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-Introduction-

The term 'room gain' is often seen used when the topic of subwoofers and the low frequency sound reproduction comes up. As I see many explanations that are lacking or incorrect, I thought I might try and clear up the topic once and for all.

In all rooms there are so-called modes. A mode consists of a function and a number; more specifically an eigenfunction and and a eigenvalue. The eigenfunction describes the shape of the mode, and the eigenvalue gives us the (angular) frequency for the mode. There are infinitely many modes in any room. They are finely distributed at higher frequencies and we rarely considered room acoustics from a modal perspective at high frequencies. However, at lower frequencies, they show up much more distinctly, and this is what we will investigate here. Rooms come in many shapes and sizes. Here, we will focus on the 'rectangular' room (cuboid, or rectangular prism, is probably more correct, but we all get the idea). While different modes can have the same eigenvalue, we will look at a design where that is not the case. So for each distinct frequency of interest, there is only one mode. Okay... One pivotal point to realize already now is that "the modes in a room exist independently of any source or excitation; they are inherently a result of the geometry of the room" (and its boundary conditions, and the fluid properties in the room). When a pressure response shows a so-called resonance, where it peaks at a certain frequency, it is likely due to a mode, but the mode is there whether you measure a peak or not. What is important is 1) how much, if at all, the modes are excited, which is determined by the placement of the source(s), and 2) where you measure the response from said source(s). Nothing you do when it comes to excitation will change the eigenfunction (shape) nor the the eigenvalue (frequency). We will see this illustrated later on. If a mode gets excited it will in principle lead to a pressure that continues to oscillate with the mode shape and frequency of the mode indefinitely. In practice however, we have losses in the room, and so it will die out fairly quickly, but we will not discuss the temporal aspect here.

-Details-

We choose the rectangular room as our prototypical room, which is probably fair considering how most rooms look. For such a room we can find the modes, assuming that the walls are hard (perfectly reflective), there is little loss in the room, and it is completely empty. You might object already that this is unrealistic, but it is better to understand the underlying theory with certain assumptions, than try to decipher what is going on in a complex room. The formulas for the modes is given in most textbooks on acoustics:

Top we have the form of each mode, second we have the modal pressure for the summation of all modes, and finally the frequency for each mode. We can see that we have an infinite sum of modes, and three integer indices to distinguish between each mode. We see that cosine functions describe the modal pressure, but remember that this pressure is not what you would measure, it is more of an abstraction that shows a solution to the homogenous wave equation (i.e. no sources) and homogenous boundary conditions (no energy is injected at the walls). Any brief excitation of the room, will lead to some or all modes oscillating forever (if no losses), satisfying the wave equation without there being any steady-state source.

If we want to, we can now calculate all modes in the room. Alternatively, we can use a room mode calculator. A goggle search gave me this one, https://amcoustics.com/tools/amroc, which is very neat as you can see the mode shape and frequencies, as well as see which muscical tone it corresponds to. So lets have a look at a room; I chose one that is (lx,ly,lz)=(length,width,height)=(4.5 m, 3 m, 2m). When we use the index notation for the summation, we will find a mode for (nx,ny,nz)=(1,0,0) at 38 Hz, for (nx,ny,nz)=(0,1,0) at 57 Hz, and on and on. I show the eigenfunction/shape below for these two modes:

Now, what will we measure in this room if we have one or more sources and a microphone? Again, we have an analytical expression that gives us the answer. (A really good resource is "The Sound Field in a Reveberation Room" by the late Finn Jacobsen.)

With this expression, we can calculate the complex sound pressure in any point for a monopole source placed in any point in the room. I have made a MATLAB script where I can have several sources with different placements, source strengths, and phases, and I can manipulate the source type so that it gives off volume acceleration, volume velocity, or volume displacement, or have a cross-over so that it switches between the three. You can also add a source which a certain phase and another source with the opposite phase and make a dipole that way, or higher order multipoles. So let's look at an example where a source is placed near a corner and the microphone placed right in the middle of the room.

The red curve is the speaker's free field response, and blue is the in room response, and the dotted lines indicate all frequencies where the room has a mode. We see that the first peak in the in room response is at 76 Hz. Hmmm... where is the mode at 38 Hz? 57 Hz? There is also a (nx,ny,nz)=(2,0,0) mode at 76 Hz, but no peak. Well, mode does not equal resonance. By positioning ourselves in the middle of the room, the modes that are symmetric will a null in the middle will not show up as resonances, even if they are excited. Conversely, we can find position for the source that will not be 'supported' by a mode. This does not mean that there will be no sound at the corresponding frequency, as I have hear some say. We can clearly see that in the result too. But what is with the slope going up as we go down in frequency? Let's us break it down in steps.

