_{1}

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Using Terahertz Optical Asymmetric Demultiplexer (TOAD) based switch we have designed all-optical parallel half adder and full adder. The approach to design this all-optical arithmetic circuit not only enhance s the computational speed but also is capable of synthesizing light as input to produce desire output. The main advantage of parallel circuit is synchronization of input which is not required. All the circuits are designed theoretically and verified through numerical simulations.

Nowadays high speed all-optical logic gates are crucial devices in optical networks because they execute essential signal processing function such as switching regeneration and header recognition processing in photonic switching nodes. A revolution has been brought about in all-optical information processing system with the help of the discovery of ultra high speed all-optical switches based on cross phase modulation (XPM) [

The basic design of TOAD based switch is shown in

where,

In the absence of a control signal, data signal (incoming signal) enters the fiber loop, pass through the SOA at different times as they counter-propagate around the loop, and experience the same unsaturated small amplifier gain G_{ss}, and recombine at the input coupler i.e. G_{ccw} ≈ G_{cw}. Then,

values can be obtained from the Equation (2). The energy of the control pulse is ten times greater than that of the incoming pulse. A filter may be used at the output of TOAD based switch to reject the control and pass the incoming pulse. The schematic diagram of TOAD based switch is shown in

A half adder circuit adds two one-bit binary numbers (A and B) and gives the output of two one-bit binary numbers, a sum (S) and a carry (Cout). The operational principle of all-optical parallel half adder is shown in

and B) [These are the light signals]. The output of the corresponding circuit is obtained from Beam Combiner (BC) as Sum and T31 as Cout. Four Cases are described below in detail.

CASE 1: When A = 0 and B = 0. Light from the Pin is incident on switch T1 and T2. As the control signal A, B is absent light emerges through the lower channel of T1, T2 as T12 = 1, T22 = 1 respectively. T21 = 0 and T22 = 1 falls on lower channel of T3 and T4 respectively and T11 = 0 and T12 = 1 falls on upper channel of T3 and T4 respectively, produces output as T31 = 0, T32 = 0, T42 = 0. Now T32 and T42 feed to BC to get output as Sum = 0. Upper output channel of switch T3 i.e., T31 produces Cout = 0.

CASE 2: When A = 0 and B = 1. Light emerges out from T42 only i.e., from lower output channel of switch T4. So, T31 and T32 become 0. Hence Sum = 1 (T42 + T32) and Cout = 0 (T31).

CASE 3: When A = 1 and B = 0. Light emerges out from T32 only i.e., from lower output channel of switch T3. So, T31 and T42 become 0. Hence Sum = 1 (T42 + T32) and Cout = 0 (T31).

CASE 4: When A = 1 and B = 1. Light emerges out from T31 only i.e., from upper output channel of switch T3. So, T32 and T42 become 0. Hence Sum = 0 (T42 + T32) and Cout = 1 (T31).

Hence, Sum =

A full adder circuit adds three one-bit binary numbers (A, B, and C) and gives the output of two one-bit binary numbers, a sum (S) and a carry (Cout). The operational principle of all-optical parallel full-adder is explained in

Depending on the state of input variables (A, B, and C) output of the corresponding circuit is obtained from BC2 for Carry Out and BC3 for Sum, respectively. The advantage of this circuit is no synchronization among the inputs are required which was compulsory for other previously designed full adders, it is a fully parallel circuit. Eight Cases are described below in detail.

