Soundton

Soundton Documentation

SOUNDTON SPEAKER PLACEMENT CALCULATOR

Author:
Andrea Cicero
AC Acustica – Acoustic Design
(c) 2023 AC Acustica (www.ac-acustica.it) and Soundton (www.soundton.com). All rights reserved.

1. INTRODUCTION

Critical listening environments such as recording studios control rooms, home theatres and hi-fi music listening rooms usually take place in small rooms, where the acoustics need to be dealt with as it strongly affects the quality of sound and the overall listening experience.
The major challenges to face when it comes to acoustic quality are [1]:
  • Reverberation: the time that the sound energy takes to decay by a certain amount of dB.
  • Frequency balance: how well-balanced the response of the room is across the entire audible range.
  • Early reflections: strong specular sound reflections coming from surfaces close to either the source or the receiver, which mix up with the direct sound and cause colouration.
These issues can be easily addressed with conventional acoustic treatment elements such as foam or mineral wool panels as well as acoustic diffusers.
However, when it comes to small rooms, the dimensions of the floor, the walls and the ceiling are comparable with the wavelength at low frequencies, therefore sound waves interact directly with the elements of the room and behave differently.
To give an example, Table 1 shows the size of half the wavelength as a function of frequency.
TABLE 1 – WAVELENGTH AS A FUNCTION OF FREQUENCY
Frequency (Hz) 20 40 60 80 100 150 200
Half Wavelength (m) Half Wavelength (m) {:[” Half “],[” Wavelength (m) “]:}\begin{array}{c}\text { Half } \\ \text { Wavelength (m) }\end{array} Half Wavelength (m) 8.58 4.29 2.86 2.14 1.72 1.14 0.86
Half Wavelength (ft) Half Wavelength (ft) {:[” Half “],[” Wavelength (ft) “]:}\begin{array}{c}\text { Half } \\ \text { Wavelength (ft) }\end{array} Half Wavelength (ft) 2.61 1.31 0.87 0.65 0.52 0.35 0.26
Frequency (Hz) 20 40 60 80 100 150 200 ” Half Wavelength (m) ” 8.58 4.29 2.86 2.14 1.72 1.14 0.86 ” Half Wavelength (ft) ” 2.61 1.31 0.87 0.65 0.52 0.35 0.26| Frequency (Hz) | 20 | 40 | 60 | 80 | 100 | 150 | 200 | | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | $\begin{array}{c}\text { Half } \\ \text { Wavelength (m) }\end{array}$ | 8.58 | 4.29 | 2.86 | 2.14 | 1.72 | 1.14 | 0.86 | | $\begin{array}{c}\text { Half } \\ \text { Wavelength (ft) }\end{array}$ | 2.61 | 1.31 | 0.87 | 0.65 | 0.52 | 0.35 | 0.26 |
From the table, it is clear that considering the dimension of an average domestic room, the region of the spectrum between 20 H z 20 H z 20Hz20 \mathrm{~Hz}20 Hz and 100 H z 100 H z 100Hz100 \mathrm{~Hz}100 Hz would be the one affected by this phenomenon, called “modal behaviour” of the room.
The modal behaviour consists of the combined effects of resonances of the room (called “modes”) caused by the matching between the room dimensions (or portions of them) with the wavelength (or portions of it) of the excitation frequency.
At a fixed listening position, room modes can easily be identified by looking at the measured frequency response curve (see Figure 1 for an example), as they cause pronounced peaks and dips.
The effects of this are not visible in the frequency domain only. Because of the “resonant” behaviour, room modes present a long decay in the time domain, as shown in the waterfall plot in Figure 1.
FIGURE 1 – MEASURED ROOM FREQUENCY RESPONSE CURVE AND THE CORRESPONDING WATERFALL PLOT AT LOW FREQUENCIES (SOFTWARE: ROOM EQ WIZARD [2]).
Moreover, modes cause the creation of areas of high and low sound pressure within the room, leading to certain frequencies sounding louder or quieter in certain parts of the room. This is strongly dependent on the source(s) and listener’s positions.
The acoustic treatment at such low frequencies is not always easy. The so-called “bass-traps” require large spaces to be installed efficiently if made of porous materials, otherwise, it is necessary to use tuned absorption devices, like membrane absorbers or Helmholtz resonators [3], which require to be engineered properly to obtain the desired absorption.
Also, low-frequency treatment often becomes an expensive and non-feasible option to pursue, especially in domestic environments where geometric and aesthetic constraints often prevail.
Therefore, the only margin of improvement resides in the quality of the loudspeakers and their location relative to the room boundaries and the listening position.
That is the reason why the Soundton Speaker Placement Calculator (SPC) has been created: it comes in handy to help in optimising the loudspeaker location given its specifics and the dimensions of the room. In the next sections, more details are provided in terms of the algorithm behind this tool, as well as some examples.

