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MAS.160 / MAS.510 / MAS.511 Signals, Systems and Information for Media Technology
Fall 2007
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M
y [ n ] = bk x [ n k ]
FIR
y [ n] = h [k ] x [n k ]
convolution
k= 0
M
k= 0
x [ n ] = Ae j� e j�ˆn
Complex exponential input
M
y [ n ] = h [ k ] Ae j� e jˆ ( nk)
k= 0
M
�
j� jˆn
jˆk
= h [ k ]e Ae e
k= 0
= H (�ˆ ) Ae e
j�
H (�ˆ )
j�ˆ n
frequency response
let
M
H (�ˆ ) = h[k]e j�ˆ k
k=0
Ex.
y [ n ] = 13 x [ n ] + 13 x [ n �1] + 13 x [ n � 2]
h [ n ] = 13 � [ n ] + 13 � [ n �1] + 13 � [ n � 2]
2
H (�ˆ ) = h[k]e j�ˆ k
k=0
= h[0]e j�ˆ 0 + h[1]e j�ˆ 1 + h[2]e j�ˆ 2
= 13 + 13 e j�ˆ + 13 e j�ˆ 2
= 13 (1+ e j�ˆ + e j 2�ˆ )
= 13 e j�ˆ (e j�ˆ +1+ e j�ˆ )
= 13 e j�ˆ (1+ 2 cos �ˆ )
FIR
1 � j�ˆ
ˆ
H (� ) = 3 e (1+ 2 cos �ˆ )
= 13 (1+ 2 cos �ˆ )(cos �ˆ � j sin �ˆ )
H (�ˆ
) =
(1+ 2 cos �ˆ )
1
1
3
Note:
H(
2�
3
) =0
H (�ˆ )
0.6
0.4
linear phase
� �ˆ
2 � 3 � ˆ < 0
=�
��ˆ + � � � ˆ < 2 � 3
�H (��ˆ ) = ��H (�ˆ )
linear phase
0.2
0
-4
�H (ˆ ) =
�ˆ
low pass filter
0.8
-3
-2
-1
0
1
2
3
4
-3
-2
-1
0
1
2
3
4
3
2
1
�H (�ˆ )0
-1
-2
-3
-4
�ˆ
principal value of phase fn
zero
Normalized vs. actual frequency
H (�ˆ ) =
Note:
1
3
(1+ 2 cos �ˆ )
H(
2�
3
1
H (�ˆ )
) =0
0.6
0.4
�ˆ =
f =?
linear phase
0.2
0
-4
2�
3
low pass filter
0.8
normalized frequency
3
actual frequency (Hz)
1
2
-3
f
fs
0
f �ˆ =�
�ˆ = 2 �f̂ = 2 �
�ˆ fs
f=
2�
-2
-3
-4
-1
� < �ˆ < �
-1
f̂ =
-2
-3
-2
0
1
�ˆ
�fs fs
=
=
2� 2
-1
0
2
zero
3
4
3
4
Nyquist
1
2
�ˆ
f
fs
�ˆ = 23� , fs = 8000Hz
f=
2�
3
8000 8000
=
Hz = 1667Hz
3
2�
Convolution / solving the difference equation
x [ n ] = 3 + 3cos(0.6�n )
y [ n ] = 13 x [ n ] + 13 x [ n �1] + 13 x [ n � 2]
input
FIR filter
sample domain
y [ n ] = 1+ cos(0.6�n ) + 1+ cos(0.6� ( n 1)) + 1+ cos(0.6� ( n 2))
= 3 + cos(0.6�n )
+ cos(0.6� ) cos(0.6�n ) + sin(0.6� ) sin(0.6�n )
+ cos(1.2� ) cos(0.6�n ) + sin(1.2� ) sin(0.6�n )
M
= 3 + 0.382cos(0.6�n 0.6� )
Multiplication in frequency domain
x [ n ] = 3 + 3cos(0.6�n )
closer �ˆ
to a zero,
the smaller the
output.
