Question
Download Solution PDFThe bit rate of the digital communication system is M kbps. The modulation used is 16 QAM. The minimum bandwidth required for ideal transmission is _________.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
In the M-Array modulation scheme, the minimum bandwidth required for ideal transmission is given by:
\({\left( {BW} \right)_{min}} = \frac{{{2R_b}}}{{{{\log }_2}N}}Hz\)
Where,
Rb = bit rate in bps
N = number of levels in M-Array scheme
Calculation:
Given that,
Bit rate = M kbps
Number of levels = N = 16
\(\therefore {\left( {BW} \right)_{{\rm{min}}}} = \frac{{{2R_b}}}{{{{\log }_2}N}}Hz = \frac{2M}{{{{\log }_2}16}}kHz\)
\( = \frac{2M}{{{{\log }_2}{2^4}}}kHz\)
\( = \frac{2M}{{4\; \times\; {{\log }_2}2}}kHz\)
\( (BW)_{min}= \frac{M}{{2}}kHz\)
So. The minimum bandwidth will be M/2 kHz, for ideal transmission.
|
For Baseband |
For Passband |
|
Binary: 1) B.W. = Rb |
Binary: 1) BW = 2 Rb |
|
Raised cosine (α) : 2) \(BW = \frac{{{R_b}}}{2}\left( {1 + \alpha } \right)\) |
Raised cosine (α) : \(2)\;BW = \frac{{2{R_b}}}{2}\left( {1 + \alpha } \right)\) = Rb (1 + α) |
|
M-ary: 1) \(B.W. = \frac{{{R_b}}}{{{{\log }_2}M}}\) |
M-ary: 1) \(B.W = \frac{{2{R_b}}}{{{{\log }_2}M}}\) |
|
Raised cosine (α): 2) \(B.W. = \frac{{{R_b}\left( {1 + \alpha } \right)}}{{2{{\log }_2}M}}\) |
Raised cosine (α) : 2) \(B.W = \frac{{{R_b}\left( {1 + \alpha } \right)}}{{{{\log }_2}M}}\) |
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