I've been studying room acoustics recently and have been looking at various models[J. Borish, JASA, 1979]. I have noticed that in many calculations of room impulse response functions there are negative numbers in the output. Does anyone here know why there would be negative numbers in the output? This, superficially at least, makes no sense to me. -Steve
Negative Numbers in Room Impulse Response
Started by ●December 5, 2003
Reply by ●December 5, 20032003-12-05
"Stephen McGovern" <smcgover@stevens.edu> wrote in message news:qwXzb.177$Pm5.159892@newshog.newsread.com...> I've been studying room acoustics recently and have been looking atvarious> models[J. Borish, JASA, 1979]. I have noticed that in many calculationsof> room impulse response functions there are negative numbers in the output. > Does anyone here know why there would be negative numbers in the output? > This, superficially at least, makes no sense to me. > -SteveIf you drop a pebble into a pond, the impulse is downward, but the waves extend above and below the surface. It's the same thing here -- after a pressure peak, the air recededs, and its inertia carries it past the equilibrium point.
Reply by ●December 5, 20032003-12-05
Hello Stephen, Think of Newton's 3rd law where for every action there is an opposite and equal reaction. When waves reflect off of solid objects, the wave undergoes a 180 degree phase shift. I.e., the pushes on the object and the object pushes back. Clay "Stephen McGovern" <smcgover@stevens.edu> wrote in message news:qwXzb.177$Pm5.159892@newshog.newsread.com...> I've been studying room acoustics recently and have been looking atvarious> models[J. Borish, JASA, 1979]. I have noticed that in many calculationsof> room impulse response functions there are negative numbers in the output. > Does anyone here know why there would be negative numbers in the output? > This, superficially at least, makes no sense to me. > -Steve > >
Reply by ●December 5, 20032003-12-05
In article qwXzb.177$Pm5.159892@newshog.newsread.com, Stephen McGovern at smcgover@stevens.edu wrote on 12/05/2003 03:40:> I've been studying room acoustics recently and have been looking at various > models[J. Borish, JASA, 1979]. I have noticed that in many calculations of > room impulse response functions there are negative numbers in the output. > Does anyone here know why there would be negative numbers in the output? > This, superficially at least, makes no sense to me.one more twist after Matt and Clay. the impulse response (or any function representing a sound signal in a medium such as air) represents the instantaneous pressure *difference* from the constant atmospheric pressure. so the total instantaneous pressure is: P(x, y, z, t) = P0 + p(x, y, z, t) P0 is about 100000 Pa (Nt/m^2) and P() is always > 0 (not even close to zero unless you are in one helluva eardrum breaking explosion somewhere). but p() is bipolar. r b-j
Reply by ●December 6, 20032003-12-06
OK, but would this still be the case if I'm dealing with a unit impulse, or in other words, a normalized Dirac function? "Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message news:180Ab.14693$zf2.1668653@news20.bellglobal.com...> "Stephen McGovern" <smcgover@stevens.edu> wrote in message > news:qwXzb.177$Pm5.159892@newshog.newsread.com... > > I've been studying room acoustics recently and have been looking at > various > > models[J. Borish, JASA, 1979]. I have noticed that in many calculations > of > > room impulse response functions there are negative numbers in theoutput.> > Does anyone here know why there would be negative numbers in the output? > > This, superficially at least, makes no sense to me. > > -Steve > > If you drop a pebble into a pond, the impulse is downward, but the waves > extend above and below the surface. It's the same thing here -- after a > pressure peak, the air recededs, and its inertia carries it past the > equilibrium point. > >
Reply by ●December 6, 20032003-12-06
So is it true that if I neglect amplitude changes as a result of distance, the walls have a reflectivity of 1, and my signal is y, then the signal under going one reflection is -y and the signal undergoing two reflections is -(-y) = y ? "Clay S. Turner" <CSTurner@WSE.Biz> wrote in message news:hn2Ab.178$IF.141@bignews4.bellsouth.net...> Hello Stephen, > Think of Newton's 3rd law where for every action there is an opposite and > equal reaction. When waves reflect off of solid objects, the waveundergoes> a 180 degree phase shift. I.e., the pushes on the object and the object > pushes back. > > Clay > > > > "Stephen McGovern" <smcgover@stevens.edu> wrote in message > news:qwXzb.177$Pm5.159892@newshog.newsread.com... > > I've been studying room acoustics recently and have been looking at > various > > models[J. Borish, JASA, 1979]. I have noticed that in many calculations > of > > room impulse response functions there are negative numbers in theoutput.> > Does anyone here know why there would be negative numbers in the output? > > This, superficially at least, makes no sense to me. > > -Steve > > > > > >
Reply by ●December 6, 20032003-12-06
