filter
Content
Tutorial on Gabor Filters
Javier R. Movellan
1
The The compl complex ex Gabor Gabor in in spac space e doma domain in
Here is the formula of a complex Gabor function in space domain g (x, y) = s(x, y) wr (x, y)
(1)
where s(x, y) is a complex sinusoidal, known as the carrier, and wr (x, y) is a 2-D Gaussian-shaped function, known as the envelop. 1.1
The complex complex sinusoid sinusoidal al carrier carrier
The complex sinusoidal is defined as follows,1 s(x, y) = exp( j exp( j (2π (2π (u0 x + v0 y ) + P ) P ))
(2)
where (u (u0 , v0 ) and P define the spatial frequency and the phase of the sinusoidal respectively. We can think of this sinusoidal as two separate real functions, conveniently allocated in the real and imaginary part of a complex function (see Figure 1).
Figure Figure 1: The real and imaginary imaginary parts parts of a com comple plex x sinuso sinusoida idal. l. The images images are 128 128 pixels. The parameters are: u0 = v0 = 1/80 cycles/pixel, cycles/pixel, P = 0 deg.
×
1
An offset constant parameter for s(x, y ) will be introduced later, to compensate the DC-component DC-component of this sinusoidal sinusoidal.. Refer to the appendix for detailed detailed explanation. explanation.
The real part and the imaginary part of this sinusoidal are Re (s(x, y)) = cos (2π(u0 x + v0 y) + P )
(3)
Im (s(x, y)) = sin (2π(u0 x + v0 y) + P )
The parameters u0 and v0 define the spatial frequency of the sinusoidal in Cartesian coordinates. This spatial frequency can also be expressed in polar coordinates as magnitude F 0 and direction ω0 : F 0
=
ω0
u0 2 + v0 2 v0 u0
(4)
= F 0 cos ω0
(5)
= tan
−1
i.e.
u0
v0 = F 0 sin ω0 Using this representation, the complex sinusoidal is s(x, y) = exp ( j (2πF 0 (x cos ω0 + y sin ω0 ) + P )) 1.2
(6)
The Gaussian envelop
The Gaussian envelop looks as follows (see Figure 2): wr (x, y) = K exp
−
π a2 (x
− x )r
2
0
+ b2 (y
− y )r
2
0
(7)
where (x0 , y0 ) is the peak of the function, a and b are scaling parameters 2 of the Gaussian, and the r subscript stands for a rotation operation3 such that (x
− x )r (y − y )r
1.3
0
= (x
− x )cos θ + (y − y )sin θ −(x − x )sin θ + (y − y )cos θ
0
=
0
0
0
(8)
0
The complex Gabor function
The complex Gabor function is defined by the following 9 parameters;
• • • • •
K
:
Scales the magnitude of the Gaussian envelop.
(a, b)
:
Scale the two axis of the Gaussian envelop.
θ
:
Rotation angle of the Gaussian envelop.
(x0 , y0 )
:
Location of the peak of the Gaussian envelop.
(u0 , v0 )
:
Spatial frequencies of the sinusoidal carrier in Cartesian coordinates. It can also be expressed in polar coordinates as ( F 0 , ω0 ).
•
P
:
Phase of the sinusoidal carrier.
Each complex Gabor consists of two functions in quadrature (out of phase by 90 degrees), conveniently located in the real and imaginary parts of a complex function. 2 3
Note that the Gaussian gets smaller in the space domain, if a and b get larger. This rotation is clockwise, the inverse of the counterclockwise rotation of the ellipse.
Figure 2: A Gaussian envelop. The image is 128 128 pixels. The parameters are as follows: x0 = y0 = 0. a = 1/50 pixels, b = 1/40 pixels, θ = 45 deg.
×
−
Figure 3: The real and imaginary parts of a complex Gabor function in space domain. The images are 128 128 pixels. The parameters are as follows: x0 = y0 = 0, a = 1/50 pixels, b = 1/40 pixels, θ = 45 deg, F 0 = 2/80 cycles/pixel, ω0 = 45 deg, P = 0 deg.
×
√
−
Now we have the complex Gabor function in space domain 4 (see Figure 3): g(x, y) =
K exp
−
π a2 (x
− x )r 0
2
+ b2 (y
exp( j (2π(u0 x + v0 y) + P ))
4
− y )r 0
2
(9)
In fact, there remains some DC component in this Gabor function. You have to compensate it to have the admissible Gabor function. Refer to the appendix.
Or in polar coordinates, g(x, y) =
K exp
−
π a2 (x
− x )r 0
2
+ b2 (y
− y )r 0
2
exp( j (2πF 0 (x cos ω0 + y sin ω0 ) + P ))
(10)
Figure 4: The Fourier transform of the Gabor filter. The peak response is at the spatial frequency of the complex sinusoidal: u p = v p = 1/80 cycles/pixel. The parameters are as follows: x0 = y0 = 0, a = 1/50 pixels, b = 1/40 pixels, θ = 45 deg, F 0 = 2/80 cycles/pixel, ω0 = 45 deg, P = 0 deg.
√
−
The 2-D Fourier transform of this Gabor 5 is as follows (see Figure 4): gˆ(u, v) =
K exp( j ( 2π (x0 (u ab
−
exp
−π
(u
− u )r 0
a2
−u )+y 0
2
+
(v
0
(v
− v )r 0
b2
2
− v )) + P )) 0
(11)
Or in polar coordinates,
5
Magnitude (ˆ g (u, v)) =
K exp ab
Phase(ˆ g (u, v)) =
−2π (x
0
2
−π (u −au )r + (v −bv )r (u − u ) + y (v − v )) + P 0
0
2
0
Refer to the appendix for detailed explanation.
0
2
0
2
(12)
2
Half-magnitude profile
The region of points, in frequency domain, with magnitude equal one-half the peak magnitude can be obtained as follows. Since the peak value is obtained for ( u, v) = (u0 , v0 ), and the peak magnitude is K/ab, we just need to find the set of points (u, v) with magnitude K/2ab. 1 K K = exp 2 ab ab or,
− − − (u
π
(u
− log 2 = −π
2
u 0 )r (v v0 )r + 2 a b2 2
−
u 0 )r (v v0 )r + a2 b2
−
2
2
(13)
(14)
or equivalently,
(u
− u )r 0
a C
2
(v
v0 )r b C
2
− +
where C =
=1 (15)
log2 = 0.46971864 π
≈ 0.5
Equation 15 is an ellipse centered at (u0 , v0 ) rotated with an angle θ with respect to the u axis. The main axis of the ellipse have length 2 a C a and 2 b C b respectively.
≈
≈
We will use the following convention: a is the length of the axis closer to ω0 , and b is the length of the axis perpendicular to the main axis 6 (See Figure 5).
(b)
(a) Theta F_0 Omega_0 Figure 5: Parameters of the Gabor kernel as reflected in the half-magnitude elliptic profile. Note that this is a figure in frequency domain.
