# Angular and Graphic Relationships for sine waves

Discussion in 'Homework Help' started by Ripplenwinder, Jan 17, 2015.

1. ### Ripplenwinder Thread Starter New Member

Jan 17, 2015
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0
I am using the book circuit analysis theory and practice and am unsure how to solve the problem in ch15 #25.

#25) A 20-khz sine wave has a value of 50 volts at t=5 micro seconds. Determine Vm and sketch the waveform.

I can't seem to figure out how to find Vm without guessing I understand how to find V knowing Vm but how might on go about finding Vm from knowing V from a given instance or position in time? Vm in this book stands for Maximum Voltage.

Here what I understand as of now
1/20000 (hz) = .00005 (T) period
(t/T)*2*pi=.62831853 (rad) , (t/T)*360=36 (deg) I don't think this step is required but......???

finding the question from the anwser 85.1 sin (125663.7061*.000005) =50.02052496V

Is there also a way of using the 1:10 relationship between the waves period and the time refernce to extract an answer?

Also I would like if you could summit any useless bit of information and equations regarding relationships within the time/frequency domain I very well just might like it.

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
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482
V=Vm*sin(2*pi*f*t)
50=Vm*sin(2*pi*20,000*0.000005)
Find Vm.

3. ### Ripplenwinder Thread Starter New Member

Jan 17, 2015
17
0
thank you for responding with a useful piece of information.

Last edited: Jan 17, 2015
4. ### Ripplenwinder Thread Starter New Member

Jan 17, 2015
17
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Do you know how to find Vm?

5. ### MrChips Moderator

Oct 2, 2009
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The question ignores the phase angle at t = 0.

v(t) = V sin(ωt + φ)

You can solve for V if you assume φ = 0.

6. ### shteii01 AAC Fanatic!

Feb 19, 2010
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482
V(t)=Vm*sin(2*pi*f*t)
You are given V(0.000005) (50), you are given f (20,000), you are given t (0.000005).
50 at time 0.000005=Vm*sin(2*pi*20,000*0.000005)

To find Vm all you have to do is:
$
Vm=\frac{V}{sin(2*pi*f*t)}
$

Just plug the eFing numbers!

7. ### Ripplenwinder Thread Starter New Member

Jan 17, 2015
17
0
yes your anwser will be zero because you are computing 50 v at its zero crossing. .00005 is the period of the wave thus 0v and .ooooo5 is 10 90 degree divisions of it's orgin (one and a half cycles) thus 0V. Because both instances occur at there zero crossing when using that formula. I know what the equation you gave is just a rewrite of Vx sin (w*t), and that equation will not solve the problem will it?

8. ### Ripplenwinder Thread Starter New Member

Jan 17, 2015
17
0
thanks I didn't see the 50/sin in the post at first the [ bracket looked messed up

9. ### WBahn Moderator

Mar 31, 2012
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4,699
My guess is that this is to be inferred by calling it a "sine" wave (as opposed to a "cosine" wave, for instance). Authors often think that the two are somehow different since they are quite distinct in a trigonometry class but forget that when talking about sinusoidal waveforms that it is purely a distinction due to an arbitrary choice of time reference (and, arguably, even in trig it is a matter of arbitrary choice of reference axis, but that is much more highly standardized than time reference).

10. ### Ripplenwinder Thread Starter New Member

Jan 17, 2015
17
0
I understand what you are talking about. I tried every different way I knew how to to try to formulate an answer to the question. I would have took me a long time until shtell01 posted 50/sin (w*t).....where I keep trying to find from 20sin(w*t) hours before I joined the form. Is there any other variations of these equations that you know that might be helpful or things for me to experement with?

11. ### Ripplenwinder Thread Starter New Member

Jan 17, 2015
17
0
I just read that you are should post you answer in the home work help section. So 50/sin(2*pi*20000*.000005)=85.06508084=Vm

12. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
It sounds like you might be well served just working on your algebra skills. That way you don't need to know a bunch of different formulas and variations on them, but rather can work with a few fundamental formulas and manipulate them to get what you need. In this case, you have

$
v(t) \; = \; V_m \sin$$2\pi f t$$
$

If we are wanting to find Vm, we just need to manipulate this equation to get Vm by itself on one side of the equation. We can do this by dividing both sides by sin(2πft) to get

$
V_m \; = \; \frac{v(t)}{\sin$$2\pi f t$$}
$

You are given that f = 20 kHz and that v(5μs) = 50V. This means that

$
V_m \; = \; \frac{50V}{\sin$$2\pi (20kHz)(5\mu s)$$}
$

the units on 2π in this case is radians/cycle and the units on Hz is cycles/second. After taking the scaling prefixes into account, we end up with

$
V_m \; = \; \frac{50V}{\sin$$0.2\pi$$} \; = \; \frac{50V}{0.5878} = 85.07V
$

Also, there is no need to report answers to 10 significant figures. As a general rule, engineering results are seldom better than 1%, or two sig figs, and so we routinely want our answers to be mathematically accurate to three sig figs. To help ensure this, intermediate results that are used in later calculations should be recorded to four or five sig figs to minimize accumulated round off errors. So the answer above would be reported as 85.1V.

13. ### Ripplenwinder Thread Starter New Member

Jan 17, 2015
17
0
Thanks again for the help. Just curious but did you figure out how to solve for Vm with just a back round in algebra (or various areas in math) or did someone show you that equation and how to apply it in order for you to know it. I do not plan on becoming an engineer nor am I attending school, I am not the best with math but most things I do work on have a much greater accuracy the 1%. Greater then 1% is the standard for audio synthesizers. I know Michael Faraday was not very well educated nor was nikola tesla very fond of math so I suppose I too can figure all this out?

14. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
Just algebra. The basic concepts of algebra is very simple -- whatever you do to one side of an equation you do to the other. As long as you do the same thing to both sides (there are some caveats such as dividing by zero not being allowed), then the two sides remain equal. So, for example, if I want to find the value of x that makes the following equation true:

5x + 20 = 70

I can subtract 20 from both sides:

5x + 20 - 20 = 70 - 20
5x = 50

Then I can divide both sides by 5:

(5x/5) = (50/5)
(5/5)x = 10
x = 10

I could also do it the other way around and divide both sides by 5 first:

(5x + 20)/5 = (70/5)
(5x)/5 + (20/5) = 14
x + 4 = 14

And now I can subtract 4 from both sides:

x + 4 - 4 = 14 - 4
x = 10

In either case I am starting with 5x+20 and asking what simple steps can I take to progressively isolate the x.

I remember in sixth grade we were first learning formulas for things like area and perimeters of simple geometric shapes and we had for the area of a circle

$
A \; = \; \pi r^2
$

This was the first time I had ever encountered this. We also knew that the radius was the distance from the center of the circuit to the edge and that the diameter was the distance from edge to edge going through the center. It was pretty obvious that the diameter was twice the radius. At one point in class I went up to the instructor really excited because I had discovered a new formula, namely

$
A \; = \; \frac{\pi d^2}{4}
$

I had NOT just substituted r=d/2 into the formula, I had discovered it independently, probably by observing the behavior of some examples. I had no idea why the 4, specifically, was there. The instructor gave me an encouraging word and then showed me how my discovery was merely the result of substituting r=d/2 into the other formula. At the same time I was disappointed at having my discovery turn out to be not very new, but I had also just had my first glimpse into a whole new mathematical world, namely algebra.

You might pick up an Algebra I text -- you can find lots of them dirt cheap, even free for Kindle, on Amazon and the like -- and start working your way through it. You will be amazed at the doors it will open for you.

15. ### Ripplenwinder Thread Starter New Member

Jan 17, 2015
17
0
Good deal! I remember some of it now seeing it again. Areas, and volumes I just went through in magnetic circuits. I have several algebra books. I have been purchasing and gathering everything I think I need already? At least in my research here is what I found and these reflect my purpose: circuits, electronics, passive and active network analysis and synthesis, electromagnetic and electromechanical machines, Laplace transformation and fourier series. My plan was to just learn the math I didn't know as I went along till I got through circuits and electronics, but I am open for suggestions. Thank you for taking time once again and I hope you didn't feel obligated to respond or help.

16. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
Oh, no one here is ever obligated to reply or help -- we do so because we want to.

You've got quite a road ahead of you if you want to start playing with transform methods. You will need a good understanding of calculus and, preferably, differential equations (Laplace transforms are merely a very powerful and handy way for solving the differential equations that describe the behavior of linear circuits). The big key to calculus is a rock solid grounding in algebra. Trig is also very useful for a lot of calc and also a lot of circuit concepts. If you can work through analytic geometry before you tackle calculus I think you will find the effort worth it. Calc I and II are pretty major courses but Calc III is not as important to understanding differential equations. However, if you want to play with electromagnetics in any depth, then the material in Calc III comes into play.

Along the way you need to get very comfortable with exponentials and logarithms and also complex numbers and complex arithmetic.

You need to be backing all of this up with the fundamental physics involved in circuits, which is usually provided well enough by a good calculus-based Physics II course.

17. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,294
482
That is standard equation for sinusoidal signals. Your particular example has phase angle of zero. The full equation is the one MrChips posted:
V=Amplitude*sin(2*pi*f*t+phase angle)

Last edited: Jan 17, 2015
18. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
Actually, MrChips posted that.

19. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,294
482
Thank you.

Sorry MrChips!

20. ### MrChips Moderator

Oct 2, 2009
12,227
3,280
@Ripplenwinder

If you are trying to get back into sine waves and trigonometry, think of a point rotating on the circumference of a circle in a counter-clockwise direction.

In the diagram above, the point P is radius a from the center of the circle.
As the point rotates in a counter-clockwise direction, the angle θ is increasing.

By trigonometry,

sin(θ) = y/a
or
y = a sin(θ)

Angular velocity ω describes how fast the angle θ is increasing. The units of ω are angle/time.
Angle is measured in radians, time in seconds.
Hence the units of angular velocity ω are radians/sec.

Now suppose the point P orbits around the center once every second, i.e. 360° in one second.
Hence the angular velocity ω = 2π rad/s.

In your question, the point rotates around the circle 20,000 times each second (20kHz).
Hence the angular velocity ω = 2πf = 2π x 20000 radians per second.

Another way of asking the question is how long would it take to go around the circle once?
The answer is 1/20 000 seconds = 1 000 000/20 000 μs = 50μs

So the next question is where would the point P be after 5μs if it began (t = 0) on the horizontal (θ = 0)
The answer is it would be 1/10 around the circle at θ = 36° or 2π/10 radians.

Now we can use
y = a sin(θ)

where θ = 36° or 2π/10 radians
y = 50
a = y/sin(θ) = 50/0.588 = 85

In summary, note that a sine wave is generated as the height y of the point P above the y=0 axis as P revolves around the center with an angular velocity ω.

Since

θ = ωt

and

ω = 2πf

we can write

y = a sin(θ)
y = a sin(ωt)
y = a sin(2πft)

Last edited: Jan 18, 2015