The simplified transfer function model from steering angle

For each of the root loci shown in Figure P8.1, tell whether or not the sketch can be a root locus. If the sketch cannot be a root locus, explain why. Give all reasons. [Section: 8.4]

Sketch the general shape of the root locus for each of the open-loop pole-zero plots shown in Figure P8.2. [Section: 8.4]

Sketch the root locus for the unity feedback system shown in Figure P8.3 for the following transfer functions: [Section: 8.4] a. G(s) = \(\frac{K(s+2)(s+6)}{s^2+8 s+25}\) b. G(s) = \(\frac{K\left(s^2+4\right)}{\left(s^2+1\right)}\) c. G(s) = \(\frac{K\left(s^2+1\right)}{s^2}\) d. G(s) = \(\frac{K}{(s+1)^3(s+4)}\)

Let G(s) = \(\frac{K\left(s+\frac{2}{3}\right)}{s^2(s+6)}\) in Figure P8.3. [Section: 8.5] a. Plot the root locus. b. Write an expression for the closed-loop transfer function at the point where the three closed-loop poles meet.

Let G(s) = \(\frac{-K(s+1)^2}{s^2+2 s+2}\) with K > 0 in Figure P8.3. [Sections: 8.5, 8.9] a. Find the range of K for closed-loop stability. b. Sketch the system's root locus. c. Find the position of the closed-loop poles when K = 1 and K = 2.

For the open-loop pole-zero plot shown in Figure P8.4, sketch the root locus and find the break-in point. [Section: 8.5]

Sketch the root locus of the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K(s+1)(s+7)}{(s+3)(s-5)}\) and find the break-in and breakaway points. Find the range of K for which the system is closed-loop stable.

The characteristic polynomial of a feedback control system, which is the denominator of the closed-loop transfer function, is given by \(s^3+2 s^2\) + (20 K + 7) s + 100 K. Sketch the root locus for this system. [Section: 8.8]

Figure P8.5 shows open-loop poles and zeros. There are two possibilities for the sketch of the root locus. Sketch each of the two possibilities. Be aware that only one can be the real locus for specific open-loop pole and zero values. [Section: 8.4]

Plot the root locus for the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K(s+2)\left(s^2+4\right)}{(s+5)(s-3)}\) For what range of K will the poles be in the right half-plane? [Section: 8.5]

For the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K\left(s^2-9\right)}{\left(s^2+4\right)}\) sketch the root locus and tell for what values of K the system is stable and unstable. [Section: 8.5]

Sketch the root locus for the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K\left(s^2+2\right)}{(s+3)(s+4)}\) Give the values for all critical points of interest. Is the system ever unstable? If so, for what range of K? [Section: 8.5]

For each system shown in Figure P8.6, make an accurate plot of the root locus and find the following: [Section: 8.5] a. The breakaway and break-in points b. The range of K to keep the system stable c. The value of K that yields a stable system with critically damped second-order poles d. The value of K that yields a stable system with a pair of second-order poles that have a damping ratio of 0.707

Sketch the root locus and find the range of K for stability for the unity feedback system shown in Figure P8.3 for the following conditions: [Section: 8.5] a. G(s) = \(\frac{K\left(s^2+1\right)}{(s-1)(s+2)(s+3)}\) b. G(s) = \(\frac{K\left(s^2-2 s+2\right)}{s(s+1)(s+2)}\)

For the unity feedback system of Figure P8.3, where G(s) = \(\frac{K(s+5)}{\left(s^2+1\right)(s-1)(s+3)}\) sketch the root locus and find the range of K such that there will be only two right-half-plane poles for the closed-loop system. [Section: 8.5]

For the unity feedback system of Figure P8.3, where G(s) = \(\frac{K}{s(s+5)(s+8)}\) plot the root locus and calibrate your plot for gain. Find all the critical points, such as breakaways, asymptotes, jo-axis crossing, and so forth. [Section: 8.5]

Given the unity feedback system of Figure P8.3, make an accurate plot of the root locus for the following: a. G(s) = \(\frac{K\left(s^2-2 s+2\right)}{(s+1)(s+2)}\) b. G(s) = \(\frac{K(s-1)(s-2)}{(s+1)(s+2)}\) Also, do the following: Calibrate the gain for at least four points for each case. Also find the breakaway points, the j\(\omega\)-axis crossing, and the range of gain for stability for each case. Find the angles of arrival for Part a. [Section: 8.5]

Given the root locus shown in Figure P8.7, [Section: 8.5] a. Find the value of gain that will make the system marginally stable. b. Find the value of gain for which the closed-loop transfer function will have a pole on the real axis at -5.

