Negative times a Negative is a Positive

Posted on 12-April-2015 by mathscinotes

Don't just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis?

— Paul Halmos on how to study mathematics
Figure 1: My Favorite Advertizing Campaign.

Figure 1: A Math Poster from My
Favorite Advertising Campaign
(Source).

On my team, I work hard to ensure that we have a non-threatening environment for questions – any questions. In fact, I often ask very basic questions in meetings so that I can make sure that I understand all the nuances of a situation. You would be amazed how often I learn things from asking questions so basic that you would think asking them would not be necessary.

Because of this approach, I frequently have non-technical people come to me and ask questions about many things, mainly remodeling and basic mathematics. I will cover the remodeling questions in other posts.

The other day I had a person come to me with the following basic mathematics question:

Why does a -1 times -1 equal a 1?

This is actually an interesting question and I last dealt with it when my kids were in school. From a mathematical standpoint, the best answer is that -1*-1 = 1 to ensure that the distributive property works, which I illustrate in Equation 1.

Eq. 1 \displaystyle 0=\text{-}1\cdot 0=\text{-}1\cdot \left( {1+\text{-}1} \right)=\text{-}1\cdot 1+\text{-}1\cdot \text{-}1=\text{-}1+1

However, this answer is not appropriate for most folks, especially schoolchildren, because they hate logical arguments and want a concrete example.

The best way to describe basic math operations is to work with analogies from their daily lives. There are numerous analogies for -1*-1=1, but for this individual the following analogy seemed to work best because they have purchased a new car.

When you buy a car, the car usually has a base price that includes a standard set of features. If you add a feature (e.g. high-end audio system), the price of the car increases. If you remove a feature (e.g. leather seats), the price of the car drops. Suppose you have selected a car with a set of features that includes price increases and decreases. What should happen to the price of a car when you remove a price reduction? It should increase. This is the same as saying that a negative times a negative is a positive.

With my kids, I used time analogies.

You hate driving, but you need to go on a long road trip to numerous locations. You assume that each destination  will add an hour of miserable driving, which you view as a negative value. If you have 10 destinations to visit, you have 10 miserable hours of driving (10*-1=-10). Suppose you remove a destination (-1*-1), now you just have 9 miserable hours of driving. Again, a negative times a negative is a positive.

Posted in General Mathematics | Comments Off on Negative times a Negative is a Positive

CML Termination Design

Posted on 11-April-2015 by mathscinotes

They're flaring gas and using diesel to fuel the pumps—it's like something Homer Simpson would do.

— North Dakota cattle rancher Vawnita Best, after seeing natural gas being flared off of wells on her property for over a year. She was simply appalled at the waste.
Figure 1: Termination Circuit.

Figure 1: Termination Circuit.

An engineer stopped by today and had forgotten how to use Mathcad to solve a pair of circuit equations to determine termination resistor values. I stepped in and helped. As I looked at the problem, I thought this might be useful to cover here as a tool demonstration.

I continue on my quest to show the usefulness of computer-algebra systems to masses. Depending on the problem at hand, I use either Mathcad or Mathematica. You only see Mathcad on this blog because I generally write on my lunch break and we only have Mathcad at work. I actually love to work with both tools, but Mathcad is better for quick calculations – it was born as an engineering tool.

Figure 1 shows the circuit in question. This circuit may look a bit strange, but it reflects the differential nature of Current-Mode Logic (CML).

We have the following requirements for this circuit:

  • V1=3.295 V
  • V2 = 1.2 V
  • VT = 1.2 V
  • RT = 50 Ω
  • R1 and R2 are related to RT by the following equation
    R_T={{R}_{1}}+\frac{1}{{\frac{1}{{{{R}_{2}}}}+\frac{1}{{{{R}_{T}}}}}}

Our mission is to select values for R1 and R2 to meet these requirements. Figure 2 shows the Mathcad solution I wrote out in less than a minute. I directly wrote out Kirchoff's voltage equations in Mathcad with no thinking. I then let Mathcad solve the system.

Figure 2: First Solution, No Thinking.

Figure 2: First Solution, No Thinking.

