[4.7 APPLICATION TO](#page-10-0) FEEDBACK AND CONTROLS

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4.7 APPLICATION TO FEEDBACK AND CONTROLS

Generally, systems are designed to produce a desired output y(t) for a given input x(t). Using the given performance criteria, we can design a system, as shown in Fig. 4.33a. Ideally, such an open-loop system should yield the desired output. In practice, however, the system characteristics change with time, as a result of aging or replacement of some components, or because of changes in the operating environment. Such variations cause changes in the output for the same input. Clearly, this is undesirable in precision systems.

Figure 4.33 (a) Open-loop and (b) closed-loop (feedback) systems.

A possible solution to this problem is to add a signal component to the input that is not a predetermined function of time but will change to counteract the effects of changing system characteristics and the environment. In short, we must provide a correction at the system input to account for the undesired changes just mentioned. Yet since these changes are generally unpredictable, it is not clear how to preprogram appropriate corrections to the input. However, the difference between the actual output and the desired output gives an indication of the suitable correction to be applied to the system input. It may be possible to counteract the variations by feeding the output (or some function of output) back to the input.

We unconsciously apply this principle in daily life. Consider an example of marketing a certain product. The optimum price of the product is the value that maximizes the profit of a merchant. The output in this case is the profit, and the input is the price of the item. The output (profit) can be controlled (within limits) by varying the input (price). The merchant may price the product too high initially, in which case, he will sell too few items, reducing the profit. Using feedback of the profit (output), he adjusts the price (input), to maximize his profit. If there is a sudden or unexpected change in the business environment, such as a strike-imposed shutdown of a large factory in town, the demand for the item goes down, thus reducing his output (profit). He adjusts his input (reduces price) using the feedback of the output (profit) in a way that will optimize his profit in the changed circumstances. If the town suddenly becomes more prosperous because a new factory opens, he will increase the price to maximize the profit. Thus, by continuous feedback of the output to the input, he realizes his goal of maximum profit (optimum output) in any given circumstances. We observe thousands of examples of feedback systems around us in everyday life. Most social, economical, educational, and political processes are, in fact, feedback processes. A block diagram of such a system, called the feedback or closed-loop system, is shown in Fig. 4.33b.

A feedback system can address the problems arising because of unwanted disturbances such as random-noise signals in electronic systems, a gust of wind affecting a tracking antenna, a meteorite hitting a spacecraft, and the rolling motion of antiaircraft gun platforms mounted on ships or moving tanks. Feedback may also be used to reduce nonlinearities in a system or to control its rise time (or bandwidth). Feedback is used to achieve, with a given system, the desired objective within a given tolerance, despite partial ignorance of the system and the environment. A feedback system, thus, has an ability for supervision and self-correction in the face of changes in the system parameters and external disturbances (change in the environment).

Consider the feedback amplifier in Fig. 4.34. Let the forward amplifier gain G = 10,000. One-hundredth of the output is fed back to the input (H = 0.01). The gain T of the feedback amplifier is obtained by [see Eq. (4.35)]

$$T = \frac{G}{1 + GH} = \frac{10,000}{1 + 100} = 99.01$$

Suppose that because of aging or replacement of some transistors, the gain G of the forward amplifier changes from 10,000 to 20,000. The new gain of the feedback amplifier is given by

$$T = \frac{G}{1 + GH} = \frac{20,000}{1 + 200} = 99.5$$

Surprisingly, 100% variation in the forward gain G causes only 0.5% variation in the feedback amplifier gain T. Such reduced sensitivity to parameter variations is a must in precision amplifiers. In this example, we reduced the sensitivity of gain to parameter variations at the cost of forward

^H Figure 4.34 Effects of negative and positive feedback.

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gain, which is reduced from 10,000 to 99. There is no dearth of forward gain (obtained by cascading stages). But low sensitivity is extremely precious in precision systems.

