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Transcript of SP05 Dig Filters 1 Classif Analysis1 Vs100609
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8/2/2019 SP05 Dig Filters 1 Classif Analysis1 Vs100609
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Signal Processing05 digitalfilters(1)seriesofexpertlectures IrR.deWild
ASTRON,32010
1
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learning
goals
2
At the end of this
lecture,
-
from
a given
transfer function
or
a given
impulse
response of a linear
system,
you
can
draw a functional
block
diagram (or
a signal
flow
diagram);
-
from
a given
block
diagram (or
a signal
flow
diagram) of a linear
system,
you
can
derive
the transfer function
and the impulse
response;
-
you
can
convert
a block
diagram from
its
direct form
into
its
transposed
form;
-
you
know
how
to adapt/modify
(in a qualitative
manner) the frequency
response of
a linear
system, by
moving
in the z-plane
the locations
of poles
and zeros;
-
you
can
classify
a digital filter according
to its
impulse
response, its
transfer function,
its
frequency
response and/or
its
functional
block
diagram;
-
you
can
identify
and analyze
comb
filters, all-pass
filters, FIR-filters
and IIR-filters;
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Contents
3
digitalfilters
classification
digitalfilters
analysis
(1)
digitalfilters analysis (2):
Digital Signal Processing
Monson H. Hayes, McGraw-Hill publ.
par. 2.3; 8 (except
8.6)
Signals
& Systems
H.P. Hsu, McGraw-Hill publ. none
see SP06
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systembehaviour four points ofview
)(
]1[))1((
1 zXz
nxTnx s
[ ])(arg)()( fHjefHfH =
)(
]1[))1((
2 fXe
nxTnx
sTfj
s
4
impulse
response
discrete-time: t = n.Ts
y[n] = h[n]
*
x[n]
(h[n]: implicit
expression)
transfer function
Y(z) = H(z)
. X(z)
H(z) = Y(z) / X(z)
frequency
response
H(f)
H(2 .f.Ts
)
complex z-plane
-
poles
& zeros
-
stability
criterium
(1/Ts
)-periodic
f-axis
-
Bode-diagram
(amplitude & phase)
-
stability
criterium
Z-Transform DTFT (Discrete-Time
Fourier
Transform)
peel-off
unit-circle
sTfj
ez
=
2
functional
structure:
block
diagram (schematic)
signal
flow
diagram (directed
graph)
)2( sss TfjTTp eez + ==
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systembehaviour analysis &synthesis
5
Filter Analysis
starting
from
a given
-
impulse
response
-
transfer function
-
functional
structure,
arrive
at the resulting
frequency
response
What
is a filter?
type of signal
processing system
frequency-specific
treatment
of
signal-amplitude
and signal-phase
Filter Synthesis
starting
from
the required
frequency
response,
arrive
at the optimally
fitting
-
impulse
response
-
transfer function
-
functional
structure
= subject ofSP06= subject ofSP05
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Contents
6
digitalfilters
classification
digitalfilters
analysis
(1)
digitalfilters analysis (2): see SP06
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digitalfilters
classification
7
according
to impulse
response
according to transferfunction
according
to systemstructure
according to frequency response
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impulse
response issues
8
relationship:
y[n]=h[n]
*x[n]
linear
difference
equation
inn(t=nTs
)
FIR (Finite Impulse Response)
IIR
(Infinite
Impulse
Response)
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impulse response overview
9
linear
digital system: y[n] = h[n] * x[n]
=
+++++=
==
M
m
m
N
m
m
M
N
mnyamnxbny
Mnyanyanya
Nnxbnxbnxbnxbny
10
21
210
][][][
][]2[]1[
][]2[]1[][][
L
L
impulse
response h[h] linear
difference
equation
in n
Z-Transform
transfer function
H(z) bi-linear
function
in z : H(z) = T(z) / N(z) ,
with
T(z) = degree-N
polynomial
in z
N(z) = degree-M polynomial in z
x[n] [ny[n] h[n]
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digitalfilters
classification
10
according
to impulse
response
according to transferfunction
according
to systemstructure
according to frequency response
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transferfunction issues
11
relationship:Y(z)=H(z).X(z) H(z) =T(z)/N(z)bi
linear function,with T(z)andN(z)polynomials inz
MA(Moving
Average)
H(z)=T(z) zplane:only zeros H(f) amplitudecharacteristic:dips
AR(AutoRegressive)
H(z)=1/N(z) zplane:only poles H(f) amplitudecharacteristic:spikes
ARMA(AutoRegressive
Moving
Average)
H(z)=T(z)/N(z) zplane:zeros &poles
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transferfunction overview
12
linear
digital system: H(z) = T(z) / N(z)
polynomials
in z:
T(z) factorization
N(z) factorization
( )
( )
=
=
=
=
==
=1
0
1
0
1
0)(M
m
m
N
m
m
M
m
m
m
N
m
m
m
pz
zz
za
zb
zH L
complex z-plane:
-
N zeros: z0
,,zM-1
-
M poles: p0
,,pN-1
stability: all poles
within unit circle!
