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).
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:I believe the equation is actually incorrect since the units do not match.
I believe the equation is actually incorrect since the units do not match.
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).
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.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.
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? <<<<
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.
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.
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,
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.
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.
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