Correct Answer - Option 4 :
\({\rm{x}}\left( { - {\rm{t}}-{{\rm{t}}_0}} \right)\)
Concept:
Unit impulse function:
It is defined as, \(\delta \left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\infty ,\;\;t = 0}\\ {0,\;\;t \ne 0} \end{array}} \right.\)
Properties:
1. \(\mathop \smallint \limits_{ - \infty }^\infty \delta \left( t \right)dt = 1\)
2. \(\delta \left( {at} \right) = \frac{1}{{\left| a \right|}}\delta \left( t \right)\)
3. x(t) δ(t – t0) = x(t0) ⇒ x(t) δ(t) = x(0)
4. \(\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right)\delta \left( {t - {t_o}} \right)dt = x\left( {{t_0}} \right)\)
5. \(\mathop \smallint \limits_{ - \infty }^T x\left( t \right)\delta \left( {t - {t_o}} \right)dt = 0\) If T < t0
6. \(\mathop \smallint \limits_{ - \infty }^\infty f\left( t \right)\delta \left( {at + b} \right)dt = \mathop \smallint \limits_{ - \infty }^\infty f\left( t \right)\frac{1}{{\left| a \right|}}\delta \left( {t + \frac{b}{a}} \right)dt\)
7. \(\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){\delta ^n}\left( {t - {t_o}} \right)dt = {\left. {\frac{{{d^n}x}}{{d{t^n}}}} \right|_{t = {t_0}}}\)
Calculation:
Using properties of impulse
\(\begin{array}{l} {\rm{x}}\left( { - {\rm{t}}} \right){\rm{*\delta }}\left( { - {\rm{t}} - {{\rm{t}}_0}} \right){\rm{\;}} = {\rm{\;x}}\left( { - {\rm{t}}} \right){\rm{*\delta }}\left( {{\rm{t}} + {{\rm{t}}_0}} \right)\\ = {\rm{\;x}}\left( { - \left( {{\rm{t}} + {{\rm{t}}_0}} \right)} \right)\\ = {\rm{\;x}}\left( { - {\rm{t}} - {{\rm{t}}_0}} \right) \end{array}\)
