APPENDIX D: FM THRESHOLDING AND RELATED ANALYSES 1

Problem D.1

a. Consider

z (t) = A cos (! 0 + ! d ) t + n (t) ; ! 0 = 2 f0 ; ! d = 2 fd

= A cos (! 0 + ! d ) t + nc (t) cos ! 0 t ns (t) sin ! 0 t

with the PSD of n (t) given by Sn (t) = N0 =2 for jf f0 j B=2 and 0 otherwise. Note
also that the PSD of the quadrature noise components is

N0 ; jf j B=2
Snc (f ) = Sns (f ) =
0; otherwise

Also, the quadrature noise components are uncorrelated. It follows that

n0c (t) cos (! 0 + ! d ) t n0s (t) sin (! 0 + ! d ) t = n0c (t) [cos ! 0 t cos ! d t sin ! 0 t sin ! d t]
0
ns (t) [sin ! 0 t cos ! d t + cos ! 0 t sin ! d t]
= n0c (t) cos ! d t n0s (t) sin ! d t cos ! 0 t
n0c (t) sin ! d t + n0s (t) cos ! d t sin ! 0 t
nc (t) cos ! 0 t ns (t) sin ! 0 t

Therefore

nc (t) = n0c (t) cos ! d t n0s (t) sin ! d t

ns (t) = n0c (t) sin ! d t + n0s (t) cos ! d t

These two equations can be solved for n0c (t) and n0s (t). The results are

n0c (t) = nc (t) cos ! d t + ns (t) sin ! d t

n0s (t) = nc (t) sin ! d t + ns (t) cos ! d t

To …nd the PSDs of n0c (t) and n0s (t) we …rst …nd their autocorrelation functions. By
de…nition,

[nc (t) cos ! d t + ns (t) sin ! d t]

Rn0c ( ) = E
[nc (t + ) cos ! d (t + ) + ns (t + ) sin ! d (t + )]
= Rc ( ) cos ! d t cos ! d (t + ) + Rs ( ) sin ! d t sin ! d (t + )

Noting that Rc ( ) = Rs ( ) and using a trig identity, we …nd that

Rn0c ( ) = Rc ( ) cos ! d
APPENDIX D: FM THRESHOLDING AND RELATED ANALYSES 2

Similarly, it can be shown that

Rn0s ( ) = Rs ( ) cos ! d

The PSDs follow by applying the modulation theorem:

1 1
Sn0c (f ) = Sn0s (f ) = Snc (f + fd ) + Snc (f fd )
8 2 2
< N0 ; jf j B=2 fd
= N0 =2; B=2 fd < jf j B=2 + fd
:
0; otherwise

b. To …nd the cross-spectral density of n0c (t) and n0s (t), we …rst …nd their cross-correlation
function:

Rn0c n0s ( ) = E n0c (t) n0s (t + )

[nc (t) cos ! d t + ns (t) sin ! d t]
= E
[ nc (t + ) sin ! d (t + ) + ns (t + ) cos ! d (t + )]
= Rc ( ) cos ! d t sin ! d (t + ) + Rs ( ) sin ! d t cos ! d (t + )

Again noting that Rc ( ) = Rs ( ) and using a trig identity, we …nd that

exp (j! d ) exp ( j! d )

Rn0c n0s ( ) = Rs ( ) sin ! d = Rs ( )
2j

Application of the frequency translation theorem results in the following for the cross-
spectral density:
j
Sn0c n0s (f ) = [Sns (f fd ) Sns (f + fd )]
2
8
< jN0 =2; B=2 fd f B=2 + fd
= jN0 =2; (B=2 + fd ) f (B=2 fd )
:
0; otherwise

For fd 6= 0 this cross-PSD is not identically zero as a sketch will show. Thus n0c (t) and
n0s (t) are not uncorrelated. However, they are when sampled at the same instant because
Rn0c n0s (0) = 0. Therefore, they are statistically independent when sampled at the same
instant since they are jointly Gaussian.

