Chapter 16: Filter Circuits

Exercises

Ex. 16.3-1
1
H (s) =
n s +1
⎛ s ⎞ 1 1250
H (s) = H ⎜ = =
n ⎝ 1250 ⎟⎠ s s + 1250
+1
1250
Problems
Section 16.3: Filters

P16.3-1
Equation 16-3.2 and Table 16-3.2 provide a third-order Butterworth low-pass filter having a
cutoff frequency equal to 1 rad/s.
1
H n (s) =
( s + 1)( s 2 + s + 1)
Frequency scaling so that ω = 2π 100=628 rad/s:
c

1 6283 247673152
H L (s) = = =
⎛ s ⎞⎛⎛ s ⎞
2
s ⎞ ( s + 628)( s + 628s + 628 ) ( s + 628)( s 2 + 628s + 394384)
2 2

⎜ + 1 ⎟ ⎜⎜ ⎜ ⎟ + + 1⎟
⎝ 628 ⎠ ⎝ ⎝ 628 ⎠ 628 ⎟⎠

P16.3-2
Equation 16-3.2 and Table 16-3.2 provide a third-order Butterworth low-pass filter having a
cutoff frequency equal to 1 rad/s and a dc gain equal to 1.

1
H ( s) =
n ( s + 1)( s 2 + s + 1)
Multiplying by 5 to change the dc gain to 5 and frequency scaling to change the cutoff
frequency to ωc = 100 rad/s:

5 5 ⋅1003 5000000
H L ( s) = = =
⎛ s ⎞ ⎛ ⎛ s ⎞2 s ⎞ ( s + 100)( s + 100 s + 100 ) ( s + 100)( s 2 + 100 s + 10000)
2 2

⎜ + 1⎟ ⎜⎜ ⎜ ⎟ + + 1⎟
⎝ 100 ⎠ ⎟
⎝ ⎝ 100 ⎠ 100 ⎠

1
P16.3-3
Use Table 16-3.2 to obtain the transfer function of a third-order Butterworth high-pass filter
having a cutoff frequency equal to 1 rad/s and a dc gain equal to 5.

5s 3
H (s) =
n ( s + 1)( s 2 + s + 1)
Frequency scaling to change the cutoff frequency to ω = 100 rad/s:
c
3
⎛ s ⎞
5⎜ ⎟
⎝ 100 ⎠ 5s 3 5s 3
H H (s) = = =
⎛ s ⎞ ⎛⎛ s ⎞
2
s ⎞ ( s + 100)( s 2 + 100 s + 1002 ) ( s + 100)( s 2 + 100 s + 10000)
⎜ + 1⎟ ⎜ ⎜ ⎟ + + 1⎟
⎝ 100 ⎠ ⎜⎝ ⎝ 100 ⎠ 100 ⎟⎠

P16.3-4
Use Table 16-3.2 to obtain the transfer function of a fourth-order Butterworth high-pass filter
having a cutoff frequency equal to 1 rad/s and a dc gain equal to 5.

5s 4
Hn ( s) = 2
( s + 0.765s + 1)( s 2 + 1.848s + 1)
Frequency scaling can be used to adjust the cutoff frequency 500 hertz = 3142 rad/s:
4
⎛ s ⎞
5⋅⎜ ⎟
⎝ 3142 ⎠ 5 ⋅ s4
HH (s) = =
⎛ ⎛ s ⎞2 ⎛ s ⎞ ⎞⎛⎛ s ⎞ ⎛ s ⎞ ⎞ ( s + 2403.6 s + 3142 )( s + 5806.4s + 3142 )
2 2 2 2 2

⎜⎜ ⎜ ⎟ + 0.765 ⎜ ⎟ + 1⎟⎟ ⎜⎜ ⎜ ⎟ + 1.848 ⎜ ⎟ + 1⎟⎟

⎝ ⎝ 3142 ⎠ ⎝ 3142 ⎠ ⎠ ⎝ ⎝ 3142 ⎠ ⎝ 3142 ⎠ ⎠

P16.3-5
First, obtain the transfer function of a second-order Butterworth low-pass filter having a dc gain
equal to 2 and a cutoff frequency equal to 2000 rad/s:

