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Frequency Shift Keying

Module by: Don Johnson

Summary: Frequency Shift Keying uses the bit to affect the frequency of a carrier sinusiod.

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A different signal set from the previous two uses the bit to affect the frequency of a carrier sinusoid.

s 0 0=A p T tsin2π f 0 t s 0 0 A p T t 2 π f 0 t (1)
s 1 0=A p T tsin2π f 1 t s 1 0 A p T t 2 π f 1 t (2)

Figure 1
Figure 1 (sig31.png)

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 . This signal set is known as frequency shift keying or FSK. 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.

Figure 2: This plot shows the FSK waveform for same bitstream used in the BPSK example.
Figure 2 (sig32.png)

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 (Figure 3).

Figure 3: The depicted decomposition of the FSK-modulated alternating bit stream into its frequency components simplifies the calculation of its bandwidth.
Figure 3 (sig33.png)

Each component can be thought of as a fixed-frequency sinusoid multiplied by a square wave of period 12T 1 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 . This bandwidth is larger than the BPSK bandwidth.

Exercise 1

Show that the bandwidth for FSK is always larger. What is the smallest it can be?

Solution

We want to show that f 1 − f 0 +1T>32T f 1 f 0 1 T 3 2 T , which means that f 1 − f 0 >12T f 1 f 0 1 2 T must be true. As we have f 0 = k 0 T f 0 k 0 T , f 1 = k 1 T f 1 k 1 T , the smallest difference between the two frequencies is 1T 1 T , giving a transmission bandwidth of 2T 2 T or 2R 2 R in terms of data rate.

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