1
0.8
H (�ˆ )
0.6
0.4
0.2
0
-4
-3
-2
-1
0
1
2
3
4
-3
-2
-1
0
1
2
3
4
3
2
�H (�ˆ )
1
0
-1
-2
-3
-4
H (�ˆ ) = 13 (1+ 2 cos �ˆ )
�H (�ˆ ) = ��ˆ
�ˆ
H (0) = 1 H (0.6 � ) = 0.127
�H (0) = 0
�H (
0.6 � ) = 0.6 �
y[n ] = 3(1) + 3(0.127) cos(0.6�n 0.6� )
= 3 + 0.382cos(0.6�n 0.6� )
h1[n]=
[1/3,1/3,1/3]
x [ n]
h2[n]=
[1/3,-1/3,1/3]
h1[n]=[1/3,1/3,1/3]
h2[n]=[1/3,-1/3,1/3]
lowpass
highpass
1
1
0.9
0.9
0.8
0.8
0.7
H 1 (�ˆ )
0.7
H 2 (�ˆ )
0.6
0.5
0.4
0.6
0.5
0.4
0.3
0.3
0.2
0.2
0.1
0
0
y[ n ]
0.1
0.5
1
1.5
2
2.5
0
0
3
0.5
�ˆ
1
1.5
2
2.5
3
3.5
�ˆ
x [ n]
y[ n ]
h1[n]* h2[n]
bandstop
h1[n]* h2[n]=[1/9,0,1/9,0,1/9]
0.35
H (�ˆ ) = H 1 (�ˆ ) � H 2 (�ˆ )
0.3
0.25
H 1 (�ˆ ) H 2 (�ˆ )
0.2
0.15
0.1
0.05
0
0
0.5
1
1.5
�ˆ
2
2.5
3
3.5
sample domain convolution
= freq domain multiplication
(and visa versa)
Ideal
High pass
�c=2.6
Ideal
Low pass
�c=.8
x [ n]
lowpass
H 1 (�ˆ )
highpass
1
1
0.8
0.8
H 2 (�ˆ )
0.6
0.6
0.4
0.4
0.2
0.2
0
0
0.5
1
1.5
2
y[ n ]
2.5
0
3
0
0.5
�ˆ
1
1.5
2
2.5
3
�ˆ
x [ n]
y[ n ]
h1[n]* h2[n]
null
1
0.9
0.8
0.7
H 1 (�ˆ ) H 2 (�ˆ )
0.6
0.5
passbands don’t overlap
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
�ˆ
2
2.5
3
Ideal
High pass
�c=0.8
Ideal
Low pass
�c=.2.6
x [ n]
lowpass
H 1 (�ˆ )
highpass
1
1
0.8
0.8
H 2 (�ˆ )
0.6
0.6
0.4
0.4
0.2
0.2
0
0
0.5
1
1.5
2
y[ n ]
2.5
0
3
0
0.5
�ˆ
1
1.5
2
2.5
3
�ˆ
h1[n]* h2[n]
x [ n]
y[ n ]
Ideal Bandpass
�c=0.8 -> 2.6
bandpass
1
0.8
H 1 (�ˆ ) H 2 (�ˆ )
0.6
passbands overlap
0.4
0.2
0
0
0.5
1
�ˆ
1.5
2
2.5
3
Ideal
Low pass
�c=.8
x [ n]
y[ n ]
Ideal
High pass
�c=2.6
lowpass
H 1 (�ˆ )
highpass
1
1
0.8
0.8
H 2 (�ˆ )
0.6
0.6
0.4
0.4
0.2
0.2
0
0
0.5
1
1.5
2
2.5
0
3
0
0.5
�ˆ
1
1.5
2
2.5
3
�ˆ
h1[n]+ h2[n]
x [ n]
y[ n ]
Ideal Bandstop
�c=0.8 -> 2.6
bandstop
1
0.8
H 1 (�ˆ ) + H 2 (�ˆ )
0.6
passbands don’t overlap
0.4
0.2
0
0
0.5
1
�ˆ
1.5
2
2.5
3
Ideal
All - pass
�[n]
x [ n]
y[ n ]
Ideal
Low pass
�c=.8
-
all-pass
H 1 (�ˆ )
lowpass
1
1
0.8
0.8
H 2 (�ˆ )
0.6
0.6
0.4
0.4
0.2
0.2
0
0
0.5
1
1.5
2
2.5
0
3
0
0.5
�ˆ
1
1.5
�ˆ
�[n]- h2[n]
Highpass
�c=0.8
x [ n]
y[ n ]
highpass
1
0.8
1� H 2 (�ˆ )
0.6
0.4
0.2
0
0
0.5
1
�ˆ
1.5
2
2.5
3
2
2.5
3
h1[n]=[1/3,1/3,1/3]
h2[n]=�[n]-h1[n]=[2/3,-1/3,-1/3]
lowpass
highpass
1.4
1.4
1.2
1.2
H 2 (�ˆ )
1
H 1 (�ˆ )
0.8
= 1� H 1 (�ˆ )
0.6
0.4
1
0.8
0.6
0.4
0.2
0.2
0
0
0.5
1
1.5
2
2.5
3
0
0
0.5
1
1.5
�ˆ
2
2.5
3
�ˆ
h5 [n]
h3[n] =h1[n]* h2[n]
h4[n] =h1[n]+ [1/3 -1/3 1/3]
=[2/3 0 2/3]
=[2/9, 1/9, 0, -2/9, -1/9]
bandpass
bandstop
1.4
1.4
H 3 (�ˆ )
= H 1 (�ˆ ) H 2 (�ˆ )
1.2
1.2
H 4 (�ˆ )
1
0.8
= H4
0.6
0.4
(�ˆ ) + H (�ˆ )
1
0.8
0.6
5
0.4
0.2
0.2
0
0
0.5
1
1.5
2
2.5
3
0
�ˆ
0
0.5
1
1.5
�ˆ
2
2.5
3
FIR low-pass filter response vs. number of taps N
N=5
N=20
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
b =[0, -0.049, 0.549, 0.549, -0.049, 0]
0
0
0.5
1
1.5
2
2.5
3
0
0
0.5
1
1.5
N=50
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0.5
1
1.5
2.5
3
N=100
1
0
2
2
2.5
3
0
0
0.5
1
More coefficients (“taps”), can achieve more ideal response;
Longer to compute, more memory, greater phase shift
1.5
2
2.5
3
Note: These aren’t L-point averagers,
fir1(), Hammingwindow, Wn=0.8
Causal FIR Filter
x [ n]
Causal FIR filter
y[ n ]
In an FIR filter, the output y at each sample n is a weighted sum
of the present input, x[n], and past inputs, x[n-1], x[n-2],…, x[n-M].
