03.Transfer functions (the s-domain)

22/10/2004 3.5 The connection between the transfer function and the impulse response 95

3.5 The connection between the transfer function and the impulse response

We can make some statements about how the transfer function impinges upon the impulse response. We made some o the cu remarks about how the impulse response contained the essential time-domain character of the system Σ = (A, b, ct, 01). We justified this in some way by stating Proposition 2.32. We will now see that the information contained in the impulse response is also directly contained in the transfer function, and further in about the simplest way possible.

3.5.1 Properties of the causal impulse response We begin by looking at the case that is the most interesting for us; the causal impulse response. Our discussions in this section will be important for the development of certain fundamental aspects of, for example, input/output stability.

We begin by making the essential connection between the impulse and the transfer function. We state the result for both the left and right causal Laplace transform of the impulse response.

3.22 Theorem For a SISO linear system Σ = (A, b, ct, 01), we have

L0++ (hΣ)(s) = L0+−(hΣ)(s) = TΣ(s)

provided that Re(s) > σmin(h+Σ).1

Proof Since h+Σ has no delta-function at t = 0 as we are supposing D = 01, we have

Thus clearly it does not matter in this case whether we use the left or right Laplace transform. We have

Now recall that if B, C Rn×n have the property that BC = CB then eB+C = eBeC. Noting that the n × n matrices −stIn and At commute, we then have

1Here is an occasion where we actually think of TΣ as a function of the complex variable s rather than as a rational function.

Since the terms in the matrix e(−sIn+A)t satisfy an inequality like (E.2), the upper limit on the integral is zero so we have

L0+−(hΣ)(s) = −ct(−sIn + A)−1b = ct(sIn − A)−1b = TΣ(s)

as claimed.

3.23 Remark Note that we ask that the feedforward term D be zero in the theorem. This can be relaxed provided that one is willing to think about the “delta-function.” The manner in which this can be done is the point of Exercise E3.1. When one does this, the theorem no longer holds for both the left and right causal Laplace transforms, but only holds for the left causal Laplace transform.

We should, then, be able to glean the behaviour of the impulse response by looking only at the transfer function. This is indeed the case, and this will now be elucidated. You will recall that if f(t) is a positive real-valued function then

The idea is that lim supt→∞ will exist for bounded functions that “oscillate” whereas limt→∞ may not exist for such functions. With this notation we have the following result.

3.24 Proposition Let Σ = (A, b, ct, 01) be a SISO linear system with impulse response h+Σ

and transfer function TΣ. Write

N(s) D(s)

where (N, D) is the c.f.r. of TΣ. The following statements hold.

(i) If D has a root in C+ then

22/10/2004 3.5 The connection between the transfer function and the impulse response 97

Proof (i) Corresponding to a root with positive real part will be a term in the partial fraction expansion of TΣ of the form

β

or

(s − σ)2 + ω2 k

with σ > 0. By Proposition E.11, associated with such a term will be a term in the impulse response that is a linear combination of functions of the form

t`eσt or t`eσt cos ωt or t`eσt sin ωt.

3.5.2 Things anticausal We will have occasion to make a few computations using the anticausal impulse response, and the causal impulse response in relation to anticausal inputs. In this section we collect the relevant results.

The first result is a natural analogue of Theorem 3.22. Since the proof is exactly the same with the exception of some signs, it is omitted.

3.25 Theorem For a SISO linear system Σ = (A, b, ct, 01) we have

L0+− (h−Σ)(s) = L0−−(h−Σ)(s) = TΣ(s),

provided that Re(s) < σmax(h−Σ).

Thus we see that there is a natural tension between whether to use the left or right causal Laplace transform. It is important to realise that the two things are di erent, and that their di erences sometimes make one preferable over the other.

