Lecture 4: Analog Filter (II)

Topics:
– Frequency Response of LTI Systems
– Filter Design using Pole-zero Placement
– Practical Filter Design Specifications
– Butterworth Filter Design
 Lecture 4: Analog Filter (II)
Topics:
– Frequency Response of LTI Systems
– Filter Design using Pole-zero Placement
– Practical Filter Design Specifications
– Butterworth Filter Design
Frequency response of an LTI system

Recap: The Laplace transform:

F (s) = L f (t)=  − f (t)e−s t dt, where s = + j.



Exercise 1: Show that L ()=1.

Recall the sampling property
L (t)= 0  (t)e−s t dt = e−s0 =1

of the unit impulse function:
− 
−
f ( ) ( )d = f (0).

Recap: Y (s) = H (s)F (s)  Y (s) = H (s)1= H (s).

f (t) = (t) Filter h(t) Conclusion: The transfer

L L function H(s) is the Laplace
transform of the impulse
F (s) =1 H(s) H (s) response h(t).
Frequency response of an LTI system

Recap:
f (t ) Filter y(t)

Y () = F ()H ()

H () , the Fourier transform of the impulse response ( h(t)  H () )

of a system is called the frequency response of the system.

Recap: Connection between the Laplace and Fourier transforms:

(Recall that F () and F ( j)
F ( j) = F (s) s= j
represent the same thing!)

The Frequency response H ( j) of an asymptotically stable system can

be obtained by substituting s= j in the system’s transfer function H(s).
Frequency response of an LTI system
Recap: Frequency response of a system

Y () = F () H () ,

f (t ) Filter y(t)
Y () = F () + H ().

Recap:

e jt
Filter y(t) = h(t)e jt
=  h( )e j(t − )d
−



jt − j jt
=e h( )e d = e H ()
−
 
H ( ) = Fourier transform of h(t )

e jt Filter y(t) = H () e jt

Frequency response of an LTI system

e jt Filter y(t) = H ( j) e jt

e− jt Filter y(t) = H (− jω) e− jt

A “practical” signal: f (t) = cos( t +  )

f (t) = cos( t +  ) = e
2

1 j( t +  ) − j( t +  )
+e 
 e j  j t  e − j  − j t
=  e +  e
 2   2 
Frequency response of an LTI system

 e j  j t  e − j  − j  t
f (t) =  e +  e
 2  2 
Applying superposition:

 e j  j t  e
− j
 − j t
y(t) =   H ( j)e +   H (− j  )e
 2  2 

= e
2

1 j( t + )
H ( j) e jH ( j) + e− j( t + ) H ( j) e− jH ( j) 

Recap: Conjugate symmetry property:

H (− j) = H ( j) ,
H (− j) = −H ( j).
Frequency response of an LTI system

y(t) = e
2

1 j( t + )
H ( j) e jH ( j) + e− j( t + ) H ( j) e− jH ( j) 
= H ( j)
2

1 j( t + +H ( j)) − j( t + +H ( j))
e +e 
= H ( j) cos( t + + H ( j))

f (t) = cos( t +  ) y(t) = H ( j) cos( t + + H ( j))

Filter
 Lecture 4: Analog Filter (II)
Topics:
– Frequency Response of LTI Systems
– Filter Design using Pole-zero Placement
– Practical Filter Design Specifications
– Butterworth Filter Design
Filter design by placement of the poles and zeros of H(s)

Recap:

F (s) H (s) Y ( s) = H ( s) F ( s)

bn s n +bn−1s n−1 +  +b1s +b0

H (s) = n
s + an−1s n−1 +  + a1s + a0

(s − z1)(s − z2 )  (s − zn )
= bn
(s − p1)(s − p2 )  (s − pn )

z1, z2, . . . , zn are the zeros of H(s) and p1, p2, . . . , pn are the poles of H(s).

