Structure and Interpretation of Signals and Systems by Edward Ashford Lee and Pravin Varaiya - HTML preview

PLEASE NOTE: This is an HTML preview only and some elements such as links or page numbers may be incorrect.
Download the book in PDF, ePub for a complete version.

W + ( ˆ

H1 + ˆ

H2) ˆ

X .

From this it is evident that if we choose

ˆ

H2 = − ˆ

H1,

then y will be a clean (noise-free) signal, equal to w. The following examples describe

real-world applications of this pattern.

Example 14.11:

A connection to the telephone network uses two wires (called

a twisted pair, consisting of tip and ring) to connect a telephone to a central

office. The central office may be, perhaps, 4 kilometers away. The two wires carry

voice signals to and from the customer premises, representing the voice signals as a

voltage difference across the two wires. Since two wires can only have one voltage

difference across them, the incoming voice signal and the outgoing voice signal

share the same twisted pair.

The central office needs to separate the voice signal from the local customer

premises (called the near-end signal) from the voice signal that comes from the

other end of the connection (called the far-end signal). The near end signal is

typically digitized (sampled and quantized), and a discrete-time representation of

the voice signal is transmitted over the network to the far end. The network itself

consists of circuits that can carry voice signals in one direction at a time. Thus, in

the network, four wires (or equivalent) are required for a telephone connection, one

wire pair for each direction.

As indicated in figure 14.4, the conversion from a two-wire to a four-wire connec-

tion is done by a device called a hybrid. A hybrid is a Wheatstone bridge, a circuit

that can separate two signals based on the electrical impedance looking into the lo-

cal twisted pair and the electrical impedance looking into the network. The design

of this circuit is a suitable topic for a text on electrical circuits.

A connection between subscribers A and B involves two hybrids, one in each sub-

scriber’s central office. The hybrid in B’s central office ideally will pass all of the

Lee & Varaiya, Signals and Systems

621

14.2. PARALLEL COMPOSITION

x = signal from A

twisted pair

Hybrid

Hybrid

at A's

at B's

CO

CO

A

B

echo

y = leakage of A's signal plus signal from B

x

H

H

2

1

w

w

2

1

y

w

echo canceller

echo path

Figure 14.4: A telephone central office converts the two-wire connection with a

customer telephone into a four-wire connection with the telephone network using

a device called a hybrid. An imperfect hybrid leaks, causing echo. An echo

canceller removes the leaked signal.

622

Lee & Varaiya, Signals and Systems

14. COMPOSITION AND FEEDBACK CONTROL

incoming signal x to B’s two-wire circuit, and none back into the network. How-

ever, the hybrid is not perfect, and some of the incoming signal x leaks through the

hybrid into the return path back to A. The signal y in the figure is the sum of the

signal from B and the leaked signal from A. A hears the leaked signal as an echo,

since it is A’s own signal, delayed by propagation through the telephone network.

If the telephone connection includes a satellite link, then the delay from one end

of the connection to the other is about 300ms. This is the time it takes for a radio

signal to propagate to a geosynchronous satellite and back. The echo traverses

this link twice: once going from A to B, and the second time coming back. Thus,

the echo is A’s own signal delayed by about 600ms. For voice signals, 600ms of

delay is enough to create a very annoying echo that can make it difficult to speak.

Humans have difficulty speaking when they hear their own voices 600ms later.

Consequently, the designers of the telephone network have put echo cancellers in

to prevent the echo from occuring.

Let ˆ

H1 be the transfer function of the hybrid leakage path. The echo canceller is

the filter ˆ

H2 placed in parallel composition with the hybrid, as shown in the figure.

The output w2 of this filter is added to the output w1 + w of the hybrid, so the signal

that actually goes back is y = w2 + w1 + w. If

ˆ

H2 = − ˆ

H1,

then y = w and the echo is cancelled perfectly. Moreover, note that as long as ˆ

H1 is

stable and causal, so will be the echo canceller ˆ

H2.

However, ˆ

H1 is not usually known in advance, and also it changes over time. So

either a fixed ˆ

H2 is designed to match a ‘typical’ ˆ

H1, or an adaptive echo canceller

is designed that estimates the characteristics of the echo path ( ˆ

H1) and changes ˆ

H2

accordingly. Adaptive echo cancellers are common in the telephone network today.

The following example combines cascade and parallel composition to achieve noise can-

cellation.