y [ n ] = b0 x [ n ] + b1 x [ n �1] + K + bM x [ n � M ]
M
y [ n ] = bk x [ n k ]
k= 0
Causal IIR filter
Infinite Impulse Response
x [ n]
y[ n ]
Causal IIR filter
For an IIR filter, the output y at each sample n is a weighted sum of the present input, x[n], past inputs, x[n-1], x[n-2],…, x[n-M],
and past outputs, y[n-1], y[n-2],…, y[n-N]..
y[n] = b0 x [n] + b1 x[n �1] +K+ bM x [n � M ]
+ a1 y[n �1] + a2 y[n � 2] +K+ a M y[n � N ]
N
M
y[n] = al y[n l ] + bk x[n k ]
l=1
k=0
N�M
proper system
IIR Impulse Response
x[n] = 1� [n
]
y[n] = 12 y[n �1] + 2x [n]
To calculate output at step n, need previous outputs. At step n=0, assume initial rest conditions. x[n]=0, y[n]=0 for n<0
y[0] = 12 y[0 �1] + 2x [0]
= 12 � 0 + 2 �1
= 2
y[1] = 12 y[1�1] + 14 x[1]
= 12 y[0] + 14 x [1]
= 12 � 2 + 14 � 0
=1
for n>0
0
y[n] = 12 y[n �1]
+ 2x [n]
= 12 y[
n �1]
n = [ �1
x [ n] = [ 0
�
2 0 1 2 3 4 K]
0 1 0 0 0 0 K]
y[n] = 12 y[n �1] + 2x [n]
y[ n ] = [ 0
0 2 1
1
2
1
4
1
8
K2 � 12 n ] = h[n]
impulse response
2
1.8
1.6
1.4
y[n]
1.2
1
TextEnd
0.8
0.6
0.4
0.2
0
1
2
3
4
5
6
n
7
8
9
10
decays exponentially,
goes on forever
“infinite impulse response”
IIR impulse response
N
M
l=0
k=0
y[n] = ak x [n l ] + bk x[n k ]
IIR filter
�1
n=0
x [ n ] = �[n] = �
�0 otherwise
Delta function
M
y[n] x=� [ n ] = h[n] � � bk� [n � k ]
k=0
Can’t read filter coefficients
off of impulse response
impulse response
y [ n] =
�
� h[n]x [n � k ]
k=��
Convolution sum:
LTI systems: FIR/IIR
IIR convolution
x[n] = x[k]� [n � k ]
�
y[n] = � x[k]h[n � k ]
k=��
We can decompose x[n]
into scaled delayed impulses.
Here assume finite length x[n].
Convolution sum:
LTI: FIR / IIR
y[n] = x[0]h[n] + x[1]h[n �1]+ x[2]h[n � 2]+ K
Each impulse has a scaled impulse response. The final output is a sum of those scaled and delayed
impulse responses. Now the impulse responses are
Infinite in length.
IIR convolution
x[n] = 2� [n] � 3� [n �1] + 2� [n � 2]
h[ n] = [ 0
n=-2
0 2 1
1
2
1
4
1
8
-1
2
3
4
0
1
�
y[n]
=
�
x[k]h[
n � k
]
K2 � 12 n ]
…
Each impulse has a scaled impulse response. The final output is
a sum of those scaled and delayed
impulse responses.