3
Half-magnitude frequency and orientation bandwidths
Frequency and orientation bandwidths of neurons are commonly measured in terms of the half-magnitude responses. Let u0 , v0 the preferred spatial frequency of a neuron. In polar coordinates this spatial frequency can be expressed as F 0 and ω0 . 6
More precisely a and b are 1.06 times the length of the respective axis
To find the half-magnitude frequency bandwidth, we probe the neuron with sinusoidal images of orientation ω0 and different spatial frequency magnitudes F . We increase F with respect to F 0 until the magnitude of the response is half the magnitude at (F 0 , ω0 ). Let’s call that value F max . We then decrease F with respect to F 0 until the magnitude of the response is half the response at ( F 0 , ω0). Call that F min . Half-magnitude frequency bandwidth is defined as follows: ∆F 1/2 = F max
− F
(16)
min
or, when measured in octaves,7 ∆F 1/2 = log2 (F max /F min )
(17)
Half-magnitude orientation bandwidth is obtained following the same procedure but playing with the orientation ω instead of the frequency magnitude F . ∆ω1/2 = ωmax In Gabor functions with θ0 follows (See Figure 6)
≈ω
0
−ω
(18)
min
the frequency bandwidth can be obtained as
∆F 1/2 = 2 a C
≈a
(19)
and the orientation bandwidth can be approximated as follows (see Figure 6) ∆ω1/2
≈ 2tan
−1
b C F 0
(20)
0.5 Delta W ~ arctan( b C / F o)
bC
Delta F = aC Fo
Figure 6: A half-magnitude profile and its relationship to the orientation and frequency bandwidths. 7
Octave is a unit used for showng the ratio, as an index of 2. k octaves = 2k × 100.0%
4
Effective spread and rms spread
The rms (which stands for root mean squares) length, rms width, and rms area of a 2-D function are defined in terms of their first and second moments: The moments of a complex function g(x, y) are defined by converting the function into a probability density (which must be always positive and integrates to 1.0) and then calculating the standard first and second moments. A common way to achieve this is as follows: From the function g(x, y) we construct the following probability density f (x, y) =
1 2 g(x, y) Z
|
(21)
|
2
where g(x, y) is the magnitude of the signal, which is always positive, and Z guarantees that f (x, y) integrates to 1.0, i.e.
|
|
+∞
Z =
+∞
−∞
−∞
2
|g(x, y)| dxdy
(22)
Once we have defined a probability density function, the standard statistical measures of location and scale follow.
2
µX = E X (x) =
f (x, y) xdxdy
f (x, y) (x
σX = E X (x
− µX )
2
=
(23) 2
− µX )
dxdy
(24)
2 with similar equations for µY and σY .
µY = E Y (y) =
f (x, y) y dxdy
f (x, y) (y
2 σY = E Y (y
And, σXY = E XY ((x
2
− µY )
=
− µX ) (y − µY )) =
f (x, y) (x
− µY )
(25) 2
dxdy
(26)
− µX ) (y − µY ) dxdy
(27)
The rms width and length are defined as the σX and σY of a rotated version of f (x, y) so that the covariance σXY of the rotated distribution be zero. Let X r , Y r represent the rotated variables for which the covariance is zero, the rms length and width are ∆X rms = ∆Y rms =
2 σX
(28)
2 σY
(29)
r
r
Similar definitions can be obtained also in the frequency domain, by working with the Fourier transform of the original complex function. ∆U rms = ∆V rms =
2 σU
(30)
2 σV
(31)
r
r
The rms area in the space and frequency domains are defined as follows: Area(XY )rms = (∆X rms ) (∆Y rms )
(32)
Area(U V )rms = (∆U rms ) (∆V rms )
(33)
Some papers work with what are known as effective length, width and areas. They are simply the rms measures multiplied by 2π
√
∆X eff =
√
2π ∆X rms
(34)
and so on. It can be shown that the following relationships hold on any 2D function with finite moments 1 (∆X rms ) (∆U rms ) (35) 4π 1 (∆Y rms ) (∆V rms ) (36) 4π and 1 Area(XY )rms Area(U V )rms (37) 16π2
≥ ≥
≥
It is easy to verify that the Gabor complex function achieves the lower limits of the uncertainty relations. For a given area in the space domain it provides the maximum possible resolution in the frequency domain, and vice-versa. It can be shown that the rms width and lengths of Gabor functions are as follows: a (38) ∆U rms = 2 π
√
∆V rms =
b √ 2 π
(39)
Tow see why, simply consider that the probability density associated with the Ga1 2 2 bor function f (x, y) = g(x, y) is Gaussian with variances equal to ∆U rms and Z 2 ∆V rms .
|
|
Moreover, from the uncertainty relations,
5
1 2a π 1 = 2b π
∆X rms =
√
(40)
∆Y rms
√
(41)
Gabor functions as models of simple cell receptive fields
? showed that the real part of complex Gabor functions fit very well the receptive
field weight functions found in simple cells in cat striate cortex. Here are some of the constraints that can be found in the literature:
• The orientation of the Gaussian ω
is rougly equivalent to the orientation of the carrier. θ0 = ω0 . The absolute deviations, “the twist” have a Median of about 10 degrees (see ?, p. 1249). 0
8
8
It’s easier to consider in the frequency domain. Note that the long axis in the frequency domain becoms the short axis in the space domain. Don’t get confused!
• In macaque V1, most cells have a spatial frequency bandwidth between 1 and 1.5 octaves. The median is about 1.4 octaves (see ?, p. 551).
• Orientation bandwidths are typically about 40 degrees (mean = 65 degrees, median = 42 degrees, mode = 30 degrees) (see ?, p. 535 and 541).
• From the three constraints mentioned above, it may be derived (see equqtion 59) that the aspect ratio is typically about 1.24, a/b ≈ 1.24. • In the area mapping the fovea, there are more kernels oriented vertically i.e.
and horizontally than oriented diagonally (about 3 to 2). (see ?, p. 537).
• Pairs of adjacent simple cells in the visual cortex of the cat are in quadrature (?). We can then put these two cells in the real and imaginary parts of a complex function and treat them as a complex Gabor receptive field.
6
Gabor functions for spatial frequency filtering
Consider a massive set of simple cell neurons with Gabor kernel functions with equal parameters except for the location parameters ( x0 , y0 ). Let all these neurons be distributed uniformly about the foveal field. Each point in the foveal field contains at least two neurons in quadrature. We can model the operation of such a set of neurons as a convolution operation (assuming a continuous and uniform distribution of filters in all the foveal locations). Since convolution in space domain is product in frequency domain, the set of Gabor functions work as bandpass frequency filters of the foveal image. The peak frequency is controlled by the spatial frequency of the sinusoidal carrier (u0 , v0 ). The half-magnitude region is controlled by the rotation θ and scale parameters a, b, of the Gaussian envelop.
7
Energy filtering
A quadrature pair (or a Hilbert Transform pair) is a set of two linear operators with the same amplitude response but phase reponses shifted by 90 degrees. Strictly speaking sine and cosine Gabor operators are not quadrature pairs because cosine phase Gabors have some DC response, whereas sine gabors do not. However, one can have quadrature Gabor pairs that look very much like sine/cosine pairs. Thus the sine and cosine Gabor pair is commonly refered to as a quadrature pair. A system that sums the square of the outputs of a quadrature pair is called an energy mechanism ( ?). Energy mechanisms have unmodulated responses to drifting sinusoidals. Complex cells in V1 are commonly modeled as energy mechanisms since they are unmodulated by drifting sinusoidals. Simple cells respond to a drifting sinusoidal with a half-wave rectified analog of the signal, suggesting that the cells are linear up to rectification. Complex cells respond to a drifting sinusoidal in an unmodulated way, as a maintained discharge. ? showed that complex receptive fields are composed of subunits. The subunits of model complex cells are model simple cells with identical amplitude response. ? have shown that behavior of complex cell to stimuli made of paris of bars flashed in sequence is consist with an energy mechanism.
