The Acoustical Channel the Transfer Function of Water Pipes

Transcription

1 The Acoustical Channel the Transfer Function of Water Pipes Supervisor Prof Werner Henkel Student Behrouz Touri 1 Introduction Because of the sever exponentially attenuation of the electro magnetic waves in impure water, the ultrasound waves are the only means of communication through water for long distance. Severe study of underwater sound propagation dates back to the first and second world war. In that age and even now, the main motivation for using underwater acoustics is to use it as the tool for monitoring the movement of underwater objects. However now a day there are many other application for underwater acoustics such as sesfloor imaging, communication using underwater modems and... Although there are lots of works done in this area it seems that there are still a long way to go comparing with wireless communication and there are lots of challenging problems in this area. These problems are distributed over wide area of sciences such as studying of wave propagation in shallow waters, non homogeneous media to engineering such as designing of efficient transceivers for special purposes or... In this project our objective was to study channel response of water filled ducts. Our motivation to do such a project was to see the possibilities of information transmission using water filled pipes. Since water pipes are frequently used in almost all houses and factories, then the possibility of ultrasound transmission using those media will lead to many application such as central energy consumption monitoring units in big houses, control of the valves or flows in factories or... In this project we first studied propagation of ultrasound waves through water and water pipes and then try to see wave propagation through a real water filled pipelines. Then we tried to explain the differences between the practical and theoretical results. Our study has its own weaknesses and we will come to this point at the end of this report. 1

2 Figure 1: Duct with rectangular cross section 2 Sound propagation in ducts Consider a medium in which p(x, y, z, t) is the related pressure scalar field at the point (x, y, z) at the time t. If there is no mean flow in point (x, y, z) then the wave equation describing such a scalar field will be 2 p c 2 2 p t 2 = 0 (1) If physical properties of medium is not time dependent and is homogeneous and linear then c will be simply a constant which describes the sound speed in the medium. Fortunately, in our experiments all of the mentioned conditions are almost satisfied. 2.1 Ducts of rectangular cross section Although our study was about the pipelines with circular cross section, studying of wave propagation in ducts with rectangular cross section may help us to understand the behavior of waves in ducts with circular cross section much more better. Suppose we have a duct with rectangular cross section with rigid boundary (which means the acoustic impedance of the boundary is infinity). More over suppose that the length of the sides of the duct are L x and L y. Now consider Cartesian coordinates for which the z-axis placed along one of the corners of the duct and is parallel to the axis of the duct and two sides with lengths L x and L y are lied on x and y axis respectively (see figure 1). Then we will have the differential equation: 2 p c 2 2 p t 2 = 0 (2) With boundary condition p x = 0 at x = 0, L x and p y = 0 at y = 0, L y. The intuitive reason for the mentioned boundary condition is that the particle 2

3 speeds should be equal to zero at the boundaries. And this means that the amplitude of pressure p should be maximum at the boundaries which means the corresponding derivatives of p should be zero at the boundaries. Another description for mentioned property is that the movement of each tiny differential volume is caused by the difference between the pressure of the both sides. Since the boundary is rigid, so we haven t particle movement normal to the boundaries which means the difference of pressure at the corresponding direction is equal to zero which itself means that the derivative of the pressure field should be zero at corresponding direction. If we put an ideal sinusoid point source operating with frequency ω 2π in the origin then the appropriate solutions to above equation are [2, 1, 4] In which p lm = A lm cos k xl x cos k ym ye j(ωt kzz), l, m Z (3) k xl = lπ L x, k ym = mπ L y, k z = ω 2 k 2 c xl kym 2 (4) The function (wave) p lm is called lmth mode. Now suppose that k lm = Then k z = ( ω c )2 klm 2. According to the equation 3, if k lm > ω c k 2 xl + k2 yl. then we will have an exponential decaying function of z which means at sufficiently large distance from the source (origin) the amplitude of the corresponding wave p lm became negligible. So, we can define the cut-off frequency ω lm = ck lm for the p lm component (mode lm). So, in order to the mode lm propagate we should have ω ω lm. Since k lm increase with l, m, so just finitely many modes can propagate through the duct. If the frequency is sufficiently low, or in other word the wave length is sufficiently high, then from the above discussion it will be obvious that the only traveling component will be 00 mode. It is worth while to mention that 00 mode, will always propagate and in for this mode we have k z = ω c Phase velocity and group velocity One should distinguish between these two velocities. Phase velocity c p is defined to be the rate of the change of phase at frequency ω along the z axis. From equation 3 we will have: c p = ω ω = = k z ( ω c )2 klm 2 1 c 1 = 2 ( k lm 2 ω ) c 1 ( ω lm ω ) 2 (5) So, it is obvious that the phase velocity of a propagating mode is greater than c. But the fact is that the energy can not propagate faster than the speed of the sound. So, what is the problem? Actually each component s speed can not determine the speed of the energy movement. That is why we should define a more descriptive expression for the speed of the wave. This expression is called 3

