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ASK Modulation

Amplitude shift keying, OOK, bandwidth, probability of error.

Mohith N
Updated: 19 March 2026
7 min read

Amplitude Shift Keying is the simplest form of digital modulation, where the amplitude of a sinusoidal carrier is switched between discrete levels to represent different binary symbols. Despite its simplicity, ASK and its special case On-Off Keying are widely used in optical fiber communications and RFID systems. Studying ASK builds the conceptual foundation for understanding more complex modulation schemes.

ASK / OOK ModulationBinary Datam(t): 1 0 1 1 0NRZ bit stream0CarrierOOK s(t)bit=1 (A=Ac)bit=0 (A=0)bit=1 (A=Ac)bit=1 (A=Ac)bit=0 (A=0)OOK: A in {0, Ac}Binary ASK special case100% modulation indexGeneral ASK: A in {A1, A2}A1 and A2 both nonzeroPartial modulation indexBW = 2Rb (null-to-null)Bandwidth proportionalto bit rate
Figure 1: OOK waveform for bit sequence 10110, showing carrier amplitude switching and key parameters.

Core Concept: ASK and OOK

In Amplitude Shift Keying (ASK), the carrier amplitude takes one of M discrete values corresponding to M possible symbols. For binary ASK (M = 2), the signal is: s(t) = Ai cos(2πfct), where Ai takes one of two values A1 or A2 for the bit duration Tb. The phase and frequency of the carrier remain constant. Only the amplitude encodes information.

On-Off Keying (OOK) is the special case where A1 = Ac and A2 = 0. Bit 1 is transmitted by sending the carrier; bit 0 is transmitted by sending nothing (zero signal). OOK is the most commonly used form of ASK in optical fiber communications because laser sources are easily turned on and off. In RF systems OOK is sensitive to amplitude noise, but optical systems have very high SNR on the link, making OOK practical.

Mathematical Expression

The general binary ASK signal is expressed as: s1(t) = Ac cos(2πfct) for bit 1 and s0(t) = 0 (or A0 cos(2πfct)) for bit 0. The transmission bandwidth using a rectangular pulse and null-to-null definition is BT = 2Rb, where Rb = 1/Tb is the bit rate. Under raised cosine filtering with rolloff factor α, BT = Rb(1 + α).

The probability of error for coherent OOK is: Pb = Q(sqrt(Eb/N0)). Note carefully that this is Q(sqrt(Eb/N0)) and not Q(sqrt(2Eb/N0)) as in BPSK. This 3 dB disadvantage arises because in OOK, half the symbols carry zero energy (the off state), so the average energy per bit is half that of BPSK for the same peak amplitude.

Practical Understanding

ASK is very sensitive to amplitude noise and channel fading. Any multiplicative disturbance in the channel directly corrupts the amplitude, making ASK unreliable in wireless fading channels. For this reason, ASK is almost exclusively used in optical communications (Gigabit Ethernet, 100G DWDM using OOK) where the channel is stable and high-speed amplitude switching is easy to implement with a Mach-Zehnder modulator.

Non-coherent detection of OOK uses an envelope detector, which finds the magnitude of the received signal and compares it to a threshold. This eliminates the need for carrier phase synchronization and reduces receiver complexity. The BER for non-coherent OOK is slightly worse than coherent detection.

Example
Given:
Bit rate Rb = 1 Mbps, OOK modulation, Eb/N0 = 10 (linear), carrier amplitude Ac = 1 V

Why this formula applies:
For coherent OOK (binary ASK), BER = Q(sqrt(Eb/N0)) because only one of two symbols carries energy.

Formula:
Pb = Q(sqrt(Eb/N0))
Bandwidth (null-to-null) BT = 2Rb

Substitution:
Pb = Q(sqrt(10)) = Q(3.162)
BT = 2 × 1 × 10^6 = 2 MHz

Calculation:
Q(3.162) ≈ 7.8 × 10^-4
If BPSK were used: Pb = Q(sqrt(2 × 10)) = Q(4.47) ≈ 3.9 × 10^-6
Difference shows 3 dB power penalty of OOK vs BPSK.

Final Answer with units:
BER of coherent OOK ≈ 7.8 × 10^-4, Transmission bandwidth = 2 MHz
Exam Tip: A very common GATE trap is confusing the BER formula for OOK and BPSK. OOK gives Pb = Q(sqrt(Eb/N0)) while BPSK gives Pb = Q(sqrt(2Eb/N0)). The factor of 2 inside the square root means BPSK requires 3 dB less power for the same BER. Always check which scheme the question refers to.
ASK Coherent Detection MechanismReceivedr(t)Multiplyx cos(2πfct)Integrateover TbCompareto thresholdDecision0 or 1OOK Signal Space (1D Constellation)s0 = 0bit 0s1 = sqrt(Eb)bit 1Threshold = sqrt(Eb)/2d = sqrt(Eb)Distance between symbols = sqrt(Eb)BPSK distance = 2 sqrt(Eb/2) = sqrt(2Eb) — larger, hence lower BER
Figure 2: Coherent ASK receiver chain and OOK one-dimensional constellation, showing why BPSK has larger symbol distance and lower BER.
  • The received signal is multiplied by the local carrier reference cos(2πfct) and integrated over one bit period. The output is sampled and compared to a threshold.
  • The threshold for OOK is set at Ac²Tb/4 (half the energy of the on-symbol) when both bits are equally likely.
  • Non-coherent detection uses an envelope detector followed by a threshold comparator, removing the need for phase reference but degrading BER slightly.
  • The 1D signal space shows s0 at origin and s1 at sqrt(Eb); the distance between symbols is sqrt(Eb), which is smaller than the BPSK distance of sqrt(2Eb), explaining the 3 dB BER disadvantage.

Quick Revision

  • ASK varies carrier amplitude to encode data; all other carrier parameters remain constant.
  • OOK is binary ASK with A2 = 0; carrier is either fully on or fully off.
  • Coherent OOK BER: Pb = Q(sqrt(Eb/N0)); coherent BPSK BER: Pb = Q(sqrt(2Eb/N0)). OOK is 3 dB worse.
  • Transmission bandwidth for rectangular pulses: BT = 2Rb (null-to-null).
  • ASK is noise-sensitive and not preferred for wireless fading channels; heavily used in optical fiber systems.
  • Common trap: OOK average energy per bit is Eb = Ac²Tb/4, not Ac²Tb/2, because the off-symbol contributes zero energy.

ASK Modulation Quiz

Test your knowledge of ASK bandwidth, OOK, and probability of error analysis.

Question 1 of 3

Q1.For Binary ASK (OOK) with bit rate Rb, the minimum (Nyquist) transmission bandwidth is: