Sine Interpret

[dr.power]

Hi guys,

I do know what is the meaning of "Vm*sin(Wt)" and how to draw it, And of course what Vm and "W" stand for. I do know that W is 2pi*F too.
But I have got this equation ""W*Vm*sin(Wt)"". Now I have a trouble to interpret it.
Any idea plz?
You can assume that Vm is 4V and F is say 50KHz.

Thanks a bunch

 

[crutschow]

It means that the amplitude of the waveform amplitude is equal to Vm * ω (W is actually ω [omega]) or Vm*2*pi*f. For your example the equation would be 2*pi*50k*sin(2*pi*50k*t).

I believe the equation is actually incorrect since the units do not match.

 

[dr.power]

Thank crutschow for your input,

It means that the amplitude of the waveform amplitude is equal to Vm * ω (W is actually ω [omega]) or Vm*2*pi*f. For your example the equation would be 2*pi*50k*sin(2*pi*50k*t).

Actually the equation is this ""-Vm*w^2*sin(t*w)"" Where F is the audio signal frequency.

I believe the equation is actually incorrect since the units do not match.

Although I am amazed about the mismatched units behinds the sine function(radian/s and Volts), But I must say that the equations are correct! Suppose that F happens to be 2KHz and Vm happens to be 4V. Then we'll get this:
-(2Pi)²*(2kHz)²*4*sin(2pi*2kHz)t. Yet I can not interpret the behind of the sine function.
2kHz and 4V do not have similar units so that I could multiple then by each other. So I have no idea what it stands for and how I deal with it?

 

[dr.power]

I believe the equation is actually incorrect since the units do not match.

You can simply reach to such an equation having W or W² in behinds of the sine function by taking one or second order derivatives of a signal like "Vm*sin(Wt)" or "Vm*cos(Wt)".

 

[MrAl]

Hi drpower,


Here are four frequency sweep plots. One is for sin(w*t), another for w*sin(w*t), and another for w^2*sin(w*t).
We allow w to increase as well as time t for all but the first plot.

As you can see, sin(w*t) is just the sine wave.
w*sin(w*t) means that as we let the frequency w increase, the amplitude also increases. The amplitude increases as a ramp.
w^2*sin(w*t) means that as we let frequency increase, the amplitude increases again, but this time it increases much faster than with just w*sin(w*t).

So the w or w^2 makes the sin(w*t) look like it got a special kind of high pass filtering where frequency amplitudes increase either as w or as w^2.

The last plot (sin(w*t) with frequency sweep) shows the sine just being swept through some frequencies. You can see there that although the frequency changes the amplitude remains constant. Compare to the 2nd and 3rd drawings, where the amplitude changes markedly.

 

[dr.power]

Hello MrAl,

Thank you very much for your reply,

As you can see, sin(w*t) is just the sine wave.
w*sin(w*t) means that as we let the frequency w increase, the amplitude also increases. The amplitude increases as a ramp.

So it says that by incarsing the frequncey of the function, the amplitude or the gain would be incarse too. Is that happen in a liner scale regarding to "w*sin(wt)"?

your plots make sense, but yet I am not able to see how "w" behinds the sine function (the first "w") affects on the function? Any mathematic/numerical explanation plz? Furthermore I do not know how/Why the said first "w"does affect on the MAGNETUDE of the sine function (w does not have the unite as the magnetude (ie voltage))?

w^2*sin(w*t) means that as we let frequency increase, the amplitude increases again, but this time it increases much faster than with just w*sin(w*t).

So for Vm=1V in the first normal sine plot, the magnetude is able to cover more than 30V for "w*sin(wt)" and more than 900V for w²*sin(wt).!
I have to take you through PM regarding to this, So that I notice how it does happen in Nonliner air!!!!

So the w or w^2 makes the sin(w*t) look like it got a special kind of high pass filtering where frequency amplitudes increase either as w or as w^2.

