Question:medium
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).
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).
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).
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Find bits per symbol from \(\log_2(16)\), convert bit rate to symbol rate, then use \(B_{min}=R_s\) for QAM.
Updated On: Jul 20, 2026
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Correct Answer: 1
Solution and Explanation
Step 1: Work out bits per symbol.
A 16-QAM constellation has 16 points, and since $2^4 = 16$, every symbol transmitted carries $4$ bits of information.
Step 2: Convert the bit rate into a symbol rate.
Given the bit rate $R_b = 4\times10^6$ bits/s, dividing by the $4$ bits carried per symbol gives the baud rate:
\[ R_s = \frac{4\times10^6}{4} = 10^6 \text{ symbols/s} \]
Step 3: Recall the Nyquist minimum bandwidth rule for QAM.
For an ideal (Nyquist, zero-ISI) pulse shape, a passband QAM signal needs a transmission bandwidth equal to the symbol rate, $B_{min}=R_s$, since the in-phase and quadrature channels share the same passband through quadrature multiplexing instead of doubling it.
Step 4: Plug in the numbers.
\[ B_{min} = 10^6 \text{ Hz} = 1 \text{ MHz} \]
Step 5: Conclude.
\[ \boxed{B_{min} = 1.00 \text{ MHz}} \]
A 16-QAM constellation has 16 points, and since $2^4 = 16$, every symbol transmitted carries $4$ bits of information.
Step 2: Convert the bit rate into a symbol rate.
Given the bit rate $R_b = 4\times10^6$ bits/s, dividing by the $4$ bits carried per symbol gives the baud rate:
\[ R_s = \frac{4\times10^6}{4} = 10^6 \text{ symbols/s} \]
Step 3: Recall the Nyquist minimum bandwidth rule for QAM.
For an ideal (Nyquist, zero-ISI) pulse shape, a passband QAM signal needs a transmission bandwidth equal to the symbol rate, $B_{min}=R_s$, since the in-phase and quadrature channels share the same passband through quadrature multiplexing instead of doubling it.
Step 4: Plug in the numbers.
\[ B_{min} = 10^6 \text{ Hz} = 1 \text{ MHz} \]
Step 5: Conclude.
\[ \boxed{B_{min} = 1.00 \text{ MHz}} \]
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