Tutorial 2 - Waveforms Learning Objective To distinguish between unidirectional and alternating waveforms. To look at different waveforms. Define instantaneous values. Key Questions What is a waveform? How do we work with AC values? The most natural waveform is the sine wave.  Sine waves are the result of circular motion.  This is obvious when you consider that an AC generator has a rotor that spins on an axis.  There are two other Physics phenomena (things that can be seen) that are very closely related to circular motion: ·         Wave motion; ·         Simple Harmonic Motion. This graph shows a sine wave: What is the peak voltage?
 Question 2 How many cycles are there?
 Question 3 What is the RMS voltage of this waveform?
 Here is the same sine wave as above, but this time with the angle in radians (rad). Since the CRO shows a voltage-time graph, it is far more convenient to place the time (t) on the horizontal axis instead of the angle.  There are two convenient time quantities that we can use: Time period (T), which is the time taken by one whole cycle. Time (t) is any time.  Period (T) is a property of the wave. Frequency (f), which is the reciprocal of the time period. We can relate both to the angular velocity: Angular velocity is the angle that is swept every second by a rotating object.  It is given the physics code w, (“omega”, a Greek lower case letter long ‘o’, ō).   Angular velocity can be linked to period: A sinusoidal alternating waveform has a frequency of 40 Hz.  Calculate: (a) The time period (b) Angular velocity.
 We can write a general equation for a sinusoidal waveform as: The equation for current is very similar:  A student looked at these equations and wrote R = R0 sin wt relate the resistance to the time.  Explain what he did wrong. A peak current of 100 mA flows in an alternating current of frequency 80 Hz. What is the angular velocity (w)? What is the instantaneous current when t = 14 ms?

### Other Waveforms

We have looked in detail at the sine wave as it’s the most common varying signal.  But it is NOT the only signal.  We can have varying signals that have a different shape, and signals that do not fall below zero.  So we need some definitions:

• Unidirectional waveforms only go in one direction.  Usually this is positive.

• Alternating waveforms go from positive to negative and back.

###### This is a Square Wave: This wave is unidirectional.

### Rectangular wave

The rectangular wave has a mark time, where the voltage is high (at +A), and a space time where it is low (0 V).  Note that:

• The negative half is not negative at all; it’s slightly above 0.  However electrical engineers still call it that.

• Not all square waves are unidirectional.  There is no reason why square waves cannot be alternating.

• Not every sine wave is alternating.  There is no reason why a sine wave cannot be unidirectional.

The mark time and the space time can be any values.  The mark to space ratio is given by:

• Ratio = mark time ÷ space time

• The duty cycle is the fraction of the cycle at which the wave is high:

• Duty cycle = mark time ÷ period (× 100 %)

• The time period is given by:

• T = mark time + space time

 Worked example A rectangular waveform has a frequency of 40 Hz and a duty cycle of 30 %. What is the time period? What is the mark time? What is the space time? What is the mark to space ratio? Answer T = 1/f = 1/40 = 0.025 s = 25 ms Mark time = 0.30 × 25 = 7.5 ms Space time = 25 – 7.5 = 17.5 ms Mark to space ratio = 7.5 ÷ 17.5 = 0.43

 Compare this wave with the square wave.

### Pulses In pulses the mark time is much shorter than the space time.

The object of these pulses is to turn on (or off) a semiconductor device, such as a counter.  The change from off to on occurs as the pulse passes.  Rising edge triggered devices turn on as the pulse goes from low to high.  Falling edge triggered devices turn on when the pulse goes from high to low.  The duration of the high signal is immaterial.

 A rectangular wave form has a mark time of 10 ms and a duty cycle of 25 %.  What is its frequency?

### Triangular Waves

Not surprisingly, these wave have a distinctive triangular shape.  They are used when electronic devices need a linear increase in voltage. Triangular wave forms are generally alternating.  The positive slope or ramp is usually of the same time duration as the negative ramp.  This particular wave is symmetrical, but we can produce triangular waves where the negative ramp is steeper than the positive ramp (or vice versa).  The slopes are linear.

### Sawtooth Waveform These wave-forms are self-descriptive. The decay or rise is almost instantaneous, while the ramp is linear.

### Complex Waveform Sine waves give pure tones, but are very dull to listen to.  Music, speech, and other sounds are much more complex.  However complex the wave-forms are, there is still a periodicity to the sounds that can be captured and measured on a CRO.  To analyse a really complex wave, we need to use a storage CRO that commits a wave to a memory.  An ordinary CRO will display the wave in real time, but to analyse the wave, it would have to be constantly played.

There are now computer-based oscilloscopes that do this job really well and have built-in functions to analyse the waves.  Alternatively there are data-logging peripherals that display the result on a computer.

