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 (SNR) per bit (\(E_b/N_0\)) 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 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) \(\approx\) Coherent FSK>BPSK \(\approx\) 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. \[ \boxed{\text{ASK}} \]