ELEN E4810: Digital Signal Processing
Topic 5:
Transform-Domain Systems
1. Frequency Response (FR)
2. Transfer Function (TF)
3. Phase Delay and Group Delay
Dan Ellis
2013-10-09
1
1. Frequency Response (FR)

Fourier analysis expresses any
signal as the sum of sinusoids
1
X(ej )ej n d
e.g. IDTFT: x[n] =
2


Sinusoids are the eigenfunctions of LSI
systems (only scaled, not ‘changed’)
Knowing the scaling for every sinusoid
fully describes system behavior
describes how a
→ frequency response system affects each
pure frequency
Dan Ellis
2013-10-09
2
Sinusoids as Eigenfunctions

IR h[n] completely describes LSI system:
x[n]
h[n]
y[n] = x[n]
h[n] =
h[m]x[n
m]
m

Complex sinusoid input i.e. x[n] = ej
y[n] =
=
y[n] =

j 0 (n m)
h[m]e
m
j 0m
j 0n
h[m]e
·
e
m
H(ej )
= |H(ej )|ej
H(ej 0 ) · x[n] = |H(ej 0 )| · ej(
0 n+
Output is sinusoid scaled by FT at
Dan Ellis
2013-10-09
0n
(
0
3
( )
0 ))
System Response from H(ej!)

If x[n] is a complex sinusoid at !0
then the output of a system with IR h[n]
is the same sinusoid scaled by |H(ej! )|
and phase-shifted by arg{H(ej! )} = µ(!0)
where H(ej!) = DTFT{h[n]}
0
0
(Any signal can be expressed as sines...)
 |H(ej!)| “magnitude response” → gain

arg{H(ej!)} “phase resp.” → phase shift
Dan Ellis
2013-10-09
4
Real Sinusoids

In practice signals are real e.g.
x[n] =A cos( 0 n + )
A j( 0 n+ )
=
e
+e
2
A j j 0n A
= e e
+ e
2
2

|X(ejω)|
j(
j
A/2
0 n+ )
e
j
0n
ω0
- ω0
ω
A j
A j
j 0 j 0n
y[n] = e H(e )e
+ e
H(e j 0 )e j 0 n
2
2
Real h[n] H(e j ) = H (ej ) = |H(ej )|e j (
y[n] = A|H(ej 0 )| cos (
Dan Ellis
0n
2013-10-09
+
+ (
0 ))
5
)
Real Sinusoids
A cos(
0n + )

h[n]
A|H(ej 0 )| cos (
0n
+
+ (
A real sinusoid of frequency !0
passed through an LSI system
with a real impulse response h[n]
has its gain modified by |H(ej!0)|
and its phase shifted by µ(!0).
Dan Ellis
2013-10-09
6
0 ))
Transient / Steady State

Most signals start at a finite time, e.g.
x[n] = ej 0 n µ[n] What is the effect?
y[n] =
=
=
n
j 0 (n m)
h[n] x[n] =
h[m]e
m= ⇥
⇥
⇥
j 0 (n m)
j 0 (n m)
h[m]e
h[m]e
m= ⇥
m=n+1
⇥
j 0 j 0n
H(e )e
( m=n+1 h[m]e j 0 m )ej 0 n
Steady state
- same as with pure sine input
Dan Ellis
Transient response
- consequence of gating
2013-10-09
7
Transient / Steady State

x[n] = ej
0n
µ[n]
⇥ y[n] = H(ej 0 )ej



transient
⇥
m=n+1 h[m]e
(
0n
j
0m
0n
FT of IR h[n]’s tail from time n onwards
zero for FIR h[n] for n ≥ N
tends to zero with large n for any ‘stable’ IR
Total output
Steady State
Transient
-40
Dan Ellis
)ej
-30
-20
-10
0
2013-10-09
10
20
30
40
time / n
8
FR example

MA filter
1
y[n] =
M
M
1
]
x[n
=0
=x[n]
1/M
h[n]
-1
1 2 3 4 5 6
H(ej ) = DTFT{h[n]}
=
h[n]e
n=
j n
1
=
M
1
1 1 e j M
=
e
=
j
M 1 e
M
Dan Ellis
2013-10-09
M
1
e
j n
n=0
j
(M 1)
2
sin(M /2)
sin( /2)
9
n
FR example

