## Introduction

I received an email yesterday from a sales engineer who was having difficulty measuring the ring voltage on one of our telephone circuits. The numbers he was getting did not agree with what my engineering group had measured. The discrepancy had to do with the shape of the ring voltage waveform and what a standard voltmeter actually measures. As with many engineering problems, developing a practically useful solution depends on coming to an agreement on definitions and determining what the instruments are really measuring. This is a harder thing than one might think and this problem provides a nice illustration of basic problem solving.

## Analysis

### Ring Voltage Specification

The ring voltage on a telephone is specified as a Root-Mean-Square (RMS) voltage. Electrical engineers like to use RMS voltages for varying waveforms because the RMS voltage can be thought of as the equivalent DC voltage with respect to producing power in the load. Many people think of RMS voltage as a form of “average” voltage, even though it really is the average of the square of the voltage.

The Wikipedia defines the RMS voltage of a periodic signal in terms of an integral (Equation 1).

 Eq. 1 ${V_{RMS}} = \sqrt {\frac{1}{T}\int\limits_0^T {v{{\left( t \right)}^2}dt} }$

where T is the period of the signal and v(t) is the function that describes the waveform.

### Ring Voltage Waveform

Figure 1 illustrates the ring voltage waveform that the sales engineer was dealing with. Trapezoidal waveforms in telephony are common.

Figure 1: Example of a Trapezoidal Telephone Ring Voltage

The voltage waveform actually goes both positive and negative, but the polarity does not matter when calculating power into a resistive load. We will work here with the positive side.

### Computing the RMS Voltage of a Trapezoidal Waveform

Equations 2-4 illustrate how to derive and expression for the RMS voltage of a trapezoidal waveform in terms of k and T, which are defined in Figure 1. This derivation assumes that the triangular portions of the trapezoid are identical.

 Eq. 2 $V_{RMS}^2 = \frac{1}{T}\int\limits_0^T {v{{\left( t \right)}^2}dt} = \frac{2}{T} \cdot \int\limits_0^k {{{\left( {\frac{{A \cdot t}}{k}} \right)}^2}dt} + \frac{1}{T} \cdot \int\limits_0^{T - 2 \cdot k} {{A^2}dt}$
 Eq. 3 $V_{RMS}^2 = 2 \cdot \frac{{{A^2} \cdot k}}{{3 \cdot T}} + \frac{{{A^2}}}{T} \cdot \left( {T - 2 \cdot k} \right) = {A^2} \cdot \left( {1 - \frac{4}{3} \cdot \frac{k}{T}} \right)$
 Eq. 4 ${V_{RMS}} = A \cdot \sqrt {1 - \frac{4}{3} \cdot \frac{k}{T}}$

### Crest Factor

Just to complicate matters, the ring voltage is actually controlled by two specifications: an RMS voltage level and a Crest Factor (CF). CF is defined shown in Equation 5.

 Eq. 5 $CF \triangleq \frac{{\left| {\max \left( {v(t)} \right)} \right|}}{{{V_{RMS}}}}$

CF is commonly used because many electronic devices are sensitive to peak-to-average ratios, which CF measures. Telephony specifications require that 1.2 ≤CF ≤ 1.6.

### Correction Factor for a Peak Detecting Meter

While there are electronic voltmeters that measure true RMS voltages for any waveform, the sales engineer in this case is using an electronic voltmeter that assumes the operator is always measuring a sinusoidal signal. It measures the peak voltage of the voltage waveform and divides that value by the square root of 2, an approach which produces the correct RMS value for a sinusoidal waveform but not a trapezoid (see Equation 6).

 Eq. 6 ${V_{Meter}} = \frac{A}{{\sqrt 2 }}$

Equations 7-8 show how to calculate a correction factor for adjusting the meters reading to give the RMS voltage for a trapezoid.

 Eq. 7 ${V_{RMS}} = \frac{{{V_{RMS}}}}{{\max \left( {\left| {v(t)} \right|} \right)}} \cdot \frac{{\max \left( {\left| {v(t)} \right|} \right)}}{{\sqrt 2 }}.\sqrt 2$
 Eq. 8 ${V_{RMS}} = \frac{{{V_{Meter}}}}{{CF}} \cdot \sqrt 2$

## A Field Example

In this particular case, the field engineer was measuring 60 V for something that Engineering said was 65 V. We can summarize the critical variables as follows:

• Trapezoid voltage peak voltage (A), is 86 V (set by the telephone ringer circuit design).
• CF is 1.33 (set by Engineering in the firmware)
• Value measured in the field, VMeter, is 60 V.

We can compute the correct RMS value by using Equation 8. This calculation is shown in Equation 9.

 Eq. 9 ${V_{RMS}} = \frac{{{V_{Meter}}}}{{CF}} \cdot \sqrt 2 \doteq 60{\text{ V}} \cdot \frac{{1.414}}{{1.333}} = 65{\text{ V}}$

This shows that the result measured by the sales engineer and my engineering group were actually the same.

I am an engineer who encounters interesting math and science problems almost every day. I am not talking about BIG math here. These are everyday problems where a little bit of math really goes a long way. I thought I would write some of them down and see if others also found them interesting.
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### 5 Responses to Measuring the Ring Voltage on a Telephone

1. Joel says:

Finally something near and dear to my heart

• mathscinotes says:

There is something strangely satisfying about working on POTS stuff. The optics stuff is always difficult to write about because it gets very deep. The telephony stuff is much more “normal” analysis.

2. Hamman says:

My first time to visit your blog – I absolutely love it !

I come from much of a same background as yourself – I have grown up in small town rural Ireland. I have had a deep passion for electronics since age five. This later developed into amateur radio, programming, science and engineering. Although I was bright at school – I too hated math and found it terrible hard work. I found it interesting but incredibly tricky and treacherous !

My mother told me that you would need to be really good at math to studying anything electrical. This disheartened me – but I carried on. I worked hard at math-gradually becoming better.

I am currently studying Electonics Engineering which now has a predominately has a software/firmware content. I selected Electronics Engineering in preference to Software Engineering as I did not want to be stuck at a PC writing code all day .
Unfortunately for the most part- this is what Electronics Engineering has became.

The analog that I know and love is only taught in the extreme basics. Graduates now have far less analog knowledge than 20 years ago. I am in my 3rd year ( doing my second last exam for the course in just over 3 hours from now!). I continue to work on analog projects in my own time ( currently developing a new unusual product which involves an analog multiplier- the reason I found your website ).

I find the math that you detail here to be excellent – you lay it out nicely and its at a standard that (at least the majority of which) I can understand .
It is with pleasure that I can now read engineering text and actually have a handle on what the maths are telling me.

I admire people like you – my aspiration is to become as proficient you.

Thank you for the wonderful blog – I will be a regualr from now on.

Best Reguards,
Hamman
Ireland

N.B I would also add to your list of analog gurus -Bob Wildar.

• mathscinotes says:

I hope things went well on your exam! Thank you for the kind comments. Like you, electronics has been a love since I first began building radios in grade school. I even did my graduate work in analog electronics. Good luck on your career.