Transcript Chapter 2-latest

Transformations of Continuous-Time Signals
Continuous time signal:
• Time is a continuous variable
• The signal itself need not be continuous .
Time Reversal
Time Reversal
compressing
Expanding
Let

2.2
Periodic Functions
X(t) is periodic if given t and T ,is there some period T > 0 such that
Or the maximum common frequency f1 , f 2
x (t )  cos(18t )  cos(12t )
f 1  9 Hz
Integer
frequency
1X 9
3X3
9X1
The common frequencies
f 2  6 Hz
1X 6
2X3
3X2
6X1
3X3 2X3
9X1 6X1
f 0  The maximum frequency
T 0  The period  (1/3) second
f 0  3 Hz
Minimum period
You can show that the period T0 of x(t) is (1/3) as follows
x (t )  cos(18t )  cos(12t )
x (t  (1/3))  cos(18 (t  (1/3))  cos(12 (t  (1/3))
 cos(18t  6 )  cos(12t  4 )
0
Since
1
cos(18t  6 )  cos(18t )cos(6 )  sin(18t )sin(6 )
 cos(18t )
Similarly
1
0
cos(12t  4 )  cos(12t )cos(4 )  sin(12t )sin(4 )
 cos(12t )
x (t  (1/3))  cos(18t )  cos(12t )  x (t )
2.3 Common Signals in Engineering
RL
i(0) is the initial current
RC

R
v (t )
C
v (t )  v (0)e

t
RC

In general
x (t )  Ce
 at
Exponential function
Exponential Signals
x(t (= C eat
C and a can be complex
Exponential Signals appear in the solution of many physical systems
Chapter 7 in EE 202
Chapter 9 in 202
Sinusoidal ?
RL and RC
Sinusoidal steady state
What that’s to do with the exponential ?
Euler’s identity
cos( ) 
e
e  j   cos( )  j sin( )
j
e
2
 j
sin( ) 
e
j
e
2j
 j
Chapter 7 in EE 201
RL and RC
Complex sinusoidal
Since e
j 0T
e
j 2 T
T
e
j 2
 10
o
Critical damped sinusoid
(The most general case)
underdamped sinusoid
2.4
Block function (window) defined as
The unit step function has the property
(1)
(2)
The Unit Impulse Function
let's look at a signal
What is its derivative ?
Define it as :
Now as   0
d
dt


 (t )
 t  0
 0 else
 (t )  
t
0
Properties of Delta Function
x (t ) (t )
(I)
x(t)
x(t)
t
0
 (t )
X
t
0

0
x (0) (t )
x (0) (t )
x (0)
t

 x(t ) (t )dt   x(0) (t )dt


0 (t )  0
0


 x(0)  (t )dt  x(0)

t
x(t)
( II )
 (t  t 0 )
x (t 0 ) (t )
x (t 0 )
0
t0
t
t
t0


x (t ) (t  t 0 )dt  x (t 0 )

Exercise



e
 t 2
 (t  10)dt  e
 (10)2
 e 100
( II )


x (l ) (t  l )dt  x (t )

x(l)
 (t  l )
x (t )
0
l
This property is known as “convolution” which will be useful in chapter 3
Note  (t )   (t )
"Even Function"
In this section we will break down special type of functions in terms of
known other simpler functions
Example
Step Function
u (t )
0 t  0
u (t )  
1 t  0
1
t
t
t
r (t )


t
u (t )dt
0


t

t 0
t 0
t
t
Ramp Function
0 t  0
r (t )  
t t  0
u (t )
Unit Step Function
1
t
Integration
r (t ) Unit Ramp Function
t
t
t
Time Shift
u (t  t 0 )
u (t )
1
1
0
t
0 t0
u (t  t 0 )
1
t 0 0
t
t
r (t  t 0 )
r (t )
0
t
0 t0
r (t  t 0 )
t 0 0
t
t
Time folding
u ( t )
u (t )
1
1
t
0
t
0
u (t  6)  u (6  t )
1
0
t
6
r ( t )
r (t )
0
t
t
0
0
r (t )  
t
t 0
t 0
u (t )
x (t )
1
1
2
0
t
0
2
t
u (t  2)
1
2
x (t) = u (t )  u (t  2)
t
Pulse Function
(t )
 1
 1
 (t) = u  t    u  t  
2

