Question

A system with x(t) and output y(t) is defined by the input-output relation:
$\text{Y(t) =}{\int }_{-\infty }^{-2t}\text{x}\left(\tau \right)\text{d}\tau$
The system will be

• A.
  casual, time-invariant and unstable
• B.
  casual, time-invariant and stable
• C.
  non-causal, time-variant and unstable
• D.
  non-causal, time-invariant and unstable

Answer 1

The correct option is C

non-causal, time-variant and unstable
$\text{First of all we will check for causality,}$
$\text{Y(t) =}{\int }_{-\infty }^{-2t}\text{x(}\tau \text{) d}\tau$
$\text{So for t = -2}$
$\text{Y(-2) =}{\int }_{-\infty }^4\text{x(}\tau \text{) d}\tau$

So output depends on future values of input along with past and present values of input so system is non-causal.

Let us find output for shifted input $x(t-t_o)$
$\text{Y(t) =}{\int }_{-\infty }^{2t}\text{x(}\tau -t_o\text{) d}\tau$
$\text{Y(t) =}{\int }_{-\infty }^{(2t-t_o)}\text{x(}\tau \text{) d}\tau$ ....(i)

$\text{Now shift the output by}{t}_o$
$\text{then, y(t-t}{}_o\text{) =}{\int }_{-\infty }^{(2t-t_o)}\text{x(}\tau \text{) d}\tau$ ....(ii)

$\text{so from eq. (i) and (ii),}$
$\text{y(t)}\ne \text{y(t - t}{}_o\text{)}$
$\text{Therefore system is time variant.}$

Non-causal and time variant is present only in option (d).