digital filter banks - the catholic university of...

Copyright © 2001, S. K. Mitra1

Digital Filter BanksDigital Filter Banks• The digital filter bank is set of bandpass

filters with either a common input or asummed output

• An M-band analysis filter bank is shownbelow


Digital Filter BanksDigital Filter Banks

• The subfilters in the analysis filterbank are known as analysis filters

• The analysis filter bank is used todecompose the input signal x[n] into a set ofsubband signals with each subbandsignal occupying a portion of the originalfrequency band

)(zHk

][nvk



• An L-band synthesis filter bank is shownbelow

• It performs the dual operation to that of theanalysis filter bank



• The subfilters in the synthesis filterbank are known as synthesis filters

• The synthesis filter bank is used to combinea set of subband signals (typicallybelonging to contiguous frequency bands)into one signal y[n] at its output

)(zFk

][nvk^


Uniform Digital Filter BanksUniform Digital Filter Banks

• A simple technique to design a class offilter banks with equal passband widths isoutlined next

• Let represent a causal lowpass digitalfilter with a real impulse response :

• The filter is assumed to be an IIRfilter without any loss of generality

)(0 zH

∑∞−∞=

−= nnznhzH ][)( 00

][0 nh

)(0 zH



• Assume that has its passband edgeand stopband edge around π/M, where Mis some arbitrary integer, as indicated below

)(0 zH

ωπ0 2π

pω sω

Mπ

pωsω



• Now, consider the transfer functionwhose impulse response is given by

where we have used the notation• Thus,

)(zHk][nhk

,][][][ 0/2

0kn

MMnkj

k Wnhenhnh −π ==

MjM eW /2π−=10 −≤≤ Mk

( ) ,][][)( 0∑∑ ∞−∞=

−∞−∞=

− == nnk

Mnn

kk zWnhznhzH10 −≤≤ Mk



• i.e.,

• The corresponding frequency response isgiven by

• Thus, the frequency response of isobtained by shifting the response ofto the right by an amount 2πk/M

),()( 0k

Mk zWHzH = 10 −≤≤ Mk

),()( )/2(0

Mkjjk eHeH π−ωω = 10 −≤≤ Mk

)(zHk)(0 zH


Uniform Digital Filter BanksUniform Digital Filter Banks• The responses of , , . . . ,

are shown below)(zHk )(zHk )(zHk



• Note: The impulse responses are, ingeneral complex, and hence doesnot necessarily exhibit symmetry withrespect to ω = 0

• The responses shown in the figure of theprevious slide can be seen to be uniformlyshifted version of the response of the basicprototype filter

][nhk|)(| ωj

k eH

)(0 zH


Uniform Digital Filter BanksUniform Digital Filter Banks• The M filters defined by

could be used as the analysis filters in theanalysis filter bank or as the synthesis filtersin the synthesis filter bank

• Since the magnitude responses of all Mfilters are uniformly shifted version of thatof the prototype filter, the filter bankobtained is called a uniform filter bank

),()( 0k

Mk zWHzH = 10 −≤≤ Mk


Uniform DFT Filter BanksUniform DFT Filter BanksPolyphase Implementation• Let the prototype lowpass transfer function

be represented in its M-band polyphaseform:

where is the -th polyphasecomponent of :

∑ −=

−= 100

M MzEzzH l ll )()(

)(zH0

l

,][][)( ∑∑ ∞=

−∞=

− +== 0 00 nn

nn znMhznezE lll

)(zEl

10 −≤≤ Ml


Uniform DFT Filter BanksUniform DFT Filter Banks

• Substituting z with in the expressionfor we arrive at the M-band polyphasedecomposition of :

• In deriving the last expression we have usedthe identity

)(zH0

kMzW

∑ −=

−−= 10

M kMM

MkMk WzEWzzH l l

ll )()(

)(zHk

1010 −≤≤= ∑ −=

−− MkzEWzM MkMl l

ll ,)(

1=kMMW



• The equation on the previous slide can bewritten in matrix form as

].... )([)(

kMM

kM

kMk WWWzH 121 −−−−=

−−−

−

−

)(

)(

)(

)(

)(

MM

M

M

M

M

zEz

zEzzEz

zE

11

22

11

0

10 −≤≤ Mk

...


