Topics: •  From simple filters to echo and reverb •  Variable delay single tap FIR filter •  Variable delay single tap IIR filter •  Plucked string filters •  Karplus - Strong plucked string models

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Introduction to ���Audio and Music Engineering

Lecture 22

Back to the simple filter …

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Y = X + a 1z −1X

1 unit delay a1

+ input X output Y

x (n − 1) a 1x (n − 1)

Multiply by z/z: zz1 + a 1z −1

1 =z + a 1z

Y = H (z)XH (z) = 1 + a 1z

−1

We want to know the magnitude of H(z). H (z) =z + a 1

z

We are interested in |H(z)| for the allowed frequencies (on unit semicircle in z-plane). z → e jω

Continued …

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but H (z) =z + a 1

z→ H (e jω ) =

e jω + a 1

e jω e jω = 1

H (e jω ) = e jω + a 1 = e jω − (−a 1 )so

This is simply the distance from the point to –a1 on the real axis. e jω

–a1

e jω

1

e jω − (−a 1 )

e jω

−a 1

c b

a

a +b =c

b =c −a

The further away is from –a1 the larger the filter response. e jω

“zero” of H(z)

What does look like?

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H (e jω )

Low pass

–a1

e jω

1

ω = 0

π

H (e jω )

ω

1

1 + a1 2

0

1 - a1 ω = π

High pass

a1

e jω

1

ω = 0

π

H (e jω )

ω

1

1 + a1 2

0

1 - a1 ω = π

ω = π4

ω = π2

ω = 3π4

ω = π4

ω = π2

ω = 3π4

Feed-forward and feed-back filters

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x (n) = [1, 0, 0, 0, 0, 0...]

Z-1 a1

+ input X output Y

feed-forward

Response is non-zero only for n = 0,1 .

y (n) = [1,a 1 , 0, 0, 0, 0...]

y (n) = x (n) + a 1x (n − 1)

impulse

response

Finite impulse response filter : FIR

feed-back

Z-1 b1

+ input X output Y

y (n) = x (n) + b 1y (n − 1)

x (n) = [1, 0, 0, 0, 0, 0...]

Response never returns to zero!

y (n) = [1,b 1 ,b 12 ,b 1

3,b 14...]

impulse

response

Infinite impulse response filter : IIR

Frequency response of feedback filter

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Z-1 b1

+ input X output Y

Y = X + b 1z −1Y

zz

multiply by

Y (1 − b 1z −1 ) = XY = 1

(1 − b 1z −1 ) X H (z) = 11 − b 1z

−1

zz

11 − b 1z −1 = z

z − b 1H (z) =

zz − b 1

H (ω ) =e jω

e jω − b 1

= 1e jω − b 1

z → e jω

b1

e jω

1

e jω − b 1e jω

-1 So the feedback filter response is proportional to the inverse of the distance from a point on the unit semicircle to the point b1. “pole” of H(z)

π

H (e jω )

ω

1

2

0 - b1

e jω

1

1/(1+b1)

1/(1-b1)

Frequency response of the feedback filter

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low pass

b1

e jω

1

ω = 0

π

H (e jω )

ω

1

2

0

ω = π

ω = π4

ω = π2

ω = 3π4

1/(1+b1)

1/(1-b1)

high pass

ω = 0ω = π

ω = π4

ω = π2

ω = 3π4

Feedback filter stability

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e jω

1 π

H (e jω )

ω

1 2

0

3 4 5

As the pole moves closer to the unit circle the response increases.

As the pole moves out past the unit circle the filter response diverges. … the filter is unstable.

Z-1 b1

+input X output

Y

y (n) = x (n) + b 1y (n − 1)

x (n) = [1, 0, 0, 0, 0, 0...]

y (n) = [1,b 1 ,b 12 ,b 1

3,b 14...]

If b1 > 1, y(n) grows without bound!

input impulse

response

Poles must remain within the unit circle to make the filter stable.

Higher order filters

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feed-forward

Z-1

-1

+ input X

Z-1

+ output Y

Y = X − z −1X + z −1z −1XY = X − z −1X + z −2X = (1 − z −1 + z −2 )X

H (z) = (1 − z −1 + z −2 )1

z2

z2 = (z2 − z + 1)z2

H (z) =(z2 − z + 1)

z2

z → e jωThe denominator is always 1.

e jω( )2 = e j2ω = 1

(z2 − z + 1) = 0Where are the zeros?

Finding the zeros of H(z)

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(z2 − z + 1) = 0

→ z = 1 ± 1 − 42 = 1

2 ± j 32

Convert to polar form

r = 1 2( )2 + 3 2( )2⎡

⎣⎢

⎤

⎦⎥

1/2

= 1 4 + 3 4⎡⎣ ⎤⎦1/2

= 1

θ = tan−1 3 21 2

⎛

⎝⎜

⎞

⎠⎟ = tan−1 3( ) = ± π 3 ±60( )

z = e + jπ 3 and z = e − jπ 3zeros are at …

e jω

1

e jω

“zeros” of H(z)

ω = 0ω = π

Finding the transfer function

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num(H (z)) = z − e jπ/3( ) z − e − jπ/3( ) = z − e jπ/3( ) ⋅ z − e − jπ/3( )

e jω

1

e jω

ω = 0ω = π

Just the product of the distances from to each of

the two zeros. e jω

π

H (e jω )

ω0 π / 3

General filters

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+ input X

output Y

a0

Z-1 a1

Z-2 a2

Z-1 b1

Z-2 b2

. . .

. . .

Y = a0 + a 1z −1 + a2z −2 + ...( )X − b 1z −1 + b2z −2 + ...( )Y

Y =a0 + a 1z −1 + a2z −2 + ...( )1 + b 1z −1 + b2z −2 + ...( ) X H (z) =

z − z 1( ) z − z2( ) z − z3( ) ...z − p 1( ) z − p2( ) z − p3( ) ...

Solve for Y You can factor the numerator and denominator

But it might not be easy!

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Finding the frequency response …

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H (ω ) =e jω − z 1( ) e jω − z2( ) e jω − z3( ) ...e jω − p 1( ) e jω − p2( ) e jω − p3( ) ...

Multiply the distances to each zero then divide by the products of the distances to each of the poles.