j.Im(z)
Re(z)
10
j
X O
O
X
X O
O
zero
pole
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digitalfilters
classification
13
according
to impulse
response
according to transferfunction
according
to systemstructure
according to frequency response
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systemstructure
issues
14
representation (visualisation):block
diagram
(block
schematic)
signal
flow
diagram(directed
graph)
canonical
structures
directform I&IItransposed
form
I&II
frequencysampled
structure
not
explicitly
treated
MA transversal form FIRonly
outside scope ladder
structure,
lattice structure
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systemstructure overview
15
[derived
from
h[n]
or
from
H(z):]
N(z)
= degree-M
polynomial
in z
AR (Auto Regressive)
part:
feed-back paths poles
T(z)
= degree-N
polynomial
in z
MA (Moving
Average)
part:
feed-forward paths zeros
H(z) =
T(z) / N(z)
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digitalfilters
classification
16
according
to impulse
response
according to transferfunction
according
to systemstructure
according to frequency response
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frequency response issues
17
relationship:
Y(f)=H(f).X(f) H(f) =T(f)/N(f)
Bodediagram: amplitude &phase characteristicsLowPass
&
HighPass
BandPass
&
BandStop
MultiBand
AllPass
(i.e. phaseshift
only)
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frequency response overview
ss
s
zjTfj
Tf
TzTfz
z
ezezfHzH s
2
1
2
1]arg[:periodic2]arg[
circleunit1
:)()(]arg[2
+
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Contents
19
digitalfilters
classification
digitalfilters
analysis
(1)
digitalfilters analysis (2): see SP06
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digitalfilters
analysis
(1)
20
functional systemstructure family tree
directform
I&II
transposed form I&II
lowpass
filterexamples
overview
according
tofrequency
response
transferfunction
&complexzplane
transfer
function
&
impulse
response
transferfunction
&frequency
response
transferfunction
&block
diagram
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block diagramversussignal flow diagram
21
see
Digital Signal
Processing
(M. Hayes): p.288 (CH 8.2)
signal
flow
diagram (graph)
directed
branch
sink
node
source
node
block
diagram (schematic)
functional
block
(sub-system)
functional
interconnection
signal
delay
signal
scaling
signal
addition
input signal
output signal
In this
lecture, we adhere
to block
diagrams
(block
schematics) !
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block schematic(1) Direct
Form I
22
Ts
z
-1
y[n]
Y(z)
x[n]
X(z)b0
Ts
z-1
Ts
z-1
Ts
z-1
Ts
z
-1
Ts
z-1
Ts
z-1
b2
b1
b4
b3
-a1
-a3
-a2
]3[]2[]1[]4[]3[]2[]1[][][ 32143210 ++++= nyanyanyanxbnxbnxbnxbnxbny
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block schematic(2a) Direct
Form II
23
Direct Form
-
I Direct Form - II : step 1 swap T(z) and 1/N(z)
Ts
z-1
y[n]
Y(z)
x[n]
X(z)b0
Ts
z-1
Ts
z-1
Ts
z-1
Ts
z-1
Ts
z-1
Ts
z-1
b2
b1
b4
b3
-a1
-a3
-a2
( ) ( )
( ) ( )]4[]3[]2[]1[][]3[]2[]1[]3[]2[]1[]4[]3[]2[]1[][][
43210321
32143210
+++++=
=+++++=
nxbnxbnxbnxbnxbnyanyanya
nyanyanyanxbnxbnxbnxbnxbny
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block schematic(2b) Direct
Form II
24
Direct Form
-
I Direct Form - II : step 2 combine delay-elements
Ts
z-1
y[n]
Y(z)
x[n]
X(z)b0
Ts
z-1
Ts
z-1
Ts
z-1
b2
b1
b4
b3
-a1
-a3
-a2
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block schematic(3a) Transposed Form I
)}}}({)({)({
)}}}}({)({)({)({)()(
3
1
2
1
1
1
4
1
3
1
2
1
1
1
0
zYazzYazzYaz
zXbzzXbzzXbzzXbzzXbzY
+++
++++=
)()()(
)()()()()()(
3
3
2
2
1
1
4
4
3
3
2
2
1
10
zYzazYzazYza
zXzbzXzbzXzbzXzbzXbzY
++++=
25
Direct Form
-
I Transposed Form - I
rewrite
the Z-domain
input-output
relationship
]3[]2[]1[
]4[]3[]2[]1[][][
321
43210
++++=
nyanyanya
nxbnxbnxbnxbnxbny
Z-Transform
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block schematic(3b) Transposed
Form
I
26
y[n]
Y(z)
x[n]
X(z)b0
Ts
z
-1
b2
b1
b4
b3
-a1
-a3
-a2
Ts
z-1
Ts
z-1
Ts
z-1
Ts
z
-1
Ts
z-1
Ts
z-1
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block schematic(4) Transposed Form II
27
Transposed
