# hnc question 3

Discussion in 'Homework Help' started by redshaw, Jul 25, 2008.

1. ### redshaw Thread Starter Member

Jul 15, 2008
12
0
hi guys

can you confirm or deny if i am correct in my working for this question

a complex waveform comprises a fundamental sinusodial voltage which has a peak value of 36V and a frequency of 400Hz together with a third harmonic component having a peak value of 12V leading the fundamental by 90 degress. Write down an expression for the instantaneous value of the complex voltage and use it to determine the value of the voltage at 3 ms from the strat of the cyle.

v=36sin400 + 12sin1200 + 90
v=23.14 + 100.39
v= 123.53V

v0.003=0.37V

2. ### blazedaces Active Member

Jul 24, 2008
130
0
Alright, you're close, but not quite:

For a frequency f a sinusoid without a phase shift would have the form PV*sin(2*Pi*f*t) where PV is the peak voltage and Pi... is obvious enough.

The phrase "leading the fundamental" I've never heard before, but 90 degrees implies it's talking about the phase, not the final voltage value, so for that sin function you would have PV*sin(2*Pi*f*t + PhaseChange). In this case, phase change would be 90, technically turning the sin into a cosine. You would not add 90 outide of the sinusoid.

Now, how exactly are you arriving at 1200 for the inside of the second sin function? I would say assume it has the same frequency as the first one (400) unless something else is specified.

I hope I didn't provide you with any faulty information.
Good luck.

-blazed

3. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,414
1,252
e = V * sin(angle)

So in this case, I would tend to think it would be

Vo = 36 + (12 * sin(-90))

since you are mixing both the primary frequency and the harmonic

4. ### dr.nosduh New Member

Mar 8, 2008
1
0
The general expression for the 400Hz fundamental and its 3rd harmonic is of the form...
v = V1sin(ωt + Ф1) + V2sin(3ωt + Ф2)

At 400Hz, ω = 2πf = 2512 rad/s
So the 3rd harmonic term 1200Hz which is 3ω = 7536 rad/s.
Ф1 = 0 as this is the reference term
Ф2 = π/2 (which is 90°)
Subsituting these values into the general expression yields...
v = 36sin(2512t) + 12sin(7536t + π/2)

At 3ms we then get...
v = (36 x 0.95) + (12 x -0.82)
v = 24.4V

Thanks,
Doc

May 21, 2015
5
0
The above solution helped me a lot to answer my homework that has another part: How to generate a triangular wave from the fundamental sine wave? Explain the formation of square wave with the help of few representative sketches

6. ### WBahn Moderator

Mar 31, 2012
18,277
4,949
This is Homework Help, not Homework Done For You.

Please refrain from simply working student's problems for them -- this seldom actually helps the student and encourages cheating.

7. ### WBahn Moderator

Mar 31, 2012
18,277
4,949
Of course it helped you a lot to answer your homework, but it probably didn't help you learn much about the concepts. But it did encourage you to ask for someone to simply work your next homework problem for you.

8. ### WBahn Moderator

Mar 31, 2012
18,277
4,949
Even though someone else already worked your problem for you, others may benefit from see where you are going wrong with your initial effort (and thanks for providing an initial attempt to work your own homework, by the way).

First, you need to learn to track units. This would have immediately highlighted two errors with your first line:

v(t) = 36V·sin(400 Hz) + 12V·sin(1200 Hz) + 90°

You have three terms and the units on the first two are volts while the units on the third are degrees. In order to add terms, they MUST have the same units, therefore you know that this equation cannot be correct and any work that proceeds from this point is guaranteed to be incorrect and a waste of time.

Second, the argument to the sine function HAS to be unitless (this is because sine is a "transcendental" function, which takes a dimensionless argument and produces a dimensionless result). So you KNOW that you are missing something since Hz has units of "cycles per second".

The sine function takes an argument that needs to be in units of "radians" (which is a dimensionless unit) and so you need to first convert the frequency from Hz to radians/second and then you need to multiply by time to get the "per second" to go away. There are 2pi radians in one cycle, so you have:

Now you need to figure out how to deal with that "leading by 90°" spec.

In general a sinusoidal waveform is described by

$
a(t) \; = \; A \sin$$\omega t + \phi$$
$

Where Φ is known as the phase angle of the waveform.

$
-\pi \le \phi \lt \pi
$

(which side gets the equality is arbitrary)

If you have two waveforms at the same frequency

$
a(t) \; = \; A \sin$$\omega t + \phi_a$$
b(t) \; = \; B \sin$$\omega t + \phi_b$$
$

Then imagine if Φb is slightly more positive than Φa. That means that waveform b(t) will reach any particular amplitude slightly before a(t) will, meaning that b(t) is "leading" a(t). The amount by which it is leading a(t) is (Φb-Φa).