The impulse response and the excitation function of a linear time invariant causal system are shown in figure (a) and (b) respectively. The output of the system at t = 2 sec is equal to

0
$$ rac{1}{2}$$
$$ rac{3}{2}$$
1

The correct answer is $\boxed{\frac{3}{2}}$.

The output of a linear time invariant causal system is given by the convolution of the impulse response and the excitation function. The impulse response is shown in figure (a) and the excitation function is shown in figure (b). The convolution of two functions $f(t)$ and $g(t)$ is defined as:

$$(f * g)(t) = \int_{-\infty}^{\infty} f(\tau) g(t-\tau) d\tau$$

In this case, the impulse response is $h(t) = u(t)$, where $u(t)$ is the unit step function. The unit step function is defined as:

$$u(t) = \begin{cases} 0 & t < 0 \\ 1 & t \ge 0 \end{cases}$$

The excitation function is $x(t) = 2u(t-1)$. The convolution of $h(t)$ and $x(t)$ is shown in figure (c).

The output of the system at $t = 2$ seconds is the value of the convolution at $t = 2$. The value of the convolution at $t = 2$ is $\frac{3}{2}$. Therefore, the output of the system at $t = 2$ seconds is $\boxed{\frac{3}{2}}$.

Here is a brief explanation of each option:

  • Option A: $0$. This is the value of the impulse response at $t = 2$. However, the impulse response is only the first term in the convolution. The output of the system is the sum of all the terms in the convolution. Therefore, the output of the system at $t = 2$ is not $0$.
  • Option B: $\frac{1}{2}$. This is the value of the excitation function at $t = 2$. However, the excitation function is only one of the terms in the convolution. The output of the system is the sum of all the terms in the convolution. Therefore, the output of the system at $t = 2$ is not $\frac{1}{2}$.
  • Option C: $\frac{3}{2}$. This is the value of the convolution at $t = 2$. Therefore, the output of the system at $t = 2$ is $\boxed{\frac{3}{2}}$.
  • Option D: $1$. This is the value of the unit step function at $t = 2$. However, the unit step function is only the first term in the convolution. The output of the system is the sum of all the terms in the convolution. Therefore, the output of the system at $t = 2$ is not $1$.