Longitudinal Impedance






In hemodynamics, longitudinal impedance has to do with impedance to oscillatory/pulsatile flow along the length of a vessel – in the longitudinal direction.   

\( \LARGE p_1+\frac{\rho}{2} v_1^2 + \rho g h_1 = p_2+\frac{\rho}{2} v_2^2 + \rho g h_2 + L_{\mu} + \rho \int_1^2  \frac{dv}{dt} ds \)

Relating this to the familiar Bernoulli equation above, we are talking about terms of the equation involving the pressure difference, the frictional losses (\(L_{\mu}\)), and the term with the integration along a streamline involving acceleration, \(dv/dt\). 

Longitudinal impedance includes a part due to the mass of fluid -- the inertance, represented by the coil or inductor in the above figure, and a part due to resistance – impedance due to friction.  Inertance and resistance were discussed in previous articles but a brief recap is in order.  We saw that the resistance is due specifically to friction caused by fluid layers sliding against one another. So resistance depends on the way the flow actually occurs - specifically the way the fluid layers slide against each other (the "velocity profile"). In the special case of Poiseuille flow of a Newtonian fluid through a circular conduit, the resistance can be calculated from specifics of the tube and the fluid viscosity.

\(\Large Z_R = R = \frac{8\mu \Delta z}{\pi r_0^4}\)

\(Z\) is the symbol being used to represent impedance, so \(Z_R\) is the impedance due to resistance alone.  We saw previously that the inertance is the thing that multiplies \(dq/dt\) in a flow equation.  If we consider for a moment the blue plug of fluid in the above figure, an expression of force equals mass times acceleration (\(\bar{F} = m\bar{a}\)) looks like this:

\(\Large \Delta p A = \rho A \Delta z \frac{dv}{dt} \)

\(\rho\) is fluid density, \(A\) is the cross sectional area, and \(\Delta p\) is the upstream pressure minus the downstream. The stuff on the left-hand side of this equation is the force due to pressure acting on the blue fluid parcel, the stuff on the right hand side is the mass of the fluid multiplied by the acceleration.  We want to replace \(dv/dt\) with \(dq/dt\) ("flow acceleration") which is done using the approximation \(q = vA\), which implies \(dv/dt = \frac{dq/dt}{A}\).

 \(\Large \Delta p = \frac{\rho \Delta z}{\pi r_0^2}   \frac{dq}{dt} \)

 \(\Large L = \frac{\rho \Delta z}{\pi r_0^2}  \)

 \(\Large \Delta p = L   \frac{dq}{dt} \) 

 \(L\) is the inertance - the thing that multiplies the rate of change of flow.

Whenever impedance comes up, we are talking about a ratio of 2 sinusoids of a specific frequency OR we're talking about a function - a whole spectrum of sinusoid ratios over a range of frequencies. Now I'm shifting to Fourier Domain and represent pressure and flow sinusoids  with capital letters, \(P\) and \(Q\).  We've seen that differentiating a sinusoid with respect to time is equivalent to shifting the oscillation by 90° and multiplying by the (temporal) angular frequency.

\(\Large \Delta P = ( j \omega L) Q\)

\(\Large \frac{\Delta P}{Q} = Z_I= j \omega L\)

The impedance, represented by \(Z\), is a ratio of pressure and flow sinusoids. \(\omega = 2 \pi f\) is the angular frequency of an oscillation when \(f\) is given in cycles per second. I'm using \(Z_I\) to represent an impedance due to inertance alone.  We've already seen that \(j \omega L\) is the impedance due to a pure inertance.

For the simple electrical analog implied by the figure above, we have 2 impedances in series arrangement with each other.  The equivalent longitudinal impedance (\(Z_L\)) for this simplified combination is simply the sum of the impedances due to resistance and inertance:

\(\Large Z_L = Z_R+Z_I\)

\(\Large Z_R = R\)

\(\Large Z_I = j\omega L\)

\(\Large Z_L = R + j\omega L\)

The above is for a finite length of tubing.  For future developments, we need to determine the longitudinal impedance per unit of tube length. To do this, we allow the length of the tubing in question to approach 0 and divide the above impedance representations by \(\Delta z\), the length of the tube.

\(\Large R'\sim \frac{8\mu}{\pi r_0^4}\)

\(\Large L' \sim \frac{\rho}{\pi r_0^2}\)

\(R'\) and \(L'\) are now the resistance and inertance per unit of length of the conduit.  \( \Delta P\) becomes \(-dP/dz\) where the negative sign comes in because \( \Delta P\) is upstream minus downstream, as opposed to the reverse.

\(\LARGE Z_L' \equiv \frac{-dP/dz}{Q} = \frac{-P_z}{Q}\)

So this is actually the definition of the longitudinal impedance, \(Z_L'\).  I've retained a prime with \(Z_L'\) to indicate that quantities are per unit of length, but will be discontinuing that notation shortly.  \(Z_L'\) is the ratio of 2 sinusoids – a negative pressure gradient sinusoid divided by a flow sinusoid of the same frequency.  \(P_z\) is a shorthand for \(dP/dz\) that's in common use.  \(Z_L'\) is complex and function of frequency  in a way that we will soon be exploring.  We've made some approximations above that will be revisited, but \(Z_L'\) can be expressed as a modulus and phase – a magnitude and angle.  Using the rules of complex numbers we have the following:

\(\Large |Z_L'| = \sqrt{R'^2+\omega^2 L'^2} \)

\(\Large \angle Z_L' = \tan^{-1}\left[\frac{\omega L'}{R'}\right]\)

The equation shows that the magnitude of the longitudinal impedance increases without bound as frequency increases.  \(\omega\) ( angular frequency ) is the aspect of the equation that depicts the acceleration rate.  It takes more and more oscillatory pressure force to produce a flow oscillation as the frequency increases.  

Once Again without Dimensions

That's it! We are done! Well, sort of.  We could use these last 2 equations to compute a reasonable approximation for the longitudinal impedance spectrum given the dimensions of the tube, viscosity, fluid density, etc. You supply these parameters and we plug them in to the equations for \(R'\), \(L'\), \(|Z_L|\), etc. and out comes the answer.    However we would have a separate plot for each set of conditions and I don't have space here to put in a plot for each of the arteries. So we'll take this opportunity to delve a little deeper by looking at a nondimensionalized version of the equations.  Engineers and physicists always do this if possible.  Notice for starters that all of the terms in the equation for \(Z_L'\) have the same physical units – impedance per unit of length.  What happens next is we divide the whole equation by \(R'\) (Poiseuille resistance per unit length).  This results in a nondimensional ratio in place of each of the original terms. Dividing \(R'\) by \(R'\) gives 1. For the other terms:

\(\Large \frac{Z_I'}{Z_R'} \sim \frac{\rho}{\pi r_0^2} \frac{\pi r_0^4}{8\mu} = \frac{\rho \omega r_0^2}{8 \mu}\)

\(\Large \frac{Z_L'}{R'} = 1 + j  \frac{\rho \omega r_0^2}{8 \mu}\)

The last equation is a nondimensionalized version of the earlier equation for \(Z_L'\).  But now the result of the equation (the \(y\) axis) is in terms of multiples of the Poiseuille resistance ( per unit length ).  At zero frequency, in other words, the impedance of every tube is 1 Poiseuille resistance.  We see that the impedance increases not with frequency per se, but has to do with tube radius, fluid density, and fluid viscosity.  We can in fact bundle this effect into 1 variable:

\(\Large \zeta \equiv  \frac{\rho \omega r_0^2}{8 \mu}\)

I've called it \(\zeta\) but don't get freaked by the Greek.  Now we can rewrite the expressions for longitudinal impedance in fully nondimensional form as follows:

\(\Large \frac{Z_L'}{Z_R'} = 1 + j \zeta \)

\(\Large \left|\frac{Z_L'}{R'}\right| = \sqrt{1 + \left[\frac{\rho \omega r_0^2}{8 \mu} \right]^2} = \sqrt{1+\zeta^2}\)

\(\Large \angle \frac{Z_L'}{R'} = \tan^{-1}\left[\frac{\rho \omega r_0^2}{8 \mu}\right] =  \tan^{-1}(\zeta)\)

By thinking about this problem in nondimensional terms, we rescale the result so that the equations are applicable to every tube.  Here are the graphs for the modulus and phase of the normalized impedance ( same graphs for every tube): 


Here's how this works:  You know the tube radius, fluid viscosity and density for your situation so you can compute the independent variable \(\zeta = \rho \omega r_0^2/(8 \mu)\).  You can read off the value of the phase directly from the lower plot (or by stuffing \(\zeta\)  into the equation.  For the modulus however, you have to multiply by \(8\mu/(\pi r_0^4)\) to convert the nondimensional result ( from plot or equation ) to one suitable for your specific tube. 


Interpreting the Math

Take a moment to consider the meaning of the equations.  At zero frequency, the modulus of the longitudinal impedance is simply the Poiseuille resistance.  The angle of the longitudinal impedance is 0 meaning that the negative pressure gradient is exactly in phase with the flow.  There are no accelerations (\(\omega = 0\)) so all of the pressure force is directed towards overcoming friction.  (It might make more sense to consider a very slowly oscillating sinusoid for which the values of the impedance modulus and phase are very close to the zero frequency values.)

With increasing oscillation frequency,   the negative  pressure gradient (pressure force oscillation) must increase in magnitude to generate a flow oscillation of a given magnitude.  Far enough away from zero frequency, we have essentially a linear increase in pressure force with frequency to generate a given flow magnitude.  for \(\omega L' >> R'\) :

\(\Large \sqrt{R'^2+\omega^2L'^2} \approx \omega L'\)

 The angle of the longitudinal impedance also changes; while in phase with the flow at low frequency, it transitions as shown so that the angle approaches +90° (\(\pi/2\) radians).  In other words the pressure force "leads" the flow by 90° in the limit for large frequency.  This should not be surprising if you remember that a net force induces an acceleration ( not a velocity).  One can view the entire longitudinal impedance spectrum as a transition to greater and greater flow acceleration with increasing frequency.  At very low frequency there is almost no net force on the fluid, i.e. almost all the pressure force is balanced by friction. At very high frequency, frictional forces are very low in comparison with the acceleration.