8
Contrast Normalization
? have shown that stimuli presented at orientations orthogonal to the optimal orientation inhibit simple cells activity. (?) have shown similar inhibitory effect
between frequency bands. These inhibitory effects may play a serve as a gain control (or contrast normalization) mechanism. ? proposes the following model of gain control in complex cells: The amplitude response of each energy mechanism is divided by the total energy at all orientations and nearby spatial frequencies: ¯i = E
E i κ+
j
(42)
E j
where κ is a positive constant to avoid zero denominators.
9
Functional Interpretations
Section in preparation:
• minimizes number of neurons needed to achieve a desired frequency resolution.
• spatially and frequency localized. • matched to “logons” likely to occur in images. • for natural images the Gabor representation is more sparse than the
δ
(pixel) representation and than the DOG representation.
10
Constructing an idealized V1
We’ll use the following simplified assumptions. They all have empirical support.
• The orientation of the complex sinusoid carrier and the Gaussian envelop are the same: ω0 = θ.
• The half-magnitude frequency bandwidth is constant for all neurons when measured in octaves.
• The half-magnitude orientation bandwidth is constant for all neurons. • The upper half-magnitude contour of a frequency band coincides with the
lower contour of the next frequency band. This is just a constrain we use for convenience.
From these assumptions above, we can derive the relationship between the parameters F 0 , a and b. From equations 17 and 19, we know that the frequency bandwidth in octaves is ∆F 1/2 = log2
F 0 + a C F 0 a C
−
where C =
(43)
log2 = 0.46971864 π
≈ 0.5
Thus, K a C
(44)
2∆F 1 2∆F + 1
(45)
a = F 0 where K a =
−
With respect to the orientation bandwidth, equation 20 tells us that tan
1 ∆ω 2
=
b = F 0
K b C
b C F 0
(46)
Thus,
where K b = tan
(47)
1 ∆ω 2
(48)
Therefore, in this model the aspect ratio of a and b is constant: λ=
a K a = b K b
(49)
Moreover, from equation 19 1 ∆F = a C = F 0 K a 2
(50)
We can now locate our frequency peaks such that the upper half-magnitude contour of one channel coincides with the lower half-magnitude contour of the the next channel. Let µi signify the peak frequency of the ith band, We know F max for band i is 1 i F max = µi + ∆F i = µi + µi K a = µi (1 + K a ) 2
(51)
and F min for band i + 1 is i+1 F max = µi+1
− 12 ∆F i
+1
= µi+1
− µi
+1
K a = µi+1 (1
− K a)
(52)
We want these two values to coincide, therefore µi+1 = µi
1 + K a 1 K a
−
(53)
Thus, the peak frequencies follow a geometric series µi = µ1 Ri−1
(54)
1 + K a 1 K a
(55)
where R= 10.1
−
Example
If we use the standard values for simple cell median bandwidths from macaque striate cortex:
• ∆F = 1.4 octaves. • ∆ω = 40 degrees.
Then,
2∆F 1 = 0.45040 2∆F + 1 1 K b = tan ∆ω = 0.36397 2 K a a = µi = 0.9589 µi C a K a λ= = = 1.23746 b K b a b = = 1.1866 µi λ 1 + K a R= = 2.6390 1 K a
−
(56)
(57)
K a =
(58) (59) (60) (61)
−
Suppose we want three frequency bands and we want the F 0 of the third band to be 0.25. Then, 0.25 µ1 = = 0.03589 (62) 2.63902 and 1 ∆F 1 = K a µ1 = 0.01617 (63) 2 Thus, the half magnitude interval 9 is (0.01973, 0.05207) The second band peaks at µ2 = µ1 R = 0.09473
(64)
and
1 ∆F 2 = K a µ2 = 0.04267 2 Thus, the half magnitude interval is (0.05207, 0.1374)
(65)
Finally, the third band peaks at µ3 = µ2 R = 0.2500
(66)
and
1 ∆F 3 = K a µ3 = 0.1126 2 Thus, the half magnitude interval is (0.1374, 0.3626)
(67)
These three Gabors cover the frequency bands of (0 .01973, 0.3626)
11
Appendix
11.1 11.1.1
Fourier transform of the Gabor function The integral of the Gaussian
Let I denote the integral ∞
I
exp
−∞
9
∞
− ≡ − 2
πx
dx =
exp
π y 2 dy
(68)
−∞
The half magnitude interval here is the frequency coverage of that Gabor in terms of 1 1 half-magnitude profile: µi − ∆F i , µi + ∆F i 2 2
Then,
∞
2
I
=
∞
− − − − − − π x2 dx
exp
−∞
−∞
∞
=
exp
∞
π x2 + y 2
exp
−∞
−∞
π y 2 dy
−
(69)
dxdy
Using a polar coordinates r and ϕ, this equation can be rewritten as: 2π
∞
I 2
=
π r2 rdrdϕ
exp
0
0
∞
=
π r2 dr
2π r exp
0
=
π r2
exp
∞
(70)
0
=1
Accordingly,
∞
I =
−
π x2 dx = 1
exp
(71)
−∞
11.1.2
Fourier transform of the Gaussian
The Fourier transform of the simple 1-D Gaussian is ∞
exp( πx2 ) exp( 2πjfx) dx
−∞
−
−
∞
− − − − − − − − − − − =
exp
2
π(x + jf )
πf 2 dx
(72)
−∞
∞
πf 2
= exp
πx
exp
2
dx
(x
−∞
≡ x + jf )
πf 2
= exp
In the same way, the Fourier transform of the simple 2-D Gaussian is ∞
∞
exp
−∞
π(x2 + y 2 ) exp( 2πjux) exp( 2πjvy) dxdy
−∞ ∞
∞
2
=
exp( πy 2 ) exp( 2πjvy) dy
exp( πx ) exp( 2πjux) dx
−∞
−∞
−
(73)
= exp( πu2 ) exp( πv 2 )
−
= exp
−
π(u2 + v 2 )
− − −
and so on. More generally, ∞
exp
−∞
π xT x exp
2πj uT x dx = exp
π uT u
−
(74)
That is, the Fourier transform of an N-dimensional Gaussian is also an Ndimensional Gaussian.
11.1.3
Several properties of the Fourier transform
Several properties of the Fourier transform will be introduced in this section. Let ˆ denote the Fourier transform of a function f (x) , i.e. f (u) ∞
−
f (x) exp
−∞
Then, ∞
f (x
−∞
= exp = exp
= exp and also, ∞
− x ) exp −2πj uT x
0
− − −
T
2πj u x0
0
−
f (x ) exp
−∞
0
T
2πj u x
dx (76)
dx
(x
≡x−x ) 0
ˆ 2πj x0 T u f (u)
−
2πj uT x dx
2πj uT A−1 Ax dx
− −
f (Ax) exp
−∞ ∞
f (x ) exp
−∞
11.1.4
(75)
− x ) exp −2πj uT (x − x )
2πj x0 u
∞
=
f (x
∞
T
−∞
=
dx
−∞
f (Ax) exp =
∞
ˆ 2πj uT x dx = f (u)
1 ˆ −T f A u A
T
−
2πj A
u
T
x
(77)
1 dx A
(x
≡ Ax)
Fourier transform of the Gabor function
Here, we have the simple Gaussian envelop and sinusoidal carrier: w (x) = exp π xT x
−
(78)
T
s (x) = exp j 2π u0 x + P Now we want a Fourier transform of the following Gabor function, g (x) = K w (A (x x0 )) (s (x) C )
−
= K w (A (x where K and C are constants.