2 8000000
HL ( s) = =
⎛ s ⎞
2
⎛ s ⎞ s + 2828s + 4000000
2

⎜ ⎟ + 1.414 ⎜ ⎟ +1
⎝ 2000 ⎠ ⎝ 2000 ⎠

Next, obtain the transfer function of a second-order Butterworth high-pass filter having a
passband gain equal to 2 and a cutoff frequency equal to 100 rad/s:

2
 2
⎛ s ⎞
2⋅⎜ ⎟
⎝ 100 ⎠ 2 ⋅ s2
HH ( s) = =
⎛ s ⎞
2
⎛ s ⎞ s 2 + 141.4s + 10000
⎜ ⎟ + 1.414 ⎜ ⎟ +1
⎝ 100 ⎠ ⎝ 100 ⎠

Finally, the transfer function of the bandpass filter is

16000000 × s 2
HB ( s) = HL (s)× HH (s) = 2
( s + 141.4s + 10000 ) ( s 2 + 2828s + 4000000 )

P16.3-6
2
⎛ 250 ⎞
⎜ s ⎟ 250000s 2
HB ( s) = 4 ⎜ 1
⎟ =
s + 2502 ⎟ ( s + 250 s + 62500 )
2
250
⎜ s2 +
2

⎝ 1 ⎠

P16.3-7
First, obtain the transfer function of a second-order Butterworth high-pass filter having a dc gain
equal to 2 and a cutoff frequency equal to 2000 rad/s:
2
⎛ s ⎞
2⎜ ⎟
⎝ 2000 ⎠ 2s 2
HL ( s) = =
⎛ s ⎞
2
⎛ s ⎞ s 2 + 2828s + 4000000
⎜ ⎟ + 1.414 ⎜ ⎟ +1
⎝ 2000 ⎠ ⎝ 2000 ⎠

Next, obtain the transfer function of a second-order Butterworth low-pass filter having a pass-
band gain equal to 2 and a cutoff frequency equal to 100 rad/s:

2 20000
HH ( s) = =
⎛ s ⎞
2
⎛ s ⎞ s + 141.4s + 10000
2

⎜ ⎟ + 1.414 ⎜ ⎟ +1
⎝ 100 ⎠ ⎝ 100 ⎠

Finally, the transfer function of the band-stop filter is

2s 2 ( s 2 + 141.4s + 10000 ) + 20000 ( s 2 + 2828s + 4000000 )

HN ( s) = HL ( s) + HH ( s) =
(s 2
+ 141.4s + 10000 )( s 2 + 2828s + 4000000 )
2s 4 + 282.8s 3 + 40000s 2 + 56560000s + 8 ⋅1010
=
( s 2 + 141.4s + 10000 )( s 2 + 2828s + 4000000)

3
P16.3-8

( )
2
⎛ ⎞
2
250 2 + 62500
⎜ s ⎟ 4 s
HN (s) = 4 − 4⎜ 1
⎟ = 2
s + 2502 ⎟ ( s + 250 s + 62500 )
2
250
⎜ s2 +
⎝ 1 ⎠

P16.3-9
2
⎛ ⎞
⎜ 2502 ⎟ 4 ⋅ 2504
HL ( s) = 4 ⎜ ⎟ =
s + 2502 ⎟ ( s + 250s + 62500 )
2
250
⎜ s2 +
2

⎝ 1 ⎠

P16.3-10
2
⎛ ⎞
⎜ s2 ⎟ 4 ⋅ s4
HH ( s) = 4 ⎜ ⎟ =
s + 2502 ⎟ ( s + 250s + 62500 )
2
250
⎜ s2 +
2

⎝ 1 ⎠

4
Section 16.4: Second-Order Filters

P16.4-1
The transfer function is
1
s L×
Cs
1 sL s
sL+
Vo ( s ) Cs s LC + 1
2
sL RC
H (s) = = = = 2 =
Vs ( s ) s L × 1 sL
+ R s LCR + s L + R s 2 +
s
+
1
Cs s LC + 1
2
RC LC
+R
1
sL+
Cs
so
1 1 ω C
K = 1 , ω 02 = and = 0 ⇒ Q = RCω 0 = R
LC RC Q L

