Transcript EE441 Data Structures (Fall 2004)

Course Outline (Tentative)
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Fundamental Concepts of Signals and Systems
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Linear Time-Invariant (LTI) Systems
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Fourier Transform & Properties …
Modulation (An application example)
Discrete-Time Frequency Domain Methods
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Response to complex exponentials
Harmonically related complex exponentials …
Fourier Integral
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Convolution integral and sum
Properties of LTI Systems …
Fourier Series
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Signals
Systems
DT Fourier Series
DT Fourier Transform
Sampling Theorem
Laplace Transform
Z Transform
Chapter II
Linear Time-Invariant (LTI) Systems
Linear Time-Invariant Systems
Important class of systems as many physical phenomena can
be modeled as such

Representation of DT Signals in Terms of Pulses
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DT unit impulse can be used to construct any DT signal
Think of a DT signal as a sequence of individual impulses
Consider
x[n]
-1
-4
...
-3 -2
...
2 3
0 1
4
n
Linear Time-Invariant Systems
x[n] is actually a sequence of time-shifted and scaled impulses
x[-2]δ[n+2]
x[0]δ[n]
x[-1]δ[n+1]
-1
-2
0
x[1]δ[n-1]
01
012
x[2]δ[n-2]
Linear Time-Invariant Systems

It is possible to re-generate an arbitrary signal by
sampling it with shifted unit impulse:
x[n] = - - - + x[-4] δ[n+4] + x[-3] δ[n+3] + x[-2] δ[n+2] +
+ x[-1] δ[n+1] + x[0] δ[n] + x[1] δ[n-1] + - - -

This is called as sifting property:
x[n] 

 x[k ] [n  k ]
k 
shifted impulse
weights
Sifting property of DT impulse
DT Unit Impulse Response
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x[n] is superposition of scaled & shifted impulses
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The output of a linear system to x[n] =



superposition of the scaled responses of the system to each of
these shifted impulses
Let hk[n] be the response of a linear system to δ[n-k].
From superposition property of linear systems, the response y[n] is the
sum of individual responses, i.e.,
x[n] 

 x[k ] [n  k ]
k 

y[n] 

 x[k ]h [n]
k 
k
DT Unit Impulse Response

In general, if the system is also time-invariant, then
hk[n], response to δ[n-k], is a time-shifted version of
h0[n], which is the response of the system to δ[n];
hk [n]  h0 [n  k ]

So, for k=0, i.e., input is δ[n], then the response of the
system is
h[n]  h0 [n]
Unit Impulse Response
h[n] is the output (response) of the LTI system
when δ[n] (impulse) is the input
Convolution Sum
Representation of LTI Systems
Recall that for an LTI system (due to superposition property):
x[n] 

 x[k ] [n  k ]

y[n] 
k 

 x[k ]h [n]
k 
k
Hence, for an LTI system, the output y[n] for the input x[n] is
y[n] 

 x[k ]h[n  k ]
k 
Convolution Sum
Superposition Sum
convolution of the sequences (signals) x[n] and h[n], and represented by:
y[n]  x[n]  h[n]
Convolution Sum
Example
Example: Consider an LTI system with impulse response h[n] and input x[n] is
applied to the system. Find the output y[n].
2
x[n]
0.5
n
0 1 2
0 1
y[n] 
h[n]
1

 x[k ]h[n  k ]  x[0]h[n  0]  x[1]h[n 1]  0.5h[n]  2h[n 1]
k 
2.5
0.5
0 1 2
2 h[n-1]
0.5 h[n]
+
1 2 3
2.5
2
0.5
0 1 2 3
y[n]
Convolution Sum
Example

