Systems: Definition
x[n]
y[n]
S
A system is a transformation from an input signal
.
x[n] into an output signal y[n]
Example: a filter
SIGNAL
s[n]

NOISE
x[n]
v[n]
y[n]  s[n]
Filter
Systems and Properties: Linearity
Linearity:
x1[n]
S
y1[n]
x2 [ n ]
S
y 2 [ n]

x[n]  a1 x1[n]  a2 x2 [n]
S
y[n]  a1 y1[n]  a2 y2[n]
Systems and Properties: Time Invariance
if
x[n]
S
y[n]
time
time
y[n  D]
x[n  D]
then
D
time
S
D
time
Systems and Properties: Stability
Bounded Output
Bounded Input
x[n]
y[n]
S
Systems and Properties: Causality
x[n]
y[n]
S
the effect comes after the cause.
Examples:
y[n]  3x[n  1]  2 x[n  2]  4 x[n  3]
y[n]  3x[n  1]  2 x[n  2]  4 x[n  3]
Causal
Non Causal
Finite Impulse Response (FIR) Filters
x[n]
Filter
y[n]
General response of a Linear Filter is Convolution:
N
y[n]  h[n] * x[n]   h[]x[n  ]
 0
Written more explicitly:
y[n]  h[0]x[n]  h[1]x[n  1]  ... h[ N ]x[n  N ]
Filter Coefficients
Example: Simple Averaging
x[n]
Filter
y[n]
1
y[n]  x[n]  x[n  1]  ...  x[n  9]
10
Each sample of the output is the average of the last ten
samples of the input.
It reduces the effect of noise by averaging.
FIR Filter Response to an Exponential
Let the input be a complex exponential x[n]  e j0n
Then the output is
N
y[n]   h[]e j0 ( n )
 0
 N
 j0 )  j0 n
   h[]e
e
  0




x[n]  e
j0n
Filter
y[n]  H 0 e j0n
Example
x[n]  e
j0n
Filter
y[n]  H 0 e
j0 n
Consider the filter
1
y[n]  x[n]  x[n  1]  ...  x[n  9]
10
j 0.1n
with input
Then
x[n]  e
 1 9  j 0.1  1 1  e j 0.1 10
 j1.4137
H 0.1     e

0
.
6392
e

 j 0.1
 10 0
 10 1  e
and the output


y[n]  0.6392e j1.4137 e j 0.1 n
Frequency Response of an FIR Filter
x[n]  e
j0n
Filter
N
H ( )   h[n]e  jn
y[n]  H 0 e j0n
    
n 0
is the Frequency Response of the Filter
Significance of the Frequency Response
If the input signal is a sum of complex exponentials…
x[n]   X k e
jk n
Filter
y[n]  Yk e
k
jk n
k
… the output is a sum is a sum of complex exponential.
Each coefficient is multiplied by the corresponding
frequency response:
Xk
Yk  H (k ) X k
Example
Consider the Filter
x[n]
defined as
y[n]
Filter
1
y[n]  x[n]  x[n  1]  ...  x[n  4]
5
Let the input be:
x[n]  3 cos(0.1 n  0.2)  2 cos(0.3 n  0.7)
Expand in terms of complex exponentials:


e
 1.0e


e
x[n]  1.5e j 0.2 e j 0.1 n  1.5e  j 0.2 e  j 0.1 n 
 j 0.7
j 0.3 n

 1.0e j 0.7
 j 0.3 n
Example (continued)
The frequency response of the filter is (use geometric
sum)
 j 5


1
1
1

e
 j
 j 4

H ( )  1  e  ...  e
 
 j 
5
5  1 e

Then




e
 H 0.2 1.0e


e
y[n]  H 0.1  1.5e j 0.2 e j 0.1 n  H  0.1  1.5e  j 0.2 e  j 0.1 n 
 j 0.7
j 0.3 n

 H  0.2  1.0e j 0.7
 j 0.3 n
with H (0.1 )  0.904e  j 0.6283 , H (0.1 )  0.904e j 0.6283
H (0.2 )  0.647e  j1.2566 , H (0.1 )  0.647e j1.2566
Just do the algebra to obtain:
y[n]  2.712cos(0.1 n  0.428)  1.294cos(0.3 n  1.956)
The Discrete Time Fourier Transform (DTFT)
Given a signal of infinite duration
x[n] with    n  
define the DTFT and the Inverse DTFT
X ( )  DTFTx[n] 

 j n
x
[
n
]
e

n  
1
x[n]  IDTFTX ( ) 
2
Periodic with period
2

jn
X
(

)
e
d


X (  2 )  X ( )
General Frequency Spectrum for a Discrete Time
Signal
Since X ( ) is periodic we consider only the
frequencies in the interval      
| X () |


FS
2
0
0
  (rad )
F
F (Hz)
S
2
*
X
(


)

X
()
If the signal x[n] is real, then
Example: DTFT of a rectangular pulse …
Consider a rectangular pulse of length N
x[n]
1
0
Then
where
N 1
N 1
X ( )   e  j n
n 0
WN ( )  e
n
1  e  j N

 WN ( )
 j
1 e
 j ( N 1) / 2
sin  N / 2
sin  / 2
Example of DTFT (continued)
x[n]
1
n
N 1
0
DTFT
WN ()
12
N
10
8
6
4
2
0
-3

-2
-1
2

N
0
1
2
N
2
3


Why this is Important
x[n]
y[n]
Filter
Recall from the DTFT
1
x[ n] 
2

j n
X
(

)
e
d


Then the output



1
jn
y[n] 
X
(

)
H
(

)
e
d

2 
Which Implies
Y ()  DTFTy[n]  H () X ()

Summary Linear FIR Filter and Freq. Resp.
x[n]
y[n]
Filter
N 1
Filter Definition:
y[n]  h[n] * x[n]   h[]x[n  ]
 0
N 1
 jn
H
(

)

h
[
n
]
e
,     
Frequency Response:

n 0
DTFT of output
Y ( )  H ( ) X ( )
Frequency Response of the Filter
x[n]
Filter
y[n]
Frequency Response:
N 1
H ( )   h[n]e  jn ,
    
n 0
We can plot it as magnitude and phase. Usually the
magnitude is in dB’s and the phase in radians.
Example of Frequency Response
Again consider FIR Filter
1
y[n]  x[n]  x[n  1]  ...  x[n  9]
10
The impulse response can be represented as a vector of
length 10
h  0.1, 0.1 ... 0.1
Then use “freqz” in matlab
freqz(h,1)
to obtain the plot of magnitude and phase.
Example of Frequency Response (continued)
Magnitude (dB)
0
-20
-40
-60
-80
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
0.9
1


Phase (degrees)
100
0
-100
-200