− x )) 0
−
exp j 2π u0 T x + P
(79)
− C
Given this function, its Fourier transform will be ∞
gˆ (u) =
g (x) exp
−∞ ∞
= K exp( jP )
−
2πj uT x dx
w (A (x
−∞
0
∞
− KC
−∞
w (A (x
0
− x )) exp −2πj uT x 0
− x )) exp −2πj (u − u )T x
dx
dx
(80)
Using the previously introduced properties, we can see gˆ (u) = K exp( jP ) exp
−
2πj x0 T (u
− u ) A1 wˆ 0
− KC exp −2πj x T u A1 wˆ K T = A exp −2πj x (u − u )
0
0
0
exp( jP ) w ˆ A−T (u
A−T u
A−T (u
0
(81)
− u ) − C exp −2πj x T u 0
−u )
0
0
w ˆ A−T u
Remember that C is a constant that compensates the DC component of the Gabor function, which means gˆ (0) must be equal to 0. Therefore, C = exp j 2π x0 T u0 + P
then
T
exp 2πj u0 (x
gˆ (u) =
K exp j A T
−
w ˆ A
w ˆ
(82)
x0 )) exp j 2π u0 T x0 + P
g (x) = K w (A (x
and
A−T u0
− − − − − − − − − − − − − − − − − − − − − − − − − − x0 )
2π x0 T (u
(u
u0 )
w ˆ
T
−
w ˆ
A
(83)
u0
u0 ) + P T
−
A
T
−
u0 w ˆ A
(84)
u
Using the following definitions:
we can get
g (x) = K exp
w (x) = exp
π xT x
w ˆ (u) = exp
π uT u
T
x0 ) AT A (x
π (x
exp 2πj u0 T (x
and
gˆ (u) =
K exp j A exp
x0 ) exp j 2π u0 T x0 + P
exp
2π x0 T (u
π (u
exp
x0 )
(85)
(86)
π u0 T A−1 A−T u0
u0 ) + P
T
u0 ) A−1 A−T (u
π u0 T A−1 A−T u0 exp
u0 )
π uT A−1 A−T u
(87)
In order to have the generalized 2-D Gabor function, consider A is a diagonal matrix and V is a rotation matrix such that a 0 cos θ sin θ D= , V = 0 b sin θ cos θ
≡ DV , where D
−
(88)
Then, admissible 2-D Gabor function is introduced: g (x, y) = K exp
−
2
π a (x
− x )r
exp(2πj (u0 (x
2
2
+ b (y
0
−x )+v 0
0
(y
2
− y )r 0
exp( j (2π (u0 x0 + v0 y0 ) + P ))
− y ))) − exp 0
− π
u0 r 2 v0 r 2 + 2 a2 b
(89)
And its Fourier transform is: K gˆ (u, v) = exp( j ( 2π (x0 (u ab
−
−u )+y 0
0
− − − − exp
exp
11.2
0
− −
u0 r 2 v0 r 2 + a2 b2
π
− v )) + P ))
u0 )r 2 (v v0 )r 2 + a2 b2
(u
π
(v
exp
π
(90) u r 2 vr 2 + 2 a2 b
Another formula of the Gabor function
In other papers, you may see another formula representation of the Gabor function. For example, in most papers, x0 = y0 = 0, P = 0. Then, g (x, y) = K exp π a2 xr 2 + b2 yr 2
−
− − −
exp(2πj (u0 x + v0 y))
K gˆ (u, v) = ab
exp
exp
π
π
(u
− u )r 0
a2
2
− − − − exp
+
u0 r 2 v0 r 2 + a2 b2
(v
π
u0 r 2 v0 r 2 + 2 a2 b
(91)
v0 )r 2 b2
exp
π
u r 2 vr 2 + 2 a2 b
(92)
Moreover, a = b σ in some paper. The rotation angle has no effect (θ = 0) in this case. g (x, y) = K exp πσ 2 x2 + y 2 (93) π 2 2 exp(2πj (u0 x + v0 y)) exp u + v 0 0 σ2 K π gˆ (u, v) = 2 exp (u u0 )2 + (v v0 )2 2 σ σ (94) π π 2 2 2 2 exp u + v exp u + v 0 0 σ2 σ2
≡
− − − − − − − − −
Then if you restrict the magnitude of spatial frequency of the sinusoidal carrier F 0 to satisfy this equation: σ2 2 2 F 0 = u0 + v0 = (95) 2π the Gabor function will be g (x, y) = K exp
−
√
πσ 2 x2 + y 2
√ − − − − − − − − exp j
gˆ (u, v) =
2πσ 2 (x cos ω0 + y sin ω0 )
K exp σ2
exp
π (u σ2 σ2 2
2
u0 ) + (v
exp
σ2 2
2
v0 )
π u2 + v 2 σ2
exp
(96)
(97)
Finally if you use K = 2πσ 2 , g (x, y) = 2πσ 2 exp
πσ 2 x2 + y2
−
√ − − − − − − − − exp j
2πσ 2 (x cos ω0 + y sin ω0 )
gˆ (u, v) = 2π exp
exp
π (u σ2 σ2 2
2
u0 ) + (v
exp
exp
v0 )
σ2 2
(98)
2
π u2 + v 2 σ2
(99)
Additionally, you can use angular frequency (ν, ξ ) instead of (u, v). Then, g (x, y) = 2πσ 2 exp
πσ 2 x2 + y 2
−
− − − − − − − − exp( j (ν 0 x + ξ 0 y))
gˆ (u, v) = 2π
1 (ν 4πσ 2
exp
σ2 2
exp
σ2 2
exp
ν 0 )2 + (ξ
ξ 0 )2
1 ν 2 + ξ 2 4πσ 2
exp
(100)
(101)
In fact, angular frequency representation can be seen in many papers. So it may be useful to have the quite general Gabor function 10 in that format: g (x, y) = K exp
π a2 xr 2 + b2 yr 2
−
− − −
exp( j (ν 0 x + ξ 0 y))
K gˆ (u, v) = ab
exp
10
1 4π
exp
1 4π
(ν
− exp 0
ν 0 r 2 ξ 0 r 2 + 2 a2 b
Only x0 = y0 = 0, P = 0 are assumed.
1 4π
ν 0 r 2 ξ 0 r 2 + a2 b2
−
− ν )r a2
−
2
+
(ξ
exp
− ξ )r
(102)
2
0
b2
1 4π
ν r 2 ξ r 2 + 2 a2 b
(103)
12
History
• The first version of this document, which was 14 page long, was written by • • • •
Javier R. Movellan in 1996. The document was made open source under the GNU Free Documentation License 1.1 on August 6 2002, as part of the Kolmogorov Project. On September 3 2002 we added the changes made by Kenta Kawamoto. These included a 7 page Appendix with sections on the Fourier transform of the Gabor function, and an altenative formula for the Gabor function. October 9, 2003. Javier R. Movellan changed the license to GFDL 1.2 and included an endorsement section. Fall 2005. Georgios Britzolakis reported a bug on equation 60.