If this were my design, I would finish off the calculation by selecting standard resistor values. I have a Mathcad program that does that for me. For this situation, the engineer working on the circuit will select the final part values.

As I thought about the circuit, I did not need to write such a complex equation for V2. Since VT = V2, no current is flowing through RT. This means I can write a simpler set of circuit equations that provide me the same solution, which I show in Figure 3. As expected, I get the same results.

Figure 2: Simpler Approach Using Circuit Insights.

Figure 3: Simpler Approach Using Circuit Insights.

My intent here is just to illustrate how useful these tools are in an engineer's daily life. I rest my case.

Postscript

The engineer that I worked with on this problem sent me his lab results. This may not mean much to most of you, but the testing verified that our modeling worked as expected. We have a properly terminated signal with the correct levels.

Figure M: Confirmation in the Lab.

Figure 4: Confirmation in the Lab.

Posted in Electronics | 1 Comment

Rate of Technological Progress

Posted on 10-April-2015 by mathscinotes

History ... is, indeed, little more than the register of the crimes, follies, and misfortunes of mankind. But what experience and history teach is this - that peoples and governments have never learned anything from history, or acted on principles deduced from it.

— Georg Wilhelm Friedrich Hegel
Figure 1: PAM Dirac, a Key Contributor to Understanding the Physics of the Electron.

Figure 1: PAM Dirac, a Key
Contributor to Understanding
the Physics of the Electron.

I saw the following LED history graphic in Machine Design magazine today. I like the graphic because it shows how technological changes often occurs – I only wish there had been some additional space for the quantum mechanical developments of the 1920s.

The basic development process is:

  • Someone notices an interesting phenomenon that people do not understand (e.g. light emitted by electrically stimulated crystals).
  • Theoreticians working in a seemingly unrelated area develop a mathematical framework to describe the related physics (e.g. Dirac, Bloch et al. developing the theory of electrons and the band theory of solids).
  • First applications of the new physics appear (e.g. solid state physics applied to the development of the transistor).
  • A period of time passes while the full impact of the new physics becomes part of the mainstream. During this time the original phenomenon is fully explained.
  • The applications folks now begin to apply their knowledge of the original phenomenon to the first mass market applications (e.g. red LED developed).
  • The technology is generalized (e.g. yellow, blue, white LED technology developed).
  • The technology is everywhere (e.g. televisions, home lighting, etc).
Figure 1: History of the LED.

Figure 1: History of the LED.

Posted in History of Science and Technology | Comments Off on Rate of Technological Progress

Prime Number Magnitudes

Posted on 9-April-2015 by mathscinotes

If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.

— Winston Churchill
Figure 1: Prime Garden (renoir_girl).

Figure 1: Prime Garden.
renoir_girl: https://www.flickr.com/photos/renoir_girl/

I am responsible for some of the authentication features in our products and these features use prime numbers. People often have basic questions on prime numbers, such as:

  • What happens if I choose the same prime number as someone else?
  • Are there enough prime numbers?

I normally give some nebulous answer like "there are more 512-bit prime numbers than there are atoms in the Universe". Recently, I saw a great response on Quora that quoted  Bruce Schneier's Applied Cryptography book on prime numbers. Here are the quotes:

If everyone needs a different prime number, won’t we run out? No. In fact, there are approximately 10^151 primes 512 bits in length or less. For numbers near n, the probability that a random number is prime is approximately one in ln n. So the total number of primes less than n is n /(ln n). There are only 10^77 atoms in the universe. If every atom in the universe needed a billion new primes every microsecond from the beginning of time until now, you would only need 10^109 primes; there would still be approximately 10^151 512-bit primes left.

What if two people accidentally pick the same prime number? It won’t happen. With over 10^151 prime numbers to choose from, the odds of that happening are significantly less than the odds of your computer spontaneously combusting at the exact moment you win the lottery.

If someone creates a database of all primes, won’t he be able to use that database to break public-key algorithms? Yes, but he can’t do it. If you could store one gigabyte of information on a drive weighing one gram, then a list of just the 512-bit primes would weigh so much that it would exceed the Chandrasekhar limit and collapse into a black hole...so you couldn't retrieve the data anyway

Schneier's responses are fantastic and I thought I would explore the numbers behind them just a bit more. I plugged some of these number into a Mathcad worksheet and also provided some links to backup material.

