Electrical Signal-To-Noise Ratio (SNR)

Optical SNR, although useful for design purposes, is not what governs the BER at the receiver. In this tutorial, we focus on the electrical SNR of the current generated when an ASE-degraded optical signal is incident on a photodetector. For simplicity of discussion, we use the configuration shown in the figure below and assume that a single optical amplifier is used before the receiver to amplify a low-power signal before it is detected. This configuration is sometimes used to improve the receiver sensitivity through optical pre-amplification.

1. ASE-Induced Current Fluctuations

When the contribution of ASE noise to the signal field E_s is included, the photocurrent generated at the receiver can be written as

where G is the amplifier gain, i_s and i_T are the current fluctuations induced by the shot noise and thermal noise, respectively, E_cp is the part of the ASE copolarized with the signal, and E_op is the orthogonally polarized part of the ASE. It is necessary to separate the ASE into two parts because only the copolarized part of the ASE can beat with the signal.

The ASE-induced current noise has its origin in the beating of E_s with E_cp and the beating of ASE with itself. To understand this beating phenomenon more clearly, notice that the ASE occurs over a broader bandwidth than the signal bandwidth Δν_s. It is useful to divide the ASE bandwidth Δν_o into M bins, each of bandwidth Δν_s, and write E_cp in the form

where φ_m is the phase of the noise component at the frequency ω_m = ω_l + m(2πΔν_s) and ω_l is the lower boundary of the filter passband. The spectral density of ASE for a lumped amplifier is given by

A form identical to that of E_cp above applies for E_op.

Using  and E_cp expression above and including all beating terms, the current I can be written in the form

where i_{sig - sp} and i_{sp - sp} represent current fluctuations resulting from signal-ASE and ASE-ASE beating, respectively, and are given by

since these two noise currents fluctuate rapidly with time, we need to find their averages and variances. It is easy to see that ⟨i_{sig - sp}⟩ vanishes. However, ⟨i_{sp - sp}⟩ has a finite value resulting from the terms in the double sum for which m = n. This average value is given by

Variances of the two noise currents can also be calculated by squaring them and averaging over time. We write the final result directly here:

where Δf is the effective noise bandwidth of the receiver. The total variance σ² of current fluctuations can be written as

where σ_T² is due to thermal noise and the shot-noise variance has an additional contribution resulting from the average, that is

where P_ASE = 2S_ASEΔν_o is the total ASE power entering the receiver.

2. Impact of ASE on SNR

We can now calculate the electrical SNR at the receiver. Noting that ⟨I⟩ = R_d(GP_s + P_ASE), the SNR is given by

The important question is whether the SNR_e is improved or degraded because of signal amplification before its detection. To answer this question, we compare the above equation with the SNR realized in the absence of optical amplifier. Setting G = 1 and P_ASE = 0, this SNR is given by i_{sig - sp}2    

Consider first the case of an ideal receiver with no thermal noise and 100% quantum efficiency so that R_d = q/hν₀. In this case, SNR without an amplifier is given by

When an optical amplifier is used, the current variance is dominated by σ²_{sig - sp}. Neglecting σ²_{sp - sp} and P_ASE in SNR_e equation above, the SNR is found to be

Using

the noise figure of an optical amplifier is found to be given by

In practice, thermal noise exceeds shot noise by a large amount, and it should be included before concluding that an optical amplifier always degrades SNR_e. Neglecting shot noise in SNR_e' expression above and retaining only the dominant term σ²_{sig - sp} in SNR_e expression, we find that

As this ratio can be made quite large by lowering P_s and increasing the amplifier gain G, electrical SNR can be improved by 20 dB or more compared with its value possible without amplification. This apparent contradiction can be understood by noting that the receiver noise, dominated by σ²_{sig - sp}, is so large in magnitude that thermal noise can be neglected in comparison to it. In other words, optical pre-amplification of the signal helps to mask the thermal noise, resulting in an improved SNR. In fact, if we retain only the dominant noise term, the electrical SNR of the amplified signal becomes

This should be compared with the optical SNR of GP_s/P_ASE. As seen in the equation above, electrical SNR is higher by a factor of Δν_o/(2Δf) under identical conditions since ASE noise contributes only over the receiver bandwidth Δf that is much narrower than the filter bandwidth Δν_o.

3. Noise Buildup in an Amplifier Chain

As mentioned earlier, long-haul lightwave systems require cascading of multiple optical amplifiers in a periodic fashion. The buildup of ASE noise is the most critical factor for such systems for two reasons. First, in a cascaded chain of optical amplifiers, the ASE accumulates over many amplifiers and degrades the SNR as the number of amplifiers increases. Second, as the level of ASE grows, it begins to saturate optical amplifiers and reduce the gain of amplifiers located further down the fiber link. The net result is that the signal level drops while the ASE level increases. Numerical simulations show that the system is self-regulating in the sense that the total power obtained by adding the signal and ASE powers (P_TOT = P_s + P_ASE) remains relatively constant. The figure below shows this self-regulating behavior for a cascaded chain of 100 amplifiers with 100-km spacing and 35-dB small-signal gain. The power launched by the transmitter is 1 mW. The other parameters are n_sp = 1.3 and G₀exp(- αL_A) = 3. The signal and ASE powers become comparable after 10,000 km. Clearly, ASE-induced gain saturation should be avoided as much as possible. We assume this to be the case in the following discussion.

To estimate the SNR associated with a long-haul lightwave system, we assume that all amplifiers are spaced apart by a constant distance L_A, and the amplifier gain G ≡ exp(αL_A) is just large enough to compensate for fiber losses in each fiber section. The total ASE power for a chain of N_A amplifiers is then given by

and can be used to find the optical SNR using SNR_o = P_in/P_ASE. To calculate the electrical SNR, we need to find the current variance at the receiver. If we assume that the receiver noise is dominated by the signal-ASE beat noise and include only this dominant contribution, the electrical SNR is found to be

This equation shows that the electrical SNR is reduced by a factor of N_A simply because ASE noise is increased by that factor. However, one should not conclude from this equation that the system performance can be improved by placing fewer amplifiers along the link. As discussed before, if we reduce the number of amplifiers, each amplifier would add more noise because it has to operate with a higher gain. The results shown in the figure below also apply here because the optical and electrical SNRs are related to each other as SNR_e/SNR_o = Δν_o/(2Δf).