3.6.1 Computing outputs for SISO linear systems in input/output form using the right causal Laplace transform We begin looking at systems that are given to us in input/output form. Thus we have a SISO linear system (N, D) in input/output form with deg(D) = n, and we are concerned with obtaining solutions to the initial value problem

for a known function u(t). We shall assume that u is su ciently di erentiable that the needed derivatives exist at t = 0+, and that u possesses a right causal Laplace transform. Then, roughly speaking, if one wishes to use Laplace transforms one determines the right causal Laplace transform L0++ (u), and then inverse Laplace transforms the function

L0++ (y)(s) = ND((ss))L0++ (u)(s)

to get y(t). This is indeed very rough, of course, because we have lost track of the initial conditions in doing this. The following result tells us how to use the Laplace transform method to obtain the solution to (3.12), properly keeping track on initial conditions.

on [0, ∞). Then the right causal Laplace transform of the solution to the initial value problem (3.12) is

where

N(s) = cnsn + cn−1sn−1 + cn−2sn−2 + · · · + c1s + c0

D(s) = sn + pn−1sn−1 + · · · + p1s + p0,

and where we take pn = 1.

Proof Note that by Corollary E.8 the right causal Laplace transform of y(k)(t) is given by

with a similar formula holding, of course, for u(k)(t). Thus, if we take the Laplace transform of the di erential equation in (3.12) and use the initial conditions given there we get the relation

100 3 Transfer functions (the s-domain) 22/10/2004

3.28 Remark The assumptions of di erentiability on the input can be relaxed as one can see by looking at the proof that one really only needs for u to satisfy the conditions of the Laplace transform derivative theorem, Theorem E.7, for a su ciently large number of derivatives. This can be made true for a large class of functions by using the definition of the Laplace transform on distributions. This same observation holds for results that follow and possess the same hypotheses.

One may, if it is possible, use Proposition 3.27 to obtain the solution y(t) by using the inverse Laplace transform. Let’s see how this works in an example.

3.29 Example We have (N(s), D(s)) = (1 − s, s2 + 2s + 2) and we take as input u(t) = 1(t)t. The initial value problem we solve is

y¨(t) + 2y˙(t) + 2y(t) = t − 1, y(0+) = 1, y˙(0+) = 0.

Taking the Laplace transform of the left-hand side of this equation gives

(s2 + 2s + 2)L0++ (y)(s) − (s + 2)y(0+) − y˙(0+) = (s2 + 2s + 2)L0++ (y)(s) − s − 2.

One may verify that (s + 2)y(0+) + y˙(0+) is exactly the expression

nk−1

X X

pksjy(k−j−1)(0+)

k=1 j=0

in the statement of Proposition 3.27 with D as given. The Laplace transform of the righthand side of the di erential equation is

(1 − s)L0++ (u)(s) + u(0+) = 1 s−2 s,

using the fact that the Laplace transform of u(t) is s12 . As with the expression involving the left-hand side of the equation, we note that −u(0+) is exactly the expression

nk−1

XX

−cksju(k−j−1)(0+)

k=1 j=0

in the statement of Proposition 3.27. Combining our work, the Laplace transform of the di erential equation gives

To compute the inverse Laplace transform we perform the partial fraction expansion for the first term to get

The inverse Laplace transform of the first term is 12 t, and of the second term is −1. To determine the Laplace transform of the third term we note that the inverse Laplace of

1

s2 + 2s + 2

is e−t sin t, and the inverse Laplace transform of

s + 1

s2 + 2s + 2

is e−t cos t. Putting this all together we have obtained the solution

y(t) = e−t 32 sin t + 2 cos t + 2t − 1

This example demonstrates how tedious can be the matter of obtaining “by hand” the solution to even a fairly simple initial value problem.

It is sometimes preferable to obtain the solution in the time-domain, particularly for systems of high-order since the inverse Laplace transform will typically be di cult to obtain in such cases. One way to do this is proposed in Section 3.6.3.

