1

CLASS 3

(Sections 1.4)

Unit impulse and unit step functions

∙ Used as building blocks to construct and represent other signals.

Discrete-time unit impulse and unit step functions:

±[n] =

⎧⎨⎩

0, n = 0

1, n ∕= 0

, and u[n] =

⎧⎨⎩

0, n < 0

1, n ≥ 0

.

−5 −4 −3 −2 −1 0 1 2 3 4 50

1

n

δ[n

]

−5 −4 −3 −2 −1 0 1 2 3 4 50

1

n

u[n

]

∙ ±[n] = u[n]− u[n− 1].

∙ u[n] =∑n

m=−∞ ±[m] =∑∞

k=0 ±[n− k].

∙ To pick the n0−th sample of a signal x[n], multiply x[t] by ±[n− n0] because

x[n]±[n− n0] = x[n0]±[n− n0].

−5 −4 −3 −2 −1 0 1 2 3 4 5

−1

0

1

n

x[n

]

−5 −4 −3 −2 −1 0 1 2 3 4 5

−1

0

1

n

x[n

]δ[

n−

2]

2

Continuous-time unit impulse and unit step functions:

u(t) =

⎧⎨⎩

0, t < 0

1, t > 0

, and ±(t) = limΔ↓0

±Δ(t) where ±Δ(t) =u(t)− u(t−Δ)

Δ.

00

1

t

u(t)

00

t

δ ∆(t)

∆

1/∆

area=1

We use the following graphical notations for ±(t) and k±(t):

00

t

δ(t)

1 area=1

00

t

kδ(t)

k area=k

∙ ±(t) = ddtu(t).

∙ u(t) =∫ t

−∞ ±(¿)d¿ =∫∞0

±(t− ¿)d¿ .

∙ To pick the value of a signal x(t) at t = t0, multiply x(t) by ±(t− t0) because

x(t)±(t− t0) = x(t0)±(t− t0) ⇒∫ ∞

−∞x(t)±(t− t0)dt = x(t0).

3

Example:

Consider a unit mass with initial velocity v(0).

If we apply the force f(t) = k±Δ(t), v(t) will be

v(t) = v(0) + k

∫ t

0

±Δ(¿)d¿, for t ≥ 0.

0t

v(t)

v(0)

v(0)+k

∆

As Δ ↓ 0, the velocity transfer from v(0) to v(0) + k will be faster.

If we apply the idealized force f(t) = k±(t), v(t) will be

v(t) = v(0) + k

∫ t

0

±(¿)d¿ = v(0) + ku(t), for t ≥ 0.

In other words, the velocity will jump from v(0) to v(0) + k instantaneously.

0t

v(t)

v(0)

v(0)+k