Topics:�•  Single “tap” feed-forward (FIR) filter�•  Single “tap” feed-back (IIR) filter�•  Frequency response and pole-zero diagrams�•  Higher order FIR filters �•  More general filters�

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Introduction to �Audio and Music Engineering �

Lecture 23�

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 …�

5�

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?�

6�

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 �

11 �

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! �

-1 �

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. �