1 L
Pick C = 1 μ F. Then L = = 1 H and R = Q = 1000 Ω
Cω02
C

P16.4-2
The transfer function is
1
I 0 (s) LC
T ( s) = =
I s ( s) s 1
s2 + +
RC LC
so
1 1 ω C
k = 1 , ω02 = and = 0 ⇒ Q = RCω0 = R
LC RC Q L

1 L
Pick C = 1μ F then L = = 25 H and R = Q = 3535 Ω
Cω02
C

1
P16.4-3
The transfer function is
1
−
R1 R C 2
T ( s) =
1 ⎛ R⎞ 1
s2 + ⎜ 2+ ⎟ s + 2 2
R C ⎝ R1 ⎠ R C

Pick C = 0.01 μ F , then

1
= ω0 = 2000 ⇒ R = 50000 = 50 k Ω
RC
ω0 1 ⎛ R⎞ R
= ⎜ 2 + ⎟ ⇒ R1 = = 8333 = 8.33 k Ω
Q RC ⎝ R1 ⎠ Q−2

P16.4-4
Pick C = 0.02 μ F. Then R1 = 40 kΩ, R2 = 400 kΩ and R 3 =3.252 kΩ.

P16.4-5
Pick C1 = C2 = C = 1 μ F . Then
106
=ω 0
R1 R2
and
1 ω R1 R1
= 0 ⇒ Q= ⇒ R2 =
R1C Q R2 Q2

106
In this case, since Q = 1, we have R 2 = R1 and R1 = = 1000 = 1 kΩ
1000

2
P16.4-6

The node equations are

R2
V0 ( s ) = Va ( s )
1
R2 +
C 2s
V0 ( s ) −Va ( s )
− C1s (Va ( s ) −Vi ( s ) ) = 0
R1
Doing a little algebra
Vo ( s ) ⎛ 1 ⎞
− V a ( s ) ⎜ + s C1 ⎟ = − s C1 V i ( s )
R1 ⎜ R1 ⎟
⎝ ⎠
Substituting for V a ( s )

Vo ( s ) ⎛ R 2 C 2 s + 1 R1 C 1 s + 1 ⎞
− − Vo ( s ) ⎜ + ⎟⎟ = s C 1 V i ( s )
R1 ⎜ R2 C 2 s R
⎝ 1 ⎠
Vo ( s )
=
( C 1 s ) R1 R 2 C 2 s
=
R1 R 2 C 1 C 2 s 2
V i ( s ) − R 2 C 2 s + ( R 2 C 2 s + 1)( R1 C 1 s + 1) R1 R 2 C 1 C 2 s 2 + R1 C 1 s + 1

The transfer function is:

V0 ( s ) s2
H ( s) = =
Vi ( s ) s 2 + s
+
1
R 2 C 2 R1 R 2 C1 C 2

1 1 ω R2
Pick C1 = C2 = C = 1 μ F. Then = ω and = 0 ⇒Q= ⇒ R1Q 2 = R 2 .
C R1 R 2 0 R 2C Q R1
1
In this case R1 = R 2 = R and = ω0 ⇒ R = 1000 Ω .
CR

3
P16.4-7
1 1
V ( s) Cs LC
T (s) = 0 = =
Vi ( s) L s + R + 1 R
s2 + s+
1
Cs L LC

When R = 25 Ω, L = 10−2 H and C = 4 ×10−6 F, then the transfer function is

25×106
H (s) =
s 2 + 2500 s + 25×106
so
ωold = 25×106 = 5000
and
ωnew 250
kf = = = 0.05
ωold 5000
The scaled circuit is

P16.4-8

The transfer function of this circuit is

4
 R2 1
s
V0 ( s ) 1+ R 2 C 2 s R 2 C 1s R1 C 2
H (s) = =− =− =−
Vi ( s ) R1 +
1 (1+ R1 C 1 s )(1+ R 2 C 2 s ) ⎛ 1
s 2 +⎜ +
1 ⎞
⎟ s+
1
C 1s ⎜ R1 C 1 R 2 C 2 ⎟ R1 R 2 C 1 C 2
⎝ ⎠