Problem D.2
The derivation here follows S. O. Rice, "Noise in FM Receivers," Proc. of the Symp. of
Time Series Analysis, M. Rosenblatt, ed., New York: Wiley (1963), pp. 395-422.
APPENDIX D: FM THRESHOLDING AND RELATED ANALYSES 3

a. We need the additional clicks due to modulation. Thus, consider

z (t) = A cos (! 0 + ! d ) t + nc (t) cos ! 0 t ns (t) sin ! 0 t

= A cos (! 0 + ! d ) t + nc (t) cos (! 0 + ! d ) t n0s (t) sin (! 0
0
+ !d) t
= A+ n0c (t) cos (! 0 + ! d ) t n0s (t) sin (! 0 + ! d ) t
, R (t) cos [(! 0 + ! d ) t + (t)]

where q
n0s (t)
R (t) = [A + n0c (t)]2 + [n0s (t)]2 and tan (t) =
A + n0c (t)
It follows from the cross-correlation function found in Problem D.1 that

Rn0c n0s (0) = E n0c (t) n0s (t) = 0

From an appropriate sketch, it follows that the probability of a counter-clockwise origin

encirclement in a small time interval is

Pcc = Pr A + n0c (t) < 0 and n0s (t) undergoes a + to zero crossing in [0; ]

The condition for n0s (t) to undergo a + to zero crossing in [0; ], from an appropriate
0 (t) dn0s (t)
sketch, is that n0s (0) > 0 and that n0s (0) + dndt
s
< 0 for all dt < 0. Therefore
t=0 t=0

dn0s (t) dn0s (t)

Pcc = Pr A + n0c (0) < 0; n0s (0) > 0; n0s (0) + < 0; <0
dt t=0 dt t=0

Thus we need the joint pdf of

dn0s (t)
X , n0c (0) ; Y , n0s (0) ; and Z ,
dt t=0

These are jointly Gaussian random variables (the derivative is a linear operation). The
APPENDIX D: FM THRESHOLDING AND RELATED ANALYSES 4

following expectations hold:

E [X] = E [Y ] = E [Z] = 0
Z 1 Z 1
E X 2
= E Y =2
Sn0c (f ) df = Sn0s (f ) df = N0 B , a
1 1
Z 1
E Z2 = jj2 f j2 Sn0s (f ) df
1
1 2
= N0 B 3 + 4 2 fd2 N0 B , b + a! 2d
3
E [XY ] = E [Y Z] = 0
dRn0c n0s ( ) d ( Rs ( ) sin ! d )
E [XZ] = = = a! d
d =0 d =0
dn0s (t) dns (t)
E [Z j X] = E j n0c (0) = E ! d nc (0) + j nc (0)
dt t=0 dt t=0
= ! d E [nc (0) j nc (0)] = ! d X

Therefore, the joint pdf of X; Y; Z is

fXY Z (x; y; z) = fZjXY (z j x; y) fX (x) fY (y)

= fZjX (z j x) fX (x) fY (y)
( " #)
1 x2 + y 2 (z + a! d )2
= p exp +
(2 )3=2 a b 2a 2b
APPENDIX D: FM THRESHOLDING AND RELATED ANALYSES 5

Thus,

Pcc = Pr [X < A; 0 < Y < Z ; all Z < 0]

Z AZ 0 Z z Z AZ 0 Z z
= fXY Z (x; y; z) dydzdx = fZj X (z j x) fX (x) fY (y) dydzdx
1 1 0 1 1 0
Z AZ 0 Z z ( " #)
1 x2 + y 2 (z + x! d )2
= p exp + dydzdx
1 1 0 (2 )3=2 a b 2a 2b
8 " # 9
Z A Z (z + x! d )2 =2b Z z exp y 2 =2a
exp x2 =2a < 0 exp =
= p p p dy dz dx; << 1
1 2 a : 1 2 b 0 2 a ;
8 9
Z A Z (z + x! d )2 =2b
exp x2 =2a < 0 z exp = x z
= p p p dz dx; = p ; & = p
1 2 a : 1 2 a 2 b ; a b
r Z 2 3
2 Z 1 exp (& + )2 =2 r
b 1 exp =2
4 5 a A2
= p p & p d& d ; = ! d ; =
2 a p2 2 0 2 b 2a
r Z "Z #
2
b 1 exp =2 1 exp u2 =2
= p p (u ) p du d ; u = & +
2 a p2 2 2
r Z "Z Z 1 #
2
b 1 exp =2 1 exp u2 =2 exp u2 =2
= p p u p du p du d
2 a p2 2 2 2
r Z " #
2 2 2 =2
b 1 exp =2 exp
= p p p Q( ) d
2 a p2 2 2
r (Z Z 1 )
b 1 exp 1 + 2 2 =2 p
2
= p
p d p
Q ( ) exp =2 d ; v = 1+ 2
2 a 2 2 2
r Z
b 1 p 1
2
= p Q 2 (1 + 2 ) p
Q ( ) exp d
2 a 1+ 2 2