2. ACOUSTICS BACKGROUND

2.1. PREDICTION OF LOW-FREQUENCY RESPONSE

As mentioned in the previous section, small rooms exhibit resonant behaviour at low frequencies. To determine the room “cut-off” frequency, below which single resonances become prominent, Schroeder’s formula is used [ 4 ] : [ 4 ] : [4]:[4]:[4]:
f s = 2000 T / V f s = 2000 T / V f_(s)=2000sqrt(T//V)f_{s}=2000 \sqrt{T / V}fs=2000T/V
where T T TTT is the reverberation time (typically 0.5 to 0.8 seconds in domestic environments) and V V VVV is the room volume in m 3 m 3 m^(3)\mathrm{m}^{3}m3. For example, a 60 m 3 / 2120 f t 3 60 m 3 / 2120 f t 3 60m^(3)//2120ft^(3)60 \mathrm{~m}^{3} / 2120 \mathrm{ft}^{3}60 m3/2120ft3 domestic room will show resonances up to about 180 H z 180 H z 180Hz180 \mathrm{~Hz}180 Hz, therefore it is expected that the low-frequency response of the room will be affected by the room dimensions up to that frequency.
In a rectangular room having dimensions w w www (width), l l lll (length) and h h hhh (height), it is possible to calculate the room modal frequencies according to the following:
f n x , n y , n z = c 2 ( n x w ) 2 + ( n y l ) 2 + ( n z h ) 2 f n x , n y , n z = c 2 n x w 2 + n y l 2 + n z h 2 f_(n_(x),n_(y),n_(z))=(c)/(2)sqrt(((n_(x))/(w))^(2)+((n_(y))/(l))^(2)+((n_(z))/(h))^(2))f_{n_{x}, n_{y}, n_{z}}=\frac{c}{2} \sqrt{\left(\frac{n_{x}}{w}\right)^{2}+\left(\frac{n_{y}}{l}\right)^{2}+\left(\frac{n_{z}}{h}\right)^{2}}fnx,ny,nz=c2(nxw)2+(nyl)2+(nzh)2
where c c ccc is the speed of sound, equal to 343 m / s 343 m / s 343m//s343 \mathrm{~m} / \mathrm{s}343 m/s or 1130 f t / s 1130 f t / s 1130ft//s1130 \mathrm{ft} / \mathrm{s}1130ft/s, and n x , n y n x , n y n_(x),n_(y)n_{x}, n_{y}nx,ny and n z n z n_(z)n_{z}nz are integer numbers from 0 to ∞ ∞ oo\infty∞.
As an example, Table 2 shows the first 20 modes for a 4.65 × 5.55 × 3.00 m ( 15.2 × 18.2 × 9.8 f t ) 4.65 × 5.55 × 3.00 m ( 15.2 × 18.2 × 9.8 f t ) 4.65 xx5.55 xx3.00m(15.2 xx18.2 xx9.8ft)4.65 \times 5.55 \times 3.00 \mathrm{~m}(15.2 \times 18.2 \times 9.8 \mathrm{ft})4.65×5.55×3.00 m(15.2×18.2×9.8ft) room.
TABLE 2 – FIRST 20 MODES FOR A 4.65 X 5.55 × 3.00 M ( 15.2 × 18.2 × 9.8 5.55 × 3.00 M ( 15.2 × 18.2 × 9.8 5.55 xx3.00M(15.2 xx18.2 xx9.85.55 \times 3.00 \mathrm{M}(15.2 \times 18.2 \times 9.85.55×3.00M(15.2×18.2×9.8 FT) ROOM
n x n x n_(x)n_{x}nx n y n y n_(y)n_{y}ny n z n z n_(z)n_{z}nz f n ( H z ) f n ( H z ) f_(n)(Hz)f_{n}(\mathrm{~Hz})fn( Hz)
0 1 0 30.9
1 0 0 36.9
1 1 0 48.1
0 0 1 57.2
0 2 0 61.8
0 1 1 65.0
1 0 1 68.0
1 2 0 72.0
2 0 0 73.8
1 1 1 74.7
2 1 0 80.0
0 2 1 84.2
1 2 1 91.9
0 3 0 92.7
2 0 1 93.3
2 2 0 96.2
2 1 1 98.3
1 3 0 99.8
0 3 1 108.9
