Yes, it's important to realize that negative feedback 'moves' the poles and zeroes from their original position in the open loop amp.
Just the poles, the zeros stay where they are.
For a simple feedback system, zeros in the forward path also occur in the closed-loop input-to-output transfer (when the forward path has zero gain for some value of s, there is nothing the feedback can do about that), zeros in the feedback path do not end up in the closed-loop transfer.
For a simple feedback system, zeros in the forward path also occur in the closed-loop input-to-output transfer (when the forward path has zero gain for some value of s, there is nothing the feedback can do about that), zeros in the feedback path do not end up in the closed-loop transfer.
You're right. A zero in forward path stays unchanged in the closed loop transfer. However, NFB introduces a pole on slightly higher freq than the zero, that tends to 'undo' the zero so the closed loop response gets flattened.
I built a demonstrator circuit for conditional stability somewhere around 1994, the schematic was something like this. It's a negative-feedback amplifier with a DC loop gain of about -16 000 000 000. Each LF356 contributes a pole at about -50 pi rad/s, C1 adds another pole at about -60 rad/s. Hence, the loop gain drops with -60 dB/decade, -270 degrees phase shift, until R2 and L1 kick in and reduce the slope to first order (-20 dB/decade, -90 degrees).
The response is well-damped (roughly third-order Butterworth at 46 kHz) when the potmeter is at its maximum. Reduce the potmeter setting and the point where the 0 dB loop gain axis is crossed gradually shifts into the region where the loop gain drops at a too high rate. The square wave response gets more and more overshoot and ringing when the loop gain is turned back until it starts to oscillate. Further reduction of the loop gain reduces the oscillation frequency. Theoretically it should get stable again when the loop gain is turned way down, but the attenuation of the potmeter was too small for that.
The root locus was supposed to be something like this. The crosses are the poles of the loop gain, the circles the zeros (which don't end up in the transfer in this circuit) and the triangles the closed-loop poles. The root locus shows how the poles move as the loop gain goes from 0 to maximum. You clearly see that two poles are in the right half plane at too low loop gain, meaning Re(p) > 0, hence an unstable circuit.
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I think you mixed up the quadrants: a positive real part indicates an unstable system, because there are terms that increase exponentially with time.
You can calculate the transfer function of an amplifier like this:
-Replace all components with linearized models consisting of ideal capacitors, ideal inductors, ideal resistors and linear controlled sources
-Connect one independent source Vin(s) or Iin(s) to the input
-Use Z = 1/sC for the impedance of an ideal capacitor and Z = sL for the impedance of an ideal inductor
-Write out and solve the network equations. If you can't think of a better way to do that, you can always use modified nodal analysis.
-Calculate the output quantity divided by the input quantity, this is the transfer function.
For a feedback amplifier, the analysis often gets more manageable and useful when you cut open the loop somehow and first calculate the loop gain, rather than the closed-loop transfer.
The equations tend to get very complicated, especially when you include collector-base, anode-grid or drain-gate capacitances. One approach is to neglect those at first and try to come up with an amplifier topology where their effect is limited. Something with cascoded stages, for example.
I frequently calculate transfer functions of filter structures, but not of complete amplifiers. The analysis just tends to get too complicated.
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Interpreting s as a differentiation-to-time operator and using V = Z I, the equation Z = sL for the impedance of an ideal inductor simply means that the voltage across an ideal inductor is the inductance times the derivative of the current to time. Similarly, the equation Z = 1/(sC) for the impedance of an ideal capacitor simply means that the current through an ideal capacitor is the capacitance times the derivative of the voltage to time.
Marcel, can you talk about how (complex?) poles (and zeros) appear in the (real?) transfer function? Obviously, there is no asymptote in a Butterworth filter transfer function. Is that because the pole is displaced into the complex s plane and only the ~skirt falls on real frequencies? Exactly how do features in the complex s plane appear in a Bode plot?
The Bode frequency response plot is the intersection of the amplitude with the vertical plane
passing through the sigma = 0 axis, where s = jw.
See figure 32-7 here.
http://www.dspguide.com/CH32.PDF
passing through the sigma = 0 axis, where s = jw.
See figure 32-7 here.
http://www.dspguide.com/CH32.PDF
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Yes, fill in s = j omega with omega = 2 pi f to find the transfer under stationary sine wave excitation. The magnitude of the transfer under stationary sine wave excitation is proportional to the product of the distances of j omega to the zeros divided by the product of the distances of j omega to the poles. You can also find the phase response by adding and subtracting the angles from the zeros and poles to j omega.
For convenience, here is the figure 32-7.
http://www.dspguide.com/CH32.PDF
The pole-zero plot (at left) is as seen from above the 3D amplitude plot.
The Bode frequency response plot (at right) is seen from the right side of the 3D amplitude plot.
And s = ( sigma + jw ).
The Bode plot is at the intersection of the 3D amplitude plot, with the vertical plane where s = jw,
which passes through the sigma = 0 axis.
That's what you do when finding the frequency response of the s-domain transfer function, first set s = jw
and so hold sigma = 0. Then there are no e ^ ( real number x t ) terms in the transfer function.
So all the exponential terms are of the form e ^ ( j wt ) which are sine and cosine terms.
Recall that Euler's formula is: e ^ ( j wt ) = cos wt + j sin wt
where w = 2 x Pi x f
and f = frequency
http://www.dspguide.com/CH32.PDF
The pole-zero plot (at left) is as seen from above the 3D amplitude plot.
The Bode frequency response plot (at right) is seen from the right side of the 3D amplitude plot.
