Describing a sinusoidal signal mathematically and it's uses.

Thread Starter

richard3194

Joined Oct 18, 2011
173
Hi. I need to get a good grasp of the basic math of how a pure sinusoidal signal is described. At the moment I am somewhat confused, not only about the math, but the ways it can be practically applied.

Let's begin with this simple equation: ν = A sin θ. I'm not sure you could say it describes the signal. What I am sure about is that if someone sets a question and asks me to calculate ν, as long as I'm told the angle (theta) I can do it. If , in a question, A is 200V (Voltage at peak: Vp) and the sine of a given angle (22.02431 °) is 0.375: ν will be + 75V. Like I say, I'm not sure this equation has much application to circuit design. At the moment all I can say is the equation is true. Certainly there is no description of the frequency of the sinusoidal waveform.

Now, lets look at ω. This is 2πf. Well, this does describe something, the frequency of the signal. In radians per second, rather than cycles per second. But says nothing about amplitude. So, I can write f = ω or f = 2πf. so, 1 Hz = 6.28 radians per second. Time is a part of the description. I'm not sure whether this is a valid equation f (t) = 2πf (t) and if it is, what is means, or whether it has any particular application.

Here we have something else again, which looks like an attempt to describe the amplitude and the frequency: I'm looking at PLL's and I see this: A sin (2πf t + θ). It does not read ν = A sin (2πf t + θ), but I guess it could do so. Also it does not read ν (t) = A sin (2πf t + θ). So, what is going on here?

I do seem to understand when there is a situation where an exam-type of question is asked. For instance: Write the expression describing a sinusoidal signal, where the period is 20 Ms, A is 200V and ν is 75V. Okay, so the frequency is 50Hz. So, we get: +75V = 200 sin (100π + 0.375). I got the sine of theta, by dividing 75 (opposite) by 200 (hypotenuse). Of course the answer (+75V) has nothing to do with the frequency part of the equation, I mean, the 100π is ignored in the calculation of the answer. Which to me seems an odd thing. Clearly, this mathematical description is of a signal that is 22.02431 ° from the start of the cycle. That is 0.384 of a radian. So, as I have written it, the worked out equation says nothing, directly, about the phase angle. Or, the point, in terms of phase angle, at which 75V is reached. I'm assuming that + 0.375 is the correct number in the equation, and not 0.384.

I'm not at all sure why someone would devise a way of calculating ν, and half of the equation has nothing to do with the means of calculating that. Of course, A sin (2πf t + θ), which I see when looking at PLL, says nothing explicitly about calculating ν. If fact it does not look like an equation as such.

And finally, I'm not sure what is going on when I see things like: ν (t). Or f (t). Thanks if you can help. Rich
 
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wayneh

Joined Sep 9, 2010
17,496
Let's begin with this simple equation: ν = A sin θ. I'm not sure you could say it describes the signal. ...as long as I'm told the angle (theta) I can do it.
That's a good start. Now you just need to express θ as a function of time.
Now, lets look at ω. This is 2πf.
This almost gets us what we need, except we still don't see a factor for time here. I think you know that θ = ωt = 2πƒt. Now we've got something. Note how the units of frequency are inverse time (per sec), so multiplying by elapsed time t gives radians.
So, I can write f = ω or f = 2πf.
Well now you're just being sloppy.
...so, 1 Hz = 6.28 radians per second.
True.
I'm not sure whether this is a valid equation f (t) = 2πf (t) and if it is, what is means, or whether it has any particular application.
More sloppiness. Is ƒ a function of time, or a constant? Frequency can vary with time but in this type of problem it's usually taken as a constant.
Here we have something else again, which looks like an attempt to describe the amplitude and the frequency: I'm looking at PLL's and I see this: A sin (2πf t + θ). It does not read ν = A sin (2πf t + θ), but I guess it could do so. Also it does not read ν (t) = A sin (2πf t + θ). So, what is going on here?
Sloppiness on the part of the author. ν (t) = A sin (2πƒ• t + θ) is the better, more precise way to present it.
I mean, the 100π is ignored in the calculation of the answer.
It's not ignored so much as dismissed. Consider, what is the sine of 100π ? What matters is only the point in the current cycle, not how many complete cycles may have occurred prior to the one of interest
 

Alec_t

Joined Sep 17, 2013
14,280
A sinusoidal waveform can be defined as V(t)=Asin(ωt+Φ), where V(t) is the instantaneous voltage at time t, A is the wave amplitude (peak voltage), ω is the angular frequency in radians and Φ is a constant phase angle.
Since ω = 2πf, where f is the frequency in Hertz (cycles per second), the waveform equation can be re-written as V(t)=Asin(2πft+Φ).
 