3 0 0 110.6
n_(x) n_(y) n_(z) f_(n)(Hz) 0 1 0 30.9 1 0 0 36.9 1 1 0 48.1 0 0 1 57.2 0 2 0 61.8 0 1 1 65.0 1 0 1 68.0 1 2 0 72.0 2 0 0 73.8 1 1 1 74.7 2 1 0 80.0 0 2 1 84.2 1 2 1 91.9 0 3 0 92.7 2 0 1 93.3 2 2 0 96.2 2 1 1 98.3 1 3 0 99.8 0 3 1 108.9 3 0 0 110.6| $n_{x}$ | $n_{y}$ | $n_{z}$ | $f_{n}(\mathrm{~Hz})$ | | :—: | :—: | :—: | :—: | | 0 | 1 | 0 | 30.9 | | 1 | 0 | 0 | 36.9 | | 1 | 1 | 0 | 48.1 | | 0 | 0 | 1 | 57.2 | | 0 | 2 | 0 | 61.8 | | 0 | 1 | 1 | 65.0 | | 1 | 0 | 1 | 68.0 | | 1 | 2 | 0 | 72.0 | | 2 | 0 | 0 | 73.8 | | 1 | 1 | 1 | 74.7 | | 2 | 1 | 0 | 80.0 | | 0 | 2 | 1 | 84.2 | | 1 | 2 | 1 | 91.9 | | 0 | 3 | 0 | 92.7 | | 2 | 0 | 1 | 93.3 | | 2 | 2 | 0 | 96.2 | | 2 | 1 | 1 | 98.3 | | 1 | 3 | 0 | 99.8 | | 0 | 3 | 1 | 108.9 | | 3 | 0 | 0 | 110.6 |
Once the modal frequencies are known, it is possible to predict the pressure frequency response curve, identified as p ( ω , r ) p ( ω , r ) p(omega,r)p(\omega, r)p(ω,r) as it is a function of the angular frequency ω = 2 π f ω = 2 π f omega=2pi f\omega=2 \pi fω=2πf and the listening position r ≡ ( x , y , z ) r ≡ ( x , y , z ) r-=(x,y,z)r \equiv(x, y, z)r≡(x,y,z) according to the modal decomposition model:
p ( r , ω ) = ∑ n x ∑ n y ∑ n z H a n ( ω ) A ω n ( r , r 0 ) ( ω 2 − ω n 2 − j 2 ω n δ n ω ) p ( r , ω ) = ∑ n x ∑ n y ∑ n z H a n ( ω ) A ω n r , r 0 ω 2 − ω n 2 − j 2 ω n δ n ω p(r,omega)=sum_(n_(x))sum_(n_(y))sum_(n_(z))H_(an)(omega)(A_(omega_(n))(r,r_(0)))/((omega^(2)-omega_(n)^(2)-j2omega_(n)delta_(n)omega))p(r, \omega)=\sum_{n_{x}} \sum_{n_{y}} \sum_{n_{z}} H_{a n}(\omega) \frac{A_{\omega_{n}}\left(r, r_{0}\right)}{\left(\omega^{2}-\omega_{n}^{2}-j 2 \omega_{n} \delta_{n} \omega\right)}p(r,ω)=∑nx∑ny∑nzHan(ω)Aωn(r,r0)(ω2−ωn2−j2ωnδnω)
where:
  • n n quad n\quad nn is the n th n th n^(“th “)\mathrm{n}^{\text {th }}nth mode index, as a function of n x , n y n x , n y n_(x),n_(y)n_{x}, n_{y}nx,ny and n z n z n_(z)n_{z}nz;
  • r 0 ≡ ( x 0 , y 0 , z 0 ) r 0 ≡ x 0 , y 0 , z 0 r_(0)-=(x_(0),y_(0),z_(0))r_{0} \equiv\left(x_{0}, y_{0}, z_{0}\right)r0≡(x0,y0,z0) is the source position;
  • H a n ( ω ) H a n ( ω ) H_(an)(omega)H_{a n}(\omega)Han(ω) is the anechoic frequency response of the sound source at the angular frequency ω ω omega\omegaω;
  • ω n = c k n = c k n x 2 + k n y 2 + k n z 2 ω n = c k n = c k n x 2 + k n y 2 + k n z 2 omega_(n)=ck_(n)=csqrt(k_(n_(x))^(2)+k_(n_(y))^(2)+k_(n_(z))^(2))\omega_{n}=c k_{n}=c \sqrt{k_{n_{x}}^{2}+k_{n_{y}}^{2}+k_{n_{z}}^{2}}ωn=ckn=cknx2+kny2+knz2 is the modal angular frequency;
  • k n i = n i π L i k n i = n i π L i quadk_(n_(i))=(n_(i)pi)/(L_(i))\quad k_{n_{i}}=\frac{n_{i} \pi}{L_{i}}kni=niπLi is the modal wavenumber corresponding to the i th i th i^(“th “)i^{\text {th }}ith dimension;