A sinusoid of a given frequency can always be resolved into a sine wave and a cosine wave of the same frequency.  This allows us to determine ( from data or from mathematical expressions )  how much of the longitudinal impedance is due to resistance and how much is due to inertance.  Consider the following 2 plots showing the negative pressure gradient sinusoid (oscillating pressure force ) and the resulting flow below it ( blue sinusoid ).  The pressure  force sinusoid was chosen semi-arbitrarily as a cosine wave for illustrative purposes and both pressure force and flow have been normalized to magnitude 1.0. It should be clear from the figure that the pressure force and the flow are not particularly in phase with each other. However the blue flow wave is readily resolved into a cosine wave (red) that is in phase with the pressure force and a sine wave (black) that lags the pressure force by 90°.  The resistive part of the longitudinal impedance at this frequency is the modulus of the negative pressure divided by the modulus of the red (in phase) part of the flow; the inertial part is the modulus of the pressure force sinusoid divided by the modulus of the black (out of phase) part of the flow.

In terms of the development above, the resistive part of \(Z'_L\):

\(\Large |Z_L| \cos[\angle Z_L]=R'\)

And the inertial part:

\(\Large|Z_L| \sin[\angle Z_L]=\omega L'\)

After all, the equations are showing that \(Z'_L = \sqrt{R'^2 + \omega^2 L'^2}\) is the hypotenuse of a right triangle with \(R'\) the side adjacent and \(\omega L'\) the side opposite of the angle in question.  At low frequency (small \(\zeta\) specifically), the flow is more nearly in phase with the negative pressure gradient and the resistive part is much greater than the inertial (flow and pressure gradient still normalized to 1.0):

\(\Large \alpha = 1\)



(red is the resistive/in-phase part of the flow; black the inertial/out of phase part) At high values of \(\zeta\), the flow is largely out of phase with the pressure force sinusoid (flow and pressure gradient still normalized to 1.0):

\(\Large \alpha = 6\)


 (red is the resistive/in-phase part of the flow; black the inertial/out of phase part)

Now we're going to add the next widget for understanding longitudinal impedance. In the development above, it was assumed that the resistance was due to a parabolic (Poiseuille) profile regardless of the oscillation frequency and that the inertance was due to a flat profile.  We've been mixing metaphors!  In fact we've already seen that the velocity profile in oscillatory flow varies between a parabolic profile at 0 frequency to a very flat profile at high frequency or, more specifically, at a high value of \(\alpha = r_0 \sqrt{\rho \omega/\mu}\).  Hey, WAIT A MINUTE!  \(\alpha\) is starting to look a heck of a lot like the thing that was called \(\zeta\) in the above!  And that's no accident. The thing we called \(\zeta\) above was the result of dividing the inertial part of the impedance by the resistive part.  That's what \(\alpha\) and \(\zeta\) are.  The fact that one is the square root of another and multiplied by a constant is rather immaterial. The important part is that they both result from the ratio of inertial to resistive forces acting on the fluid.  We see this sort of thing over and over again in physics and engineering analysis.

We already figured out the longitudinal impedance as it depends on velocity profile in a previous article; I won't put you through it again but here are the results:  

\(\Large \frac{-P_z(j\omega)}{Q(j\omega)} = \frac{\rho \omega}{j \pi r_0^2} \frac{1}{\left[1-F_{10} \right]}\)

\(\Large M'_{10} e^{j\varepsilon'_{10}} =1- F_{10} = \left[1-\frac{2 J_1(j^{3/2}\alpha)}{j^{3/2}\alpha J_0(j^{3/2}\alpha)}\right]\) 

\( \frac{-P_z}{Q}\) is the definition of the longitudinal impedance; the stuff on the right hand side of the equation shows how we might compute it from information about the tube and fluid within.  Womersley also decided to express \(1-F_{10}\) in terms of a modulus and phase, \(M'_{10}\) and \(\varepsilon'_{10}\) as shown.  Here are plots of \(1-F_{10}\) in terms of \(M'_{10}\) and \(\varepsilon'_{10}\) to help you get the feel for them:


HERE is a table of values pertaining to Womersley's analysis. 



The foregoing gives us a mathematical approximation of the longitudinal impedance.  This turns out to be a difficult thing to measure, either in vitro or in vivo because of the difficulty in measuring the pressure gradient.  However let's take a little time to get acquainted with the meaning – the consequences associated with the equations and graphs.


Physical factors affecting inertance and resistance



Effect of velocity profile and alpha on resistance

Effect of velocity profile and alpha on inertance

Videos of velocity profiles in action



Fully normalized plots, \(\alpha^2\) x-axis





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