2003-10-07Dan Ellis 1

ELEN E4810: Digital Signal Processing

Topic 5: Transform-Domain Systems

1. Frequency Response (FR)

2. Transfer Function (TF)

3. Phase Delay and Group Delay

2003-10-07Dan Ellis 2

1. Frequency Response (FR) Fourier analysis expresses any

signal as the sum of sinusoids e.g. IDTFT:

Sinusoids are the eigenfunctions of LSI systems (only scaled, not ‘changed’)

Knowing the scaling for every sinusoid fully describes system behavior

frequency response

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x n[ ] = X e jω( )e

jωn dω−π

π∫

describes how a system affects eachpure frequency

2003-10-07Dan Ellis 3

Sinusoids as Eigenfunctions IR h[n] completely describes LSI system:

Complex sinusoid input i.e.

Output is sinusoid scaled by FT at

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= h k[ ]x n − k[ ]∀k

∑x[n] h[n] y[n] = x[n] h[n]

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x n[ ] = e jω0n

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y n[ ] = h k[ ]ejω0 n−k( )

k∑

= h k[ ]e− jω0k ⋅e jω0n

k∑

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⇒ y n[ ] = H e jω0( ) ⋅x n[ ]

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=H e jω0( ) ⋅ej ω0n+θ ω0( )( )

H(ej) = H(ej)ej()

2003-10-07Dan Ellis 4

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)} = () where H(ej) = DTFT{h[n]}

(Any signal can be expressed as sines...) H(ej) “magnitude response” gain arg{H(ej)} “phase resp.” phase shift

2003-10-07Dan Ellis 5

In practice signals are real e.g.

Real h[n] H(e-j) = H*(ej) = H(ej)e-j()

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x n[ ] = A cos ω0n +φ( )

= A2 e j ω0n+φ( ) + e− j ω0n+φ( )

( )

= A2 e jφe jω0n + A

2 e− jφe− jω0n

Real Sinusoids

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⇒ y n[ ] = A2 e jφ H e jω0( )e

jω0n + A2 e− jφ H e− jω0( )e

− jω0n

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⇒ y n[ ] = A H e jω0( ) cos ω0n +φ +θ ω0( )( )

-

X(ej)

2003-10-07Dan Ellis 6

Real Sinusoids

A real sinusoid of frequency

passed through an LSI system with a real impulse response h[n]has its gain modified by H(ej) and its phase shifted by ().

Acos(0n+) h[n] H(ej0)Acos(0n++(0))

2003-10-07Dan Ellis 7

Transient / Steady State Most signals start at a finite time e.g.

What is the effect?

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x n[ ] = e jω0nμ n[ ]

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y n[ ] = h n[ ]∗x n[ ] = h k[ ]ejω0 n−k( )

k=−∞

n∑

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= h k[ ]ejω0 n−k( )

k=−∞

∞∑ − h k[ ]e

jω0 n−k( )

k=n

∞∑

= H e jω0( )ejω0n − h k[ ]e− jω0k

k=n

∞∑( )e

jω0n

Steady state- same as with pure sine input

Transient response- consequence of gating

2003-10-07Dan Ellis 8

Transient / Steady State

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

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x n[ ] = e jω0nμ n[ ]

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⇒ y n[ ] = H e jω0( )ejω0n − h k[ ]e− jω0k

k=n

∞∑( )e

jω0ntransient

2003-10-07Dan Ellis 9

FR example MA filter

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y n[ ] = 1M

x n − l[ ]l =0

M −1∑

= x n[ ]∗h n[ ]n-1 1 2 3 4 5 6

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H e jω( ) = DTFT h n( ){ }

= h n[ ]e− jωn

n=−∞

∞∑ = 1

M e− jωn

n=0

M −1∑

=1M

1 − e− jωM

1 − e− jω=

1M

e− jω M −1( ) /2 sin Mω /2( )sin ω /2( )

2003-10-07Dan Ellis 10

FR example MA filter:

Response to ...

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H e jω( ) = 1

M e− jω M −1( ) /2 sin Mω /2( )sin ω /2( )

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⇒ H e jω( ) = 1

Msin Mω /2( )sin ω /2( )

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( )=− M −1( )ω2 +π ⋅r

(jumps at sign changes:r=M/2)

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x n[ ] = e jω0n + e jω1n

2003-10-07Dan Ellis 11

FR example MA filter input0 = 0.1 H(ej0) ~ 4/5 ej0

1 = 0.5 H(ej1) ~ -1/5 ej1

output€

x n[ ] = e jω0n + e jω1n

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y n[ ] = H e jω0( )ejω0n + H e jω0( )e

jω1n

2003-10-07Dan Ellis 12

2. Transfer Function (TF)

Linking LCCDE, ZT & Freq. Resp...

