Selected Topics in Applied Signal Processing - Digital Signal Processing: An Introduction with MATLAB and Applications - page 154

¼

/ ð k Þ
/ ð k þ 1 Þ

u k
v k

x k ¼

Then, using Eq. 3.40 one has:

¼

u k þ 1
v k þ 1

/ ð k þ 1 Þ
/ ð k þ 2 Þ

x k þ 1 ¼

¼ ¼ g ð x k Þ

v k
2v k u k þ K 2 sin ð u k Þ rK 2 sin ð v k Þ

f ð x k Þ
h ð x k Þ

¼

The problem is now expressed in a form which can be analyzed with fixed point
analysis:

and ;

u
v

g ð x Þ¼ f ð x Þ
h ð x Þ

g(x) = x with x ¼ where f(x) = v and h ð x Þ¼

2v u þ K 2 sin ð u Þ rK 2 sin ð v Þ:
As there is a function of two variables, one should use the Jacobian (instead of
the derivative):
g 0 ð x Þ¼ of = ou

¼

of = ov 0 1

oh = ou oh = ov 1 þ K 2 cos ð u Þ 2 rK 2 cos ð v Þ

v ¼

ð mod 2p Þ , at which:

u

0
0

The fixed point is x ¼

:

0 1

g 0 ð x Þ¼

1 þ K 2 2 rK 2

For this system to have a fixed point as above, one should have:

j k i j \1 8 i

(x * ), i.e., roots of |k I - L| = 0. Hence we where k i are the eigenvalues of L = g 0

get:

r [ 1

and

4
r þ 1

0\K 2 \

which define the locking range of SDPLL2.

3.4.2.5 PM Demodulation Using the SDPLL

Using Eq. 3.29 for the 1st-order SDPLL yields:

/ ð k Þ¼ h ð k Þ x 0 X y ð i Þ¼ h ð k Þ K 1 X

k 1 k 1

sin ½ / ð i Þ ð 3 : 41 Þ

i ¼ 0 i ¼ 0