100 μ F 500 μ F
Pick km = 1000 so that the scaled capacitances will be = 0.1 μ F and = 0.5 μ F.
1000 1000
Before scaling ( R1 = 20 Ω, C 1 =100 μ F, R 2 = 10 Ω and C2 = 500 μ F )

-100 s
H (s) =
s + 700 s +105
2

After scaling ( R1 = 20000 Ω = 20 kΩ, C1 = 0.1 μ F, R 2 =10000 Ω =10 kΩ, C 2 = 0.5 μ F)

−100s
H (s) =
s + 700 s +105
2

P16.4-9
This is the frequency response of a bandpass filter, so

ω0
k s
Q
H (s) =
ω0
s2 + s +ω 0 2
Q
From peak of the frequency response

ω 0 = 2π × 10 × 10 6 = 62.8 × 106 rad/s and k =10 dB = 3.16

Next
ω0
= BW = (10.1× 106 − 9.9 × 106 ) 2π = (0.2 × 106 )2π = 1.26 × 106 rad/s
Q

So the transfer function is

3.16(1.26)106 s (3.98)106 s
H (s) = =
s 2 + (1.26)106 s + 62.82 ×1012 s 2 + (1.26)106 s + 3.944×1015

5
P16.4-10
1
R2 ×
(a) V0 (ω ) Z j jω C 2
H (ω ) = = − 2 where Z 1 = R1 − and Z 2 =
Vs (ω ) Z1 ω C1 R2 +
1
jω C 2
jω R 2C 1 1 1
∴ H (ω ) = − where ω1 = , ω2 =
⎛ jω ⎞ ⎛ jω ⎞ R1 C 1 R2 C 2
⎜⎜ 1+ ⎟⎟ ⎜ 1+ ⎟
⎜ ω2 ⎟
⎝ ω1 ⎠ ⎝ ⎠

1 1
(b)
ω1 = , ω2 =
R1 C 1 R2 C 2

(c)
jω R 2C 1 ω R 2C 1 R2
H (ω ) = − = = ω 1 R 2C 1 =
⎛ jω ⎞ ω R1
⎜⎜ 0+ ⎟⎟ (1+0 ) ω1
⎝ ω 1 ⎠

P16.4-11

Voltage division:
1
Cs
V0 ( s ) = V ( s ) , ⇒ V1 ( s ) = (1 + s R C )V0 ( s )
1 1
R+
Cs
KCL:
V1 − Vs V −V
+ 1 0 + (V1 −V0 ) nC s = 0
mR R
Combining these equations gives:

⎡ 1 sC 2 ⎤ V
V0 ⎢ +s C + + s n R C2 ⎥ = s
⎣m R m ⎦ mR

6
Therefore
V0 (ω ) 1 1
H (ω ) = = =
Vs (ω ) 1+ s ( m +1) R C + n m R 2 C 2 s 2 ⎛ω ⎞
2
ω
1−⎜ +j
⎜ ω 0 ⎟⎟ Qω0
⎝ ⎠
1 mn
where ω 0 = and Q =
m n RC m +1

P16.4-12

1 R2 1
R2 − s
V0 ( s ) C 2s R 2 C 2 s +1 R1 C 2
H (s) = =− = =
VS ( s ) 1 R1 C 1 s +1 ⎛ 1 1 ⎞
R1 + s 2 +⎜ + ⎟ s+
1
C s C1 s ⎜ R1 C 1 R 2 C 2 ⎟ R1 R 2 C 1 C 2
1 ⎝ ⎠

Substituting the element values R1 = 100 Ω , R 2 = 200 Ω , C1 = 0.02 μ F and C 2 = 50 pF we

determine the center frequency and bandwidth of the filter to be

1
ω0 = = 70.7 k rad sec = 2π (11.25 kHz )
R1 R 2 C 1 C 2
ω0 1 1
BW = = + = 150 k rad s =2π ( 23.9 kHz )
Q R1 C 1 R 2 C 2

