Room Shape Reconstruction
With a Single Mobile Acoustic Sensor
Fangrong Peng, Tiexing Wang, Biao Chen
Electrical Engineering and Computer Science
Syracuse University
GLOBALSIP 2015
GLOBALSIP 2015
Room Shape Reconstruction
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Room Shape Reconstruction with Acoustic Sensors
Related Works
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“Can one hear the shape of a room: The 2-d polygonal case”
[I. Dokmanic, Y. Lu, etc., ICASSP 2010]
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“Acoustic echoes reveal room shape” [I. Dokmanic, R.
Parhizkar, etc., PNAS 2013]
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“L1 regularized room modeling with compact microphone
array” [D. Ba, F. Ribeiro, etc., ICASSP 2010]
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“3d room geometry estimation from measured impulse
responses” [S. Tervo and T. Tossavainen, ICASSP 2012]
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etc.
GLOBALSIP 2015
Room Shape Reconstruction
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Room Shape Reconstruction with Acoustic Sensors
Previous Works
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multiple acoustic stimuli / multiple microphones are used
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I
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complicated deployment
topology of the device array required
higher-order echoes are required
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often significantly weaker and of poor resolution
Question
Can we use a single co-located loudspeaker/microphone and solely
first-order echoes to reconstruct the room shape?
GLOBALSIP 2015
Room Shape Reconstruction
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Image Source Model (ISM)
Convex Polygonal Room with N = 4 Walls
y
S3
W3
R3
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co-located source and receiver at O
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first-order images S1 - S4
reflecting points Ri
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second-order images S12 , S21 , etc.
reflecting path O - R12 - R21 - O
S2
W2
W4
O
x
R21
S4
S12
R12
W1
S1
S21
GLOBALSIP 2015
Room Shape Reconstruction
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Mapping between ISM and Room Shape
Room Impulse Response
ideal room impulse response (RIR)
h(t) =
Ñ
X
an · δ(t − tn )
(1)
n=0
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the number of received echoes Ñ, typically Ñ ≥ N
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interesting parameter: time of arrival (TOA) tn
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the distance between source and the n-th ‘wall’
rn =
c · tn
,
2
(2)
where c is the speed of sound
GLOBALSIP 2015
Room Shape Reconstruction
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Mapping between ISM and Room Shape
Retrieve TOA
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probing signal s(t) used to excite echoes
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received signal r (t) = s(t) ∗ h(t)
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s(t) usually chosen to be broadband signals, e.g., chirp
signals, s.t.
s(t) ∗ s(t) ≈ δ(t),
then
s(t) ∗ r (t) ≈ h(t)
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TOA tn ’s can be obtained by picking the peeks
GLOBALSIP 2015
Room Shape Reconstruction
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Mapping between ISM and Room Shape
Co-located loudspeaker and microphone with one measurement
y
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ri as the distance between the
source and the i-th wall
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W1 W2 W3 and W1 W20 W3 have
the same set of ri
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one set of first-order TOA
measurements is not sufficient
W2′
W3
r2
r3
x
O
r1
W2
W1
Question
How many sets of TOA measurements are sufficient to retrieve a
triangle?
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Room Shape Reconstruction
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Identifiability by First-order Echoes
Lemma 1
For any given triangle T , suppose there are three interior points
that are not on a straight line, and they are able to receive the
first-order echoes from all walls. Let R be the distance matrix with
Rij being the distance between the i-th point and the j-th wall.
Then R uniquely defines the triangle with probability 1.
Note
1. We assume that the measurement points are within a
substantially small range, such that the echoes arrive in the
same order for all points.
2. Denote rj as the j-th column of R.
GLOBALSIP 2015
Room Shape Reconstruction
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Retrieve a Triangle
Co-located loudspeaker and microphone with three measurements
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measurement taken at Oi :
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walls:
(xi , Ri1 ), i = 1, 2, 3
y
W1 : y = 0
W2 : y = a2 · x + b2
W3 : y = a3 · x
y = a2 x + b 2
y = a3 x
R22
R12
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O2
W3
R13
O1
R11
O3
W2
x
O
W1
GLOBALSIP 2015
R32
constraints:
|a2 xi − Ri1 + b2 |
p
= Ri2 , i = 1, 2, 3
a22 + 1
|a3 xi − Ri1 |
p
= Ri3 , i = 1, 2, 3
a32 + 1
Room Shape Reconstruction
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Retrieve a Triangle
Solution
s
a3 = ±
2BC
2
D − B2 − C 2
Ba3
p
,
B + kC a32 + 1
p
kA a32 + 1
p
b2 =
,
B + kC a32 + 1
p
Ri1 + k a32 + 1 · Ri3
xi =
a3
a2 =
I
− 1, (3)
two candidates for a3
reconstructed distances between
Oi to W2 using two candidates:
(4)
+
Di2
=
q
(A + BRi3 − CRi1 ) a32 + 1
s
2
q
(Ba3 )2 + B + C a32 + 1
=
q
(A + BRi3 − CRi1 ) a32 + 1
−s
2
q
(Ba3 )2 + B − C a32 + 1
(5)
−
Di2
(6)
k = sign(a3 ), A, B, C, D are parameters
depending on R
GLOBALSIP 2015
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+
−
one of Di2
and Di2
matches
Ri2
Room Shape Reconstruction
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Triangle Recovery Algorithm
1
2
3
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set k = 1
calculate a3 , a2 , b2 and xi ’s using Eqs. (3)-(6)
calculate the reconstructed distances d2 = [D12 , D22 , D32 ]T
if d2 6= r2
set k = −1 and reconstruct the triangle again
GLOBALSIP 2015
Room Shape Reconstruction
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Retrieve a Convex Polygon
Extended Triangle Recovery Algorithm
Sequentially apply triangle recovery algorithm on R:
1. apply the triangle recovery algorithm on {r1 , r2 , r3 }
2. apply the triangle recovery algorithm on {r1 , rj , r3 }, j = 4 · · · N
using the a3 and xi ’s from last step
Problem
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parallel edges exist in {r1 , r2 , r3 }?
