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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 frequenciesthan those occupied by the signal.

We use analog communication techniques for analog message signals, like music, speech, and television. Transmission andreception of analog signals using analog results in an inherently noisy received signal (assuming the channel addsnoise, 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 onamplitude modulation (AM). The amplitude modulated message signal has the form
x t A c 1 m t 2 f c t
where f c is the carrier frequency and A c the carrier amplitude . Also, the signal's amplitude is assumed to be less than one: m t 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 f c W f c W , assuming the signal's bandwidth is W Hz (see the figure ). The carrier frequency is usually much larger than the signal's highestfrequency: f c W , which means that the transmitter antenna and carrier frequency are chosen jointly during the design process.

The AM coherent receiver along with the spectra of key signals is shown for the case of a triangular-shaped signalspectrum. The dashed line indicates the white noise level. Note that the filters' characteristics — cutoff frequency andcenter frequency for the bandpass filter — must be match to the modulation and message parameters.

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

m ^ t LPF x t 2 f c t LPF A c 1 m t 2 f c t 2
Because of our trigonometric identities, we know that
2 f c t 2 1 2 1 2 2 f c t
At this point, the message signal is multiplied by a constant and a sinusoid at twice the carrier frequency. Multiplication bythe constant term returns the message signal to baseband (where we want it to be!) while multiplication by the double-frequencyterm yields a very high frequency signal. The lowpass filter removes this high-frequency signal, leaving only the basebandsignal. Thus, the received signal is
m ^ t A c 2 1 m t

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

The signal-related portion of the transmitted spectrum is given by X f 1 2 M f f c 1 2 M f f c . Multiplying at the receiver by the carrier shifts thisspectrum to f c and to f c , and scales the result by half.

1 2 X f f c 1 2 X f f c 1 4 M f 2 f c M f 1 4 M f 2 f c M f 1 4 M f 2 f c 1 2 M f 1 4 M f 2 f c
The signal components centered at twice the carrier frequency are removed by the lowpass filter, while the baseband signal M f emerges.

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

Questions & Answers

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suppose the production function is given by ( L, K)=L¼K¾.assuming capital is fixed find APL and MPL. consider the following short run production function:Q=6L²-0.4L³ a) find the value of L that maximizes output b)find the value of L that maximizes marginal product
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Source:  OpenStax, Fundamentals of electrical engineering i. OpenStax CNX. Aug 06, 2008 Download for free at http://legacy.cnx.org/content/col10040/1.9
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