Graphing the response of an underdamped circuit

[paulmdrdo]

Homework Statement

Graph the response v(t)

Relevant Equations

SEE THE ATTACHED PHOTO

I'm having difficulty as to how I would produce an approximately graph of the response just by hand. I was able to determine the first maximum by taking the derivative of the response and setting to zero and I'm stuck. How do I know the succeding minimum and maximum of this response? TIA.

 

[The Electrician]

The derivative of the response is periodic with the same period as v(t), so the next zero will be one period after the first zero, and so on for the rest of them.

Just how extreme is "by hand"? Are you able to use a calculator to calculate values of the expression for a few values of t? The graph will be zero at t=0 and again after one period of the sine wave. You could calculate a few points in between and plot them.

You can see that the exponent in the exponential is 2t. That will decay rapidly, almost vanishing in the time of one cycle of the sine wave.

 

[LvW]

Sin SQRT[(2)*t] ?
Time t in deg or rad?

 

[Master1022]

I'm having difficulty as to how I would produce an approximately graph of the response just by hand. I was able to determine the first maximum by taking the derivative of the response and setting to zero and I'm stuck. How do I know the succeding minimum and maximum of this response? TIA.
View attachment 242162
View attachment 242161

With these $y = A\exp(-kt) sin(t)$ (whether it is $sin$ or $cos$), if you are looking for a 'rough' response, I think an easy way is to first sketch your $Ae^{-kt}$ portion of the graph (above and below the t-axis) and this is the boundaries of your graph (because the trig part can vary from 0 to 1, so it will never go outside of this 'envelope'). Then you can think of a couple different points (e.g. $0, \frac{\pi}{2}, \pi, etc...$ and label those points on the graph). For example, I would think:
- for t = 0, sin(0) = 0 so y = 0
- for t = $\frac{\pi}{2}$, sin(pi/2) = 1, so the we will be at the value of $y = Ae^{-kt}$
- for t = pi, sin(pi) = 0, so y = 0 again
- for t = $\frac{3\pi}{2}$, sin(3pi/2) = -1, so we will be at the value of $y = \mathbf{-}Ae^{-kt}$

Then just connect the dots with a wave. I think this way is easier to do by hand when you just want a sketch. Hope that is of some use. Also note that the time period of the wave-forms/ oscillations remains constant despite the decreasing amplitude.

For the maximum and minimum, you could differentiate to confirm.