Finish

3.6.2 Computing outputs for SISO linear systems in input/output form using the left causal Laplace transform As mentioned in the preamble to this section, the right causal Laplace transform is the more useful than its left brother for solving initial value problems, by virtue of its encoding the initial conditions in a convenient manner. However, it is possible to solve these same problems using the left causal Laplace transform. To do so, since the initial conditions at t = 0− are all zero (we always work with causal inputs and outputs), it turns out that the correct way to encode the initial conditions at t = 0+ is to add delta-function inputs. While this is not necessarily the recommended way to solve such equations, it does make precise the connection between the left and right causal transforms in this case. Also, it allows us to better understand such things as the impulse response.

Let us begin by considering a proper SISO linear system (N, D) in input/output form. For this system, let us consider inputs of the form

where u0 : R → R has the property that u0, u(1)0 , . . . , u(deg(0 N)−1) are continuous on [0, ∞) and that u(deg(0 N)) is piecewise continuous on [0, ∞). Thus we allow inputs which satisfy the derivative rule for the right causal Laplace transform (cf. Theorem E.7), and which additionally allows a delta-function as input. This allows us to consider the impulse response as part of the collection of problems considered. Note that by allowing a delta-function in our input, we essentially mandate the use of the left causal Laplace transform to solve the equation. Since we are interested in causal outputs, we ask that the output have initial conditions

where, as usual, n = deg(D).

Let us first state the solution of the problem just formulated.

D ddt y(t) = N ddt u(t)

with initial conditions (3.15) has the form y(t) = y0(t) + dδ(t) where d = and where y0 is the solution of the initial value problem

D ddt y0(t) = N ddt u0(t), y0(0+) =, y0(1)(0+) =, . . . , y0(n−1)(0+) =,

where

N(s) = cnsn + cn−1sn−1 + cn−2sn−2 + · · · + c1s + c0

D(s) = sn + pn−1sn−1 + · · · + p1s + p0.

Proof Taking the left causal Laplace transform of the di erential equation, using Corollary E.8, gives

Since the initial conditions at t = 0− for y are all zero, we may express y = y0 + y1 where

D(s)L0+−(y0)(s) = N(s)L0+−(u0)(s)

and

n

X

D(s)L0+−(y1)(s) = cjcsj.

j=0

3.6.3 Computing outputs for SISO linear systems in input/output form using the causal impulse response To see how the Laplace transform method is connected with solving equations in the time-domain, we make the observation, following from Theorem 3.22, that the Laplace transform of the impulse response hN,D is the transfer function TN,D in the case where (N, D) is strictly proper. Therefore, by Exercise EE.5, the inverse Laplace

Let us address the question of which initial value problem of the form (3.12) has the expression (3.16) as its solution. That is, let us determine the proper initial conditions to obtain the solution (3.16).

[0, ∞). Then the integral (3.16) is the solution to the initial value problem (3.12) provided that the initial values y(0+), y(1)(0+), . . . , y(n−1)(0+) are chosen as follows:

(i) let y(0+) = 0;

Proof From Proposition 3.27 it su ces to show that the initial conditions we have defined are such that

nk−1

X X

pksjy(k−j−1)(0+) − cksju(k−j−1)(0+) = 0, k = 1, . . . , n − 1.

k=1 j=0

This will follow if the coe cient of each power of s in the preceding expression vanishes. Starting with the coe cient of sn−1 we see that y(0+) = 0. The coe cient of sn−2 is then

determined to be

y(1)(0+) + pn−1y(0+) − cn−1u(0+),

Let’s see how we may use this to obtain a solution to an initial value problem in the time-domain using the impulse response hN,D.

Proof That every solution of (3.12) can be expressed in the form of (3.17) follows from Proposition 2.32. It thus su ces to show that the given solution satisfies the initial conditions. However, by the definition of yh and by Proposition 3.31 we have

Let’s see how this works in a simple example.

3.33 Example (Example 3.29 cont’d) We take (N(s), D(s)) = (1 − s, s2 + 2s + 2) so the di erential equation is

y¨(t) + 2y˙(t) + 2y(t) = u(t) − u˙(t).

