The Transmission Line

The video depicts sinusoidal pressure and flow waves of a single frequency traveling through a compliant tube in accordance with Transmission Line Theory.  This article starts with the mathematics of transmission lines showing the reason for a specific relationship between the pressure and flow waves – the physical determinants of the characteristic impedance.  It also shows the determinants of propagation through the line ( propagation coefficient ).  SKIP THE MATH to skip the derivation of transmission line behavior but proceed to the interpretation of the result.

In the previous article it was shown that pressure and flow waves arise in the compliant conduits of the circulation due to straightforward considerations of the physics.  It was seen previously that the longitudinal impedance stems from a consideration of the forces due to pressure and viscous friction acting to accelerate a parcel of fluid in a conduit.  This time we're working in the Fourier domain from the start. $$Z_L$$ is the longitudinal impedance:

$$\Large Z_L = \frac{-dP/dz}{Q} = R'(j\omega)+j\omega L'(j\omega)$$

$$P$$ and $$Q$$ represent sinusoidal oscillations of pressure and flow at a specific angular frequency, $$\omega$$.   $$R'$$ and $$L'$$ represent the resistance and inertance per unit of tube length respectively. It has already been seen that the values of $$R'$$ and $$L'$$ are frequency dependent through a non-dimensional flow parameter, $$\alpha = r_0 \sqrt{\rho \omega/\mu}$$.  However we are still working with a linear system and one frequency does not interact with another. I won't show the dependence explicitly in the following, i.e. $$(j\omega)$$ is omitted after $$R'$$ and $$L'$$. Consequently this equation is essentially a spectrum of the equations, with specific values for $$R'$$ and $$L'$$  at each frequency.

For our purposes the transverse impedance is due principally to a the compliance of the tube:

$$\Large Z_T = \frac{P}{-dQ/dz} = \frac{1}{j\omega C' (+G')}$$

It may not be obvious at a glance, but $$-dQ/dz$$ is quite literally the rate of change of tube cross sectional area ( $$dA/dt$$ ). This was shown in the previous article and we'll revisit it again.  $$C'$$ is the compliance per unit of tube length ( $$dA/dP$$ ) so the above could be rearranged to:

$$\Large \frac{dA}{dP} \frac{dP}{dt}= \frac{dA}{dt}$$

i.e. it's a direct expression of the chain rule of calculus.  $$G'$$ is shown here in parentheses for completeness if you care to check this discussion against some other source on transmission line theory.  $$G'$$ is the conductance per unit of tube length and has units of reciprocal resistance (per unit of length).  Expressing it that way makes the equations easier to read but it wouldn't have to be done that way.  In essence the conductance allows fluid to leak out through the wall of the tube, with pressure driving the flow through the wall.  Of course this is relevant to hemodynamics also and it's sometimes used as a means to represent smaller arteries that siphon off some of the flow as it passes through a larger one, such as the aorta.

We've already eliminated the time variable from the equations by working in the Fourier domain so these are now coupled ordinary differential equations,  technically speaking. To solve this problem, we multiply the first equation by $$Q$$ and the second by $$dQ/dz$$:

$$\Large \frac{-dP}{dz} = (R'+j\omega L')Q$$

$$\Large (j\omega C' (+G'))P = \frac{-dQ}{dz}$$

Differentiating the first of these 2 with respect to $$z$$, the flow can be eliminated:

$$\Large \frac{d^2P}{dz^2} = (R'+j\omega L') \frac{dQ}{dz}$$

$$\Large \frac{d^2P}{dz^2} / (R'+j\omega L') = \frac{dQ}{dz} = - (j\omega C' (+G')) P$$

$$\Large \frac{d^2P}{dz^2} = (R'+j\omega L') (j\omega C' (+G')) P$$

All of THIS stuff  $$(R'+j\omega L') (j\omega C' (+G'))$$ is just a number ( a complex one ) and in fact you'll notice that it is equal to $$Z_L/Z_T$$ which is easier to write.  The solution to this differential equation is:

$$\Large P = P_a e^{-\gamma z}+ P_r e^{+\gamma z}$$

where

$$\Large \gamma \equiv \sqrt{(R'+j\omega L') (j\omega C' (+G'))} = \sqrt{Z_L/Z_T}$$

$$P_a$$ and $$P_r$$ are arbitrary (complex) constants that we are free to choose to fit the circumstances of our problem. They turn out to represent pressure waves traveling antegrade ($$P_a$$) and retrograde ($$P_r$$). We can go through this whole process again to solve for the flow, $$Q$$ and we find that we get the very same equation.