LCCDE:

Take ZT:

Hence:

or:

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dk y n − k[ ]k=0

N∑ = pk x n − k[ ]

k=0

M∑

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dk z−kY z( )k

∑ = pk z−k X z( )k

∑

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Y z( ) =pk z−k

k∑

dk z−k

k∑

X z( )

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Y z( ) = H z( )X z( )

Transferfunction

H(z)

2003-10-07Dan Ellis 13

Transfer Function (TF) Alternatively, ZT

Note: same H(z)

e.g. FIR filter, h[n] = {h0, h1,... hM-1}

pk=hk, d0=1, DE is

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y n[ ] = h n[ ]∗x n[ ]

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Y z( ) = H z( )X z( )

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=pkz−k∑dkz−k∑

h n[ ]z−n

n∑

⎧ ⎨ ⎪

⎩ ⎪

... if systemhas DE form

... from IR

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1⋅y n[ ] = hk x n − k[ ]k=0

M −1∑

2003-10-07Dan Ellis 14

Transfer Function (TF) Hence, MA filter:

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y n[ ] = 1M

x n − l[ ]l =0

M −1∑

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⇒ h n[ ] =1M 0 ≤ n < M

0 o.w.

⎧ ⎨ ⎩

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H z( ) = 1M z−n

n=0

M −1∑

=1 − zM

M 1 − z−1( )

=zM −1

M ⋅zM −1 z −1( ) z-plane

Re{z}

Im{z}

zM=1 i.e. M roots of 1@ z=ejr/M

pole @ z=1cancels

ROC?

2003-10-07Dan Ellis 15

TF example y[n] = x[n-1] - 1.2 x[n-2] + x[n-3]

+ 1.3 y[n-1] - 1.04 y[n-2] + 0.222 y[n-3]

factorize:

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⇒ H z( ) =Y z( )X z( )

=z−1 −1.2z−2 + z−3

1 −1.3z−1 +1.04z−2 − 0.222z−3

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H z( ) =z−1 1 −ζ 0z−1

( ) 1 −ζ 0*z−1

( )

1− λ 0z−1( ) 1− λ 1z

−1( ) 1− λ 1

*z−1( )

0 = 0.6+j0.80 = 0.31 = 0.5+j0.7

2003-10-07Dan Ellis 16

TF example

Poles i ROC causal ROC is z > maxi includes u.circle stable

Re{z}

Im{z}

0

1

1

0

0€

H z( ) =z−1 1−ζ 0z−1

( ) 1−ζ 0*z−1

( )

1 − λ 0z−1( ) 1 − λ 1z

−1( ) 1− λ 1

*z−1( )

2003-10-07Dan Ellis 17

TF FR DTFT H(ej) = ZT H(z)z = ej

i.e. Frequency Response is Transfer Function eval’d on Unit Circle

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H z( ) =p0 1 −ζ k z−1

( )k=1

M∏

d0 1 − λ k z−1( )k=1

N∏

=p0z−M z −ζ k( )

k=1

M∏

d0z−N z − λ k( )k=1

N∏

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⇒ H e jω( ) =

p0

d0

e jω N−M( )e jω −ζ k( )k=1

M∏

e jω − λ k( )k=1

N∏

factor:

2003-10-07Dan Ellis 18

TF FR

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H e jω( ) =

p0

d0

e jω N−M( )e jω −ζ k( )k=1

M∏

e jω − λ k( )k=1

N∏

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⇒ H e jω( ) =

p0

d0

e jω −ζ kk=1

M∏

e jω − λ kk=1

N∏

Magnituderesponse

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( )=argp0

d0

⎧ ⎨ ⎩

⎫ ⎬ ⎭+ω N − M( )

+ arg e jω −ζ k{ }k=1

M∑ − arg e jω − λ k{ }k=1

N∑

i, i are TF roots on z-plane

Phaseresponse

2003-10-07Dan Ellis 19

FR: Geometric Interpretation Have

On z-plane:€

H e jω( ) =

p0

d0

e jω N−M( )e jω −ζ k( )k=1

M∏

e jω − λ k( )k=1

N∏

Constant/linear part 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 ofall these vectors

Im{z}

eji

eji

Re{z}

ej

2003-10-07Dan Ellis 20

FR: Geometric Interpretation Magnitude H(ej) is product

of lengths of vectors from zeros divided by product of lengths of vectors from poles

Phase () is sum of angles of vectors from zeros minus sum of angles of vectors from poles

