Question

Which of the following modulation scheme gives the maximum probability of error?

• A. ASK
• B. FSK
• C. PSK
• D. QPSK

Hint:

Probability of error (\(P_e\)) depends on the modulation scheme and the signal-to-noise ratio (\(E_b/N_0\)).
Generally, for coherent detection in AWGN: \(P_{e,BPSK} = Q(\sqrt{2E_b/N_0})\) \(P_{e,coherent FSK} = Q(\sqrt{E_b/N_0})\) \(P_{e,coherent ASK (OOK)} = Q(\sqrt{E_b/N_0})\) (if average power of bit '1' is used for \(E_b\)) or \(Q(\sqrt{2E_b/N_0})\) (if \(E_b\) is average energy per bit over 0 and 1).
BPSK generally offers better error performance than ASK and coherent FSK for the same \(E_b/N_0\). ASK is often the most susceptible.
\(Q(x) = \frac{1}{\sqrt{2\pi}}\int_x^\infty e^{-t^2/2} dt\) is the tail probability of a standard normal distribution. Larger argument \(x\) means smaller \(Q(x)\) (better performance).

Answer 1

The Correct Option is A

Comparing common digital modulation schemes in terms of probability of error (\(P_e\)) for a given signal-to-noise ratio (\(E_b/N_0\)) in an Additive White Gaussian Noise (AWGN) channel: 

• ASK (Amplitude Shift Keying) / OOK (On-Off Keying):
  • The error probability for coherent ASK (OOK) is \(P_e = Q(\sqrt{E_b/N_0})\) or \(P_e = Q(\sqrt{2E_b/N_0})\) depending on the definition of \(E_b\).
  • For non-coherent envelope detection, performance is worse.
  • ASK is generally the most susceptible to noise because information is encoded in the amplitude, which is directly affected by noise.
• FSK (Frequency Shift Keying):
  • Coherent FSK: \(P_e = Q(\sqrt{E_b/N_0})\).
  • Non-coherent FSK: \(P_e = \frac{1}{2}e^{-E_b/(2N_0)}\).
  • Coherent FSK has similar performance to coherent ASK (OOK). Non-coherent FSK is worse than coherent FSK but can be better than non-coherent ASK.
• PSK (Phase Shift Keying) / BPSK (Binary PSK):
  • For coherent BPSK, \(P_e = Q(\sqrt{2E_b/N_0})\).
  • BPSK is generally more robust (lower \(P_e\) for a given \(E_b/N_0\)) than ASK and non-coherent FSK.
• QPSK (Quadrature Phase Shift Keying):
  • For QPSK, the bit error rate (BER) can be similar to BPSK for the same \(E_b/N_0\), i.e., \(P_b \approx Q(\sqrt{2E_b/N_0})\), but it transmits two bits per symbol, making it more bandwidth efficient.
  • Symbol error rate is higher, but bit error rate is comparable to BPSK.

General ranking from worst (highest \(P_e\)) to best (lowest \(P_e\)) for a given \(E_b/N_0\), using common coherent detection schemes:

• Non-coherent ASK/FSK > Coherent ASK (OOK) ≈ Coherent FSK > BPSK ≈ QPSK (BER)

Among the options typically implying coherent detection if not specified:

• ASK generally has the highest probability of error for a given \(E_b/N_0\) compared to PSK and FSK (coherent versions).
• QPSK (a form of PSK) offers good performance.

 

Therefore, ASK is the modulation scheme that usually gives the maximum probability of error among these choices under comparable conditions.

Final Answer:

ASK