7
P16.4-13

Va −V0 ⎫
C 1 s (Va −Vs ) + = 0⎪ −
1
s
R1 ⎪ V0 ( s ) R2 C 2
⎬ ⇒ H ( s )= =
1 1
Va ⎪ V ( s ) s2 + s+
− −C 2 s V0 = 0 s
R2 ⎪⎭ R1 C 1 R1 R 2 C 1 C 2

Comparing this transfer function to the standard form of the transfer function of a second order
bandpass filter and substituting the element values R1 = 1 kΩ , R 2 = 100 Ω , C1 = 1 μ F and
C 2 = 0.1 μ F gives:
1
ω0 = = 104 rad sec
R1 R 2 C 1 C 2
1
BW = = 103 rad/sec
R1 C 1
ω0
Q= = 10
BW

8
P16.4-14
Node equations:

Vc ( s) −Vs ( s) V ( s) −V0 ( s )
a C sVc ( s) + + c =0
R R
V ( s ) −Vc ( s )
s (Vs ( s ) −V0 ( s) ) + s
C
=0
a R

Solving these equations yields the transfer function:

⎛2 ⎞ 1 1
s 2 +⎜ + a ⎟ s+
V0 ( s ) ⎝ a ⎠ RC ( RC)
2

H ( s) = =
Vs ( s ) ⎛ 2⎞ 1 1
s 2 +⎜ ⎟ s+
⎝ a ⎠ RC (RC)
2

1
We require 105 = . Pick C = 0.01μ F then R = 1000 Ω . Next at s = j ω 0
RC
2
+a
a2
H (ω 0 ) = a = 1+
2 2
a
The specifications require
a2
201 = H (ω 0 ) = 1 + ⇒ a = 20
2

9
P16.4-15
Node equations:

Va Va −V0
+ =0
R2 R1
Va −Vb
C sVa + =0
R
V −V V −V
C s (Vb −V0 ) + b s + b a = 0
R R

Solving the node equation yields:

⎛ R1 ⎞ 1
⎜⎜1+ ⎟⎟ 2 2
V0 ( s ) ⎝ R2 ⎠ R C
=
Vs ( s ) ⎛ R ⎞ 1 1
s 2 +⎜ 2− 1 ⎟ s+ 2 2
⎜ R2 ⎟ R C RC
⎝ ⎠
1 1
ω0 = = = 41.67 k rad sec
RC (1.2×10 ) ( 20×10−9 )
3

10
Section 16.5: High-Order Filters

P16.5-1
This filter is designed as a cascade connection of a Sallen-Key low-pass filter designed as
described in Table 16.4-2 and a first-order low-pass filter designed as described in Table 16.5-2.

Sallen-Key Low-Pass Filter:

MathCad Spreadsheet (p16_5_1_sklp.mcd)

c
The transfer function is of the form H(s) =----------------- .
s^2 + bs + a

2
Enter the transfer function coefficitents: a := 628 b := 628
ω0
Determine the Filter Specifications: ω0 := a Q := ω0 = 628 Q=1
b
−6
Pick a convenient value for the capacitance: C := 0.1⋅ 10

1 1 4 4
Calculate resistance values: R := A := 3 − R = 1.592 × 10 R⋅ ( A − 1) = 1.592 × 10
C⋅ ω0 Q

Calculate the dc gain. A=2

1
First-Order Low-Pass Filter:

MathCad Spreadsheet (p16_5_1_1stlp.mcd)

-k
The transfer function is of the form H(s) =-------- .
s+p
Enter the transfer function coefficitents: p := 628 k := 0.5p
−6
Pick a convenient value for the capacitance: C := 0.1⋅ 10
1 1 4 4
Calculate resistance values: R2 := R1 := R1 = 3.185 × 10 R2 = 1.592 × 10
C⋅ p C⋅ k

P16.5-2
This filter is designed as a cascade connection of a Sallen-key high-pass filter, designed as
described in Table 16.4-2, and a first-order high-pass filter, designed as described in Table 16.5-
2.