Now, consider what happens when we add (instead of subtract) the signal fed back to the input. Such addition means the sign on the feedback connection is + instead of − (which is same as changing the sign of H in Fig. 4.34). Consequently,

$$T = \frac{G}{1 - GH}$$

If we let G = 10,000 as before and H = 0.9×10−4, then

$$T = \frac{10,000}{1 - 0.9(10^{4})(10^{-4})} = 100,000$$

Suppose that because of aging or replacement of some transistors, the gain of the forward amplifier changes to 11,000. The new gain of the feedback amplifier is

$$T = \frac{11,000}{1 - 0.9(11,000)(10^{-4})} = 1,100,000$$

Observe that in this case, a mere 10% increase in the forward gain G caused 1000% increase in the gain T (from 100,000 to 1,100,000). Clearly, the amplifier is very sensitive to parameter variations. This behavior is exactly opposite of what was observed earlier, when the signal fed back was subtracted from the input.

What is the difference between the two situations? Crudely speaking, the former case is called the negative feedback and the latter is the positive feedback. The positive feedback increases system gain but tends to make the system more sensitive to parameter variations. It can also lead to instability. In our example, if G were to be 111,111, then GH = 1, T = ∞, and the system would become unstable because the signal fed back was exactly equal to the input signal itself, since GH = 1. Hence, once a signal has been applied, no matter how small and how short in duration, it comes back to reinforce the input undiminished, which further passes to the output, and is fed back again and again and again. In essence, the signal perpetuates itself forever. This perpetuation, even when the input ceases to exist, is precisely the symptom of instability.

Generally speaking, a feedback system cannot be described in black and white terms, such as positive or negative. Usually H is a frequency-dependent component, more accurately represented by H(s); hence it varies with frequency. Consequently, what was negative feedback at lower frequencies can turn into positive feedback at higher frequencies and may give rise to instability. This is one of the serious aspects of feedback systems, which warrants a designer’s careful attention.

4.7-1 Analysis of a Simple Control System

Figure 4.35a represents an automatic position control system, which can be used to control the angular position of a heavy object (e.g., a tracking antenna, an anti-aircraft gun mount, or the position of a ship). The input θ*ⁱ* is the desired angular position of the object, which can be set at any given value. The actual angular position θ*^o* of the object (the output) is measured by a potentiometer whose wiper is mounted on the output shaft. The difference between the input θ*ⁱ*

(set at the desired output position) and the output θ*^o* (actual position) is amplified; the amplified output, which is proportional to θ*ⁱ* − θo, is applied to the motor input. If θ*ⁱ* − θ*^o* = 0 (the output being equal to the desired angle), there is no input to the motor, and the motor stops. But if θ*^o* = θi, there will be a nonzero input to the motor, which will turn the shaft until θ*^o* = θi. It is evident that by setting the input potentiometer at a desired position in this system, we can control the angular position of a heavy remote object.

The block diagram of this system is shown in Fig. 4.35b. The amplifier gain is K, where K is adjustable. Let the motor (with load) transfer function that relates the output angle θ*^o* to the motor input voltage be G(s) [for a starting point, see Eq. (1.32)]. This feedback arrangement is identical to that in Fig. 4.18d with H(s) = 1. Hence, T(s), the (closed-loop) system transfer function relating the output θ*^o* to the input θi, is

$$\frac{\Theta_o(s)}{\Theta_i(s)} = T(s) = \frac{KG(s)}{1 + KG(s)}$$

From this equation, we shall investigate the behavior of the automatic position control system in Fig. 4.35a for a step and a ramp input.

STEP INPUT

If we desire to change the angular position of the object instantaneously, we need to apply a step input. We may then want to know how long the system takes to position itself at the desired angle, whether it reaches the desired angle, and whether it reaches the desired position smoothly (monotonically) or oscillates about the final position. If the system oscillates, we may want to know how long it takes for the oscillations to settle down. All these questions can be readily answered by finding the output θo(t) when the input θi(t) = u(t). A step input implies instantaneous change in the angle. This input would be one of the most difficult to follow; if the system can perform well for this input, it is likely to give a good account of itself under most other expected situations. This is why we test control systems for a step input.