As input we take u(t) = t and as initial conditions we take y(0+) = 1 and y˙(0+) = 0. We first obtain the homogeneous solution yh(t). We first determine the initial conditions, meaning we need to determine the initial conditions y˜0 and y˜1 from Proposition 3.31. These we readily determine to be y˜0 = 0 and y˜1 = c1u(0+) − p1y˜0 = 0. Thus yh should solve the initial value problem

y¨h(t) + 2y˙h(t) + 2yh(t) = 0, yh(0+) = 1, y˙h(0+) = 0.

Recall how to solve this equation. One first determines the roots of the characteristic polynomial that in this case is s2 + 2s + 2. The roots we compute to be −1 ±i. This gives rise to two linearly independent solutions to the homogeneous di erential equation, and these can

be taken to be

y1(t) = e−t cos t, y2(t) = e−t sin t.

Any solution to the homogeneous equation is a sum of these two solutions, so we must have yh(t) = C1y1(t)+C2y2(t) for appropriate constants C1 and C2. To determine these constants we use the initial conditions:

yh(0+) = C1 = 1

y˙h(0+) = −C1 + C2 = 0,

from which we ascertain that C1 = C2 = 1 so that yh(t) = e−t(cos t + sin t).

We now need the impulse response that we can compute however we want (but see Example 3.41 for a slick way to do this) to be

hN,D(t) = e−t(2 sin t − cos t).

Now we may determine

In practice, one does not—at least I do not—solve simple di erential equations this way. Rather, I typically use the “method of undetermined coe cients” where the idea is that after solving the homogeneous problem, the part of the solution that depends on the right-hand side is made a general function of the same “form” as the right-hand side, but with undetermined coe cients. One then resolves the coe cients by substitution into the di erential equation. This method can be found in your garden variety text on ordinary di erential equations, for example [Boyce and Diprima 2000]. A too quick overview is given in Section B.1.

3.6.4 Computing outputs for SISO linear systems Next we undertake the above constructions for SISO linear systems Σ = (A, b, c, D). Of course, we may simply compute the transfer function TΣ and proceed using the tools we developed above for systems in input/output form. However, we wish to see how the additional structure for state-space systems comes into play. Thus in this section we consider the initial value problem

3.34 Lemma Let y(t) be the output for the SISO linear system Σ = (A, b, ct, D) subject to the input u(t) and the initial condition x(0+) = x0: thus y(t) is defined by (3.18). Then there exists unique functions y1, y2, y3 : [0, ∞) → R satisfying:

(i) y1(t) satisfies

x˙ (t) = Ax(t), x(0+) = x0 y1(t) = ctx(t);

(ii) y2(t) satisfies

x˙ (t) = Ax(t) + bu(t), x(0+) = 0 y1(t) = ctx(t);

(iii) y3(t) satisfies

y3(t) = Du(t);

(iv) y(t) = y1(t) + y2(t) + y3(t).

Proof First note that y1(t), y2(t), and y3(t) are indeed uniquely defined by the conditions of (i), (ii), and (iii), respectively. It thus remains to show that (iv) is satisfied. However, this is a straightforward matter of checking that y1(t) + y2(t) + y3(t) as defined by the conditions (i), (ii), and (iii) satisfies (3.18).

The idea here is that to obtain the output for (3.18) we first look at the case where D = 01 and u(t) = 0, obtaining the solution y1(t). To this we add the output y2(t), defined again with D = 01, but this time with the input as the given input u(t). Note that to obtain y1 we use the given initial condition x0, but to obtain y2(t) we use the zero initial condition. Finally, to these we add y3(t) = Du(t) to get the actual output.

Our objective in this section is to provide scalar initial value problems whose solutions are y1(t), y2(t), and y3(t). These are easily implemented numerically. We begin by determining the scalar initial value problem for y1(t).