### AC Values

We can determine the value of any changing wave-form for any instant of time.  Clearly we could use a voltage time graph. And it doesn’t take a genius to read a voltage value off at a particular time.  However it makes sense to have a standard range of values for any waveform.  Let’s look at them:

###### Instantaneous values

These are the values at any moment of time.  Instead of using capitals for the codes, some books use lower case letters, e.g. v for V, i for I, etc.  In these notes, I will use V0 for peak voltage and V for instantaneous voltage.

###### Peak values

These are the maximum values or amplitude of the waveform.  We will represent these from now on as V0, I0 etc.  You will see variations in textbooks.  Bird writes them as Vm, Im, etc.  The peak to peak value is the difference between the highest value (most positive) and the lowest value (most negative). We have seen how we can work out the peak voltage by finding out the peak-to-peak voltage and dividing it by 2.  This only works is the waveform is symmetrical about the zero line.  If there is an offset, i.e. the waveform is displaced more to one side of the zero line than the other, we need to find the difference between the maximum and the minimum, then divide that by 2.

###### Average values

This is the average value measured over half a cycle.  A whole cycle for an alternating symmetrical waveform would be zero.  This is given by this relationship:

Average value = area under the graph ÷ length of the base For a sine wave, the average value is given by:

Average value = (2/p) × maximum value

To 3 significant figures, 2/p = 0.637

If we know the function of the graph, we can work out the area of the graph by using calculus integration.  But if it’s a complex waveform with no obvious mathematical function, you have simply to count the squares under the graph.  It’s tedious but gives you an answer that is approximate, but pretty close.

Question 9

The peak value of an alternating sinusoidal waveform is 15 V.  What is the average value?
###### Form Factor

The form factor is the ratio of the rms value to the average value for a particular alternating current.  We can write it as:

Form factor = RMS value ÷ average value

Here are some form factors: Question 10 Show that the form factor for a sine wave has a value of about 1.1
Question 11 Show that the form factor for a square wave is 1.
###### Effective value

The effective voltage for an alternating voltage is the voltage that would give the same heating effect as a direct current.  We have seen this before as the rms voltage.  In these notes I will write Vrms and Irms.

For a sinusoidal wave-form we can write: For other wave-forms, it’s not as simple as that.  We have to take a number of readings and use a more complex looking formula: So for our complex wave form, we would take some values: So our formula becomes: The more intervals chosen, the closer the value gets to the true value.

The following data are currents taken at intervals from an alternating waveform:

 Time (ms) 10 20 30 40 50 60 Current (A) 0.75 1.35 1.45 1.25 1.12 0.34

What is the effective current?
###### Peak Factor

The peak factor is the ratio of the maximum value to the rms value for a particular alternating current.  We can write it as:

Peak factor = maximum value ÷ rms value Worked Example

These data give the instantaneous values of voltage and time for a half cycle of an alternating voltage.

 Time (ms) 0 0.5 1 1.5 2 2.5 3 3.5 4 Voltage (V) 0 7 14 36 40 56 68 3 0

Plot these data on a graph and find:

a)      (a) The frequency of the supply;

b)      (b) The peak value;

c)      (c) The average value;

d)      (d) The rms value;

e)      (e) The form factor. Answer a)      Frequency = 1 ÷ (4.0 × 10-3) = 250 Hz   b)      Peak value = 70 V   c)      To get the voltages for this, we will sample every 0.25 ms and read the intermediate values. We are using the mid-ordinate rule Area = (0.25 × 10-3) × (2 + 5 + 9 + 12 + 20 + 30 + 39 + 44 + 51 + 62 + 70 + 50 + 20) = (0.25 × 10-3) × 452 = 0.113 V s. Average value = area ÷ base = 0.113 ÷ 4.0 × 10-3 = 28.25 V   d)     RMS value = Ö((22 + 52 + 92 + 122 + 202 + 302 + 392 + 442 + 512 + 622 + 702 + 502 + 202 + 0 + 0)) (15) = Ö(19256 ÷ 15) = Ö1284 = 35.8 V   e)      Form factor = RMS value ÷ average value = 35.8 V ÷ 28.25 V = 1.27

That was a rather tedious calculation which required a fair amount of number-crunching.  A spreadsheet program such as Excel would certainly help.  Note that the smoothed curve dropped slightly below zero.  The negative values were ignored for the average value.  They were counted as 0 points for the RMS value.

The problem with these data was the uncertainty between 3.5 ms and 4.0 ms.  Was the instantaneous value, V­, at 0 volts or was it 1.5 V?  The data don’t tell us. Question 13 Show that the peak factor for a sine wave has a value of about 1.4