Form
-
I
Transposed Form - II:
combine delay-elements
y[n]
Y(z)
x[n]
X(z)b0
Ts
z-1
b2
b1
b4
b3
-a1
-a3
-a2
Ts
z-1
Ts
z
-1
Ts
z-1
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block schematic(5a) FrequencySampled Form
28
=
=
+=
=
==1
0
][][)(Nn
n
nn
n
n znhznhzHFIR-type
filter:
h[n] = h(nTs
) Ts
-sampled
&
NTs
-finite
substitute
Inverse-DFT-expression
of h[n]
into
FIR-expression
of H(z):
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
==
1
0 1
2
1
0 1
2
2
1
0 1
2
2
1
0
1
0
2
1
0
1
0
21
0
1
][1
1
1][
1
1
1][
1][
1
][
1
][)(
Nk
kk
Nj
NNk
kk
Nj
Nkj
Nk
k
k
N
j
NNk
Nj
Nk
k
Nn
n
nnk
Nj
Nn
n
nNk
k
nkN
jNn
n
n
ze
kH
N
z
ze
zekH
N
ze
zekH
NzekH
N
zekHNznhzH
parallelserial
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block schematic(5b) FrequencySampled Form
=
=
=
1
0
220
00 ][)()(Nn
n
nTfjfj senheHfH
=
=
=
=
==
1
0
2
0
1
0
20 0
0
][][)(Nn
n
nkN
jNn
n
nTNT
kj
s
s
enhkHenTh
NT
kH
ss
29
Discrete-Time
Fourier
Transform: h[n] = h(nTs
) Ts
-sampled
(& finite)
(DTFT)
H(f) (1/Ts
)-periodic
& f-continuous
frequency-sampled
approximation
of one
period
of H(f)
Discrete Fourier
Transform: h[n] = h(nTs
) Ts
-sampled
& (NTs
)-limited
(~ period)
(DFT)
H[k] = H(k/NTs
) (1/Ts
)-periodic
& (1/NTs
)-sampled
=
=
=
=
+
=
=
1
0
2
0
1
0
2
0
00
][1
][1
)(Nk
k
nkN
jNk
k
TnNT
kj
s
s ekHN
nheNT
kH
NTnh
ss
Inverse Discrete Fourier
Transform
(apply
duality
property):
see DA05
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block schematic(5c) FrequencySampled Form
30
x[n]
X(z)
1 / N
Ts
z-1
H[0]
y[n]
Y(z)
Ts
z-1
1
H[1]
Ts
z-1
exp
( j2
/ N )
H[N-1]
Ts
z-1exp ( j2(N-1)/ N )
H[k]
Ts
z-1
exp ( j2.k/ N )
N x
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digitalfilters
analysis
(1)
31
functional systemstructure family tree
directform
I&II
transposed form I&II
lowpass
filterexamples
overview
according
tofrequency
response
transferfunction
&complexzplane
transfer
function
&
impulse
response
transferfunction
&frequency
response
transferfunction
&block
diagram
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lowpass filterexamples:overview (1)( )
( )
=
=
=
=
==
=1
0
1
0
1
0)(M
m
m
N
m mMN
M
m
m
m
N
m
m
m
pz
zz
z
za
zb
zH L
32
transfer function complex z-planeH(z) = T(z) / N(z)
M zeros: z0
,,zM-1N poles: p0
,,pN-1
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lowpass filterexamples:overview (2)
33
T(z)order N(z)order filtertype(impulse
response/zplane
/frequency
response)
N=M M=N comb
filter:
type1
N=M
notches,
equidistant
on
unit
circle
|z|
=1
type2
M=Npeaks,equidistant
on
unitcircle
|z|=1
N=M M=N allpass
filter:
N=Mpolezero
pairs:geometric
inversew.r.t.unitcircle
0 M IIRfilter:
Mpoles,within
unitcircle|z|=1
M=1 1pole: real &pos./neg.;[complex outside scope]
M=2 2poles:2xreal ;1x2foldreal;complexconjugate pair
N 0 FIRfilter:
N
zeros;
N
=even/odd
h[n]=geometric
sequence
(H(z):nonlinear
phase)
h[n]=even/odd
function H(z): linear phase 4FIRtypes
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lowpass
filterexamples
comb
filters
allpass filters
FIRfilters
IIRfilters
evaluation
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lowpass filterexamples comb filter intuitive
35
block
diagram
Ts
z-1
1
Ts
z-1
Ts
z-1
- 1
x[n]
X(z)
N delay-units time-shift: = N.Ts [s]
y[n]
Y(z)
harmonic
response
constructive
combining
at output, if
x[n N] and x[n] are in counter-phase
( i.e. for
= , 3, 5,
[rad] )
destructive
combining
at output, if
x[n N] and x[n] are in phase
( i.e. for
= 2, 2, 2,
[rad] )
x[n -N]
X(z) . z-N
l f l l
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lowpass filterexamples comb filter 2types
( )( )N
NNNNNN
z
zzzzzzzH
=++++== )1(12211 111)( L
36
linear
digital system: H(z) = T(z) / N(z)
comb
filter
type 1
1 pole
N-fold
at z = 0
N zeros
equidistant
on
circle
|z| =
( < 1)
N notches (dips) equidistant on unit circle |z| = 1
1 -
impulse
response
2 -
transfer function
3 -
block
schematic
comb
filter
type 2
1 zero
N-fold
at z = 0
N poles
equidistant
on
circle
|z| =
( < 1)
N peaks (spikes) equidistant on unit circle |z| = 1
( )( ) NNN
NNNNz
z
zzzzzzH
=
++++=
=
)1(12211 11
1
1
1)(
L
M. Hayes: pp.331
exercise
type
1 : see
next
slides
type 2 : do-it-yourself!
