Digital Filters

1
Digital Filters

• Well decide on some desired spectral
  characteristics
• Well use tools or mathematics to create a filter
  that is as close as possible

2
Why we may design filters

• Low pass filters
• Remove high frequency noise
• Create muffled sounds
• Separate bass
• Blur images
• High pass filters
• Remove rumble, wind noise
• Edge enhancement
• Others
• Band pass filters
• Tone controls

3
Well start by analyzing filters

• We want to understand what happens
• What we are usually interested in
• Magnitude frequency response

4
Digital Filters and Rates

• An interesting thing about digital signals
• Playing at different rates simply scales
• A 1000 Hz sine wave at fs 8000 is identical to
  a 0.125 Hz sine wave at fs1
• What we really care about is that the sine wave
  is 1/8 the sample rate!

5
Normalization

• Well assume a sample rate of 1 sample per second
  (or 2p radians per second) for all future
  mathematics!
• What!?!?

Normalized Time
6
Normalized sample rate

• 44,100 no more!
• Suppose I want to generate a filter that
  emphasizes 440 Hz with fs44100
• This is equivalent to a filter that emphasizes
  440/44100 at fs 1 or 2p radians/second
• Normalization
• Were only interested in what happens from
  frequency 0 to frequency 0.5

7
Examples

• Suppose we have a filter with a magnitude
  response of 0.78 at 0.0226 Hz (assuming fs1)
• What would be the equivalent frequency if we ran
  the filter at fs44100?
• 0.0226 44100 997Hz

TT
8
Now were only interested in integral delays

• yt xt a1xt-1
• ejwta1ejw(t-1)
• H(w) 1 a1e-jw
• H(w) (1 a12 2a1cos(w))1/2
• w ranges from 0 to p

x
y

a1
Delay 1
9
Frequency response for a11
H(w) (1 a12 2a1cos(w))1/2
yt xt a1xt-1
2
Low pass response!
1
Amplitude
0
0
0.5
Percentage of fs
10
Frequency response for a1 -1
H(w) (1 a12 2a1cos(w))1/2
yt xt a1xt-1
2
High pass response!
1
Amplitude
0
0
0.5
Percentage of fs
11
More general

• More general equation for delay of 1
• yt a0xt a1xt-1
• If input is ejwt
• xta0a1e-jw
• Well use z ejw as an operator for delay
• yt a0a1z-1 xt

12
Some conventions

• When z ejw
• Let X represent the entire sequence xt
• a0X implies every sample xt gets multiplied by a0
• Let Y represent the entire sequence yt
• Well use
• Y a0a1z-1 X
• H(z) a0a1z-1 Transfer Function or Equation
• Y H(z) X

13
Flowgraphs with z
x
y

3
z-1

• yt xt 3xt-1
• H(z) 13z-1

x
y

3
TT
5
z-1

• yt xt 3xt-1 5xt-2
• H(z) 13z-15z-2

z-2
14
Example Transfer Functions

• yt a0xt a1xt-1
• H(z) a0a1z-1
• yt xt - xt-1 xt-2
• H(z) 1-z-1z-2
• yt xt 17xt-25
• H(z) 117z-25

• Transfer Functions

TT
15
How to determine frequency response

• Step 1 Determine transfer function
• Step 2 Get rid of negative exponents
• Step 3 Factor numerator, determine zeros,
  multiplicand
• Step 4 Plot the zeros of the numerator on the
  z-plane
• Step 5 For every frequency, take the products
  of the distances to the zeros times any
  multiplicand

16
Step 1 Determine the transfer function

• yt xt 0.5xt-1
• H(z) 10.5z-1

17
Step 2 Get rid of negative exponents

• H(z) 10.5z-1 (z 0.5)/z

Multiply by z/z
18
Step 3 Factor numerator, determine zeros

• H(z) (z 0.5)/z
• This is zero for z -0.5

19
Step 4 Plot the zeros of the numerator on the
z-plane

• We call all z values that result in a zero value
  the zeros of the transfer function
• We can plot those on a unit circle

z-0.5
20
Step 5 For every frequency, take products of
distances to the zeros

• Frequencies are points on the circle
• Distance to the points is the gain of the filter

f0.2 e0.2(2p)j cos(0.2(2p))jsin(0.2(2p))
0
0.5
z-0.5
21
A more complex filter

• yt xt xt-1 xt-2
• H(z) 1z-1z-2
• (z2z1)/z2
• Binomial theorem time!!

22
Two zeros!

• For any frequency ejw, take product of distances
  to all zeros!

0.33
0
0.5
23
Frequency response
24
Handling any multiplied gain

• When factoring, you must get into this form
• a(zz1)(zz2)(zzn)/zn
• a is a gain multiplied by the filter

25
Another more complex filter

• yt 2xt xt-1 2xt-2
• H(z) 2z-12z-2
• (2z2z2)/z2
• 2(z20.5z1)/z2
• Binomial theorem again

26
Two zeros!

• For any frequency ejw, take product of distances
  to all zeros the gain value (2)!

0
0.5
27
Frequency response