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MAS.160 / MAS.510 / MAS.511 Signals, Systems and Information for Media Technology
Fall 2007
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Causal FIR filter
x [ n]
Causal FIR filter
y[ n ]
Q:What is the definition of an FIR filter?
Causal FIR filter
x [ n]
Causal FIR filter
y[ n ]
Q:What is the definition of an FIR filter?
A: 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].
Causal FIR filter
x [ n]
Causal FIR filter
y[ n ]
Q:What is the formula for an FIR filter?
Causal FIR filter
x [ n]
Causal FIR filter
y[ n ]
Q:What is the formula for an FIR filter?
y [ n ] = b0 x [ n ] + b1 x [ n 1] + K + bM x [ n M ]
M
y [ n ] = � bk x [ n k ]
k= 0
�
n x[n
]
x n =� n
0
[ ] [ ] Causal FIR filter
2
b0 = 1,b1 = 3,b2 = 1
1
0
1
0
y[n]
=
?
1
0
h[n]
=
?
2
0
y[1
] =
?
3
0
1
0.9
0.8
0.7
x[n]
0.6
0.5
TextEnd
0.4
0.3
0.2
0.1
0
-2
-1
0
1
2
n
3
4
5
y[n] =
?
�y[n]�
� n x[n]
�
�
0 �
x n = � n
�
0
[ ] [ ] Causal FIR filter h[n]
=
y[n]
x[
n ]=� [ n ]
2
� 0 �
b
=
1,b
=
3,b
=
1
1
0
1
2
0 �
�
b 1
�
�
0
1
�b 3 �
1
0 �
�
y[n] = 1x[n] + 3x[n 1] +1x[n 2]
1
b
�
�
2
0
h[n] = y[n] = 1� [n] + 3� [n 1] +1� [
n 2]
� 0 ��
3
0 h[n] = bn
0
1
2
h[1] = 1� [1] + 3� [11] + 1� [1 2]
h[1] = 1� [1] + 3� [0] +1� [1]
0.9
0.8
h[1] = 1(0) + 3(1) +1(0)
0.7
x[n]
0.6
0.5
TextEnd
h[1] = 3
0.4
0.3
0.2
3
h[1]=b1
2.5
2
y[n]
1
1.5
TextEnd
h[0]=b0
h[2]=b2
1
0.5
0.1
0
-2
-1
0
1
2
n
3
4
5
0
-2
-1
0
1
2
n
3
4
5
x [ n]
Causal FIR filter
b0 ,b1 ,b2
h[n]
M
y [ n ] = � bk x [ n k ]
y[ n ]
weighted sum of
delayed inputs
k= 0
h[n] = bn
M
y[n] = � h[k]x [n k ]
k=0
y[n] = h[n]* x[n]
Convolution of
impulse response
and input
�
n
0
1
2
3
4
5
x [n
] x[n]
=
sin(
2� � 0.33n)u[n]
0
0.88
0.84
0.06
0.90
0.81
1
0.8
0.6
0.4
x[n]
0.2
0
TextEnd
-0.2
-0.4
-0.6
-0.8
-1
0
0.5
1
1.5
2
2.5
n
3
3.5
4
4.5
5
Causal FIR filter
b0 = 1,b1 = 3,b2 = 1
h[n] = [1, 3, 1]
y[ 3
] =
?
y[n]
=
?