Calculations

number of primes number of atoms mass of the universe
Posted in General Mathematics, software | Comments Off on Prime Number Magnitudes

Temperature Sensing and a Current Ratio of 17

Posted on 8-April-2015 by mathscinotes

Mathematics is the part of physics where experiments are cheap.

Introduction

Figure 1: My Idea of a Thermometer.

Figure 1: My Idea of a Thermometer.

One of the most common diagnostic functions requested for an electronic system is to measure its own temperature. We want to know the hardware temperature when a problem occurs because many system characteristics are affected by temperature and temperature may give us a clue to the problem's root cause.

Today, I was looking at the specification for a temperature measurement integrated circuit (TMP75B) and I noticed that it has a very common block diagram – two switched current sources of different drive levels feeding a transistor's base-emitter junction.

As I looked at this block diagram, I recalled reading that the current ratio of the current sources is normally set to 17. You don't often see the number 17 randomly appear in electronics, so there must be a reason. Let's do a little mathematical experimentation to see if we can figure out why.

Background

Block Diagram

Figure 2 shows a block diagram of the TMP75B. Notice the two current sources that drive the base-emitter junction of a diode-connected, bipolar transistor. This is a very common architecture for an integrated circuit-based, temperature sensor.

Figure 2: Block Diagram of TI TMP75B.

Figure 2: Block Diagram of TI's TMP75B.

Ratio Quote

Here is a quote from an ON Semiconductor specification that mentions that the current ratio is typically set at 17.

Setting Ic1 as a fixed multiple, N, of Ic2 gives:

Eq. 1 \displaystyle \Delta {{V}_{{BE}}}=\frac{{{{k}_{b}}\cdot T}}{q}\cdot \ln \left( N \right)

This is the equation used internally in 2−current ON Semiconductor devices to calculate temperature based on the difference in Vbe measurements. The typical value used for N is 17.

I have seen 17 used as the current ratio for other switched, dual-current source temperature sensor circuits, but I do not recall where.

Definitions

Before we proceed to the Analysis section, we need to define a few terms.

Analog-to-Digital Converter (ADC)
An ADC is a device that converts a continuous physical quantity (usually voltage) to a digital number that represents the quantity's amplitude (Wikipedia).
ADC Count Number (NADC)
The number of discrete binary numbers (sometimes called codes) available to represent a range of analog input, like 0 V to 1 V. The ADC count number is normally a power of 2 (Wikipedia).
LSB Voltage (δVBE)
The minimum change in voltage required to guarantee a change in the output code level is called the Least Significant Bit (LSB) voltage (Wikipedia).

Analysis

Approach

First, I will derive Equation 1. We can use Equation 1 to derive Equation 2, which is very useful for designers because it relates the temperature resolution to the current ratio and ADC resolution.

Eq. 2 \displaystyle \ln \left( N \right)\cdot {{N}_{{ADC}}}=\frac{{{{V}_\text{Ref}}\cdot q}}{{{{k}_{b}}\cdot \delta T}}

where

  • kb is Boltzmann's constant.
  • δT is temperature resolution, which is smallest temperature change the system can measure.
  • q is elementary charge.
  • VRef is the ADC reference voltage. This represents the high-end analog voltage that the ADC will convert properly.

Setup

Figure 3 shows the how I defined the critical constants in the analysis to follow. I also show how we can compute δT knowing the positive temperature range and maximum positive ADC value.

Figure 3: Analysis Setup and Temp Sensor Requirements.

Figure 3: Analysis Setup and Temperature Sensor Requirements.

Derivation of Equation 1

We can derive Equation 1 as shown in Figure 4.

Figure 4: Derivation of Equation 1.

Figure 4: Derivation of Equation 1.

Derive Equation 2

Figure 5 shows how we can derive Equation 2. This is the key equation needed to explain the appearance of N= 17.