Example: Consider LTI system with h[n]
2
0.5
x[n]
h[n]
1
n
0 1 2
0 1

y[n]=
 x[k] h[n-k]
k  
Consider h[n-k] as a time-reversed and shifted version of h[k]
If we plot h[-k] then we can obtain h[n-k] simply by shifting to
right (by n) if n>0, or to left if n<0
Convolution Sum
Example
2

x[k]
0.5
0

k
1
1
h[n-k] , n<0
k
n-2 n-1
n
Flip h[k]
By shifting h[k] for all possible values
of n, pass it through x[n].
for n  0, y[n]   x[k ]h[n  k ]  0
k
for n  0, y[0]   x[k ]h[0  k ]  0.5
0
k
k
-2
-1
h[0-k]
0
for n  2, y[2]   x[k ]h[2  k ]  0.5  2  2.5
k
k
k
-1
0
h[1-k]
1
1
k
2.5
2
2
h[3-k]
n-2 n-1 n
h[n-k] , n>3
2.5
2
0.5
3
k
0
k
h[2-k]
k
0 1
for n  3, y[3]   x[k ]h[3  k ]  2
for n  3, y[n]   x[k ]h[n  k ]  0
k
0
for n  1, y[1]   x[k ]h[1  k ]  0.5  2  2.5
y[n]=0
0 1 2 3
y[n]
Convolution Sum
Example
Ex:
Consider x[n]=  n u[n] for 0< <1 and h[n]=u[n]. Find y[n].
y[n]=
 x[k] h[n-k]
k
1
x[k]=  k u[k]
h[-k]
---
k
k
0 1 2
0
Convolution Sum
Example
for n<0
x[k]h[n-k]=0
for  k
for n≥0
1
x[k]=  k u[k]
k
x[k]h[n-k]≠0
k
0 1
0 1 2
for n≥0
h[n-k]
--
for
0≤k≤n
n
Convolution Sum
Example
 k , 0  k  n
x[k]h[n-k] = 
0 , o/w
n
y[n]   =
k
k 0
1   n1
1
 1   n1 
y[n]  
 u[n]
 1 
for n  0
Convolution Sum
Example
Ex: Consider an LTI system with input x[n]=2n u[-n] and h[n]=u[n]. Find y[n].
1
1
1
1
---
1
2
x[k]=2k
4
u[-k]
h[n-k]
---
8
1
16
k
k
0
-4 -3 -2 -1 0
n
for n≥0, x[k]h[n-k] has nonzero samples for k≤0
0
y[n] =

k  

0

x[k]h[n-k] =
k  
2k
1
=  
k 0  2 
k
=
1
1
1
2
=2
Convolution Sum
Example
for n<0, x[k]h[n-k] has nonzero samples for k≤n
y[n] =
 x[k]h[n-k] = 
k  
k  

1
  
l 0  2 

n
n
l n

2k
1
=  
l  n 2 
l
l
1
n
 2     2n 1
l 0  2 
2
1
, n0
2
y[n]   n1
, n0
2
1
1
1
---
2
4
8
1
16
-5 -4 -3 -2 -1 0 1 2
y[n]
---
Representation of CT Signals in terms
of Impulses
Consider a pulse or “staircase” approximation,
xˆ (t ) , to a CT signal x(t)
x(t)
Similar to DT case, xˆ (t ) can be
expressed as a linear combination
of delayed pulses
---Δ
Δ 2Δ
t
kΔ
0
1
 , 0t 
Define   (t )   
0,
o/w
δΔ(t)
1

0 Δ
t
Representation of CT Signals in
Terms of Impulses
Since   (t ) has unit amplitude, we can represent
of shifted and scaled rectangular pulses…
xˆ (t ) 

 x(k)
k  
As Δ
0,
xˆ (t )

xˆ (t )
in terms
(t  k)
approaches to x(t)

x(t )  lim
 0
 x(k) (t  k)

k  

x(t )   x( ) (t   )d (sifting property)

Representation of CT Signals in
Terms of Impulses
Other derivations for this ??
Recall:
x(t ) (t   )  x( ) (t   )
Integrate both sides over  :




 x(t ) (t   )d   x( ) (t   )d




x(t )   (t   )d 
 x( ) (t   )d
1

x(t )   x( ) (t   )d

CT Unit Impulse Response &
Convolution Integral
Let us define
hˆk (t )
yˆ (t ) 
as the response of a linear system to
  (t  k)

ˆ (t )
x
(
k

)
h

k
k  
As Δ  0 yˆ (t) approaches to y(t)

y(t) = lim
 0
 x(k)hˆ
k  

Hence,
y (t ) 
k
(t ) (note as   0   (t  k)   (t  )
 x( )h (t )d

hence
hˆk (t )  h (t )
)
CT Unit Impulse Response &
Convolution Integral
By time-invariance
h (t )  h0 (t   )
and therefore,
h(t )  h0 (t )
impulse response!
 x( )h(t   )d
Convolution
integral

y (t ) 

y (t )  x(t )  h(t )
An LTI system is completely characterized by its impulse response
(i.e., h[n] in DT, h(t) in CT)
Convolution Integral
Example
 at
Let x(t )  e u (t ) , a>0 be the input to an LTI system with
h(t) = u(t). Find y(t)!
1
x ( )
1
h( )