k=��
y[n] = 2 � h[n] � 3� h[n �1]+ 2 � h[n � 2]
n>2
� 0
�
=
�
�
�
�
0
0
4
0
0 �6 �3
0
4
2
1
1
2
1
4
1
8
�3
2
�3
4
�3
8
0
4
2
1
1
2
�4
2
1
1
2
1
4
K2 � 12 n�1 �
�
1 n�2
K� 3� 2 �
K2 � 12 n�3 �
n�2
�
1
K2
[ 2 ] �
K[ ]
1
2
n �3
[
2
12
2
� 3 12 + 2]
IIR convolution
x[n]
5
0
TextEnd
h[n]
-5
-2
5
0
2*h[n]
-3*h[n-1]
2*h[n-2]
2
4
6
8
10
0
2
4
6
8
10
0
2
4
6
8
10
0
2
4
6
8
10
0
2
4
6
8
10
0
2
4
n
6
8
10
TextEnd
-5
-2
10
y[n]
0
0
TextEnd
-10
-2
10
0
-10
-2
10
TextEnd
0
-10
-2
10
TextEnd
0
TextEnd
-10
-2
[
y[n] = 0 0 4 �4 2 1 1
2
1
4
K2 � [
]
1
2
n�2
]
IIR convolution
x[n] = 2� [n] � 3� [n �1] + 2� [n � 3]
h[ n] = [ 0
0 2 1
1
2
1
4
1
8
y[n] = h[n]* x[n]
y[n] = 0 0 4 �4 2 1
[
1
2
K2 � 12 n
]
1
4
K2 � [
4
3
2
y[n]
1
0
TextEnd
-1
-2
-3
-4
-2
0
2
4
n
6
8
10
]
1
2
n�2
]
Frequency response
�
y[n]
=
�
h[ k ] x [n � k]
convolution
k=0
x [ n ] = Ae j� e j�ˆn
�
y[n]
=
�
h[ k ] Ae
j�
e
Complex exponential input
j�ˆ ( n�k )
k=0
(
)
�
=
�
h[ k ]e� j�ˆ k Ae j� e j�ˆ n
k =0
= H (�ˆ ) Ae e
j�
j�ˆ n
frequency response
let
�
H (�ˆ
) =
� h[k]e
j�ˆ k
k=0
This sum must converge for
H (�ˆ ) to exist.
H (�ˆ ) is the Discrete Time
Fourier Transform of the impulse
function h[n]
Frequency response
FIR
IIR
M
y [ n ] = bk x [ n k ]
k= 0
M
h[ k ] = bk� [n k ]
k=0
M
H (�ˆ ) = h[k]e
N
M
l=1
�
k=0
y[n] = al y[n l ] + bk x[n k
]
h[ k ] � � bk� [n � k ]
k=0
x
j�ˆ k
k=0
M
H (�ˆ ) = bk e j�ˆ k
Argh!
�
H (�ˆ ) = � h[k]e j�ˆ k
k=0
k=0
Easy to go from difference
equation to frequency response
because h[n] finite length and
h[n] = [b0, b1,…].
Tough to go from diff.eqn.
to freq. response because
h[n] infinite length,
h[n]=f(a ,bk) is complicated,
and H (�ˆ ) may be unbounded.
l
temporal space - n
complex frequency space- z
N
M
y[n] = al y[n l ] + bk x[n k ]
l=1
�
k=0
h[ k ] � � bk� [n � k ]
z-transform
M
H ( z) =
k=0
Argh!
@#! road block
�
H (�ˆ ) = � h[k]e j�ˆ k
bk z
M
�( z zzi )
k
k=0
N
1 ak z k
=
k=1
i=0
N
�
(z z pi )
i=0
Fourier
transform
k=0
frequency space - �
Hurray!
H (� ) = H (e j� ) = H ( z) z=e j�
The frequency response is
H(z) evaluated on unit circle
Benefits of z-plane and z-transforms:
1. Get around road block by using z-plane and z-transforms.
Compute system function from diff.eq. coefficients, then evaluate on the unit
circle to find the frequency response.
2. z-plane (pole/zeros) will tell us if system stable and frequency response exists.
3. By using z-transforms, solution to diff.eq goes from solving convolution in
n-space to solving algebraic equations in z-domain (easier).
And lots more…!
Frequency response
�
y[n]
=
�
h[ k ] x [n � k]
convolution
k=0
x [ n ] = Ae j� e j�ˆn
�
y[n]
=
�
h[ k ] Ae
j�
e
Complex exponential input
j�ˆ ( n�k )
k=0
(
)
�
=
�
h[ k ]e� j�ˆ k Ae j� e j�ˆ n
k =0
= H (�ˆ ) Ae e
j�
j�ˆ n
frequency response
let
�
H (�ˆ
) =
� h[k]e
j�ˆ k
k=0
This sum must converge for
H (�ˆ ) to exist.