MA filter:
1
H(e ) =
e
M
j
j
(M 1)
2
sin(M /2)
sin( /2)
6
4
2
0
1 sin(M /2)
j
H(e ) =
M sin( /2)
(M 1)
( )=
+ ·r
2
6
4
2
0
0
-
2
(jumps at sign changes:
r= M!/2π )

Response to
Dan Ellis
x[n] = ej
2013-10-09
0n
+ ej
0
1n
2
(M = 5)
...
10
FR example
6
4
2
0
6
4


MA filter
j
input x[n] = e
= 0.1
= 0.5
0
1

2
0
0
0n
+e
j
1n
-
H(ej 0 ) 0.8ej 0
H(ej 1 ) ( )0.2ej
output y[n] = H(ej 0 )ej
0n
0
2
2
1
+ H(ej 1 )ej
1n
2
1
x[n]
0
y[n]
-1
-2
-20
Dan Ellis
0
20
40
2013-10-09
60
80
n
11
M
2. Transfer Function (TF)
Linking LCCDE, ZT & Freq. Resp...
N

N
LCCDE:
dk y[n
k] =
k=0

Take ZT:
k]
k=0
k
dk z
k

Hence: Y (z) =

or:
Dan Ellis
pk x[n
Y (z) =
Y (z) =
P
p z
Pk k
k dk z
X(z)
k
k
k
X(z)
H(z)X(z)
2013-10-09
k
pk z
Transfer
function
H(z)
12
Transfer Function (TF)



Alternatively, y[n] = h[n] x[n]
ZT → Y (z) = H(z)X(z)
Note: same H(z) =
P
P pk z
dk z
... if system
has DE form
k
k
n
n h[n]z
... from IR
e.g. FIR filter, h[n] = {h0, h1,... hM-1}
M
pk=hk, d0=1, DE is 1 · y[n] =
Dan Ellis
2013-10-09
1
hk x[n
k=0
13
k]
Transfer Function (TF)

Hence, MA filter:
1
y[n] =
M
H(z) =
=
=
1
M
M 1
=0
M 1
=0
z
zM
M ·z M
] ⇥ h[n] =
1
1 (z 1)
0
Im{z}
n
1 z M
M (1 z 1 )
(ignore poles
at z=0)
Dan Ellis
x[n
0 n M
otherwise
1
M
zM=1 i.e.
M roots of 1
@ z=ej2πr/M
|H(ej!)|

Re{z}
1
ROC?
pole @ z=1
cancels
2013-10-09
z-plane
14
TF example

y[n]
=
x[n 1] 1.2x[n 2] + x[n 3]
+ 1.3y[n 1] 1.04y[n 2] + 0.222y[n
Y (z)
H(z) =
=
X(z)
1

H(z) =
z
1.3z
1.2z 2 + z 3
1 + 1.04z 2
0.222z
3]
1
3
factorize:
z
(1
Dan Ellis
(1
0z
1 )(1
0z
1
)(1
0z
1 )(1
1z
1
2013-10-09
)
1z
1
1)
ζ0 = 0.6+j0.8
λ0 = 0.3
λ1 = 0.5+j0.7
→ ...
15
TF example
H(z) =
z
(1
(1
0z
1 )(1
0z
1
)(1
0z
1 )(1
1z
1
ζ0 = 0.6+j0.8
λ0 = 0.3
λ1 = 0.5+j0.7

Poles ∏i → ROC

causal → ROC is |z| > max|∏i|

includes u.circle → stable
Dan Ellis
2013-10-09
)
1z
1
1)
Im{z}
λ1
×
λ0 ×
λ*1
ζ0
Re{z}
1
×
16
ζ ∗0
TF → FR

DTFT H(ejω) = ZT H(z)|z = ejω
i.e. Frequency Response is
Transfer Function eval’d on Unit Circle
1   k z ) p0 z  (z   k )
(
k=1
k=1
H (z) =
=
N
N
1
N
d0  (1   k z ) d0 z  (z   k )
k=1
k=1
M
j
e



