Math Problem Statement

32-level FSK is used with fc = 5000 kHz and fd = 25kHz. Excess bandwidth factor r= 0.55 . What is the bandwidth efficiency in bps/Hz?

Solution

In a Frequency Shift Keying (FSK) system, the bandwidth efficiency (measured in bps/Hz) is a key metric that indicates how effectively the system uses its available bandwidth.

To calculate the bandwidth efficiency, we need to consider the following steps:

1. Determine the Data Rate:

The number of bits transmitted per symbol is calculated as:

n=log2(M)n = \log_2(M)

where M=32M = 32 (the number of levels in the M-ary FSK system).

So,

n=log2(32)=5 bits per symboln = \log_2(32) = 5 \text{ bits per symbol}

The symbol rate (baud rate) can then be written as:

Rs=RbnR_s = \frac{R_b}{n}

where RbR_b is the bit rate and RsR_s is the symbol rate. Since we are interested in bandwidth efficiency, we focus on how much bandwidth is required to transmit the symbols.

2. Calculate the Bandwidth Required:

The bandwidth of an M-ary FSK system is calculated using the following formula, considering the excess bandwidth factor rr:

B=(1+r)×M×fdB = (1 + r) \times M \times f_d

where:

  • BB is the bandwidth.
  • r=0.55r = 0.55 (excess bandwidth factor).
  • M=32M = 32 (number of levels).
  • fd=25 kHzf_d = 25 \text{ kHz} (frequency deviation).

Now, substituting the values:

B=(1+0.55)×32×25 kHzB = (1 + 0.55) \times 32 \times 25 \text{ kHz} B=1.55×32×25 kHzB = 1.55 \times 32 \times 25 \text{ kHz} B=1.55×800 kHzB = 1.55 \times 800 \text{ kHz} B=1240 kHz=1.24 MHzB = 1240 \text{ kHz} = 1.24 \text{ MHz}

3. Calculate the Bandwidth Efficiency:

The bandwidth efficiency η\eta (in bps/Hz) is given by:

η=RbB\eta = \frac{R_b}{B}

Since each symbol represents 5 bits and the system uses M=32M = 32 symbols, we assume the bit rate RbR_b is proportional to the symbol rate. The number of bits per second is:

Rb=5×fd=5×25 kHz=125 kbpsR_b = 5 \times f_d = 5 \times 25 \text{ kHz} = 125 \text{ kbps}

Finally, the bandwidth efficiency:

η=RbB=125 kbps1240 kHz0.1 bps/Hz\eta = \frac{R_b}{B} = \frac{125 \text{ kbps}}{1240 \text{ kHz}} \approx 0.1 \text{ bps/Hz}

Final Answer:

The bandwidth efficiency is approximately 0.1 bps/Hz.


Would you like more details or have any questions about the calculation?

Here are 5 related questions you might find helpful:

  1. How does increasing the number of levels in FSK affect bandwidth efficiency?
  2. What is the role of the excess bandwidth factor in FSK systems?
  3. Can bandwidth efficiency improve without increasing the data rate?
  4. How does FSK compare to PSK or QAM in terms of bandwidth efficiency?
  5. What factors influence the choice of fcf_c and fdf_d in an FSK system?

Tip: Increasing the modulation order (e.g., moving from 32-FSK to 64-FSK) can improve data rates but requires more bandwidth. Always balance the trade-offs between bandwidth efficiency and system complexity.

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Math Problem Analysis

Mathematical Concepts

Digital Communications
Frequency Shift Keying (FSK)
Bandwidth Efficiency

Formulas

n = log2(M)
B = (1 + r) * M * fd
η = Rb / B

Theorems

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Suitable Grade Level

Undergraduate - Communication Systems