Question
Download Solution PDFIn amplitude shift keying, transmission bandwidth is equal to
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFIn amplitude shift keying, transmission bandwidth is equal to Base bandwidth.
Additional Information
The bandwith of differnet passband modulation scheme is as shown:
|
Modulation Scheme |
Bandwidth |
|
ASK and PSK |
2Rb (Same as baseband BW) |
|
FSK |
fH - fL + 2Rb where fH is the higher cut-off frequency, fL is the lower cut-off frequency. |
|
DSB-FC & DSB-SC |
2fm Where fm is the frequency of the message signal. |
|
SSB-SC |
fm |
|
VSB-SC |
fm + fv where fv is the vestige frequency |
|
NBFM (β < 1) |
2fm |
|
WBFM (β < 1) |
2fm(1 + βf) Where βf is the modulation index of WBFM, given by βf = Δf/fm, where Δf is the frequency deviation. |
|
PM |
2fm(1 + βp) Where βp is the modulation index of pm, given by βp = Kp × Am, where Kp is amplitude sensitivity of PM signal, Am is the amplitude of message signal. |
|
M - array PSK |
2Rb/n Where n is the number of bits. |
|
M – array QAM |
For rectangular pulse, 2Rb/n For raised cosine signal, Rb(1 + α)/n, where α is the roll-off. |
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