Complex Number Sine Wave Problem

[badsanta010]

Homework Statement

Two alternating voltages are given by:

V1= 12 sin 200pi t
V2= 18 sin 200pi t + pi / 3

i) Plot each wave form on the same axis for one complete cycle
ii) Add both together and plot the resultant waveform on the same axis
iii) Using complex numbers, confirm ii) graphical answer

Homework Equations

2pi f t ( to work out frequency)

Cycle time = 1/f

The Attempt at a Solution

Hi.

Any help would be greatly appreciated

Sorry I am unsure how to type the correct symbols hopefully you can understand what I mean easily enough.

So far I have completed questions 1 and 2 but its the final questions I am stuck on.

For question 1 I have:

Created a table with the following columns and done all the calculations needed.

V1

t (time)
200pi t
sin (200pi t) Using radians for this calculation
12 sin (200pi t)

V2

t (time)
200pi t
pi/3
sin(200pi t + pi/3) Using radians for this calculation
18 sin(200pi t +pi/3)

Using the completed tables I then plotted the two waves on the same graph.

For questions 2 I have:

Added the two graphs together by simply adding the final columns ( 12 sin(200pi t) + 18 sin(200pi t + pi/3) and plotted these on the same graph.

So I now have 3 graphs on the same axis and that's where I am stuck. I don't understand what it means by confirming graphical answer using complex numbers.

I know that V1 and V2 can be converted into complex number form but I don't understand how this can be used to prove the graphical answer?? A push in the right direction would be great.

V1 = 12 < 0 = 12 + 0j
V2 = 18< 60 = 9 + 15.58j

Please ask if you need any more information or don't understand anything.

Thanks

Any help would be great.

 

[HallsofIvy]

I presume that you have learned that
$$e^{i\theta}= cos(\theta)+ i sin(\theta)$$
and from that
$$sin(ax)= \frac{e^{ix}- e^{-ix}}{2i}$$

so that
$$V1= 12\frac{e^{200\pi i t}- e^{-200\pi i t}}{2i}= -6i\left(e^{200\pi i t}- e^{-200\pi i t}\right)$$
and
$$V2= 18\frac{e^{200\pi i t+ i\pi/3}- e^{-200\pi i t+ i\pi/3}}{2i}$$
$$= -9ie^{i\pi/3}\left(e^{200i\pi t}- e^{-200i\pi t}\right)$$

And, of course,
$$e^{i\pi/3}= cos(\pi/3)+ i sin(\pi/3)= \frac{1}{2}+ i\frac{\sqrt{3}}{2}$$

Add those.

 

[badsanta010]

Thanks for the reply.

Sorry no doesn't look familiar. Have looked through my notes nothing either. I can kind of follow what you have done so I will give it a go and see how I get on.

Thanks