Javier R. Movellan
1
The The compl complex ex Gabor Gabor in in spac space e doma domain in
Here is the formula of a complex Gabor function in space domain g (x, y) = s(x, y) wr (x, y)
(1)
where s(x, y) is a complex sinusoidal, known as the carrier, and wr (x, y) is a 2-D Gaussian-shaped function, known as the envelop. 1.1
The complex complex sinusoid sinusoidal al carrier carrier
The complex sinusoidal is defined as follows,1 s(x, y) = exp( j exp( j (2π (2π (u0 x + v0 y ) + P ) P ))
(2)
where (u (u0 , v0 ) and P define the spatial frequency and the phase of the sinusoidal respectively. We can think of this sinusoidal as two separate real functions, conveniently allocated in the real and imaginary part of a complex function (see Figure 1).
Figure Figure 1: The real and imaginary imaginary parts parts of a com comple plex x sinuso sinusoida idal. l. The images images are 128 128 pixels. The parameters are: u0 = v0 = 1/80 cycles/pixel, cycles/pixel, P = 0 deg.
×
1
An offset constant parameter for s(x, y ) will be introduced later, to compensate the DC-component DC-component of this sinusoidal sinusoidal.. Refer to the appendix for detailed detailed explanation. explanation.
The real part and the imaginary part of this sinusoidal are Re (s(x, y)) = cos (2π(u0 x + v0 y) + P )
(3)
Im (s(x, y)) = sin (2π(u0 x + v0 y) + P )
The parameters u0 and v0 define the spatial frequency of the sinusoidal in Cartesian coordinates. This spatial frequency can also be expressed in polar coordinates as magnitude F 0 and direction ω0 : F 0
=
ω0
u0 2 + v0 2 v0 u0
(4)
= F 0 cos ω0
(5)
= tan
−1
i.e.
u0
v0 = F 0 sin ω0 Using this representation, the complex sinusoidal is s(x, y) = exp ( j (2πF 0 (x cos ω0 + y sin ω0 ) + P )) 1.2
(6)
The Gaussian envelop
The Gaussian envelop looks as follows (see Figure 2): wr (x, y) = K exp
−
π a2 (x
− x )r
2
0
+ b2 (y
− y )r
2
0
(7)
where (x0 , y0 ) is the peak of the function, a and b are scaling parameters 2 of the Gaussian, and the r subscript stands for a rotation operation3 such that (x
− x )r (y − y )r
1.3
0
= (x
− x )cos θ + (y − y )sin θ −(x − x )sin θ + (y − y )cos θ
0
=
0
0
0
(8)
0
The complex Gabor function
The complex Gabor function is defined by the following 9 parameters;
• • • • •
K
:
Scales the magnitude of the Gaussian envelop.
(a, b)
:
Scale the two axis of the Gaussian envelop.
θ
:
Rotation angle of the Gaussian envelop.
(x0 , y0 )
:
Location of the peak of the Gaussian envelop.
(u0 , v0 )
:
Spatial frequencies of the sinusoidal carrier in Cartesian coordinates. It can also be expressed in polar coordinates as ( F 0 , ω0 ).
•
P
:
Phase of the sinusoidal carrier.
Each complex Gabor consists of two functions in quadrature (out of phase by 90 degrees), conveniently located in the real and imaginary parts of a complex function. 2 3
Note that the Gaussian gets smaller in the space domain, if a and b get larger. This rotation is clockwise, the inverse of the counterclockwise rotation of the ellipse.
Figure 2: A Gaussian envelop. The image is 128 128 pixels. The parameters are as follows: x0 = y0 = 0. a = 1/50 pixels, b = 1/40 pixels, θ = 45 deg.
×
−
Figure 3: The real and imaginary parts of a complex Gabor function in space domain. The images are 128 128 pixels. The parameters are as follows: x0 = y0 = 0, a = 1/50 pixels, b = 1/40 pixels, θ = 45 deg, F 0 = 2/80 cycles/pixel, ω0 = 45 deg, P = 0 deg.
×
√
−
Now we have the complex Gabor function in space domain 4 (see Figure 3): g(x, y) =
K exp
−
π a2 (x
− x )r 0
2
+ b2 (y
exp( j (2π(u0 x + v0 y) + P ))
4
− y )r 0
2
(9)
In fact, there remains some DC component in this Gabor function. You have to compensate it to have the admissible Gabor function. Refer to the appendix.
Or in polar coordinates, g(x, y) =
K exp
−
π a2 (x
− x )r 0
2
+ b2 (y
− y )r 0
2
exp( j (2πF 0 (x cos ω0 + y sin ω0 ) + P ))
(10)
Figure 4: The Fourier transform of the Gabor filter. The peak response is at the spatial frequency of the complex sinusoidal: u p = v p = 1/80 cycles/pixel. The parameters are as follows: x0 = y0 = 0, a = 1/50 pixels, b = 1/40 pixels, θ = 45 deg, F 0 = 2/80 cycles/pixel, ω0 = 45 deg, P = 0 deg.
√
−
The 2-D Fourier transform of this Gabor 5 is as follows (see Figure 4): gˆ(u, v) =
K exp( j ( 2π (x0 (u ab
−
exp
−π
(u
− u )r 0
a2
−u )+y 0
2
+
(v
0
(v
− v )r 0
b2
2
− v )) + P )) 0
(11)
Or in polar coordinates,
5
Magnitude (ˆ g (u, v)) =
K exp ab
Phase(ˆ g (u, v)) =
−2π (x
0
2
−π (u −au )r + (v −bv )r (u − u ) + y (v − v )) + P 0
0
2
0
Refer to the appendix for detailed explanation.
0
2
0
2
(12)
2
Half-magnitude profile
The region of points, in frequency domain, with magnitude equal one-half the peak magnitude can be obtained as follows. Since the peak value is obtained for ( u, v) = (u0 , v0 ), and the peak magnitude is K/ab, we just need to find the set of points (u, v) with magnitude K/2ab. 1 K K = exp 2 ab ab or,
− − − (u
π
(u
− log 2 = −π
2
u 0 )r (v v0 )r + 2 a b2 2
−
u 0 )r (v v0 )r + a2 b2
−
2
2
(13)
(14)
or equivalently,
(u
− u )r 0
a C
2
(v
v0 )r b C
2
− +
where C =
=1 (15)
log2 = 0.46971864 π
≈ 0.5
Equation 15 is an ellipse centered at (u0 , v0 ) rotated with an angle θ with respect to the u axis. The main axis of the ellipse have length 2 a C a and 2 b C b respectively.
≈
≈
We will use the following convention: a is the length of the axis closer to ω0 , and b is the length of the axis perpendicular to the main axis 6 (See Figure 5).
(b)
(a) Theta F_0 Omega_0 Figure 5: Parameters of the Gabor kernel as reflected in the half-magnitude elliptic profile. Note that this is a figure in frequency domain.
