12: polyphase filters - imperial college london · 12: polyphase filters 12: polyphase filters...

12: Polyphase Filters


• Heavy Lowpass filtering

• Maximum DecimationFrequency

• Polyphase decomposition

• Downsampled PolyphaseFilter

• Polyphase Upsampler

• Complete Filter

• UpsamplerImplementation

• DownsamplerImplementation

• Summary

DSP and Digital Filters (2016-9045) Polyphase Filters: 12 – 1 / 10

Heavy Lowpass filtering







• Complete Filter



• Summary


Filter Specification:Sample Rate: 20 kHzPassband edge: 100 Hz (ω1 = 0.03)Stopband edge: 300 Hz (ω2 = 0.09)








• Complete Filter



• Summary


Filter Specification:Sample Rate: 20 kHzPassband edge: 100 Hz (ω1 = 0.03)Stopband edge: 300 Hz (ω2 = 0.09)Passband ripple: ±0.05 dB (δ = 0.006)








• Complete Filter



• Summary


Filter Specification:Sample Rate: 20 kHzPassband edge: 100 Hz (ω1 = 0.03)Stopband edge: 300 Hz (ω2 = 0.09)Passband ripple: ±0.05 dB (δ = 0.006)Stopband Gain: −80 dB (ǫ = 0.0001)








• Complete Filter



• Summary



This is an extreme filter because the cutoff frequency is only 1% of theNyquist frequency.








• Complete Filter



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Symmetric FIR Filter:








• Complete Filter



• Summary




Symmetric FIR Filter:Design with Remez-exchange algorithmOrder = 360

0 1 2 3

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0

M=360

ω (rad/s)








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Symmetric FIR Filter:Design with Remez-exchange algorithmOrder = 360

0 1 2 3

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0

M=360

ω (rad/s)0 0.05 0.1

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0

ω1

ω2

ω (rad/s)

Maximum Decimation Frequency







• Complete Filter



• Summary


If a filter passband occupies only a small fractionof [0, π], we can downsample then upsamplewithout losing information.

0 1 2 3-60

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0

ω1

ω2 ω








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• Summary



Downsample: aliased components at offsets of2πK

are almost zero because of H(z)

0 1 2 3-60

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0

ω1

ω2 ω








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• Summary




are almost zero because of H(z)Upsample: Images spaced at 2π

Kcan be

removed using another low pass filter

0 1 2 3-60

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0

ω1

ω2 ω

0 1 2 3-60

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0ω = 2π /4 K = 4

ω








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Kcan be

removed using another low pass filterTo avoid aliasing in the passband, we need

2πK

− ω2 ≥ ω1 ⇒ K ≤ 2πω1+ω2

0 1 2 3-60

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0

ω1

ω2 ω

0 1 2 3-60

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0ω = 2π /4 K = 4

ω

0 1 2 3-60

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0ω = 2π /7 K = 7

ω








• Complete Filter



• Summary





Kcan be


2πK

− ω2 ≥ ω1 ⇒ K ≤ 2πω1+ω2

0 1 2 3-60

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0

ω1

ω2 ω

0 1 2 3-60

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0ω = 2π /4 K = 4

ω

0 1 2 3-60

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0ω = 2π /7 K = 7

ω

Centre of transition band must be ≤ intermediate Nyquist freq, πK








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• Summary





Kcan be


2πK

− ω2 ≥ ω1 ⇒ K ≤ 2πω1+ω2

0 1 2 3-60

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0

ω1

ω2 ω

0 1 2 3-60

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0ω = 2π /4 K = 4

ω

0 1 2 3-60

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0ω = 2π /7 K = 7

ω


We must add a lowpass filter to remove the images:








• Complete Filter



• Summary





Kcan be


2πK

− ω2 ≥ ω1 ⇒ K ≤ 2πω1+ω2

0 1 2 3-60

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0

ω1

ω2 ω

0 1 2 3-60

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0ω = 2π /4 K = 4

ω

0 1 2 3-60

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0ω = 2π /7 K = 7

ω


We must add a lowpass filter to remove the images:

Passband noise = noise floor at output of H(z) plus 10 log10 (K − 1) dB.

Polyphase decomposition







• Complete Filter



• Summary


For our filter: original Nyquist frequency = 10 kHz and transition bandcentre is at 200 Hz so we can use K = 50.








• Complete Filter



• Summary



We will split H(z) into K filters each of order R − 1.








• Complete Filter



• Summary



We will split H(z) into K filters each of order R − 1. For convenience,assume M + 1 is a multiple of K (else zero-pad h[n]).








