1
ME360  Solved Examples for Homework #8
Problem 1.
1)
Derive 1st, 2nd, and 3rd order Butterworth denominator polynomials for
ω
=
ω
c
=3
.
5
π
rad/s. Find the poles
of the polynomials for each case, draw splane map with pole location for each case, comment on the distribution of
poles in the splane. Plot the magnitude spectral densities of the corresponding three low pass filters on the same graph
and comment on their relative filtering properties. Plot the phase responses of these three filters on the same graph
and comment on their relative phase distortion.
First order:
T
(
s
)=
1
as
+1
=
⇒

T
(
jω
)

2
=

1
jaω
+1

2
=
1
1+(
aω
)
2
=
1
1+
a
2
ω
2

T
(
jω
)

ω
=3
.
5
π
=
1
√
2
=
⇒
1
√
1+(3
.
5
πa
)
2
=
1
√
2
=
⇒
3
.
5
πa
=1=
⇒
a
=
1
3
.
5
π
So,
T
(
s
)=
1
as
+1
=
1
s/
3
.
5
π
+1
=
3
.
5
π
s
+3
.
5
π
pole:
s
=
−
3
.
5
π
.I
t
l
i
e
so
nac
i
r
c
l
eo
f
R
=3
.
5
π
in the splane.
15
10
5
0
5
10
15
10
5
0
5
10
Second order:
T
(
s
)=
1
as
2
+
bs
+1
=
⇒

T
(
jω
)

2
=

1
−
aω
2
+
jbω
+1

2
=
1
(1
−
aω
2
)
2
+
b
2
ω
2
=
1
a
2
ω
4
+(
b
2
−
2
a
)
ω
+1
thus,
b
2
−
2
a
=0
and

T
(
jω
)

ω
=3
.
5
π
=
1
√
2
=
⇒
1
√
1+(3
.
5
π
)
4
a
2
=
1
√
2
=
⇒
(3
.
5
π
)
4
a
2
=1=
⇒
a
=
1
(3
.
5
π
)
2
b
2
=2
a
=
⇒
b
=
√
2
3
.
5
π
So,
T
(
s
)=
1
as
2
+
bs
+1
=
1
s
2
/
(3
.
5
π
)
2
+
s
√
2
/
(3
.
5
π
)+1
=
(3
.
5
π
)
2
s
2
+3
.
5
π
√
2
s
+(3
.
5
π
)
2
poles:
s
=3
.
5
π
∗
(
−
√
2
2
±
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15
10
5
0
5
10
15
10
5
0
5
10
Third order:
T
(
s
)=
1
as
3
+
bs
2
+
cs
+1
=
⇒

T
(
jω
)

2
=

1
−
jaω
3
−
jbω
2
+
jcω
+1

2
=
1
(1
−
bω
2
)
2
+(
cω
−
aω
3
)
2
=
1
a
2
ω
6
+(
b
2
−
2
ac
)
ω
4
+(
c
2
−
2
b
)
ω
2
+1
thus,
c
2
−
2
b
=0
,b
2
−
2
ac
=0
.
and

T
(
jω
)

ω
=3
.
5
π
=
1
√
2
=
⇒
1
√
1+(3
.
5
π
)
6
a
2
=
1
√
2
=
⇒
(3
.
5
π
)
6
a
2
=1=
⇒
a
=
1
(3
.
5
π
)
3
c
2
=2
b
=
⇒
c
4
=4
b
2
=8
ac
=
⇒
c
3
=8
a
=
⇒
c
=
2
3
.
5
π
b
=
c
2
2
=
2
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 Spring '08
 Staff
 SEPTA Regional Rail, Lowpass filter, Butterworth Filter, Vout Vin

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