6.14 Frequency shift keying

In frequency-shift keying (FSK), the bit affects the frequency of a carrier sinusoid.

The frequencies $f_0$ , $f_1$ are usually harmonically related to the bit interval. In the depicted example, $f_0=\frac{3}{T}$ and $f_1=\frac{4}{T}$ . As can be seen from the transmitted signal for our example bit stream ( [link] ), the transitions at bit interval boundaries are smoother than thoseof BPSK .

To determine the bandwidth required by this signal set, we again consider the alternating bit stream. Think of it as two signalsadded together: The first comprised of the signal $s_0(t)$ , the zero signal, $s_0(t)$ , zero, etc. , and the second having the same structure but interleaved with the first and containing $s_1(t)$ ( [link] ).

Each component can be thought of as a fixed-frequency sinusoid multiplied by a square wave of period $2T$ that alternates between one and zero. This baseband square wave has the same Fourier spectrum as our BPSK example, but with theaddition of the constant term $c_0$ . This quantity's presence changes the number of Fourier series terms required for the 90% bandwidth: Now we needonly include the zero and first harmonics to achieve it. The bandwidth thus equals, with $f_0< f_1$ , $f_1+\frac{1}{2T}-f_0-\frac{1}{2T}=f_1-f_0+\frac{1}{T}$ . If the two frequencies are harmonics of the bit-interval duration, $f_0=\frac{k_0}{T}$ and $f_1=\frac{k_1}{T}$ with $k_1> k_0$ , the bandwidth equals $\frac{k_1+-k_0+1}{T}$ . If the difference between harmonic numbers is $1$ , then the FSK bandwidth is smaller than the BPSK bandwidth. If the difference is $2$ , the bandwidths are equal and larger differences produce a transmission bandwidthlarger than that resulting from using a BPSK signal set.