The passband gain of the Sallen key stage is 2 and the passband gain of the first-order stage is
2.5 So the overall passband gain is 2 × 2.5 = 5

Sallen-Key High-Pass Filter:

2
MathCad Spreadsheet (p16_5_2_skhp.mcd)
A s^2
The transfer function is of the form H(s) =----------------- .
s^2 + bs + a

Enter the transfer function coefficitents: a := 10000 b := 100

ω0
Determine the Filter Specifications: ω0 := a Q := ω0 = 100 Q=1
b
−6
Pick a convenient value for the capacitance: C := 0.1⋅ 10

1 1 5 5
Calculate resistance values: R := A := 3 − R = 1 × 10 R⋅ ( A − 1) = 1 × 10
C⋅ ω0 Q

Calculate the passband gain. A=2

First-Order High-Pass Filter:

MathCad Spreadsheet (p16_5_2_1sthp.mcd)

-ks
The transfer function is of the form H(s) =-------- .
s+p
Enter the transfer function coefficitents: p := 100 k := 2.5
−6
Pick a convenient value for the capacitance: C := 0.1⋅ 10
1 5 5
Calculate resistance values: R1 := R2 := k⋅ R1 R1 = 1 × 10 R2 = 2.5 × 10
C⋅ p

3
P16.5-3
This filter is designed as a cascade connection of a Sallen-key low-pass filter, a Sallen-key high-
pass filter and an inverting amplifier.

Sallen-Key Low-Pass Filter:

MathCad Spreadsheet (p16_5_3_sklp.mcd)

c
The transfer function is of the form H(s) =----------------- .
s^2 + bs + a

Enter the transfer function coefficitents: a := 4000000 b := 2828

ω0 3
Determine the Filter Specifications: ω0 := a Q := ω0 = 2 × 10 Q = 0.707
b
−6
Pick a convenient value for the capacitance: C := 0.1⋅ 10

1 1 3 3
Calculate resistance values: R := A := 3 − R = 5 × 10 R⋅ ( A − 1) = 2.93 × 10
C⋅ ω0 Q

Calculate the dc gain. A = 1.586

4
Sallen-Key High-Pass Filter:

MathCad Spreadsheet (p16_5_3_skhp.mcd)

c s^2
The transfer function is of the form H(s) =----------------- .
s^2 + bs + a

Enter the transfer function coefficitents: a := 10000 b := 141.4

ω0
Determine the Filter Specifications: ω0 := a Q := ω0 = 100 Q = 0.707
b
−6
Pick a convenient value for the capacitance: C := 0.1⋅ 10

1 1 5 4
Calculate resistance values: R := A := 3 − R = 1 × 10 R⋅ ( A − 1) = 5.86 × 10
C⋅ ω0 Q

Calculate the passband gain. A = 1.586

1.6×106
Amplifier: The required passband gain is = 4.00 . An amplifier with a gain equal to
141.4×2828
4.0
= 1.59 is needed to achieve the specified gain.
2.515

5
P16.5-4
This filter is designed as the cascade connection of two identical Sallen-key bandpass filters:

Sallen-Key BandPass Filter:

MathCad Spreadsheet (p16_5_4_skbp.mcd)

cs
The transfer function is of the form H(s) =----------------- .
s^2 + bs + a

Enter the transfer function coefficitents: a := 62500 b := 250

ω0
Determine the Filter Specifications: ω0 := a Q := ω0 = 250 Q=1
b
−6
Pick a convenient value for the capacitance: C := 0.1⋅ 10

1 1
Calculate resistance values: R := A := 3 −
C⋅ ω0 Q

4 4 4
R = 4 × 10 2⋅ R = 8 × 10 R⋅ ( A − 1) = 4 × 10

Calculate the pass-band gain. A⋅Q = 2

6
P16.5-5
This filter is designed using this structure:

Sallen-Key Low-Pass Filter:

MathCad Spreadsheet (p16_5_5_sklp.mcd)

c
The transfer function is of the form H(s) =----------------- .
s^2 + bs + a

Enter the transfer function coefficitents: a := 10000 b := 141.4

ω0
Determine the Filter Specifications: ω0 := a Q := ω0 = 100 Q = 0.707
b
−6
Pick a convenient value for the capacitance: C := 0.1⋅ 10