0
0

y (t ) 
 x( )h(t   )d

Convolution Integral
Example
h( )
h(t   )


t
0
h(t   )
0
for t>0

0
For t<0,
t>0
t
x( )h(t   )  0
for t<0
y(t)=0
e  a , 0    t
x( )h(t   )  
 0 , o/w
Convolution Integral
Example
t
for t>0
y(t )   e
0
 a
1 a
d   e
a
1
y (t )  (1  e  at )u (t )
a
1
a
y (t )
t
t
0
1
 (1  e at )
a
Properties of Convolution and
LTI Systems
1) Commutative Property:
Convolution is a commutative operation in both DT and CT.

x[n]  h[n]  h[n]  x[n]   h[k ]  x[n  k ]


x(t )  h(t )  h(t )  x(t )   h( )x(t   )d
Easily shown
by substituting
variables

Practical use ??
(take h[k], time-reverse and shift x[k] whenever it’s easier)
Properties of Convolution and
LTI Systems
2) Distributive Property:
Convolution distributes over addition.
x[n]  (h1[n]  h2[n])  x[n]  h1[n]  x[n]  h2[n]
x(t )  (h1 (t )  h2 (t ))  x(t )  h1 (t )  x(t )  h2 (t )
Practical interpretation ??
x
h1
+
h2
(parallel interconnection of systems)
y
x
h1+h2
y
Properties of Convolution and
LTI Systems
( x1[n]  x2[n])  h[n]  x1[n]  h[n]  x2[n]  h[n]
( x1 (t )  x2 (t ))  h(t )  x1 (t )  h(t )  x2 (t )  h(t )
Practical ??
- (response to sum of two inputs = sum of responses to inputs individually)
- Break a complicated convolution into several simpler ones (useful !!)
Properties of Convolution and
LTI Systems
3) Associative Property:
x[n]  (h1[n]  h2[n])  ( x[n]  h1[n])  h2[n]
x(t )  (h1 (t )  h2 (t ))  ( x(t )  h1 (t ))  h2 (t )
x[n]
h1[n]
h2[n]
y[n]
: (x[n]* h1[n])* h2[n]
by associativity
x[n]
h1[n]*h2[n]
y[n]
x[n]
h2[n]*h1[n]
y[n]
x[n]
h2[n]
h1[n]
y[n]
Result 1
by commutativity
Result 2
Properties of Convolution and
LTI Systems
Results:
Impulse response of cascade LTI systems is the convolution of their
individual impulse responses
 Overall system response is independent of the order of the systems in
cascade !

4) LTI Systems with & w/o Memory:
h[n]  0 for
If 
h(t )  0 for
n0
t 0
then the LTI system is memoryless !
(look at the convolution sum

Hence,
 h[n]  K [n] 


 h(t )  K (t ) 
y[n]   h[k ]x[n  k ] )
k
 y[n]  Kx[n] 


 y (t )  Kx(t ) 
memoryless
Properties of Convolution and
LTI Systems
5) Invertibility of LTI Systems
x(t)
x(t)
h(t)
y(t)
h1(t)
Identity System
δ(t)
w(t)=x(t)
x(t)
if the system is invertible then h(t )  h1 (t )   (t )
h[n]  h1[n]   [n]
 h1 (t ) 
 is the impulse response of the LTI inverse system.
and 
 h1[n] 
Properties of Convolution and LTI Systems
Example
Consider LTI system of time-shift:
y(t)=x(t-t0)
Impulse response of the system is
h(t)=δ(t-t0)
y (t )  x(t )  h(t )
x(t  t0 )  x(t )   (t  t0 )
Convolution of a signal with a
shifted impulse shifts the signal
Inverse: shift the output back!
h1 (t )   (t  t0 )
h(t )  h1 (t )   (t  t0 )   (t  t0 )   (t )
Properties of Convolution and LTI Systems
Example
Consider an LTI system with h[n]=u[n]
y[n] 