CASE 1: When A = 0, B = 0 and C = 0. Light from the Pin is incident on switch T1, T2, T3. As the control signal A, B, C is absent light emerges through the lower channel of T1, T2 and T3 as T12 = 1, T22 = 1 and T32 = 1 respectively. T21 = 0, T22 = 1 and T31 = 0 falls on lower channel of T4, T5 and T8 respectively and T11 = 0, T12 = 1 and Pin falls on upper channel of T4, T5 and T8 respectively, produces output as T41 = 0, T42 = 0, T52 = 0 and T81 = 0. Now T42 and T52 feed to BC1 to get output as X = 0. Again, X is applied on upper and lower channel of switch T6 and T7 as input respectively, also T41 is applied on lower channel of T9 and T81 is applied on lower and upper channel of switch T6 and T7 as input respectively and Pin is applied on upper channel of T9 Hence, T91 = 0. Switch T6 produces output as X2 = 0 from its upper

channel and X3 = 0 from its lower channel. Switch T7 produces output X4 = 0 from its upper channel and X5 = 0 from its lower channel. Now X3 and X5 is combined using BC3 to produce output Sum = 0. X2 and T91 is combined using BC2 to produce Cout = 0.

CASE 2: When A = 0, B = 0 and C = 1. Light emerges out from X5 only i.e., from lower output channel of switch T7. So, T91, X3 and X2 becomes 0. Hence Sum = 1 (X3 + X5) and Cout = 0 (T91 + X2).

CASE 3: When A = 0, B = 1 and C = 0. Light emerges out from X3 only i.e., from lower output channel of switch T6. So, T91, X2 and X5 becomes 0. Hence Sum = 1 (X3 + X5) and Cout = 0 (T91 + X2).

CASE 4: When A = 0, B = 1 and C = 1. Light emerges out from X2 only i.e., from upper output channel of switch T6. So, T91, X3 and X5 becomes 0. Hence Sum = 0 (X3 + X5) and Cout = 1 (T91 + X2).

CASE 5: When A = 1, B = 0 and C = 0. Light emerges out from X3 only i.e., from lower output channel of switch T6. So, T91, X2 and X5 becomes 0. Hence Sum = 1 (X3 + X5) and Cout = 0 (T91 + X2).

CASE 6: When A = 1, B = 0 and C = 1. Light emerges out from X2 only i.e., from upper output channel of switch T6. So, T91, X3 and X5 becomes 0. Hence Sum = 0 (X3 + X5) and Cout = 1 (T91 + X2).

CASE 7: When A = 1, B = 1 and C = 0. Light emerges out from T91 only i.e., from upper output channel of switch T9. So, X2, X3 and X5 becomes 0. Hence Sum = 0 (X3 + X5) and Cout = 1 (T91 + X2).

CASE 8: When A = 1, B = 1 and C = 1. Light emerges out from X5 and T91 i.e., from lower output channel of switch T7 and upper output channel of switch T9 respectively. So, X3 and X2 becomes 0. Hence Sum = 1 (X3 + X5) and Cout = 1 (T91 + X2).

In this way addition of any three bit number can be done with this circuit.

The parameters used in this simulation are taken from the literature survey of different research papers [_{ss}) = 30 dB, gain recovery time of SOA (τ_{e}) = 90 ps, saturation energy of the SOA (E_{sat}) = 700 fJ, eccentricity of the loop (T_{asym}) = 30 ps, line-width enhancement factor (α) = 6, full width at half maximum of control pulse (σ) = 6 ps, bit period (T_{c}) = 100 ps, and a control pulse energy (E_{cp}) = 70 fJ so that the operational conditions are satisfied. The simulated input and output waveforms of half adder are shown in

In this paper, we have reported parallel all-optical half adder and full adder. Here, in this proposed scheme, the significant advantage is that the proposed circuit can perform addition operations, which are all-optical in nature. This theoretical model has been verified through numerical simulation.

The authors are grateful to Technical Education Quality Improvement Programme (TEQIP) phase II by National Project Implementation Unit (Approval No.-CEMK/ TEQIP-lI/R&D/Project/15-16/03) for providing the grant for this work.

Gayen, D.K. (2016) Optical Arithmetic Operation Using Optical Demultiplexer. Circuits and Systems, 7, 3485-3493. http://dx.doi.org/10.4236/cs.2016.711296