First, let's us choose another position of the microphone/head, here I took (xmic,ymic,zmic)=(3.5,2,1.5).

We see many more modes being excited now, as we are not placed in a null for the lower modes. Second, let us put some damping into the equations (slight damping will not affect the discussion we have had so far):

Thirdly, the source is set to have a flat volume acceleration as a function of frequency. This is equivalent to how most speakers behave; point a laser at a driver in its pass band area (i.e. where it is supposed to play music), measure the velocity, convert to acceleration, and you should see a flat amplitude response for acceleration. (This is a good test btw. to give to acoustic engineers. They will often think that it must be flat amplitude for the displacement that gives a flat pressure amplitude). So the source is 'perfect' in that it has infinite bandwidth from DC and upwards. We need to consider the roll-off of a speaker. A closed sub will roll-off in a second order manner, so that the pressure response resembles that of a second-order high-pass function. And a ported speaker will roll-off with a fourth-order response. So if we assume that we have a closed sub in a corner, and assume that it behaves as a second-order high pass filter with a cut-off frequency of 35 Hz (I think the Linkwitz Thor has this without the external filter applied; I had that setup once and it was so nice), we finally get this result:

Wow... We have a flat in room response for a sub with finite bandwidth. Why is that? Well, we have actually forgotten one thing; the indices (nx,ny,nz) do not start at (1,0,0), (0,1,0), or (0,0,1); they start at (0,0,0)! We have a mode at 0 Hz that pressurizes the room the same in all positions, so no gradients, and it satisfies the wave equation as discussed already. Room gain is simply the effect of having an eigenmode at DC! With the increase in resulting pressure as the excitation frequency gets closer and closer to the modal frequency being second-order (see the denominator in the equation higher up), and the output from a closed sub goes down by an order of two, we can get a resulting pressure which is flat towards DC. So while we do not have sound at 0 Hz, since there are no oscillations, the effect of any mode extends away from it. Also, for a ported subwoofer we will not be able to have this flat response, because of its higher order roll-off, but the in room response will still be modified by the mode. If you actually do the calculations, be it analytically or numerically, the 'zeroth' mode will pop right out:

Does all of the mathematics make sense from a physical standpoint? Luckily yes: Imagine a bike pump. Put your finger on the output valve. Squeeze. This is the situation you have in your room at low frequencies, it simply acts as an acoustic compliance. The impedance of the compliance is 1/(i*2*pi*Ca), where f is the frequency. And acoustic impedance is pressure divided by volume velocity(!). We can rearrange and get that pressure is equal to volume displacement(!) divided by a constant (the acoustic compliance), and so pressure is directly proportional to displacement, and displacement is exactly what our closed sub can deliver below its resonance frequency. [If you prefer, you can just think of Hooke's law in structural mechanics for a spring; the force is the stiffness times the displacement, and since pressure and force are linked simply via an area, we can again see that a constant displacement across a (low) frequency range, will give a constant force/pressure]. And what about the ported sub? When the driver moves inwards at low frequencies below the driver resonance, air will just be pushed out of the port, and so there is no net volume displacement; there is an acoustic short-circuit, just as you have for a dipole. So again it makes physical sense that we cannot take the same advantage of room gain as for the closed enclosure case.

-Closing remarks-

We have been considering ideal conditions with low loss rectangular rooms and probably more importantly no leakage. In practice there will be leakage and so we can not go flat to DC. But still the analysis explains what is going on, and in cars the effect is also very important to consider (I have a story there for another time).

-Conclusion-

Room gain is not some esoteric effect that the acoustic text books do not cover; it is actually a simple subset of known modal analysis combined with loudspeaker characteristics. For a sealed room we will have flat low frequency pressure response for a transducer with flat volume displacement, and a sealed sub gives us exactly that below its operating range.

- About me -

René (no need for Dr title, please), BSEE, MSc (Physics), PhD (Microacoustics), FEM and BEM simulations specialist in/for loudspeaker, hearing aid, and consultancy companies. Own company Acculution, blog at acculution.com/blog

The term 'room gain' is often seen used when the topic of subwoofers and the low frequency sound reproduction comes up. As I see many explanations that are lacking or incorrect, I thought I might try and clear up the topic once and for all.

In all rooms there are so-called modes. A mode consists of a function and a number; more specifically an eigenfunction and and a eigenvalue. The eigenfunction describes the shape of the mode, and the eigenvalue gives us the (angular) frequency for the mode. There are infinitely many modes in any room. They are finely distributed at higher frequencies and we rarely considered room acoustics from a modal perspective at high frequencies. However, at lower frequencies, they show up much more distinctly, and this is what we will investigate here. Rooms come in many shapes and sizes. Here, we will focus on the 'rectangular' room (cuboid, or rectangular prism, is probably more correct, but we all get the idea). While different modes can have the same eigenvalue, we will look at a design where that is not the case. So for each distinct frequency of interest, there is only one mode. Okay... One pivotal point to realize already now is that "the modes in a room exist independently of any source or excitation; they are inherently a result of the geometry of the room" (and its boundary conditions, and the fluid properties in the room). When a pressure response shows a so-called resonance, where it peaks at a certain frequency, it is likely due to a mode, but the mode is there whether you measure a peak or not. What is important is 1) how much, if at all, the modes are excited, which is determined by the placement of the source(s), and 2) where you measure the response from said source(s). Nothing you do when it comes to excitation will change the eigenfunction (shape) nor the the eigenvalue (frequency). We will see this illustrated later on. If a mode gets excited it will in principle lead to a pressure that continues to oscillate with the mode shape and frequency of the mode indefinitely. In practice however, we have losses in the room, and so it will die out fairly quickly, but we will not discuss the temporal aspect here.

-Details-

We choose the rectangular room as our prototypical room, which is probably fair considering how most rooms look. For such a room we can find the modes, assuming that the walls are hard (perfectly reflective), there is little loss in the room, and it is completely empty. You might object already that this is unrealistic, but it is better to understand the underlying theory with certain assumptions, than try to decipher what is going on in a complex room. The formulas for the modes is given in most textbooks on acoustics:

Top we have the form of each mode, second we have the modal pressure for the summation of all modes, and finally the frequency for each mode. We can see that we have an infinite sum of modes, and three integer indices to distinguish between each mode. We see that cosine functions describe the modal pressure, but remember that this pressure is not what you would measure, it is more of an abstraction that shows a solution to the homogenous wave equation (i.e. no sources) and homogenous boundary conditions (no energy is injected at the walls). Any brief excitation of the room, will lead to some or all modes oscillating forever (if no losses), satisfying the wave equation without there being any steady-state source.

If we want to, we can now calculate all modes in the room. Alternatively, we can use a room mode calculator. A goggle search gave me this one, https://amcoustics.com/tools/amroc, which is very neat as you can see the mode shape and frequencies, as well as see which muscical tone it corresponds to. So lets have a look at a room; I chose one that is (lx,ly,lz)=(length,width,height)=(4.5 m, 3 m, 2m). When we use the index notation for the summation, we will find a mode for (nx,ny,nz)=(1,0,0) at 38 Hz, for (nx,ny,nz)=(0,1,0) at 57 Hz, and on and on. I show the eigenfunction/shape below for these two modes:

Now, what will we measure in this room if we have one or more sources and a microphone? Again, we have an analytical expression that gives us the answer. (A really good resource is "The Sound Field in a Reveberation Room" by the late Finn Jacobsen.)

With this expression, we can calculate the complex sound pressure in any point for a monopole source placed in any point in the room. I have made a MATLAB script where I can have several sources with different placements, source strengths, and phases, and I can manipulate the source type so that it gives off volume acceleration, volume velocity, or volume displacement, or have a cross-over so that it switches between the three. You can also add a source which a certain phase and another source with the opposite phase and make a dipole that way, or higher order multipoles. So let's look at an example where a source is placed near a corner and the microphone placed right in the middle of the room.

The red curve is the speaker's free field response, and blue is the in room response, and the dotted lines indicate all frequencies where the room has a mode. We see that the first peak in the in room response is at 76 Hz. Hmmm... where is the mode at 38 Hz? 57 Hz? There is also a (nx,ny,nz)=(2,0,0) mode at 76 Hz, but no peak. Well, mode does not equal resonance. By positioning ourselves in the middle of the room, the modes that are symmetric will a null in the middle will not show up as resonances, even if they are excited. Conversely, we can find position for the source that will not be 'supported' by a mode. This does not mean that there will be no sound at the corresponding frequency, as I have hear some say. We can clearly see that in the result too. But what is with the slope going up as we go down in frequency? Let's us break it down in steps.