3.35 Lemma Let Σ = (A, b, ct, 01) be a SISO linear system and let y1(t) be the output determined by condition (i) of Lemma 3.34. Then y1(t) solves the initial value problem

D ddt y1(t) = 0, y1(0+) = ctx0, y1(1)(0+) = ctAx0, . . . , y1(n−1)(0+) = ctAn−1x0, where (N, D) is the c.f.r. for TΣ.

106 3 Transfer functions (the s-domain) 22/10/2004

Proof Note that y1(t) = cteAtx0. Taking Laplace transforms and using Exercise EE.4 gives

Thus we have

D(s)L0++ (y)1(s) = ctadj(sIn − A)x0,

and since the right-hand side is a polynomial of degree at most n − 1 (if deg(D) = n), by (3.13) this means that there are some initial conditions for which y1(t) is a solution to

3.36 Lemma Let Σ = (A, b, ct, 01) be a SISO linear system, let u: [0, ∞) → R possess a right causal Laplace transform, and assume that u, u(1), . . . , u(deg(N)−1) are continuous on [0, ∞) and that u(deg(N)) is piecewise continuous on [0, ∞). If y2(t) is the output determined by condition (ii) of Lemma 3.34, then

using Theorem 3.22, Exercise EE.5, and the fact that (N, D) is the c.f.r. of TΣ. This means

= hΣ(t)u(t) +

0

Thus y(1)(0+) = hΣ(0+)u(0+). We may proceed, using mathematical induction if one wishes to do it properly, to derive

Now we observe that a simple computation given hΣ(t) = cteAtb demonstrates that

h(Σk)(0+) = ctAkb,

The above two lemmas give us the “hard part” of the output for (3.18)—all that remains is to add y3(t) = Du(t). Let us therefore summarise how to determine the output for the system (3.18) from a scalar di erential equation.

3.37 Proposition Let Σ = (A, b, ct, D) be a SISO linear system, let u: [0, ∞) → R possess a right causal Laplace transform, and assume that u, u(1), . . . , u(deg(N)−1) are continuous on [0, ∞) and that u(deg(N)) is piecewise continuous on [0, ∞). If y(t) is the output defined by (3.18) for the input u(t) and initial state x0, then y(t) is the solution of the di erential equation

L0+(y)(s) = c (sIn − A)− b + D L0+(u)(s),

modulo initial conditions. The initial conditions for y are derived as in Lemma 3.36, using the fact that we now have

Now admittedly this is a lengthy result, but with any given example, it is simple enough to apply. Let us justify this by applying the result to an example.

3.38 Example (Example 3.29 cont’d) We take Σ = (A, b, ct, D) with

Thus this is simply the state-space version of the system we dealt with in Examples 3.29 and 3.33. As input we take again u(t) = t, and as initial state we select x0 = (45 , 15 ). Following Proposition 3.37 we compute

ctx0 = 1, ctAx0 + ctbu(0+) = 0,

and we compute the transfer function to be

1 − s

TΣ(s) = s2 + 2s + 2.

Therefore, the initial value problem given to us by Proposition 3.37 is

y¨(t) + 2y˙(t) + 2y(t) = t − 1, y(0+) = 1, y˙(0+) = 0.

Well now, if this isn’t the same initial value problem encountered in Examples 3.29 and 3.33! Of course, the initial state vector x0 was designed to accomplish this. In any case, we may solve this initial value problem in the manner of either of Examples 3.29 and 3.33, and you will recall that the answer is

y(t) = e−t(32 sin t + 2 cos t) + 2t − 1.

3.6.5 Formulae for impulse, step, and ramp responses In this section we focus on developing initial value problems for obtaining some of the basic outputs for SISO systems. We provide this for both SISO linear systems, and those in input/output form. Although the basic definitions are made in the Laplace transform domain, the essential goal of this section is to provide initial value problems in the time-domain, and in so doing provide the natural method for numerically obtaining the various responses. These responses can be obtained by various control packages available, but it is nice to know what they are doing, since they certainly are not performing inverse Laplace transforms!