4 Group velocity. Simply group velocity defined as the velocity of progress of the center of gravity of a group of waves that differ somewhat in frequency [4]. Another definition is the speed of the peak of the traveling wave. Exact analytical derivation of the group velocity is usually not always possible. Using some simplifying assumptions the phase velocity of the l, m component in 3 will be equal to the: c g = c 1 ( ω lm ω )2 (6) In order to have an intuitive filling about the above expression lets rewrite 3 as: p lm = 1 4 A lm e j(ωt±k xlx±k ymy k zz) (7) ± In above equality ± means sum over every combination of ± in the expression 6. Now lets consider for example e j(ωt+k xlx+k ymy k zz). The constant phase plane for the mentioned component is (ωt+k xl x+k ym y k z z) = α. This means that in time t 0 all of the points satisfying (ωt 0 +k xl x+k ym y k z z) = α will have the same phase. But the normal vector to this plane is n = (k xl, k ym, k z ). So, for this component, the energy propagate in the n = (k xl, k ym, k z ) direction. For other terms of the 6 we have the same argument. These are nothing else than several reflecting components from the wall (see figure 2). Now, if the energy of the sound propagate with the speed of c in the direction n, then the velocity of this component along the direction of the duct (zaxis) will be equal to the ck z = c 1 ( ω lm kxl 2 + k2 ym + kz 2 ω )2 (8) 2.2 Ducts of circular cross section The concepts and principles of wave propagation along the ducts with circular cross section is the same as ducts with rectangular cross section. Consider a circular cross section duct with radius r 0 centered at origin and positioned along the z-axis in R 3. Again we should solve the same differential equations 2 but now the boundary conditions should be satisfy in the region r = r 0 in the cylindrical coordinates. In order to avoid the details here we will just mention the solution of the new problem. In this case all of the functions are written in the cylindrical coordinates. The solution of the above equation is the summation of the terms p lm (θ, r, z) = A lm J m (k lm r) cos(mθ)e j(ωt k zz) In which l, m N {0} and J m (r) is the first kind Bessel function of mth order. As it was mentioned before the derivative in the normal direction to the walls of the solutions should be zero in boundaries. Which means: J m (k lm r) r=r0 = 0 J r m(k lm r 0 ) = 0 (10) 4 (9)

5 So if the l th zero of the m-th order Bessel function is located in z lm then k lm = z lm r0. Since the Bessel functions can not be expressed in terms of elementary functions, there are famous tables for the l + 1th zero of mth order Bessel function. You can find such a table in Appendix A. An important fact is that the 00 mode which is for m = 0, l = 0 always propagate because the derivative of the J 0 function is zero at the origin so k 00 = 0. So, the 00 mode will always propagate. In the equation 9 k z = ( ω c )2 k 2 lm (11) Again we see that in order for lm mode to propagate, k z should be a real number. So if we define ω lm = ck lm then ω lm will be the lower cut-off frequency for the lmth mode. Exactly as the case of rectangular cross section ducts, the phase velocity and c group velocity of the lm mode will be c p = and c ω 1 ( lm ω ) 2 g = c 1 ( ω lm ω ) 2 respectively. Generally speaking, in any such a transmission line the group velocity is equal to v g = ( k z ω ) 1 (12) Which in our case: v g = ( ( ω ω c )2 klm 2 ) 1 = ( 2ω 1 c 2 ) 1 (13) 2 ( ω c )2 klm 2 = c 1 ( ck lm ω )2 = c 1 ( ω lm ω )2 (14) Which is the same as the one we claim. So, as it can be seen form the table of the Appendix A, the modes will propagate in the order of 00, 01, 02, 10, 03,... 3 Practical results In our experiment, we investigated the transfer function of cylindrical pipelines which are the same as cylindrical ducts. Our experiment was mainly on 1-meter and 2-meters pieces of copper pipelines with radius r 0 = 1.15cm. In figure 3 you can find the building blocks of our experiment and how they were placed next to each other. Simply, the output of the network analyzer will pass through the amplifier and then it will be fed to the piezoelectric transmitter which is placed in one 1 Here we used l + 1th zero instead of lth zero because we wanted to have the index set l starts with 0. 5