(
k)
p
j

NM
j
(
)
k=1
0
 H (e ) = e
N
j
d0
k=1 (e   k )
factor:
Dan Ellis
p0 
M
1
2013-10-09
M
M
17
TF → FR
p0 j
j
H(e ) = e
d0
(N
M)
p0
H(e ) =
d0
⇥(⌅) = arg
+
p0
d0
M
⇧
k=1
Dan Ellis
⌅
ej
+ ⌅ · (N
arg e
j
k
k
(ej
M
j
e
k=1
N
j
|e
k=1
j
⇤
M
k=1
N
k=1
k)
Magnitude
response
k
k|
M)
⇥
2013-10-09
N
⇧
ζk, λk are
TF roots
on z-plane
Phase
response
arg e
j
⇤k
k=1
18
⇥
FR: Geometric Interpretation
M
j
e



(
k)
p
j

NM
j
(
)
k=1
0
 Have H e
( )= d e
N
j
0
k=1 (e   k )
Constant/
linear part

On z-plane:
Im{z}
×
ej!−∏i
×
Dan Ellis
ej!−≥i
ej!
Re{z}
×
Product/ratio of terms
related to poles/zeros
Each (ej! - ∫) term corresponds
to a vector from pole/zero ∫ to
point ej! on the unit circle
Overall FR is product/ratio of
all these vectors
2013-10-09
19
FR: Geometric Interpretation

Im{z}
≥i
∏i×
ej!−∏i
ej!−≥i
ej!

×
Re{z}
×
Dan Ellis
Magnitude |H(ej!)| is product of
lengths of vectors from zeros
divided by product of lengths
of vectors from poles
Phase µ(!) is sum of angles of
vectors from zeros
minus sum of angles of
vectors from poles
2013-10-09
20
FR: Geometric Interpretation

Magnitude and phase of a single zero:
Imaginary Part
1
0.6
0.4
0.8
0.5
0.2
1
0
0.0
0
magnitude
2
phase
-0.5
-0.8
-0.6
-1
-1

-0.5
/2
-0.2
0
-0.4
0
0.5
Real Part
- /2
1
-
0
0.5
1.5
2
Pole is reciprocal mag. & negated phase
Dan Ellis
2013-10-09
21
FR: Geometric Interpretation

(
(
1
Imaginary Part
)(
)(
j0.3 
 j0.3 
Multiple
z  0.8e
z  0.8e
poles, H (z ) =
j0.3 
 j0.3 
zeros:
z  0.9e
z  0.9e
0.6
0.4
0.8
0.5
0.2
1
0
0.0
0
magnitude
2
phase
-0.5
-0.8
-0.6
-1
-1
Dan Ellis
-0.2
-0.5
0
-0.4
0
0.5
Real Part
/2
- /2
1
-
2013-10-09
0
0.5
1.5
22
2
V
)
)
Geom. Interp. vs. 3D surface

3D magnitude surface for same system
Full surface
Dan Ellis
Showing ROC
2013-10-09
23
Geom. Interp: Observations

Roots near unit circle
→ rapid changes in magnitude & phase




zeros cause mag. minima (= 0 → on u.c.)
poles cause mag. peaks (→ 1÷0=∞ at u.c.)
rapid change in relative angle → phase
Pole and zero ‘near’ each other
cancel out when seen from ‘afar’;
affect behavior when z = ej! gets ‘close’
Dan Ellis
2013-10-09
24
M
PEZdemo
Filtering


Idea: Separate information in frequency
with constructed H(ej!)
e.g. x [n] = A cos(1n) + B cos( 2 n)
interested
in this part

Construct a filter:
|H(ej!1)|
∼1
|H(ej!2)| ∼ 0

don’t care about
this part
B/
2
X(ej!)
H(ej!)
A/
2
−!2 −!1
B/
2
“filtered
out”
!
A/
2
!1
!2
Then y[n] = h[n]  x [n]  A cos(1n +  (1 ))
Dan Ellis
2013-10-09
25
Filtering example Æ

x[n]
Consider
Ø
z
filter ‘family’:
z
Æ
3 pt FIR filters
with h[n] = {Æ Ø Æ}
Frequency Response:
-1
-1

+
y[n]
Æ
-2 -1
Ø
Æ
1 2 3 4
n
( ) = n h[n]e =  + e + e
 j
j
 j
 j
= e ( +  (e + e )) = e ( + 2 cos  )
can set Æ and Ø
j
to obtain desired
 H (e ) =  + 2 cos 
|H(e )| ...
H e
j
 jn
 j
2 j
j!
Dan Ellis
2013-10-09
26
Filtering example (cont’d)



