Tutorial 3 - Sinusoidal Waveforms

Learning Objective

Use the idea of a rotating phase vector.

Recognise and use V = V₀ sin wt calculate instantaneous values.

Work out the resultant from two phasors.

Modify the equation to take into account the phase angle.

Key Question

What is a Phasor?

What happens if there is a phase angle?

What happens if there are two waves?

Phasor and Algebraic Representation of Sine Waves

Phasors

Alternating currents can be represented by phase vectors or phasors.  A phasor is representation of a sinusoidal wave form as a rotating vector. 

Sinusoidal motion was very closely linked with circular motion and simple harmonic motion.  Phasors are particularly useful when you have two or more alternating currents that are a fixed amount out of phase.  This happens when we put reactive components like capacitors and inductors into an electrical circuit.

In our study of phasors, we will assume that:

·         The amplitude, A, remains constant;

·         The angular velocity, w, remains constant;

·         The phase relationship, f, remains constant.

(The curious looking symbol, f, is “phi”, a Greek lower-case letter ‘f’ or ‘ph’.  It is the physics code for phase angle.  We will look at that soon.)

Let’s look at a phasor that represents a sinusoidal waveform.  It has frequency f, and an angular velocity of w rad s^-1.

This diagram is showing how there is a rotating vector that is projected onto a moving piece of paper.  It traces a sine wave.

The rotating vector is turning at a constant angular velocity of w rad s^-1.  By convention it turns anticlockwise.  By convention, the zero point is the 3 o’clock position.  Angles are measured in radians.

The vector starts on the line OA.

Let us consider the situation after a time t.  The vector has moved to position OB and has passed through an angle q (as q = wt).  The perpendicular line BC is a projection of the vector onto the line OA. 

By simple trigonometry, we can write:

We can rearrange this to give:

OA is the voltage.  So we can give the instantaneous voltage at any time as:

 

Adding a phase difference

In the last piece of theory, we started off with the phase vector starting at 0 rad.  Starting at the beginning is, after all, a very good place to start.  But we don’t have to start from zero.  We can start at any point we like.

Suppose we started off with the vector at f radians from the zero point.  After t seconds, the phasor would have rotated through an angle q = wt.

 

So the total angle would be wt + f.

 

If we started below the zero point our total angle would become wt – f.

 

So we can rewrite the expression for instantaneous voltage at any time and any phase angle as:

 

 

 

These relationships work with angles in radians.  Therefore it is important to convert any angles that are in degrees to radians.  You must ensure that your calculator is set to radians, not degrees.

 

Worked Example A sinusoidally alternating voltage is given by this relationship:   V = 75 sin(200 pt – 0.25)   a.       What is the amplitude? b.      What is the periodic time? c.       What is the instantaneous voltage 3.7 ms after the AC passes through the zero point going in the positive direction?

Answer a.       Amplitude occurs when sin(200 pt – 0.25) = 1        V0 = 75 × 1 = 75 V

b.      Angular velocity = 200 p rad/s.        w = 2p/T Þ T = 2p ÷ 200 p = 0.01 s = 10 ms

c.       V = 75 sin(200 p × 3.7 × 10-3 – 0.25)        V = 2.71 V

 

Combining two wave forms

In many electronic systems, it is possible to have alternating currents that are not in phase, but a fraction of a cycle apart.  For example, in a capacitor, the current is 90^o or p/2 rad ahead of the voltage.  But we could have a phase difference of any angle.  We have just been working with the relationship:

Like any wave, electrical wave-forms can interact to superpose.  This means that the waves add together vectorially. 

• If the waves are in the same direction, they add up.  

• If they are in opposite directions they take away.

Here we have two low frequency AC waves that are interacting.

Wave 1 is ahead of Wave 2 by 0.25 rad.  We say that it leads by 0.25 rad.  If it were behind by 0.25 rad, it would lag by 0.25 rad.

We also see that the there is a resultant wave.  This would be the current you would measure if you were using a multimeter.  The resultant lags Wave 1 by about 0.12 rad, and leads Wave 2 by about 0.13 rad.

Maths Note: In the formulae, all angles must be in radians.  However when dealing with the angular relationships between phasors, it is perfectly correct to express the angles in degrees.  I have done this throughout my arguments as it's much easier.  You do need to convert into radians before putting any phase difference into the formula.

Using Phasors

This is a much easier way of going about things.  We can go about it by accurate drawing, or by finding the resultant by calculation.

Accurate Drawing

Whichever method we use, we draw the phasor at t = 0.  Consider two alternating voltages, V₁ and V₂ that have a phase difference of f.

Question 4

Which voltage is leading?

Now we use a vector parallelogram to work out where the resultant will go:

Let’s have a look at this by accurate drawing.  You will need:

·         Graph paper;

·         A sharp pencil;

·         A ruler.