Im{z}

eji

eji

Re{z}

ej

i

i

2003-10-07Dan Ellis 21

FR: Geometric Interpretation Magnitude and phase of a single zero:

Pole is reciprocal mag. / negated phase

2003-10-07Dan Ellis 22

FR: Geometric Interpretation Multiple

poles / zeros:

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H z( ) =z − 0.8e j0.3π

( ) z − 0.8e− j0.3π( )

z − 0.9e j0.3π( ) z − 0.9e− j0.3π

( )

M

2003-10-07Dan Ellis 23

Geom. Interp. vs. 3D surface 3D magnitude surface for same system

2003-10-07Dan Ellis 24

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’

2003-10-07Dan Ellis 25

Filtering Idea: Separate information in frequency

with constructed H(ejw) e.g.

Construct a filter:H(ej) H(ej)

Then

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x n[ ] = A cos ω1n( ) + B cos ω2n( )interested in this part

don’t care aboutthis part

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y n[ ] = h n[ ]∗x n[ ] ≈ A cos ω1n +θ ω1( )( )

H(ej)

X(ej) “filteredout”

2003-10-07Dan Ellis 26

Filtering example 3 pt FIR filter h[n] = mix of 1 = 0.1 rad/samp

and 2 = 0.4 rad/samp

= =

n-1 1 2 3 4-2

to removei.e. make

H(ej) = 0

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H e jω( ) = h n[ ]e− jωn

∀n∑ = α + βe− jω +αe−2 jω

= e− jω β +α e jω + e− jω( )( ) = e− jω β + 2α cosω( )

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⇒ H e jω( ) = β + 2α cosω want = 1 at = 0.4

= 0 at = 0.1 ...

M

2003-10-07Dan Ellis 27

Filtering example Filter

IR

Freq.resp

input/output

2003-10-07Dan Ellis 28

3. Phase- and group-delay For sinusoidal input x[n] = cos0n,

we saw

i.e.

or

where is phase delay

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y n[ ] = H e jω0( ) cos ω0n +θ ω0( )( )gain phase shift

or time shift

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cos ω0 n +θ ω0( )

ω0

⎛

⎝ ⎜

⎞

⎠ ⎟

⎛

⎝ ⎜

⎞

⎠ ⎟

cos ω0 n − τ p ω0( )( )( )

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τ p ω( ) =−θ ω( )

ω

subtraction so positive τp meansdelay (causal)

2003-10-07Dan Ellis 29

Phase delay example For our 3pt filter:

i.e. 1 sample delay (at all frequencies)(as observed)

n-1 1 2 3 4-2

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H e jω( ) = e− jω β + 2α cosω( )

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⇒ ( )=−

⇒ τ p ω( ) = −−ωω

⎛ ⎝ ⎜

⎞ ⎠ ⎟= +1

2003-10-07Dan Ellis 30

Group Delay Consider a modulated carrier

e.g. x[n] = A[n]·cos(cn) with A[n] = Acos(mn) and m << c

Delay of envelope and carrier may differ

2003-10-07Dan Ellis 31

Group Delay

So: x[n] = Acos(mn)·cos(cn)

Now:

Assume H(ej) around cm

but (c-m) = l ; (c+m) = u ...

c

X(ej)

c+m

c-m

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=A2 cos ωc −ωm( )n + cos ωc +ωm( )n[ ]

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y n[ ] = h n[ ]∗x n[ ]

=A2

H e j ωc −ωm( )( )cos ωc −ωm( )n

+H e j ωc +ωm( )( )cos ωc +ωm( )n

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

2003-10-07Dan Ellis 32

Group Delay

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y n[ ] = A2

cos ωc −ωm( )n +θ l[ ]

+cos ωc +ωm( )n +θ u[ ]

⎛

⎝ ⎜

⎞

⎠ ⎟

= A cos ωcn +θ u +θ l

2

⎛ ⎝ ⎜

⎞ ⎠ ⎟⋅cos ωmn +

θ u −θ l

2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

phase shiftof carrier

phase shiftof envelope

2003-10-07Dan Ellis 33

Group Delay If (c) is locally linear i.e.

(c+) = (c) + S,

Then carrier phase shift

so carrier delay , phase delay

Envelope phase shift

delay group delay

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S =dθ ω( )

dω ω=ωc

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l +θ u

2=θ ωc( )

€

− c( )

ωc

= τ p

€

u −θ l

2= ωm ⋅S

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τ g ωc( ) = −dθ ω( )

dω ω=ωc

2003-10-07Dan Ellis 34

Group Delay

If () is not linear around c, A[n] suffers “phase distortion” correction...

c

τp, phasedelay

τg, groupdelay

()

n