# 6.11: Modulated Communication

- Page ID
- 1863

Learning Objectives

- An introduction to frequency and amplitude modulation.

Especially for wireless channels, like commercial radio and television, but also for wireline systems like cable television, an analog message signal must be **modulated**: The transmitted signal's spectrum occurs at much higher frequencies than those occupied by the signal.

Note

We use analog communication techniques for analog message signals, like music, speech, and television. Transmission and reception of analog signals using analog results in an inherently noisy received signal (assuming the channel adds noise, which it almost certainly does).

The key idea of modulation is to affect the amplitude, frequency or phase of what is known as the **carrier** sinusoid. Frequency modulation (FM) and less frequently used phase modulation (PM) are not discussed here; we focus on amplitude modulation (AM). The amplitude modulated message signal has the form

\[x(t)=A_{c}(1+m(t))\cos (2\pi f_{c}t)\]

where **f _{c}** is the

**carrier frequency**and

**A**the

_{c}**carrier amplitude**. Also, the signal's amplitude is assumed to be less than one:

\[\left | m(t) \right |< 1\]

From our previous exposure to amplitude modulation (see the Fourier Transform example), we know that the transmitted signal's spectrum occupies the frequency range

\[\left [ f_{c}-W,f_{c}+W \right ]\]

assuming the signal's bandwidth is **W **Hz (see Fig. 6.11.1). The carrier frequency is usually much larger than the signal's highest frequency: **f _{c }>> W**, which means that the transmitter antenna and carrier frequency are chosen jointly during the design process.

Ignoring the attenuation and noise introduced by the channel for the moment, reception of an amplitude modulated signal is quite easy (see [link]). The so-called **coherent** receiver multiplies the input signal by a sinusoid and lowpass-filters the result (Fig. 6.11.1).

\[\widehat{m}(t)=LPF(x(t)\cos (2\pi f_{c}t))\]

\[\widehat{m}(t)=LPF(A_{c}(1+m(t))\cos ^{2}(2\pi f_{c}t))\]

Because of our trigonometric identities, we know that

\[\cos ^{2}(2\pi f_{c}t)=\frac{1}{2}(1+\cos (2\pi 2f_{c}t))\]

At this point, the message signal is multiplied by a constant and a sinusoid at twice the carrier frequency. Multiplication by the constant term returns the message signal to baseband (where we want it to be!) while multiplication by the double-frequency term yields a very high frequency signal. The lowpass filter removes this high-frequency signal, leaving only the baseband signal. Thus, the received signal is

\[\widehat{m}(t)=\frac{A_{c}}{2}(1+m(t))\]

Exercise \(\PageIndex{1}\)

This derivation relies solely on the time domain; derive the same result in the frequency domain. You won't need the trigonometric identity with this approach.

**Solution**

The signal-related portion of the transmitted spectrum is given by

\[X(f)=\frac{1}{2}M(f-f_{c})+\frac{1}{2}M(f+f_{c})\]

Multiplying at the receiver by the carrier shifts this spectrum to **f _{c}** and to

**-f**, and scales the result by half.

_{c}\[\frac{1}{2}X(f-f_{c})+\frac{1}{2}X(f+f_{c})=\frac{1}{4}(M(f-2f_{c})+M(f))+\frac{1}{4}(M(f+2f_{c})+M(f))\]

\[\frac{1}{2}X(f-f_{c})+\frac{1}{2}X(f+f_{c})=\frac{1}{4}M(f-2f_{c})+\frac{1}{2}M(f)+\frac{1}{4}M(f+2f_{c})\]

The signal components centered at twice the carrier frequency are removed by the lowpass filter, while the baseband signal **M(f) **emerges.

Because it is so easy to remove the constant term by electrical means—we insert a capacitor in series with the receiver's output—we typically ignore it and concentrate on the signal portion of the receiver's output when calculating signal-to-noise ratio.

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