In frequency-shift keying (FSK), the bit
affects the frequency of a carrier sinusoid.
s
0
t=A
p
T
tsin2π
f
0
t
s
0
t
A
p
T
t
2
f
0
t
(1)
The frequencies
f
0
f
0
,
f
1
f
1
are usually harmonically related to the bit interval. In the
depicted example,
f
0
=3T
f
0
3
T
and
f
1
=4T
f
1
4
T
. As can be seen from the transmitted signal for our
example bit stream (Figure 2), the
transitions at bit interval boundaries are smoother than those
of BPSK.
To determine the bandwidth required by this signal set, we again
consider the alternating bit stream. Think of it as two signals
added together: The first comprised of the signal
s
0
t
s
0
t
, the zero signal,
s
0
t
s
0
t
, zero, etc., and the second having
the same structure but interleaved with the first and containing
s
1
t
s
1
t
(Figure 3).
Each component can be thought of as a fixed-frequency sinusoid
multiplied by a square wave of period
2T
2
T
that alternates between one and zero. This baseband square wave
has the same Fourier spectrum as our BPSK example, but with the
addition of the constant term
c
0
c
0
. This quantity's presence changes the number of
Fourier series terms required for the 90% bandwidth: Now we need
only include the zero and first harmonics to achieve it. The
bandwidth thus equals, with
f
0
<
f
1
f
0
f
1
,
f
1
+12T−(
f
0
−12T)=
f
1
−
f
0
+1T
f
1
1
2
T
f
0
1
2
T
f
1
f
0
1
T
. If the two frequencies are harmonics of the
bit-interval duration,
f
0
=
k
0
T
f
0
k
0
T
and
f
1
=
k
1
T
f
1
k
1
T
with
k
1
>
k
0
k
1
k
0
, the bandwidth equals
k
1
+−
k
0
+1T
k
1
k
0
1
T
. If the difference between harmonic numbers is
11, then the FSK bandwidth is
smaller than the BPSK bandwidth. If the
difference is 22, the bandwidths are
equal and larger differences produce a transmission bandwidth
larger than that resulting from using a BPSK signal set.
"Electrical Engineering Digital Processing Systems in Braille."