Question

A wireless digital transmission scheme is using \(16\)-QAM over an additive white Gaussian noise channel and a maximum-likelihood receiver. Consider the information bit rate from source to be \(4\times10^{6}\) bits per second.

The minimum transmission bandwidth (in MHz) of the modulated signal necessary for optimum recovery of information at the receiver is (rounded off to two decimal places).

Hint:

Find bits per symbol from \(\log_2(16)\), convert bit rate to symbol rate, then use \(B_{min}=R_s\) for QAM.

Answer 1

Correct Answer: 1

Step 1: Find how many bits each 16-QAM symbol carries.
\(16\)-QAM has \(16\) distinct constellation points, and
\[ 16=2^4 \]
so each symbol carries
\[ k=\log_2(16)=4\text{ bits} \]

Step 2: Find the symbol rate.
The information bit rate is
\[ R_b=4\times10^6\text{ bits/s} \]
Each symbol packs \(4\) bits, so the symbol (baud) rate is
\[ R_s=\frac{R_b}{k}=\frac{4\times10^6}{4}=1\times10^6\text{ symbols/s} \]

Step 3: Recall the minimum bandwidth needed for a QAM signal.
For QAM, the baseband in-phase and quadrature pulse streams are each shaped to just fit in a bandwidth of \(R_s/2\) (satisfying the Nyquist criterion for zero inter-symbol interference at the minimum possible bandwidth). When these two streams are placed on quadrature carriers to build the passband QAM signal, the two sidebands combine so that the total transmission bandwidth required is exactly
\[ B_{min}=R_s \]

Step 4: Substitute the symbol rate.
\[ B_{min}=1\times10^6\text{ Hz}=1\text{ MHz} \]

Final Answer:
\[ \boxed{B_{min}=1.00\text{ MHz}} \]