 2
1
1

2
0
t
1
2
(t )
Chang of Scale
Consider the pulse function (t)
1
1

2
Plot (2t ) ?

1
Since  (t ) = 
0

1
t 
2
1
2
else
t
0
1
2

1
Since  (2t ) = 
0

2
1
1/ 4
0
 2t 
1

2
else
 (2t )
1/ 4
1
t
1
4
t 
1
4
r (t )  r (t  2)
r (t )  r (t  2)
2.5
2.0  0.0  2
2.1  0.1  2
2.5  0.5  2
2.1
2
r (t )
2.1
0.1
1
2
They will cancel
each other
t
t
0.5
r (t  2)
r (t )  r (t  2)
t
2
In general
r (t )  r (t  m )
t
m
x (t )
We want to express or decompose this function to singularities functions
2
0
4
2
t
First we cancel the uprising ramp after t =2 with subtracting downward ramp
Which will result in horizontal line segment
2
r (t )
r (t  2)
Therefore we need to subtract another ramp
r (t  2)
0
2
However what we need is a downward ramp
4
t
r (t )  2r (t  2)
2
r (t )
r (t  2)
r (t )  2r (t  2)
r (t  2)
t
However the downward ramp will continue indefinitely
2
Therefore we need to cancel this part of downward ramp
r (t  4)
0
To cancel this part of downward ramp
We add an upward ramp at t = 4
t
2
4
x (t)=r (t )  2r (t  2)  r (t  4)
x (t )
x (t) = r (t ) 2r (t  2)  r (t  4)
2
0
2
4
t
Example We will break f (t) into simpler singular functions
To cancel an upward ramp we subtract down ward ramp
f (t)  3u(t) r(t) r(t 1) 5u(t  2)
Let f (t) be the input to a system
y(t) ?
system
We seek the output y(t)
f (t)  3u(t)  r(t)  r(t 1) 5u(t  2)
system
y4 (t)
y3(t)
y2 (t)
y1(t)
y(t)  y1(t)  y2(t)  y3(t)  y4(t)
if the system linear as will be shown later
Input
Output
The relations between cause and effect (Input/output) are expressed as equations
Output
Input
Examples
Electric Heater
Car
Speed
paddle
Speed
A another example of a physical system is a voltage amplifier such as that used
in public-address systems
amplifier
block diagram
Representation of a general system
y(t) = T[x(t)]
where the notation T[x(t)] indicates a transformation or mapping
This notation T[.] does not indicate a function
that is, T[x(t)] is not a mathematical function into which we substitute
x(t) and directly calculate y(t).
The explicit set of equations relating the input x(t) and the output
y(t) is called the mathematical model, or simply, the model, of the
system



  Mathmatical Model

Properties of Continuous  Time Systems
y(t )  S[ x(t )]
x(t )
S[ ]
y(t  to )
time shift
S[ x(t  to )]
x(t  to )
x(t )
time shift
S[ ]
t
Is the capacitor is time-invariant ? VC (t ) 1  i( )d
C 
t to
VC (t  to )  1  i( )d
C 
t
i(t )
S[ ]
VC (t )  S[i(t )]  1  i( )d
C 
time shift
t
i(t  to )
i(t )
S[ ]
time shift
S[i(t  to )]  1  i(  to )d
C 
t
The capacitor is time-invariant if S[i(t  to )]  VC (t  to ) 1  i(  to )d
C 
t
OR
t to
1 i(  t )d  1 i( )d
o


C 
C 
both output of the block diagram are equal
both output of the block diagram are equal
The capacitor is time-invariant
are equal
The resistor is time-invariant
are not equal
is time-variant
t
Is z(t ) =  x( )d

time-invariant ?
are they equal ?
let  =  to
S[ x(t  to )]=
t to
 x( )d
 z(t  to )
time-invariant
t
Is z(t ) =  x( )d
time-invariant ?
0
t
z(t  to ) =  x( )d
0
are they equal ?
t
S[ x(t  to )] =  x(  to )d
0
let  =  to
S[ x(t  to )]=
t to

t
o
x( )d
t
  x( )d
0
 z(t  to )
time-variant
Examples of Linear systems
 c[ax1 (t )  bx2 (t )]  cax1 (t )  cbx2 (t )  cy1 (t )  cy2 (t )
Examples of Nonlinear systems
We can stop here and imply the system is non linear
violates Additivity due to the cross  terms
To satisfy homogeneity  S[cx(t)]  cy(t)  cx(t)  ca
 cx(t)  ca
We can stop here and imply the system is non linear
To check additivity
S[ x1(t)] = x1(t)  a
x1(t)
x2(t)
x1(t)  x2(t)
S[ ]
S[ x2(t)] = x2(t)  a
S[ x1(t)  x2(t)] = x1(t)  x2(t)  a
 S[ x1(t)]+S[ x2(t)]