H (�ˆ ) is the Discrete Time
Fourier Transform of the impulse
function h[n]
System function
�
y[n]
=
�
h[ k ] x [n � k ]
convolution
k=��
x [ n] = z n
y[n]
=
power sequence
of complex numbers
�
�
h[ k ] z
( n�k )
k=��
(
�
=
�
h[ k ] z
k =��
�k
)z
n
= H ( z) z n
= H ( z) e j�H ( z) z n
�
H (z) = � h[ k ] z � k
system function
k=��
let
�
H (z) = � h[ k ] z � k
k=��
scaled and shifted
power sequence
LTI
z
characteristic functions
of LTI systems
n
1
0.8
Re( z
n
)
0.6
0.4
z=e
0.2
0
z n +(z * )n
2
j 216�
-0.2
-0.4
sinusoid
-0.6
-0.8
-1
0
5
10
15
20
25
30
35
1
Re( z
n
) z n +(z * )n
2
0.5
z = 0.9e
0
j 216�
damped sinusoid
-0.5
0
5
10
15
20
25
30
35
1
0.9
0.8
0.7
0.6
z
n
z = 0.8
0.5
0.4
0.3
exponentials
0.2
0.1
0
0
5
10
15
20
25
30
35
z-plane and sample responses zn
Im(z)1
2
0
1
0
-2
0
5
1
-1
1
1
0
0
-1
0
-1
5
0
0
5
-1
0
0
5
1
0
-1
0
0
5
-1
1
0
0
5
Re(z)
0.5
-1
5
1
1
1
0
0
5
1
0
0
1
-1
0
-1
0
5
-1
-1
0
5
0
5
0
5
zn
z-plane
1
Im(z)
high freq
z =1
low freq
DC, �=0
0
-1
0
5
1
1
Re(z)
z = e ± j�ˆ
0
-1
z = �1
unit circle
sinusoids
0
5
1
Nyquist sampled
sinusoid
0
-1
0
5
zn
z-plane
1
± j
�4
z = 0.5e
0
-1
high freq
Im(z)
0
5
0
5
0
5
low freq
1
± j
�2
z = 0.5e
0
-1
0.5
Re(z)
z = 0.5e
|z|<1
Damped sinusoids
1
± j 34�
0
-1
1
z = �0.5
0
-1
0
5
Nyquist sampled
damped sinusoid
zn
z-plane
1
Im(z)
z =1
0
-1
10
z = 0.6
1
0
-1
10
Re(z)
z = 0.4
5
5
0
-1
0
5
0
5
+Re(z) axis
exponential
decay
1
z=0
0
-1
Im(z)
z = �0.5
1
-1
1
-Re(z) axis
0
Re(z)
0
5
�s 2
2 samples/cycle
1
z = �1
0
-1
0
5
Nyquist sampled
signals
zn
z-plane
Im(z)
1
z = ±1j
0
-1
0
5
0
5
0
5
1
1
Re(z)
z = ± j0.5
0
-1
Im(z) axis
every other
sample 0
1
z=0
0
-1
Im(z)
|z| > 1 unstable
2
1
z = ± j1.1
Re(z)
0
-2
0
5
|z| < 1 stable
System function
�
y[n]
=
�
h[ k ] x [n � k ]
convolution
k=��
x [ n] = z n
y[n]
=
power sequence
of complex numbers
�
�
h[ k ] z
( n�k )
k=��
(
�
=
�
h[ k ] z
k =��
�k
)z
n
= H ( z) z n
= H ( z) e j�H ( z) z n
�
H (z) = � h[ k ] z � k
system function
k=��
let
�
H (z) = � h[ k ] z � k
k=��
scaled and shifted
power sequence
LTI
z-transform
z
x[n] � X(z) =
�
� x[k]z
definition
�k
k=��
Ex.
impulse
x[n ] = � [ n ]
z
�
X(z) =
�
�� [ k ] z
�k
k=��
= z�0
Ex.
y[n] = h[n]� x[n]
= h[n]� �[n]
=1
Y (z) = H(z) �1
= H(z)
H(z) is the z-transform
of the impulse response
h[n]
y[n] = h[n]
H (z) = � h[ k ] z � k
c z
c z
�
k=��
LTI
Ex.
n0 sample delayed impulse
x[n] = � [ n � n 0 ]
n0th power of z-1
X(z) =
�
�
�k
�
k
�
n
z
[
]
�
0
k=��
=z
Ex. finite response
�n 0
= (z
)
�1 n 0
8
6
4
x[n]
2
0
TextEnd
-2
-4
-6
-8
-1
0
1
2
3
4
5
6
n
x[n] = 1� [ n ] � 2� [ n �1] + 5� [ n � 2] � 7� [ n � 3]
cz
X(z) = 1� 2z�1 + 5z�2 � 7z�3
7
8
9
10
Ex.
y [ n ] = 13 x [ n ] + 13 x [ n �1] + 13 x [ n � 2]
FIR
h [ n ] = 13 � [ n ] + 13 � [ n �1] + 13 � [ n � 2]
2
H ( z) = h[k]zk
k= 0
= h[0]z 0 + h[1]z1 + h[2]z2
= 13 + 13 z1 + 13
z2
=
1
3
(z )
1 0
+
1
3
( z ) + (z )
1 1
2
1
3
1 2
polynomial in z-1
2
h[n] = bk� [ n k ] � H ( z) = bk zk
k= 0
sequence
�
n-domain (sample space)
FIR only!
k= 0
polynomial
z-domain (complex freq space)
2
z
+ z +1
1
1 �1
1 �2
H ( z) = 3 + 3 z + 3 z =
3z
2
Ex.
y[n] = H ( z) z n
H ( z) = 0
num=0
y[n] = 0
(
)
z = 12 1± j 3 = e
H ( z) = �
denom=0
z2 + z + 1 = 0
y[n] = �
z 2 = 0
z = 0, 0
± j 2�
zeros
3
roots of numerator
poles
roots of denominator
5
4
3
|H(z)|
2
1
2
0
2
1
2
1
0
Im(z)
0
-1
-1
-2
-2
Re(z)
FIR filters only have zeros
on unit circle, and poles
are either at 0 or �.