�
n
0
1
2
3
4
5
x [n
] x[n] =
sin(
2� � 0.33n)u[n]
Causal FIR filter
0
b0 = 1,b1 = 3,b2 = 1
0.88
h[n] = [1, 3, 1]
0.84
M
M
bk
x [
n k ]
=
�
h[k]x[
n k
]
0.06
y[n]
=
k
�
=0
k =
0
0.90
y[n]
=
1x[n] + 3x[
n 1
]
+
1x[n 2]
0.81 y[ 3
] =
1x[ 3
] + 3x[ 3 1]
+
1x[
3 2
]
y[ 3
] =
1x[ 3] + 3x
[2] +
1x[1]
y[n
] 0
0.88
1.78
1.72
0.13
1.84
�
n
0
1
2
3
4
5
2
y[ 3
] =
1(0.06
)
+ 3(0.84
)
+
1(
0.88
)
1.5
1
0.8
0.6
0.5
y[n]
y[ 3] =
1.72
1
0
TextEnd
0.4
-0.5
x[n]
0.2
0
TextEnd
-1
-0.2
-0.4
-1.5
-0.6
-0.8
-1
-2
0
0.5
1
1.5
2
2.5
n
3
3.5
4
4.5
5
0
0.5
1
1.5
2
2.5
n
3
3.5
4
4.5
5
Graphical Convolution
y[n] = h[n]* x[n] = x[n]* h[n]
M
�
k=0
k=��
sum
y[n] = � h[k]x [n k ] = � x[k]h[n � k ]
multiply
flip
shift
1
1
0.5
x[n]
0.5
x[n]
0
TextEnd
0
TextEnd
-0.5
-0.5
-1
-2
-1
0
1
2
3
4
5
-1
-2
n
-1
0
1
h[1]=b1
2.5
4
5
2
3
4
5
b1=3
TextEnd
1
h[0]=b0
h[-n]
2
h[2]= b2
0.5
0
-2
-1
0
1
2
3
4
5
1
b2=1
TextEnd
0
-2
n
b0=1
-1
0
1
n
1
y[n] = x [n 2]h[n (n 2)]+ x[n 1]h[n (n 1)]+ x[n]h[n n]
0.5
y[n]
h[-n]
3
3
flip
2
1.5
2
n
3
0
TextEnd
y[n] = h[0]x [n] + h[1]x[n 1] + h[2]x [n 2
]
-0.5
-1
-2
-1
0
1
2
n
3
4
5
Graphical Convolution
sum
0
y[n] = � x[k]h[n k ]
k= M
multiply
flip
shift
n=0
1
0.5
x[n]
0
0
0
TextEnd
0
-0.5
-1
-2
-1
0
1
2
3
4
5
2
3
4
5
2
3
4
5
n
3
h[-n]
flip/
shift by n
b1=3
2
1
b0=1
TextEnd
multiply
b2=1
0
-2
-1
0
sum
1
n
1
y[n]
0.5
0
0
TextEnd
-0.5
-1
-2
-1
0
1
n
y[0] = x [2]h[2]+ x [1]h[1]+ x[0]h[0] = (0)1+ (0) 3+ (0)1 = 0
Graphical Convolution
sum
0
y[n] = � x[k]h[n k ]
k= M
flip
multiply
shift
n=1
1
0.88
x[n]
0.5
0
0
TextEnd
0
-0.5
-1
-2
-1
0
1
2
3
4
5
2
3
4
5
2
3
4
5
n
3
h[-n]
2
b1=3
TextEnd
1
b0=1
multiply
b2=1
0
-2
-1
0
1
sum
n
2
0.88
1
y[n]
flip/
shift by n
0
TextEnd
-1
-2
-2
-1
0
1
n
y[1] = x [1]h[2]+ x [0]h[1]+ x[1]h[0] = (0)1+ (0) 3+ (0.88)1 = 0.88
Graphical Convolution
sum
0
y[n] = � x[k]h[n k ]
k= M
flip
multiply
shift
n=2
1
0.88
x[n]
0.5
0
0
TextEnd
-0.5
-0.84
-1
-2
-1
0
1
2
3
4
5
n
3
h[-n]
2
b1=3
TextEnd
1
b0=1
multiply
b2=1
0
-2
-1
0
1
2
n
2
sum
3
4
5
3
4
5
1.78
1
y[n]
flip/
shift by n
0
TextEnd
-1
-2
-2
-1
y[ 2 ] = x [ 0 (
0
1
2
n
)]h[2]+ x[1]h[1]+ x[2]h[0]
= (0)1+ (0.88) 3+ (0.84 )1 = 1.78
Graphical Convolution
sum
0
y[n] = � x[k]h[n k ]
k= M
flip
multiply
shift
n=3
1
0.88
x[n]
0.5
-0.06
0
TextEnd
-0.5
