written 6.1 years ago by | • modified 5.0 years ago |
Similar Question
Explain AM.
- Mathematical equation
- AM waveform
- An amplitude and power spectrum
- Modulator coefficient
- Transmission power
written 6.1 years ago by | • modified 5.0 years ago |
Similar Question
Explain AM.
written 6.1 years ago by | • modified 6.1 years ago |
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.
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,
$eAM = (Ec + Em \cos wmt ) \cos wct \\ ∴Ec=\big[1+\frac{Em}{Ec} \coswmt \big] \coswct \\ → \text{Modulation index} m=\frac{Em}{Ec} \\ ∴eAM=Ec (1+m \coswmt ) \coswct \\ ∴ eAM = Ec \cos wct + m Ec \cos wmt \cos wct \\ \text{Compare by,} \\ 2 \cos A \cos B= \cos (A+B) + \cos (A-B)$
Bandwidth of AM wave
B.W is obtained by subtracting of highest and lowest frequency. component in frequency. spectrum.
$∴B.W= fusb – flsb = (fc + fm)-(fc-fm) \\ =2fm$