#poles=#zeros
“extra” zero/poles are at z=�.
system response |H(z)|
pz plot
1
5
0.8
4
0.6
0.4
3
Imaginary part
|H(z)|
2
1
2
0.2
2
0
-0.2
2
TextEnd
-0.4
0
2
-0.6
1
2
-0.8
1
Im(z) 0
0
-1
-1
-2
-1
Re(z)
-1
-2
-0.5
0
Real part
0.5
1
1
0.9
frequency response
H (� ) = H (e j� ) = H ( z) z=e j�
0.8
0.7
|H(ej�)|
0.6
0.5
0.4
0.3
The frequency response is
H(z) evaluated on unit circle
0.2
0.1
0
-3
-2
-1
0
�
1
2
3
Ex.
N
M
l=1
k=0
N
M
l=1
k=0
y[n] = al y[n l ] + bk x[n k ]
IIR
h[n] = al h[n l ] + bk� [n k ]
N
M
H (z) = al H (z)z + bk z k
l
l=1
(
k=0
N
H (z) 1 al z
l=1
M
H ( z) =
l
bk z
)
M
= bk z k
k=0
k
k=0
N
1 ak z k
freqz(b,a)
k=1
sequence
�
n-domain (sample space)
polynomial
z-domain (complex freq space)
Equivalent ways to represent the system
1
2
N
M
y[n] = al y[n l ] + bk x[n k ]
l=1
�
k=0
b1
x[n]
b0
a1
unit delay
+ +
difference equation
x[n]=�[n]
c
c
inspection
M
3
bk z
z
h[n] = y[n] x[ n ]=� [ n ] � H ( z) =
k=0
N
z = e
6
=
k=1
4
j�
c
system function
polynomial
H (� ) = H (e j� ) = H ( z) z=e j�
frequency response
y[n]
M
k
1 ak z k
impulse response
sequence
block diagram
�( z zzi
)
i=0
N
�( z z pi )
i=0
pole-zero
locations
5
All poles must be inside
unit circle for H (�)
to converge and the system
to be stable.
(FIR filter always stable)
IIR filter
difference equation
y [ n ] = x [ n ] + y [ n �1] � 0.5y [ n � 2]
1
0.8
0.6
M
Imaginary part
0.4
0.2
2
0
-0.2
H ( z) =
2
TextEnd
bk z k
k=0
N
1 ak z k
k=1
-0.4
-0.6
-0.8
-1
-1
-0.5
0
Real part
0.5
1
1
z2
H ( z) =
=
1� z �1 + 0.5z �2 z 2 � z + 0.5
M
=
=
frequency response
H (� ) = H (e j� ) = H ( z) z=e j�
�( z � zzi )
i=0
N
�
(z � z pi )
i=0
z�z
(z � 0.5 � j0.5)(z � 0.5 + j0.5)
zeros: 0, 0
poles: 0.5±j0.5
zeros: 0, 0
poles: 0.5±j0.5
1
0.8
IIR filter
0.6
M
Imaginary part
0.4
0.2
2
0
-0.2
H ( z) =
2
TextEnd
=
-0.4
-0.6
-0.8
-1
-1
-0.5
0
Real part
0.5
1
1
z2
=
� 2
�1
�2
�1
�2
1� z + 0.5z
1� z + 0.5z z
z2
= 2
z � z + 0.5
frequency response
H (� ) = H (e j� ) = H ( z) z=e j�
1
�( z � zzi )
i=0
N
�
(z � z pi )
i=0
z�z
(z � 0.5 � j0.5)(z � 0.5 + j0.5)
z2
1
= 2
=
z � z + 0.5 1� z �1 + 0.5z �2
M
H ( z) =
bk z
k
k=0
N
1 ak z k
k=1
difference equation
y [ n ] = x [ n ] + y [ n �1] � 0.5y [ n � 2]
zeros: 0, 0
poles: 0.5±j0.5
1
0.8
IIR filter
0.6
z2
H ( z) =
( z � 0.5 � j0.5)( z � 0.5 + j0.5)
Imaginary part
0.4
0.2
2
0
-0.2
2
z2
1
= 2
=
z � z + 0.5 1� z�1 + 0.5z�2
TextEnd
-0.4
-0.6
H (�ˆ ) = H ( z) z= e j�ˆ
-0.8
-1
-1
-0.5
0
Real part
0.5
y [ n ] = x [ n ] + y [ n �1] � 0.5y [ n � 2]
1
Magnitude Response (dB)
10
1
5
0.8
0
0.6
-5
TextEnd
-10
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
Normalized frequency (Nyquist == 1)
0.6
0.8
1
100
Phase (degrees)
0.4
y[n]
0.2
50
0
0
-0.2
-50
-100
-1
TextEnd
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
Normalized frequency (Nyquist == 1)
0.6
0.8
1
-0.4
0
2
4
6
8
10
n
12
14
16
18
20
zeros: 0, 0
poles: 0.5±j0.5
1
0.8
IIR filter
0.6
z2
H ( z) =
( z � 0.5 � j0.5)( z � 0.5 + j0.5)
Imaginary part
0.4
0.2
2
0
-0.2
2
z2
1
=
= 2
z � z + 0.5 1� z�1 + 0.5z�2
TextEnd
-0.4
-0.6
H (�ˆ ) = H ( z) z= e j�ˆ
-0.8
-1
-1
-0.5
0
Real part
0.5
1
y [ n ] = x [ n ] + y [ n �1] � 0.5y [ n � 2]
1
To go from p-z map to
impulse response:
Look at locations of
poles. Closer poles are
to zeros, the weaker
their response.