l f l l
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lowpass filterexamples comb filter type1(1a)
37
( )( )N
NNNNNN
z
zzzzzzzH
=++++== )1(12211 111)( L
transfer function
complex Z-plane
comb
filter
type 1
1 pole
N-fold
at z = 0
N zeros
equidistant
on
circle
|z| =
( < 1)
N notches (dips) equidistant on unit circle |z| = 1
NOTE
evaluation slides comb & IIR = FIR
l fil l
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lowpass filterexamples comb filter type1(1b)N
NN
z
zzH
=)(
38
transfer function
complex Z-plane
][1
2 radN+
n=2
0
n=1
n=N-1
O O
O
n=3
O
Re(z)
jj.Im(z)
1
n=0(=N)OXN x
l fil l
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lowpass filterexamples comb filter type1(2)
sNTfjN efH= 21)(
NNzzH
= 1)(
][][][ Nnnnh N =
39
transfer function
impulse
response
frequency
response
t = nTs
[s]0
NTs
N
.(t NTs
)
(t)
NjnjNNT
NT
nj
N
ss
NnjN
NT
NT
nj
N
ss
eee
NT
nH
NT
nf
eeNT
nHNT
nf
ss
s
s
+===
++=
===
=
+
111:for
111:for
22
21
21
2
2
21
maxima
minima
l fil l
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lowpass filterexamples comb filter type1(3)sNTfjN efH
= 21)(
40
transfer function
frequency
response
f [Hz]
1 / Ts1 / NTs0
= 1
0 <
< 1
e.g. N = 5
1 + N
1 -
N
exercise
Applying
the duality
property
of the Fourier
transform,
which
case do you
get
when
you
swap t-
and f-domain?
l filt l
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lowpass filterexamples comb filter type1(4)
41
NNzzH
= 1)(
transfer function
block
diagram
Ts
z-1
1
Ts
z-1Ts
z-1
N
x[n]
X(z)
y[n]
Y(z)
x[n -N]
X(z) . z-N
N
delay-units
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lowpass
filterexamples
comb
filters
allpass filters
FIRfilters
IIRfilters
evaluation
l filt l
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lowpass filterexamples allpass filter (1a)
( ) ( )
( )
( ) ( )
( ) 1,,1,0;
:zeros
1,,1,0;
:poles
2
1
2
2
2
==
==
==
==
+
+
+
+
Nneerz
eererz
Nneerz
eererz
nN
jj
njNjNjN
nN
jj
njNjNjN
L
L
( )( ) NNj
NjN
NN
NN
zer
erz
z
zzH
=
=
11
)()(
43
transfer function
complex Z-plane
N zeros
equidistant
on
circle
|z| =
r < 1
1 -
impulse
response2 -
transfer function
3 - block schematic
N x pole-zero pair: geometric inverse (with respect to unit circle)
N poles
equidistant
on
circle
|z| = r-1
> 1
jj
erer
==
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lowpass filterexamples allpass filter (1b)( )( ) NN
NN
NNj
NjN
NN
NN
z
z
zer
erz
z
zzH
=
=
=
)(1
11
)()(
44
][1
2rad
N+
n=2
r
0
n=1
n=N-1
O
O
O
n=3
O
Re(z)
X
j
j.Im(z)
1/r
n=0(=N)O
X
XXX
transfer function
complex Z-plane
1
jj erer ==
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lowpass filterexamples allpass filter (2)
][)(][][][ nxNnxNnyny NN =
NN
NN
z
zzH
=
1
)()(
NN
N
NN
N
NN
NN
zz
z
z
zzH
=
=
1
)(
11
)()(
45
transfer function
Impulse
response
][][
][)(][][
2nuNnu
nuNnunh
NN
NNN
=
=
implicit
formulation
of impulse
response ( in terms
of x[n] and y[n] )
explicit
formulation
of impulse
response (= h[n] )
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lowpass filterexamples allpass filter (3)( )( ) NNj
NjN
NN
NN
zer
erz
z
zzH
=
=
11
)()(
46
exercises
1
prove that
the amplitude
characteristic
is flat
( i.e. H(f) = 1 for
all f );
2
compute
the phase
characteristic
for
f = 0 ,
1/4Ts
, 1/2Ts
, 3/4Ts
, 1/Ts ;
transfer function
frequency
response
( )
( )s
s
s
s
NTfjNj
NjNTfj
NTfjN
NNTfj
eer
ere
e
efH
=
=
2
2
2
2
11
)()(
(peel-off
the unit circle)
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lowpass filterexamples allpass filter (4)
47
NN
NN
z
zzH
=
1
)()(
transfer function
block diagram
Ts
z-1
1
Ts
z-1
Ts
z-1
(*)N
x[n]
X(z)
y[n]
Y(z)
x[n -N]
X(z).z-N
N
delay-units
Ts
z-1
N
N
delay-units
y[n -N]
Y(z).z-N
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lowpass
filterexamples
comb
filters
allpass filters
FIRfilters
IIRfilters
evaluation
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lowpass filterexamples FIR geom.seq.&lin.phase
=
=
+=
=
==Nn
n
nn
n
n znhznhzH0
][][)(
)(1
11)(
11
1
)1(1221
0
=
=++++==
++
++
=
=
zz
z
z
zzzzzzH
N
NNNNNN
Nn
n
nnL
49
linear
phase:
restrictions
on
h[n] 4 types
h[n] sampled
rectangular
pulse
| H(f) | periodic-sinc
(sin(Nx
/ Nsinx
)
arg{H(f)}
linear
function
of f
finite
geometric
sequence:
h[n] = n
, n = 0,1,,N ; 0 <
< 1
=
=
=Nn
n
nzzH
0
)(
special case: uniform weighting
of N successive
samples
N
1h[n]
1
exercise
ForN=4
and N=5, derive
z-plane, impulse
response,
frequency
response, and
block
schematic.