Given the unity feedback system of Figure P8.3, where G(s) = \(\frac{K(s+2)}{s(s+1)(s+3)(s+5)}\) do the following: [Section: 8.5] a. Sketch the root locus. b. Find the asymptotes. c. Find the value of gain that will make the system marginally stable. d. Find the value of gain for which the closed-loop transfer function will have a pole on the real axis at -0.5.

For the unity feedback system of Figure P8.3, where G(s) = \(\frac{K(s+\alpha)}{s(s+3)(s+6)}\) find the values of \(\alpha\) and K that will yield a second-order closed-loop pair of poles at \(-1 \pm j 100\). [Section: 8.5]

For the unity feedback system of Figure P8.3, where G(s) = \(\frac{K(s-1)(s-2)}{s(s+1)(s+2)}\) sketch the root locus and find the following: [Section: 8.5] a. The breakaway and break-in points b. The j\(\omega\)-axis crossing c. The range of gain to keep the system stable d. The value of K to yield a stable system with second order complex poles, with a damping ratio of 0.5.

For the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K(s+10)(s+20)}{(s+30)\left(s^2-10 s+100\right)}\) do the following: [Section: 8.7] a. Sketch the root locus. b. Find the range of gain, K, that makes the system stable. c. Find the value of K that yields a damping ratio of 0.707 for the system's closed-loop dominant poles. d. Find the value of K that yields closed-loop critically damped dominant poles.

For the system of Figure P8.8(a), sketch the root locus and find the following: [Section: 8.7] a. Asymptotes b. Breakaway points c. The range of K for stability d. The value of K to yield a 0.7 damping ratio for the dominant second-order pair To improve stability, we desire the root locus to cross the jo-axis at j 5.5. To accomplish this, the open-loop function is cascaded with a zero, as shown in Figure P8.8 (b). e. Find the value of \(\alpha\) and sketch the new root locus. f. Repeat Part c for the new locus. g. Compare the results of Part c and Part f. What improvement in transient response do you notice?

Sketch the root locus for the positive-feedback system shown in Figure P8.9. [Section: 8.9]

Root loci are usually plotted for variations in the gain. Sometimes we are interested in the variation of the closed-loop poles as other parameters are changed. For the system shown in Figure P8.10, sketch the root locus as \(\alpha\) is varied. [Section: 8.8]

Given the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K}{(s+1)(s+2)(s+3)}\) do the following problem parts by first making a second-order approximation. After you are finished with all of the parts, justify your second-order approximation. [Section: 8.7] a. Sketch the root locus. b. Find K for 20% overshoot. c. For K found in Part b, what is the settling time, and what is the peak time? d. Find the locations of higher-order poles for K found in Part b. e. Find the range of K for stability.

For the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K\left(s^2-2 s+2\right)}{(s+2)(s+4)(s+5)(s+6)}\) do the following: [Section: 8.7] a. Sketch the root locus. b. Find the asymptotes. c. Find the range of gain, K, that makes the system stable. d. Find the breakaway points. f. Find the location of higher-order closed-loop poles when the system is operating with 25% overshoot. g. Discuss the validity of your second-order approximation. h. Use MATLAB to obtain the closed-loop step response to validate or refute your second-order approximation.

The unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K(s+2)(s+3)}{s(s+1)}\) is to be designed for minimum damping ratio. Find the following: [Section: 8.7] a. The value of K that will yield minimum damping ratio b. The estimated percent overshoot for that case c. The estimated settling time and peak time for that case d. The justification of a second-order approximation (discuss) e. The expected steady-state error for a unit ramp input for the case of minimum damping ratio

For the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K(s+2)}{s(s+6)(s+10)}\) find K to yield closed-loop complex poles with a damping ratio of 0.55. Does your solution require a justification of a second-order approximation? Explain. [Section: 8.7]

For the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K(s+\alpha)}{s(s+2)(s+6)}\) find the value of \(\alpha\) so that the system will have a settling time of 2 seconds for large values of K. Sketch the resulting root locus. [Section: 8.8]

For the unity feedback system shown in Figure 8.3, where G(s) = \(\frac{K(s+5)}{\left(s^2+8 s+25\right)(s+1)^2(s+\alpha)}\) design K and \(\alpha\) so that the dominant complex poles of the closed-loop function have a damping ratio of 0.5 and a natural frequency of 1.2 rad/s.