0.8
0.6
0.4
y[n]
0.2
0
-0.2
-0.4
0
2
4
6
8
10
n
12
14
16
18
20
zeros: 0, 0
poles: 0.5±j0.5
1
0.8
IIR filter
0.6
z2
H ( z) =
( z � 0.5 � j0.5)( z � 0.5 + j0.5)
Imaginary part
0.4
0.2
2
0
-0.2
2
z2
1
= 2
=
z � z + 0.5 1� z�1 + 0.5z�2
TextEnd
-0.4
-0.6
H (�ˆ ) = H ( z) z= e j�ˆ
-0.8
-1
-1
-0.5
0
Real part
0.5
y [ n ] = x [ n ] + y [ n �1] � 0.5y [ n � 2]
1
Magnitude Response (dB)
10
5
0
-5
TextEnd
-10
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
Normalized frequency (Nyquist == 1)
0.6
0.8
1
-0.4
-0.2
0
0.2
0.4
Normalized frequency (Nyquist == 1)
0.6
0.8
1
Phase (degrees)
100
50
0
-50
-100
-1
TextEnd
-0.8
-0.6
From zp map to freq response:
Go around unit circle from DC
to Nyquist. Response goes down
near zeros, goes up near poles.
Closer poles are to zeros, the weaker
their effect.
z-transform
x[n] =
�
� x[k]�[n � k ] � X(z) =
k=��
sequence
�
n-domain (sample space)
�
�k
x[k]z
�
k=��
polynomial
z-domain (complex freq space)
M
h[n ] � H ( z) =
impulse response
sequence
b z
k
k
k= 0
N
1 ak zk
system function
k=1
polynomial
H(z) is the z-transform of the impulse response h[n].
LTI
z-transform
x[n] =
�
� x[k]�[n � k ] � X(z) =
k=��
sequence
�
n-domain (sample space)
�
�k
x[k]z
�
k=��
polynomial
z-domain (complex freq space)
Why’re we interested?
M
y [ n] = h [k ] x [n k ]
convolution sum
Y (z) = H(z)X(z)
polynomial multiplication
k= 0
z-transform
x[n] � X(z) =
�
�k
x[k]z
�
k=��
impulse
x[n] = � [ n ]
�
X(z) =
�
�� [ k ] z
�k
k=��
= z�0
=1
Ex.
y[n] = h[n]� x[n]
= h[n]� �[n]
c z
Y (z) = H(z) �1
= H(z)
c z
y[n] = h[n]
H(z) is the z-transform
of the impulse response
h[n]
LTI
n0th power of z-1
n0 sample delayed impulse
x[n] = � [ n � n 0 ]
X(z) =
�
�
�k
�
k
�
n
z
[
]
�
0
k=��
=z
Ex. finite response
�n 0
= (z
)
�1 n 0
8
6
4
x[n]
2
0
TextEnd
-2
-4
-6
-8
-1
0
1
2
3
4
5
6
n
x[n] = 1� [ n ] � 2� [ n �1] + 5� [ n � 2] � 7� [ n � 3]
cz
X(z) = 1� 2z�1 + 5z�2 � 7z�3
7
8
9
10
Infinite signals
�
x[n]u[n] � X(z) = � x[k]z�k
right-sided signals
k= 0
x[n] = a u[n]
n
�
�
X(z) = � a k z�k
k= 0
�
= � ( az
)
�1 k
k= 0
= 1+ az + ( az
�1
) + (az ) K
�1 2
geometric series
�1 3
Infinite signals
x[n] = a n u[n]
�
�
X(z) = � a k z�k
k= 0
= 1+ az + ( az
�1
geometric series
X ( z) = 1+ az + ( az
�1
- az X ( z) = � az
�1
�1
) + (az ) K
�1 2
�1 3
) + (az ) K
� ( az ) � ( az ) � ( az ) K
�1 2
�1 3
�1 2
�1 3
�1 4
�1
1�
az
(
) X (z) = 1
1
X ( z) =
1� az�1
az�1 < 1
or
z > a
region of convergence
Infinite series:
x[n] = a n u[n]
IIR filter
�
X(z) = � a k z�k
�
k= 0
x[n]=0 n<0
right sided
1
X ( z) =
1� az�1
z >
a
region of convergence
Finite series:
x[n] = a n ( u[n � M] � u[n � N]) � X(z) =
az )
(
X ( z) =
�1 M
� ( az
1� az�1
lim
X(z) = N � M
z�a
N 1
FIR filter
a z
k k
k= M
)
�1 N
all z
region of convergence
Ex.
x[n] = 1u[ n ]
X(z) =
X(z) = ?