Example 14.12:

Consider a microphone in a noisy environment. For example,

a traffic helicopter might be used to deliver live traffic reports over the radio, but

the (considerable) background noise of the helicopter would be highly undesirable

Lee & Varaiya, Signals and Systems

623

14.3. FEEDBACK COMPOSITION

w

w 1

w 4

H 1

x

y

H 2

H 3

w

w

2

3

Figure 14.5: Traffic helicopter noise cancellation/equalization problem.

on the radio. Fortunately, the background noise can be cancelled. Referring to

Figure 14.5, suppose that w is the announcer’s voice, x is the engine noise, and

ˆ

H1 represents the acoustic path from the engine noise to the microphone. The

microphone picks up both the engine noise and the announcer’s voice, producing

the noisy signal w4. We can place a second microphone somewhere far enough from

the announcer so as to not pick up much of his or her voice. Since this microphone

is in a different place, say on the back of the announcer’s helmet, the acoustic path

is different, so we model that path with another transfer function ˆ

H2. To cancel the

noise, we design a filter ˆ

H3. This filter needs to equalize (invert) ˆ

H2 and cancel ˆ

H1.

That is, its ideal value is

ˆ

H3 = − ˆ

H1/ ˆ

H2.

Of course, as with the equalization scenario, we have to ensure that this filter re-

mains stable. Once again, in practice, it is necessary to make the filter adaptive.

14.3

Feedback composition

Consider the feedback composition in figure 14.6. It is a composition of two systems

with transfer functions ˆ

H1 and ˆ

H2. We assume that these systems are causal and that ˆ

H1

and ˆ

H2 are proper rational polynomials in z or s. The regions of convergence of these

two transfer functions are those suitable for causal systems (the region outside the largest

624

Lee & Varaiya, Signals and Systems

14. COMPOSITION AND FEEDBACK CONTROL

x

e

w

y

H 1

H 2

!

H

H =

1 H 2

1 + H 1 H 2

Figure 14.6: Negative feedback composition of two LTI systems with transfer func-

tions H1 and H2.

circle passing through a pole, for discrete time, and the region to the right of the pole with

the largest real part, for continuous-time).

In terms of Laplace or Z transforms, the signals in the figure are related by

ˆ

Y = ˆ

H ˆ ˆ

2H1E ,

and

Ê = ˆ

X − ˆ

Y .

Notice that, by convention, the feedback term is subtracted, as indicated by the minus

sign adjacent to the adder (for this reason, this composition is called negative feedback).

Combining these two equations to elimintate Ê, we get

ˆ

Y = ˆ

H ˆ

1H2( ˆ

X − ˆ

Y ),

which we can solve for the transfer function of the composition,

ˆ

Y

ˆ

H ˆ

H

ˆ

1

2

H =

=

.

(14.1)

ˆ

X

1 + ˆ

H ˆ

1H2

This is often called the closed-loop transfer function, to contrast it with the open-loop

transfer function, which is simply ˆ

H ˆ

1H2. We will assume that this resulting system is

causal, and that the region of convergence of this transfer function is therefore determined

by the roots of the denominator polynomial, 1 + ˆ

H ˆ

1H2.

Lee & Varaiya, Signals and Systems

625

14.3. FEEDBACK COMPOSITION

The closed-loop transfer function is valid as long as the denominator 1 + ˆ

H ˆ

1H2 is not

identically zero (that is, it is not zero for all s or z in C – it may be zero some s or z in C).

This is sufficient for the feedback loop to be well-formed, although in general, this fact

is not trivial to demonstrate (Exercise 8 considers the easier case where ˆ

H ˆ

1H2 is causal

and strictly proper, in which case the system ˆ

H ˆ

1H2 has state-determined output). We will

assume henceforth, without comment, that the denominator is not identically zero.

Feedback composition is useful for stabilizing unstable systems. In the case of cascade

and parallel composition, a pole of the composite must be a pole of one of the components.

The only way to remove or alter a pole of the components is to cancel it with a zero. For

this reason, cascade and parallel composition are not effective for stabilizing unstable

systems. Any error in the specification of the unstable pole location results in a failed

cancellation, which results in an unstable composition.

In contrast, the poles of the feedback composition are the roots of the denominator 1 +

ˆ

H ˆ

1H2, which are generally quite different from the poles of ˆ

H1 and ˆ

H2. This leads to the

following important conclusion:

The poles of a feedback composition can be different from the poles of its component

subsystems. Consequently, unstable system can be effectively and robustly stabilized by

feedback.

The stabilization is robust in that small changes in the pole or zero locations do not result

in the composition going unstable. We will be able to quantify this robustness.