As I told if you plz kindly could give me a mathematic or at least an example by the numerical values it would be very ok to completely understand it.
By considering w and w^2 as a high pass filter can you plz let me know if they obey the 3dB or 6dB/octave as normall high pass filters too?


Another question.
is there any meaning for something like ""w*Vm"" (wihout ans sine function in front of it) where w is 2pi*F and Vm is a peak volatge? If so IS it interpretable?

Thanks a lot

 

[MrAl]

Hello again,


Let me ask you a question...

Say we have a low pass filter made from a single resistor and a single capacitor. We analyze the output with a (peak) unit sinusoidal input wave.
The transfer function is:
1/(s+1)

so the amplitude is:
1/sqrt(w^2+1)

The output will be a sine wave, so the amplitude of the sine wave output for four different frequencies is:

0.1Hz, ampl=0.8467
0.2hz, ampl=0.6227
0.4Hz, ampl=0.3697
0.8Hz, ampl=0.1951

So what do these four distinct waves look like?

Well, since the wave for 0.1Hz has an amplitude of 0.8467 the output wave therefore is a sinusoid with peak of 0.8467, so the relative form is:
0.8467*sin(2*pi*0.1*t).

What about the 0.2Hz wave?
That is a sinusoid with peak 0.6227. The form is 0.6227*sin(2*pi*0.2*t).

Ditto for the other two waves with their respective amplitudes.

How did we know this?
Because we found that the amplitudes were going to be:
1/sqrt(w^2+1)

times the input, or:
Vout=(1/sqrt(w^2+1))*sin(w*t)

Now this is not an exact representation of the output, because we did not include the phase shift. However, if we look at this on a scope and sync at the zero degree point on the sinusoid, we'll see this wave just as above. It makes sense to interpret it this way then when the phase is not important in our problem.

Now here comes the question for you...

You see that we have 1/sqrt(w^2+1) multiplying the sine wave. This definitely is a function that has 'w' in the denominator.

My question for you then is this:

>>>> Do you see any reason why we can NOT interpret the factor 1/sqrt(w^2+1) as an amplitude of the sinusoid? <<<<

 

[dr.power]

Hello again,


Let me ask you a question...

Say we have a low pass filter made from a single resistor and a single capacitor. We analyze the output with a (peak) unit sinusoidal input wave.
The transfer function is:
1/(s+1)

so the amplitude is:
1/sqrt(w^2+1)

The output will be a sine wave, so the amplitude of the sine wave output for four different frequencies is:

0.1Hz, ampl=0.8467
0.2hz, ampl=0.6227
0.4Hz, ampl=0.3697
0.8Hz, ampl=0.1951

So what do these four distinct waves look like?

Well, since the wave for 0.1Hz has an amplitude of 0.8467 the output wave therefore is a sinusoid with peak of 0.8467, so the relative form is:
0.8467*sin(2*pi*0.1*t).

What about the 0.2Hz wave?
That is a sinusoid with peak 0.6227. The form is 0.6227*sin(2*pi*0.2*t).

Ditto for the other two waves with their respective amplitudes.

How did we know this?
Because we found that the amplitudes were going to be:
1/sqrt(w^2+1)

times the input, or:
Vout=(1/sqrt(w^2+1))*sin(w*t)

Now this is not an exact representation of the output, because we did not include the phase shift. However, if we look at this on a scope and sync at the zero degree point on the sinusoid, we'll see this wave just as above. It makes sense to interpret it this way then when the phase is not important in our problem.

Now here comes the question for you...

You see that we have 1/sqrt(w^2+1) multiplying the sine wave. This definitely is a function that has 'w' in the denominator.

My question for you then is this:

>>>> Do you see any reason why we can NOT interpret the factor 1/sqrt(w^2+1) as an amplitude of the sinusoid? <<<<

Hi MrAl,

Thanks a lot as usual.

Actually I know what do you mean by your above example. But you knew that Vm is equal to 1/sqrt(w^2+1), in your case.