see SP04
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lowpass filterexamples FIR geometric sequence (1a)
50
)(1
11)(
11
1
)1(1
221
=
=++++=
++
++
zz
z
z
zzzzzH N
NNNN
NNL
nN
j
ez 12
+=
transfer function
complex Z-plane
1 -
impulse
response
2 -
transfer function
3 -
block
schematicNOTE
evaluation slides comb & IIR = FIR
N+1 zeros:
1 pole:1 pole
(N-fold):
N zeros:
1 pole
(N-fold):
z = 0
z =
(n=0,1,2,,N)
nNjez 12
+=
z = 0
(n=1,2,,N)
zero-pole
cancellation
for
z=
finite
geometric
sequence: h[n] = n
, n = 0,1,,N see SP04
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lowpass filterexamples FIR geometric sequence (1b)
51
][1
2rad
N+
n=2
0
n=1
n=N
O
O
O
n=3
O
Re(z)X .
N x
j.
j.Im(z)
transfer function
complex Z-plane)()(11
=
++
zz
zzH
N
NN
j
1
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lowpass filterexamples FIR geometric sequence (2)][]2[]1[][][
2
NnnnnnhN
++++= L
sss NTfjNTfjTfj eeefH ++++= 222221)( L
NNzzzzH
++++= L2211)(
52
transfer function
impulse
response
frequency
response
t = nTs
(t) .(t Ts
) N.(t NTs
)
0 NTsTs
maxima & minima ???
finite
geometric
sequence
h[n] = n
n = 0,1,,N
0 <
< 1
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lowpass filterexamples FIR geometric sequence (3a)sss NTfjNTfjTfj
eeefH
++++=
22222
1)(L
NNzzzzH
++++= L2211)(
53
transfer function
frequency
response
|H(f)|
f [Hz]1/NTs+1/2Ts-1/2Ts 0
H(z): N zeros
on
circle
|z|=
H(f): N notches
(dips)
on
unit circle
|z|=1
N notches
oddNif,for
evenNiffor1
1
21
21
221
1
ss
ss
NTN
NTN
NTN
T
N
f
f
+
+
=
==
=
+
minglobal
(1-N+1)/(1-) = global
max
for
f = 0
0 <
< 1
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54
transfer function
frequency
response
lowpass filterexamples FIR geometric sequence (3b)0 <
< 1
min
for
f = 1/2Ts
:
(1-8)/(1+)
=
1
1
:0for
8
fmax
7 notchesexample:
N = 7 (odd)
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55
transfer function
frequency
response
)1(1
1)0(
:0for1
1
1
1
1
1
1
)1(
:1)N,1,2,(nfor2-N
1
1
1
1
1
1
)1(
:N),1,2,(nfor1-N
21
11
2
1
11
2
21
)1(2
)1()1(21
21
21
1
2
1
1
2
21
)1(2
)1()1(
21
21
21
NN
Njn
Nj
N
Njn
Nj
jnjN
TTN
nj
TNTN
n
jN
s
s
nNj
N
nNj
njN
TTN
nj
TNTN
nj
N
s
s
H
f
eeee
ee
e
e
TN
nH
NT
nf
ee
e
e
e
TN
nH
NTnf
ss
ss
ss
ss
++++=
=
=
+=
=
=
+
+
=+
=
=
=
=
+
==
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
L
L
L
maximumglobal
maximalocal
(notches)minima
s
s
Tfj
TNfjN
e
efH
++
=
2
)1(21
1
1)(
lowpass filterexamples FIR geometric sequence (3c)1
turn resp. N+1
turns
on
circles
with
radii
resp. N+1
( circlein complex z-plane
with
radius
)
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lowpass filterexamples FIR geometric sequence (4a)
56
1
)1(1221
1
1
1)(
++
=++++= z
z
zzzzH
NNNN
L
Ts
z-1
y[n]
Y(z)
1
Ts
z-1
Ts
z-1
N2
transversal
structure
(tapped
delay
line; MA only)
x[n]X(z)
transfer function
block
diagram
x[n -N]X(z) . z-N
[N delay-units]
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lowpass filterexamples FIR geometric sequence (4b)
57
1
)1(1221
1
1
1)(
++
=++++= z
z
zzzzH
NNNN
L
Ts
z-1
y[n]
Y(z)
1
Ts
z-1
Ts
z-1
-
N+1 -
cascade of
MA structure
and
AR structure
x[n]
X(z)
[N+1
delay-units]
Ts
z-1
transfer function
block
diagram
correct:
IF
MA-structure
THEN
FIR-behaviour
incorrect: IF FIR-behaviourTHEN MA-structure
x[n (N+1)]
X(z).z-(N+1)
y[n -1]
Y(z).z-1
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lowpass filterexamples FIR linear phase (0a)
)()()()()()(
)()()()()()(
:proof
)()()()(odd)(
)()()()(even)(
:therevisiting
fHfHjfHfHjfHfH
fHjfHfHfhfhth
fHfHththth
fHfHththth
oddevenoddeven
oddevenoddeven
==+=
+=+=
==
==
propertiessymmetryFourier
=
=
+=
=
==Nn
n
nn
n
nznhznhzH
0
][][)(
58
H(f)
with
linear
phase which restrictions to h[n] ???
