Uses of IIR filters IIR filters can be unstable. What if the pole - - PowerPoint PPT Presentation

▶

Jan 13, 2024 388 likes •609 views

Uses of IIR filters IIR filters can be unstable. What if the pole is on the unit circle? Interesting use of IIR filter IIR filters can be unstable. What if the pole is on the unit circle? a = [1 a1 1] with (-2<a1<2)

SLIDE 1

Uses of IIR filters

IIR filters can be unstable.
What if the pole is on the unit

circle?

SLIDE 2

Interesting use of IIR filter

IIR filters can be unstable.
What if the pole is on the unit

circle?

a = [1 a1 1] with (-2<a1<2)
What does the impulse response look

like?

Possible poles for bp=3.

SLIDE 3

BP=5, a=(1/32)*[32,-31,32] =[1,-0.96875,1]

SLIDE 4

Filtering 1320 kHz at 125 MHz

We need a very narrow band filter to

recover the 10 kHz signal.

A direct filter requires too many

taps.

Decimation: taking every Nth sample

allows reduction in the frequency range and # of taps.

By 2, signals in the range 31.25 to 62.5

are aliased into 31.26 to 0 MHz.

When you decimate by 2 a frequency at 1/2 the sampling frequency appears as a constant (f=0).

SLIDE 5

Filtering 1320 kHz at 125 MHz

Use a filter to precondition the

signal prior to decimation.

Filter the signal to reduce the

energy that is aliased from the (31.25 to 62.5 band) into the (0 to 31.25 band) and on.

Half Band Filter: Design problem...

When you decimate by 2 a frequency at 1/2 the sampling frequency appears as a constant (f=0).

SLIDE 6

Half Band FIR/IIR Filter

Design a Filter with 4 poles and 4

zeros with the following constraints.

Start with a constrained mathematical

model.

Efficient half-band filter:
All poles are on the y-axis and have

magnitude < 1.

All zeros are on the unit circle between

pi/2 and –pi/2.

Final transfer function takes on the form:

H(z) = (1 + b01z-1 + 1)(1 + b11z-1 + 1) (1 + a02z-2)(1 + a12z-2)

bX1 are between 0 and 2
aX2 are between 0 and 1

SLIDE 7

Coefficients

b01,b02 = 0.750,1.750
a02,a12 = 0.125,0.5625

SLIDE 8

Block Diagrams

z-1 z-1 b01 z-1 z-1 b11

z-1 z-1 a02

+ - subtraction

z-1 z-1 a12

+ - subtraction

SLIDE 9

FIR Cascade

z-1 z-1 3 z-1 z-1 7

x4 <<<2 x4 <<<2 x4 <<<2 x4 <<<2 bp is increased by 4.

3 z-1 z-1

x4 <<<2 x4 <<<2

z-1 7 z-1 z-1

x4 <<<2 x4 <<<2 bp is increased by 4. x8-1 x4-1

SLIDE 10

IIR Cascade

9 z-2

bp is increased by 7.

1 z-2

x8 <<<3 correct bp >>>3 x16 <<<4 correct bp >>>4

SLIDE 11

FIR/IIR Cascade

9 z-2 1 z-2

x8 <<<3 correct bp >>>3 x16 <<<4 correct bp >>>4

3 z-1 z-1

x4 <<<2 x4 <<<2

z-1 7 z-1 z-1

x4 <<<2 x4 <<<2 x8-1 x4-1 Total Width increase by 22. BP net is increased by 11. 1,.75,1 = 2.75 = 3 bits. BP is increased by 2. Total = 5 1,1.75,1 = 3.75 = 3 bits BP is increased by 2. Total = 5 Sum(.125^(0:inf)) = 2. BP is increased by 3. Total = 5. Sum(.5625^(0:inf)) = 3. BP is increased by 4. Total = 7. Total BP = 11. Coefficients ... ceil(log2(sum(f1×f2×f3×f4)))+1=6 14 input BP is 11 Coefficients are 6 =31 (we will use 32)

SLIDE 12

Impulse Response

SLIDE 13

Comparison to Simulation

SLIDE 14

Decimate by 2 without LPF

SLIDE 15

Decimate by 2 with LPF

SLIDE 16

FIR/IIR Cascade

9 z-2

bp is increased by 7.

1 z-2

x8 <<<3 correct bp >>>3 x16 <<<4 correct bp >>>4

3 z-1 z-1

x4 <<<2 x4 <<<2

z-1 7 z-1 z-1

x4 <<<2 x4 <<<2 bp is increased by 4. x8-1 x4-1 bp net is increased by 11.

z-1 ↓2

SLIDE 17

FIR/IIR Cascade

9 z-1

bp is increased by 7.

1 z-1

x8 <<<3 correct bp >>>3 x16 <<<4 correct bp >>>4

3 z-1 z-1

x4 <<<2 x4 <<<2

z-1 7 z-1 z-1

x4 <<<2 x4 <<<2 bp is increased by 4. Decimate here x8-1 x4-1 bp net is increased by 11.

z-1 ↓2

SLIDE 18

Stage 1: Before Decimation

SLIDE 19

Stage 2: After Decimation and cascading

Again 3 more times

SLIDE 20

Stage 4: After decimation by 16

125/16 is 7.8125

SLIDE 21