Modes of Response - Each chapter heading in the contents table is clickable and will take you

Let us examine the response of a plant to a step input. If is a unit step input, i.e.

, then the response is

. (2.18)

s = –3 K₂ = –2A

G s 

K_i r_i s= p_i

K_i

s = z₁ p₁ p₁–z₁

r₁

s = –1 G s  s = –2.01

G s  r₁s p– ₂s p– ₃s p– _n+r₂s p– ₁s p– ₃s p– _n + s p– ₁

 s p– ₂s p– _n

---

r₁+r₂++r_n

  sⁿ^–¹+lower order terms in s

m = n–1 sⁿ^–¹

r_i

i=1 n

 ⁼ ^A

sⁿ^–¹ r_i

i=1 n

 ⁼ ⁰

m = n

U s  U s  = 1s Y s 

Y s  G s 1 ---s

Sec. 2.6 Modes of Response 33 This can be expressed in partial fraction form as:

. (2.19)

Using a table of inverse Laplace transforms yields the time response to the step input,

. (2.20)

Note that

• the response comprises a steady-state component and transient terms; is the response after all the transients components have decayed away;

• , is the amplitude at time zero of the transient terms, ;

• if the input step-size were increased by a factor , all the terms in (2.19) and (2.20) are multiplied by the same factor;

• the form of the transient response is determined by the n roots of the characteristic polynomial, , or by the ^{n poles}of .

• if and are a complex-conjugate pole pair, i.e. , then the associated residues are also a complex-conjugate pair, .

For the present time, assume all roots of have negative real parts. In general, there will be real and complex roots of of the form,

for the real root,

and and for the complex pair .

Associated with each real root , there is a term in the partial fraction expansion .

Taking the inverse transform of the latter term, as in (2.19) - (2.20), results in a term in the time-domain response . If is negative, this response is a monotonically, exponential- ly-decaying mode¹.

Likewise, for the complex pair, and, there are terms in the partial fraction expansion of the form

1. See a note on the term ‘mode’ in Section 3.5.2.

Y s  K₀ ---s K₁

s p– ₁ --- K₂

s p– ₂

---  K_n s p– _n ---

+ + + +

y t  = K₀+K₁e^p¹^t+K₂e^p²^t++K_ne^pⁿ^t

y t  K₀ K₀

K_i, i0 K_ie^pⁱ^t

y t 

Q s  G s 

p₁ p₂ p₂ = p₁

K₂ = K₁ Q s 

Q s  p_i = _i i^th

p_k = _k–j_k p_k₊₁ = _k+j_k k, k+1 p_i = _i

K_is–_i

K_ie^ⁱ^t _i

p_k = _k–j_k p_k₊₁ = _k+j_k

34 Control systems techniques Ch. 2

, where and are complex conjugates.

Correspondingly, there will be a term in the time-domain response of the form

. If is negative, this response is an oscillatory, exponentially-decaying, sinusoidal mode. In power systems analysis the term “mode” usually refers to a broader set of properties that characterize the physical behaviour of the natural system responses. Other modal characterisations include whether it is an electro-mechanical mode, for example, or a controller mode, etc.

The significance of the above analysis is that it reveals, by a simple examination of the poles of the transfer function, the nature of the transient response in the time domain. The real part of the pole, , measured in Neper/s (Np/s), indicates how rapidly the modes in the response decay away. The imaginary part, , of a complex pole pair is the frequency in rad/

s of the damped sinusoidal oscillation. These results provide another valuable short-cut in linear analysis: there is no need to solve the differential equations to determine the nature of the transients in the response.

Let us illustrate some of the important concepts outlined above by means of examples. They are intended to provide some useful insights into the dynamic behaviour of systems.

Example 7

In the following cases, find the time-response of the plant to a step input of magnitude A units.

Case 1. The plant is described by the second-order differential equation , where u(t) and y(t) are the input and output signals, respectively.

Assuming initial conditions are all zero, the plant transfer function is found by replacing the differential operator ^p by the Laplace operator ^s, i.e.

For a step input, . The response can be found using a partial fraction expansion, i.e.

Note that the sum of the residues in this case is zero since . Using the inverse Laplace transform tables, the response in the time domain is found to be

. K_k

s–_k–j_k

--- K_k^* s–_k+j_k ---

+ K_k K_k^*

Ae^^k^tsin_kt+_k _k

_k

_k

p²y t +5py t +6y t  = 6pu t +6u t 

Y s  U s 

--- G s  6s+1 s²+5s+6

--- 6s+1 s+2

 s+3 ---

= =

U s  = A s

Y s  G s U s  6A s +1 s s +2s+3 --- A

---s 3A s+2

 

--- 4A s+3

 

--- – +

= = =

n m– 2

 

y t  = A+3Ae^–^2t–4Ae^–^3t

Sec. 2.6 Modes of Response 35

Note that the coefficients of the modes and , 3A and 4A respectively, are the initial amplitudes of the transient response and are of comparable magnitude.

Case 2. Assume that the zero in the previous plant transfer function lies at Np/s instead of at Np/s. The coefficient of the mode is then zero. (Note that the pole at Np/s still exists in the plant; this aspect is considered later.) This case shows that pole-zero cancellation causes the amplitude of mode to become small or negligible.

