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ESE Electronics 2011 Paper 1: Official Paper

Option 2 : \(\displaystyle\int_{-\infty}^\infty f(t) \ \delta(\tau) \ d\tau =1\)

CT 1: Current Affairs (Government Policies and Schemes)

54993

10 Questions
10 Marks
10 Mins

__Concept:__

**Unit Impulse function:**

A continuous-time unit impulse function δ(t), also called a Dirac delta function is defined as:

δ (t) = ∞ , t = 0

= 0, otherwise

**The unit impulse function is represented by an arrow with the strength of ‘1’ which represents its area.**

\(\mathop \smallint \limits_{ - \infty }^\infty δ \left( t \right)dt = 1\)

**Properties of Delta function:**

**Scaling Property:**

\(δ \left( {at} \right) = \frac{1}{{\left| {at} \right|}}δ \left( t \right)\)

**Multiplication Property:**

X(t).δ(t – t_{0}) = x(t_{0})δ(t – t_{0})

**Sampling Property:**

\(\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right)δ \left( {t - {t_0}} \right)dt = x\left( {{t_0}} \right)\)

**Important Expressions:**

X(t).δ(t) = x(0)δ(t)

\(\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right)δ \left( t \right)dt = x\left( 0 \right)\)

**Explanation:**

From the concept part we can find that:

**option(1) is true**

From option (2)

\(\displaystyle\int_{-\infty}^\infty f(t) \ δ(\tau) \ d\tau =1\)

**This option is false**

As

\(\displaystyle\int_{-\infty}^\infty \ δ(\tau) \ d\tau =1\)

**Both options (3) and (4) follow delta function properties,**

**Hence they are true.**

__Important Points__

**Differentiation Property:**

\(\mathop \smallint \limits_{{t_1}}^{{t_2}} x\left( t \right){δ ^n}\left( {t - {t^0}} \right)dt = {\left( { - 1} \right)^n}x\left( {{t_0}} \right)\)

**Even Signal property:**

δ(-t) = δ(t)

**Convolution Property**:

\(\mathop \smallint \limits_{ - \infty }^\infty x\left( \tau \right)\delta \left( {t - \tau } \right)d\tau = x\left( t \right)\)

**Integration Property:**

\(\mathop \smallint \limits_{{t_1}}^{{t_2}} x\left( t \right)\delta \left( t \right)dt = x\left( 0 \right)\) for (t_{1} < t_{2})

0, otherwise.