1
CLASS 3
(Sections 1.4)
Unit impulse and unit step functions
∙
Used as building blocks to construct and represent other signals.
1
u[n]
δ[n]
Discrete-time unit impulse and unit step functions:
⎧
⎧




⎨ 0, 𝑛 = 0
⎨ 0, 𝑛 < 0
𝛿[𝑛] =
,
and
𝑢[𝑛] =
.




⎩ 1, 𝑛 ∕= 0
⎩ 1, 𝑛 ≥ 0
0
−5 −4 −3 −2 −1 0 1 2 3 4 5
n
1
0
−5 −4 −3 −2 −1 0 1 2 3 4 5
n
∙
𝛿[𝑛] = 𝑢[𝑛] − 𝑢[𝑛 − 1].
∙
𝑢[𝑛] =
∙
To pick the 𝑛0 −th sample of a signal 𝑥[𝑛], multiply 𝑥[𝑡] by 𝛿[𝑛 − 𝑛0 ] because
∑𝑛
𝑚=−∞ 𝛿[𝑚] =
∑∞
𝑘=0
𝛿[𝑛 − 𝑘].
x[n]
1
0
−1
−5 −4 −3 −2 −1 0 1 2 3 4 5
n
x[n]δ[n−2]
𝑥[𝑛]𝛿[𝑛 − 𝑛0 ] = 𝑥[𝑛0 ]𝛿[𝑛 − 𝑛0 ].
1
0
−1
−5 −4 −3 −2 −1 0 1 2 3 4 5
n
2
Continuous-time unit impulse and unit step functions:
⎧


⎨ 0, 𝑡 < 0
𝑢(𝑡) − 𝑢(𝑡 − Δ)
.
𝑢(𝑡) =
,
and
𝛿(𝑡) = lim 𝛿Δ (𝑡) where 𝛿Δ (𝑡) =
Δ↓0

Δ

⎩ 1, 𝑡 > 0
1
δ∆(t)
u(t)
1/∆
0
area=1
0
0
t
0∆
t
We use the following graphical notations for 𝛿(𝑡) and 𝑘𝛿(𝑡):
1
0
area=1
area=k
kδ(t)
δ(t)
k
0
0
t
0
t
𝑑
𝑢(𝑡).
𝑑𝑡
∙
𝛿(𝑡) =
∙
𝑢(𝑡) =
∙
To pick the value of a signal 𝑥(𝑡) at 𝑡 = 𝑡0 , multiply 𝑥(𝑡) by 𝛿(𝑡 − 𝑡0 ) because
∫𝑡
−∞
𝛿(𝜏 )𝑑𝜏 =
∫∞
0
𝛿(𝑡 − 𝜏 )𝑑𝜏 .
∫
𝑥(𝑡)𝛿(𝑡 − 𝑡0 ) = 𝑥(𝑡0 )𝛿(𝑡 − 𝑡0 )
∞
⇒
𝑥(𝑡)𝛿(𝑡 − 𝑡0 )𝑑𝑡 = 𝑥(𝑡0 ).
−∞
3
Example:
Consider a unit mass with initial velocity 𝑣(0).
If we apply the force 𝑓 (𝑡) = 𝑘𝛿Δ (𝑡), 𝑣(𝑡) will be
∫
𝑡
𝑣(𝑡) = 𝑣(0) + 𝑘
𝛿Δ (𝜏 )𝑑𝜏,
for 𝑡 ≥ 0.
0
v(t)
v(0)+k
v(0)
0 ∆
t
As Δ ↓ 0, the velocity transfer from 𝑣(0) to 𝑣(0) + 𝑘 will be faster.
If we apply the idealized force 𝑓 (𝑡) = 𝑘𝛿(𝑡), 𝑣(𝑡) will be
∫
𝑡
𝑣(𝑡) = 𝑣(0) + 𝑘
𝛿(𝜏 )𝑑𝜏 = 𝑣(0) + 𝑘𝑢(𝑡),
for 𝑡 ≥ 0.
0
In other words, the velocity will jump from 𝑣(0) to 𝑣(0) + 𝑘 instantaneously.
v(t)
v(0)+k
v(0)
0
t