Question

The input x(t) and the corresponding output y(t) of a system are related by
$y\left(t\right)=\underset{-\infty }{\overset{5t}{\int }}x\left(\tau \right)d\tau$
The system is

• A. time variant and causal
• B. time variant and non causal
• C. time invariant and causal
• D. time invariant and non causal

Answer 1

The correct option is B time variant and non causal

$y\left(t\right)=\underset{-\infty }{\overset{5t}{\int }}x\left(\tau \right)d\tau$

$\text{For t = 1,}$

$y\left(1\right)=\underset{-\infty }{\overset{5}{\int }}x\left(\tau \right)d\tau$

Here output depends on future values of x(t), so system is noncausal.

Now for shifted input $t-t_0$, output is
$y^{\prime }\left(t\right)=\underset{-\infty }{\overset{5t}{\int }}x(\tau -t_0)d\tau =\underset{-\infty }{\overset{5t}{\int }}x(\tau -t_0)d\tau$

$\tau -t_0={\tau }^{\prime }$

$\Rightarrow y^{\prime }t=\underset{-\infty }{\overset{(5t-t_0)}{\int }}x\left({\tau }^{\prime }\right)d{\tau }^{\prime }$

$andy(t-t_0)=\underset{-\infty }{\overset{(5t-t_0)}{\int }}x\left(\tau \right)d\tau \ne y^{\prime }\left(t\right)$

So system is time variant.