M.Hayes: pp.189+190
H.P.Hsu: pp.333+334
put restrictions
on
the (anti)symmetry
of
h[n]
!!!
1 -
impulse
response
2 -
transfer function
3 -
block
schematic
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lowpass filterexamples FIR linear phase (0b)
fNTffNTfff
eefHefHfHfH
nNhnhnhh
ss
NTfjfjfj s
~)(2)()(
)()()()(
][][][)symmetric"("even[n]
2)()(
====
==
59
So: h[n] (anti-)symmetric arg{H(f)} ~ f
fNTffNTfff
eeefHefHfHfH
nNhnhnhh
ss
NTfjfjjfj s
~)(2)()(
)()()()(
][][][)symmetric"-anti("odd[n]
2
2)()(
==
==
==
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low pass filterexamples FIR
linear
phase
(0c)
60
Only
digital
(FIR-)filters can
realize
an
exact linear
phase
!
4linearphase
FIRfilter
types
Neven
centre:n=
N/2
Nodd
centre:n=
(N
1)/2
,n=
(N+1)/2
h[n]even:
h[Nn]= h[n]
h[(N/2)m]= h[(N/2)+m]
h[0] = h[N]
h[1] = h[N 1]
....
h[N/2] = h[N/2]trivial
h[(N1)/2m]= h[(N+1)/2 +m]
h[0] = h[N]
h[1] = h[N 1]
....
h[(N1)/2] = h[(N+1)/2]
h[n]odd:
h[Nn]=
h[n]
h[(N/2)m]= h[(N/2)+m]
h[0] =
h[N]
h[1] =
h[N 1]
....
h[(N/2)1] =
h[(N/2)+1]
h[N/2] =
h[N/2] =0
h[(N1)/2m]= h[(N+1)/2 +m]
h[0] =
h[N]
h[1] =
h[N 1]
....
h[(N1)/21]=
h[(N+1)/2+1]
h[(N1)/2] =
h[(N+1)/2]
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low pass filterexamples FIR
linear
phase
(1a)
N
nn
zNnhznhznhnh
znhzHznhzH
=
=
+
++++=
===
][]2[]1[][
][)(][)(
21L
Nn
0n
th :FIRorderN
61
H(z) = degree-N
polynomial
in z equation H(z)=0 has N solutions:
H(z) = 0 z = z0, z1, , zN-1
transfer function
complex Z-plane
=
=
=
=
=
=
===1
000
)(][][)(Nl
l
l
NNn
n
nNNNn
n
nzzzznhzznhzH
location
of the N zeros
z0
,z1
, , zN-1
in the complex z-plane:
-
not
necessarily
equidistant
-
not
necessarily
on
(unit) circle;
-
appearing
as
complex conjugate
pairs (w.r.t. real
axis)
and/or
as reciprocal
pairs (w.r.t. unit circle);-
appearing
as conjugate
reciprocal
quadruples;
N+1 terms N factors
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low pass filterexamples FIR
linear
phase
(1b)
NzNnhznhznhnhzH ++++= ][]2[]1[][)( 21 L
62
transfer function
complex Z-plane
][0 rad
r0
O
O
O
r0
O
Re(z)X .
N x
j.r
j.Im(z)
j
1
compare with geometric-sequence FIR!
O
O
N
zeros
1/r0
][0 rad
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low pass filterexamples FIR
linear
phase
(2a)
63
transfer function
impulse
response
N evenN odd
h[n]
even
h[n]
odd
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low pass filterexamples FIR
linear
phase
(2b)
64
transfer function
impulse
response
frequency
response
h[n]even
h[n]
odd
N odd N even
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low pass filterexamples FIR
linear
phase
(3a)
65
amplitude:
1
main
lobe,
N-2
side
lobes
of unequal
width;
phase: linear
with
f ( incl. phase-jumps
of 2
[rad] )
transfer function
frequency
response
=
=
=Nn
n
nTfj senhfH0
..2][)(
h[n] =1
, for
n=0,1,,N
= 0 , for other n
N oddN even
example
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low pass filterexamples FIR
linear
phase
(3b)
66
example: N = 6 z-plane: 3 x 2 zeroscomplex conjugate
pairs (w.r.t. real
axis)
reciprocal
pairs (w.r.t. unit circle)
transfer function
frequency
response
=
=
=Nn
n
nTfj senhfH0
..2][)(
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low pass filterexamples FIR
linear
phase
(4)
67
=
=
=Nn
n
nznhzH
0
][)(
= MA-part of
- direct form I & II- transposed form I & II
transfer function
block
diagram
Ts
z-1
y[n]
Y(z)
h[n2]
Ts
z-1
Ts
z-1
transversal
structure
(tapped
delay
line; MA only)
x[n]
X(z)
x[n -N]
X(z) . z-N
[N delay-units]
h[n1] h[nN]h[n]
compare with geometric-sequence FIR!
