**A unit impulse function**δ(t) can be considered a narrow pulse (of any shape) occurring at zero time such that area under the pulse is unity and the time for which the pulse occurs tends to zero. In the limit t → 0, the pulse reduces to a unit impulse δ(t). consider a narrow rectangular pulse of width A and height 1/A units, so that the area under the pulse = 1 as shown in the Fig. 1(a).

Now if we go on reducing width A and maintain the area as unity then the height 1/A will go on increasing. Ultimately when A → 0, 1/A → ∞ and it results the pulse of infinite magnitude. It may then be called an impulse of magnitude unity and it is denoted by δ(t). it is not possible to draw an impulse function on paper, hence it is represented by a vertical arrow at t = 0 as shown in the Fig.1(b).

If in the above example the area under the narrow pulse is maintained at K units while the period of pulse is reduced, it is called to be an impulse of magnitude 'K' and is denoted by K δ(t), as shown in the Fig1(c).

An important property of impulse function is that, if it is multiplied by any function and integrated then the result is the value of the function at t =0.

So response C(s) can be determined for any input once T(s) is determined.

**Note**: The equation {C(s) = R(s) . T(s)} is applicable only on Laplace domain and cannot be used in time domain. The equation {c(t) = r(t)} is not at all valid in time domain.

Now consider that input be unit impulse i.e.

r(t) = δ(t) = unit impulse input

r(t) = δ(t) = unit impulse input

**.**R(s) = L{δ(t)} = 1^{.}.
Substituting in above,

C(s) = 1

Now c(t) = L

C(s) = 1

**.**T(s) = T(s)Now c(t) = L

^{-1}{C(s)} = L^{-1}^{}{T(s)} = T(t)
Thus we can say that for impulse input, impulse response C(s) equals the transfer function T(s). So impulse response is c(t) = T(t) as C(s) = T(s) hence we can conclude that,

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