Frequency
( j − z1)( j − z2 )  ( j − zn )
response: H ( j) H ( j) = bn
0

( j − p1)( j − p2 )  ( j − pn )
Filter design by placement of the poles and zeros of H(s)

Im
( j − z1)( j − z2 )  ( j − zn ) j
H ( j) = bn
( j − p1)( j − p2 )  ( j − pn )
r1
1
z1 0 Re
Filter design by placement of the poles and zeros of H(s)

Im
( j − z1)( j − z2 )  ( j − zn ) j
H ( j) = bn
( j − p1)( j − p2 )  ( j − pn )
r1e j1 r1
1
z1 0 Re
Filter design by placement of the poles and zeros of H(s)

Im
( j − z1)( j − z2 )  ( j − zn ) j
H ( j) = bn d1
( j − p1)( j − p2 )  ( j − pn ) p1 1 r2
r1
1 2
d2
(r1e j1 )(r2e j2 )  (rne jn )
= bn 2
z1 0 z2 Re
(d1e j1 )(d2e j2 )  (dne jn ) p2

r1r2  rn j[(1+2 ++n )−(1+2++n )]

= bn e
d1d2  dn

r1r2  rn product of the distances of the zeros to j

H ( j) = bn = bn
d1d2  dn product of the distances of the poles to j

H ( j) = (1 +2 ++n ) −(1 +2 ++n )

= sum of the zero angles to j − sum of the pole angles to j
Filter design by placement of the poles and zeros of H(s)

Gain enhancement by a pole

Consider the hypothetical case of a single pole at −𝑎 +𝑗𝜔0.
The amplitude response at a specific value of 𝜔, 𝐻(𝑗𝜔) , is found by measuring the
length of the line that connects the pole to the point 𝑗𝜔.
If the length of the above mentioned line is 𝑑, then 𝐻(𝑗𝜔) is proportional to 1/𝑑 .

K
H ( j ) = Im
d
j0 H ( j )
• As 𝜔 increases from zero, 𝑑 decreases
progressively until 𝜔 reaches the value 𝜔0 and,
d j
therefore, 1/𝑑 increases.
• As 𝜔 increases beyond 𝜔0, 𝑑 increases − 0
progressively. Re
• Therefore, the peak of 𝐻(𝑗𝜔) occurs at 𝜔0. As α 0 0 
d'
becomes smaller, i.e., as the pole moves closer
to the imaginary axis the gain enhancement at
𝜔0 becomes more prominent (𝑑 becomes very
small.)
Filter design by placement of the poles and zeros of H(s)

Gain enhancement by a pole

Consider the hypothetical case of a single pole at −𝑎 +𝑗𝜔0.
The amplitude response at a specific value of 𝜔, 𝐻(𝑗𝜔) , is found by measuring the
length of the line that connects the pole to the point 𝑗𝜔.
If the length of the above mentioned line is 𝑑, then 𝐻(𝑗𝜔) is proportional to 1/𝑑 .

K
H ( j ) = Im
d
j0 H ( j )
• In conclusion, we can enhance a gain at a
frequency 𝜔0 by placing a pole opposite the
d j
point 𝑗𝜔0 .
• The closer the pole is to 𝑗𝜔0, the higher is the − 0 Re
gain at 𝜔0 and furthermore, the enhancement is
more prominent around 𝜔0. 0 0 
d'
• In the extreme case of α = 𝟎 (pole on the
imaginary axis) the gain at 𝝎𝟎 goes to infinity.
• Recall that poles must lie on the left half of the
𝑠 −plane.
Filter design by placement of the poles and zeros of H(s)

Gain enhancement by a pole

In a real system, a complex pole at − α +𝑗𝜔0 must be accompanied by its
conjugate pole − α −𝑗𝜔0.

The amplitude response at a specific value of 𝜔, 𝐻(𝑗𝜔) , is found by measuring the

length of the two lines that connect the poles to the point 𝑗𝜔.
If the lengths of the above mentioned lines are
𝑑, 𝑑′ then 𝐻(𝑗𝜔) = 𝐾/𝑑𝑑′ Im
j0 H ( j )
We can see graphically that the presence of d j
the conjugate pole does not affect
substantially the behaviour of the system − 0 Re
around 𝜔0. This is because as we move
around 𝝎𝟎, 𝒅′ does not change 0 0 
d'
dramatically
Filter design by placement of the poles and zeros of H(s)

Gain enhancement by a pole

In a real system, a complex pole at − α +𝑗𝜔0 must be accompanied by its
conjugate pole − α −𝑗𝜔0.

