# Solution to the 1D wave equation for a finite length plane wave tube

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• jeremiahrose99
In summary: I also see the material properties of density and speed of sound.In summary, the conversation discusses the acoustics of a finite plane-wave tube with arbitrary impedances at both ends. The 1D wave equation for pressure inside the tube is given, and the general solution for a semi-infinite tube is provided as the sum of one incident and one reflected wave. The textbooks also provide a model for the impedance at any point using the definition of specific acoustic impedance and some calculus. It is mentioned that the impedance for the entire tube is determined by the termination impedance at one end. However, there is confusion about whether this is enough to determine both complex amplitudes A and B, and if there are additional reflections in a finite
jeremiahrose99
TL;DR Summary
I'm trying to understand the acoustics of a finite plane-wave tube terminated by arbitrary impedances at both ends.
Hi there! This is my first post here - glad to be involved with what seems like a great community!

I'm trying to understand the acoustics of a finite plane-wave tube terminated by arbitrary impedances at both ends. So far all of the treatments I've managed seem only to address a different problem: that of a semi-infinite tube with an impedance at one end.
The 1D wave equation for pressure inside the tube is:
$$\frac{d^2p}{dx^2}=\frac{1}{c^2}\frac{d^2p}{dt^2}$$
The general solution for the semi-infinite tube is given (by all the textbooks I've read) as the sum of one incident and one reflected wave:
$$p(x,t) = Ae^{j (\omega t-k x)} + Be^{j (\omega t+k x)}$$
Where A and B are complex amplitudes containing both the amplitude and phase information for each sinusoid. Using the definition of specific acoustic impedance and some calculus, the textbooks arrive at a model for the (complex) impedance at any point: $$Z_s(x) = \frac{p(x,t)}{-u(x,t)} = \rho_0 c \frac{A e^{j k x} + B e^{-j k x}}{A e^{j k x} - B e^{-j k x}}$$
Then, if I define the termination of the pipe to be of impedance Z0 at x=0, I can eliminate A and B from the above equation and with some manipulation arrive at:
$$\frac{Z_s(x)}{\rho_0 c} = \frac{\frac{Z_0}{\rho_0 c} + i \tan{k x}}{1 + i \frac{Z_0}{\rho_0 c}\tan{k x}}$$
This equation says that the impedance for the entire tube is completely and uniquely determined by the termination impedance at one end. This seems to exclude the possibility of a second arbitrary impedance further along the tube, which says to me that these results cannot apply to a finite tube with two arbitrary impedances (one at each end).

What, then, is the solution to the 1D wave equation for a finite tube and how to I find it? Is it possible to obtain similar models to above for the impedance in such a tube?

jeremiahrose99 said:
The general solution for the semi-infinite tube is given (by all the textbooks I've read) as the sum of one incident and one reflected wave:
$$p(x,t) = Ae^{j (\omega t-k x)} + Be^{j (\omega t+k x)}$$
Do you understand the physical picture in this region of the tube? Do you see why these two terms are here for this specific scenario?

jeremiahrose99 said:
the impedance for the entire tube is completely and uniquely determined by the termination impedance at one end
Are you sure of this statement? I also see the material properties of density and speed of sound.

jeremiahrose99 said:
Summary:: I'm trying to understand the acoustics of a finite plane-wave tube terminated by arbitrary impedances at both ends.

Hi there! This is my first post here - glad to be involved with what seems like a great community!

I'm trying to understand the acoustics of a finite plane-wave tube terminated by arbitrary impedances at both ends. So far all of the treatments I've managed seem only to address a different problem: that of a semi-infinite tube with an impedance at one end.
The 1D wave equation for pressure inside the tube is:
$$\frac{d^2p}{dx^2}=\frac{1}{c^2}\frac{d^2p}{dt^2}$$
The general solution for the semi-infinite tube is given (by all the textbooks I've read) as the sum of one incident and one reflected wave:
$$p(x,t) = Ae^{j (\omega t-k x)} + Be^{j (\omega t+k x)}$$
Where A and B are complex amplitudes containing both the amplitude and phase information for each sinusoid. Using the definition of specific acoustic impedance and some calculus, the textbooks arrive at a model for the (complex) impedance at any point: $$Z_s(x) = \frac{p(x,t)}{-u(x,t)} = \rho_0 c \frac{A e^{j k x} + B e^{-j k x}}{A e^{j k x} - B e^{-j k x}}$$
Then, if I define the termination of the pipe to be of impedance Z0 at x=0, I can eliminate A and B from the above equation and with some manipulation arrive at:
$$\frac{Z_s(x)}{\rho_0 c} = \frac{\frac{Z_0}{\rho_0 c} + i \tan{k x}}{1 + i \frac{Z_0}{\rho_0 c}\tan{k x}}$$
This equation says that the impedance for the entire tube is completely and uniquely determined by the termination impedance at one end.
I am not an expert but I don't understand this. You have one boundary condition and you say that this determines both A and B. So you have one equation for two unknowns, and you clearly cannot determine both of them. So there has to be more input that just that single boundary condition.

