# 6.15: Frequency Shift Keying

Learning Objectives

- 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.

\[s_{0}(t)=A_{P_{T}}(t)\sin (2\pi f_{0}t)\]

\[s_{1}(t)=A_{P_{T}}(t)\sin (2\pi f_{1}t)\]

Fig. 6.15.1

The frequencies **f _{0}**,

**f**are usually harmonically related to the bit interval. In the depicted example,

_{1}\[f_{0}=\frac{3}{T}\; and\; f_{1}=\frac{4}{T}\]

As can be seen from the transmitted signal for our example bit stream (Fig. 6.15.2) the transitions at bit interval boundaries are smoother than those of BPSK.

*Fig. 6.15.2 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 signals added 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

*Fig. 6.15.3 The depicted decomposition of the FSK - modulated alternating bit stream into its frequency components simplifies the calculation 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 the addition of the constant term **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}\[f_{1}+\frac{1}{2T}-\left ( f_{0}-\frac{1}{2T} \right )=f_{1}-f_{0}+\frac{1}{T}\]

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 bandwidth larger than that resulting from using a BPSK signal set.

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