Differentiating the second of the 2 wrt $$z$$ allows the pressure to be eliminated:

$$\Large -Z_T \frac{d^2Q}{dz^2} = \frac{dP}{dz}$$

$$\Large \frac{d^2Q}{dz^2} = (R'+j\omega L') (j\omega C' (+G'))Q$$

$$\Large Q = Q_a e^{-\gamma z} + Q_r e^{+\gamma z}$$

$$Q_a$$ and $$Q_r$$ are 2 more arbitrary constants, except that they're not independent of $$P_a$$ and $$P_b$$.  A better way to go after $$Q$$ is to replace the first equation on the page with the solution we've already found for $$P$$.

$$\Large -\frac{dP}{dz} = -(-P_a \gamma e^{-\gamma z} + P_r \gamma e^{+\gamma z}) = P_a \sqrt{\frac{Z_L}{Z_T}} e^{-\gamma z} - P_r \sqrt{\frac{Z_L}{Z_T}} e^{+\gamma z} =Z_L Q$$

Now we'll define the characteristic impedance $$Z_0$$ as:

$$\Large Z_0 \equiv \sqrt{Z_L Z_T} = \sqrt{ \frac{R'+j\omega L'}{j\omega C' (+G')}}$$

That way it isn't so messy when we divide the previous equation by $$Z_L$$:

$$\Large Q = \frac{1}{Z_0} (P_a e^{-\gamma z} - P_r e^{+\gamma z})$$

What we have here is the flow $$Q$$ written in terms of the antegrade and retrograd pressure waves.  The measured pressure results from the summation  of the antegrade and retrograde pressure waves.  However the measured flow results from the difference; the retrograde flow wave is subtracted from the antegrade.  Also we have these relationships:

$$\Large P_a = Z_0 Q_a$$

$$\Large P_r =-Z_0 Q_r$$

$$Z_0(j\omega)$$ is the frequency dependent ratio of pressure to flow, equal to

$$\Large Z_0 \equiv \sqrt{Z_L Z_T} = \sqrt{ \frac{R'+j\omega L'}{j\omega C' (+G')}}$$

It is the same spectrum everywhere in the tube, i.e. as long as the physical determinants do not change ($$R'$$, $$L'$$, $$C'$$, $$G'$$).

The video at the top of the article shows the pressure and flow waves traveling in the transmission line when there is essentially no resistance.  We are starting here with the situation of no reflected waves as the latter complicate things considerably. The viscosity parameter of the model was set as close to 0 as technically feasible. In that case we have a characteristic impedance:

$$\Large Z_0 = \sqrt{ \frac{R'+j\omega L'}{j\omega C'}} \approx \sqrt{ \frac{j\omega L'}{j\omega C'}} = \sqrt{\frac{L'}{C'}}$$

This is just a real number.  Furthermore, with viscosity essentially 0 the inertance isn't very frequency dependent; we simply have a flat velocity profile and $$L' = \rho/A$$.  That means that the pressure and flow waves are simply scaled multiples of each other at all frequencies, with $$\sqrt{L'/C'}$$  the single valued ratio for $$P/Q$$.  If we plot time varying pressure against flow for the above model, we get a hysteresis loop which looks like this:

We get the same hysteresis loop at every location in the tube since the magnitude of the pressure and flow waves don't change ( no friction ).  The slope of this line has physical units of pressure/flow; it's the characteristic impedance.  It appears as a line rather than a loop because pressure and flow are perfectly in phase when there's no friction.  That's the result of having $$Z_0$$ with no imaginary part ( phase = 0 ).

The only thing different about the next video in comparison with the top of the page is that some viscosity has been added to the model fluid.

This obviously shows attenuation of the traveling wave as a result of having some viscosity in the fluid.  The characteristic impedance is no longer a real number:

$$\Large Z_0 = \sqrt{\frac{R'+j\omega L'}{j\omega C'}}$$