To integrate the last integral, consider

Z 1 Z 1
xQ (Cx) exp 2
x =2 dx = uvj1
B vdu
B B

integrated by parts with

Z 1 exp t2 =2 exp C 2 x2
u = Q (Cx) = p dt; du = C p dx
Cx 2 2
2
dv = x exp x =2 dx; v = exp x2 =2
APPENDIX D: FM THRESHOLDING AND RELATED ANALYSES 6

Thus, by parts
Z 1 Z
1
x2 x2 1
x2 exp C 2 x2 =2
xQ (Cx) exp dx = Q (Cx) exp C exp p dx
D 2 2 B D 2 2
Z
D2 1 exp 1 + C 2 x2 =2
= Q (CD) exp C p dx
2 D 2
Z 1
D2 C exp w2 =2
= Q (CD) exp p p p dw
2 1 + C 2 D 1+C 2 2
D2 C p
= Q (CD) exp p Q D 1 + C2
2 1 + C2
p
Therefore, Pcc with the above integral substituted with C = and D = 2 , becomes
r Z 1
b 1 p 2
Pcc = p Q 2 (1 + ) 2 Q ( ) exp d
2 a p
1+ 2 2
r 8 < 1
p
2)
9
=
b p
1+ 2
Q 2 (1 +
= h p p i
2 a: Q 2 exp ( ) p1+ 2 Q 2 (1 + 2 ) ;
r
b 1+ 2 p p
= p Q 2 (1 + 2 ) Q 2 exp ( )
2 a 1+ 2
np p p o
= 1 + 2Q 2 (1 + 2 ) Q 2 exp ( )

where
!d fd
= =
2
r s
1 2 3
1 b 1 3 ( N0 B ) B
= = = p
2 a 2 N0 B 2 3
A 2 A 2
= =
2a 2N0 B

b. It follows that the probability of a clockwise origin encirclement is given by this

APPENDIX D: FM THRESHOLDING AND RELATED ANALYSES 7

expression but with fd replaced by fd ( = ). Taking this into account,

Pc p p p
= 1 + 2Q 2 (1 + 2 ) + fd Q 2 exp ( )
p p h p i
= 1 + 2Q 2 (1 + 2 ) + fd 1 Q 2 exp ( )
p p p
= 1 + 2Q 2 (1 + 2 ) fd Q 2 exp ( ) + fd exp ( )
Pcc
= + fd exp ( )
Pcc
> ; fd > 0

Therefore
1
mod = (Pc + Pcc )
Pcc A2
= 2 + fd exp ; fd > 0
2N0 B
To make this applicable to square wave modulation, replace fd by jfd j to get
h p p p i
sw mod = 2 1 + 2Q 2 (1 + 2 ) fd Q 2 exp ( ) + jfd j exp ( )
, +
where
h p p p i
= 2 1+ 2Q 2 (1 + 2) fd Q 2 exp ( ) (reduces to (D.23) for fd = 0)
2 s 0vu !1 3
fd 2 u fd 2
fd p
= 24 1+ Q t2
@ 1+ A fd Q 2 exp ( )5

2 s 0v
u !1 3
B fd 2 u fd 2
fd p
= 24 p 1 + 12 Q @t2 1 + 12 A fd Q 24 exp ( )5
2 3 B B B

2fd p p
= 2 p Q 8 fd Q 6 exp ( ) ; B = 2fd
3
0 s 1 0s 1
4fd 2
A A 2
A A A2 A2
= p Q @2 2fd Q @ 3 exp ; =
3 N0 B N0 B 2N0 B 2N0 B

A2
= jfd j exp
2N0 B
Typical results for v and are shown below versus SNR.
APPENDIX D: FM THRESHOLDING AND RELATED ANALYSES 8

Average encirclements per second, ν and δν, for fd = 5 Hz

0
10

-2
10

-4
10
ν, δν

-6
10

-8
10

ν
-10
δν
10
-15 -10 -5 0 5 10
A2/(2N0B) in dB
APPENDIX D: FM THRESHOLDING AND RELATED ANALYSES 9

Average encirclements per second, ν and δν, for fd = 10 Hz

0
10

-2
10

-4
10
ν, δν

-6
10

-8
10

ν
-10
δν
10
-15 -10 -5 0 5 10
A2/(2N0B) in dB