And s = ( sigma + jw ).
The Bode plot is at the intersection of the 3D amplitude plot, with the vertical plane where s = jw,
which passes through the sigma = 0 axis.
That's what you do when finding the frequency response of the s-domain transfer function, first set s = jw
and so hold sigma = 0. Then there are no e ^ ( real number x t ) terms in the transfer function.
So all the exponential terms are of the form e ^ ( j wt ) which are sine and cosine terms.
Recall that Euler's formula is: e ^ ( j wt ) = cos wt + j sin wt
where w = 2 x Pi x f
and f = frequency
Unfortunately, figure 32-7 has a couple zeros that ~hide the effect of the poles on the Bode plot. Zeros may happen in amplifier compensation, but I am mostly interested in dominant pole compensation and in crossover filters. Any zeros not typically used in these applications distract and confuses the issue.
And you could also talk about the poles(?) in a high pass filter vs a low-pass filter.
And you could also talk about the poles(?) in a high pass filter vs a low-pass filter.
Here's a nice visualization with two poles.
This is a good discussion of active LP and HP filters.
See eqns. 1-6 for LP and 1-64 for HP. The numerator is the only difference between them.
This is a good discussion of active LP and HP filters.
See eqns. 1-6 for LP and 1-64 for HP. The numerator is the only difference between them.
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That's only true for some special cases, including Butterworth and Linkwitz-Riley filters and any second-order filter.
In general, when you look at the frequency response on a logarithmic frequency scale, you can change a low-pass into a high-pass by mirroring the response in a line at some frequency fm. That boils down to changing f into fm2/f.
Using s = j omega, the s in the low-pass transfer has to be replaced with -omegam2/s. For a filter with real-valued inputs and outputs, the response is symmetrical around zero frequency, so you might as well drop the minus sign. Low- to high-pass conversions are usually done on filters normalized to unity radian bandwidth, so then you can use omegam = 1 and replace s with 1/s.
To be continued, I have a train to catch now.
In general, when you look at the frequency response on a logarithmic frequency scale, you can change a low-pass into a high-pass by mirroring the response in a line at some frequency fm. That boils down to changing f into fm2/f.
Using s = j omega, the s in the low-pass transfer has to be replaced with -omegam2/s. For a filter with real-valued inputs and outputs, the response is symmetrical around zero frequency, so you might as well drop the minus sign. Low- to high-pass conversions are usually done on filters normalized to unity radian bandwidth, so then you can use omegam = 1 and replace s with 1/s.
To be continued, I have a train to catch now.
Suppose you have an all-pole low-pass transfer function Hlp(s) = 1/(ansn + an - 1sn - 1 + ... + a1s + 1). When you then replace s by 1/s and multiply the numerator and denominator with sn, you get n extra zeros in the origin, but also swap the order of the coefficients of the denominator polynomial. The poles only stay the same if those coefficients happen to stay the same when you swap their order.
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If you would leave the poles as is and just add n zeros in the origin, you would tilt the response by adding a +20n dB/decade slope to it. (The distance from a zero in the origin to j omega is omega, so proportional to frequency. Expressed in decibel per decade, that's 20 dB/decade. With n such zeros, you add 20n dB/decade.)
That works fine for a Butterworth response, but for for example Chebyshev, it would mean that the passband ripples remain at the frequency they used to lie at, but that has turned into the stopband after low- to high-pass transformation. For example, a 1 dB Chebyshev response would turn into a high-pass response with 1 dB ripples in the stopband and a big hump at the start of the passband.
That works fine for a Butterworth response, but for for example Chebyshev, it would mean that the passband ripples remain at the frequency they used to lie at, but that has turned into the stopband after low- to high-pass transformation. For example, a 1 dB Chebyshev response would turn into a high-pass response with 1 dB ripples in the stopband and a big hump at the start of the passband.
So, a high pass filter has zero(s) at zero Hz and a lowpass has zero(s) at infinity??? ...that tilt the pole skirts towards the passband? I suppose you wouldn't see the zero(s) at infinity on a s-plain graph? And that would be 20dB/decade/zero??
When you have a low-pass transfer function of the form Hlp(s) = 1/(ansn + an - 1sn - 1 + ... + a1s + 1) and you fill in s = j omega, you will find that for large omega, the first term of the denominator dominates and the magnitude of the transfer becomes inversely proportional to the nth power of frequency, corresponding to -20 n dB/decade.
As the magnitude of the transfer tends to zero with the nth power when the frequency tends to infinity, some people like to refer to that as n zeros at infinity. Looking at it that way has the advantage that the number of zeros stays the same when you transform a low-pass into a high-pass or the other way around. Others just find it confusing and say that the low-pass filter has no zeros at all. Those are just two mathematical abstractions, it depends on whether you prefer extended complex numbers or normal complex numbers.
In the pole-zero plot of the low-pass filter, you just see the n solutions of ansn + an - 1sn - 1 + ... + a1s + 1 = 0, that is, the n poles.
As the magnitude of the transfer tends to zero with the nth power when the frequency tends to infinity, some people like to refer to that as n zeros at infinity. Looking at it that way has the advantage that the number of zeros stays the same when you transform a low-pass into a high-pass or the other way around. Others just find it confusing and say that the low-pass filter has no zeros at all. Those are just two mathematical abstractions, it depends on whether you prefer extended complex numbers or normal complex numbers.
In the pole-zero plot of the low-pass filter, you just see the n solutions of ansn + an - 1sn - 1 + ... + a1s + 1 = 0, that is, the n poles.