Calculate Rise Time for Sinusoidal Signal at f_0

[VinnyCee]

Algebraically calculate the Rise Time value for the sinusoidal signal at an arbitrary frequency f_0 (expressed in Hz). Assume that the signal has zero average, that is

V_{min}\,=\,-V_{max}

Express the Rise Time in units of the period T\,=\,\frac{1}{f_0}.


MY WORK SO FAR:

\left(V_{min}\,+\,0.9\,V_{ppk}\right)\,-\,\left(V_{min}\,+\,0.1\,V_{ppk}\right)

V_{ppk}\,=\,V_{max}\,-\,V_{min}

RISE\,=\,0.8\,V_{max}\,-\,0.8\,V_{min}

RUN\,=\,T

SLOPE equals RISE over RUN, but I don't know how to relate the two things to get an equation for the answer.

The problem goes on to ask us to calculate the Rise Time value in msec for the sinusoidal signal at 500Hz. But I need the equation in order to solve it. Please help!

 

[Astronuc]

The rise time of a pulse is the time taken to go from 10% to 90% of its final value.

So given a simple sin wave sin\omega t, one wants to find the time that it takes the signal to go from t₁ = 0.1 to t₂ = 0.9 if the signal goes from 0 - 1, or t₁ = -0.8 to t₂ = 0.8 if the signal goes from -1 to 1.

Now how to find t?

If one knows the value of the function, e.g. 0.1, then

sin\omega t\,=\,0.1, or

\omega t\,=\,sin^{-1}(0.1), or

t\,=\,\frac{sin^{-1}(0.1)}{\omega}

One then needs to know the relationship between f and \omega .

 

[VinnyCee]

Angular frequency, \omega, is equal to 2\,\pi times the frequency f(in Hertz).

So the Rise Time for a -1 to 1 sinusoidal signal is T\,=\,\frac{sin^{-1}(-0.8)}{2\,\pi\,f_0}? What about the positive one? Should I be subtracting it from the one just stated?

 

[berkeman]

Sorry if I missed it, but the term rise time must be qualified with what fraction of the total waveform amplitude the rise time is measured over.

For example, it is common with trapezoidal logic signals to refer to the "10 to 90 rise time", which means that you measure the time it takes during the rising transition for the signal to go from 10% of the final value to 90% of the final value.

Was something like this mentioned in the original problem statement?

 

[VinnyCee]

Yes, its exactly what is stated earlier in the lab manual. The problem itself does not specify this though, it is implied.

Does the T equation that I have above look right for the 10 to 90 rise time in terms of f_0?

 

[berkeman]

Angular frequency, \omega, is equal to 2\,\pi times the frequency f(in Hertz).

So the Rise Time for a -1 to 1 sinusoidal signal is T\,=\,\frac{sin^{-1}(-0.8)}{2\,\pi\,f_0}? What about the positive one? Should I be subtracting it from the one just stated?

I'm not sure I understand your approach, but just calculate what fraction of a period it takes for the sine wave to go from zero to 0.8 in amplitude. You'll get something a little less than a radian, I believe. Then multiply by 2 to get the total 10-90 rise time of the sine wave (from -0.8 to + 0.8). Since the answer is in terms of radians at the frequency of any sine wave, that shows that the frequency does not matter.

 

[VinnyCee]

So the T above is the correct answer?

 

[VinnyCee]

Algebraically calculate the Rise Time value for the sinusoidal signal at an arbitrary frequency f_0 (expressed in Hz). Assume that the signal has zero average, that is

V_{min}\,=\,-V_{max}

Express the Rise Time in units of the period T\,=\,\frac{1}{f_0}.

Using your algebraic results, calculate the Rise Time in \mu s for the sinusoidal signal at 50 kHz.


Please help! I have NO IDEA what the problem is trying to get at. Please help!