-0.84
-1
-2
-1
0
1
2
3
4
5
n
3
h[-n]
2
b0=1
b2=1
TextEnd
1
0
-2
-1
0
1
2
multiply
3
4
5
4
5
n
2
sum
1
y[n]
flip/
shift by n
b1=3
0
TextEnd
-1
-2
-2
-1.71
-1
0
1
2
n
3
y[ 3] = x [1]h[2]+ x[2]h[1]+ x[3]h[0] = (0.88)1+ (0.84 ) 3+ (0.06)1 = 1.72
Graphical Convolution
0
y[n] = � x[k]h[n k ]
k= M
y[n] = x[k]* h[k]
1
x[n]
0.5
0
TextEnd
-0.5
-1
-2
-1
0
1
2
3
4
5
2
3
4
5
2
3
4
5
n
3
b1
h[n]
2
b0
b2
TextEnd
1
0
-2
-1
0
1
n
2
y[n]
1
0
TextEnd
-1
-2
-2
-1
0
1
n
Graphical convolution by decomposition
1. Remember impulse response
x[n] = �[n]
causal FIR filter
h[n] = b0� [n] + b1� [n 1] + b2� [n 2]
b0 ,b1 ,b2
h[n]
1
3
h[1]=b1
0.9
2.5
0.8
0.7
2
0.5
TextEnd
y[n]
x[n]
0.6
1.5
TextEnd
h[0]=b0
0.4
h[2]=b2
1
0.3
0.2
0.5
0.1
0
-2
-1
0
1
2
n
3
4
5
0
-2
-1
0
1
2
n
3
4
5
Graphical convolution by decomposition
2. Decompose input into sum of scaled delayed impulses
Input
Input as impulses
x[n] = 0.88� [n 1] 0.84� [n 2] 0.06� [n 3]
x[n] = sin(2 � � 0.33n)u[n]
+ 0.90� [n 4 ] 0.81� [n 5]
x[1]
2
0
1
TextEnd
-2
-2
2
0.8
x[2]
0.6
0
-1
0
1
2
3
4
5
2
3
4
5
2
3
4
5
2
3
4
5
2
3
4
5
n
TextEnd
0.4
-2
-2
2
x[3]
0
TextEnd
-0.2
x[4]
-0.6
-0.8
-1
-2
0
-1
0
1
2
n
3
4
0
0
0
1
n
-1
0
1
n
TextEnd
-2
-2
2
5
-1
TextEnd
-2
-2
2
-0.4
x[5]
x[n]
0.2
-1
0
1
n
TextEnd
-2
-2
-1
0
1
n
Graphical convolution by decomposition
3. find impulse responses to each impulse
Input as impulses
Impulse responses
h[n] = b0� [n] + b1� [n 1] + b2� [n 2]
x[3]
x[4]
x[5]
2
x1[n 2] = 0.84�[n 2]
-1
x[1]h[n-1]
3
4
5
0
1
2
4
5
x1[n 3] = 0.06�[n 3]
n
-1
0
1
2
3
4
TextEnd
-1
0
1
2
3
4
TextEnd
-2
-2
-1
0
1
2
n
FIR
3
4
5
2
0
-2
2
0
-2
-2
5
n
0
-2
-2
5
n
2
-2
x[2]h[n-2]
3
TextEnd
-2
-2
2
0
1
n
-2
-2
2
0
0
TextEnd
-2
-2
2
0
-1
x[3]h[n-3]
x[2]
-2
-2
2
0
x1[n 1] = 0.88�[n 1]
TextEnd
x[4]h[n-4]
0
2
0
-2
-2
x[5]h[n-5]
x[1]
2
2
0
-2
-2
y1[n]
= 0.88(h[n 1])
TextEnd
-1
0
1
2
y2 [n] = 0.84 (h[n 2])
3
4
5
6
7
3
4
5
6
7
3
4
5
6
7
3
4
5
6
7
3
4
5
6
7
n
TextEnd
-1
0
1
2
y3 [n] = 0.06(h[n 3])
n
TextEnd
-1
0
1
2
y4 [n] = 0.06(h[n 4])
n
TextEnd
-1
0
1
2
y5 [n] = 0.81(h[n 5])
TextEnd
-1
0
1
n
2
n
Graphical convolution by decomposition
3. sum impulse responses to get total response
Input as impulses
Impulse responses
3
4
5
x[2]h[n-2]
n
-1
0
1
2
3
4
5
n
-1
0
1
2
3
4
-1
0
1
2
3
4
TextEnd
-2