Dr_Nate said:
Do you understand the physical picture in this region of the tube? Do you see why these two terms are here for this specific scenario?

Thanks for the reply Dr Nate. Yes I'm pretty sure I understand it - it's just the sum of the incident and the reflected wave. There can be no others because there are no reflections coming from the infinite end. But for a finite tube, I imagine there would be new reflections of different phases happening all the time, so I don't know what the solution would be for that...

Dr_Nate said:
Are you sure of this statement? I also see the material properties of density and speed of sound.

I just meant that there are no other variables within my control (I can change the termination impedance, but I can't change the speed of sound), and that there is no room in the equation for a second terminating impedance, meaning the result doesn't work for a finite tube. Does that make sense?

jeremiahrose99 said:
Thanks for the reply Dr Nate. Yes I'm pretty sure I understand it - it's just the sum of the incident and the reflected wave. There can be no others because there are no reflections coming from the infinite end. But for a finite tube, I imagine there would be new reflections of different phases happening all the time, so I don't know what the solution would be for that...
It looks like you understand what I was asking.
jeremiahrose99 said:
I just meant that there are no other variables within my control (I can change the termination impedance, but I can't change the speed of sound), and that there is no room in the equation for a second terminating impedance, meaning the result doesn't work for a finite tube. Does that make sense?

Thanks for your replies Dr Nate. Yes, the impedance of the semi-infinite tube alaso varies with position and wavenumber. However, the question I asked is about a different case, a finite tube, with impedances at both ends. Can you help me with that?

jeremiahrose99 said:
Thanks for your replies Dr Nate. Yes, the impedance of the semi-infinite tube alaso varies with position and wavenumber. However, the question I asked is about a different case, a finite tube, with impedances at both ends. Can you help me with that?
So this is pretty much the same as an electrical transmission line, right? The same equations apply, so the same solutions apply. Are you wanting to drive the CW (or pulsed?) input waveform into the transmission line via the source impedance at one end? And then solve for the waveform distribution versus time?

That is very do-able. It's easier in the pulsed input case, but still do-able in the CW input case.

berkeman said:
So this is pretty much the same as an electrical transmission line, right? The same equations apply, so the same solutions apply. Are you wanting to drive the CW (or pulsed?) input waveform into the transmission line via the source impedance at one end? And then solve for the waveform distribution versus time?

That is very do-able. It's easier in the pulsed input case, but still do-able in the CW input case.

Yes! I haven't fully gotten my head around electric-acoustical impedance analogies yet, but I do know that a finite acoustic pipe can be modeled as a finite electrical transmission line.

Looking for a continuous (steady-state?) solution. I was hoping for a general solution for arbitrary impedances at each end, but perhaps something more specific would be easier so I will describe the exact setup for you:

I have a pipe with a driver at one end, open at the other end. The open end has a complex radiation (load) impedance Zmr. The driven end has an input impedance Zm0 (due to Zmr, reflection interference, and friction in the pipe) and a source impedance Zmd (due to the mechanical resistance, inertia and compliance of the driver). The driving "voltage" in your analogy is the mechanical force acting on the driver (there are at least three different definitions of acoustic impedance: I am using the "mechanical" impedance definition which is basically force divided by velocity).

Trying to get solutions for the time/space evolution of the pressure waveform in the pipe, analytical approximations for resonant frequencies in the pipe, and a model for the impedance at different points in the pipe.

berkeman

## 1. What is the 1D wave equation for a finite length plane wave tube?

The 1D wave equation for a finite length plane wave tube is a mathematical representation of the propagation of a wave through a tube with a finite length. It is described by the equation: ∂²u/∂t² = c²∂²u/∂x², where u is the displacement of the wave, t is time, x is the position along the tube, and c is the wave speed.

## 2. What are the boundary conditions for the solution to the 1D wave equation for a finite length plane wave tube?

The boundary conditions for the solution to the 1D wave equation for a finite length plane wave tube are that the displacement of the wave must be zero at both ends of the tube. This represents the fact that the wave is confined within the tube and cannot escape.

## 3. How is the solution to the 1D wave equation for a finite length plane wave tube derived?

The solution to the 1D wave equation for a finite length plane wave tube is derived using the method of separation of variables. This involves assuming a solution of the form u(x,t) = X(x)T(t) and substituting it into the wave equation. This results in two ordinary differential equations, which can be solved to obtain the general solution.

## 4. What is the significance of the solution to the 1D wave equation for a finite length plane wave tube?

The solution to the 1D wave equation for a finite length plane wave tube is significant because it allows us to understand the behavior of waves in confined spaces. It can be used to model and predict the propagation of sound waves in musical instruments, as well as electromagnetic waves in optical fibers.

## 5. Are there any limitations to the solution of the 1D wave equation for a finite length plane wave tube?

Yes, there are some limitations to the solution of the 1D wave equation for a finite length plane wave tube. It assumes that the tube is perfectly rigid and has no losses, which may not be the case in real-world situations. Additionally, it only applies to one-dimensional waves, so it cannot be used to model more complex wave phenomena.

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