  • A ω n ( r , r 0 ) = j ρ Q ω c 2 p n ( r ) p n ( r 0 ) A ω n r , r 0 = j ρ Q ω c 2 p n ( r ) p n r 0 A_(omega_(n))(r,r_(0))=j rho Q omegac^(2)p_(n)(r)p_(n)(r_(0))A_{\omega_{n}}\left(r, r_{0}\right)=j \rho Q \omega c^{2} p_{n}(r) p_{n}\left(r_{0}\right)Aωn(r,r0)=jρQωc2pn(r)pn(r0) is the amplitude factor for the n th n th n^(“th “)\mathrm{n}^{\text {th }}nth mode, with j j jjj the imaginary unit, ρ ρ rho\rhoρ density of air, Q Q QQQ is the source volume velocity, p n ( r ) = cos ⁡ ( k n x x ) cos ⁡ ( k n y y ) cos ⁡ ( k n z z ) p n ( r ) = cos ⁡ k n x x cos ⁡ k n y y cos ⁡ k n z z p_(n)(r)=cos(k_(n_(x))x)cos(k_(n_(y))y)cos(k_(n_(z))z)p_{n}(r)=\cos \left(k_{n_{x}} x\right) \cos \left(k_{n_{y}} y\right) \cos \left(k_{n_{z}} z\right)pn(r)=cos⁡(knxx)cos⁡(knyy)cos⁡(knzz) is the gain (or mode shape) due to the listener’s position, p n ( r 0 ) = cos ⁡ ( k n x x 0 ) cos ⁡ ( k n y y 0 ) cos ⁡ ( k n z z 0 ) p n r 0 = cos ⁡ k n x x 0 cos ⁡ k n y y 0 cos ⁡ k n z z 0 p_(n)(r_(0))=cos(k_(n_(x))x_(0))cos(k_(n_(y))y_(0))cos(k_(n_(z))z_(0))p_{n}\left(r_{0}\right)=\cos \left(k_{n_{x}} x_{0}\right) \cos \left(k_{n_{y}} y_{0}\right) \cos \left(k_{n_{z}} z_{0}\right)pn(r0)=cos⁡(knxx0)cos⁡(knyy0)cos⁡(knzz0) is the gain (or mode shape) due to the source position
  • δ n = c β ω n ( ϵ n x w + ϵ n y l + ϵ n z h ) δ n = c β ω n ϵ n x w + ϵ n y l + ϵ n z h delta_(n)=(c beta)/(omega_(n))((epsilon_(n_(x)))/(w)+(epsilon_(n_(y)))/(l)+(epsilon_(n_(z)))/(h))\delta_{n}=\frac{c \beta}{\omega_{n}}\left(\frac{\epsilon_{n_{x}}}{w}+\frac{\epsilon_{n_{y}}}{l}+\frac{\epsilon_{n_{z}}}{h}\right)δn=cβωn(ϵnxw+ϵnyl+ϵnzh) is the damping factor of the n th n th n^(“th “)\mathrm{n}^{\text {th }}nth mode, with β β beta\betaβ the average admittance of the room boundaries, ϵ n i ϵ n i epsilon_(n_(i))\epsilon_{n_{i}}ϵni equal to 1 if the mode is grazing in the i th i th i^(“th “)\mathrm{i}^{\text {th }}ith direction, otherwise equal to 2.
The obtained quantity will be an array of complex pressure values, one for each frequency bin f f fff considered. To obtain the in-room SPL curve in d B d B dB\mathrm{dB}dB, it is necessary to calculate the following:
S P L ( ω , r ) = 10 log ⁡ ( | p ( ω , r ) | 2 p r e f 2 ) S P L ( ω , r ) = 10 log ⁡ | p ( ω , r ) | 2 p r e f 2 SPL(omega,r)=10 log((|p(omega,r)|^(2))/(p_(ref)^(2)))S P L(\omega, r)=10 \log \left(\frac{|p(\omega, r)|^{2}}{p_{r e f}^{2}}\right)SPL(ω,r)=10log⁡(|p(ω,r)|2pref2)
where p ref = 2 ⋅ 10 − 5 P a p ref = 2 ⋅ 10 − 5 P a p_(“ref “)=2*10^(-5)Pap_{\text {ref }}=2 \cdot 10^{-5} \mathrm{~Pa}pref =2⋅10−5 Pa