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• Summary




Example: M = 399, K = 50








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• Summary




Example: M = 399, K = 50⇒ R = M+1K

= 8








• Complete Filter



• Summary




Example: M = 399, K = 50⇒ R = M+1K

= 8

H(z) =∑M

m=0 h[m]z−m








• Complete Filter



• Summary




Example: M = 399, K = 50⇒ R = M+1K

= 8

H(z) =∑M

m=0 h[m]z−m

=∑K−1

m=0 h[m]z−m +∑K−1

m=0 h[m+K]z−(m+K) + · · · [R terms]








• Complete Filter



• Summary




Example: M = 399, K = 50⇒ R = M+1K

= 8

H(z) =∑M

m=0 h[m]z−m

=∑K−1

m=0 h[m]z−m +∑K−1

m=0 h[m+K]z−(m+K) + · · · [R terms]

=∑R−1

r=0

∑K−1m=0 h[m+Kr]z−m−Kr








• Complete Filter



• Summary




Example: M = 399, K = 50⇒ R = M+1K

= 8

H(z) =∑M

m=0 h[m]z−m

=∑K−1

m=0 h[m]z−m +∑K−1

m=0 h[m+K]z−(m+K) + · · · [R terms]

=∑R−1

r=0


=∑K−1

m=0 z−m

∑R−1r=0 hm[r]z−Kr

where hm[r] = h[m+Kr]

=∑K−1

m=0 z−mHm

(

zK)








• Complete Filter



• Summary




Example: M = 399, K = 50⇒ R = M+1K

= 8

H(z) =∑M

m=0 h[m]z−m

=∑K−1

m=0 h[m]z−m +∑K−1

m=0 h[m+K]z−(m+K) + · · · [R terms]

=∑R−1

r=0


=∑K−1

m=0 z−m



=∑K−1

m=0 z−mHm

(

zK)

Example: M = 399, K = 50, R = 8h3[r] = [h[3], h[53], · · · , h[303], h[353]]








• Complete Filter



• Summary




Example: M = 399, K = 50⇒ R = M+1K

= 8

H(z) =∑M

m=0 h[m]z−m

=∑K−1

m=0 h[m]z−m +∑K−1

m=0 h[m+K]z−(m+K) + · · · [R terms]

=∑R−1

r=0


=∑K−1

m=0 z−m



=∑K−1

m=0 z−mHm

(

zK)

Example: M = 399, K = 50, R = 8h3[r] = [h[3], h[53], · · · , h[303], h[353]]

This is a polyphase implementation of the filter H(z)

Downsampled Polyphase Filter







• Complete Filter



• Summary


H(z) is low pass so we downsampleits output by K without aliasing.








• Complete Filter



• Summary



The number of multiplications per inputsample is M + 1 = 400.








• Complete Filter



• Summary




Using the Noble identities, we can movethe resampling back through the addersand filters. Hm(zK) turns into Hm(z)at a lower sample rate.








• Complete Filter



• Summary





We still perform 400 multiplications butnow only once for every K inputsamples.








• Complete Filter



• Summary





We still perform 400 multiplications butnow only once for every K inputsamples.

Multiplications per input sample = 8 (down by a factor of 50 ,) but v[n] hasthe wrong sample rate (/).

Polyphase Upsampler







• Complete Filter



• Summary


To restore sample rate: upsample andthen lowpass filter to remove images

Polyphase Upsampler







• Complete Filter



• Summary



We can use the same lowpass filter,H(z), in polyphase form:∑K−1

m=0 z−m


Polyphase Upsampler







• Complete Filter



• Summary




m=0 z−m


This time we put the delay z−m afterthe filters.

Polyphase Upsampler







• Complete Filter



• Summary




m=0 z−m


This time we put the delay z−m afterthe filters.Multiplications per output sample = 400

Polyphase Upsampler







• Complete Filter



• Summary




m=0 z−m



Using the Noble identities, we can movethe resampling forwards through thefilters. Hm(zK) turns into Hm(z) at alower sample rate.

Polyphase Upsampler







• Complete Filter



• Summary




m=0 z−m



Using the Noble identities, we can movethe resampling forwards through thefilters. Hm(zK) turns into Hm(z) at alower sample rate.

Multiplications per output sample = 8(down by a factor of 50 ,).

Complete Filter







• Complete Filter



• Summary


The overall system implements:

Complete Filter







• Complete Filter



• Summary


The overall system implements:Need an extra gain of K to compensate for the downsampling energy loss.