The amplitude response at a specific value of 𝜔, 𝐻(𝑗𝜔) , is found by measuring the

length of the two lines that connect the poles to the point 𝑗𝜔.
If the lengths of the above mentioned lines are
𝑑, 𝑑′ then 𝐻(𝑗𝜔) = 𝐾/𝑑𝑑′ Im
j0 H ( j )
We can see graphically that the presence of d j
the conjugate pole does not affect
substantially the behaviour of the system − 0 Re
around 𝜔0. This is because as we move
around 𝝎𝟎, 𝒅′ does not change 0 0 
d'
dramatically
Filter design by placement of the poles and zeros of H(s)

Gain suppression by a zero

Consider a real system with a pair of complex conjugate zeros at −𝑎+𝑗𝜔0 and
−𝑎−𝑗𝜔0
The amplitude response at a specific value of 𝜔, 𝐻(𝑗𝜔) is again found by measuring
the length of the two lines that connect the zeros to the point 𝑗𝜔.

If the lengths of the above mentioned lines are 𝑟, 𝑟′ then 𝐻(𝑗𝜔) = 𝐾𝑟𝑟′.

In that case, the minimum of 𝐻(𝑗𝜔) occurs at 𝜔0.

As 𝑎 becomes smaller, i.e., as the zero Im

moves closer to the imaginary axis, the j0 H ( j )
gain suppression at 𝜔0 becomes more r j
prominent.

In the extreme case of 𝒂 = 𝟎 (zero on the − 0 Re

imaginary axis) the gain at 𝝎𝟎 goes to 0 0 
'
zero. r
Filter design by placement of the poles and zeros of H(s)

Phase response by a pole and a zero

Response to a pole

∠𝐻(𝑗𝜔) = − 𝜃1 +𝜃2 .

Response to a zero
∠𝐻(𝑗𝜔) = 𝜙1 +𝜙2
Filter design by placement of the poles and zeros of H(s)

Lowpass filters
c
Fist-order lowpass filter: H (s) =
s +c
0

Im If d is the distance from the pole −ωc to a

n =1 point jω:
d j c
H ( j ) = H(0) = 1.
− c 0 Re d

As ω increases, d increases and |H(jω)| decreases monotonically with ω, as N = 1.

This is clearly a lowpass filter with gain enhanced in the vicinity of ω = 0.
Filter design by placement of the poles and zeros of H(s)

Lowpass filters
| H ( j ) |
ideal (n = )
1 n =1
n=2
n = 10 n=4

0 c n=8 

Im Im
jc Butterworth filter n = 5
0
jc

0 Re 0 Re

− j c − j c
Filter design by placement of the poles and zeros of H(s)
Bandpass filters Im
| H ( j ) |

In the case of a bandpass j0 ideal

filter, we need enhanced
gain over the entire
passband 0 Re
0 0 
− j 0

Notch (bandstop) filters

Im
A practical second- | H ( j ) |
j0
order notch filter (or  = 87
bandstop filter) to 1
obtain zero gain at a   = 80
frequency ω = ω0 can  = 60
0 Re
be realized with two ideal
poles and two zero 0 0 
− j 0
 Lecture 4: Analog Filter (II)
Topics:
– Frequency Response of LTI Systems
– Filter Design using Pole-zero Placement
– Practical Filter Design Specifications
– Butterworth Filter Design
Practical filter specifications

Gp = minimum passband gain

0

Gs = maximum stopband gain

0

| H ( j ) | | H ( j ) |

1 1
Gp Gp

Gs Gs
0 p s  0  s1  p1  p 2 s 2 
Lowpass filter Bandpass filter
Practical filter specifications
| H ( j ) | | H ( j ) |

1 1
Gp Gp

Gs Gs
0 s p  0  p1  s1 s 2  p 2 
Highpass filter Bandstop filter

Decibels
G Ĝ (dB)
Gˆ (dB) = 20 log10 G 1/10 -20
1/2 ≈ -6
1/√2 ≈ -3
1 0
√2 ≈3
2 ≈6
10 20
 Lecture 4: Analog Filter (II)
Topics:
– Frequency Response of LTI Systems
– Filter Design using Pole-zero Placement
– Practical Filter Design Specifications
– Butterworth Filter Design
Butterworth filters
The amplitude response |H(j ω)| of an n-th order Butterworth
lowpass filter is

1 At ω = 0 the DC gain is H (0) = 1

H ( j) =
H (c ) = 1/ 2 = −3 dB
2n
 
1+ 
 c  c = 3 dB-cutoff frequency
| H ( j ) | = 3 dB bandwidth
ideal (n = )
1 n =1
n=2
n = 10 n=4

0 c n=8 
Butterworth filters
Normalised Butterworth lowpass filter: with ωc = 1

1
H ( j ) =
1+ 2n

H ( j)
ideal (n = )
1 n =1
0.707 n=2
n = 10 n=4

0 1 n=8 
Butterworth filters
1
Normalised Butterworth lowpass filter: H ( j ) =
1+ 2n

Recap: Conjugate symmetry property: H (− j) = H ( j) ,

H (− j) = −H ( j).