Figure 5: Relationship Between ADC Resolution and Current Ratio.

Figure 5: Relationship Between ADC Resolution and Current Ratio.

Numerical Example

In Figure 6, I assume that the voltage used for the ADC reference voltage as 1 V (a guess). I can then evaluate the right-hand side of Equation 2 to determine N for various ADC count values.

Figure 6: Evaluation of Equation 2.

Figure 6: Evaluation of Equation 2.

As you can see in Figure 6, a current ratio of 17 appears when we use an ADC with 16 bits. The other ratios are large, which would be difficult to control accurately. So we may have found where ratio of 17 comes from.

Dynamic Range

One issue remains – the TMP75B only reports back 12 bits of temperature information, yet I am assuming a 16-bit ADC. What limits the amount of information that can be returned?

Figure 7 investigates that issue and, like many engineering issues, the problem has to do with dynamic range. While we need a 16-bit converter to get the temperature measurement resolution we want, the input voltage does not vary enough to use all 16-bits of the ADC.

Figure 7: Dynamic Range of Temperature Reading.

Figure 7: Dynamic Range of Temperature Reading.

The analysis in Figure 7 shows that there is only 12 bits of information in the data returned from the ADC because ΔVBE is a small voltage and will only use a fraction of the 16-bit ADC's dynamic range.

Conclusion

I came up with a theory as to where the current ratio of 17 came from. It may not be totally correct for each part, but it is a reasonable working theory given my lack of internal part details.

Posted in Electronics | Comments Off on Temperature Sensing and a Current Ratio of 17

Reasonable Budget Plan for a Young Person

Posted on 7-April-2015 by mathscinotes

It's clearly a budget. It's got a lot of numbers in it.

- George W. Bush

In my role as an annoying father, I regularly talk to my sons about financial matters and this post summaries a recent discussion on budgeting. It's the last thing on their minds, but I wouldn't be a good dad if I didn't prepare them for the financial shock that is independent living. One reasonable budget plan for a young person is called the "60% solution" and its percentages are based on gross income.

Depending on their career path after university, young people could be making good money. However, even those in medical professions don't always have the greatest financial situation. Young doctors or physicians may well want to check out LeverageRX for advice on loans, mortgages, and more to set them on the right path to financial success. These days, everyone is struggling with finances. Be it managing the expenses, paying loans, medical bills, saving up for retirement, and more, youngsters do not realize where their money is going.

But then, the Internet also puts this generation at a greater advantage. Along with managing their studies, they could earn additional income through a number of ways. Besides the usual part-time job, they can gain from a multitude of online gigs -- title loan affiliate programs, freelancing for content marketing companies and much more. The idea is to explore all possibilities and opportunities out there.

Having explained that, let's return to our main discussion on how to plan the budget. The plan is simple and can be summarized as follows:

  • 60% to committed expenses

    I think of this category as fixed expenses (e.g. housing, food, insurance, phone) and taxes. A typical Renter will spend most of their income on rent and utilities, hence why this section is so big.

  • 10% to retirement.

    I might argue for 15%, but 10% is a good place to start when you are young. The earlier you start saving for retirement, the better your retiremnt will be, even if it feels a long way off.

  • 10% to short-term savings

    Life often presents us with unexpected bills, which I call "knowable unknowables". This category includes things like a transmission failure on a relatively new car.

  • 10% to long-term savings/debt money

    You know over time you will need to replace your car, upgrade your house, and pay off debt.

  • 10% for personal enjoyment

    The stuff that makes life fun. You can stretch this one a little bit, but not too much. After all, there's no point in going through the years saving all of your money only to not have enjoyed your life!

This approach is discussed in detail at a number of web sites, like:

Here is a video with Sharon Epperson that summarizes the approach.

Posted in Financial | Comments Off on Reasonable Budget Plan for a Young Person

Computing Percent Differences

Posted on 6-April-2015 by mathscinotes

Information consists of the differences that make a difference.

— Edward Tufte

I have recently been computing a lot of percentage differences – mainly in variance calculations. I have been using the formula that I was taught in 7th grade, namely

\displaystyle \Delta\%=\frac{N-O}{\text{O}}

where

  • N is an updated or new number.
  • O is the original number.
  • Δ% is the percentage change.