First, let's us choose another position of the microphone/head, here I took (xmic,ymic,zmic)=(3.5,2,1.5).

We see many more modes being excited now, as we are not placed in a null for the lower modes. Second, let us put some damping into the equations (slight damping will not affect the discussion we have had so far):

Thirdly, the source is set to have a flat volume acceleration as a function of frequency. This is equivalent to how most speakers behave; point a laser at a driver in its pass band area (i.e. where it is supposed to play music), measure the velocity, convert to acceleration, and you should see a flat amplitude response for acceleration. (This is a good test btw. to give to acoustic engineers. They will often think that it must be flat amplitude for the displacement that gives a flat pressure amplitude). So the source is 'perfect' in that it has infinite bandwidth from DC and upwards. We need to consider the roll-off of a speaker. A closed sub will roll-off in a second order manner, so that the pressure response resembles that of a second-order high-pass function. And a ported speaker will roll-off with a fourth-order response. So if we assume that we have a closed sub in a corner, and assume that it behaves as a second-order high pass filter with a cut-off frequency of 35 Hz (I think the Linkwitz Thor has this without the external filter applied; I had that setup once and it was so nice), we finally get this result:

Wow... We have a flat in room response for a sub with finite bandwidth. Why is that? Well, we have actually forgotten one thing; the indices (nx,ny,nz) do not start at (1,0,0), (0,1,0), or (0,0,1); they start at (0,0,0)! We have a mode at 0 Hz that pressurizes the room the same in all positions, so no gradients, and it satisfies the wave equation as discussed already. Room gain is simply the effect of having an eigenmode at DC! With the increase in resulting pressure as the excitation frequency gets closer and closer to the modal frequency being second-order (see the denominator in the equation higher up), and the output from a closed sub goes down by an order of two, we can get a resulting pressure which is flat towards DC. So while we do not have sound at 0 Hz, since there are no oscillations, the effect of any mode extends away from it. Also, for a ported subwoofer we will not be able to have this flat response, because of its higher order roll-off, but the in room response will still be modified by the mode. If you actually do the calculations, be it analytically or numerically, the 'zeroth' mode will pop right out:

Does all of the mathematics make sense from a physical standpoint? Luckily yes: Imagine a bike pump. Put your finger on the output valve. Squeeze. This is the situation you have in your room at low frequencies, it simply acts as an acoustic compliance. The impedance of the compliance is 1/(i*2*pi*Ca), where f is the frequency. And acoustic impedance is pressure divided by volume velocity(!). We can rearrange and get that pressure is equal to volume displacement(!) divided by a constant (the acoustic compliance), and so pressure is directly proportional to displacement, and displacement is exactly what our closed sub can deliver below its resonance frequency. [If you prefer, you can just think of Hooke's law in structural mechanics for a spring; the force is the stiffness times the displacement, and since pressure and force are linked simply via an area, we can again see that a constant displacement across a (low) frequency range, will give a constant force/pressure]. And what about the ported sub? When the driver moves inwards at low frequencies below the driver resonance, air will just be pushed out of the port, and so there is no net volume displacement; there is an acoustic short-circuit, just as you have for a dipole. So again it makes physical sense that we cannot take the same advantage of room gain as for the closed enclosure case.

-Closing remarks-

We have been considering ideal conditions with low loss rectangular rooms and probably more importantly no leakage. In practice there will be leakage and so we can not go flat to DC. But still the analysis explains what is going on, and in cars the effect is also very important to consider (I have a story there for another time).

-Conclusion-

Room gain is not some esoteric effect that the acoustic text books do not cover; it is actually a simple subset of known modal analysis combined with loudspeaker characteristics. For a sealed room we will have flat low frequency pressure response for a transducer with flat volume displacement, and a sealed sub gives us exactly that below its operating range.

- About me -

René (no need for Dr title, please), BSEE, MSc (Physics), PhD (Microacoustics), FEM and BEM simulations specialist in/for loudspeaker, hearing aid, and consultancy companies. Own company Acculution, blog at acculution.com/blog

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