We saw in (Reference) that the transfer function of a linear timeinvariant 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 ssplane where the transfer function is either nonexistent (for a pole) or zero (for a zero). These points can be plotted in the ssplane 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 jΩβkjΩβk have the same magnitude and phase as a vector from the zero βkβk to the jΩjΩ axis in the complex ssplane. Likewise, the quantities jΩαkjΩαk have the same magnitude and phase as vectors from the pole αkαk to the jΩjΩ axis.
Example 3.1 Consider a firstorder 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 firstorder 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 secondorder 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 polezero 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 ssplane, 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 highfrequency sinusoids used in wireless communications.