Uniform DFT Filter BanksUniform DFT Filter Banks• All M equations on the previous slide can

be combined into one matrix equation as

• In the above D is the DFT matrixM D 1−

MM ×

−−−

−

−

)(

)(

)(

)(

)(

MM

M

M

M

M

zEz

zEzzEz

zE

11

22

11

0

...

=

−−−−−−

−−−−

−−−−

− 21121

1242

121

1

210

1

11

1111

)()()(

)(

)(

)(

)()()(

MM

MM

MM

MMMM

MMMM

M WWW

WWWWWW

zH

zHzHzH

......

......

...

...

...

...

......


Uniform DFT Filter BanksUniform DFT Filter Banks• An efficient implementation of the M-band

uniform analysis filter bank, morecommonly known as the uniform DFTanalysis filter bank, is then as shown below



• The computational complexity of an M-banduniform DFT filter bank is much smaller thanthat of a direct implementation as shownbelow



• For example, an M-band uniform DFTanalysis filter bank based on an N-tapprototype lowpass filter requires a total of

multipliers• On the other hand, a direct implementation

requires NM multipliers

NMM +22log


Uniform DFT Filter BanksUniform DFT Filter Banks• Following a similar development, we can

derive the structure for a uniform DFTsynthesis filter bank as shown below

Type I uniform DFT Type II uniform DFTsynthesis filter bank synthesis filter bank



• Now can be expressed in terms of

• The above equation can be used todetermine the polyphase components of anIIR transfer function

−−−

−

−

)(

)(

)(

)(

)(

MM

M

M

M

M

zEz

zEzzEz

zE

11

22

11

0

M1=

− )(

)()()(

zH

zHzHzH

M 1

210

D......

)( Mi zE

)(zH0


Nyquist FiltrsNyquist Filtrs• Under certain conditions, a lowpass filter

can be designed to have a number of zero-valued coefficients

• When used as interpolation filters thesefilters preserve the nonzero samples of theup-sampler output at the interpolator output

• Moreover, due to the presence of thesezero-valued coefficients, these filters arecomputationally more efficient than otherlowpass filters of same order


LLthth-Band Filters-Band Filters• These filters, called the Nyquist filters or

Lth-band filters, are often used in single-rateand multi-rate signal processing

• Consider the factor-of-L interpolator shownbelow

• The input-output relation of the interpolatorin the z-domain is given by

L][nx ][ny)(zH][nxu

)()()( LzXzHzY =


LLthth-Band Filters-Band Filters

• If H(z) is realized in the L-band polyphaseform, then we have

• Assume that the k-th polyphase componentof H(z) is a constant, i.e., :

∑ −=

−= 10 )()( L

iL

ii zEzzH

α=)(zEk

)(...)()()( 1)1(

11

0L

kkLL zEzzEzzEzH −−−− +++=

)(...)( 1)1(

1)1( L

LLL

kk zEzzEz −

−−+

+− +++


LLthth-Band Filters-Band Filters• Then we can express Y(z) as

• As a result,

• Thus, the input samples appear at the outputwithout any distortion for all values of n,whereas, in-between output samplesare determined by interpolation

∑−

=

−− +α=1

0)()()()(

LLLLk zXzEzzXzzY

ll

l

k≠l

][][ nxkLny α=+

)1( −L


LLthth-Band Filters-Band Filters• A filter with the above property is called a

Nyquist filter or an Lth-band filter• Its impulse response has many zero-valued

samples, making it computationallyattractive

• For example, the impulse response of anLth-band filter for k = 0 satisfies thefollowing condition

=][Lnh otherwise,0

0, =α n


LLthth-Band Filters-Band Filters• Figure below shows a typical impulse

response of a third-band filter (L = 3)

• Lth-band filters can be either FIR or IIRfilters


LLthth-Band Filters-Band Filters• If the 0-th polyphase component of H(z) is a

constant, i.e., then it can be shownthat

(assuming α = 1/L)• Since the frequency response of is

the shifted version of ,the sum of all of these L uniformly shiftedversions of add up to a constant