-2
-1
0
1
FIR
2
3
4
TextEnd
-1
1
2
3
4
5
6
7
3
4
5
6
7
3
4
5
6
7
3
4
5
6
7
3
4
5
6
7
n
0
-2
TextEnd
-1
0
1
2
n
2
0
-2
TextEnd
-1
0
1
2
n
2
0
-2
TextEnd
-2
5
0
2
-2
5
n
0
-2
-2
5
n
2
-2
x[3]h[n-3]
x[3]
x[4]
x[5]
2
TextEnd
-2
-2
2
0
1
TextEnd
-2
-2
2
0
0
TextEnd
-2
-2
2
0
-1
x[4]h[n-4]
x[2]
-2
-2
2
0
x[1]h[n-1]
TextEnd
x[5]h[n-5]
0
-1
0
1
2
n
2
0
-2
-2
TextEnd
-1
0
1
2
n
n
sum
total response
2
y[n]
x[1]
2
0
TextEnd
-2
-2
-1
0
1
2
3
n
4
5
6
7
Graphical convolution by decomposition
x[n]
1
Input
0
TextEnd
-1
-2
-1
0
1
2
3
4
5
n
Impulse response
h[n]
2
0
TextEnd
-2
-2
-1
0
1
2
3
4
5
6
7
3
4
5
6
7
3
4
5
6
7
3
4
5
6
7
3
4
5
6
7
3
4
5
6
7
3
4
5
6
7
Impulse responses
x[1]h[n-1]
n
2
0
-2
TextEnd
x[2]h[n-2]
-2
-1
x[3]h[n-3]
2
0
-2
TextEnd
-1
0
1
2
n
2
0
-2
TextEnd
-2
x[4]h[n-4]
1
n
-2
-1
0
1
2
n
2
0
-2
TextEnd
-2
x[5]h[n-5]
0
2
-1
0
1
2
n
2
0
-2
-2
TextEnd
-1
0
1
2
n
total response
y[n]
2
0
TextEnd
-2
-2
-1
0
1
2
n
Synthetic polynomial multiplication
n
-2
-1
0
1
2
3
4
5
x[n] 0
0
0
0.88 -0.84 -0.06 0.90 -0.81
h[n]
1
3
1
===============================================
0
0
0
0.88 2.64 0.88
-0.84 -2.52 -0.84
-0.06 -0.18 -0.06
0.9
2.7
-0.81
===============================================
y[n] 0
0
0
0.88 1.80 -1.7 -0.12 1.83
try h1[n]* h2[n]
h1[n]=[1/3,1/3,1/3]
h2[n]=[1/3,-1/3,1/3]
Impulse response
M
y [ n ] = � bk x [ n k ]
FIR filter
k= 0
�1
n =
0
x [ n ] =
�[n]
=
�
Delta function
�
0 otherwise
M
y [ n ] x=
� [n ]
=
h[n]
=
�
bk� [ n k
]
k= 0
impulse response
y [ n] =
�
� h[n]x [n � k ]
k=��
convolution sum
LTI: FIR, IIR
Frequency response
M
y [ n ] =
�
h [ k ] x [ n k ]
convolution
k= 0
x [ n ] = Ae j� e j�ˆn
Complex exponential input
�ˆ = �Ts
M
y [ n ] =
�
h [ k ] Ae j� e j�ˆ ( nk)
k= 0
M
�
� j� j�ˆn
j�ˆk
=
�
h [ k ]e Ae e
� k= 0
= H (�ˆ ) Ae j� e j�ˆn
H (�ˆ )
frequency response
let
M
H (�ˆ
) =
�
h[k]e
j�ˆk
k= 0
M
y [ n] = � h [k ] x [n k ]
convolution
k= 0
x [ n ] = Ae j� e j�ˆn
y [ n ] = H (�ˆ ) Ae e
j�
j�ˆn
complex exponential input
M
H (�ˆ ) = � h[k ]e j�ˆk
k= 0
frequency response
complex
j � +�H (�ˆ )) j�ˆn
y [ n ] = H (�ˆ ) Ae (
e
output same frequency
as input, but amplitude scaled
and a phase shift
LTI: FIR & IIR
� n x [n]�
�
�x n = � n
1 �[ ] [ ]
�0
�1
0 �
�
�
0 �
�2
�3
0 �
�
�
0 �
y[ n ] = ?