14.3.1

Proportional controllers

In control applications, one of the two systems being composed, say ˆ

H2, is called the plant.

This is a physical system that is given to us to control. Its transfer function is determined

by its physics. The second system being composed, say ˆ

H1, is the controller. We design

this system to get the plant to do what we want. The following example illustrates a

simple strategy called a proportional controller or P controller.

Example 14.13: For this example we take as the plant the simplified continuous-

time helicopter model of example 12.2,

1

˙

y(t) =

w(t).

M

626

Lee & Varaiya, Signals and Systems

14. COMPOSITION AND FEEDBACK CONTROL

Here y(t) is the angular velocity at time t and w(t) is the torque. M is the moment

of inertia.

We have renamed the input w (instead of x) because we wish to control the heli-

copter, and the control input signal will not be the torque. Instead, let’s define the

input x to be the desired angular velocity. So, to get the helicopter to not rotate, we

provide input x(t) = 0.

Let us call the impulse response of the plant h2, to conform with the notation in

Figure 14.6; it is given by

∀ t ∈ R, h2(t) = u(t)/M,

where u is the unit step. The transfer function is

ˆ

H2(s) = 1/(Ms),

with

RoC(h) = {s ∈ C | Re(s) > 0}.

ˆ

H2 has a pole at s = 0, so this is an unstable system.

As a compensator we can simply place a gain K in a negative feedback composition,

as shown in Figure 14.7. The intuition is as follows. Suppose we wish to keep the

helicopter from rotating. That is, we would like the output angular velocity to be

zero, y(t) = 0. Then we should apply an input of zero, x(t) = 0. However, the

plant is unstable, so even with a zero input, the output diverges (even the smallest

non-zero initial condition or the smallest input disturbance will cause it to diverge).

With the feedback arrangement in Figure 14.7, if the output angular velocity rises

above zero, then the input is modified downwards (the feedback is negative), which

will result in a negative torque being applied to the plant, which will counter the

rising velocity. If the output angular velocity drops below zero, then the torque will

be modified upwards, which again will tend to counter the dropping velocity. The

output velocity will stabilize at zero.

To get the helicopter to rotate, for example to execute a turn, we simply apply a

non-zero input. The feedback system will again compensate so that the helicopter

will rotate at the angular velocity specified by the input.

The signal e is the difference between the input x, which is the desired angular

velocity, and the output y, which is the actual angular velocity. It is called the error

signal. Intuitively, this signal is zero when everything is as desired, when the output

angular velocity matches the input.

Lee & Varaiya, Signals and Systems

627

14.3. FEEDBACK COMPOSITION

error

torque

x

e

w

y

H

1

1= K

H =

2

desired

Ms

!

angular

angular

velocity

velocity

K / M

H = s + K / M

Figure 14.7: A negative feedback proportional controller with gain K.

K = 0

!

!2

2

K =

K = +2

K = !2

K = !

root locus

Figure 14.8: Root locus of the helicopter P controller.

A compensator like that in example 14.13 and Figure 14.7 is called a proportional controller or P controller. The input w to the plant is proportional to the error e. The

objective of the control system is to have the output y of the plant track the input x as

closely as possible. I.e., the error e needs to be small. We can use (14.1) to find the

transfer function of the closed-loop system.

Example 14.14: Continuing with the helicopter of example 14.13, the closed loop

system transfer function is

K ˆ

H(s)

K/M

ˆ

G(s) =

=

.

(14.2)

1 + K ˆ

H(s)

s + K/M

628

Lee & Varaiya, Signals and Systems

14. COMPOSITION AND FEEDBACK CONTROL

which has a pole at s = −K/M. If K > 0, the closed loop system is stable, and

if K < 0, it is unstable. Thus, we have considerable freedom to choose K. How

should we choose its value?

As K increases from 0 to ∞, the pole at at s = −K/M moves left from 0 to −∞. As

K decreases from 0 to −∞, the pole moves to the right from 0 to ∞. The locus of the

pole as K varies is called the root locus, since the pole is a root of the denominator

polynomial.

Figure 14.8 shows the root locus as a thick gray line, on which are marked the

locations of the pole for K = 0, ±2, ±∞. Since there is only one pole, the root locus

comprises only one ‘branch’. In general the root locus has as many branches as

the number of poles, with each branch showing by the movement of one pole as K

varies.

