# 6.14 Frequency shift keying

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Frequency Shift Keying uses the bit to affect the frequency of a carrier sinusoid.

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

${s}_{0}(t)=A{p}_{T}(t)\sin (2\pi {f}_{0}t)$
${s}_{1}(t)=A{p}_{T}(t)\sin (2\pi {f}_{1}t)$

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 . This plot shows the FSK waveform for same bitstream used in the BPSK example .

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] ). The depicted decomposition of the FSK-modulated alternating bit stream into its frequency components simplifies thecalculation of its bandwidth.

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.

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