For the step input θi(t) = u(t), i(s) = 1/s and

$$\Theta_o(s) = \frac{1}{s}T(s) = \frac{KG(s)}{s[1+KG(s)]}$$

Let the motor (with load) transfer function relating the load angle θo(t) to the motor input voltage be G(s) = 1/(s(s+8)). This yields

$$\Theta_o(s) = \frac{\frac{K}{s(s+8)}}{s\left[1 + \frac{K}{s(s+8)}\right]} = \frac{K}{s(s^2 + 8s + K)}$$

Let us investigate the system behavior for three different values of gain K. For K = 7,

$$\Theta_o(s) = \frac{7}{s(s^2 + 8s + 7)} = \frac{7}{s(s+1)(s+7)} = \frac{1}{s} - \frac{\frac{7}{6}}{s+1} + \frac{\frac{1}{6}}{s+7}$$

Figure 4.35 (a) An automatic position control system. (b) Its block diagram. (c) The unit step response. (d) The unit ramp response.

and

$$\theta_o(t) = \left(1 - \frac{7}{6}e^{-t} + \frac{1}{6}e^{-7t}\right)u(t)$$

This response, illustrated in Fig. 4.35c, shows that the system reaches the desired angle, but at a rather leisurely pace. To speed up the response let us increase the gain to, say, 80.

For K = 80,

$$\Theta_o(s) = \frac{80}{s(s^2 + 8s + 80)} = \frac{80}{s(s + 4 - j8)(s + 4 + j8)}$$ $$= \frac{1}{s} + \frac{\frac{\sqrt{5}}{4}e^{j153^\circ}}{s + 4 - j8} + \frac{\frac{\sqrt{5}}{4}e^{-j153^\circ}}{s + 4 + j8}$$

and

$$\theta_o(t) = \left[1 + \frac{\sqrt{5}}{2}e^{-4t}\cos{(8t + 153^\circ)}\right]u(t)$$

This response, also depicted in Fig. 4.35c, achieves the goal of reaching the final position at a faster rate than that in the earlier case (K = 7). Unfortunately the improvement is achieved at the cost of ringing (oscillations) with high overshoot. In the present case, the percent overshoot (PO) is 21%. The response reaches its peak value at peak time tp = 0.393 second. The rise time, defined as the time required for the response to rise from 10% to 90% of its steady-state value, indicates the speed of response.† In the present case tr = 0.175 second. The steady-state value of the response is unity so that the steady-state error is zero. Theoretically it takes infinite time for the response to reach the desired value of unity. In practice, however, we may consider the response to have settled to the final value if it closely approaches the final value. A widely accepted measure of closeness is within 2% of the final value. The time required for the response to reach and stay within 2% of the final value is called the settling time ts. ‡ In Fig. 4.35c, we find ts ≈ 1 second (when K = 80). A good system has a small overshoot, small tr and ts and a small steady-state error.

A large overshoot, as in the present case, may be unacceptable in many applications. Let us try to determine K (the gain) that yields the fastest response without oscillations. Complex characteristic roots lead to oscillations; to avoid oscillations, the characteristic roots should be real. In the present case, the characteristic polynomial is s² +8s+K. For K > 16, the characteristic roots are complex; for K < 16, the roots are real. The fastest response without oscillations is obtained by choosing K = 16. We now consider this case.

For K = 16,

$$\Theta_o(s) = \frac{16}{s(s^2 + 8s + 16)} = \frac{16}{s(s+4)^2} = \frac{1}{s} - \frac{1}{s+4} - \frac{4}{(s+4)^2}$$

and

$$\theta_o(t) = [1 - (4t + 1)e^{-4t}]u(t)$$

This response also appears in Fig. 4.35c. The system with K > 16 is said to be underdamped (oscillatory response), whereas the system with K < 16 is said to be overdamped. For K = 16, the system is said to be critically damped.