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lowpass
filterexamples
comb
filters
allpass filters
FIRfilters
IIRfilters
evaluation
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low pass filterexamples IIR order1&order2(1)
=
=+++=
z
z
zzzzH
1
221
1
11)( L
69
linear
digital system: H(z) = T(z) / N(z)
IIR filter: T(z) = 1, N(z) = order-M
polynomial
in z
M poles, within unit circle
M = 1
1 pole: real, pos./neg. ; [complex]M = 2
2 poles: 2 x real
; complex conjugate
pair; xxx
M = 1
M = 2)1)(1(
11)(
1
2
1
1
2
2
1
10
=
++=
zpzpzazaazH
M.H. Hayes, Dig. Signal
Processing:
CH.2 pp. 79 (S.P. 2.10)
M.H. Hayes, Dig. Signal
Processing:
CH.8 pp. 185+186CH.2 pp. 79 (S.P. 2.10)
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IIR order1&order2(2)
70
IIR-filter:
order 1 1 pole-
1 x real
& positive
-
1 x real
& negative
order 2 2 poles-
2 x real
-
1 x real
& 2-fold
-
2 complex conjugate
transfer function
impulse
response
complex Z-plane
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low pass filterexamples IIR order1(1a)
==+++=
z
z
zzzzH 1221
1
1
1)(L
71
see SP04 transfer functioncomplex Z-plane
NOTE evaluation slides comb & IIR = FIR
infinite
geometric
sequence: h[n] = n
, n = 0,1,,N
1 -
impulse
response
2 -
transfer function
3 -
block
schematic
1 zero: z = 0
1 pole: z = real
stability -1 < < +1
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p p IIR order1(1b)
=
=
z
z
zzH
11
1)(
72
0
stability: |z| = r < 1
ORe(z)X
X = r . exp[j2.f0
.Ts
]
1
1||
:circleunit2
=
=
z
ez sTfj
j
single real
pole:
single complex pole:
transfer function
complex Z-plane
sTr
=
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p p IIR order1(1c)
=
=
z
z
zzH
11
1)(
73
( )( ) ( ) ( )( ) ( ) ( ){ }( ) ( ) ( ){ } LL
L
L
=+++++
+++++=
=++++=
==
====
33221
33221
3332221
1
3sin2sinsin0
3cos2coscos1
1
1
1)(
)2argandwith(
zrzrzr
zrzrzr
zerzerzer
zerzH
Trer
jjj
j
s
j
j
f0
single complex
pole
z = |H(f)|: not even; single notch at f0 = /(2Ts) [Hz]
h[n]: complex (I-part and Q-part)
= S.S.B.-system outside scope of Theme 8
transfer function
complex Z-plane
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p p IIR order1(2a)L+++=
]2[]1[][][
2
nnnnh
74
real: |H(f)| not even
0 <
< 1 h[n] monotonously decreasing
exp(2.f0
.Ts
) = +1 f0 = 0 [d.c.]
-1 <
< 0
h[n] alternatingly decreasingexp(j2.f0
.Ts
) = -1 f0 = 1/(2Ts) [Nyquist rate]
complex: |H(f)| not even
= r.exp(j), 0 < r < 1 h[n] complex: oscillating & decreasing
exp(j2.f0
.Ts
) = exp(j)
f0 = /(2Ts) = single notch
1,2
= r.exp(j), 0 < r < 1 h[h] real: oscillating & decreasing IIR, order 2
cos(2.f0
.Ts
) = cos() f0 = /(2Ts) = complexconjugate pair
of notches
transfer function
impulse
response
1
221
1
11)(
=+++=
zzzzH
L
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p p IIR order1(2b)
]1[][][]2[]1[][][2
+=+++= nynxnynxnxnxny L
75
transfer function
impulse
response
( )1
221
1
)()(1)(
=+++=
z
zXzXzzzY
L
< 1 stable
> 1 instable
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p p IIR order1(2c)
76
transfer function
impulse
response
frequency
response
( )1
221
1
)()(1)(
=+++=
z
zXzXzzzY
L
lowpass filterexamples
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p p IIR order1(3a)
=
=
=+++=
z
z
z
zzzH
1
221
1
1
1)( L
=
=
=+++=
s
s
s
ss
Tfj
Tfj
Tfj
TfjTfj
e
e
e
eefH
2
2
2
2222
1
1
1)( L
77
transfer function
frequency
response
construction rule:
H(z) = bi-linear function of z ifzon circle, thenH(z)on circle
sTfj
ez
=2
lowpass filterexamples
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p p IIR order1(3b)
78
transfer function
frequency
response
lowpass filterexamples
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p p IIR order1(4)1
221
1
11)(
=+++=
zzzzH
L
79
Ts
z-1
y[n]
Y(z)
x[n]
X(z)
-
transfer function
block
schematic
y[n -N]
Y(z).z-NAR structure
lowpass filterexamples
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p p IIR order2(1a)
2
2
11
0
20
2
11