3.39 Definition Let (N, D) be a SISO linear system in input/output form. The step re-

transform of the unit slope ramp input u(t) = t1(t). Of course, we could define the response to an input u(t) = tk1(t) for k ≥ 2 by noting that the Laplace transform of such an input is

k! . Indeed, it is a simple matter to produce the general formulas following what we do in

sk+1

this section, but we shall not pursue this level of generality here.

We now wish to produce the scalar initial value problems whose solutions are the impulse, step, and ramp responses Fortunately, the hard work has been done already, and we essentially have but to state the answer.

3.40 Proposition Let (N, D) be a proper SISO linear system in input/output form with

D(s) = sn + pn−1sn−1 + · · · + p1s + p0

N(s) = cnsn + cn−1sn−1 + · · · + c1s + c0,

and let ΣN,D = (A, b, ct, D) be the canonical minimal realisation of (N, D). The following statements hold.

(i)If (N, D) is strictly proper then the impulse response hN,D is the solution to the initial value problem

D ddt hN,D(t) = 0, hN,D(0+) = ctb, h(1)N,D(0+) = ctAb, . . . , h(N,Dn−1)(0+) = ctAn−1b.

(ii)The step response 1N,D(t) is the solution to the initial value problem

D ddt 1N,D(t) = c0, 1N,D(0+) = D, 1(1)N,D(0+) = ctb, . . . , 1(N,Dn−1)(0+) = ctAn−2b.

Proof Let us look first at the impulse response. Since the Laplace transform of hN,D is the transfer function TN,D we have

D(s)L0++ (hN,D)(s) = N(s).

As we saw in the course of the proof of Proposition 3.27, the fact that the right-hand side of this equation is a polynomial in s of degree at most n − 1 (if deg(D) = n) implies that hN,D is a solution of the di erential equation D ddt hN,D(t) = 0. The determination of the initial conditions is a simple matter of di erentiating hN,D(t) = cteAtb the required number of times, and evaluating at t = 0.

The last two statements follow from Lemma 3.36 choosing u(t) = 1(t) for the step response, and u(t) = t1(t) for the ramp response.

The impulse response is generalised for proper systems in Exercise E3.1. One can, I expect, see how the proposition gets generalised for inputs like u(t) = tk1(t). Let’s now determine the impulse, step, and ramp response for an example so the reader can see how this is done.

3.41 Example We take (N(s), D(s)) = (1 − s, s2 + 2s + 2), the same example we have been using throughout this section. We shall merely produce the initial value problems that give the step and ramp response for this problem, and not bother with going through the details of obtaining the solutions to these initial value problems. The canonical minimal realisation is ΣN,D = (A, b, ct, D) where

We readily compute ctb = −1 and ctAb = 3.

For the impulse response, we solve the initial value problem

We obtain the solution by looking for a solution of the form est, and so determine that s should be a root of s2 + 2s + 2 = 0. Thus s = −1 ± i, and so our homogeneous solution will be a linear combination of the two linearly independent solutions y1(t) = e−t cos t and y2(t) = e−t sin t. That is,

hN,D(t) = C1e−t cos t + C2e−t sin t.

To determine C1 and C2 we use the initial conditions. These give

hN,D(0+) = C1 = −1

h˙ N,D(0+) = − C1 + C2 = 3.

Solving gives C1 = −1 and C2 = 2, and so we have

hN,D(t) = e−t(2 sin t + cos t).

For the step response the initial value problem is

In like manner, the initial value for the ramp response is

Doing the requisite calculations gives

You may wish to refer back to this section on computing step responses later in the course of reading this book, since the step response is something we shall frequently write down without saying much about how we did it.

3.42 Remark Those of you who are proponents of the Laplace transform will wonder why one does not simply obtain the impulse response or step response by obtaining the inverse Laplace transform of the transfer function or the transfer function multiplied by 1s . While this is theoretically possible, in practice it is not really a good alternative. Even numerically, determining the inverse Laplace transform is very di cult, even when it is possible.