2 1
H ( j)H (-j) = H ( j) =
1+ 2n

Substitute s= j :
1
H (s)H (-s) = 2n
s
1+   =0
 j

The poles of H (s)H (-s) satisfy: s 2n = − j 2n

Butterworth filters
Normalised Butterworth lowpass filter:

The poles of H (s)H (-s) satisfy: s 2n = − j 2n

Im Im
j
  /2

−1 0 Re 0 Re

−1 = e j (2k −1) , k integer j = e j / 2

 s = − j = −1 e
2n 2n
( )
j / 2 2n
= e j (2k −1)e jn = e j (2k −1+n)

j (2k +n−1)
 sk = e 2n
, k = 1, 2,  , 2n
Butterworth filters
Normalised Butterworth lowpass filter:
Im

Poles of H (s)H (-s) :


j (2k +n−1)  /n
sk = e 2n
, k = 1, 2,  , 2n
−1 0 1 Re

Since we want H (s) stable, we select the

poles on the LHP (i.e., for k = 1, 2,  , n):

j (2k +n−1)  
sk = e 2n
= cos (2k + n −1) + jsin (2k + n −1), k = 1, 2,  , n,
2n 2n
1
and H (s) is given by: H (s) =
(s − s1)(s − s2 )(s − sn )
Butterworth filters
Normalised Butterworth lowpass filter:

For example, for n = 4, the poles are :

Butterworth filters
Normalised Butterworth lowpass filter:
Im
For example, for n = 4, the poles are :
n=4
-0.3827  j0.9239  /4
 /4
-0.9239  j0.3827 −1 0 Re

1
H (s) =
(s + 0.3827 − j0.9239)(s + 0.3827 + j0.9239)
1

(s + 0.9239− j0.3827)(s + 0.9239+ j0.3827)
1
= 2
(s + 0.7654s +1)(s 2 +1.8478s +1)
1
= 4
s + 2.6131s3 +3.4142s 2 + 2.6131s +1
Butterworth filters
Normalised Butterworth lowpass filter:

In general:
1 1
H (s) = = n
Bn (s) s + an−1s n−1 ++ a1s +1

Bn (s) = Butterworth polynomial of nth order → Tables!

Tables 7.1 and 7.2 on
Page 509 of Lathi’s textbook.

Frequency scaling: The normalised filter H (s) has 3 dB bandwidth c =1.

 s 
For a different c : H (s) = H  
 c  1
H (s) =
In the previous example, for c =10: s 4 + 2.6131s3 +3.4142s 2 + 2.6131s +1

 s  1
H (s) = H   = 4 3 2
 10   s   s   s   s 
  + 2.6131  +3.4142  + 2.6131  +1
 10   10   10   10 
10,000
= 4
s + 26.131s3 +341.42s 2 + 2,613.1s +10,000
Butterworth filters
Determination of the filter order n

| H ( j ) | 1
H ( j) = 2n
1  
1+ 
 c 
Gp

Gˆ (dB) = 20 log10 G
Gs
0 p s 
Gain in dB
Lowpass filter

minimum passband gain

   2n 
0


2n
  −Gˆ /10
Gˆ p = 20 log10 H ( j p ) = −10 log10 1+ p     p  =10 p −1
  c    c 
maximum stopband gain
0
   2n    
2n

Gˆ s = 20 log10 H ( js ) = −10 log10 1+ s     s  =10−Gs /10 −1

ˆ

  c    c 
Butterworth filters
Determination of the filter order n

 p 
2n
ˆ
  =10−Gp /10 −1
 c   s 
2n
10 −Gˆ s /10
−1
   = −Gˆ /10
   10 p −1
 p
2n
 s  ˆ
  =10−Gs /10 −1
 c 

 n=
( ˆ
)( −Gˆ /10
log10 10−Gs /10 −1 / 10 p )
−1
2 log10 (s / p )

 p 
2n
ˆ p
  =10−Gp /10 −1  c =
 c  10 
−Gˆ p /10
−1 
1/ 2n

Alternatively:
2n
 s  ˆ s
  =10−Gs /10 −1  c =

 c 10 
−Gˆ s /10
−1 
1/ 2n