I recently discovered that this formula fails miserably when dealing with negative quantities – I had never considered what happens when the O variable is negative.

Consider the case when O = -4 and N = -2. This should reflect a positive improvement of 50%, but instead the sign is negative.

\displaystyle \Delta\%=\frac{-2--4}{-4} = \frac{2}{-4}=-50\%

Looks like I have been computing percent differences incorrectly since 7th grade. Here is how I am going to compute percent differences going forward.

\displaystyle \Delta\%=\frac{N-O}{\left| O \right|}

I did a bit of googling and some folks prefer to the use the sign instead of the absolute value function in their percent difference formula (example).

Posted in Financial | Comments Off on Computing Percent Differences

Lighthouse Visual Ranges

Posted on 5-April-2015 by mathscinotes

Extraordinary claims require extraordinary evidence.

— Carl Sagan

Introduction

Figure 1: Flat Point Lighthouse in Nova Scotia.

Figure 1: Flat Point Lighthouse in Nova Scotia.

It is Easter and I have a terrible cold − I am not going to see family this year. Instead, I am going to lay here and write for a while.

Yesterday, I received a question from a reader who was puzzled by a web page written by a flat earther that presented a seemingly rational argument in favor of the flat earth position. In a nutshell, the flat earther's argument says that to see a lighthouse at long distance on spherical Earth would mean that you would have to be able to see around the horizon, which they claim is not possible. Therefore, the Earth must be flat. I may be mischaracterizing their argument, so you may want to visit web sites that go into the details of the flat earth rationale. Of course, I argue that refraction can and does literally allow you to see "around" the horizon.

The questioner's request was to help him understand the fallacies in the arguments being made. I generally avoid these discussions and simply refer people who write me on this subject to wikis and blogs that are focused on these topics. For example, in the case of the flat earthers, the RationalWiki does a good job of debunking their arguments. However, I am in bed and feeling cranky, so here I go.

While I will not argue about non-falsifiable arguments, I can discuss the parameters critical to computing lighthouse ranges and how nominal lighthouse ranges are computed in a standard table of lighthouse ranges. As a side topic in Appendix A, I review one of flat earther calculations and with my own calculation show that refraction and tidal water level changes can easily explain the ability to see a lighthouse over-the-horizon. You do not need to assume that the Earth is flat.

Background

I have done a fair amount of optical range modeling in the Earth's atmosphere. Truth be told, this is all related to my interest in battleships and optical fire control systems. However, I did write a blog post that referenced lighthouses and the material presented here is based on that post.

Analysis

Formula

Starting Point

Equation 1 gives the distance one can see from an elevated point (e.g. lighthouse or battleship mast) to a point on the horizon, assuming a nominal level of refraction.

Eq. 1 \displaystyle s\left[ {\text{km}} \right]=3.86\cdot \sqrt{{h\left[ m \right]}}

where

    • s is the distance along the Earth in kilometers.
    • h is the height above the sea in meters.

The weakness of this formula is that the front coefficient (3.86) assumes a specific lapse rate, which describes the atmospheric temperature variation with height. Variations in the lapse rate can produce enormous increases or decreases in the visual range of a lighthouse. Variations in the lapse coefficient can even explain atmospheric ducts that can guide optical signals (and other electromagnetic signals) over very long ranges.

Also, the height of the lighthouse above the sea varies with time (i.e. tides) and weather (e.g. storm surge). Most official tables of lighthouse ranges are based on the mean high water level about the lighthouse.

Model Representation

Figure 2 shows how the visual range of a lighthouse depends on both the height of the lighthouse and the viewer. Figure 2 is actually taken from a post on battleships, but the geometry is identical for both cases. This means that I must apply Equation 1 twice:

  • Compute the range from the lighthouse to the horizon.
  • Compute the range from the viewer to the horizon.
  • Add the ranges together.

Figure 2: Figure from My Battleship Work --- Also True for Viewing Lighthouses.