Regarding to my equation, I have no idea what is the meaning of "w*Vm" or "w²*Vm".
We both knew before from somewhere that ""w²*Vm" is nonlinear gain, But I just wanted to see if there is any mathematical relationship between the "w" behinds the sine function and Vm (thats the w factor which affects on Vm so that creates the amplitude of the function "w*Vm*sin(wt)", and it is w² factor which affects on Vm so that it creates final nonlinear amplitude in "w²*Vm*sin(wt)").

We know for instance that "V=R*I".
"R" has the unit of ohms, And "I" has the unit of Ampere, But It is not confusing to me when somebody tells that R*I is an voltage because it is defined to be volatge before, And of course (in contrast to something like w²*Vm) I am informed about that definition.
I myself have no idea if there is a definition for "w²*Vm" as as the amplitude or not.
Anyway If you think that it has some sort of problems to to be explained (specially for a beginner like me), We can give up for the REASON. Your plots at post #5 are clear enough and make sense what is happening to the frequency and amplitude of a sine wave when there is factor of w or w² behinds it. It was good if we could have a numerical example regarding to "w*Vm*sin(wt)" and "w²*Vm*sin(wt)" too.
But I do not like to bother you anymore. As I told your plots make a lot of sense.

Thanks a bunch for everything.

 

[MrAl]

Hello again,


"Actually I know what do you mean by your above example. But you knew that Vm is equal to 1/sqrt(w^2+1), in your case."

Yes but that multiplies sin(w*t):

Vout=1/sqrt(w^2+1)*sin(w*t)

and now you want to call 1/sqrt(w^2+1) equal to "Vm", but what is Vm here and what is sin(w*t) here?
If Vm has unit of volts, how did we get that? And if sin(w*t) has unit of volts, how can we multiply Vm*sin(w*t)=volts, we would get volts^2.

For the following:
V=R*i
dV/di=R


How about this:
v=d(w*t)/dt * sin(w*t)

Does that make more sense to you?
'w' is not a function of 't' although it is a function of frequency. When we take the derivative of w*t we see that we get a gain that changes with frequency. For f=1, the gain is 2*pi, for f=2, the gain is 4*pi, etc. For every unit frequency we get a gain of 2*pi.

 

[dr.power]

Hello again,


"Actually I know what do you mean by your above example. But you knew that Vm is equal to 1/sqrt(w^2+1), in your case."

Yes but that multiplies sin(w*t):

Vout=1/sqrt(w^2+1)*sin(w*t)

and now you want to call 1/sqrt(w^2+1) equal to "Vm", but what is Vm here and what is sin(w*t) here?
If Vm has unit of volts, how did we get that? And if sin(w*t) has unit of volts, how can we multiply Vm*sin(w*t)=volts, we would get volts^2.

For the following:
V=R*i
dV/di=R


How about this:
v=d(w*t)/dt * sin(w*t)

Does that make more sense to you?
'w' is not a function of 't' although it is a function of frequency. When we take the derivative of w*t we see that we get a gain that changes with frequency. For f=1, the gain is 2*pi, for f=2, the gain is 4*pi, etc. For every unit frequency we get a gain of 2*pi.

Thank you very much MrAl.

Ok plz let's have an exampl eso that see what is happening, Maybe I could solve the problem by the below example (I hope).
In my example I choice:
Vm=3V
F=2KHz.
V(t)= Vm*sin(2pi*F)= 3*sin(2pi*2KHz)t= 3*sin(12566.3)t.

Now suppose we get this:
V(t)= 2pi*F*Vm*sin(2pi*F)t= 12566.3*3*sin(12566.3)t= 37698.9*sin*(12566.3)t.

Now if we get this:
V(t)= (2pi*F)^2*Vm*sin(2pi*F)t= (12566.3)^2*3*sin(12566.3)t= 473735687.1*sin(12566.3)t.

Of course the above calculations show that if F raises then the amplitude will raise too.

Now I want to know if I have done it just right? I multiplied the W and w^2 by Vm so that get a new Vm ,

To be honest I just memorized what "Vm*sin(wt)" is, and what is its waveform and how to draw it.
I also can memorize "W*Vm*sin(wt) just as your saying. Everything would be solved just if you confirm my above three calculations.

