Question

The output y(t) of a system is related to its input x(t) as
$y\left(t\right)={\int }_0^{t}x(\tau -2)d\tau$
where, x(t) = 0 and y(t) = 0 for t $\le$ 0. The transfer function of the system is

• A. $\frac{1}{s}$
• B. $\frac{(1-e^{-2s})}{s}$
• C. $\frac{e^{-2s}}{s}$
• D. $\frac{1}{s}-e^{-2s}$

Answer 1

The correct option is C $\frac{e^{-2s}}{s}$

Put $x\left(t\right)=\delta \left(t\right)$

$y\left(t\right)=h\left(t\right)={\int }_0^t\delta (\tau -2)d\tau =u(t-2)$

So, $H\left(s\right)=\frac{e^{-2s}}{s}$
Alternatively,
$y\left(t\right)={\int }_0^{t}x(\tau -2)d\tau$

$Y\left(s\right)=\frac{1}{s}\times e^{-2s}$

$\therefore \frac{Y\left(s\right)}{X\left(s\right)}=\frac{e^{-2s}}{s}$