The input x(t) and the output y(t) of a system are related as

$y(t)=e^{-t}{\int }_{-\mathrm{\infty }}^t{e}^{\tau }x(\tau )d\tau ,-\mathrm{\infty }<t<\mathrm{\infty .}$

The system is

The correct option is Linear and time-invariant.

To understand why the system is both linear and time-invariant, let us analyze its properties step-by-step.

The relationship between the input $x(t)$ and the output $y(t)$ is given by:

$$y(t)=e^{-t}{\int }_{-\infty }^t{e}^{\tau }x(\tau )d\tau$$

We can rewrite this expression by moving the term $e^{-t}$ inside the integral, since the integration is with respect to $\tau$:

$$y(t)={\int }_{-\infty }^t{e}^{-(t-\tau )}x(\tau )d\tau$$

This is in the exact form of a convolution integral, defined as:

$$y(t)={\int }_{-\infty }^{\infty }h(t-\tau )x(\tau )d\tau$$

where $h(t)$ is the impulse response of the system. Comparing the two equations, we identify the impulse response of the system as:

$$h(t)=e^{-t}u(t)$$

where $u(t)$ is the unit step function, which accounts for the upper limit of integration being $t$ instead of $\infty$.

1. Linearity:
A system represented by a convolution integral is always linear because integration is a linear operator. If we apply an input $x_1(t)$, the output is $y_1(t)$. If we apply $x_2(t)$, the output is $y_2(t)$. Applying $a{x}_1(t)+b{x}_2(t)$ results in $a{y}_1(t)+b{y}_2(t)$, satisfying the principle of superposition and homogeneity. Therefore, the system is linear.

2. Time-Invariance:
Any continuous-time system whose input-output relation is expressed as a convolution with a fixed impulse response $h(t)$ is time-invariant. To verify, let us shift the input by $t_0$, i.e., $x_d(t)=x(t-t_0)$:
$$y_d(t)={\int }_{-\infty }^t{e}^{-(t-\tau )}x(\tau -t_0)d\tau$$ Let us perform a change of variables by setting $\lambda =\tau -t_0$. Thus, $d\lambda =d\tau$. The limits of integration change from $\tau =-\infty$ to $\lambda =-\infty$, and from $\tau =t$ to $\lambda =t-t_0$. Substituting these values gives: $$y_d(t)={\int }_{-\infty }^{t-t_0}{e}^{-(t-(\lambda +t_0))}x(\lambda )d\lambda$$ Simplifying the exponent: $$y_d(t)={\int }_{-\infty }^{t-t_0}{e}^{-((t-t_0)-\lambda )}x(\lambda )d\lambda =y(t-t_0)$$ Since the shifted input produces an output that is identically shifted in time, the system is time-invariant.

Thus, the system is both linear and time-invariant.