BR-549

Joined Sep 22, 2013
4,928
Let's say I gave you that task of plotting a 6" circle on graph paper. You would take a compass or a radius and sweep one turn from the origin. A 3" radius will make a 6" circle. X-Y graph.

Now.......how long did it take to make that circle or rotation? How would you graph that? So...they replace the X axis for a time axis. And plot Y against time....instead of X.

A time graphed rotation or circle. Not plane graphed.
 

Thread Starter

richard3194

Joined Oct 18, 2011
173
This is what I think the situation is:

Student says to teacher : I need something so I can deal with undistorted or pure sinusoidal signals. Teacher takes a piece of paper and writes on it: A sin (2πf t + θ) and says mess with this. Student looks at what teacher wrote and thinks, what much good is this, it's not even a formula or equation. I seem to recognise things in it though.

Student says to teacher: I recognize things in what you wrote, but I cannot make head nor tail of it. It hardly seems relevant, given it's not an equation. What do you mean by mess with it? Teacher says: Well, you have got to use your intelligence and make something of what I gave you - I do assure you that I have given you what you need. Okay, let me start you off.

1) You can use what I gave you to calculate the amplitude of a sinusoidal signal at a point in time (t). For example you can write ∨ (t) = A sin (2πf t + θ). Student says: You can do that? Teacher says: Sure. But, let me tell you, you the value of ν is not a matter of the 2πf t terms. Which means that effectively you are working on this ν (t) = A sin θ. In this particular case you can choose to omit the 2πf t terms. But, if you have the data you can include the angular frequency 2πf i.e. (ω). If you do, then you are saying something about frequency as well as amplitude

(NB: I think I understand ν (t), i.e. voltage at time t. I'm not actually sure about the t term right after the 2πf term. Why is that in there?)

2...
 
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MrChips

Joined Oct 2, 2009
30,708
In my opinion, the teacher has provided you with proper guidance.

A sin (2πf t + θ) has a lot of information in it as it stands.

Even if the teacher failed to explicitly state

v(t) = A sin (2πf t + θ)

you should have been able to figure this out on your own.

What grade, level, year of schooling are you in? High school, college, university?
What course is this? Math, physics, electricity, electronics, general science?

Do you understand the meaning and origins of the following equation?

y = r sin ( θ )

If so, explain in your own words.
 

Thread Starter

richard3194

Joined Oct 18, 2011
173
Please note that I'm not studying any course. The bit with the teacher, it's fictitious, but, I've made up as a method of showing my thinking.

Also:

* When I post I'm thinking of others who may see this thread in time to come. So, I'm not neccesarily posting everything I may know.

* It is said, I should have figured this out: v(t) = A sin (2πf t + θ). Well, you know, I've never come across an equation where some terms are uneccessary in order to calculate the answer. And that is why I started this thread. I thought I could see, that if I ignore the 2πf t terms I could write, at least, v = A sin (θ), or maybe v (t) = A sin (θ). I need to put some work in regarding this t term. I mean, I'm not entirely sure you can write v (t) = A sin (θ), but pretty sure I can write v = A sin (θ).
 

MrChips

Joined Oct 2, 2009
30,708
Hello Rich, Ok. There are some gaps in your knowledge base. I am willing to help you fill those gaps but it will require some effort on your part. I will not give you the straight answers. You have to provide your own. It would help me if I knew your level of education or age.

What do the following terms mean to you?

sin (60)
sin (π/2)

cos (180)
cos (2π)

tan (90)
tan (π/4)

Explain in your own words.
 

Janis59

Joined Aug 21, 2017
1,834
The class number 10 ?
Theme - harmonic oscillations.

Or class number 12 ?
Theme - power electricity.

Anyway, the Giancoli `Physics` 7th international edition is best ever seen about most of prost physics themes. All other books it is well knotted that newcomer may not distinguish where is rope beginning and where is the very end.
 

