and $\eta$ is viscosity. The mean signal value is the expected displacement for a given combination of modulus and viscosity. These bounds offer a lower limit on performance for any unbiased estimator. It is an optimistic estimate because it does not fully account for signal decorrelation associated with radiation force induced displacement.

The performance bound for the fixed-viscosity estimator is simpler as it has only one free parameter, yielding:

$$C_{\mu \mu } \ge {\raise0.7ex\hbox{${\sigma_{x}^{2} }$} \!\mathord{\left/ {\vphantom {{\sigma_{x}^{2} } {\left( {\frac{{\partial m^{t} }}{{\partial \theta_{\mu } }}\frac{\partial m}{{\partial \theta_{\mu } }}} \right)}}}\right.\kern-0pt} \!\lower0.7ex\hbox{${\left( {\frac{{\partial m^{t} }}{{\partial \theta_{\mu } }}\frac{\partial m}{{\partial \theta_{\mu } }}} \right)}$}}.$$

(44)

The lower bound on the variance of modulus estimates for the fixed-viscosity algorithm is smaller than that for the free-viscosity estimator. This is not surprising because the fixed-viscosity estimator has only one degree of freedom. It should be noted that the total error for the fixed-viscosity algorithm may actually be higher however because of bias resulting from differences between the actual and assumed viscosity.