For the unity feedback system shown in Figure 8.3, where G(s) = \(\frac{K}{s(s+3)(s+4)(s+8)}\) do the following: [Section: 8.7] a. Sketch the root locus. b. Find the value of K that will yield a 10% overshoot. c. Locate all nondominant poles. What can you say about the second-order approximation that led to your answer in Part b? d. Find the range of K that yields a stable system.

Repeat Problem 32 using MATLAB. Use one program to do the following: a. Display a root locus and pause. b. Draw a close-up of the root locus where the axes go from -2 to 0 on the real axis and -2 to 2 on the imaginary axis. c. Overlay the 10 % overshoot line on the close-up root locus. d. Select interactively the point where the root locus crosses the 10% overshoot line, and respond with the gain at that point as well as all of the closed-loop poles at that gain. e. Generate the step response at the gain for 10% overshoot.

For the unity feedback system shown in Figure 8.3, where G(s) = \(\frac{K\left(s^2+4 s+5\right)}{\left(s^2+2 s+5\right)(s+3)(s+4)}\) do the following: [Section: 8.7] a. Find the gain, K, to yield a 1 -second peak time if one assumes a second-order approximation. b. Check the accuracy of the second-order approximation using MATLAB to simulate the system.

For the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K(s+2)(s+3)}{\left(s^2+2 s+2\right)(s+4)(s+5)(s+6)}\) do the following: [Section: 8.7] a. Sketch the root locus. b. Find the j \(\omega\)-axis crossing and the gain, K, at the crossing. c. Find all breakaway and break-in points. d. Find angles of departure from the complex poles. e. Find the gain, K, to yield a damping ratio of 0.3 for the closed-loop dominant poles.

Repeat Parts a through c and e of Problem 35 for [Section: 8.7] G(s) = \(\frac{K(s+4)}{s(s+1)(s +2)(s+10)}\)

For the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K}{(s+3)\left(s^2+4 s+5\right)}\) do the following: [Section: 8.7] a. Find the location of the closed-loop dominant poles if the system is operating with 15% overshoot. b. Find the gain for Part a. c. Find all other closed-loop poles. d. Evaluate the accuracy of your second-order approximation.

For the system shown in Figure P8.11, do the following: [Section: 8.7] a. Sketch the root locus. b. Find the jo-axis crossing and the gain, K, at the crossing. c. Find the real-axis breakaway to two-decimal-place accuracy. d. Find angles of arrival to the complex zeros. e. Find the closed-loop zeros. f. Find the gain, K, for a closed-loop step response with 30% overshoot. g. Discuss the validity of your second-order approximation.

Sketch the root locus for the system of Figure P8.12 and find the following: [Section: 8.7] a. The range of gain to yield stability b. The value of gain that will yield a damping ratio of 0.707 for the system’s dominant poles c. The value of gain that will yield closed-loop poles that are critically damped

Repeat Problem 39 using MATLAB. The program will do the following in one program: a. Display a root locus and pause. b. Display a close-up of the root locus where the axes go from -2 to 0.5 on the real axis and -2 to 2 on the imaginary axis. c. Overlay the 0.707 damping ratio line on the close-up root locus. d. Allow you to select interactively the point where the root locus crosses the 0.707 damping ratio line, and respond by displaying the gain at that point as well as all of the closed-loop poles at that gain. The program will then allow you to select interactively the imaginary-axis crossing and respond with a display of the gain at that point as well as all of the closed-loop poles at that gain. Finally, the program will repeat the evaluation for critically damped dominant closed-loop poles. e. Generate the step response for the critically damped case.