3
Half-magnitude frequency and orientation bandwidths
Frequency and orientation bandwidths of neurons are commonly measured in terms of the half-magnitude responses. Let u0 , v0 the preferred spatial frequency of a neuron. In polar coordinates this spatial frequency can be expressed as F 0 and ω0 . 6
More precisely a and b are 1.06 times the length of the respective axis
To find the half-magnitude frequency bandwidth, we probe the neuron with sinusoidal images of orientation ω0 and different spatial frequency magnitudes F . We increase F with respect to F 0 until the magnitude of the response is half the magnitude at (F 0 , ω0 ). Let’s call that value F max . We then decrease F with respect to F 0 until the magnitude of the response is half the response at ( F 0 , ω0). Call that F min . Half-magnitude frequency bandwidth is defined as follows: ∆F 1/2 = F max
− F
(16)
min
or, when measured in octaves,7 ∆F 1/2 = log2 (F max /F min )
(17)
Half-magnitude orientation bandwidth is obtained following the same procedure but playing with the orientation ω instead of the frequency magnitude F . ∆ω1/2 = ωmax In Gabor functions with θ0 follows (See Figure 6)
≈ω
0
−ω
(18)
min
the frequency bandwidth can be obtained as
∆F 1/2 = 2 a C
≈a
(19)
and the orientation bandwidth can be approximated as follows (see Figure 6) ∆ω1/2
≈ 2tan
−1
b C F 0
(20)
0.5 Delta W ~ arctan( b C / F o)
bC
Delta F = aC Fo
Figure 6: A half-magnitude profile and its relationship to the orientation and frequency bandwidths. 7
Octave is a unit used for showng the ratio, as an index of 2. k octaves = 2k × 100.0%
4
Effective spread and rms spread
The rms (which stands for root mean squares) length, rms width, and rms area of a 2-D function are defined in terms of their first and second moments: The moments of a complex function g(x, y) are defined by converting the function into a probability density (which must be always positive and integrates to 1.0) and then calculating the standard first and second moments. A common way to achieve this is as follows: From the function g(x, y) we construct the following probability density f (x, y) =
1 2 g(x, y) Z
|
(21)
|
2
where g(x, y) is the magnitude of the signal, which is always positive, and Z guarantees that f (x, y) integrates to 1.0, i.e.
|
|
+∞
Z =
+∞
−∞
−∞
2
|g(x, y)| dxdy
(22)
Once we have defined a probability density function, the standard statistical measures of location and scale follow.
2
µX = E X (x) =
f (x, y) xdxdy
f (x, y) (x
σX = E X (x
− µX )
2
=
(23) 2
− µX )
dxdy
(24)
2 with similar equations for µY and σY .
µY = E Y (y) =
f (x, y) y dxdy
f (x, y) (y
2 σY = E Y (y
And, σXY = E XY ((x
2
− µY )
=
− µX ) (y − µY )) =
f (x, y) (x
− µY )
(25) 2
dxdy
(26)
− µX ) (y − µY ) dxdy
(27)
The rms width and length are defined as the σX and σY of a rotated version of f (x, y) so that the covariance σXY of the rotated distribution be zero. Let X r , Y r represent the rotated variables for which the covariance is zero, the rms length and width are ∆X rms = ∆Y rms =
2 σX
(28)
2 σY
(29)
r
r
Similar definitions can be obtained also in the frequency domain, by working with the Fourier transform of the original complex function. ∆U rms = ∆V rms =
2 σU
(30)
2 σV
(31)
r
r
The rms area in the space and frequency domains are defined as follows: Area(XY )rms = (∆X rms ) (∆Y rms )
(32)
Area(U V )rms = (∆U rms ) (∆V rms )
(33)
Some papers work with what are known as effective length, width and areas. They are simply the rms measures multiplied by 2π
√
∆X eff =
√
2π ∆X rms
(34)
and so on. It can be shown that the following relationships hold on any 2D function with finite moments 1 (∆X rms ) (∆U rms ) (35) 4π 1 (∆Y rms ) (∆V rms ) (36) 4π and 1 Area(XY )rms Area(U V )rms (37) 16π2
≥ ≥
≥
It is easy to verify that the Gabor complex function achieves the lower limits of the uncertainty relations. For a given area in the space domain it provides the maximum possible resolution in the frequency domain, and vice-versa. It can be shown that the rms width and lengths of Gabor functions are as follows: a (38) ∆U rms = 2 π
√
∆V rms =
b √ 2 π
(39)
Tow see why, simply consider that the probability density associated with the Ga1 2 2 bor function f (x, y) = g(x, y) is Gaussian with variances equal to ∆U rms and Z 2 ∆V rms .
|
|
Moreover, from the uncertainty relations,
5
1 2a π 1 = 2b π
∆X rms =
√
(40)
∆Y rms
√
(41)
Gabor functions as models of simple cell receptive fields
? showed that the real part of complex Gabor functions fit very well the receptive
field weight functions found in simple cells in cat striate cortex. Here are some of the constraints that can be found in the literature:
• The orientation of the Gaussian ω
is rougly equivalent to the orientation of the carrier. θ0 = ω0 . The absolute deviations, “the twist” have a Median of about 10 degrees (see ?, p. 1249). 0
8
8
It’s easier to consider in the frequency domain. Note that the long axis in the frequency domain becoms the short axis in the space domain. Don’t get confused!
• In macaque V1, most cells have a spatial frequency bandwidth between 1 and 1.5 octaves. The median is about 1.4 octaves (see ?, p. 551).
• Orientation bandwidths are typically about 40 degrees (mean = 65 degrees, median = 42 degrees, mode = 30 degrees) (see ?, p. 535 and 541).
• From the three constraints mentioned above, it may be derived (see equqtion 59) that the aspect ratio is typically about 1.24, a/b ≈ 1.24. • In the area mapping the fovea, there are more kernels oriented vertically i.e.
and horizontally than oriented diagonally (about 3 to 2). (see ?, p. 537).
• Pairs of adjacent simple cells in the visual cortex of the cat are in quadrature (?). We can then put these two cells in the real and imaginary parts of a complex function and treat them as a complex Gabor receptive field.
6
Gabor functions for spatial frequency filtering
Consider a massive set of simple cell neurons with Gabor kernel functions with equal parameters except for the location parameters ( x0 , y0 ). Let all these neurons be distributed uniformly about the foveal field. Each point in the foveal field contains at least two neurons in quadrature. We can model the operation of such a set of neurons as a convolution operation (assuming a continuous and uniform distribution of filters in all the foveal locations). Since convolution in space domain is product in frequency domain, the set of Gabor functions work as bandpass frequency filters of the foveal image. The peak frequency is controlled by the spatial frequency of the sinusoidal carrier (u0 , v0 ). The half-magnitude region is controlled by the rotation θ and scale parameters a, b, of the Gaussian envelop.
7
Energy filtering
A quadrature pair (or a Hilbert Transform pair) is a set of two linear operators with the same amplitude response but phase reponses shifted by 90 degrees. Strictly speaking sine and cosine Gabor operators are not quadrature pairs because cosine phase Gabors have some DC response, whereas sine gabors do not. However, one can have quadrature Gabor pairs that look very much like sine/cosine pairs. Thus the sine and cosine Gabor pair is commonly refered to as a quadrature pair. A system that sums the square of the outputs of a quadrature pair is called an energy mechanism ( ?). Energy mechanisms have unmodulated responses to drifting sinusoidals. Complex cells in V1 are commonly modeled as energy mechanisms since they are unmodulated by drifting sinusoidals. Simple cells respond to a drifting sinusoidal with a half-wave rectified analog of the signal, suggesting that the cells are linear up to rectification. Complex cells respond to a drifting sinusoidal in an unmodulated way, as a maintained discharge. ? showed that complex receptive fields are composed of subunits. The subunits of model complex cells are model simple cells with identical amplitude response. ? have shown that behavior of complex cell to stimuli made of paris of bars flashed in sequence is consist with an energy mechanism.
