Frequency response of feedback amplifiers
• In previous context of discussion for feedback amplifiers, it
is assumed that open-loop gain and feedback ratio are
independent of frequency.
• But open-loop gain of real amplifiers is a function of
frequency. Magnitude response drops off and phase shift
increases at high frequencies.
• When feedback is applied to the open-loop amplifier,
undesirable frequency response (also transient response)
can result.
• Considering frequency dependence, the closed-loop gain
of a feedback amplifier should be re-formatted as function
A( s)
of Laplace variable S as follows: A f ( s) 
1  A( s)  ( s)
• The zeros and poles of the above transfer function are the
key to understand the behavior of the feedback amplifier
Transient response in terms of pole location
• Remember from the Circuit Analysis Course, the mathematical form of
the transient response is related to the location of the poles in the
complex domain.
Suppose   0, then a pole at s  - (on the negative real axis) results in
the transient response of the form exp(- t), which eventually decays to 0.
On the other hand, a pole s   (on the positive real axis) results in a transient
response of the form exp( t), which goes to infinity.
• Obviously, we do not want to have poles on the positive real axis,
because the transient response eventually drives the amplifier into
voltage limits, resulting in nonlinear distortion.
In circuit analysis, exponetial transient terms are written in the form exp(-t/  ),
where  is called the time constant.
For a pole at s  - , the time constant is   1/ .
• Approximately within 5 time constant, the amplitude of exp terms
decays to negligible value compared to initial amplitude.
• The greater the distance of the pole from the origin, the faster the
transient response decays
Transient response in terms of pole location
For circuit tr ansfer functions, complex poles occur in conjugate paris. So, a pole at
s  -  j always has a conjugate s  - - j
A pair of complex poles incurs a transient terms of the form exp(- t)(Acos t  Bsin t),
where A and B are constants that depend on the details of the circuit and initial conditions .
If  is larger tha n  , the transient dies to an insignific ant amplitude within one cycle of
sinusoid. On the other hand, if  is much smaller th an  , many oscillatio ns occur before
the amplitude becomes small. This is called ringing and should be avoided in amplifiers
for pulse - type signals.
Poles in the right half of the complex plane at s    j (  0) lead to transient terms
of the form e t ( A cos t  B sin t ). This is even worse because they grow in amplitude
until reaching output limit.
As a conclusion , in amplifier design, for stability, the complex poles must fall in the left
plane of the complex plane. Furthermor e, to poles must have    to avoid ringing.
Transient response in terms of pole location
Frequent response in terms of pole location
Bode magnitude contributi ons for poles at various locations in the complex plane
are shown in the following plot. Notice that complex poles with  much less than
 display a sharp gain peak.
Desired pole location
Often, amplifiers are required to
have nearly constant gain for a
given range of frequency and the
gain is required to roll off at
higher frequencie s. Considerin g
both trans ient response and
frequency response in terms of
pole locations, the desired pole
locations for most amplifiers are
with  45 o of the negative real axis.
Poles in this region gives transient
response decaying faster and
frequency response showing no
excessive gain peaks.
Effects of feedback on pole: one pole I
• Negative feedback has dramatic effects on pole locations
of amplifiers (OpAmp), which in turn affects transient
response and frequency response of the amplifiers
• First, considering a one-pole (or dominant pole) amplifier,
the open loop gain is of the form
A( s) 
A0
, where A0 is open  loop gain and f b is open loop break frequency.
1  s / 2f b
• With feedback in this amplifier, then the closed-loop gain
A f ( s) 
A f ( s) 
A0 /(1  s /( 2f b ))
A( s)

, assuming  is constant.
1  A( s )  ( s ) 1  A0 /(1  s /( 2f b ))
A0 f ( s )
1  s /( 2f bf )
will be A0 f 
, where closed - loop dc gain and break frequency
A0
, f bf  f b (1  A0  )
1  A0 
A0 f f bf  A0 f b , this is the gain - bandwidth produce often, this is a specificat ion for an amplifier.
Using different  , we can change the gain and bandwith t o design needs.
Effects of feedback on pole: one pole II
• Transient and frequency response of feedback amplifiers are related
to the pole location, so we need to consider how the pole would
change as feedback ratio  changes
• For single-pole (dominant-pole) amplifier, the pole for the closed-loop
gain becomes s  2f bf  2f b (1  A0  )
• So, the above pole is still on the negative real axis, but moves further
from the origin as  increases
• Real amplifiers usually have more than one pole. However, sometimes
one pole is much closer to the origin than others (called dominant
pole). In that case, we can ignore the rest of the poles.
• Summary: for one-pole amplifier, the feedback back leads to a smooth
roll-off of the frequency response and fast decaying of the transient
response without ringing
Example of a one-pole amplifier:
Effects of feedback on pole: two pole I
• Now, considering a two-pole amplifier, the open loop gain
is of the form
A( s ) 
A0
, where A0 is open  loop gain at DC
(1  s / 2f b1 )(1  s / 2f b 2 )
and f b1 and f b 2 are two open-loop br eak frequency (assumed to be real values)
• Again assume feedback ratio is constant (not a function of
frequency) and evaluate the poles of the closed loop
transfer function
we need to solve equation 1  A(s)  0
s 2  s(2f b1  2f b 2 )  (1  A0  )4 2 f b1 f b 2  0, the roots are then
1
1
s  - (2f b1  2f b 2 ) 
(2f b1  2f b 2 ) 2  16 2 f b1 f b 2 (1  A0  )
2
2
• For the poles, as  increases, the poles move together until
they meet at the point in the middle. Then, further increase
causes the poles to become complex, moving away from the
real axis along the vertical line across the meeting point. (the
path followed by the poles is called a root locus)
Effects of feedback on pole: two pole II
• Usually, feedback amplifiers are
designed so that A0  is much
larger than unity, which is usually
necessary to achieve gain
stabilization, impedance control,
nonlinear distortion reduction etc.
• From the root locus, it can be
seen that a too large value of A0 
might move the poles outside the
desirable region of the s-plane
(the 45 degree negative axis). In
that case, undesirable frequency
response peaks and transient
ringing occurs.
Example of a two-pole amplifier:
Transient response
Frequency response
Effects of feedback on pole: three poles
• An amplifier with three or more poles can be analyzed using the
same method as in the one pole and two poles case, but math
analysis get much more complicated.
• Qualitatively, in three poles case, feedback can cause the poles to
move even to the right half of the complex plane, thus making the
amplifier instable.
• Example root lotus for 3 poles and 4 poles are shown below:
Example of a two-pole amplifier:
Transient response