Thread Starter

richard3194

Joined Oct 18, 2011
173
Hi. Can I just mention one thing to get of my chest. I'm a bit dissapointed, that enginers use the term phase offset, in the case where we are calculating the amplitude of the phasor or wave cycle in terms of the fraction of time elapsed relative to the origin. Yet, at some point (t) a fraction of the period, the phase of the signal has moved on from zero, but it does not feel like a phase offset. Phase offset does seem entirely appropriate when the origin of one phasor is some time before or after a reference phasor's point of origin. OK, just felt I wanted to say this. :)

For example: Phase offset ν (t) = A sin θ
 
Last edited:

MrChips

Joined Oct 2, 2009
30,708
Hi. Can I just mention one thing to get of my chest. I'm a bit dissapointed, that enginers use the term phase offset, in the case where we are calculating the amplitude of the phasor or wave cycle in terms of the fraction of time elapsed relative to the origin. Yet, at some point (t) a fraction of the period, the phase of the signal has moved on from zero, but it does not feel like a phase offset. Phase offset does seem entirely appropriate when the origin of one phasor is some time before or after a reference phasor's point of origin. OK, just felt I wanted to say this. :)
Phase offset is ok, perhaps a bit redundant.
Phase alone is ok, generally referring to a phase angle.
Offset alone is ok, generally referring to a phase angle θ or Φ, or time difference Δt, in this context.
 

Thread Starter

richard3194

Joined Oct 18, 2011
173
Phase offset is ok, perhaps a bit redundant.
Phase alone is ok, generally referring to a phase angle.
Offset alone is ok, generally referring to a phase angle θ or Φ, or time difference Δt, in this context.
What about phase progression, without to word offset?
 

Thread Starter

richard3194

Joined Oct 18, 2011
173
What about it?
Can you provide a context in which the expression is used?
Actually, thinking about it, the context is shown clearly by what one is messing with, which helps.

phase offset ν (t) = A sin θ

I guess the above tells you the context, that is, you are dealing with one phasor. And the engineer knows that. So, "phase offset" would be understood to be the point in the phase at some angle or time after the origin. Actually the phase offset in this context would be equal to θ. But, v (t) is not an angle, it's a voltage at some point in the phase. Which complicates my thinking. I'm not entirely sure now that it's correct to write:

phase offset ν (t) = A sin θ

But, where you are messing with two phasors, "phase offset" would clearly mean the offset between phasors. That is, the difference between the two phasors at the points where a periodic signal, on it's upward slope, passes through the origin.
 

MrChips

Joined Oct 2, 2009
30,708
An equation such as

ν (t) = A sin θ

implies that A and θ are constant values unless stated otherwise.

Hence

v(t) = A sin θ = constant
 

Thread Starter

richard3194

Joined Oct 18, 2011
173
θ is an angle. So, I'm quite happy think of this as a phase offset.

ν (t) is an instananeous voltage.

So, I'm thinking this is wrong: phase offset ν (t) = A sin θ. That is saying - the phase offset (ν (t)) equals A times the sine of θ.

And this is correct: ν (t) = A sin θ phase offset. That is saying ν (t) equals A times the sine of the phase offset (θ).
 

MrChips

Joined Oct 2, 2009
30,708
Neither is correct.

v(t) = A sin θ = constant

v(t) does not change with time assuming that A and θ are both constant values.

In this case, v(t) is a constant DC value with zero frequency. There is zero phase information.
 

Thread Starter

richard3194

Joined Oct 18, 2011
173
Neither is correct.

v(t) = A sin θ = constant

v(t) does not change with time assuming that A and θ are both constant values.

In this case, v(t) is a constant DC value with zero frequency. There is zero phase information.
I think you are misunderstanding my latest train of thought, which is the phrase "phase offset". My thought has not been on what is a constant and what is not.

I'm just saying that θ is the phase offset. And yes, it's constant. It's in degrees in this case. The sine of which is a ratio.

So, I'm just saying: The sine of phase offset = ν (t) / A.
 

WBahn

Joined Mar 31, 2012
29,976
I think you are misunderstanding my latest train of thought, which is the phrase "phase offset". My thought has not been on what is a constant and what is not.

I'm just saying that θ is the phase offset. And yes, it's constant. It's in degrees in this case. The sine of which is a ratio.

So, I'm just saying: The sine of phase offset = ν (t) / A.
If v(t) is a time-varying voltage, then v(t)/A is NOT a constant.

θ is the phase of the sinusoids at t=0. Hence, sin(θ) = v(0)/A.
 
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