two parallel and one perpendicular edges are recovered
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all convex polygons can be recovered?
think about rectangles: every three edges have a pair of
parallel ones
GLOBALSIP 2015
Room Shape Reconstruction
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Ambiguity between Rectangles and Parallelograms
Rectangle and Parallelogram
In a parallelogram (including rectangle),
y
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for any source Oi ,
H
L
L
O
x
Ri1 + Ri3
=
H
Ri2 + Ri4
=
L
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exist other parallelograms also
satisfying the condition
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extended triangle recovery algorithm
recovers all parallelograms as their
corresponding rectangles
Lemma 2
For any given rectangle R, there is an infinite number of
parallelograms having the same distance matrix R with R,
regardless of the number of measurement points.
GLOBALSIP 2015
Room Shape Reconstruction
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Retrieve a Convex Polygon
R contains only first-order echoes
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define V(R) = {rj , j = 1 · · · N}
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define V 0 (R) as the largest subset of V(R) s.t. no parallel
edges recovered from V 0 (R)
Algorithm
1. find V 0 (R)
2. apply the extended triangle recovery algorithm on V 0 (R)
3. add the parallel edges from V(R) − V 0 (R)
Question
How does it work if higher-order information is present in R?
GLOBALSIP 2015
Room Shape Reconstruction
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Retrieve a Convex Polygon
R̃ contains first-order and some higher-order echoes
Two cases of the starting set {r̃1 , r̃2 , r̃3 } of V 0 (R̃):
1. contains higher-order information, then two cases
1.1 no valid triangle is formed, i.e., the recovered a3 is not a real
number
1.2 a valid yet wrong triangle is formed, then no edges recovered
later from V 0 (R̃) can pass the distance match check
rearrange V 0 (R̃) until {r̃1 , r̃2 , r̃3 } only contains first-order TOAs
2. contains only first-order information, then
2.1 edges recovered later from the rj ’s corresponding to higherorder information can pass the distance match check
Question
Can all convex polygons be reconstructed if first-order mixed with
high-order echoes are used?
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Room Shape Reconstruction
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Recoverability of Polygons
Lemma 3
Suppose a polygon has N edges, including M pairs of parallel ones.
The polygon can be correctly and uniquely recovered either when
N − M ≥ 4, or M ≥ 3.
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Room Shape Reconstruction
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Polygon Recovery Algorithm
1 find V 0 (R̃)
2 if M == 0 or M ≥ 3
3
apply the extended triangle recovery algorithm on V 0 (R̃)
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if dk 6= r̃ for all k = 4 · · · Ñ
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rearrange V 0 (R̃) and go to step 3
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else the algorithm is done
7 if M == 1 or M == 2
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apply step 3-6 on V 0 (R̃)
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if step 6 is done
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add the edges corresponding to V(R̃) − V 0 (R̃), done
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else if step 3 is executed exceeding a maximum no. of times
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if M == 1
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more information is required to recover the shape, done
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else if M == 2
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recover a rectangle, done
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Room Shape Reconstruction
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Polygon Reconstruction
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0
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-5
0
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0
(a)
10
5
5
0
0
0
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(c)
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(b)
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(d)
Figure: original (blue) and recovered (dash purple/black) polygons: (a)
an irregular quadrilateral, (b) a rotation, (c) a reflection, (d) a hexagon
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Room Shape Reconstruction
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Conclusion
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a co-located loudspeaker and microphone is used to
reconstruct a room shape
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only first-order RIR are required
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three sets of measurements taken at three non-collinear points
inside the room are sufficient to recover a wide class of
polygonal shapes
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additional information is needed to distinguish the ambiguity
between rectangle and parallelogram
GLOBALSIP 2015
Room Shape Reconstruction
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Thank you!
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Room Shape Reconstruction
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