8
Contrast Normalization
? have shown that stimuli presented at orientations orthogonal to the optimal orientation inhibit simple cells activity. (?) have shown similar inhibitory effect
between frequency bands. These inhibitory effects may play a serve as a gain control (or contrast normalization) mechanism. ? proposes the following model of gain control in complex cells: The amplitude response of each energy mechanism is divided by the total energy at all orientations and nearby spatial frequencies: ¯i = E
E i κ+
j
(42)
E j
where κ is a positive constant to avoid zero denominators.
9
Functional Interpretations
Section in preparation:
• minimizes number of neurons needed to achieve a desired frequency resolution.
• spatially and frequency localized. • matched to “logons” likely to occur in images. • for natural images the Gabor representation is more sparse than the
δ
(pixel) representation and than the DOG representation.
10
Constructing an idealized V1
We’ll use the following simplified assumptions. They all have empirical support.
• The orientation of the complex sinusoid carrier and the Gaussian envelop are the same: ω0 = θ.
• The half-magnitude frequency bandwidth is constant for all neurons when measured in octaves.
• The half-magnitude orientation bandwidth is constant for all neurons. • The upper half-magnitude contour of a frequency band coincides with the
lower contour of the next frequency band. This is just a constrain we use for convenience.
From these assumptions above, we can derive the relationship between the parameters F 0 , a and b. From equations 17 and 19, we know that the frequency bandwidth in octaves is ∆F 1/2 = log2
F 0 + a C F 0 a C
−
where C =
(43)
log2 = 0.46971864 π
≈ 0.5
Thus, K a C
(44)
2∆F 1 2∆F + 1
(45)
a = F 0 where K a =
−
With respect to the orientation bandwidth, equation 20 tells us that tan
1 ∆ω 2
=
b = F 0
K b C
b C F 0
(46)
Thus,
where K b = tan
(47)
1 ∆ω 2
(48)
Therefore, in this model the aspect ratio of a and b is constant: λ=
a K a = b K b
(49)
Moreover, from equation 19 1 ∆F = a C = F 0 K a 2
(50)
We can now locate our frequency peaks such that the upper half-magnitude contour of one channel coincides with the lower half-magnitude contour of the the next channel. Let µi signify the peak frequency of the ith band, We know F max for band i is 1 i F max = µi + ∆F i = µi + µi K a = µi (1 + K a ) 2
(51)
and F min for band i + 1 is i+1 F max = µi+1
− 12 ∆F i
+1
= µi+1
− µi
+1
K a = µi+1 (1
− K a)
(52)
We want these two values to coincide, therefore µi+1 = µi
1 + K a 1 K a
−
(53)
Thus, the peak frequencies follow a geometric series µi = µ1 Ri−1
(54)
1 + K a 1 K a
(55)
where R= 10.1
−
Example
If we use the standard values for simple cell median bandwidths from macaque striate cortex:
• ∆F = 1.4 octaves. • ∆ω = 40 degrees.
Then,
2∆F 1 = 0.45040 2∆F + 1 1 K b = tan ∆ω = 0.36397 2 K a a = µi = 0.9589 µi C a K a λ= = = 1.23746 b K b a b = = 1.1866 µi λ 1 + K a R= = 2.6390 1 K a
−
(56)
(57)
K a =
(58) (59) (60) (61)
−
Suppose we want three frequency bands and we want the F 0 of the third band to be 0.25. Then, 0.25 µ1 = = 0.03589 (62) 2.63902 and 1 ∆F 1 = K a µ1 = 0.01617 (63) 2 Thus, the half magnitude interval 9 is (0.01973, 0.05207) The second band peaks at µ2 = µ1 R = 0.09473
(64)
and
1 ∆F 2 = K a µ2 = 0.04267 2 Thus, the half magnitude interval is (0.05207, 0.1374)
(65)
Finally, the third band peaks at µ3 = µ2 R = 0.2500
(66)
and
1 ∆F 3 = K a µ3 = 0.1126 2 Thus, the half magnitude interval is (0.1374, 0.3626)
(67)
These three Gabors cover the frequency bands of (0 .01973, 0.3626)
11
Appendix
11.1 11.1.1
Fourier transform of the Gabor function The integral of the Gaussian
Let I denote the integral ∞
I
exp
−∞
9
∞
− ≡ − 2
πx
dx =
exp
π y 2 dy
(68)
−∞
The half magnitude interval here is the frequency coverage of that Gabor in terms of 1 1 half-magnitude profile: µi − ∆F i , µi + ∆F i 2 2
Then,
∞
2
I
=
∞
− − − − − − π x2 dx
exp
−∞
−∞
∞
=
exp
∞
π x2 + y 2
exp
−∞
−∞
π y 2 dy
−
(69)
dxdy
Using a polar coordinates r and ϕ, this equation can be rewritten as: 2π
∞
I 2
=
π r2 rdrdϕ
exp
0
0
∞
=
π r2 dr
2π r exp
0
=
π r2
exp
∞
(70)
0
=1
Accordingly,
∞
I =
−
π x2 dx = 1
exp
(71)
−∞
11.1.2
Fourier transform of the Gaussian
The Fourier transform of the simple 1-D Gaussian is ∞
exp( πx2 ) exp( 2πjfx) dx
−∞
−
−
∞
− − − − − − − − − − − =
exp
2
π(x + jf )
πf 2 dx
(72)
−∞
∞
πf 2
= exp
πx
exp
2
dx
(x
−∞
≡ x + jf )
πf 2
= exp
In the same way, the Fourier transform of the simple 2-D Gaussian is ∞
∞
exp
−∞
π(x2 + y 2 ) exp( 2πjux) exp( 2πjvy) dxdy
−∞ ∞
∞
2
=
exp( πy 2 ) exp( 2πjvy) dy
exp( πx ) exp( 2πjux) dx
−∞
−∞
−
(73)
= exp( πu2 ) exp( πv 2 )
−
= exp
−
π(u2 + v 2 )
− − −
and so on. More generally, ∞
exp
−∞
π xT x exp
2πj uT x dx = exp
π uT u
−
(74)
That is, the Fourier transform of an N-dimensional Gaussian is also an Ndimensional Gaussian.
11.1.3
Several properties of the Fourier transform
Several properties of the Fourier transform will be introduced in this section. Let ˆ denote the Fourier transform of a function f (x) , i.e. f (u) ∞
−
f (x) exp
−∞
Then, ∞
f (x
−∞
= exp = exp
= exp and also, ∞
− x ) exp −2πj uT x
0
− − −
T
2πj u x0
0
−
f (x ) exp
−∞
0
T
2πj u x
dx (76)
dx
(x
≡x−x ) 0
ˆ 2πj x0 T u f (u)
−
2πj uT x dx
2πj uT A−1 Ax dx
− −
f (Ax) exp
−∞ ∞
f (x ) exp
−∞
11.1.4
(75)
− x ) exp −2πj uT (x − x )
2πj x0 u
∞
=
f (x
∞
T
−∞
=
dx
−∞
f (Ax) exp =
∞
ˆ 2πj uT x dx = f (u)
1 ˆ −T f A u A
T
−
2πj A
u
T
x
(77)
1 dx A
(x
≡ Ax)
Fourier transform of the Gabor function
Here, we have the simple Gaussian envelop and sinusoidal carrier: w (x) = exp π xT x
−
(78)
T
s (x) = exp j 2π u0 x + P Now we want a Fourier transform of the following Gabor function, g (x) = K w (A (x x0 )) (s (x) C )
−
= K w (A (x where K and C are constants.
