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This note set introduces the concept of continuous-time signals, focusing on unit step functions and their relationship with unit ramp functions. the definition of continuous-time signals, the unit step function, and the relationship between the two. It also explains how the unit ramp function can be derived from the unit step function through integration.
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EECE 301
Signals & Systems
Prof. Mark Fowler
Ch. 1 Intro C-T Signal Model Functions on Real Line System Properties D-T Signal Model Functions on Integers
LTICausalEtc
Ch. 2 Diff EqsC-T System Model Differential Equations^ D-T Signal Model Difference Equations Zero-State Response Zero-Input Response^ Characteristic Eq.
Ch. 2 ConvolutionC-T System Model^ Convolution Integral^ D-T System Model^ Convolution Sum
Ch. 3: CT
Fourier Signal
Models Fourier Series Periodic Signals Fourier Transform (CTFT)^ Non-Periodic Signals
New
System
Model
New Signal^ Models
Ch. 5: CT
Fourier System
Models Frequency Response Based on Fourier Transform New
System
Model
Ch. 4: DT
Fourier Signal
ModelsDTFT (for “Hand” Analysis)^ DFT & FFT (for Computer Analysis)
New Signal^ Model^ PowerfulAnalysis Tool
Ch. 6 & 8: LaplaceModels for CT^ Signals
&^ Systems Transfer Function New^ System
Model Ch. 7: Z Trans.Models for DT Signals
&^ Systems Transfer Function
New
System Model
Ch. 5: DT
Fourier System
Models Freq. Response for DT^ Based on DTFT New
System
Model
The arrows here show conceptual flow between ideas. Note the parallel structure between
the pink blocks (C-T Freq. Analysis) and the blue blocks (D-T Freq. Analysis).
u ( t ) 1
t
Note:
A step of height
A^ can be made from
Au
( t )
These are common textbook signals but are also common test signals, especially in control systems.
t^ = 0
Depends on
t^ value
⇒function of
t :^ f
( t )
- Write unit step as a function of
- Integrate up to
- How does area change as
t^ changes?
i.e., Find Area
Area =
f ( t )
“Running Integral of step = ramp” t ∫ ∞−
d u
t = ∫ ∞−
λ λ)(
)(
1 )( )(
tr t t d u tf
t
== ⋅=
∞−
λ λ ∫ ∞− =^
t
d u tr
λ λ)(
)(
we have:
u ( t ) 1
r ( t ) 1
t
⎧ ⎨ ⎩
< =^
0 , 0
0 , 1 )(
t^ t
tdr dt
⎧ ⎨ ⎩
≥ <
=^
0 , 0
0 ,
)(
t t t
tr
tdr dt
tu
)(
)( =
u ( t ) 1
t
u ( t -2)
t
u ( t +0.5)
t
Other Names: Delta Function,
Dirac Delta Function
Infinite heightZero widthUnit
area
a pulse with:
δ( t
δ( t
0 any for , 1 )(
0 any for , 0 )(
=
≠
= ∫ −
dtt
t
t
t
δ( t
Caution
… this is NOT the vertical axis… it is the delta function!!!
δ ( t
0
) ( ) ( )( 0 0
0
0
∀
= −
ε
δ ε t ε t
tf dt t t tf
t
f ( t )
t^0
f ( t )^0
t
t^0
f ( t )^0 )^0
δ ( t – t
As long as the integral’s limits surround the“location” of the delta… otherwise it returns zero
? ) (^5). 2 ()
2 sin( 0
=
−
∫^
dt
t t^ δπ
0 ) (^5). 2 ()
2 sin( 0
=
−
∫^
dt
t t^ δπ
Step 1
: Find variable of integration:
t
Step 2
: Find the argument of
t^ – 2.
Step 3
: Find the value of the variable of integration that causes the argument of
) to go to zero:
t^ = 2.
Step 4
: If value in Step 3 lies inside limits of integration… No! Otherwise… “return” zero…
t π
(^ − t δ
? ) 4 (^3) ( ) 3 )(
7 sin(^4
2
=
−
∫ −
dt t
tt
δ
ω
(^
)ω
δ
ω^
(^3) / 4 sin (^26). 6
) 4 (^3) ( ) 3 )(
7 sin(^4
2
−
=
−
∫ −
dt t
t t
Step 1
: Find variable of integration:
τ
Step 2
: Find the argument of
τ^ + 4
Step 3
: Find the value of the variable of integration that causes the argument of
) to go to zero:
τ^ = –
Step 4
: If value in Step 3 lies inside limits of integration… Yes! Take everything that is multiplying
): (1/3)sin(
ωτ/3)(
τ/3 – 3)
2
…and evaluate it at the value found in step 3:
(1/3)sin(–4/
ω)(–4/3 – 3)
2 = 6.26sin(–4/
ω) Because of this… handleslightly differently!
Step 0
: Change variables: let
τ^ = 3
t^ Î
d τ
dt^ Î
limits:
τL^
τL^
21 12
2
τ τδ
τ ωτ^
u ( t )^1
Derivative = 0
Derivative = 0 Derivative = “
∞ ” (“Engineer Thinking”)
δ ( t
t
x ( t )
... ...
When we say “Period” we almost always mean “Fundamental Period”
x ( t )
x ( t^