Tutorials - Digital Signal Processing: An Introduction with MATLAB and Applications - page 283

From Tables we have: d ð t Þ¼ 1 =ð p t Þ ; an from (A) we have: y(t) =-cos
(x o t).
Since H is a linear system, we have: z ð t Þ¼ y ð t Þþ d ð t Þ¼ cos ð x o t Þþ 1 =ð p t Þ:
(C) z ð t Þ¼ x ð t Þþ j Hf x ð t Þg:

Using an approach similar to that in (A) above, we have
Hf cos ð x o t Þg¼ sin ð x o t Þ:
Hence ; z ð t Þ¼ cos ð x o t Þþ jsin ð x o t Þ¼ e jx o t :
Now from Tables we have:
X ð f Þ¼ 1

2 d ð f f o Þþ 1

2 d ð f þ f o Þ ;

Z ð f Þ¼ d ð f f o Þ:

(D) Using the analytic signal, the negative part of spectrum is removed, while the
positive part is scaled by 2. This indicates that the use of the analytic signal
will lead to spectrum economy.

Tutorial 49

Q: Consider a first-order sinusoidal DPLL (SDPLL) under noise-free conditions,
center frequency f o = 1 Hz, and input signal x(t) = sin (6t + p/4).
1. Find the system equation in terms of the digital filter gain G 1 .
2. Find the steady-state phase error / ss in terms of G 1 .
3. Find the range of G 1 that ensures locking on the above incoming frequency.
4. Choose a value for G 1 inside the locking range and find the corresponding / ss .
5. Assuming t(0) = 0, plot x(t) with the first three DCO pulses for the abovevalue
of G 1 .
Solution:
1. From Tables we find the system equation as follows:

/ ð k þ 1 Þ¼ / ð k Þ x : G 1 sin ½ / ð k Þþð x x o Þ T o ð 1 Þ

We have: x = 6 rad/s, x o = 2p rad/s, and T o = 1s.
Hence, the system equation will be given by:

/ ð k þ 1 Þ¼ / ð k Þ 6G 1 sin ½ / ð k Þ 0 : 28 ð 2 Þ

2. From (2) we have:

/ ss ¼ / ss 6G 1 sin ½ / ss 0 : 28 ) / ss ¼ sin 1 ð 0 : 047 = G 1 Þ

ð 3 Þ

3. Locking conditions are as follows:
Condition 1: From (3) we have:

j 0 : 047 = G 1 j \1 ) G 1 [ 0 : 047

ð 4 Þ