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Consider signal \(x\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1,}&{\left| t \right| \le 2}\\ {0,}&{\left| t \right| > 2} \end{array}} \right.\) Let δ(t) denote the unit impulse (Dirac-delta) function. The value of the integral \(\mathop \smallint \limits_0^5 2x\left( {t - 3} \right)\delta \left( {t - 4} \right)\) dt is
1. 2
2. 1
3. 0
4. 3

1 Answer

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Correct Answer - Option 1 : 2

\(x\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1,}&{\left| t \right| \le 2}\\ {0,}&{\left| t \right| > 2} \end{array}} \right.\)

\(\mathop \smallint \limits_0^5 2x\left( {t - 3} \right)\delta \left( {t - 4} \right)dt\)

From the property of unit impulse function,

\(\smallint x\left( t \right)\delta \left( {t - {t_0}} \right)dt = x\left( {{t_0}} \right)\)

\(= \mathop \smallint \limits_0^5 2x\left( {t - 3} \right)\delta \left( {t - 4} \right)dt\)

= 2x(4 – 3) = 2x(1)

As, x(1) = 1, the integral value = 2

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