− x )) 0
−
exp j 2π u0 T x + P
(79)
− C
Given this function, its Fourier transform will be ∞
gˆ (u) =
g (x) exp
−∞ ∞
= K exp( jP )
−
2πj uT x dx
w (A (x
−∞
0
∞
− KC
−∞
w (A (x
0
− x )) exp −2πj uT x 0
− x )) exp −2πj (u − u )T x
dx
dx
(80)
Using the previously introduced properties, we can see gˆ (u) = K exp( jP ) exp
−
2πj x0 T (u
− u ) A1 wˆ 0
− KC exp −2πj x T u A1 wˆ K T = A exp −2πj x (u − u )
0
0
0
exp( jP ) w ˆ A−T (u
A−T u
A−T (u
0
(81)
− u ) − C exp −2πj x T u 0
−u )
0
0
w ˆ A−T u
Remember that C is a constant that compensates the DC component of the Gabor function, which means gˆ (0) must be equal to 0. Therefore, C = exp j 2π x0 T u0 + P
then
T
exp 2πj u0 (x
gˆ (u) =
K exp j A T
−
w ˆ A
w ˆ
(82)
x0 )) exp j 2π u0 T x0 + P
g (x) = K w (A (x
and
A−T u0
− − − − − − − − − − − − − − − − − − − − − − − − − − x0 )
2π x0 T (u
(u
u0 )
w ˆ
T
−
w ˆ
A
(83)
u0
u0 ) + P T
−
A
T
−
u0 w ˆ A
(84)
u
Using the following definitions:
we can get
g (x) = K exp
w (x) = exp
π xT x
w ˆ (u) = exp
π uT u
T
x0 ) AT A (x
π (x
exp 2πj u0 T (x
and
gˆ (u) =
K exp j A exp
x0 ) exp j 2π u0 T x0 + P
exp
2π x0 T (u
π (u
exp
x0 )
(85)
(86)
π u0 T A−1 A−T u0
u0 ) + P
T
u0 ) A−1 A−T (u
π u0 T A−1 A−T u0 exp
u0 )
π uT A−1 A−T u
(87)
In order to have the generalized 2-D Gabor function, consider A is a diagonal matrix and V is a rotation matrix such that a 0 cos θ sin θ D= , V = 0 b sin θ cos θ
≡ DV , where D
−
(88)
Then, admissible 2-D Gabor function is introduced: g (x, y) = K exp
−
2
π a (x
− x )r
exp(2πj (u0 (x
2
2
+ b (y
0
−x )+v 0
0
(y
2
− y )r 0
exp( j (2π (u0 x0 + v0 y0 ) + P ))
− y ))) − exp 0
− π
u0 r 2 v0 r 2 + 2 a2 b
(89)
And its Fourier transform is: K gˆ (u, v) = exp( j ( 2π (x0 (u ab
−
−u )+y 0
0
− − − − exp
exp
11.2
0
− −
u0 r 2 v0 r 2 + a2 b2
π
− v )) + P ))
u0 )r 2 (v v0 )r 2 + a2 b2
(u
π
(v
exp
π
(90) u r 2 vr 2 + 2 a2 b
Another formula of the Gabor function
In other papers, you may see another formula representation of the Gabor function. For example, in most papers, x0 = y0 = 0, P = 0. Then, g (x, y) = K exp π a2 xr 2 + b2 yr 2
−
− − −
exp(2πj (u0 x + v0 y))
K gˆ (u, v) = ab
exp
exp
π
π
(u
− u )r 0
a2
2
− − − − exp
+
u0 r 2 v0 r 2 + a2 b2
(v
π
u0 r 2 v0 r 2 + 2 a2 b
(91)
v0 )r 2 b2
exp
π
u r 2 vr 2 + 2 a2 b
(92)
Moreover, a = b σ in some paper. The rotation angle has no effect (θ = 0) in this case. g (x, y) = K exp πσ 2 x2 + y 2 (93) π 2 2 exp(2πj (u0 x + v0 y)) exp u + v 0 0 σ2 K π gˆ (u, v) = 2 exp (u u0 )2 + (v v0 )2 2 σ σ (94) π π 2 2 2 2 exp u + v exp u + v 0 0 σ2 σ2
≡
− − − − − − − − −
Then if you restrict the magnitude of spatial frequency of the sinusoidal carrier F 0 to satisfy this equation: σ2 2 2 F 0 = u0 + v0 = (95) 2π the Gabor function will be g (x, y) = K exp
−
√
πσ 2 x2 + y 2
√ − − − − − − − − exp j
gˆ (u, v) =
2πσ 2 (x cos ω0 + y sin ω0 )
K exp σ2
exp
π (u σ2 σ2 2
2
u0 ) + (v
exp
σ2 2
2
v0 )
π u2 + v 2 σ2
exp
(96)
(97)
Finally if you use K = 2πσ 2 , g (x, y) = 2πσ 2 exp
πσ 2 x2 + y2
−
√ − − − − − − − − exp j
2πσ 2 (x cos ω0 + y sin ω0 )
gˆ (u, v) = 2π exp
exp
π (u σ2 σ2 2
2
u0 ) + (v
exp
exp
v0 )
σ2 2
(98)
2
π u2 + v 2 σ2
(99)
Additionally, you can use angular frequency (ν, ξ ) instead of (u, v). Then, g (x, y) = 2πσ 2 exp
πσ 2 x2 + y 2
−
− − − − − − − − exp( j (ν 0 x + ξ 0 y))
gˆ (u, v) = 2π
1 (ν 4πσ 2
exp
σ2 2
exp
σ2 2
exp
ν 0 )2 + (ξ
ξ 0 )2
1 ν 2 + ξ 2 4πσ 2
exp
(100)
(101)
In fact, angular frequency representation can be seen in many papers. So it may be useful to have the quite general Gabor function 10 in that format: g (x, y) = K exp
π a2 xr 2 + b2 yr 2
−
− − −
exp( j (ν 0 x + ξ 0 y))
K gˆ (u, v) = ab
exp
10
1 4π
exp
1 4π
(ν
− exp 0
ν 0 r 2 ξ 0 r 2 + 2 a2 b
Only x0 = y0 = 0, P = 0 are assumed.
1 4π
ν 0 r 2 ξ 0 r 2 + a2 b2
−
− ν )r a2
−
2
+
(ξ
exp
− ξ )r
(102)
2
0
b2
1 4π
ν r 2 ξ r 2 + 2 a2 b
(103)
12
History
• The first version of this document, which was 14 page long, was written by • • • •
Javier R. Movellan in 1996. The document was made open source under the GNU Free Documentation License 1.1 on August 6 2002, as part of the Kolmogorov Project. On September 3 2002 we added the changes made by Kenta Kawamoto. These included a 7 page Appendix with sections on the Fourier transform of the Gabor function, and an altenative formula for the Gabor function. October 9, 2003. Javier R. Movellan changed the license to GFDL 1.2 and included an endorsement section. Fall 2005. Georgios Britzolakis reported a bug on equation 60.
