Download The natural response is

Lecture 22
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
j
s-plane

Figure 22-1
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
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
H (s)  50(s  3)(s2  4s  8)
s(s  4)(s  2s  2)
(22-1)
The poles are at s=0, -1, -1+j1, -1-j1; the zeroes are at s =-3, -2+j2, -2j2, respectively. The pole-zero plot for the function is shown in Fig.
22-2.
j
s-plane
o X
2
1
X o
-4
-3
-2
-1
X
0
-1
o X

-2
Figure 22-2
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.
Example 22-1
Given the circuit in Fig. 22-3, find the natural response of the
capacitor voltage vc(t).
i
2H
+
vs
+_
vL
ic
-
+
iR
4
1
F
4
vc
-
Fig. 22-3
Applying KCL at the capacitor node gives,
i  vc  1 dvc
4 4 dt
(22-2)
Writing KVL around the outer loop we get,
vs  vc  2 di
dt
Substituting (22-2) into (22-3) gives,
(22-3)
d 2vc  dvc  2v  2v
c
s
dt 2 dt
(22-4)
Solution of this equation when vs=0 is the natural response. As was
done previously, we get the natural response by assuming a solution
of the form Aest. Doing so, we obtain two values of s that satisfy the
equation
s2  s  2  0
(22-5)
Using the quadratic formula, we find
s 1  j 7
2
2
(22-6)
In other words, the natural response vn(t) has the form
1 Ae 2
vn (t)  Ae
1
2
st
st
(22-7)
Where,
1
7
s1   1  j 7 and s1    j
2
2
2
2
Let us now redraw the circuit given in Fig. 22-3 in phasor domain in
terms of complex impedance.
I
2s
Ic
+ VL -
Vs
+
IR
4 4
+_
s
Vc
-
Fig. 22-4
Writing KCL at the capacitor node, we get
Vc Vs  Vc  Vc  0
2s
4 4/ s
(22-8)
Simplify to get,
(s2  s  2)Vc  2Vs
(22-9)
Thus,
Vc 
2Vs
s2  s  2
(22-10)
Alternatively, you can find the impedance of RC combination and
apply voltage division principle to get Vc in terms of input voltage Vs.
Observe that the poles of the expression for Vc in (21-10) are
precisely the roots of the characteristic equation (22-5). Hence if you
know the transfer function and the poles, you can write the natural
response given in (22-7) right away. The constants A1 and A2 depend
on the initial values of the inductor current and capacitor voltage.
Determination of these for particular initial conditions is a different
issue.
Example 22-2
Find the natural response for the output voltage vo in the circuit of Fig.
22-5.
3
1H
+
vg
3
+_
vo
1F
1H
Fig. 22-5
Re-draw Fig. 22-5 in phasor notation and in terms of complex
impedance in terms of s.
s
3
+
Vg
3
+_
s
1
s
Vo
-
Fig. 22-6
The impedance of the parallel combination is,
Z p  1 ( 3 s )
s 3  s 1/ s
The total impedance seen by the voltage source is then,
(22-11)
ZT  3  s  Z p
( 22-12)
Applying the voltage divider principle,
Vo 
Zp
V
ZT g
(22-13)
Simplify to get,
1
Vg
s 2  3s  2
1
=
V
(s+1)(s+2)
Vo 
(22-14)
The natural response von(t) is then written as,
t  A e2t (22-15)
von (t )  Ae
1
2
Again, the constants A1 and A2 depend on the initial conditions of the
circuit elements.
The Complete Response
The complete response is obtained by summing the natural response
with the forced response. In the previous lecture you have seen how
we can find the forced response for complex frequency s. Now if we
add the natural response obtained through the examination of the
poles of the transfer (or network) function the complete response, say
for voltage, is given as
v(t )  v f (t )  vn (t )
Let us illustrate this though an example.
Example 22-3
Consider the circuit shown in Fig. 22-7. For this circuit,
a) Find
H (s) 
I1(s)
Vg (s)
b) Find the complete response i1(t) if vg= 2 e-t cost
12
6
i
vg
+_
i1
3H
2H
Fig. 22-7
The corresponding diagram in terms of phasors and complex
impedance is given in Fig.22-8.
12
6
I
vg
+_
I1
3s
Fig. 22-8
The total impedance seen by the source is
2s
ZT 12  3s(6  2s)
3s  6  2s
(22-16)
The total current is then,
I
Vg
 2 5s  6
V
ZT 6s  78s  72 g
(22-17)
By current divider principle,
I1 
=
5s  6
3s
Vg (
)
6s2  78s  72 5s  6
(22-18)
s
Vg
2(s2 13s 12)
Since in terms of phasor
Vg  20oV and s=-2+j1,
Current I1 in phasor domain is,
s
20o
2(s 1)(s 12)
-2+j1
2.23153.43o
=