Note that in principle, the same transfer function as the closed-loop transfer function can

be achieved by a cascade composition. But as in example 14.1, the resulting system is

not robust, in that even the smallest change in the pole location of the plant can cause

the system to go unstable (see problem 6). The feedback system, however, is robust, as

shown in the following example

Example 14.15: Continuing with the P controller for the helicopter, suppose that

our model of the plant is not perfect, and its actual transfer function is

1

ˆ

H2(s) =

,

M(s − ε)

for some small value of ε > 0. In that case, the closed loop transfer function is

K/M

ˆ

H(s) =

,

s − ε + K/M

which has a pole at s = ε − K/M. So the feedback system remains stable so long as

ε < K/M.

Lee & Varaiya, Signals and Systems

629

14.3. FEEDBACK COMPOSITION

In practice, when designing feedback controllers, we first quantify our uncertainty about

the plant, and then determine the controller parameters so that under all possible plant

transfer functions, the closed-loop system is stable.

Example 14.16: Continuing the helicopter example, we might say that ε < 0.5. In

that case, if we choose K so that K/M > 0.5, we would guarantee stability for all

values of ε < 0.5. We then say that the proportional feedback controller is robust

for all plants with ε < 0.5.

We still have a large number of choices for K. How do we select one? To understand

the implications of different choices of K we need to study the behavior of the output y

(or the error signal e) for different choices of K. In the following examples we use the

closed-loop transfer function to analyze the response of a proportional controller system

to various inputs. The first example studies the response to a step function input.

Example 14.17:

Continuing the helicopter example, suppose that the input is a

step function, ∀t, x(t) = au(t) where a is a constant and u is the unit step. This

input declares that at time t = 0, we wish for the helicopter to begin rotating with

angular velocity a. The closed-loop transfer function is given by (14.2), and the

Laplace transform of x is ˆ

X (s) = a/s, from table 13.3, so the Laplace transform of

the output is

K/M

ˆ

Y (s) = ˆ

G(s) ˆ

X (s) =

· a

s + K/M s

Carrying out the partial fraction expansion, this becomes

−a

a

ˆ

Y (s) =

+

.

s + K/M

s

We can use this to find the inverse Laplace transform,

∀t, y(t) = −ae−Kt/Mu(t) + au(t).

The second term is the steady-state response yss, which in this case equals the

input. So the first term is the tracking error ytr, which goes to zero faster for

larger K. Hence for step inputs, the larger the gain K, the smaller the tracking error.

630

Lee & Varaiya, Signals and Systems

14. COMPOSITION AND FEEDBACK CONTROL

In the previous example, we find that the error goes to zero when the input is a step

function. Moreover, the error goes to zero faster if the gain K is larger than if it is smaller.

In the following example, we find that if the input is sinusoidal, then larger gain K results

in an ability to track higher frequency inputs.

Example 14.18: Suppose the input to the P controller helicopter system is a sinu-

soid of amplitude A and frequency ω0,

∀ t ∈ R, x(t) = A(cosω0t)u(t).

We know that the response can be decomposed as y = ytr + yss. The transient re-

sponse ytr is due to the pole at s = −K/M, and so it is of the form

∀ t ∈ R, ytr(t) = Re−Kt/M,

for some constant R. The steady-state response is determined by the frequency

response at ω0. The frequency response is

K/M

ω ∈ R,

H(ω) = ˆ

H(iω) =

,

iω + K/M

with magnitude and phase given by

|

K/M

H(ω)| =

, ∠H(ω) = −tan−1 ωM .

[ω2 + (K/M)2]1/2

K

So the steady-state response is

∀t, yss(t) = |H(ω0)|Acos(ω0t + ∠H(ω0)).

Thus the steady-state output is a sinusoid of the same frequency as the input but

with a smaller amplitude (unless ω0 = 0). The larger ω0 is, the smaller the output

amplitude. Hence, the ability of the closed-loop system to track a sinusoidal input

decreases as the frequency of the sinusoidal input increases. However, increasing

K reduces this effect. Thus, larger gain in the feedback loop improves its ability to

track higher frequency sinusoidal inputs.

In addition to the reduction in amplitude, the output has a phase difference. Again,

if ω0 = 0, there is no phase error, because tan−1(0) = 0. As ω0 increases, the phase

lag increases (the phase angle decreases). Once again, however, increasing the gain

K reduces the effect.

Lee & Varaiya, Signals and Systems

631

14.3. FEEDBACK COMPOSITION

The previous two examples suggest that large gain in the feedback loop is always better.

For a step function input, it causes the transient er