�
�k
x[k]z
�
We know:
k=��
�
= �z
�
� (az )
�k
k= 0
k= 0
�
= � (z
)
1
=
1� z�1
1
=
1� az�1
z>a
�1 k
k= 0
�1 k
let
a =1
z >1
roc
Ex. x[n] = cos(�ˆ n)u[ n ]
�
� x[k]z
X(z) =
�k
k=��
�
=�
=
=
(
�
= � cos(ˆ k ) z�k
k= 0
e jˆk + e � jˆk
2
k= 0
�
1
2
k= 0
)z
jˆk �k
e
� z +
�
1
2
� (e
)
jˆ �1 k
z
k= 0
let a = e
X(z) = ?
j�ˆ
We know:
�
�k
� (az )
k= 0
�
1
2
+
�1 k
� jˆk �k
e
� z
1
=
1� az�1
z>a
k= 0
�
1
2
roc
� (e
)
� jˆ �1 k
z
k= 0
let a = e
1
1
1
1
=2
+2
j�ˆ �1
1� e z
1� e� j�ˆ z�1
� j�ˆ
intersection of roc’s
z > e j�ˆ � e� j�ˆ
z >1
Ex. x[n] = cos(�ˆ n)u[ n ]
X(z) =
�
�
k=��
k= 0
X(z) = ?
Cont.
�k
�k
ˆ
=
cos
k
z
x[k]z
� ( )
�
1
1
1
=
+2
j�ˆ �1
1� e z
1� e� j�ˆ z
�1
z
z
1
1
=2
+2
j�ˆ
� j�ˆ
z�e
z�e
1
2
� j�ˆ
j�ˆ
2z
�
e
�
e
= 12 z 2
z � (e� j�ˆ + e j�ˆ ) z + 1
z � cos(�ˆ )
=z 2
z � 2cos(�ˆ )z + 1
z >1
Table of z-transforms
signal
�[n � k]
u[n]
n 2 u[n]
� cos(�n ) u[n]
n
� n sin(�n ) u[n]
z-transform
ROC
z�k
all z
z
z �1
z >1
z( z + 1)
( z �1)
3
z( z � � cos � )
z 2 � (2� cos� ) z + � 2
z� sin �
z 2 � (2� cos� ) z + � 2
z >1
z>�
z>�
z-transform properties
a1 x1 [ n ] + a2 x 2 [ n ] � a1 X1 ( z) + a2 X 2 ( z)
linearity/superposition
x [ n � m] � z�m X ( z)
sample shift
h [ n ] � x [ n ] � H ( z) X
( z)
sample domain convolution
z domain multiplication
dX(z)
nx [ n ] � �z
dz
a n x [ n ] � X ( az )
multiply by a ramp
z domain differentiation
multiply by an exponential
z domain scaling
Ex.
X ( z) = ?
x[n] = na n u[n]
We already know
1
a u[n] �
1� az�1
n
dY (z)
ny [ n ] � �z
dz
and
d � 1 � d � z �
= �
�
�
�1 �
dz �1� az � dz � z � a �
=
na n u[ n ] � �z
��
�a
( z � a)
2
=
az�2
(1� az )
�1 2
az
�1
(1� az )
�1 2
�az�2
(1� az )
�1 2
Next:
Solve difference equation using z-transforms.
Convolution problem in the temporal domain
becomes an algebra problem in the z-domain.
Need to know how to convert from z-domain
back to temporal domain (inverse z-transform).
y [ n ] = 13 x [ n ] + 13 x [ n �1] + 13 x [ n � 2]
h [ n ] = 13 � [ n ] + 13 � [ n �1] + 13 � [ n � 2]
x[n] = cos(�ˆ n )u[ n ]
y [ n ] = h [ n ] � x [ n ] � Y (z) = H ( z) X ( z) 2
+ z +1
z
1
1 �1
1 �2
H ( z) = 3 + 3 z + 3 z =
2
3z
z � cos(�ˆ )
X(z) = z 2
z � 2cos(�ˆ )z + 1
z 2
+ z +
1
�
z � cos(�ˆ )
�
Y (z) =
� z
2
�
2
3z
�
z
� 2cos(�ˆ )z +
1
�
2
�
z � cos(�ˆ ) �
�
z
+ z +1 �
�1
y[n] = Z � z
2
2
ˆ
�
3z
�
z
� 2 cos(� )z
+
1
��