^† Delay time td, defined as the time required for the response to reach 50% of its steady-state value, is another indication of speed. For the present case, td = 0.141 second.

^‡ Typical percentage values used are 2 to 5% for ts.

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There is a trade-off between undesirable overshoot and rise time. Reducing overshoots leads to higher rise time (sluggish system). In practice, a small overshoot, which is still faster than the critical damping, may be acceptable. Note that percent overshoot PO and peak time tp are meaningless for the overdamped or critically damped cases. In addition to adjusting gain K, we may need to augment the system with some type of compensator if the specifications on overshoot and the speed of response are too stringent.

RAMP INPUT

If the anti-aircraft gun in Fig. 4.35a is tracking an enemy plane moving with a uniform velocity, the gun-position angle must increase linearly with t. Hence, the input in this case is a ramp; that is, θi(t) = tu(t). Let us find the response of the system to this input when K = 80. In this case, i(s) = 1/s2, and

$$\Theta_o(s) = \frac{80}{s^2(s^2 + 8s + 80)} = -\frac{0.1}{s} + \frac{1}{s^2} + \frac{0.1(s - 2)}{s^2 + 8s + 80}$$

Use of Table 4.1 yields

$$\theta_o(t) = \left[ -0.1 + t + \frac{1}{8}e^{-8t}\cos\left(8t + 36.87^\circ\right) \right] u(t)$$

This response, sketched in Fig. 4.35d, shows that there is a steady-state error er = 0.1 radian. In many cases such a small steady-state error may be tolerable. If, however, a zero steady-state error to a ramp input is required, this system in its present form is unsatisfactory. We must add some form of compensator to the system.

EXAMPLE 4.26 Step and Ramp Responses of Feedback Systems Using MATLAB

Using the feedback system of Fig. 4.18d with G(s) = K/(s(s + 8)) and H(s) = 1, determine the step response for each of the following cases: (a) K = 7, (b) K = 16, and (c) K = 80. Additionally, find the unit ramp response when (d) K = 80.

Example 4.21 computes the transfer functions of these feedback systems in a simple way. In this example, the conv command is used to demonstrate polynomial multiplication of the two denominator factors of G(s). Step responses are computed by using the step command. (a–c)

H = tf(1,1); K = 7; G = tf([K],conv([1 0],[1 8])); Ha = feedback(G,H);

^&gt;> H = tf(1,1); K = 16; G = tf([K],conv([1 0],[1 8])); Hb = feedback(G,H);

^&gt;> H = tf(1,1); K = 80; G = tf([K],conv([1 0],[1 8])); Hc = feedback(G,H);

^&gt;> clf; step(Ha,‘k-’,Hb,‘k—‘,Hc,‘k-.’);

^&gt;> legend(‘K = 7’,‘K = 16’,‘K = 80’,‘Location’,‘best’);

Figure 4.36 Step responses for Ex. 4.26.

(d) The unit ramp response is equivalent to the integral of the unit step response. We can obtain the ramp response by taking the step response of the system in cascade with an integrator. To help highlight waveform detail, we compute the ramp response over the short time interval of 0 ≤ t ≤ 1.5.

t = 0:.001:1.5; Hd = series(Hc,tf([1],[1 0])); >> step(Hd,‘k-’,t); title(‘Unit Ramp Response’);

DESIGN SPECIFICATIONS

Now the reader has some idea of the various specifications a control system might require. Generally, a control system is designed to meet given transient specifications, steady-state error specifications, and sensitivity specifications. Transient specifications include overshoot, rise time, and settling time of the response to step input. The steady-state error is the difference between

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the desired response and the actual response to a test input in steady state. The system should also satisfy a specified sensitivity specifications to some system parameter variations, or to certain disturbances. Above all, the system must remain stable under operating conditions. Discussion of design procedures used to realize given specifications is beyond the scope of this book.