( ) =  + 2 cos 
h[n] = {Æ Ø Æ}  H e
j
Consider input as mix of sinusoids
at !1 = 0.1 rad/samp
want to remove
and !2 = 0.4 rad/samp
i.e. make H(ej!2) = 0
Solve H(ej ) = |⇥ + 2 cos ⇤|
⇥
1 ⇤ = ⇤1 = 0.1
=
0 ⇤ = ⇤2 = 0.4
Ø = -12.46, Æ = 6.76 ...
Dan Ellis
2013-10-09
27
Filtering example (cont’d)
15

Filter
IR
10
5
0
-5
-10
-15
0
5
10
15
20
25
30
35
40
45
n
20

Freq.
resp
10
dB
0
-10
-20
0
0.2
0.4
0.6
0.8
/ rad
1
2

input/
output
1
0
-1
-2
0
Dan Ellis
20
40
60
80
2013-10-09
100
120
140
160
180
28
n
3. Phase- and group-delay

For sinusoidal input x[n] = cos!0n,
(
we saw y[n] = H e
j 0
) cos( 0n +  ( 0 ))
gain

   ( 0 ) 
i.e. cos 0  n +

 0 
 
or

cos( 0 (n   p ( 0 )))
 ( )
where  p ( ) =

Dan Ellis
2013-10-09
phase shift
or time shift
subtraction so
positive øp means
delay (causal)
is phase delay
29
Phase delay example

For our 3pt filter:
( )=e
H e

j
 j
Æ
-2 -1
Ø
Æ
1 2 3 4
( + 2 cos  )
n
  ( ) = 
  
  p ( ) =   = +1
 
i.e. 1 sample delay (at all frequencies)
(as observed)
Dan Ellis
2013-10-09
30
Group Delay

Consider a modulated carrier
e.g. x[n] = A[n]·cos(!cn)
with A[n] = Acos(!mn) and !m << !c
1
0.5
0
0.5
1
0
Dan Ellis
50
100
150
200
2013-10-09
250
300
350
400
31
X(ej!)
Group Delay

So: x[n] =
Now:
=
!c
-!m
A cos( m n) · cos( c n)
A
m )n + cos( c +
2 [cos( c
y[n] =
=

h[n]
A
2
H(ej(
+H(ej(
m )n]
x[n]
c
m)
) cos(
c
c+
m)
) cos(
c
m )n
+
Assume |H(ej!)| ∼ 1 around !c±!m
but µ(!c-!m) = µl ; µ(!c+!m) = µu ...
Dan Ellis
2013-10-09
!c
+!m
!c !
32
m )n
Group Delay
j(
=
y[n]
|H(ej!)| ∼ 1
µ(!c-!m) = µl
µ(!c+!m) = µu
A
2
=
= A cos
H(e
+H(ej(
A
2
c
m)
) cos(
c
c+
m)
) cos(
c
cos[(
+ cos[(
cn +
+
u+ l
2
phase shift
of carrier
Dan Ellis
2013-10-09
+
m )n
m )n+ l ]
c
c
m )n
m )n+ u ]
· cos
mn +
u
l
2
phase shift
of envelope
33
!c
Group Delay
µ(!)
If µ(!c) is locally linear i.e.
" µ(!c+¢!) = µ(!c) + S¢!,

µl
!
S
µu
!c–!m !c+!m
d ( )
S=
d  = c
l + u
 Then carrier phase shift
=  ( c )
2
 ( c )
=  p , phase delay
so carrier delay 
c
u  l
 Envelope phase shift
= m  S
2
d ( )
→ delay  g ( c ) = 
group delay
d  = c
Dan Ellis
2013-10-09
34
Group Delay
Envelope
(group) delay
1
0.5
Carrier
(phase) delay
0
-0.5
-1
0
50
100
150
200
!c
øp, phase
delay

250
300
350
400
n
!
øg, group
delay
µ(!)
If µ(!) is not linear around !c, A[n] suffers
“phase distortion” → correction...
Dan Ellis
2013-10-09
35
M