Lighthouse Formula

Figure 3 shows my lighthouse formula. I also grabbed some data from an old UK publication on nominal lighthouse viewing ranges for comparison.

Figure M: My Lighthouse Formula.

Figure 3: My Lighthouse Formula.

This formula also reveals another source of variation − the height of the observer above the sea. This will vary since the height of the deck of each ship is different. The standard for computation is 15 feet, but that can vary dramatically by the type of ship. For example, the flat earther web page in question here assumed a 24-foot high deck. To duplicate the UK table values, I needed to use 15 feet for the deck height.

Publication from Lighthouse Manufacturer

Results

Figure 4 shows that Equation 1 from my blog post produces the same result as that specified for nominal lighthouse ranges. So I believe that my model is reasonable.

Figure M: My Lighthouse Ranges Versus an Old Specification.

Figure 4: My Lighthouse Ranges Versus an Old Specification.

Conclusion

Reported lighthouse ranges are a strongly dependent of the assumptions made in their calculation.  You do not need to assume the Earth is flat to explain how lighthouse range estimates can vary so much from a pure spherical model. The variation naturally occurs because of changes in water level and temperature.

Appendix A: Worked Flat Earther Example

I will work one of the flat earther examples. My intent is to first show how they performed their calculation. I have highlighted the statement "not an uncommon thing", which is a very imprecise statement. In general, lighthouse ranges are stated in a conservative manner that ensures that they are almost always visible at the stated range. This statement indicates that the ranges quoted here are NOT common.

The distance across St. George's Channel, between Holyhead and Kingstown Harbour, near Dublin, is at least 60 statute miles. It is not an uncommon thing for passengers to notice, when in, and for a considerable distance beyond the centre of the Channel, the Light on Holyhead Pier, and the Poolbeg Light in Dublin Bay. The Lighthouse on Holyhead Pier shows a red light at an elevation of 44 feet above high water; and the Poolbeg Lighthouse exhibits two bright lights at an altitude of 68 feet; so that a vessel in the middle of the Channel would be 30 miles from each light; and allowing the observer to be on deck, and 24 feet above the water, the horizon on a globe would be 6 miles away. Deducting 6 miles from 30, the distance from the horizon to Holyhead, on the one hand, and to Dublin Bay on the other, would be 24 miles. The square of 24, multiplied by 8 inches, shows a declination of 384 feet. The altitude of the lights in Poolbeg Lighthouse is 68 feet; and of the red light on Holyhead Pier, 44 feet. Hence, if the earth were a globe, the former would always be 316 feet and the latter 340 feet below the horizon!

Figure 5 shows how I duplicated the flat earther's calculations. There are some issues with their calculation:

  • purely geometric, no refraction.
  • assumes a 44-foot Hollyhead lighthouse (Wikipedia states its height as 70-feet ) and a 68-foot Poolbeg lighthouse (reference states 66-feet )
  • Lighthouse heights are normally stated with respect to the high-water level, but I am unclear what the flat earthers are using.
Figure M: Worked As a Flat Earther.

Figure 5: Worked As a Flat Earther.

Figure 6 shows the geometry that I will be assuming here.

Figure 6: Basic Ship Geometry.

Figure 6: Basic Ship Geometry.

Figure 7 shows that the lights could easily seen at low tide and with reasonable atmospheric conditions.

Figure M: Refracton Makes the Lighthouses Visible on a Spherical Earth.

Figure 7: Refraction Makes the Lighthouses Visible on a Spherical Earth.

Posted in optics | 17 Comments

Earth With All the Ice Melted

Posted on 4-April-2015 by mathscinotes

Peace is that glorious moment in history when everyone stands around reloading.

— Thomas Jefferson

Last night, one of my sons and I discussed using R to process geographic data. Related to this topic, we also discussed how to visualize the impact of global sea level rise on certain countries. This topic came up because we had seen a report on television that Venice is experiencing more acqua alta – high water – events.

While digging around the Internet, I found the following video that does a good job showing what would happen to the continental outlines if all the ice on Earth melted, which would raise sea level by ~70 meters.

National Geographic also has a good article on the same topic.