2.2. A FIGURE OF MERIT FOR ACOUSTIC QUALITY

To compute the “goodness” of the calculated in-room response at a given listener’s position, a figure of merit needs to be introduced.
In this specific case, it is appropriate to measure how strong the colouration provided by the room to the frequency response is at the listener’s position, by comparing it with their anechoic responses, which is the noroom most representative scenario.
Therefore, the following single-figure metric Δ Δ Delta\DeltaΔ is defined:
Δ ( r ) = ∑ i = 1 m [ S P L ( ω m , r ) − H a n ( ω m ) ] 2 m Δ ( r ) = ∑ i = 1 m S P L ω m , r − H a n ω m 2 m Delta(r)=sqrt((sum_(i=1)^(m)[SPL(omega_(m),r)-H_(an)(omega_(m))]^(2))/(m))\Delta(r)=\sqrt{\frac{\sum_{i=1}^{m}\left[S P L\left(\omega_{m}, r\right)-H_{a n}\left(\omega_{m}\right)\right]^{2}}{m}}Δ(r)=∑i=1m[SPL(ωm,r)−Han(ωm)]2m
where m m mmm is the number of frequency bins considered. Please note that, in presence of more than one sound source, H a n ( ω ) H a n ( ω ) H_(an)(omega)H_{a n}(\omega)Han(ω) is the sum of the anechoic responses. Also, because of the definition of S P L ( ω , r ) , Δ S P L ( ω , r ) , Δ SPL(omega,r),DeltaS P L(\omega, r), \DeltaSPL(ω,r),Δ varies with the listener’s position across the room, while maintaining the same sound source(s) position(s).
The definition of Δ Δ Delta\DeltaΔ recalls the one of standard deviation in statistics, which quantifies how a certain measured experimental set of data drifts from the expected value. The bigger Δ Δ Delta\DeltaΔ, the more the room has an impact on the frequency response.
Some plots demonstrating how such metric works are shown in the examples in section 4.

3. Soundton Speaker Placement Calculator (SPC)

Soundton SPC is a web app designed to optimise the position of the elements of a sound system to achieve the best frequency response based on room dimensions.
Despite the tool being minimal in its user interface and simple to use, its calculation core performs the acoustic simulation as described in the previous section. More details on the user interface are given in the next paragraph.

3.1. USER INTERFACE

The user interface of the SPC web app is shown in Figure 2.
FIGURE 2 – THE USER INTERFACE OF THE SOUNDTON SPC WEB APP.
All the input parameters are located on the left sidebar:
  • Room dimensions: length, width, and height.
  • Listener’s height above floor level.
  • Units (S.I. or Imperial).
  • Speaker Layout: choose a loudspeaker type (generic or commercial) for each speaker slot, up to 11 loudspeakers.
In the central panel, a plan view and a front view of the room are displayed, where it is possible to drag the active loudspeakers in. The simulation results are shown in the form of a colour map in the plan view. The colourmap range is fixed so that it is possible to compare different scenarios in terms of room ratios, loudspeaker types and position, etc.
Finally, on the right sidebar, it is possible to find information on the selected loudspeaker, a legend for the colour map and an X , Y X , Y X,YX, YX,Y indicator for the mouse location on the plan view, which helps to locate better the desired listening position.
Other accessory features include:
  • “Rotate Speakers”, not relevant for the acoustic simulation as all loudspeakers are omnidirectional at low frequencies.
  • “Mirror Stereo Pair”, quickly sets the position of the Right speaker in the Stereo pair according to the Left loudspeaker.
  • “Reset Default Position”, to restore the initial loudspeaker positions.

3.2. THE ALGORITHM

The diagram in Figure 3 summarises the SPC web-app calculation algorithm.
FIGURE 3 – SPC ALGORITHM FLOWCHART.
Starting from room dimensions, a 0.40 × 0.40 m ( 1.3 × 1.3 f t ) 0.40 × 0.40 m ( 1.3 × 1.3 f t ) 0.40 xx0.40m(1.3 xx1.3ft)0.40 \times 0.40 \mathrm{~m}(1.3 \times 1.3 \mathrm{ft})0.40×0.40 m(1.3×1.3ft) resolution grid of receivers is constructed in the plan view, at a given listener’s height above the floor.
The anechoic frequency response is calculated as the sum of the anechoic responses of all active loudspeakers. This parameter is independent of the loudspeaker layout, and it is the same for all the receivers in the grid previously created, representing the reference against which the goodness of the in-room response at each receiver is calculated.
The in-room response at each receiver is then computed using the modal decomposition model, as defined in section 2.1, taking as input parameters the receivers’ coordinates, the sources’ coordinates and the overall anechoic response. In terms of absorption in the room, a default coefficient of 10 % 10 % 10%10 \%10% is assumed.
Finally, the figure of merit Δ Δ Delta\DeltaΔ is calculated at each receiver, and then plotted in the plan view as a colour map.

4. EXAMPLES

In this section, two examples of how a user can benefit from the SPC web app are described.

