Transfer Functions of Interconnections

Transfer Functions of Interconnections

Key Ideas

• It is common practice to build complex dynamical system models as interconnections of simpler subsystems, where some inputs of some subsystems are declared equal to some outputs of others.

• For an interconnection of LTI subsystems, its transfer function H(s) can be found by assuming that every signal in the interconnection is a complex multiple of the same complex exponential e^{st}. This leads to a system of linear equations whose solution gives H(s).

• In many cases the process simplifies to a sequence of additions or multiplications of transfer functions, using the rules for series, parallel, and feedback (Black's formula) connections.

• The task of designing stable interconnections is frequently aided by the Routh-Hurwitz criterion, which checks whether all roots of a polynomial have negative real part without computing the roots explicitly.

Series Connection

In a series connection, the output of the first subsystem feeds directly into the input of the second:

The overall transfer function (from u to y) is

To see this: assuming u(t)=Ue^{st}, w(t)=We^{st}, y(t)=Ye^{st} gives W=H_1(s)U and Y=H_2(s)W, so Y=H_2(s)H_1(s)U.

Python example:

import control as ct
s = ct.tf('s')
H1, H2 = 1/s, 1/(s+1)
H = H2*H1      # H = 1/(s^2+s)

Parallel Connection

In a parallel connection, both subsystems share the same input and their outputs are summed:

The overall transfer function is

Python example (continued):

H = H2 + H1    # H = (2s+1)/(s^2+s)

Feedback Connection and Black's Formula

In a feedback connection, a portion of the output is fed back and compared with the reference input:

The subsystem with transfer function P(s) has input e and output y; the subsystem with transfer function K(s) has input y and output w; and e = r - w. Black's formula gives the closed-loop transfer function from r to y:

This is memorized as “forward gain divided by one plus total loop gain.” To prove it: assuming r(t)=Re^{st}, e(t)=Ee^{st}, y(t)=Ye^{st}, w(t)=We^{st}, we get Y=P(s)E, W=K(s)Y, and E=R-W. Solving for Y gives Y = \frac{P(s)}{1+K(s)P(s)}R.

Python example (continued):

G12 = ct.feedback(H1, H2)   # H1/(1+H2*H1) = s/(s^2+s+1)
G21 = ct.feedback(H2, H1)   # H2/(1+H1*H2) = (s+1)/(s^2+s+1)

Note the result depends on the order of arguments.

A Worked Example: Arm Position Controlled by Torque with Friction

A model familiar in 6.3100 describes the propeller arm: control torque u(t) drives the arm position \theta(t) in the presence of linear friction. Introducing intermediate signals—arm velocity w(t), friction torque v(t) = bw(t), and net torque f(t) = u(t) - v(t)—we can identify four subsystems:

• \dot\theta = w: transfer function H_1(s) = 1/s.

• M\dot w = f: transfer function H_2(s) = M/s.

• v = bw: transfer function H_3(s) = b.

• Summation: f = u - v.

Using Black's formula, the transfer function from u to w is H_{u\to w}(s) = \frac{M/s}{1+bM/s} = \frac{M}{s+bM}. Then the series connection with 1/s gives the overall transfer function from u to \theta:

Routh-Hurwitz Criterion

It is often useful to know whether all roots of a polynomial have negative real parts without computing the roots. The Routh-Hurwitz criterion gives necessary and sufficient conditions for polynomials with positive leading coefficient:

• p(s)=s+p_0 is stable if and only if p_0 > 0.

• p(s)=s^2+p_1 s+p_0 is stable if and only if p_0 > 0 and p_1 > 0.

• p(s)=s^3+p_2 s^2+p_1 s+p_0 is stable if and only if p_0 > 0, p_1 > 0, p_2 > 0, and p_2 p_1 > p_0.

Example: Stabilizing a Double Integrator with PID

The block diagram below describes a PID controller attempting to stabilize the plant P(s) = 1/s^2:

By Black's formula the closed-loop transfer function is

The characteristic polynomial is s^3 + K_d s^2 + K_p s + K_i. By the degree-3 Routh-Hurwitz criterion, stability requires K_i > 0, K_p > 0, K_d > 0, and K_d K_p > K_i.

Beware of Unstable Zero/Pole Cancellations

Consider the feedback loop with K(s) = \frac{s}{s+1} and plant P(s) = 1/s^2. Black's formula gives

whose poles s = -\tfrac{1}{2} \pm j\tfrac{\sqrt{3}}{2} both have negative real parts—so the system looks stable.

However, consider the transfer function from a plant disturbance input d (injected between the controller and the plant) to the output y:

which has a pole at s = 0—the system is actually unstable with respect to disturbances.

What happened? The controller zero at s=0 cancelled the plant pole at s=0 in H_{r\to y}, masking the instability. This is an unstable zero/pole cancellation. It can be proved that cancelling any zero or pole with \operatorname{Re}(s) \ge 0 is guaranteed to produce an unstable closed-loop system—it is an absolute no-no in feedback control design.

Python example:

import control as ct
s = ct.tf('s')
H1 = s/(s+1)
H2 = 1/(s*s)
H  = H2*H1              # H = s/(s^3+s^2)  -- no cancellation performed
G  = ct.feedback(H, 1)  # G = s/(s^3+s^2+s) -- instability visible

Python's control module correctly refuses to cancel the zero/pole, so the instability is visible in G(s).

Questions

Question 1: Stabilizing a Double Integrator with a First-Order Controller

Consider the feedback system with first-order controller K(s) = c\,\dfrac{s+b}{s+a} and plant P(s) = 1/s^2:

By Black's formula the closed-loop characteristic polynomial is q(s) = s^3 + a s^2 + c s + cb. Apply the degree-3 Routh-Hurwitz conditions: cb > 0, c > 0, a > 0, and a \cdot c > cb (i.e., a > b).

Which combination of conditions on a, b, c yields a stable closed-loop system?

-- a > b > 0 and c > 0 b > a > 0 and c > 0 a, b, c > 0 (any positive values) a > b and c > 0 (no sign constraint on b)

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Which Routh-Hurwitz condition is the one that requires the strict inequality a > b (rather than just a, b > 0)?

The condition a > b comes from:

-- p_0 = cb > 0 p_1 = c > 0 p_2 = a > 0 p_2 p_1 > p_0, i.e., ac > cb

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Question 2: Unstable Zero/Pole Cancellation

Consider the standard feedback diagram with plant P(s) = \dfrac{s-1}{s^2} and a desired closed-loop transfer function H_{r\to y}(s) = \dfrac{1}{s+1}.

Solving H = \frac{PK}{1+PK} for K gives K = \frac{H}{(1-H)P}. Substituting:

The open-loop product is K(s)P(s) = \dfrac{s}{s-1}\cdot\dfrac{s-1}{s^2} = \dfrac{1}{s}, where the factor (s-1) cancels. Since \operatorname{Re}(1) > 0, this is an unstable zero/pole cancellation.

The unstable cancellation between K(s) and P(s) occurs at s =

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Why does this make the controller K(s) = \frac{s}{s-1} unacceptable?

-- Cancelling a factor with Re(s)>0 masks an unstable mode; the closed-loop system is actually unstable (visible from the disturbance transfer function) The controller gain is too large at high frequencies The plant P(s) has no zeros, so no cancellation is possible The desired H_{r \rightarrow y} has poles in the right half-plane

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