Suppose that arrows are shot at a target and land at random
distances from the target centre. The horizontal and vertical
components of these distances are formed into a 2-D random
error vector. If each component of this error vector is an
independent variable with zero-mean Gaussian pdf of variance
σ2
σ
2
, calculate the pdf's of the radial magnitude and the
phase angle of the error vector.
Let the error vector be
X1
X1
and
X2
X2
each have a zero-mean Gaussian pdf given by
fx=12πσ2e−x22σ2
f
x
1
2
σ
2
x
2
2
σ
2
(3)
Since
X1
X1
and
X2
X2
are independent, the 2-D pdf of
XX is
fX
x1
x2
=f
x1
f
x2
=12πσ2e−
x1
2+
x2
22σ2
fX
x1
x2
f
x1
f
x2
1
2
σ
2
x1
2
x2
2
2
σ
2
(4)
In polar coordinates
x1
=rcosθ
x1
r
θ
and
x2
=rsinθ
x2
r
θ
To calculate the radial pdf, we substitute
r=
x1
2+
x2
2
r
x1
2
x2
2
in the above 2-D pdf to get:
Prr<R<r+δr=∫rr+δr∫−ππ
fX
x1
x2
RdθdR
r
R
r
δ
r
R
r
r
δ
r
θ
fX
x1
x2
R
(5)
where
∫rr+δr∫−ππ
fX
x1
x2
RdθdR≃δr∫−ππ12πσ2e−r22σ2rdθ=1σ2re−r22σ2δr
R
r
r
δ
r
θ
fX
x1
x2
R
δ
r
θ
1
2
σ
2
r
2
2
σ
2
r
1
σ
2
r
r
2
2
σ
2
δ
r
Hence the radial pdf of the error vector is:
fR
r=limit
δr
→
0
Prr<R<r+δrδr=1σ2re−r22σ2
fR
r
δ
r
0
r
R
r
δ
r
δ
r
1
σ
2
r
r
2
2
σ
2
(6)
This is a
Rayleigh distribution with variance =
2σ2
2
σ
2
(these are two components of
XX, each with variance
σ2
σ
2
).
The 2-D pdf of XX
depends only on rr and not on
θθ, so the angular pdf of
the error vector is constant over any
2π
2
interval and is therefore
fΘ
θ=12π
fΘ
θ
1
2
so that
∫−ππ
fΘ
θdθ=1
θ
fΘ
θ
1