To indicate this, let us consider a concrete example. We take (N(s), D(s)) = (s2 + 3s + 1, s5 + 5s4 + 10s3 + 20s2 + 10s + 5). The impulse response and step response were generated numerically and are shown in Figure 3.9. Both of these were generated in two ways: (1) by

computing numerically the inverse Laplace transform, and (2) by solving the ordinary differential equations of Proposition 3.40. In Table 3.1 can be seen a rough comparison of the time taken to do the various calculations. Obviously, the di erential equation methods are far more e cient, particularly on the step response. Indeed, the inverse Laplace transform methods will sometimes not work, because they rely on the capacity to factor a polynomial.

3.7 Summary

This chapter is very important. A thorough understanding of what is going on here is essential and let us outline the salient facts you should assimilate before proceeding.

1.You should know basic things about polynomials, and in particular you should not hesitate at the mention of the word “coprime.”

2.You need to be familiar with the concept of a rational function. In particular, the words “canonical fractional representative” (c.f.r.) will appear frequently later in this book.

3.You should be able to determine the partial fraction expansion of any rational function.

4.The definition of the Laplace transform is useful, and you ought to be able to apply it when necessary. You should be aware of the abscissa of absolute convergence since it can once in awhile come up and bite you.

5.The properties of the Laplace transform given in Proposition E.9 will see some use.

6.You should know that the inverse Laplace transform exists. You should recognise the value of Proposition E.11 since it will form the basis of parts of our stability investigation.

7.You should be able to perform block diagram algebra with the greatest of ease. We will introduce some powerful techniques in Section 6.1 that will make easier some aspects of this kind of manipulation.

8.Given a SISO linear system Σ = (A, b, ct, D) you should be able to write down its transfer function TΣ.

9.You need to understand some of the features of the transfer function TΣ; for example, you should be able to ascertain from it whether a system is observable, controllable, and whether it is minimum phase.

10.You really really need to know the di erence between a “SISO linear system” and a “SISO linear system in input/output form.” You should also know that there are relationships between these two di erent kinds of objects—thus you should know how to determine ΣN,D for a given strictly proper SISO linear system (N, D) in input/output form.

11.The connection between the impulse response and the transfer function is very important.

12. You ought to be able to determine the impulse response for a transfer function (N, D) in input/output form (use the partial fraction expansion).

Exercises

The impulse response hN,D for a strictly proper SISO linear system (N, D) in input/output form has the property that every solution of the di erential equation D ddt y(t) = N ddt u(t)

can be written as

Z t

y(t) = yh(t) + hN,D(t − τ)u(τ) dτ

0

where yh(t) is a solution of D ddt yh(t) = 0 (see Proposition 3.32). In the next exercise, you will extend this to proper systems.

E3.1 Let (N, D) be a proper, but not necessarily strictly proper, SISO linear system in input/output form.

(a) Show that the transfer function TN,D for (N, D) can be written as

where δ(t) satisfies

Z ∞

δ(t − t0)f(t) dt = f(t0)

−∞

for any integrable function f. Is δ(t) really a map from R to R?

E3.2 Determine the transfer function from rˆ to yˆ for the block diagram depicted in Figure E3.1.

R4(s)

R5(s) R6(s)

Figure E3.1 A block diagram with three loops

At various time throughout the text, we will want to have some properties of real rational functions as s becomes large. In the following simple exercise, you will show that this notion does not depend on the sense in which s is allowed to go to infinity.

E3.3 Let (N, D) be a proper SISO linear system in input/output form.

(a) Show that the limit

lim TN,D(Reiθ)

R→∞

is real and independent of θ (−π, π]. Thus it makes sense to write

lim TN,D(s),

s→∞

and this limit will be in R.

(b)Show that the limit in part (a) is zero if (N, D) is strictly proper.

(c)If (N, D) is not strictly proper (but still proper), give an expression for the limit from part (a) in terms of the coe cients of D and N.

