Question: Explain the concept of amplitude modulation
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Explain AM.

  1. Mathematical equation
  2. AM waveform
  3. An amplitude and power spectrum
  4. Modulator coefficient
  5. Transmission power

Subject: Computer Engineering

Topic: Electronic Circuits and Communication Fundamentals

Difficulty: Medium / High

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modified 11 months ago  • written 11 months ago by gravatar for Sayali Bagwe Sayali Bagwe2.2k
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Amplitude Modulation or DSB-FC modulation –

AM is process of changing the amplitude of a high frequency. sinusoidal carrier signal in proportion with the instantaneous value of modulating signal.

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Principle:

In AM the instaneous amplitude of sinusoidal high frequency. carrier is changed in proportion to the instantaneous amplitude of the modulating signal.

AM Envelope: DSB-FC also called conventional AM The shape of the modulated waveform is called as envelope. The repetition of modulating sig and the envelop shape is identical to the shape of modulating signal.

Mathematical representation (AM)

Let modulating Signal be,

$Em= Em \cos wmt$

em – instantaneous amp

Em – peak amplitude

$wm = 2πfm$ and fm = frequency. of mod. Signal

Let carrier sig at much higher frequency than the mod sig. , The instantaneous carrier sig. ec ;

$Ec = Ec \cos wct$

Ec – peak carrier amplitude

$wc = 2π fc , fc$ – carrier frequency

Am wave is expressed as

$eAM = A cos (2πfct)$

A= Envelope of AM wave

Where,

A → instantaneous value of envelope. The mod signal either adds or gets subtracted from peak carrier amp , i.e. , Ec.

$∴A=Ec+em \\ =Ec+Em cos⁡(2πfmt ) \\ ∴AM wave ; \\ eAM=A \cos⁡(2πfct) \\ =[Ec+Em \cos⁡(2πfmt) ] \cos⁡(2πfct) \\ ∴eAM=Ec \big[1+\frac{Em}{Ec} \cos⁡(2πfmt) \cos⁡(2πfct) \big] \\ →let m= \frac{Em}{Ec}→ \text{modulation index} \\ ∴eAM=Ec [1+m \cos⁡(2πfmt ) ] \cos⁡(2πfct)$

This expression represents the time domain representation of AM signal.

Frequency spectrum of AM

We Know,

$eAM = (Ec + Em \cos wmt ) \cos wct \\ ∴Ec=\big[1+\frac{Em}{Ec} \cos⁡wmt \big] \cos⁡wct \\ → \text{Modulation index} m=\frac{Em}{Ec} \\ ∴eAM=Ec (1+m \cos⁡wmt ) \cos⁡wct \\ ∴ eAM = Ec \cos wct + m Ec \cos wmt \cos wct \\ \text{Compare by,} \\ 2 \cos A \cos B= \cos (A+B) + \cos (A-B)$

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Bandwidth of AM wave

B.W is obtained by subtracting of highest and lowest frequency. component in frequency. spectrum.

$∴B.W= fusb – flsb = (fc + fm)-(fc-fm) \\ =2fm$

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modified 11 months ago  • written 11 months ago by gravatar for Sayali Bagwe Sayali Bagwe2.2k
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