(-2+j1+1)(-2+j1+12) 1.41135o10.055.71o
I1 
(22-19)
=0.1612.7o
The forced response as a function of time is then,
i1 f  0.16e2t cos(2t 12.7o ) A (22-20)
From examination of the poles of function I1, the natural response is
t  A e12t A
I1n (t )  Ae
1
2
The complete response is then
t  A e12t  0.16e2t cos(2t 12.7o ) A (22-21)
I1n (t )  Ae
1
2
The constants A1 and A2 are to be determined from the initial
conditions.
Pole Locations and Natural Response
Consider the general form of the transfer function given in the
previous lectures
K (s - z1)(s  z2 ).......(s  zm )
H (s)  Vo 
Vin (s  p1)(s  p2 ).......(s  pn )
(22-22)
Here, z1, z2, ....zm are the zeroes and p1, p2, ......pn are the poles. The
poles and zeroes can be either real or complex, but the complex
poles and zeroes occur in conjugate pairs. Let us consider the natural
response for the voltage vo for different locations of the poles in the
complex s-plane.
Case I: Pole at the origin
If the poles are real and distinct, the natural response for voltage vo(t)
is
p1t
p2t
(22-23)
vn (t)  Ae

A
e
 .............  Ane pnt
1
2
If the pole pi is at the origin of the s-plane, that is, pi=0, then the term
of the natural response corresponding to it is Ai e0=Ai , a constant.
The location of the pole in s-plane and corresponding time response
is shown in Fig. 22-9.
j
vn
s-plane
Ai
X

t
(b)
(a)
Figure 22-9
Case II: Distinct real poles
The location of pole pi in the s-plane and corresponding time
responses are shown in Fig. 22-10. It can be observed that if pi is in
the left half s-plane, the corresponding time response is decaying,
while the response grows if the pole is in the right half of the plane.
j
vn
s-plane
X
Pi
Ai

t
(a)
(b)
j
vn
s-plane
Ai
X
Pi

t
(c)
(d)
Figure 22-10
Case III: Complex conjugate poles
If pi is a complex number, then there is a pole pj=pi*. We have seen in
earlier chapters that when there is complex conjugate pair of roots
σ±jω, the corresponding term in the natural response can be obtained
in the form Aeσt cos(ωt+θ). The same reasoning applies for poles also.
The location of the poles and the corresponding time responses are
shown in Fig. 22-11.
Note that if the pair of poles are in the left half of the s-plane (that is,
σ<0), this term is a damped sinusoid. If the pair of poles are on the
right half of the s-plane, the term is an increasing sinusoid.
j
vn
s-plane
X
Ai

t
X
(a)
(b)
j
s-plane
X
vn
Ai

t
X
(c)
(b)
Figure 22-11
SELF-TEST (22)
1. Draw the pole-zero locations for the transfer function
H (s)  152s  45s
s  6s  8
2
are given in the following 3 diagrams. State the correct answer.
X
o
X
-4
-3
-2
a)
X
X
o
-4
-2
3
b)
c)
X
o
X
2
3
4
Ans: (a)
2. The transfer function for a circuit is
H (s)  3s(s  3)
(s 1)(s  2)
Write the equation for the natural response is (A and B are
constants)
a) A e-2t + B e-3t
2t  Bet
b) Ae
1
c) A e2t + B e3t
d) A e2t + B et
Ans: (b)
3. The poles of a transfer functions are
-2,
-5, 1+ j1 , 1– j1
The natural response is
a) Stable (decaying)
b) Unstable (growing)
c) Constant
Ans: (b)