# Dispersion Power Penalty Modeling (Part 3)

#### Deriving Equation 2

Equation 2 is derived from Equation 7 by noting the following items.

• A true normal pulse has infinite length, so we cannot have a high speed data system that sends true normal pulses.
• A common choice is to select a bit time that will contain 95% of the bit energy at the point of transmission.
• If you set the bit time equal to ±2$\sigma_0$ about the signal peak ($4 \cdot \sigma_0$ total), the bit time will contain 95% of the normal pulse's energy.
• This choice is much more conservative than the choice in Equation 1, which can be shown to only have 31% of the pulse energy contained in the bit time.

Given these assumptions, we can state Equation 10 directly.

Eq. 10      $\sigma_0 = \frac{T}{4} = \frac{1}{4 \cdot B}$

We can substitute Equation 10 into Equation 7 to obtain Equation 2.

$P{P_D} = 10 \cdot \log \left( {\sqrt {1 + {{\left( {\frac{{D \cdot L \cdot {\sigma _\lambda }}}{{\frac{1}{{4 \cdot B}}}}} \right)}^2}} } \right)$

$P{P_D} = 5 \cdot \log \left( {1 + {{\left( {4 \cdot B \cdot D \cdot L \cdot {\sigma _\lambda }} \right)}^2} } \right)$

#### Deriving Equation 3

Equation 3 is derived from Equation 7 by noting the following items.

• Select a bit time that will contain 95% of the signal energy within the bit time at the point of reception.
• If you set the bit time equal to $\pm 2 \cdot \sigma_0$ about the signal peak ($4 \cdot \sigma_0$ total), the bit time will contain 95% of the normal pulse's energy.

This derivation is a bit more complicated than the previous derivations because we need to calculate $\sigma_0$ given that the signal $\sigma$ at distance L has been stretched by the factor k, which is computed using the following equation.

$k = \sqrt {1 + {{\left( {\frac{{D \cdot L \cdot {\sigma _\lambda }}}{{{\sigma _0}}}} \right)}^2}}$

Given the factor k, we can compute $\sigma_0$ as follows.

$\sigma\left(\text{Distance = L}\right) = \frac{1}{{4 \cdot B}} \Rightarrow {\sigma _0} = \frac{1}{{4 \cdot B \cdot k}}$

Substituting this equation into Equation 7 gives us the following.

${k^2} = 1 + {\left( {\frac{{D \cdot L \cdot {\sigma _\lambda }}}{{\frac{1}{{4 \cdot B \cdot k}}}}} \right)^2} = 1 + {\left( {4 \cdot B \cdot D \cdot L \cdot {\sigma _\lambda } \cdot k} \right)^2}$

This equation can be solved for the k to yield

$k = \sqrt {\frac{1}{{1 - {{\left( {4 \cdot B \cdot D \cdot L \cdot {\sigma _\lambda }} \right)}^2}}}}$

This equation does not only represent the elongation of the pulse – it also represents the reduction in amplitude of the pulse. Therefore, we simply convert this equation to dB to get Equation 3.
$P{P_D} = 10 \cdot \log \left( {\sqrt {\frac{1}{{1 - {{\left( {4 \cdot B \cdot D \cdot L \cdot {\sigma _\lambda }} \right)}^2}}}} } \right) = - 5 \cdot \log \left( {1 - {{\left( {4 \cdot B \cdot D \cdot L \cdot {\sigma _\lambda }} \right)}^2}} \right)$

This equation is very commonly used. It is different from the Equations 1 and 2 in that it contains a singularity. This singularity occurs because there are distances beyond which the dispersion is so severe that 95% of the bit energy cannot be contained in a bit time at the receiver. This characteristic has confused many people because they are not used to thinking about ISI at the receiver.

#### Deriving Equation 4

Equation 4 is Equation 3 after applying a Taylor series approximation for the reciprocal of the square root. The approximation is shown below and is only valid for small x.

$\frac{1}{{\sqrt {1 - {x^2}} }} \approx 1 + \frac{1}{2} \cdot {x^2}$

This approximation can used to simplify the $PP_D$ a bit.

$PP_D = 10 \cdot \log \left( {\frac{1}{\sqrt {1 - {\left( 4 \cdot B\cdot D \cdot L \cdot \sigma_\lambda \right)}^2}} }\right) \approx 10 \cdot \log \left( {1 + \frac{1}{2} \cdot {{\left( {4 \cdot B \cdot D \cdot L \cdot {\sigma _\lambda }} \right)}^2}} \right)$

This gives us Equation 4. It is important to remember that this equation is only valid when the dispersion power penalty is small. I do not view this as a serious restriction because I would not deploy a system with a large dispersion power penalty.

### Conclusion

I know this was long, but it will be useful. Here is what was accomplished:

• The commonly used dispersion power penalty formulas were put in one place.
• Their underlying assumptions were examined in detail.
• It was demonstrated that they all come from the same basic equation.

My next blog posts will be shorter and less specialized.

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### 2 Responses to Dispersion Power Penalty Modeling (Part 3)

1. mruffini says:

Thanks for your post. The definition of a power penalty thought it's usually the increase in power at the receiver to obtain the same BER compared to the case without the impairment.
It seems here that power penalty is only related to the loss of power due to pulse broadening, while the effect of ISI is not taken into account. Is this the reason why usually the maximum dispersion power penalty allowed is 2 dB?

• mathscinotes says:

Hi Marco,

This blog post was outcome of a discussion I had with several optics folks on the various equations used to model dispersion. The models I covered are the most common and they only model pulse broadening. All of the engineers in our little discussion group have made extensive measurements of dispersion and had found that the equations were really only useful at low power penalties. We only use the equations in this blog post to tell us whether dispersion is significant or not. We never allow a PON to be deployed with significant dispersion power penalty. For example, my company has never deployed a PON with more than 1 dB of dispersion power penalty.

At one time, I had a salesman that used one of these equations to recommend that we increase our power penalty budget, reduce our PON split, and run the PONs longer distance. We took a look at this in the lab and found that PONs with a computed power penalty significantly greater than 1 dB turned out to have erratic performance. The spectral width of the laser varies randomly and the worst-case power penalty proved difficult to predict. My customers need rock solid performance and a PON with variable BER characteristics was not acceptable.

I see you work at Trinity College. I was just there with my wife in September. I had a great time and I plan to spend more time in Ireland.

mathscinotes