AC Theory

Thread Starter

Biggsy100

Joined Apr 7, 2014
88
So I have an alternating voltage described with the following expression:

V= 100sin(100*Pi*t) + 60sin(300*pi*t - Pi/2) + 30sin(600*pi*t + pi/2)

I need to find the:

-Amplitude of Fundamental
-Frequency of Fundamental
-Amplitude of Harmonic components
-The Phase Angle
-The Instantaneous Value of v 5.12 ms from the start of cycle.

As I understand each element from the expression is equivalent to a value above (?) or do I have to break the expression up?
 

Veracohr

Joined Jan 3, 2011
711
As I understand each element from the expression is equivalent to a value above (?) or do I have to break the expression up?
The amplitudes are right there in the equation. The frequencies are there too, but you first need to know the relationship between frequency in Hertz, and the angular frequency as used in sine wave equations. In other words, there's some simple math needed to get the frequencies in Hertz out of that equation. Do you know what a fundamental frequency is, and what harmonics are?

To get the instantaneous value, just substitute the time you're given for t in the equation.

The phase angle I'm not sure about. Being a complex waveform, I don't know where people would consider the 'beginning' of the wave to be. I'm sure someone else will have an answer.
 

Thread Starter

Biggsy100

Joined Apr 7, 2014
88
Right...so, I know the fundamenatls will be the lowest of the expression, in this case 100, 60 and 30? I can see that there are 3 elements to this expression which I assume will repeat process?
 

Thread Starter

Biggsy100

Joined Apr 7, 2014
88
Not sure where to start with this? I know the fundamental is ultimatley double, so in this expression we have 50, 150 and 300 HZ (Yes/no?)

The instantaneous value of V can be obtained by substituting the given value (5.12ms) for t in the expression?

This was my idea; 100 sin(2/Pi x 50 x 5.12ms + pi/2)?
 
Last edited:

MrAl

Joined Jun 17, 2014
7,500
Hi,

Sine and cosine waves are usually written in the form:
V=A*sin(w*t+TH)
or
V=A*cos(w*t+TH)

and w is the angular frequency equal to 2*pi*f so we get:

V=A*sin(2*pi*f*t+TH)

where
A is the amplitude,
f is the frequency in Hertz,
t is time in seconds,
TH is the phase shift (often the greek letter 'theta').

The fundamental frequency is the lowest frequency of the wave.

See if you can take it from here...
 

ericgibbs

Joined Jan 29, 2010
9,832
hi 100,
Consider this part of your full equation. V= 100sin(100*Pi*t)

Inside the brackets Let t =1 Second . So 100 * ∏ = 314.2 [radians]

To get the Freq, you know that there are 2*∏ rads per cycle [ or 360 degree revolution]

So 314.2/[2*∏] = 50Hz
So you should now be able to calculate Vinst at 5.12mS..................

Work out the Freq for the other parts using the same method:
60sin(300*pi*t - Pi/2) + 30sin(600*pi*t + pi/2)

The ∏/2 is a 90 degree phase shift, so when calculating the Freq, ignore the ∏/2.

If you need to plot the individual waveforms, include the 90deg phase, its either +90deg or -90deg, you figure it out.

Do you follow OK.?

E
 

ericgibbs

Joined Jan 29, 2010
9,832
The part that is confusing me now is the phase part, would it simply be entered as 1,2,3?
hi 100,
Sorry I don't understand your post.???:confused:

The ∏/2 is the Phase Shift of the Freq in your equation.
(300*pi*t - Pi/2).

Pi/2 = 1.57 Rads, So 1.57/2*Pi = 0.25 of a Cycle, so thats 0.25 *360 = 90 Degs. [always consider the SIGN of the Phase shift]

E
 

Thread Starter

Biggsy100

Joined Apr 7, 2014
88
so first I calculate the numbers in the brackets...

314.2/(2 x pi) = 50 hz which gives me the frequency..0.020ms?

314.2 x 0.005 = 100 which will give the instantaneous value?
 

ericgibbs

Joined Jan 29, 2010
9,832
so first I calculate the numbers in the brackets...

314.2/(2 x pi) = 50 hz which gives me the frequency..0.020ms?

314.2 x 0.005 = 100 which will give the instantaneous value?
hi 100,
No, its as explained in my PM's

Vo= 100 * sin( 314.2 *.005)

Vo = 100 * sin(1.57) ; Use your calculator set to Radians

Vo = 100 *1 = 100v
 

Thread Starter

Biggsy100

Joined Apr 7, 2014
88
Ok put the question another way, I think I maybe thinking to much on the overall expression. I understand now that they all equal 100. So I am confused of the diffrentiation ?
 

ericgibbs

Joined Jan 29, 2010
9,832
ok,

100sin(100*pi*0.005) =100 ; No

60sin(300*pi*0.005)=-60 ? ;No

30sin(600*pi*0.0050= 0? ; No
This is your original equation.!
V= 100sin(100*Pi*t) + 60sin(300*pi*t - Pi/2) + 30sin(600*pi*t + pi/2)

I have explained a number of times how to solve this simple problem, I am sorry but you are totally ignoring what is being posted.

Which engineering course are you currently studying.?
 
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