�4
�> 5
0 �
system
�y[n]�
h[n] = y[n] x[ n ]=� [ n ]� 1/ 3 �
�
�
� 0 �
�
�
1/
3
�
�
� 0 �
�
�
1/
3
�
�
� 0 �
Ex.
h[n] = 13 � [n] + 13 � [n 2] + 13 � [n 4 ]
FIR
y[n] = 13 x [n] + 13 x[n 2] + 13 x [n 4 ]
4
H (�ˆ ) = � h[k]e j�ˆ k
k=0
= h[0]e j�ˆ 0 + h[2]e j�ˆ 2 + h[4]e j�ˆ 4
= 13 + 13 e j�ˆ 2 + 13 e j�ˆ 4
= 13 (1+ e j�ˆ 2 + e j 2�ˆ 4 )
= 13 e j�ˆ 2 (e j�ˆ 2 +1+ e j�ˆ 2 )
= 13 e j�ˆ 2 (1+ 2 cos 2�ˆ )
Also try by inspection
if bk’s symmetric, then factor out
e j�ˆ ( M /2) where M is the order of
the filter. This leaves complex
conjugate paired exponentials
to transform into trigonometric
functions (cosines/sines).
H (�ˆ ) = 13 e j�ˆ 2 (1+ 2 cos 2�ˆ )
1
H (�ˆ ) = 13 (1+ 2 cos 2�ˆ )
Note:
H(
) =0
H( �) = 0
�
3
0.5
H (�ˆ )
1
3
(1+ 2 cos 2�ˆ )
0
zeros
-0.5
-1
2
3
-3
-2
-1
0
1
2
3
-3
-2
-1
0
1
2
3
3
�H (�ˆ ) = 2�ˆ
2
linear phase
�H (�ˆ ) = �H (�ˆ )
1
�H (�ˆ ) 0
-1
-2
-3
�ˆ
principal value of phase fn
if not, add multiples of 2 �
� < �H (�ˆ ) < �
Want positive magnitudes, H (�ˆ ) � 0
so absorb negative sign into phase by adding
an additional � at each zero
H (�ˆ ) = 13 e j 2�ˆ (1+ 2 cos 2�ˆ
)
1
H (�ˆ
) =
(1+ 2 cos 2�ˆ )
1
3
Note:
H ( �3 ) = 0
H(
2�
3
) =0
H (�ˆ )
1
3
band stop filter
(1+ 2 cos 2�ˆ )
0
zero
-0.5
-1
-3
-2
-1
0
1
2
3
-3
-2
-1
0
1
2
3
3
�H (�ˆ ) = �2�ˆ
� �2�ˆ
�
= � �2�ˆ + �
��2�ˆ + 2 �
�
0.5
2
linear phase
0 � �ˆ < � 3
� 3 � �ˆ < 2 � 3
2 � / 3 � �ˆ < �
phase odd function
�H (�ˆ ) = �H (�ˆ
)
principal value of phase fn
� < �H (�ˆ ) < �
1
�H (�ˆ ) 0
-1
-2
-3
�ˆ
Linear Phase
delay of n0 sample periods
y[n] = x [n n0 ]
H (�ˆ ) =
e j�ˆ n0
�H (�ˆ ) = n0�ˆ
H (�ˆ ) = 1
FIR filters are linear phase if
the coefficients are symmetric
linear phase
3
y = sin(2 �� (t + nTs )
)
2
higher frequencies need larger phase shifts
than lower frequencies to achieve same time
delay
= sin(2 ��t + 2 ��nTs )
1
0
-1
�H (�ˆ ) = �2�ˆ
� �2�ˆ
�
= � �2�ˆ + �
��2�ˆ + 2 �
�
linear phase,
2 sample delay
0 � �ˆ < � 3
� 3 � �ˆ < 2 � 3
2 � / 3 � �ˆ < �
-3
-2
-3
-2
= sin(2 ��t + � )
� = 2 �Ts n�
-1
0
1
2
3
-1
0
1
2
3
3
2
�H (�ˆ )
1
0
-1
-2
-3
�ˆ
Linear Phase
“It turns out that, within very generous
tolerances, humans are insensitive to
[audio] phase shifts. …”- Floyd E. Toole,
PhD
Vice President Acoustical Engineering
Harman International Industries, Inc.
“For data transmission, a nonlinear
phase delay causes intersymbol
interference which increases error
rate, particulary if the signal-to­
noise ration is poor” - Digital Signal
Processing in Communication
Systems By Marvin E. Frerking
“These are the pulse responses of each of the filters.The pulse response is nothing more than a positive
going step response followed by a negative going step response. The pulse response is used
here because it displays what happens to both the rising and falling edges in a signal.
Here is the important part: zero and linear phase filters have left and right edges that look the same,
while nonlinear phase filters have left and right edges that look different.
