We saw in (Reference) that the transfer function of a linear time-invariant system is given by
H
(
s
)
=
Y
(
s
)
X
(
s
)
H
(
s
)
=
Y
(
s
)
X
(
s
)
(1)If we assume that H(s)H(s) is a rational function of ss then we can write
H
(
s
)
=
(
s
-
β
1
)
(
s
-
β
2
)
⋯
(
s
-
β
q
)
(
s
-
α
1
)
(
s
-
α
2
)
⋯
(
s
-
α
p
)
H
(
s
)
=
(
s
-
β
1
)
(
s
-
β
2
)
⋯
(
s
-
β
q
)
(
s
-
α
1
)
(
s
-
α
2
)
⋯
(
s
-
α
p
)
(2)where β1,β2,...,βqβ1,β2,...,βq are the zeros, and α1,α2,...,αpα1,α2,...,αp are the poles of H(s)H(s). The poles and zeros are points in the ss-plane where the transfer function is either non-existent (for a pole) or zero (for a zero). These points can be plotted in the ss-plane with “××" representing the location of a pole and a “∘∘" representing the location of a zero.
Since
H
j
Ω
=
H
(
s
)
s
=
j
Ω
H
j
Ω
=
H
(
s
)
s
=
j
Ω
(3)we have
H
j
Ω
=
j
Ω
-
β
1
j
Ω
-
β
2
⋯
j
Ω
-
β
q
j
Ω
-
α
1
j
Ω
-
α
2
⋯
j
Ω
-
α
p
H
j
Ω
=
j
Ω
-
β
1
j
Ω
-
β
2
⋯
j
Ω
-
β
q
j
Ω
-
α
1
j
Ω
-
α
2
⋯
j
Ω
-
α
p
(4)and
∠
H
j
Ω
=
∑
k
=
1
q
∠
j
Ω
k
-
β
k
-
∑
l
=
1
p
∠
j
Ω
l
-
α
l
∠
H
j
Ω
=
∑
k
=
1
q
∠
j
Ω
k
-
β
k
-
∑
l
=
1
p
∠
j
Ω
l
-
α
l
(5)Each of the quantities jΩ-βkjΩ-βk have the same magnitude and phase as a vector from the zero βkβk to the jΩjΩ axis in the complex ss-plane. Likewise, the quantities jΩ-αkjΩ-αk have the same magnitude and phase as vectors from the pole αkαk to the jΩjΩ axis.
Example 3.1 Consider a first-order lowpass filter with transfer function
H
(
s
)
=
1
s
+
2
H
(
s
)
=
1
s
+
2
(6)Then
H
j
Ω
=
1
j
Ω
+
2
H
j
Ω
=
1
j
Ω
+
2
(7)The quantity jΩ+2jΩ+2 has the same magnitude and phase as a vector from the pole at s=-2s=-2 to the jΩjΩ axis in the complex plane. The magnitude of the frequency response is the inverse of the magnitude of this vector. The length of jΩ+2jΩ+2 increases as ΩΩ increases thereby making HjΩHjΩ decrease as one would expect of a lowpass filter.
Example 3.2 Consider a first-order highpass filter with transfer function
H
(
s
)
=
s
s
+
2
H
(
s
)
=
s
s
+
2
(8)Then
H
j
Ω
=
j
Ω
j
Ω
+
2
H
j
Ω
=
j
Ω
j
Ω
+
2
(9)When Ω=0Ω=0, HjΩ=0HjΩ=0, but as ΩΩ approaches infinity, the two vectors jΩjΩ and jΩ+2jΩ+2 have equal lengths, so the magnitude of the frequency response approaches unity.
Example 3.3
Let's now look at the transfer function corresponding to a second-order filter
H
(
s
)
=
s
+
1
(
s
+
1
+
j
5
)
(
s
+
1
-
j
5
)
H
(
s
)
=
s
+
1
(
s
+
1
+
j
5
)
(
s
+
1
-
j
5
)
(10)Or
H
j
Ω
=
j
Ω
+
1
j
Ω
+
1
+
j
5
j
Ω
+
1
-
j
5
H
j
Ω
=
j
Ω
+
1
j
Ω
+
1
+
j
5
j
Ω
+
1
-
j
5
(11)The pole-zero plot for this transfer function is shown in Figure 1. The corresponding magnitude and phase of the frequency response are shown in Figure 2.
The two poles are at s=-3±j5s=-3±j5 and the zero is at s=-1s=-1. When the frequency gets close to either one of the poles, the frequency response magnitude increases since the lengths of one of the vectors jΩ±j5jΩ±j5 is small.
In the previous example we saw that as the poles get close to the jΩjΩ axis in the ss-plane, the frequency response magnitude increases at frequencies that are close to the poles. In fact, when a pole is located directly on the jΩjΩ axis, the filter's frequency response magnitude becomes infinite, and will begin to oscillate. If a zero is located on the jΩjΩ axis, the filters frequency response magnitude will be zero.
Example 3.4
Suppose a filter has transfer function
H
(
s
)
=
s
s
2
+
25
H
(
s
)
=
s
s
2
+
25
(12)The two poles are at s=±j5s=±j5 and there is a zero at s=0s=0. From (Reference), the impulse response of the filter is h(t)=cos(5t)u(t)h(t)=cos(5t)u(t). This impulse response doesn't die out like most of the impulse responses we've seen. Instead, it oscillates at a fixed frequency. The filter is called an oscillator. Oscillators are useful for generating high-frequency sinusoids used in wireless communications.