2,1 222
4
:poles a
aa
a
aaaa
p
=
=80
)()()()(1
1
)1)(1(11)(
2121
2
2
2
21
1
21
1
2
1
1
2
2
1
10
ppzppz
z
zppzpp
zpzpzazaazH
++=
++=
=
=++
=
a0
, a1
, a2
: all real
a0
1
a1
, a2 : bounded
by
stability
criteria
transfer function
complex Z-plane
1 -
impulse
response
2 -
transfer function
3 -
block
schematic
lowpass filterexamples
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IIR order2(1b)
221)cos2(1
1)(
+=
zrzrzH
cos2;
22222
12
12
1122222
211
2
21211
21
21
2
21
21
12
21
1212
2
1
212
2
1122
2
1112
2
1
=+=====
+
=
=
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IIR order2(1c)
21
2
aa
r == 21
2
aa
r ==
82
sTf
=
2
0Re(z)X
1
j
j.Im(z)
Re(z)
j.Im(z)
j
1X
X
X
0
case A ( B):two real poles
case C ( B):pair ofcomplexconjugate poles
transfer function
complex Z-plane)()()(
2121
2
2
21
2
0
2
ppzppz
z
azaza
zzH
++=
++=
p1p1
p2
p2
lowpass filterexamples
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IIR order2(2)( )]2[]1[][
1)(
210
= nxanxanxa
ny
83
transfer function
impulse
response
2
2
1
10
)()( ++
=zazaa
zXzY
a0 1
a1 cos
a2 a
[rad]
= 2.f0
.Ts1/f0
.= 2.Ts
/
lowpass filterexamples
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IIR order2(3)
84
)()()(
2121
2
2
21
2
0
2
ppzppz
z
azaza
zzH
++=
++= transfer function
frequency
response
lowpass filterexamples
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IIR order2(4)
85
y[n]
Y(z)
-a1
-a2
Ts
z-1
Ts
z
-1
x[n]
X(z)
Ts
z
-1
Ts
z-1
-a1
-a2
x[n]
X(z)
y[n]
Y(z)
AR-structure, direct form AR-structure,
transposed
form
2r.cos
r2
exercise: how to make this filter behave like an oscillator?
transfer function
block
schematic
1;1
)( 022
1
10
=++
=
azazaa
zH
low pass filter examples
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lowpass
filterexamples
comb
filters
allpass filters
FIRfilters
IIRfilters evaluation
evaluation
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comparison:IIRversusFIR(1)
87
filteringperformance:passbandstopband
range
transitionband rolloffphase linearity stability &settling time effectoffeedbackon signal error accumulation
sensitivity
tocoefficientvariations
filterorder(number ofpoles) size ofblock schematic(amount ofcircuitry)
evaluation
comparison: IIR versus FIR (2)
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comparison:IIRversusFIR(2)
88
evaluation
comb filter & IIR filter = FIR filter (1)
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{ }{ }L
LL
L
++++++=
=++++=
=
==
++
)1(1
)1(1
)1(1221
1
1
1
11
NNNN
NNNN
NN
NN
zz
zz
zzz
zz
(z)H(z)H(z)H IIR1comb1FIRN
NN z= 1(z)Hcomb1
L+++=
=
221
11
11 zz
z
(z)HIIR1
combfilter
&IIRfilter
=FIRfilter
(1)
evaluation
comb filter & IIR filter = FIR filter (2)
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the cascade of a type 1
comb-filter
of order N with
an
IIR-filter
of order 1
yields
a finite
geometric
sequence
FIR-filter
of order N
{ }{ }L
LL
L
+++
+++++==++++=
==
+
)]1([][
][)]1([]1[][)]1([]2[]1[][
][][][
1
1
12
11
NnNn
NnNnnnNnnnn
nhnhnh
NN
NN
N
IIRcombFIRN
][][][1 NnnnhN
comb =
L
+++= ]2[]1[][][2
1 nnnnhIIR
order-N
FIR
= order-1 IIR
minus N-delayed
order-1 IIR
combfilter
&IIRfilter
=FIRfilter
(2)
evaluation
comb filter & IIR filter = FIR filter (3)
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t = nTs0 NTs
(t)
combfilter
&IIRfilter
=FIRfilter
(3)
0 NTs
N.(t NTs
)
0
NTs
N.(t NTs
)
(t)
t = nTs
t = nTs
IIR + delayed IIR(parallel system) comb * IIR(cascaded system)
N.(t NTs
)
.(t Ts
)
t = nTs
(t).(t Ts
)
*
t = nTs
(t) .(t Ts
) N.(t NTs
)
0 NTsTs
evaluation
comb filter & IIR filter = FIR filter (4)
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Ts
z-1
y[n]
Y(z)1
Ts
z-1
Ts
z-1
-
N
-
x[n]
X(z)Ts
z-1
Ts
z-1
y[n]
Y(z)
1
Ts
z-1
Ts
z-1
N-1
2
x[n]
X(z)
combfilter
&IIRfilter
=FIRfilter
(4)
=
x[n -N]
X(z).z-N
y[n -1]
Y(z).z-1
x[n -(N-1)]
X(z).z-(N-1)
exercises
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exercises
see
reference
book: Solved
Problems