A corporation has a medium with a 1-MHz bandwidth (lowpass). The corporation needs to create 10 separate independent channels each capable of sending at least 10 Mbps. The company has decided to use QAM technology. What is the minimum number of bits per baud for each channel? What is the number of points in the constellation diagram for each channel? Let d = 0.
The correct answer and explanation is :
To solve this problem, we need to figure out the minimum number of bits per baud for each channel and the number of points in the constellation diagram when using Quadrature Amplitude Modulation (QAM) technology.
Given Information:
- Bandwidth of the medium = 1 MHz
- The corporation needs to create 10 separate independent channels.
- Each channel must support a minimum data rate of 10 Mbps.
- The modulation technology used is QAM.
- The symbol rate (baud rate) of the medium is determined by the bandwidth, and we will assume Nyquist’s criterion for simplicity.
- ( d = 0 ) means no noise, simplifying the situation.
Step 1: Baud Rate Calculation
According to Nyquist’s theorem, the maximum baud rate (symbols per second) for a given bandwidth is twice the bandwidth. In this case, the bandwidth is 1 MHz, so the maximum baud rate (in symbols per second) is:
[
\text{Baud rate} = 2 \times \text{Bandwidth} = 2 \times 1 \, \text{MHz} = 2 \, \text{Mbaud}
]
Since there are 10 independent channels, the bandwidth per channel is:
[
\text{Bandwidth per channel} = \frac{\text{Total Bandwidth}}{\text{Number of Channels}} = \frac{1 \, \text{MHz}}{10} = 0.1 \, \text{MHz}
]
So, the baud rate per channel is:
[
\text{Baud rate per channel} = 2 \times 0.1 \, \text{MHz} = 0.2 \, \text{Mbaud}
]
Step 2: Bits per Baud Calculation
Each channel must support a minimum data rate of 10 Mbps. The data rate ( R ) for a given modulation scheme is related to the baud rate and the number of bits per symbol (or bits per baud) ( k ) as follows:
[
R = \text{Baud rate} \times k
]
Rearranging to solve for ( k ) (the number of bits per baud):
[
k = \frac{R}{\text{Baud rate}} = \frac{10 \, \text{Mbps}}{0.2 \, \text{Mbaud}} = 50 \, \text{bits per baud}
]
Thus, the minimum number of bits per baud per channel is 50 bits per baud.
Step 3: Number of Points in the Constellation Diagram
For QAM, the number of points in the constellation diagram is ( 2^k ), where ( k ) is the number of bits per baud. Since ( k = 50 ), the number of points in the constellation diagram is:
[
2^k = 2^{50}
]
This is the number of unique symbols that can be transmitted per baud, and it shows the density of the constellation diagram.
Conclusion
- The minimum number of bits per baud for each channel is 50 bits per baud.
- The number of points in the constellation diagram for each channel is ( 2^{50} ), which represents an extremely high-density modulation scheme.
Explanation
This calculation demonstrates how bandwidth, symbol rate, and modulation affect data rate. By dividing the available bandwidth among the channels, calculating the necessary baud rate, and determining the number of bits per baud, the company can ensure that each channel meets the required data rate. However, it’s important to note that real-world considerations like noise and interference will likely affect the practical deployment of such a high-order QAM scheme.