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AM, FM, and Digital Modulated Systems Chap. 5
(Noise moves the received signal vector to a new location that might correspond to a different
signal level.) However, we know that R certainly has to be less than C, the channel capacity
(Sec. 1–9), if the errors are to be kept small. Consequently, using Eq. (1–10), we require that
where
h 6 h max
h max = log 2 a1 + S N b
(5–108a)
(5–108b)
5–11 MINIMUM-SHIFT KEYING AND GMSK
Minimum-shift keying (MSK) is another bandwidth conservation technique that has been
developed. It has the advantage of producing a constant-amplitude signal and, consequently,
can be amplified with Class C amplifiers without distortion. As we will see, MSK is equivalent
to OQPSK with sinusoidal pulse shaping [for h i (t)].
DEFINITION. Minimum-shift keying (MSK) is continuous-phase FSK with a minimum
modulation index (h = 0.5) that will produce orthogonal signaling.
First, let us show that h = 0.5 is the minimum index allowed for orthogonal continuousphase
FSK. For a binary 1 to be transmitted over the bit interval 0 6 t 6 T b , the FSK signal
would be s 1 (t) = A c cos (v 1 t + u 1 ), and for a binary 0 to be transmitted, the FSK signal would
be s 2 (t) = A c cos (v 2 t + u 2 ), where u 1 = u 2 for the continuous-phase condition at the switching
time t = 0. For orthogonal signaling, from Eq. (2–77), we require the integral of the product of
the two signals over the bit interval to be zero. Thus, we require that
T b
s 1 (t)s 2 (t) dt = A 2 c cos (v 1 t + u 1 ) cos (v 2 t + u 2 )dt = 0
L0
L 0
This reduces to the requirement that
A c
2
T b
2 c sin [1v 1 - v 2 2T b + (u 1 - u 2 )] - sin (u 1 - u 2 )
v 1 - v 2
+ A c 2
2 c sin [(v 1 + v 2 )T b + (u 1 + u 2 )] - sin (u 1 + u 2 )
v 1 + v 2
d = 0
d
(5–109a)
(5–109b)
The second term is negligible, because v 1 + v 2 is large, † so the requirement is that
sin [2ph + (u 1 - u 2 )] - sin (u 1 - u 2 )
2ph
= 0
(5–110)
† If v 1 + v 2 is not sufficiently large to make the second term negligible, choose f c = 1 2 m>T b = 1 2 mR,
where m is a positive integer. This will make the second term zero (f 1 = f c - ∆F and f 2 = f c + ∆F).