 x[k ]u[n  k ]
; as u[n-k] = 0 for n-k<0
k  
y[n] 
n
 x[k ]
accumulator
k  
LTI inverse:
y[n]=x[n]-x[n-1]
(1st difference)
h1[n] = δ[n]-δ[n-1]
Check if really inverse:
h[n]*h1[n] = u[n]*(δ[n]-δ[n-1])
= u[n]*δ[n]-u[n]*δ[n-1]
= u[n]-u[n-1] = δ[n]
Properties of Convolution and
LTI Systems
6) Causality for LTI Systems:
y[n] 

 x[k ]h[n  k ]
; the output y[n] must not depend on x[k] for k>n
k  
h[n-k] should be zero for k>n
h[n]=0
for n<0
h(t)=0
for t<0
Practical intuition ??
(impulse response must be zero before the impulse occurs:
initial rest)
For causal LTI systems, convolution sum becomes:
y[n] 
n

k  
k 0
 x[k ]h[n  k ]   h[k ]x[n  k ]
Properties of Convolution and
LTI Systems

In CT: h(t)=0
becomes:
y(t ) 
for t<0
causal; convolution integral
t


0
 x( )h(t  )d   h( ) x(t  )d
REMARK: Causal Signal ! (zero for n<0, t<0)
Properties of Convolution and
LTI Systems
7) Stability for LTI Systems:
x[n]  B, n
Consider
y[n] 


 h[k ]x[n  k ]
k  

 h[k ] x[n  k ]
k  

 B  h[k ]
<B
n
k  
if the impulse response is absolutely summable

 h[k ]  
k  
then the system is stable!
Properties of Convolution and
LTI Systems

In CT: if the impulse response is absolutely integrable

 h( ) d  

then the system is stable!
Example: Shift h(t)=δ(t-t0)


 h( ) d    (  t ) d  1
0

stable!

Accumulator: h[n]=u[n]


 u[k ]   u[k ]  
k  
k 0
unstable!
Unit Step Response
Another signal used quite often to describe the behaviour of LTI
systems corresponds to the output when x[n]=u[n] or x(t)=u(t)
s[n]  u[n]  h[n]
  h[k ]u[n  k ] 
k
n
 h[k ]
k  
h[n]  s[n]  s[n  1]
In continuous-time:
s(t )  u (t )  h(t )

  h( )u (t   )d

: the step response of a DT
LTI system is the running sum
of its impulse response
Unit Step Response
t
s (t )   h( )d

: the step response of a CT LTI system is the
running integral of its impulse response
The impulse response is the first derivative of the unit step response
ds (t )
h(t ) 
 s ' (t )
dt
Causal LTI Systems Described by
Differential and Difference Equations

CT systems for which the input and output are
related through a linear constant-coefficient
differential equation,


e.g., RC circuit
DT systems for which the input and output are
related through a linear constant-coefficient
difference equation,

e.g., bank account
Linear Constant-Coefficient
Differential Equations
Consider
dy (t )
 2 y (t )  x(t )
dt
Implicit representation of input-output relationship
Complete solution has two parts:
- particular solution, yp(t)
- homogeneous solution, yh(t)
y (t )  y p (t )  yh (t )
• yh(t) is the solution to x(t)=0, i.e., dy (t )  2 y (t )  0
• yp(t) depends on x(t)
dt
Linear Constant-Coefficient
Differential Equations
For x(t)=Ke3tu(t)  yp(t)=Ye3t
K
Substitute yp(t) and try  Y 
5
for t>0
K 3t
 yp(t) = e , t>0
5
yh(t)=Aest  Asest+2Aest=0  s=-2  yh(t)=Ae-2t
K 3t
2t
 y(t)= e  Ae
, t>0
5
Linear Constant-Coefficient
Differential Equations

K 3t
y(t)  e  Ae 2t , t  0
5
We need initial conditions.
 For causal LTI systems, we assume initial rest, y(0)=0
K
 substitute: A  
5


K 3t
 2t
y(t)

e

e
u(t)

