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UNIVERSITY OF CALIFORNIA, SAN DIEGO
Department of Electrical and Computer Engineering
BangSup Song
1
ECE163
Lecture #1: Feedback Amplifier Stability
v
i
i
i
Network
v
i
i
i
Network
v
o
i
o
Oneport network
Twoport network
Fig. 1.1: Drivingpoint and transfer concepts.
DrivingPoint and Transfer Functions
Consider any networks with one port or two ports as shown in Fig. 1.1.
On each port,
we can define its terminal voltage and the current flowing into the terminal.
In the one
port case shown on the left side, the ratio of the terminal voltage to the current can be
defined.
Rs
()
=
v
i
s
i
i
s
,
and
Gs
=
i
i
s
v
i
s
.
(
1
.
1
)
The former is drivingpoint resistance, and the latter one is drivingpoint conductance.
Their units are
±
and 1/
±
, respectively.
That is, if the input and output are referred to the
same port, the term “drivingpoint” is used.
On the other hand, in the twoport case
shown on the right side, the following four ratios can be defined.
A
v
s
=
v
o
s
v
i
s
,
A
i
s
=
i
o
s
i
i
s
,
=
v
o
s
i
i
s
,
and
=
i
o
s
v
i
s
,
(1.2)
where A
v
(s) and A
i
(s) are unitless transfer functions defined as voltage and current
gains, respectively.
The latter two definitions are the same as in (1.1) for the oneport
network, but they are called as transresistance and transconductance, respectively.
The
term “trans” is now used since the input and output ports are different.
In steadystate smallsignal analysis, impedances of reactive components such as
inductor and capacitor are frequencydependent as sL and 1/sC, where s is the complex
frequency of j
²
.
The unit of
²
is rad/sec.
Note that the angular frequency
²
is defined
as the amount of angle rotation per second, while the ordinary frequency unit in Hertz
(Hz) is defined as the number of rotations per second.
Since one rotation of a vector
covers an angle of 2
³
radian, there is a 2
³
difference between
²
and f like
²
= 2
³
f.
In
general, all these transfer functions can be represented as a ratio of two polynomial
functions N(s) and D(s) in steady state.
Let’s consider a general transfer function H(s) as
follows.
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View Full DocumentUNIVERSITY OF CALIFORNIA, SAN DIEGO
Department of Electrical and Computer Engineering
BangSup Song
2
ECE163
Hs
()
=
Ns
Ds
=
b
1
s
m
+
b
2
s
m
±
1
+
.....
+
b
m
a
1
s
n
+
a
2
s
n
±
1
+
.....
+
a
n
.
(
1
.
3
)
The solutions of D(s) = 0 are poles, and those of N(s) = 0 are zeros.
These poles and
zeros affect the magnitude and phase responses of the transfer function.
For any
networks to be (exponentially) stable, poles should be placed in the open left half
complex splane.
That is, the real part of the poles should be negative like Re [poles] < 0.
Frequency Response
If s is replaced by j
±
, the steadystate frequency response of a transfer function is
obtained.
That is, the transfer function is defined as a complex function H(j
±
) at
frequency
±
.
The complex function H(j
±
) can be represented by either of the following
rectangular or polar form.
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 Spring '08
 staff
 Amplifier

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