Lecture 20
In this lecture the following material is covered:
•
We explore the complex frequency (
s
) plane
•
Natural response of a circuit is obtained from pole
location
•
Examples are worked to find the complete response
(forced and natural) through complex frequency domain
analysis
•
Demonstrate how the location of poles of the poles can
affect the natural response
•
Finally, a test on the material of this chapter will be
conducted
.
The Complex Frequency (s) Plane
The complex frequency plane or the s-plane is sketched in Fig.22-1.
Recollect that,
s
j
σ
ω
=
+
s-plane
Figure 22-1
j
ω
σ
The horizontal (x) axis of the plane is real part
σ
and the vertical (y) axis being the
imaginary part
ω
. Also, observe that both
σ
and
ω
are frequencies; one being real part and
the other imaginary part of complex frequency. From the several of examples solved you
may have noticed that response of a circuit depends both on
σ
as well as
ω
. The first term

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
controls how quickly the transients (or oscillations) will die down or grow. The second
term controls the frequency of the oscillations in the response
.
If we denote a pole by a symbol ‘x’ and a zero by a symbol ‘
Ο
’, then
depicting the poles and zeroes of a function in the s-plane is referred
to as
pole-zero plot
. Consider for example the following function,
2
2
50(
3)(
4
8)
( )
(
4)(
2
2)
s
s
s
H s
s s
s
s
+
+
+
=
+
+
+
(22-1)
The poles are at
s
=0, -1, -1+j1, -1-j1; the zeroes are at
s
=-3, -2+j2, -2-
j2,
respectively. The pole-zero plot for the function is shown in Fig.
22-2.
s-plane
-1
-2
-3
-4
0
-1
-2
1
2
X
o
X
X
X
o
o
Figure 22-2
j
ω
σ
1`
Natural Response from Transfer (or, Network) Function
In earlier chapters we learnt how to find the natural response of a
circuit. We write the circuit equations in terms of differential equations
and solve for the response by setting the input to zero. To refresh are
our memory, let us consider an example.