B
bishshoy007
Guest
For this circuit given in the figure below :
![new.png new.png](https://www.edaboard.com/data/attachments/23/23328-86604a13e9c3e10a49de99a50263be57.jpg)
(Click to Enlarge)
Given that the initial voltage \[v(0-) = 4 V\], find out \[{i }_{c }(t)\], i.e current through the capacitor for \[0^- < t < \infty\] .
I have tried out the sum in two ways. First, I found out a differential node equation in \[v(t)\], i.e
\[v'[t] + v[t] = \delta (t)-e^{-t}u(t)\]
and solve it to get :
\[ v(t) = \left(5e^{-t}-t e^{-t}\right)u(t)\],
where \[u(t)\] is the UnitStep function. Also the required initial conditions was obtained as \[v(0+) = 5 V\], which I believe is perfectly all right. Next to find out \[{i}_{c}(t)\], I use :
\[{i}_{c}(t) = C \frac{\mathrm{d} v(t) }{\mathrm{d} t } = 5\delta (t) - 6e^{-t}u(t) + t e^{-t}u(t) \]
Now in procedure 2, which was done by a friend of mine, he found out \[{i}_{c}(t)\] by writing
\[\delta (t)-e^{-t}u(t) = i_c(t)+i_R(t)\]
and
\[i_R(t)=\int _{0^-}^ti_c(t)dt+v(0-)\].
Combining both equations and transforming into the Laplace equivalent, he arrived at
\[I_c(s) = \frac{s^2-4s-4}{(s+1)^2} \]
whose inverse would be
\[i_c(t) = \delta (t) - 6e^{-t}u(t) + t e^{-t}u(t) \].
Here lies the problem. The coefficient of the delta functions do not match.
So the two answers are different. :-? Where is the mistake ? I have tried the problem several times from scratch but still I am unable to find it out and this problem is bugging my sleep for many days. Please help me. I repeat the problem is that in the answers the strengths of the delta functions or the coefficients of the delta functions are different.
![new.png new.png](https://www.edaboard.com/data/attachments/23/23328-86604a13e9c3e10a49de99a50263be57.jpg)
(Click to Enlarge)
Given that the initial voltage \[v(0-) = 4 V\], find out \[{i }_{c }(t)\], i.e current through the capacitor for \[0^- < t < \infty\] .
I have tried out the sum in two ways. First, I found out a differential node equation in \[v(t)\], i.e
\[v'[t] + v[t] = \delta (t)-e^{-t}u(t)\]
and solve it to get :
\[ v(t) = \left(5e^{-t}-t e^{-t}\right)u(t)\],
where \[u(t)\] is the UnitStep function. Also the required initial conditions was obtained as \[v(0+) = 5 V\], which I believe is perfectly all right. Next to find out \[{i}_{c}(t)\], I use :
\[{i}_{c}(t) = C \frac{\mathrm{d} v(t) }{\mathrm{d} t } = 5\delta (t) - 6e^{-t}u(t) + t e^{-t}u(t) \]
Now in procedure 2, which was done by a friend of mine, he found out \[{i}_{c}(t)\] by writing
\[\delta (t)-e^{-t}u(t) = i_c(t)+i_R(t)\]
and
\[i_R(t)=\int _{0^-}^ti_c(t)dt+v(0-)\].
Combining both equations and transforming into the Laplace equivalent, he arrived at
\[I_c(s) = \frac{s^2-4s-4}{(s+1)^2} \]
whose inverse would be
\[i_c(t) = \delta (t) - 6e^{-t}u(t) + t e^{-t}u(t) \].
Here lies the problem. The coefficient of the delta functions do not match.
So the two answers are different. :-? Where is the mistake ? I have tried the problem several times from scratch but still I am unable to find it out and this problem is bugging my sleep for many days. Please help me. I repeat the problem is that in the answers the strengths of the delta functions or the coefficients of the delta functions are different.
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