6 Figure 2: Constant phase planes for 01 mode end of the pipe. Then the wave will propagate through the pipe and at the other side, the receiver transducer will receive the signal. Then the signal will go through the buffer circuit and finally it will go to the input of the Network analyzer. Then network analyzer will simply divide the Fourier transform of the transmitted wave by the Fourier transform of the received signal. And finally, depend on the operator s selected mode for displaying data, the phase, magnitude or... of the result will be shown in the display of the network analyzer. Network analyzer out Amplifier Trandsucer (Transmitter) Pipe Network analyzer in Buffer Transducer (Receiver) Figure 3: Building blocks of our experiment 6

7 3.1 Amplifier and buffer circuits We used a simple two stage amplifier with an approximate gain of 20dB. The first stage is nothing more than a single one stage common-emitter amplifier. The second stage is just a simple type-a, power amplifier. Figure 4: Layout of the amplifier designed circuit The buffer circuit is also a simple type A power amplifier with output resistance of approximately 47Ω. In figure5 you can see the designed circuit. 3.2 Transducers In our experiments we used water proof piezoelectric transducers for transmitting and receiving sound waves. As it is provided by the manufacture of this transducers 2, the nominal center frequency of these transducers is in 40 khz. In figure 6 you can see the frequency response of the transducers provided in the company s data sheet[3]. 3.3 Results Experiment with 1-meter piece In figure 7 you can see the frequency response of the whole scheme for 1 meter piece. As you can see the peak is in roughly 30 khz, which is about 10 khz less than what is claimed by the manufacturer of the transducers. This is mainly due to the differences of the pressure of the manufacturer measurements in 2 Murata 7

8 Figure 5: Layout of the designed buffer circuit Figure 6: Sound pressure level diagram of the transducers 8

9 air and our measurements in water. It is important to keep in mind that these transducers are built in order to operate in air, so the data sheets measurements are done in the air condition. Later we will see that the response of the air filled pipes which approves mentioned description. Another amazing fact is that when the frequency goes up, the frequency response also goes up. This is some how contradicting to our intuition and also the data sheet of used transducers. This is due to the structure of these transducers. The body of those transducers are made of black Aluminum which is not obvious unless you cut them into pieces. At the other one of the two string of wires going inside the transducers connected directly to the body of the transducers. So, part of the signal will be transmitted by electro magnetic waves through the water. One may say that those components should not travel through the water. This is true but not for short range. At the other hand part of those components will reach easily to the surface of the pipes, which are made of coppers, and then it will travel through the surface. Fortunately and by the chance in early stages of our experiments we figured out that if we connect the surface of the pipes to the ground then a noise with constant component in each frequency will disappear (which later, as I told, we find out that it is actually signal component traveling through the surface of the pipe). So, the high frequency components are mainly due to the electro magnetic components. In figure 7 you can see the time domain impulse response of the whole structure with 1-meter piece. In this figure, the vertical lines show the estimated arrival time of the related components, for example the green line shows the approximate expected arrival time for the acoustic component traveling through the water. As you can see, there is a negligible component traveling through the copper using longitudinal wave. The rest is more or less due to the water. The peak of the water component come a bit slower than the speed of the sound. Actually this is really slower and it is not just kind of latency. We will explain and deal with this problem later on in our discussion about the 2-meters piece meters piece Actually, since 1-meter piece is not long enough, there may be some higher modes components which are not negligible in 1 meter or even some electrical component traveling through water. So, we studied the wave propagation briefly in the 1-meter piece and now we will study the propagation in the 2-meters piece in more details. The frequency domain response of the air filled 2-meters piece is shown in figure 8. As you can see the peak value is in roughly 40 khz which is the same as in manufacturer s catalog and it approves our reasoning about the 30 khz peak in the water filled pipe. Here, in contrast to 1-meter piece, higher frequency components does not rise up. This also approve our description for higher frequency component in 1-meter piece. In 2-meters piece, because of the higher distance between receiver and transmitter, electro magnetic components attenuated much more than the 1-meter piece. In figure 8 the time-domain response of the 2-meters piece is shown. There, 9

