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- 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?
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?