4.1. COMPARE LISTENER’S POSITIONS WITHIN THE ROOM

The first example is about comparing the response at different listening positions while keeping the room dimensions and the speaker layout unvaried.
The room is 5.20 m 5.20 m 5.20m5.20 \mathrm{~m}5.20 m (17.0 ft) long, 4.00 m 4.00 m 4.00m4.00 \mathrm{~m}4.00 m (13.0 ft) wide, and 2.70 m ( 8.5 f t ) 2.70 m ( 8.5 f t ) 2.70m(8.5ft)2.70 \mathrm{~m}(8.5 \mathrm{ft})2.70 m(8.5ft) high. The listener’s height is set to 0.85 m 0.85 m 0.85m0.85 \mathrm{~m}0.85 m (2.78 ft). Table 3 shows the parameters related to the loudspeaker layout.
TABLE 3 – LOUDSPEAKER LAYOUT PARAMETERS.
Loudspeaker Loudspeaker Type Loudspeaker Type {:[” Loudspeaker “],[” Type “]:}\begin{array}{c}\text { Loudspeaker } \\ \text { Type }\end{array} Loudspeaker Type Distance from left wall Distance from left wall {:[” Distance from left “],[” wall “]:}\begin{array}{c}\text { Distance from left } \\ \text { wall }\end{array} Distance from left wall Distance from the front wall Distance from the front wall {:[” Distance from the “],[” front wall “]:}\begin{array}{c}\text { Distance from the } \\ \text { front wall }\end{array} Distance from the front wall Height above floor
Left Bookshelf 1.20 m / 3.9 f t 1.20 m / 3.9 f t 1.20m//3.9ft1.20 \mathrm{~m} / 3.9 \mathrm{ft}1.20 m/3.9ft 1.04 m / 3.4 f t 1.04 m / 3.4 f t 1.04m//3.4ft1.04 \mathrm{~m} / 3.4 \mathrm{ft}1.04 m/3.4ft 0.80 m / 2.6 f t 0.80 m / 2.6 f t 0.80m//2.6ft0.80 \mathrm{~m} / 2.6 \mathrm{ft}0.80 m/2.6ft
Right Bookshelf 2.80 m / 9.2 f t 2.80 m / 9.2 f t 2.80m//9.2ft2.80 \mathrm{~m} / 9.2 \mathrm{ft}2.80 m/9.2ft 1.04 m / 3.4 f t 1.04 m / 3.4 f t 1.04m//3.4ft1.04 \mathrm{~m} / 3.4 \mathrm{ft}1.04 m/3.4ft 0.80 m / 2.6 f t 0.80 m / 2.6 f t 0.80m//2.6ft0.80 \mathrm{~m} / 2.6 \mathrm{ft}0.80 m/2.6ft
Subwoofer Subwoofer 12 ′ ′ 12 ′ ′ 12^(”)12^{\prime \prime}12′′ 3.46 m / 11.3 f t 3.46 m / 11.3 f t 3.46m//11.3ft3.46 \mathrm{~m} / 11.3 \mathrm{ft}3.46 m/11.3ft 0.48 m / 1.6 f t 0.48 m / 1.6 f t 0.48m//1.6ft0.48 \mathrm{~m} / 1.6 \mathrm{ft}0.48 m/1.6ft 0.18 m / 0.6 f t 0.18 m / 0.6 f t 0.18m//0.6ft0.18 \mathrm{~m} / 0.6 \mathrm{ft}0.18 m/0.6ft
Loudspeaker ” Loudspeaker Type ” ” Distance from left wall ” ” Distance from the front wall ” Height above floor Left Bookshelf 1.20m//3.9ft 1.04m//3.4ft 0.80m//2.6ft Right Bookshelf 2.80m//9.2ft 1.04m//3.4ft 0.80m//2.6ft Subwoofer Subwoofer 12^(”) 3.46m//11.3ft 0.48m//1.6ft 0.18m//0.6ft| Loudspeaker | $\begin{array}{c}\text { Loudspeaker } \\ \text { Type }\end{array}$ | $\begin{array}{c}\text { Distance from left } \\ \text { wall }\end{array}$ | $\begin{array}{c}\text { Distance from the } \\ \text { front wall }\end{array}$ | Height above floor | | :—: | :—: | :—: | :—: | :—: | | Left | Bookshelf | $1.20 \mathrm{~m} / 3.9 \mathrm{ft}$ | $1.04 \mathrm{~m} / 3.4 \mathrm{ft}$ | $0.80 \mathrm{~m} / 2.6 \mathrm{ft}$ | | Right | Bookshelf | $2.80 \mathrm{~m} / 9.2 \mathrm{ft}$ | $1.04 \mathrm{~m} / 3.4 \mathrm{ft}$ | $0.80 \mathrm{~m} / 2.6 \mathrm{ft}$ | | Subwoofer | Subwoofer $12^{\prime \prime}$ | $3.46 \mathrm{~m} / 11.3 \mathrm{ft}$ | $0.48 \mathrm{~m} / 1.6 \mathrm{ft}$ | $0.18 \mathrm{~m} / 0.6 \mathrm{ft}$ |