1 1 5 4
Calculate resistance values: R := A := 3 − R = 1 × 10 R⋅ ( A − 1) = 5.86 × 10
C⋅ ω0 Q

Calculate the dc gain. A = 1.586

7
Sallen-Key High-Pass Filter:

MathCad Spreadsheet (p16_5_5_skhp.mcd)

c s^2
The transfer function is of the form H(s) =----------------- .
s^2 + bs + a

Enter the transfer function coefficitents: a := 4000000 b := 2828

ω0 3
Determine the Filter Specifications: ω0 := a Q := ω0 = 2 × 10 Q = 0.707
b
−6
Pick a convenient value for the capacitance: C := 0.1⋅ 10

1 1 3 3
Calculate resistance values: R := A := 3 − R = 5 × 10 R⋅ ( A − 1) = 2.93 × 10
C⋅ ω0 Q

Calculate the passband gain. A = 1.586

Amplifier: The required gain is 2, but both Sallen-Key filters have passband gains equal to 1.586.
2
The amplifier has a gain of = 1.26 to make the passband gain of the entire filter equal to 2.
1.586

8
P16.5-6
This filter is designed as the cascade connection of two identical Sallen-key notch filters.

Sallen-Key Notch Filter:

MathCad Spreadsheet (p16_5_6_skn.mcd)

c(s^2 + a)
The transfer function is of the form H(s) =----------------- .
s^2 + bs + a

Enter the transfer function coefficitents: a := 62500 b := 250

ω0
Determine the Filter Specifications: ω0 := a Q := ω0 = 250 Q=1
b
−6 −7
Pick a convenient value for the capacitance: C := 0.1⋅ 10 2⋅ C = 2 × 10

1 1
Calculate resistance values: R := A := 2 −
C⋅ ω0 2⋅ Q

4 R 4 4
R = 4 × 10 = 2 × 10 R⋅ ( A − 1) = 2 × 10
2
Calculate the pass-band gain. A = 1.5

4
Amplifier: The required passband gain is 4. An amplifier having gain equal to = 1.78
(1.5)(1.5)
is needed to achieve the required gain.

9
P16.5-7
V1 ( s ) R1 R1 C s
H a (s) = = =
Vs ( s )
(a) Voltage division gives:
R1 +
1 1 + R1 C s
Cs
V2 ( s ) Ls
H b (s) = =
V1 ( s )
(b) Voltage division gives:
R2 + L s
V2 ( s ) R1 || ( R 2 + L s ) Ls
(c) Voltage division gives: H c (s) = = ×
Vs ( s ) 1
+ R1 || ( R 2 + L s ) R 2 + L s
Cs
R1 × ( R 2 + L s )
V2 ( s ) R1 + ( R 2 + L s ) Ls
Doing some algebra: H c (s) = = ×
Vs ( s ) 1 R1 × ( R 2 + L s ) R2 + L s
+
C s R1 + ( R 2 + L s )
R1 R 2 C s + R1 L C s 2 Ls
= ×
R1 R 2 C s + R1 L C s + R1 + R 2 + L s R 2 + L s
2

R1 C s ( R 2 + L s ) Ls
= ×
R1 L C s + ( R1 R 2 C + L ) s + R1 + R 2
2
R2 + L s
R1 L C s 2
=
R1 L C s 2 + ( R1 R 2 C + L ) s + R1 + R 2

(d) 1
H c ( s ) ≠ H a ( s ) × H b ( s ) because the R 2 , L s voltage divider loads the , R1 voltage
Cs
divider.

10
P16.5-8
100 20
H (s) = ×
⎛ s ⎞⎛ s ⎞ ⎛ s ⎞⎛ s ⎞
⎜1 + ⎟ ⎜1 + ⎟ ⎜1 + ⎟ ⎜1 + ⎟
⎝ 200 π ⎠ ⎝ 20, 000 π ⎠ ⎝ 20 π ⎠ ⎝ 4000 π ⎠
2000
=
⎛ s ⎞⎛ s ⎞⎛ s ⎞⎛ s ⎞
⎜1 + ⎟ ⎜1 + ⎟ ⎜1 + ⎟ ⎜1 + ⎟
⎝ 20 π ⎠ ⎝ 200 π ⎠ ⎝ 4000 π ⎠ ⎝ 20, 000 π ⎠

P16.5-9
(a) The transfer function of each stage is
1
R2 ×
Cs
1 1 R2 R2
R 2 || R2 +
Cs Cs 1+ R2 C s R1
H i (s) = − =− =− =−
R1 R1 R1 1+ R2 C s

The specification that the dc gain is 0 dB = 1 requires R 2 = R1 .