10 H(f) (db) Frequency (Hz) x x Impulse response Water Copper long First reflect Air Copper transversal Voltage (v) Time (t) x 10 3 Figure 7: Magnitude of frequency response of the water filled 1 meter piece (top) and time domain response of the water filled 1 meter piece(bottom)

11 several vertical lines are indicator for the expected arrival time of each mode. The expected arrival time is computed using the formulas for group velocity in previous part and also using the table of Appendix A. A quick look to the diagram shows that the dominant mode is not the 00 mode which travels with the speed of the sound (green line indicates the arrival time of 00 mode). The spread shape of the impulse response is mainly due to the bandwidth of the transducers. As it is shown in the frequency response of the transducers, the bandwidth of the transducers are less than 5 khz. So, we should see a spread shape in time domain. Unfortunately, we haven t access to the impulse response of the transducers so we can not exclude the effects of the transducers in our impulse response and this is the most important defect of our study. If our explanation is correct then another question rises. Why some modes propagate better than the other modes, specially why 00 mode is not the dominant mode? I think this is because of the transducers beam pattern. In 9 each constant coefficient A lm determines the power carried by lm mode. Those coefficient will be determined by the boundary condition imposed by the source. If the source is ideal point source, then there should be a delta Dirac function generator at the center of the pipe. If so, then all of the A lm will be equal to one. More precisely p lm s in 9 are functions which are some how pairwise orthogonal to each other which means: 2π 0 0 p lm (θ, r, z)p l m (θ, r, z)rdθdr = 0 (15) for (l, m) (l, m ). So, if our source is placed in z = 0 and p(θ, r) is its pressure scalar field, in order to find A lm, it is sufficient to compute: 2π 0 0 J m (k lm r) cos(mθ)p(θ, r, t = 0)dθdr = 0 (16) So, the factor that determines which mode propagate better is the pattern of the transducers which we don t have access to its details. One may think that maybe this is because of the imperfectness of the boundary conditions. I don t believe so because the acoustic impedance of the Copper, is more than 200 times of the air one. So, at least in air it more or less satisfies the boundary conditions. Even if the boundary conditions is not ideal, the solution to the wave equation 2 for circular boundaries and finite impedance in the boundaries is the same as 9. The only things that would be changed compare with our case, are the constants k lm. In figure 9 you can see the time domain response of water filled 2-meters piece. At the very first look, it seems that the only major components are mostly due to the longitudinal component of copper and the 00th mode from water. But, by the chance we measured the same channel after a day and we see more or less very different channel than the previous one which is placed in the bottom of the same figure. In the top figure the peak is more or less carried by the 00 mode but in the other one the peak comes with some slower component. As I told before, this peak is due to slower mode and it is not due to the latency because as you can see the reflected component 11

12 H(F) (db) Frequency (Hz) x 10 4 Voltage (v) 2.5 x Time (t) Figure 8: The magnitude of frequency response of the air filled 2 meters pipe (top) and the impulse response of the air filled 2 meter pipe (bottom)

