How do i analyze the following waveform to find out the voltage at t=6us?

Question

How do I analyze the following waveform to find out the voltage at \$t=6\mu s\$?

Answer 1

The integrals you've been told about already are the general approach. That will always work. But you have very simply shaped curves. So you can take a huge shortcut. And I'm pretty sure you are supposed to be able to figure it out. It's otherwise silly make-work to set up integrals over six separate segments.

So just look at the averages for each segment. The averages are very, very easy to work out since the shapes of those curves are trivial. So, just enumerate them for each successive \$1\:\mu\textrm{s}\$ period:

$$\begin{array}{rllr} 1. & \overline{i_1}=5\:\textrm{mA} & \overline{i_2}=2.5\:\textrm{mA}\\ 2. & \overline{i_1}=5\:\textrm{mA} & \overline{i_2}=5\:\textrm{mA}\\ 3. & \overline{i_1}=-5\:\textrm{mA} & \overline{i_2}=2.5\:\textrm{mA}\\ 4. & \overline{i_1}=5\:\textrm{mA} & \overline{i_2}=-2.5\:\textrm{mA}\\ 5. & \overline{i_1}=5\:\textrm{mA} & \overline{i_2}=-2.5\:\textrm{mA}\\ 6. & \overline{i_1}=-5\:\textrm{mA} & \overline{i_2}=2.5\:\textrm{mA} \end{array}$$

Just sum all that up and multiply it by your time period (\$6\:\mu\textrm{s}\$) and you have the number of Coulombs (\$Q\$) on the capacitor at the end of the period. Assuming the capacitor starts with no charge on it, the voltage then just falls out from the basic \$V_C=\frac{Q}{C}\$.

Answer 2

Just the outline of the solution:

By KCL the currents of the two current sources add up at the upper node and go into the capacitor. Therefore to obtain \$i_c=i_1+i_2\$ you must be able to add those two waveforms graphically.

Once you have the graph of \$i_c\$, you should integrate that waveform from \$t=0\$ to \$t=6\mu s\$, then divide by \$C=20nF\$. This will give you \$v_c(6\mu s)\$, assuming the capacitor is discharged at \$t=0\$ (i.e., assuming \$v_c(0)=0\$), otherwise the problem is not well specified.