Complete Filter







• Complete Filter



• Summary



Filtering at downsampled rate requires 16 multiplications per input sample(8 for each filter). Reduced by K

2 from the original 400.

Complete Filter







• Complete Filter



• Summary





H(ejω) reaches −10 dB at thedownsampler Nyquist frequency of π

K.

0 0.05 0.1

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0

ω1

ω2π/50

ω (rad/s)

Complete Filter







• Complete Filter



• Summary






K.

Spectral components > πK

will be aliaseddown in frequency in V (ejω).

0 0.05 0.1

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0

ω1

ω2π/50

ω (rad/s)

Complete Filter







• Complete Filter



• Summary






K.



For V (ejω), passband gain (blue curve)follows the same curve as X(ejω).

0 0.05 0.1

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0

ω1

ω2π/50

ω (rad/s)

0 1 2 3-80

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0

ω1

ω (downsampled)

Complete Filter







• Complete Filter



• Summary






K.



For V (ejω), passband gain (blue curve)follows the same curve as X(ejω).Noise arises from K aliased spectralintervals.

0 0.05 0.1

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0

ω1

ω2π/50

ω (rad/s)

0 1 2 3-80

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0

ω1

ω (downsampled)

Complete Filter







• Complete Filter



• Summary






K.



For V (ejω), passband gain (blue curve)follows the same curve as X(ejω).Noise arises from K aliased spectralintervals.Unit white noise in X(ejω) gives passbandnoise floor at −69 dB (red curve) eventhough stop band ripple is below −83 dB(due to K − 1 aliased stopband copies).

0 0.05 0.1

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0

ω1

ω2π/50

ω (rad/s)

0 1 2 3-80

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0

ω1

ω (downsampled)

Upsampler Implementation







• Complete Filter



• Summary


We can represent the upsamplercompactly using a commutator.Sample y[n] comes from Hk(z)where k = n mod K .

[“@f ” indicates the sample rate]








• Complete Filter



• Summary




H0(z) comprises a sequence of7 delays, 7 adders and 8 gains.








• Complete Filter



• Summary





We can share the delays betweenall 50 filters.








• Complete Filter



• Summary






We can also share the gains andadders between all 50 filters anduse commutators to switch thecoefficients.








• Complete Filter



• Summary






We can also share the gains andadders between all 50 filters anduse commutators to switch thecoefficients.

We now need 7 delays, 7 adders and 8 gains for the entire filter.

Downsampler Implementation







• Complete Filter



• Summary


We can again use a commutator.The outputs from all 50 filters areadded together to form v[i].








• Complete Filter



• Summary



We use the transposed form ofHm(z) because this will allow usto share components.








• Complete Filter



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We can sum the outputs of thegain elements using anaccumulator which sums blocksof K samples.

w[i] =∑

K−1

r=0u[Ki− r]








• Complete Filter



• Summary





Now we can share all thecomponents and usecommutators to switch the gaincoefficients.

w[i] =∑

K−1

r=0u[Ki− r]








• Complete Filter



• Summary





Now we can share all thecomponents and usecommutators to switch the gaincoefficients.

We need 7 delays, 7 adders, 8gains and 8 accumulators in total.

w[i] =∑

K−1

r=0u[Ki− r]

Summary







• Complete Filter



• Summary


• Filtering should be performed at the lowest possible sample rate◦ reduce filter computation by K◦ actual saving is only K

2 because you need a second filter

◦ downsampled Nyquist frequency ≥ max (ωpassband) +∆ω2

Summary







• Complete Filter



• Summary





• Polyphase decomposition: split H(z) as∑K−1

m=0 z−mHm(zK)

◦ each Hm(zK) can operate on subsampled data◦ combine the filtering and down/up sampling

Summary







• Complete Filter



• Summary






m=0 z−mHm(zK)


• Noise floor is higher because it arises from K spectral intervals thatare aliased together by the downsampling.

Summary







• Complete Filter



• Summary






m=0 z−mHm(zK)



• Share components between the K filters◦ multiplier gain coefficients switch at the original sampling rate◦ need a new component: accumulator/downsampler (K : Σ)

Summary







• Complete Filter



• Summary






m=0 z−mHm(zK)



• Share components between the K filters◦ multiplier gain coefficients switch at the original sampling rate◦ need a new component: accumulator/downsampler (K : Σ)

For further details see Harris 5.

12: polyphase filters - imperial college london · 12: polyphase filters 12: polyphase filters...

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