13 which appears later, came exactly at 3 times the place where peak appears which means that is a slower mode. Another major difference is that the first impulse response is very dispersed but the second one is compacter. What is the cause of such difference? We think that this difference is due to the solved air in the water. We did the first measurement immediately after filling the pipe with water. As we know there are considerable amount of air solved in water inside the water pipelines. But when the water remain still the air inside the water dissolves. When there are lots of air bubble inside the water then the media looks like optical fibers which are made of impure glasses. When water has lots of solved air in it then the wave travels several time and with several angles from one material to another material with different acoustical impedances (air and water) and this cause dispersion of the waves. That is why in the first diagram we see that the 00 component is stronger than the rest, because the 00 mode travel less distance than the other components so it will suffer from the dispersion less than the other modes. Our description approved by figure 10. In that figure, we put some water still for 2 days and then poor the water in the pipeline slowly such that air doesn t solve in it very much. As you can see this is not like the immediately measured channel response. However it not identical to the second one but it is identical to the first one as well. At the other hand in the second figure we see some reflected components, but in the first figure we can not see such reflections. This is also in agree with our discussion. Comparing the air filled pipe and water filled pipe, we see that in air filled the tail of the time domain response is much more than water filled one. This is because of the difference of the sound speed in air and in water. As you can see in order to lmth mode propagates, k z = ( w c )2 klm 2 should be real. So, when c decreases, the higher order modes can travel (k lm is independent of c). So, that is why we see much more wider response in air-filled pipe and water-filled pipe. 3.4 Regular notches In figure7 we see that there are some regular notches. Even those notches are more regular in figure 11. Figure 11, is the frequency response of 2-meters water-filled piece after one day. What can cause such regular notches? Regular notches in frequency domain means standing wave in time domain which is caused by the summation of the wave and its reflections. As you can see in figure 9(bottom) we have a nice reflected component. This reflected component is not so nice in figure 7 (time domain response). And that is why we see more regular notches in figure 11 than figure 7. In figure 12 you can see the zoomed version of the figure Conclusion and suggestions for further studies In this project, we studied the transfer function of water pipes. Unfortunately, our study was more descriptive than analytic, because in order to have more 13

14 4 x Impulse response Water Copper long First reflect Copper transversal 1 Voltage (v) Time (t) 4 x Impulse response Water Copper long First reflect Copper transversal 1 Voltage (v) Time (t) Figure 9: The time domain impulse response of the 2-meters water-filled pipe (top) and the corresponding response after 1 day (bottom)

15 1.5 2 x Impulse response Water Copper long First reflect Copper transversal 0.5 Voltage (v) Time (t) Figure 10: Impulse response of the slowly filled pipe 15

16 0 50 H(f) (db) Frequency (Hz) x 10 4 Figure 11: Frequency domain response of water-filled 2-meters piece after 2 days 16

17 52 54 H(f) (db) Frequency (Hz) x 10 4 Figure 12: Zooming in on notches 17

18 analytical results we should have the impulse responses of the transducers, which we didn t have access to them. The main result of our study was that these channels are time variant and solved air, or any other material, can cause dramatic changes in the channel response. One of our suggestion for further investigations is the study of the impacts of solved air on transfer function of the such channels. The other result of our study is that in order to have shorter impulse response, we should move to the lower frequency which means our transducers resonant frequency should be less than the cut-off frequency of the secondpropagating mode. Which results in possibility of transmission in very limited bandwidth. Even worse, the cut off frequency of the second propagating mode in our standard pipes, is less than 30 khz which is still in hearing spectrum of some animals like dogs. The other important thing is the pattern of the transducers. If the transducers are not properly designed, then the energy at the input of the transmitter, will be probably injected to the higher modes and as we illustrated those modes will be attenuated in far distances from the source. We studied impulse response of the pipes with 1-meter and 2-meters length. But actually in order to be more realistic, one should study the behavior of the channel for longer distances. In our study the water was still and this is also far from the real application. So, it is worthwhile to study the wave propagation in pipes with moving water (or media) in them. Another interesting topic in this direction is to investigate the wave propagation over the surface of the pipes. Actually possibilities of communication through the copper has lots of advantages. For example, the speed of the sound in copper is more than two times the speed of the sound in water. And although, the influence of the moving media inside the pipes, should be less than its influence on acoustic waves traveling through the water. Appendix A Here is the table of the lth zero of mth order Bessel functions derivative for m = 0, 1, 2, 3, 4 and l = 1, 2, 3, 4 [1]: l = J J J J J

19 Appendix B Here is the table of the sound speed and the density of some used materials 3 : Material Sound speed(m/s) Density(gm/cm 3 ) Air (20 C) Water (25 C) Copper longitudinal Copper transversal References [1] M. L. Crocker. Handbook of Acoustics. John Wiley & Sons, [2] J. V. Sanders L. E. Kinsler, A. R. Frey. A. B. Coppens. Fundamentals of acoustics. John Wiley & Sons, third edition, [3] Murata Manufacturing Company, [4] K. U. Ingard P. M. Morse. Theoretical Acoustics. McGraw-Hill, From 19