Figure 4 shows the app user interface with the above setup. The dots indicate the two listening positions under analysis, which have coordinates shown in Table 4.
FIGURE 4 – SOUNDTON SPC WEB-APP USER INTERFACE FOR THE CASE STUDY.
TABLE 4 – RECEIVER COORDINATES.
Receiver Distance from left wall ( m ) ( m ) (m)(\mathbf{m})(m) Distance from the front wall (m)
M 1 M 1 M1\mathrm{M} 1M1 2.40 m / 7.8 f t 2.40 m / 7.8 f t 2.40m//7.8ft2.40 \mathrm{~m} / 7.8 \mathrm{ft}2.40 m/7.8ft 2.80 m / 9.2 f t 2.80 m / 9.2 f t 2.80m//9.2ft2.80 \mathrm{~m} / 9.2 \mathrm{ft}2.80 m/9.2ft
M 2 M 2 M2\mathrm{M} 2M2 2.00 m / 6.6 f t 2.00 m / 6.6 f t 2.00m//6.6ft2.00 \mathrm{~m} / 6.6 \mathrm{ft}2.00 m/6.6ft 4.40 m / 14.4 f t 4.40 m / 14.4 f t 4.40m//14.4ft4.40 \mathrm{~m} / 14.4 \mathrm{ft}4.40 m/14.4ft
Receiver Distance from left wall (m) Distance from the front wall (m) M1 2.40m//7.8ft 2.80m//9.2ft M2 2.00m//6.6ft 4.40m//14.4ft| Receiver | Distance from left wall $(\mathbf{m})$ | Distance from the front wall (m) | | :—: | :—: | :—: | | $\mathrm{M} 1$ | $2.40 \mathrm{~m} / 7.8 \mathrm{ft}$ | $2.80 \mathrm{~m} / 9.2 \mathrm{ft}$ | | $\mathrm{M} 2$ | $2.00 \mathrm{~m} / 6.6 \mathrm{ft}$ | $4.40 \mathrm{~m} / 14.4 \mathrm{ft}$ |
From the modal decomposition model, described in section 2.1, it is clear that the frequency response has a strong dependence on either the source or the receiving position. In this example, the term p n ( r ) p n ( r ) p_(n)(r)p_{n}(r)pn(r) varies between the two listening positions, while the rest of the equation remains the same.
According to the obtained colour map (see Figure 4), it is expected that position M1 will exhibit a frequency response closer to the anechoic response when compared to M 2 M 2 M2M 2M2. This is confirmed by the frequency response curves calculated at both listening positions, plotted in Figure 5.
Despite the large peaks and dips in both responses, due to the assumption of very little absorption at low frequency in the room, It is clear to see that the green curve (M1) presents less deviation from the anechoic response compared to the red one (M2). This is further confirmed by the calculated Δ Δ Delta\DeltaΔ values, for which the “the less, the better” rule applies, equal to 8.3 at position M1 and 13.1 at position M2.
FIGURE 5 – CALCULATED FREQUENCY RESPONSE CURVES AT M1 AND M2, COMPARED WITH THE OVERALL ANECHOIC RESPONSE OF THE LOUDSPEAKERS.
This example, therefore, demonstrates the ability of the tool to provide insight into the best listening position given a fixed loudspeaker layout.