Many applications cannot tolerate the left and right edges looking different. One example is the
display of an oscilloscope, where this difference could be misinterpreted as a feature of the signal being
measured. Another example is in video processing. Can you imagine turning on your TV to
find the left ear of your favorite actor looking different from his right ear?”
http://www.dspguide.com/ch19/4.htm
FREQZ Z-transform digital filter frequency response.
When N is an integer, [H,W] = FREQZ(B,A,N) returns the N-point frequency
vector W in radians and the N-point complex frequency response vector H
of the filter B/A:
-1
-nb
jw B(z) b(1) + b(2)z + .... + b(nb+1)z
H(e) = ---- = ---------------------------­
-1
-na
A(z) 1 + a(2)z + .... + a(na+1)z
given numerator and denominator coefficients in vectors B and A.
<snip>
FREQZ(B,A,...) with no output arguments plots the magnitude and
unwrapped
phase of B/A in the current figure window.
-1
H (�ˆ ) = 13 e2 j�ˆ (1+ 2 cos 2�ˆ ) = 13 + 13 e j�ˆ 2 + 13 e j�ˆ 4
z 1 = e j�ˆ
+ 13 z 2 + 13 z 4
H (�ˆ ) =
1
1
3
>> freqz([1/3,0,1/3,0,1/3],[1])
b(1) = 13 ,b(2) = 0,b(3) = 13 ,b(4) = 0,b(5) = 13
a(1) = 1
H (�ˆ ) = 13 e j 2�ˆ (1+ 2 cos 2�ˆ ) = 13 + 13 e j�ˆ 2 + 13 e j�ˆ 4
>> freqz([1/3,0,1/3,0,1/3],[1])
Bode Plot (frequency response curve, amp and phase)
Magnitude Response (dB)
0
Magnitude response is plotted
on a logarithmic scale.
-10
-20
decibels (dB) = 20log10(|H|)
-30
-40
-50
-60
TextEnd
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Normalized frequency (Nyquist == 1)
0.8
0.9
1
0.3
0.4
0.5
0.6
0.7
Normalized frequency (Nyquist == 1)
0.8
0.9
1
150
Phase (degrees)
100
50
0
-50
TextEnd
-100
-150
0
0.1
0.2
Note: In this plot the
normalized frequency goes
�
from DC to Nyquist (�= � ),
so this is just one side.
We normally
plot from
�
-�< � < �.
Remember:
For real filter coefficients,
magnitude is an even function;
phase is an odd function
Superposition and the frequency response
x [ n ] = 3 + 3cos(0.6�n )
y[n] = 13 x [n] + 13 x[n 2] + 13 x [n 4 ]
input
FIR filter
sample domain
y[n] = 1+ cos(0.6 �n) +1+ cos(0.6 � (n 2)) +1+ cos(0.6 � (n 4 ))
= 3+ cos(0.6 �n)
+ cos(0.6 �n) cos(1.2 � ) + sin(0.6 �n) sin(1.2 � )
+ cos(0.6 �n) cos(2.4 � ) + sin(0.6 �n) sin(
2.4 � )
= 3+ [1+ cos(1.2 � ) + cos(2.4 � )] cos(
0.6 �n)
+ [sin(1.2 � ) + sin(2.4 � )] sin(0.6 �n
)
= 3+ A cos(0.6 �n) + Bsin(
0.6 �n)
= 3+ A 2 + B 2 cos(0.6 �n + tan 1 (B / A))
= 3+ 0.618 cos(0.6 �n 0.2 � )
frequency domain
x [ n ] = 3 + 3cos(0.6�n )
closer �ˆ
to a zero,
the smaller the
output.
3
2
H (�ˆ )
1
0
-1
-3
-2
-1
0
1
2
3
-3
-2
-1
0
1
2
3
3
2
�H (�ˆ )
1
0
-1
-2
-3
H (�ˆ ) =
1
3 (1+ 2 cos 2�ˆ )
�H (�ˆ ) = 2�ˆ + �
�ˆ
H (0) = 1
H (0.6 � ) = 0.206
�H (0) = 0 �H (2 � 0.6 � + � ) = 0.2 �
y[n ] = 3(1) + 3(0.206) cos(0.6 �n 0.2 � )
= 3+ 0.618 cos(0.6 �n 0.2 � )
chirp
3
2
H (�ˆ )
1
0
-1
-3
-2
-1
0
1
2
3
-3
-2
-1
0
1
2
3
3
2
�H (�ˆ )
1
0
-1
-2
-3
�ˆ