E3.4 For the SISO linear system Σ = (A, b, ct, 01) with

for σ R and ω > 0, determine TΣ.

E3.5 For the SISO linear systems connected in series in Exercise E2.1, determine the transfer function for the interconnected SISO linear system (i.e., for the A, b, c, and D determined in that exercise).

E3.6 For the SISO linear systems connected in parallel in Exercise E2.2, determine the transfer function for the interconnected SISO linear system (i.e., for the A, b, c, and D determined in that exercise).

E3.7 For the SISO linear systems connected in the negative feedback configuration of Exercise E2.3, we will determine the transfer function for the interconnected SISO linear system (i.e., for the A, b, c, and D determined in that exercise).

(a) Use Lemma A.2 to show that

where

U= (sIn2 − A2) + b2ct1(sIn1 − A1)−1b1ct2 −1b2ct1(sIn1 − A1)−1.

(b)Use your answer from part (a), along with Lemma A.3, to show that

TΣ(s) = (1 + TΣ2 (s)TΣ1 (s))−1TΣ2 (s)TΣ1 (s).

(c) Thus we have

TΣ(s) = TΣ1 (s)TΣ2 (s) .

1 + TΣ1 (s)TΣ2 (s) Could you have deduced this otherwise?

(d)Use the computed transfer function to determine the characteristic polynomial for the system. (Thus answer Exercise E2.3(b).)

E3.8 Let Σ = (A, b, ct, D) be a SISO linear system with transfer function TΣ. Show that TΣ is proper, and is strictly proper if and only if D = 01.

E3.9 Consider the two polynomials that are not coprime:

P1(s) = s2 + 1, P2(s) = s3 + s2 + s + 1.

Answer the following questions:

(a)Construct two SISO linear system Σi = (Ai, bi, cti, 01), i = 1, 2, each with state dimension 3, and with the following properties:

1.Σ1 is controllable but not observable;

2.Σ2 is observable but not controllable;

3.TΣ (s) = P1(s) , i = 1, 2.

i P2(s)

P2(s)

why not.

Hint: Use Theorem 2.41.

E3.10 Consider a SISO linear system (N, D) in input/output form, and suppose that N has a real root z > 0 with multiplicity one. Thus, in particular, (N, D) is nonminimum phase. Let u(t) = 1(t)ezt.

(a)Show that there exists a unique set of initial conditions y(0), y(1)(0), . . . , y(n−1)(0)

for which the solution to the di erential equation D(ddt )y(t) = N(ddt )u(t), with these initial conditions, is identically zero.

Hint: Think about what Proposition 3.27 says about the method of Laplace transforms for solving ordinary di erential equations.

(b)Comment on this in light of the material in Section 2.3.3.

For the following three exercises we will consider “open-circuit” and “short-circuit” behaviour of circuits. Let us make the definitions necessary here in a general setting by talking about a SISO linear system Σ = (A, b, ct, D). A pair of functions (u(t), y(t)) satisfying

x˙ (t) = Ax(t) + bu(t)

y(t) = ctx(t) + Du(t)

is open-circuit for Σ if y(t) = 0 for all t and is short-circuit for Σ if u(t) = 0 for all t.

E3.11 Consider the circuit of Exercise E1.7 with the output of Exercise E2.16.

(a)Compute the transfer function for the system.

(b)Determine all open-circuit pairs (u(t), y(t)).

(c)Determine all short-circuit pairs (u(t), y(t)).

(d)Comment on the open/short-circuit behaviour in terms of controllability, observability, and zero dynamics.

E3.12 Consider the circuit of Exercise E1.8 with the output of Exercise E2.17.

(a)Compute the transfer function for the system.

(b)Determine all open-circuit pairs (u(t), y(t)).

(c)Determine all short-circuit pairs (u(t), y(t)).

(d)Comment on the open/short-circuit behaviour in terms of controllability, observability, and zero dynamics.