We will consider two voltages of 20 V and 30 V that are 30^o out of phase.  The 20 V vector lags the 30 V vector.

You need to choose a scale.  Let’s do 1 cm = 2 V as there is room on the A4 paper.  Yes, I have used a black biro so that it stands out better on the graph paper.

The 20 V vector is at the zero position, so that it goes from left to right.  We can see that below:

Now we need to do some construction lines to make it a parallelogram, so that we can draw the resultant vector:

Now we can measure the final vector using a ruler.  The distance is 24.2 cm, and using the scale 1 cm = 2 V, our voltage is 48.4 V.  Using a protractor, we can measure the phase angle as well.  The 30 V vector leads the resultant by 12^o.

Before we put the numbers into of these equations, we must convert the angle from degrees into radians.

 Angle (rad) = (angle (degrees) × p) ÷ 180

Question 5

What is the phase difference between the resultant voltage and the 20 V vector?

What is the phase difference between the resultant voltage and the 20 V vector?

Now let’s apply a formula to these vectors in the form of:

Let’s make the frequency 1 Hz.  So the angular velocity is 2p rad/s.  The relationship for the 20 V vector is easy, as it’s at the 0 point when t = 0.  The equation for this vector is:

V₁ = 20 sin (2pt)

Now we need to write the equation for V₂.

Question 6

Why is v₂ = 30 sin (2pt + 30) wrong?

Question 7

Show that 30^o = 0.524 rad

So our relationship for V₂ is:

V₂ = 30 sin (2pt + 0.524)

Mathematically the equation can be left in surd form:

V₂ = 30 sin (2pt + p/6)

Relative to V₁, we can write an equation for the resultant voltage:

V_R = 48.4 sin (2pt + 0.314)

We can also derive an expression for the resultant relative to V₂.

 V_R = 48.4 sin (2pt – 0.209)

Calculation

The problem with accurate drawing this that it is anything but accurate.  It is much less error-prone to do the working out of the resultant by calculation.  Let us suppose we had our 20 V and 30 V vectors with a 90^o (p/2 rad) phase difference.  The phasor diagram would be this:

The resultant can simply be worked out using Pythagoras.  We can write equations for the instantaneous voltages.

Question 8

Show that the resultant voltage is about 36 V

Question 9

Write down the expression for the 30 V vector (V₂) relative to 20 V (V₁)

We can use trigonometry to work out the phase angle relative to V₁.  Use the tan function

Angle = tan^-1(30/20) = +0.983 rad (= 56.3^o)

The plus sign tells us that the resultant phasor is leading.

Question 10

Work out the phase angle of the resultant relative to V₂.  State whether the resultant is leading or lagging V₂.

What do we do if the phasors are not at 90 degrees?

Pythagoras only works if the angles are at 90^o.   However any vector can be resolved into vertical and horizontal components.

Let’s try this method on the problem we have looked at previously.

We can see the horizontal and vertical components for V₂.  The vector V₁ is horizontal, so has no vertical component.

We can also show the vertical and horizontal components for the resultant voltage V_R.  We will use a separate diagram to show this.

We can see easily that the horizontal component is given by:

The vertical component is simply:

This is because V₁ sin q = 0

So let’s put some numbers in:

Resolving horizontally:

Resolving vertically:

We can use Pythagoras to calculate V_R:

So our resultant voltage is actually 48.4 V

Now we can use the phase angle.  Remember that tan = opposite ÷ adjacent.

f = tan^-1 0.326 = 18.1^o

Subtracting phase vectors

In the example above, we added the phase vectors V₁ and V₂.    What do we do if we want to subtract them?  Instead of V₁ + V₂, we do V₁ – V₂.

The minus sign means that the vector V₂ is going the opposite direction.

We can put on the vertical and horizontal components for V₂:

And for V_R:

In this diagram we have angle f which is the phase angle.  We also are using f' (phi-prime) which is the angle we will use with our resolutions:

f = f' + 90

The symbol f' is not in italics to make it clearer which term we are talking about.

As before we have the vertical components equal as V₁ has no vertical component.

The horizontal components add up:

Notice that for V_R the sine and cosine are the opposite way round to what we had before.  It won’t affect the result.  This way, it’s easier to work out f'.  The minus sign indicates that the term V₂ cos q is in the opposite direction to V₁.

Let’s put the numbers in.  Resolving vertically we can write:

Resolving horizontally:

Now we can work out v_R by Pythagoras.  (Note that the square of a negative number is positive):

Now we can work out f'.

Now, eventually, we can work out the phase angle:

f = 90 + 21.7 = 111.7^o = 112^o to 3 significant figures

To put this phase angle into an equation, we have to convert it into radians. 

In this case f = 1.95 rad, lagging.

We can write an expression for this:

Question 11

Two alternating currents have instantaneous values of current given by the expressions:

Work out by calculation of phase vectors an expression for I_R = I₁ + I₂.