Pole-zero cancellation, or close cancellation, is sometimes used in control system design.

However, it should be used with caution because the mode is then only partially ob- servable - or even unobservable - in the output.

Case 3. Let the plant transfer function be:

. The time response to the step input is

In this case the pole at Np/s is much closer to the origin of the complex s-plane than that at Np/s. The response of the fast mode at a time equal to its time constant, i.e. s, is about 37% of its initial amplitude (see Section 2.8.1). The response of the slow mode at s is about 82% of its initial amplitude. In this case the contribution to the overall response of the fast mode rapidly diminishes with time.

Thus the response of the slow mode with a time constant of 1/2 s, dominates until it itself decays after a time equal to four time constants (2 s).

Case 4. Furthermore, it can be shown that the initial amplitude of the response of the fast mode becomes smaller as the pole at is moved further into the left-half of the s- plane relative to the location of the pole of the slow mode. For example, for the transfer function:

the step response is .

Note that the initial amplitude – and therefore the time response – of the fast mode is almost negligible in comparison to the slower mode.

e^–^2t e^–^3t

s = –2

s = –1 e^–^2t

s = –2

e^–^2t

Y s  U s 

--- G s  20

s²+12s+20

--- 20 s+2

 s+10 ---

= =

U s  = A s

y t  = A–1.25Ae^–^2t+0.25Ae^–^10t s = –2

s = –10 e^–^10t

t = T = 1 10

e^–^2t t = 1 10

s = –10

Y s  U s 

--- 200

s+2

 s+100 ---

y t  = A–1.0204Ae^–^2t+0.0204Ae^–^100t 0.0204A

 

36 Control systems techniques Ch. 2 Case 5. Let the plant transfer function be:

The time response to the step input is .

Although the poles in this case are the same as those in Case 3, the initial amplitude of the fast mode is relatively much larger. Due to this mode the overall response of the plant is much faster, although it settles in a time determined by the slow mode.

The significance of the results illustrated in Cases 3 and 5 is that the slower modes tend to dominate the response. Case 5 reveals that the placement of a zero at an appropriate position can speed up the response of a sluggish system. In this case the location of the zero at Np/s relatively close to the pole Np/s diminishes the effect of the slow mode. This concept is commonly used in classical control system design for speeding up the response of a sluggish system.

Case 6. The plant transfer function of Case 1 is modified so that the damping ratio of its second-order poles is less than one, say,

There are a pair of complex poles at . The response to a step of magnitude results in the following partial fraction expansion

Replacing each term by its inverse Laplace transform, we find the closed-form expression for the oscillatory response is:

The denominator of the transfer function, , is of the form . (The following results are described in Section 2.8.2.1). The undamped natural frequency is

Y s  U s 

--- G s  20s+1 s²+12s+20

--- 20s+1 s+2

 s+10 ---

= =

y t  = A+1.25Ae^–^2t–2.25Ae^–^10t

e^–^10t

s = –1 s = –2

Y s  U s 

--- 6s+1 s²+2s+6

--- 6 s+1 s+1

 ²+5 ---

= 6s+1

s+ +1 j 5

 s+1– 5j  ---

s_{1 2}_ = –1j 5 A

Y s  A 1

---s 1+j 5 2s+1+j 5 ---

– 1–j 5

2s+1–j 5 --- –

y t  A 1 1+j 5e^–^¹⁺^j ⁵^t ---2

– 1–j 5e^–^¹^–^j ⁵^t

---2 –

A Ae^–^t 5 e^j ^5t–e^–^j ^5t ---j2

 

 

  e^j ^5t+e^–^j ^5t ---2

 

 

 

– +

A Ae+ ^–^t 5sin 5t–cos 5t

A1+ 6e^–^tsin 5t–24.1

s²+2s+6 s²+2_ns+_n²

Sec. 2.7 Block diagram representations 37

rad/s. The oscillatory mode thus has a damping ratio ; the damping constant is Np/s. If this were a rotor mode having a frequency of oscillation of rad/s, it would be considered well damped.

As has been pointed out earlier, the form of the mode can be derived by inspection of the poles of the transfer function without having to solve for the time response to a step input of the system described by the second-order differential equation,

. ---

Let us emphasize some important results:

• The response to an input of a linear system consists of a steady-state response and a transient response.

• The steady-state term bears a direct relationship to the input function (e.g. doubling the amplitude of the input signal doubles the amplitude of the response).

• The transient terms are determined by the initial magnitude of the input function (at time t(0⁺). However the transient response has a form which is characteristic of the system, and may by identified with the position of the poles of the transfer function; these poles are. the zeros of the characteristic equation.

• The concept not only of modes, poles and zeros, together with information provided through the partial fraction expansion, provide important engineering short-cuts for predicting the characteristic response of the plant in the time domain. By inspection of the factorised denominator of the plant transfer function - thus revealing the pole positions - we can ascertain: whether the plant responses will contain monotonic or oscillatory (i.e. sinusoidal) components, how rapidly transients decay away, and the frequency of any oscillation.

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