# Playing with poles and zeros

#### Article By : Sergio Franco Use an op amp to control the location of poles and zeros in the complex plane, and use the natural response of a circuit to illustrate the effect of pole/zero location.

In a previous blog  I wrote: During my academic career I have observed that one of the most difficult subjects to teach as well as to learn is systems theory. All those poles and zeros may make sense in class, but once the student tries to relate them to a physical circuit in the lab, there seems to be a chasm between the blackboard and the bench. In the present blog I’ll try to develop a physical feel for poles and zeros, using an op amp to control their location in the complex plane, and using the natural response of a circuit to illustrate the effect of pole/zero location.

The natural response of a one-port
Consider the passive linear one-port of Figure 1, consisting of resistors, capacitors, and inductors. Figure 1
(a) A passive one-port, and (b) its natural (or source-free) open-circuit response vn(t).

If we apply a test current I(s), the one-port will yield a voltage V(s) such that V(s) = Z(s)I(s), where I(s) and V(s) are the Laplace transforms of the applied current and the resulting voltage, and s is the complex frequency, in sec– 1. Mother Nature is such that the impedance Z(s) takes on the form of a rational function of s, that is, the ratio of a numerator polynomial N(s) and a denominator polynomial D(s), The roots of the equation N(s) = 0 are called the zeros of Z(s), and are denoted as z1, z2,… The roots of the equation D(s) = 0 are called the poles of Z(s), and are denoted as p1, p2,… Collectively, poles and zeros are referred to as roots, or also as critical frequencies. As an example, the impedance has a zero at s = 0 and a complex-conjugate pole pair at s = –3 ± j4. Expressing it in terms of its roots, The zero and pole designations stem from the fact if we plot the magnitude |Z(s)| versus s, the resulting curve appears as a tent pitched on the s plane and such that it touches the s plane at the zeros, and its height becomes infinite at the poles. Figure 2
Magnitude plot of Z(s) = (10 Ω)s/(s2 + 6s + 25). (The profile obtained by calculating |Z| on the imaginary axis provides an indication of the ac response of the one-port.)

To develop a physical feel for poles , note that if we apply a current I(s) with s close to a pole pk, we can achieve a given voltage V(s) with a rather small I(s). The closer s is to pk, the smaller I(s) for a given V(s). In the limit spk the one-port will supply a nonzero voltage even with zero applied current, or an open circuit (see Figure 1b)! This voltage is called the natural response or also the source-free response because the one-port produces it by utilizing the energies stored internally in its capacitors and inductors. These energies get dissipated in the resistors, and in the case of a passive one-port, they decay with time exponentially. In fact, systems theory predicts the following expression for the natural response: (2)

where a1, a2,,… are suitable coefficients (in volts) that depend on the stored energies, and the poles of Z(s) are the inverses of the time-constants appearing in the exponents.

What about the zeros of Z(s)? Refer now to Figure 3, representing the dual situation of Figure 1. The applied signal is now a voltage V(s) and the response is a current I(s) = [1/Z(s)]V(s), indicating that the zeros of Z(s) are now the poles of 1/Z(s). By dual reasoning, in the limit szk the one-port will supply a nonzero current even with zero applied voltage, or a short circuit (see Figure 3b)! This current is called the natural response or also the source-free response because the one-port produces it by utilizing the energies stored internally in its capacitors and inductors. Systems theory predicts the following expression for the natural short-circuit current response (3)

where b1, b2,… are suitable coefficients (in Amperes) that depend on the stored energies, and the zeros of Z(s) are the inverses of the time-constants appearing in the exponents. Figure 3
(a) A passive one-port, and (b) its natural (or source-free) short-circuit response in(t).

Summarizing, the natural responses of a one-port are governed by the roots of its impedance Z(s): the poles govern the open-circuit voltage response vn(t), and the zeros govern the short-circuit current response in(t). In a way, the roots are like the DNA of the one-port. As an example, consider the one-port of Figure 4. If at t = 0 the voltage across the capacitor is 9 V, positive at the top, what are its natural responses for t > 0? By inspection, the impedance presented by the one-port is Clearly, we have z1 = –1/(R1C) = –1/(10 ms) and p1 = –1/[(R1 + R2)C] = –1/(30 ms). Moreover, we have a1 = [20/(10 + 20)]9 = 6 V and b1 = 9/10 = 0.9 mA. Consequently,  Figure 4
Finding (a) the open-circuit and (b) the short-circuits natural responses.

Single pole control
In the circuit of Figure 5a the impedance between the node denoted as vn and ground is Z(s) = R||(1/sC) = R/(1 + sRC), so the circuit has a pole at s = –1/(RC) = –1/(1 ms). Assuming vn(0) = 1 V, we thus have: (4) Figure 5
(a) Basic circuit, and (b) the same circuit, but with provision for controlling its pole.

No matter how we choose the values of R and C, the pole of this circuit will always be negative. We wish to find ways to control it so as to drive it to zero or make it even positive. Figure 5b shows a circuit that will do the job. The non-inverting amplifier senses vn and outputs the voltage (1 + R2/R1)vn = (1 + k)vn, where R2 represents the portion of the potentiometer between its left end and the wiper. With the given component values, varying the wiper from the left end to the right end will cause k to vary over the range 0 < k < 2. Now, the voltage across R3 is (1 + k)vnvn, or kvn, positive at the right, indicating that R3 will source to C the current kvn/R3. Given that R sinks out of C the current vn/R, the net current out of C is thus iC = vn[1/R + 1/(–R3/k)], indicating that C sees R in parallel with a negative resistance of –R3/k, for a net equivalent resistance of Req = R||(–R3/k)]. Expanding gives With the given component values we have Req = (10 kΩ)/(1 – k), so the pole location is now at s = –1/(ReqC) = –(1–k)/(1 ms), and Equation (4) therefore changes to (7)

Let us discuss circuit operation as a function of the wiper setting, using also the PSpice circuit of Figure 6 to visualize the ensuing natural response types.

• With the wiper all the way to the left (k = 0), the voltage drop across R3 is zero, so R3 carries zero current and C is discharging via R with a time constant of 1 ms, as in Equation (4).
• As we move the wiper to the right, R3 now sources current to C. So long as this current is less than that sunk by R, C will still be discharging exponentially, but at a slower rate than when k = 0.
• With the wiper midway (k = 1), the current sourced by R3 equals that sunk by R, for a net capacitor current of zero, so the capacitor voltage remains constant with time.
• Moving the wiper further to the right (k > 1) makes the sourced current prevail over the sunk current, so now C charges up exponentially to yield a diverging response. The response will grow until the op amp saturates. Figure 6
PSpice circuit to display the natural responses for different values of k, assuming the capacitor is initially charged at 1 V.

Figure 7 depicts pole location as a function of k. Figure 7
Locus described by the pole as a function of k.

[Continue reading on EDN US: Pole pair control]

Sergio Franco is an author and (now emeritus) university professor.

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