Kind regards

 

[MrAl]

Hi,

Yes that looks right, as long as you multiplied correctly

If we have a voltage v=-cos(w*t) and we take the first time derivative we get:
dv/dt=w*sin(w*t)

and notice that if we multiply both sides by time we cancel the first 'w' and again get volts.

In most systems there is some other limiting factor such as the voltage of the supply that powers the circuit, and this prevents an infinite response. In the application of ultrasonic transmission in air we'd have to find out what is the limiting factor in air.

 

[dr.power]

Hi Al,
Thank you.

Hi,

Yes that looks right, as long as you multiplied correctly

If I could not do a multiplection, then I should go to hell.

If we have a voltage v=-cos(w*t) and we take the first time derivative we get:
dv/dt=w*sin(w*t)

and notice that if we multiply both sides by time we cancel the first 'w' and again get volts.

Why the first W would be canceled out by multiplying by t?
Is that due this fact that F is 1/t, so it would be "2pi*t/t"?
Now that I am thinking more to w, it seems a bit more clear that assume the W somehow as a matter of time, So W can be multiplied by anything yet let the said thing to maintain its overall uint.

In most systems there is some other limiting factor such as the voltage of the supply that powers the circuit, and this prevents an infinite response. In the application of ultrasonic transmission in air we'd have to find out what is the limiting factor in air.

I myself guess that the limiting factor is the "saturation" of air.
Actaully I think the "w^2" factor behinds the sine function is a donation for our job due to the nasty response (limited bandwidth) of the output sensors.
We'll talk more in this regard later.

 

[MrAl]

Hi again,

What would you say happens to cause the air to 'saturate' ? Would this be due to the inertia of the air particles you think?

Here's another example:
V=Vm*cos(w*t) [instantaneous voltage]
dV/dt=-Vm*w*sin(w*t) [rate of change of instantaneous voltage with time]

For Vm*cos(w*t) the w is angular frequency and t is time, w*t is just angle. Vm could be volts peak.

For dV/dt, that's change in voltage per unit time, and we can rewrite:
dV/dt=-Vm*2*pi/Tp*sin(w*t)
where Tp is the period.
So we have on each side informally:
volts/time=volts/time

 

[dr.power]

Hi again,

What would you say happens to cause the air to 'saturate' ? Would this be due to the inertia of the air particles you think?

Hi,

I guess that maybe the imaginary SPRINGS which hold the air molecules connected to each other Are able to change their position somewhat (they are not able to move infinitely when disturbed by voice).

 

[MrAl]

Hi,


Well, my quick question was not thought out. Inertia does not eat up power in a system it only stores it, but resistance does. So it's the thermoviscous attenuation effect of the air that eats up some of the power. Also, thermoviscous attenuation is frequency dependent.

 

[dr.power]

Hi,


Well, my quick question was not thought out. Inertia does not eat up power in a system it only stores it, but resistance does. So it's the thermoviscous attenuation effect of the air that eats up some of the power. Also, thermoviscous attenuation is frequency dependent.

What is thermoviscous really? I guess that the friction bettwen th eair molecules is the other reason...

 

[MrAl]

Hi,

That's it yes, and the faster the particles move the more friction and heat and that means more loss so the response is worse at higher frequencies.
We ignore that in many cases for almost the same reason we dont consider the distance a listener is from an audio speaker unless the system is going to be used in a large room or outside.

 

[dr.power]

Hi,

That's it yes, and the faster the particles move the more friction and heat and that means more loss so the response is worse at higher frequencies.
We ignore that in many cases for almost the same reason we dont consider the distance a listener is from an audio speaker unless the system is going to be used in a large room or outside.

Actually I do not know why the higher frequncies would be absorbed much faster than the lower ones. Any idea plz?

Plz delete several pf your PMs so that I could sent you 1-2 PM's.,..

 

[MrAl]

Hi,

More friction.

My box is filled up again?