"Stephen McGovern" <smcgover@stevens.edu> wrote in message news:jFrAb.300$Pm5.249066@newshog.newsread.com...> So is it true that if I neglect amplitude changes as a result of distance, > the walls have a reflectivity of 1, and my signal is y, > then the signal under going one reflection is -y > and the signal undergoing two reflections is -(-y) = y ? > >Hello Stephen, Yes this is absolutely true. Think about using hollow cylinders as resonators. If an end is closed, the wave will reflect with a 180 degree phase shift. If an end is open the wave will reflect without a phase shift. This not only occurs with sound waves, but also occurs with E-M waves as well. If the terminating impedance is lower than the characteristic impedance of the propagation channel, the reflection is inverted. And likewise if the termination impedance is higher, the reflection is noninverted. Clay
Reply by ●December 6, 20032003-12-06
Clay S. Turner wrote:> "Stephen McGovern" <smcgover@stevens.edu> wrote in message > news:jFrAb.300$Pm5.249066@newshog.newsread.com... > >>So is it true that if I neglect amplitude changes as a result of distance, >>the walls have a reflectivity of 1, and my signal is y, >>then the signal under going one reflection is -y >>and the signal undergoing two reflections is -(-y) = y ? >> >> > > > Hello Stephen, > Yes this is absolutely true. Think about using hollow cylinders as > resonators. If an end is closed, the wave will reflect with a 180 degree > phase shift. If an end is open the wave will reflect without a phase shift. > This not only occurs with sound waves, but also occurs with E-M waves as > well. If the terminating impedance is lower than the characteristic > impedance of the propagation channel, the reflection is inverted. And > likewise if the termination impedance is higher, the reflection is > noninverted. > > ClayTo be cutesy, one could say that it always inverts and always doesn't. In actuality, either the pressure inverts, or the velocity. With E-M waves, either E inverts or M does. When neither inverts, the wave keeps going. If one could arrange that both invert, the wave would also keep going. When the wave reflects, either does, but not both. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●December 7, 20032003-12-07
"Stephen McGovern" <smcgover@stevens.edu> wrote in message news:8trAb.298$Pm5.247664@newshog.newsread.com...> OK, but would this still be the case if I'm dealing with a unit impulse,or> in other words, a normalized Dirac function?Since what we usually call the "room impulse response" includes a transducer, then the easiest way to deduce the existence of negative content is to recognize that the response cannot pass DC, unless the room is sealed and the transducer can permanently add or remove air, which aren't designed to do. So even without reflections, the response will include about as much negative content as positive, because everything has to add up to zero. Room reverberations then just stir that up. Now, if you were to make your unit impulse by suddenly adding a bit of pressurized air in the middle of the room, and you measured the response with a real pressure sensor instead of a microphone, then you may indeed read a mostly positive response. Now, there will likely still be some negative content, because the wall reflections don't preserve high frequencies up to infinity, and many kinds of low-pass filters cross zero in the ringing part of their impulse responses, but there doesn't need to be enough to balance out the positive parts of the response. And finally, if you were to measure the room's response to really all-positive stimuli that are bandlimited to exclude very high frequencies, you could certainly get an all-positive result.
Reply by ●December 7, 20032003-12-07
Clay S. Turner wrote:> "Stephen McGovern" <smcgover@stevens.edu> wrote in message > news:jFrAb.300$Pm5.249066@newshog.newsread.com...>>So is it true that if I neglect amplitude changes as a result of distance, >>the walls have a reflectivity of 1, and my signal is y, >>then the signal under going one reflection is -y >>and the signal undergoing two reflections is -(-y) = y ?> Yes this is absolutely true. Think about using hollow cylinders as > resonators. If an end is closed, the wave will reflect with a 180 degree > phase shift. If an end is open the wave will reflect without a phase shift. > This not only occurs with sound waves, but also occurs with E-M waves as > well. If the terminating impedance is lower than the characteristic > impedance of the propagation channel, the reflection is inverted. And > likewise if the termination impedance is higher, the reflection is > noninverted.I think you have to be careful with what you mean by signal. For the EM wave, such as a signal on a transmission line, for a shorted line the voltage is inverted and the current is not. For an open ended line, the current is inverted and the voltage is not. The first time I saw this demonstrated, the instructor was trying to make the analogy to closed and open ended tubes (cylindrical resonators). After the analogy came out wrong measuring the voltage on the line, the next lecture he came back with a current probe on the oscilloscope to make the analogy right. For sound waves, it is much easier to build a pressure sensitive detector than an air velocity detector. A closed end tube is a pressure antinode and velocity node. A shorted transmission line is a voltage node and current antinode. For electromagnetic fields on a perfect conductor, the E field is inverted, and the B field is not. For a sound wave on a solid wall, pressure is not inverted. The velocity is more complicated, because it is a longitudinal wave and not a transverse wave. -- glen