1
The specification of a break frequency of 1000 rad/s requires = 1000 .
R2 C
Pick C = 0.1 μ F . Then R 2 = 10 kΩ so R1 = 10 kΩ .

(b) ⎛
2
⎞
−1 −1 1 1
H (ω ) = × ⇒ H (10, 000 ) = ⎜ ⎟ = = −40.1 dB
ω ω ⎝ + 2
⎠ 101
1+ j 1+ j 1 10
1000 1000

11
Section 16.7 How Can We Check…?

P16.7-1
ω0 100
ω 0 = 10000 = 100 rad s and = 25 ⇒ Q = = 4 ≠ 5
Q 25
This filter does not satisfy the specifications.

P16.7-2
ω0 100 75
ω 0 = 10000 = 100 rad s , = 25 ⇒ Q = = 4 and k= = 3
Q 25 25

This filter does satisfy the specifications.

P16.7-3
ω0 20 600
ω 0 = 400 = 20 rad s , = 25 ⇒ Q = = 0.8 and k = = 1.5
Q 25 400

This filter does satisfy the specifications.

P16.7-4
ω0 25 750
ω 0 = 625 = 25 rad s , = 62.5 ⇒ Q = = 0.4 and k = = 1.2
Q 62.5 625

This filter does satisfy the specifications.

P16.7-5
ω0 12
ω 0 = 144 = 12 rad/s and = 30 ⇒ Q = = 0.4
Q 30

This filter does not satisfy the specifications.

1
PSpice Problems
SP 16-1

1
SP 16-2

2
SP 16-3

3
SP 16-4

4
SP 16-5

5
SP 16-6

6
SP 16-7

Vs 7 0 ac 1
R1 7 6 200
R2 7 1 100
R3 7 2 50
L1 6 5 10m
L2 1 3 10m
L3 2 4 10m
C1 5 0 1u
C2 3 0 1u
C3 4 0 1u

.ac dec 100 100 10k

.probe
.end

SP 16-8

Vs 1 0 ac 1
R1 1 2 100

7
 C1 2 3 0.2u
R2 3 4 200k
C2 3 4 50p
Xoa5 3 0 4 FGOA
.subckt FGOA 1 2 4
*nodes listed in order - + o
Ri 1 2 500k
E 3 0 1 2 100k
Ro 4 3 1k
.ends FGOA
.ac dec 100 1k 100k
.probe
.end

SP 16-9

Vs 1 0 ac 1
L1 1 2 2.5m
Rw 2 0 8
C2 1 3 34.82u
L2 3 4 0.364m
Rmr 4 0 8
C3 1 5 5u
Rt 5 0 8
.ac dec 100 10 100k
.probe
.end

Bw=4.07k - 493 HZ ∼ 3600HZ

8
Design Problems

DP 16.1
s
−
V0 ( s ) RC
=
V1 ( s ) 2 2
s+ s+
R3 C R R3 C 2

2 ω 2
2π (100.103 ) = ω 0 = 2
and 2π (10.103 ) = BW = 0 =
R R3 C Q R3 C

2 2
C = 100 pF is specified so R3 = = 318 kΩ and R = = 1.6 kΩ
(100×10 −12
) (2π ×10×10 )
3
R3C 2ω 02

DP 16.2

1
DP 16.3
Choose ω1 = 0.1 , ω 2 = 2 , ω3 = 5 , ω 4 = 100 rad s . The corresponding Bode magnitude plot
is:

( ω)
2 2
1+ s ⎛1+ s ⎞
⎜ ω3 ⎟⎠
H (s) = 1 ⎝

(1+ sω ) (1+ sω )
2 2

2 4

Minimum gain is − 46.2 dB at f min = 0.505 Hz