The voltage on capacitor is given by

$$V_c=\dfrac{1}{C} \int i(t)dt$$

and the current is

$$i(t)=0.4 \sin(2t-\frac{\pi}{4})u(t)$$

when u(t) is unit step function. May you show how to integrate the expression, how to find V, how to regard the step function while integrating?

• consider using laplace to determine response to a step – JonRB Feb 23 '14 at 21:28
• May you show how to integrate more or less step by step? – user40987 Feb 23 '14 at 21:32
• Consider: What does the step function look like from -infinity to 0? What does it look like from 0 to infinity? How can you use this to simplify the integrand? – Theran Feb 23 '14 at 21:40

First thing to note is that the step function is defined as

$$u(t) = \left.\begin{cases} 0 & t<0\\ 1 & t>0\\ \end{cases}\right\}$$

so

$$V_c = \frac{1}{C}\int_{-\infty}^{t_f} i(t) dt$$

$$V_c = \frac{1}{C}\int_{-\infty}^{0} 0.4\sin (2t - \frac{\pi}{4} )u(t) dt + \frac{1}{C}\int_0^{t_f} 0.4\sin (2t - \frac{\pi}{4} )u(t) dt$$

$$V_c = \frac{0.4}{C}\int_0^{t_f} \sin (2t - \frac{\pi}{4} ) dt$$

So all the step function does is set all values of t that are less than zero to zero.

You then continue to compute the integral in the normal way, so:

$$V_c = \frac{0.4}{C} \left[ \frac{1}{2} \cos(2t - \frac{\pi}{4})\right]_0^{t_f}$$

• @ KillaKem Why does the integral need to be considered at first place from -infinity? Doesn't time starts at 0? – user40987 Feb 24 '14 at 3:28
• Starting the integral at -infinity is the general case, it works if the function is zero for t < 0 and if the function is not zero for t < 0, that is why we start from -infinity.In your example because we know t is time we know the system is not defined for t < 0, so we can just start the integral at t=0 immediately.But the point of the function u(t) is to make it absolutely clear that the integral starts at t = 0, so anyone evaluating the integral knows they are working with a system that is zero or all values below 0 (this is called a causal system). – KillaKem Feb 24 '14 at 12:25

First, write your first equation more correctly as so

$$v_C(t) = \frac{1}{C}\int_{-\infty}^t i_C(\tau)\, d\tau$$

Now, because of the presence of the unit step we have

$$v_C(t) = \frac{1}{C}\int_{-\infty}^t 0.4 \sin(2\tau-\frac{\pi}{4})u(\tau)\, d\tau = \frac{1}{C}\int_0^t 0.4 \sin(2\tau-\frac{\pi}{4})\, d\tau$$

In other words, the unit step in the integrand has the effect of changing the lower limit of integration to $\tau = 0$.

The step function simply allows you to start integrating at $t = 0$ and know that $V_c = 0,\ t \leq 0$.

The integral itself should now be straightforwardly solved by using the primitive function and inner derivative.