α=)(0 zE

∑ −= =α=1

0 1)(Lk

kL LzWH

)( kLzWH

)( )/2( LkjeH π−ω )( ωjeH

)( ωjeH


Half-Band FiltersHalf-Band Filters

• An Lth-band filter for L = 2 is called a half-band filter

• The transfer function of a half-band filter isthus given by

with its impulse response satisfying)()( 2

11 zEzzH −+α=

=]2[ nh otherwise,0

0, =α n



• The condition

reduces to (assuming α = 0.5)

• If H(z) has real coefficients, then

• Hence

)()( 21

1 zEzzH −+α=

1)()( =−+ zHzH

)()( )( ω−πω =− jj eHeH

1)()( )( =+ ω−πω jj eHeH



• and add upto 1 for all θ

• Or, in other words, exhibits asymmetry with respect to the half-bandfrequency π/2, hence the name “half-bandfilter”

)( )2/( θ−πjeH )( )2/( θ+πjeH

)( ωjeH


Half-Band FiltersHalf-Band Filters• Figure below illustrates this symmetry for a

half-band lowpass filter for which passbandand stopband ripples are equal, i.e.,and passband and stopband edges aresymmetric with respect to π/2, i.e.,

sp δ=δ

π=ω+ω sp



• Attractive property: About 50% of thecoefficients of h[n] are zero

• This reduces the number of multiplicationsrequired in its implementation significantly

• For example, if N = 101, an arbitrary Type 1FIR transfer function requires about 50multipliers, whereas, a Type 1 half-bandfilter requires only about 25 multipliers



• An FIR half-band filter can be designedwith linear phase

• However, there is a constraint on its length• Consider a zero-phase half-band FIR filter

for which , with• Let the highest nonzero coefficient be h[R]

][*][ nhnh −α= 1|| =α



• Then R is odd as a result of the condition

• Therefore R = 2K+1 for some integer K• Thus the length of h[n] is restricted to be of

the form 2R+1 = 4K+3 [unless H(z) is aconstant]

=]2[ nh otherwise,0

0, =α n


Design of Linear-PhaseDesign of Linear-PhaseLLthth-Band Filters-Band Filters

• A lowpass linear-phase Lth-band FIR filtercan be readily designed via the windowedFourier series approach

• In this approach, the impulse responsecoefficients of the lowpass filter are chosenas where is theimpulse response of an ideal lowpass filterwith a cutoff at π/L and w[n] is a suitablewindow function

][][][ nwnhnh LP ⋅= ][nhLP



• Now, the impulse response of an ideal Lth-band lowpass filter with a cutoff atis given by

• It can be seen from the above that

∞≤≤∞−ππ= n

nLnnhLP ,)/sin(][

Lc /π=ω

...,2,for0][ LLnnhLP ±±==



• Hence, the coefficient condition of the Lth-band filter

is indeed satisfied• Hence, an Lth-band FIR filter can be

designed by applying a suitable windoww[n] to

=][Lnh otherwise,0

0, =α n

][nhLP



• There are many other candidates for Lth-band FIR filters

• Program 10_8 can be used to design an Lth-band FIR filter using the windowed Fourierseries approach

• The program employs the Hammingwindow

• However, other windows can also be used



• Figure below shows the gain response of ahalf-band filter of length-23 designed usingProgram 10_8

0 0.2 0.4 0.6 0.8 1

-80

-60

-40

-20

0

ω/π

Gai

n, d

B



• The filter coefficients are given by

• As expected, h[n] = 0 for

;0]10[]10[;002315.0]11[]11[ ==−−==− hhhh;0]8[]8[;005412.0]9[]9[ ==−==− hhhh

;0]6[]6[;001586.0]7[]7[ ==−−==− hhhh

;0]4[]4[;003584.0]5[]5[ ==−==− hhhh

;0]2[]2[;089258.0]3[]3[ ==−−==− hhhh;5.0]0[;3122379.0]1[]1[ ===− hhh

10,8,6,4,2 ±±±±±=n

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