5
Linear Constant-Coefficient
Differential Equations
In general, Nth order linear constant-coefficient differential
equation is
d k y(t) M
d k x(t)
ak
 bk

k
k
dt
dt
k 0
k 0
N
 has particular solution + homogeneous solution
d k y(t)
ak
0

k
dt
k 0
 initial (auxiliary) conditions are necessary.
N
 for causal LTI systems, initial rest:
dy( 0 )
d N 1 y( 0 )
y( 0 ) 
   
0
N 1
dt
dt
Linear Constant-Coefficient
Differential Equations
Assume an LTI system described by Nth order linear constantcoefficient differential equation as follows
d k y(t)
 x(t )

k
k 0 dt
N
 In this case, we find that h(t) (impulse response) satisfies
d k h(t)
0

k
k 0 dt
 with the initial (auxiliary) conditions (recall that for
causal LTI systems, initial rest):
N
dh( 0 )
d N 2 h( 0 )
d N 1h( 0 )
h( 0 ) 
   
 0,
 1 for t  0 and h(t )  0 for t  0
N 2
N 1
dt
dt
dt
Check Auxiliary text 2 (Yuksel) for proof.
Linear Constant-Coefficient
Differential Equations
Example: Consider the LTI system described by
y ''(t )  7 y '(t )  12 y(t )  et u (t )
Let us find the impulse response first. h(t) satisfies
h ''(t )  7h '(t )  12h(t )  0 with h(0)  0, h '(0)  1
The characteristic equation is:
s 2  7 s  12  0  Satisfied for s=-3,
and s=-4
c1e3t  c2e4t , for t  0
h(t )  
0
, t0

When the initial conditions are used, the coefficients are determined as
h(0)  0  c1  c2

 c1  1, c2  1
h '(0)  1  3c1  4c2 
h(t )  (e3t  e4t )u(t )
You can then obtain the output by using convolution
Linear Constant-Coefficient
Difference Equations
In DT:
N
a
k 0
M
k
y[n  k]  bk x[n  k]
k 0
N
1 M

Rearrange it, then: y[n] 
 bk x[n  k]  ak y[n - k] 
a0  k 0
k 1



Recursive operation as the output at time n is a
function of previous values of input and output.
We need initial conditions: y[-1], ..., y[-N]
Linear Constant-Coefficient
Difference Equations
As a special case, N=0, then
M 
bk 
y[n]     x[n  k]

k  0  a0 
(non-recursive (LTI system),
No need for initial conditions)
 Its impulse response:
 bn

,
0

n

M


h[n]   a0

0 ,

otherwise


 This is a finite duration impulse response. Such systems
are called finite impulse response (FIR) systems.
Linear Constant-Coefficient
Difference Equations
Example: Consider the difference equation
y[n]  x[n] 
1
y[n  1]
2
1
y[n]  y[n  1]  x[n]
2
 recursive; initial conditions are needed

Assume x[n]  Kδ[n] and initial rest, i.e., x[n]=0, y[n]=0 for n≤-1:
1
y[1]  K
2
1
1
y[1]  x[1]  y[0]  K
2
2
2
1
1
y[2]  x[2]  y[1]    K
2
2
y[0]  x[0] 



n
n
1
1
1
h[n]

  u[n]
y[n]  x[n]  y[n - 1]    K  for K=1,
2
2
2
For N≥1:
Infinite
Impulse
Response
(IIR)
Block Diagram Representations


Used to help in understanding and implementation of
systems
DT Case:

First-order difference equation
y[n]+ay[n-1]=bx[n] (addition, multiplication, delay)
x2[n]
x1[n]
+
x1[n]+ x2[n]
x[n]
y[n]=-ay[n-1]+bx[n]:
x[n]
b
a
ax[n]
x[n]
D
y[n]
+
D
-a
y[n-1]
x[n-1]
Block Diagram Representations
CT Case: First-order differential equation
dy(t)
1 dy(t) b
 ay(t)  bx(t)  y(t)  
 x(t)
dt
a dt
a
x(t)
b/a
y(t)
+
D
-1/a
Block Diagram Representations
In real implementations, differentiators are difficult to implement and
extremely sensitive to error & noise.
t
dy(t)
 bx(t) - ay(t)  y(t)   (bx( ) - ay( ))d
dt
-
(For initial rest, take integral from -∞ to t)
x(t)
t
∫
 x( ) d
-
x(t)
b
+
∫
-a
OP-AMP !
y(t)