Posted in General Science | Comments Off on Earth With All the Ice Melted

Return Loss of a Glass-Air Interface

Posted on 3-April-2015 by mathscinotes

...the acquisition of knowledge is hard ... Our minds are prone to illusions, fallacies, and superstitions ....

— Steven Pinker

Introduction

Figure 1: Return Measurement Example.

Figure 1: Return Loss Measurement Example.

I received an email today about a deployment issue that involved the reflection of light from unterminated connectors. When light travels down a fiber and encounters a change in the index of refraction, part of the energy will reflect back toward the transmitter because of a phenomenon called Fresnel reflection, which I define below (source).

Fresnel reflection is something we all encounter in our daily life. It is what causes the reflections we see in our windows. Here is a quote from an optics text.

When a ray of light strikes a change of refractive index, and is approaching at an angle close to normal, most of the light passes straight through....  Most of the light but not all. A very small proportion is reflected back off of the boundary. We have seen this effect with normal window glass. Looking at a clean window we can see two images. We can see the scene in front of us and we can also see the faint reflection of what is behind us. Light therefore is passing through the glass and is also being reflected off the surface.

Background

Definitions

We quantify the size of this return "echo" using a parameter called return loss. Let's begin by precisely defining both Fresnel reflection and return loss. Note the 4% transmission loss (aka reflected power) figure highlighted below – I will be explaining where this comes from later.

Fresnel Reflection
In optics, the reflection of a portion of incident light at a discrete interface between two media having different refractive indices. Fresnel reflection occurs at the air-glass interfaces at the entrance and exit ends of an optical fiber. Resultant transmission losses, on the order of 4% per interface, can be reduced considerably by the use of index-matching materials. The coefficient of reflection depends upon the refractive index difference, the angle of incidence, and the polarization of the incident radiation. For a normal ray, the fraction of reflected incident power is given by

\displaystyle R=\frac{{{{{\left( {{{n}_{1}}-{{n}_{2}}} \right)}}^{2}}}}{{{{{\left( {{{n}_{1}}+{{n}_{2}}} \right)}}^{2}}}}

where R is the reflection coefficient and n1 and n2 are the respective refractive indices of the two media.

Return Loss
In telecommunications, return loss is the loss of power in the signal returned/reflected by a discontinuity in a transmission line or optical fiber. This discontinuity can be a mismatch with the terminating load or with a device inserted in the line. It is usually expressed as a ratio in decibels (dB):

\displaystyle RL(\text{dB})=10{{\log }_{{10}}}\frac{{{{P}_{\text{i}}}}}{{{{P}_{\text{r}}}}}

where RL(dB) is the return loss in dB, Pi is the incident power and Pr is the reflected power.

Connector Types

Figure 2 shows common ways of polishing the end of a fiber. Fiber-To-The-Home system usually use the Angle Polished Connector (APC) type because we do not want the reflections to interfere with our analog video transmission, which are VERY sensitive to reflections. The Ultra-Polished Connectors (UPC) are used for most other applications.

Figure 2: Common Fiber-Optic Connector Polishing.

Figure 2: Common Fiber-Optic Connector Polishing.

Usage Example

I usually see the return loss expressed in terms of dB. For example, the following quote is from an ADC document on connectors

When connectors are unmated — such as unused ports in an FTTP distribution frame—the return loss for APC connectors is -65 dB or greater, compared to UPC connectors that will be in the neighborhood of 14 dB. This is an important consideration for building today’s FTTX architectures.

I will show that the 4% and 14 dB numbers actually represent the same value, which I will demonstrate next.

Analysis

Figure 2 shows how you compute the reflection percentage and the return loss for the fiber/air interface that you encounter with an unterminated, Ultra-Polished Connector (UPC).

Figure 2: Return Loss and Reflection Percentage Calculation.

Figure 2: Return Loss and Reflection Percentage Calculation.

Wikipedia Return Loss Discussion Index of Refraction of Air Index of Refraction for Various forms of Silicon Dioxide

Conclusion

Just a short calculation, but I wanted to document it so that people can see where the 4% and 14 dB numbers come from.

Posted in Fiber Optics | 2 Comments