4.2. Compare Different SubWoOfer Positions

The next example is similar to the previous one, however, the source position is now changed, while keeping fixed the listening position. This represents a typical scenario in Hi-Fi setups where it is necessary to find a suitable location for the subwoofer to provide the best response at the selected listening position.
Room dimensions and the listener’s height remain the same as in the previous example. In this case, two scenarios, P 1 P 1 P1P 1P1 and P 2 P 2 P2P 2P2, are analysed, between which the position of a single subwoofer is changed from corner to mid-room, as shown in Table 5.
TABLE 5 – SUBWOOFER COORDINATES.
Scenario Loudspeaker Type Loudspeaker Type {:[” Loudspeaker “],[” Type “]:}\begin{array}{c}\text { Loudspeaker } \\ \text { Type }\end{array} Loudspeaker Type Distance from the left wall Distance from the left wall {:[” Distance from the “],[” left wall “]:}\begin{array}{c}\text { Distance from the } \\ \text { left wall }\end{array} Distance from the left wall Distance from the front wall Distance from the front wall {:[” Distance from the “],[” front wall “]:}\begin{array}{c}\text { Distance from the } \\ \text { front wall }\end{array} Distance from the front wall Height above floor
P 1 P 1 P1\mathrm{P} 1P1 Subwoofer 12″ 3.82 m / 12.5 f t 3.82 m / 12.5 f t 3.82m//12.5ft3.82 \mathrm{~m} / 12.5 \mathrm{ft}3.82 m/12.5ft 0.18 m / 0.6 f t 0.18 m / 0.6 f t 0.18m//0.6ft0.18 \mathrm{~m} / 0.6 \mathrm{ft}0.18 m/0.6ft 0.18 m / 0.6 f t 0.18 m / 0.6 f t 0.18m//0.6ft0.18 \mathrm{~m} / 0.6 \mathrm{ft}0.18 m/0.6ft
P 2 P 2 P2\mathrm{P} 2P2 Subwoofer 12 ′ ′ 12 ′ ′ 12^(”)12^{\prime \prime}12′′ 2.00 m / 6.6 f t 2.00 m / 6.6 f t 2.00m//6.6ft2.00 \mathrm{~m} / 6.6 \mathrm{ft}2.00 m/6.6ft 0.78 m / 2.6 f t 0.78 m / 2.6 f t 0.78m//2.6ft0.78 \mathrm{~m} / 2.6 \mathrm{ft}0.78 m/2.6ft 0.18 m / 0.6 f t 0.18 m / 0.6 f t 0.18m//0.6ft0.18 \mathrm{~m} / 0.6 \mathrm{ft}0.18 m/0.6ft
Scenario ” Loudspeaker Type ” ” Distance from the left wall ” ” Distance from the front wall ” Height above floor P1 Subwoofer 12″ 3.82m//12.5ft 0.18m//0.6ft 0.18m//0.6ft P2 Subwoofer 12^(”) 2.00m//6.6ft 0.78m//2.6ft 0.18m//0.6ft| Scenario | $\begin{array}{c}\text { Loudspeaker } \\ \text { Type }\end{array}$ | $\begin{array}{c}\text { Distance from the } \\ \text { left wall }\end{array}$ | $\begin{array}{c}\text { Distance from the } \\ \text { front wall }\end{array}$ | Height above floor | | :—: | :—: | :—: | :—: | :—: | | $\mathrm{P} 1$ | Subwoofer 12″ | $3.82 \mathrm{~m} / 12.5 \mathrm{ft}$ | $0.18 \mathrm{~m} / 0.6 \mathrm{ft}$ | $0.18 \mathrm{~m} / 0.6 \mathrm{ft}$ | | $\mathrm{P} 2$ | Subwoofer $12^{\prime \prime}$ | $2.00 \mathrm{~m} / 6.6 \mathrm{ft}$ | $0.78 \mathrm{~m} / 2.6 \mathrm{ft}$ | $0.18 \mathrm{~m} / 0.6 \mathrm{ft}$ |
The chosen listening position is located at 3.20 m ( 10.5 f t ) 3.20 m ( 10.5 f t ) 3.20m(10.5ft)3.20 \mathrm{~m}(10.5 \mathrm{ft})3.20 m(10.5ft) from the left wall and 2.40 m ( 7.8 f t ) 2.40 m ( 7.8 f t ) 2.40m(7.8ft)2.40 \mathrm{~m}(7.8 \mathrm{ft})2.40 m(7.8ft) from the front wall. The output of the calculations carried out with the Soundton SPC web app is shown in Figure 6.
In this case, the term p n ( r 0 ) p n r 0 p_(n)(r_(0))p_{n}\left(r_{0}\right)pn(r0), described in section 2.1, is the only one varying in the equation between the two scenarios. The colour map for scenario P1 shows a yellow tile at the listening position, while the same position is in green for the P2 scenario. Therefore, it is expected that the frequency response curve for the first will deviate more from the anechoic response of the subwoofer.
In the same fashion as done in the previous example, the frequency response curves at the listening position are plotted for each scenario, as well as the corresponding Δ Δ Delta\DeltaΔ values (see Figure 7). It is clear that the green curve (P2), with Δ = 7.6 Δ = 7.6 Delta=7.6\Delta=7.6Δ=7.6, is closer to the anechoic response of the subwoofer, while the presence of strong peaks on the red curve ( Δ = 12.0 ) ( Δ = 12.0 ) (Delta=12.0)(\Delta=12.0)(Δ=12.0) makes P 1 P 1 P1\mathrm{P} 1P1 an unfavourable scenario for the subwoofer positioning.
FIGURE 6 – SOUNDTON SPC WEB-APP USER INTERFACE FOR THE CASE STUDY (P1 ON THE LEFT, P2 ON THE RIGHT).
FIGURE 7 – CALCULATED FREQUENCY RESPONSE CURVES FOR P1 AND P2 SCENARIOS, COMPARED WITH THE ANECHOIC RESPONSE OF THE SUBWOOFER.

5. REFERENCES

[1] M. Kleiner and J. Tichy, Acoustics of small rooms. Boca Raton, Florida ; CRC Press, 2014.
[2] J. Mulcahy, “Room EQ Wizard.” https://www.roomeqwizard.com/ (accessed Feb. 02, 2023).
[3] T. J. Cox and P. D’Antonio, Acoustic Absorbers and Diffusers. Routledge Taylor & Francis, 2009. doi: 10.4324 / 9781482266412 10.4324 / 9781482266412 10.4324//978148226641210.4324 / 978148226641210.4324/9781482266412.
[4] H. Kuttruff, Room Acoustics, 5th ed. London, 2002. doi: 10.1201/9781482286632.