What is the bandwidth efficiency for FSK, ASK, PSK and QPSK for a bit error rate of 10-7 on a channel with an SNR of 12 dB?

1. Break Down the Problem

To determine the bandwidth efficiency for different modulation schemes (FSK, ASK, PSK, and QPSK) at a given bit error rate (BER) and signal-to-noise ratio (SNR), we will need to:

1. Calculate the bandwidth efficiency for each modulation scheme.
2. Use the appropriate formulas for each modulation type based on the given SNR and BER.

2. Relevant Concepts

The bandwidth efficiency (η) can be defined as: $\eta = \frac{R_b}{B}$ Where:

• $R_b$ = Bit rate (in bits/second)
• $B$ = Bandwidth (in Hertz)

The relationship between SNR and BER for different modulation schemes can be given by specific formulas:

• For PSK and QPSK: $\text{BER} \approx \frac{1}{2} \operatorname{erfc}\left(\sqrt{\frac{S}{N}}\right)$

• For FSK:

• For ASK:

For a specific SNR (in linear scale): $\text{SNR}_{\text{linear}} = 10^{\frac{\text{SNR}_{\text{dB}}}{10}} = 10^{\frac{12}{10}} = 15.85$

3. Analysis and Detail

1. PSK and QPSK: Assuming a BER of $10^{-7}$:

   • Calculate $SNR$ in linear scale.
   • Use the inverse of the BER formula to find the required operating conditions.

2. FSK: The bandwidth is typically more significant than for PSK and QPSK.

3. ASK: Similar principles apply as in FSK, but has lower efficiency due to square law demodulation.

From known values:

• For PSK: $\eta_{PSK} = 1 \text{ bit/Hz}$

• For QPSK: $\eta_{QPSK} = 2 \text{ bits/Hz}$

• For FSK: $\eta_{FSK} \approx 0.5 \text{ bits/Hz}$

• For ASK: $\eta_{ASK} \approx 1 \text{ bit/Hz}$

4. Verify and Summarize

• For SNR of 12 dB, bandwidth efficiency for each modulation:
  • FSK: ~0.5 bits/Hz
  • ASK: ~1 bit/Hz
  • PSK: ~1 bit/Hz
  • QPSK: ~2 bits/Hz

Final Answer

• FSK: 0.5 bits/Hz
• ASK: 1 bit/Hz
• PSK: 1 bit/Hz
• QPSK: 2 bits/Hz