Given the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K(s+z)}{s^2(s+10)}\) do the following: [Section: 8.7] a. If z = 2, find K so that the damped frequency of oscillation of the transient response is 5 rad/s. b. For the system of Part a, what static error constant (finite) can be specified? What is its value? c. The system is to be redesigned by changing the values of z and K. If the new specifications are %OS = 4:32% and \(T_s\) = 0:8 s, find the new values of z and K.

Given the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K}{(s+1)(s+3)(s+6)^2}\) find the following: [Section: 8.7] a. The value of gain, K, that will yield a settling time of 4 seconds b. The value of gain, K, that will yield a critically damped system

You are given the unity-feedback system of Figure P8.3, where G(s) = \(\frac{K(s+0.02)}{s^2(s+4)(s+10)(s+25)}\) Use MATLAB to plot the root locus. Use a closeup of the locus (from -5 to 0 and -1 to 6) to find the gain, K, that yields a closed-1oop unit-step response, c(t), with 20.5 % overoot and a settling time of \(T_s\) = 3 seconds. Mark on the time response graph all other relevant characteristics, such as the peak time, risetime, and final steady-state value.

Let G(s) = \(\frac{K(s-1)}{(s+2)(s+3)}\) in Figure P8.3. [Section: 8.7]. a. Find the range of K for closed-loop stability. b. Plot the root locus for K > 0. c. Plot the root locus for K < 0. d. Assuming a step input, what value of K will result in the smallest attainable settling time? e. Calculate the system's \(e_{s s}\) for a unit step input assuming the value of K obtained in Part d. f. Make an approximate hand sketch of the unit step response of the system if K has the value obtained in Part d.

Given the unity feedback system shown in Figure P8.3, where G(s) = \(\frac{K}{s(s+1.25)(s+8)}\) evaluate the pole sensitivity of the closed-loop system if the second-order, underdamped closed-loop poles are set for [Section: 8.10] a. \(\zeta\) = 0.591 b. \(\zeta\) = 0.456 c. Which of the two previous cases has more desirable sensitivity?

Repeat Problem 3 but sketch your root loci for negative values of K. [Section: 8.9]

Figure P8.13 shows the block diagram of a closed loop control of a linearized magnetic levitation system (Galvão, 2003). Assuming A = 1300 and \(\eta\) = 860, draw the root locus and find the range of K for closed-loop stability when: a. G(s) = K; b. G(s) = \(\frac{K(s+200)}{s+1000}\)

The simplified transfer function model from steering angle \(\delta(s)\) to tilt angle \(\varphi(s)\) in a bicycle is given by G(s) = \(\frac{\varphi(s)}{\delta(s)}\) = \(\frac{a V}{b h} * \frac{s+\frac{V}{a}}{s^2-\frac{g}{h}}\) In this model, h represents the vertical distance from the center of mass to the floor, so it can be readily verified that the model is open-loop unstable. (Åström, 2005). Assume that for a specific bicycle, a = 0:6 m, b = 1:5 m, h = 0:8 m, and g = 9:8 m/sec. In order to stabilize the bicycle, it is assumed that the bicycle is placed in the closed-loop configuration shown in Figure P8.3 and that the only available control variable is V, the rear wheel velocity. a. Find the range of V for closed-loop stability. b. Explain why the methods presented in this chapter cannot be used to obtain the root locus. c. Use MATLAB to obtain the system’s root locus.

A technique to control the steering of a vehicle that follows a line located in the middle of a lane is to define a look-ahead point and measure vehicle deviations with respect to the point. A linearized model for such a vehicle is \(\left[\begin{array}{c} \dot{V} \\ \dot{r} \\ \dot{\psi} \\ \dot{Y}_g \end{array}\right]=\left[\begin{array}{cccc} a_{11} & a_{12} & -b_1 K & \frac{b_1 K}{d} \\ a_{21} & a_{22} & -b_2 K & \frac{b_2 K}{d} \\ 0 & 1 & 0 & 0 \\ 1 & 0 & U & 0 \end{array}\right]\left[\begin{array}{c} V \\ r \\ \psi \\ Y_g \end{array}\right]\) where V = vehicle's lateral velocity, r = vehicle's yaw velocity, \(\psi\) = vehicle's yaw position, and \(Y_g\) = the y-axis coordinate of the vehicle's center of gravity. K is a parameter to be varied depending upon trajectory changes. In a specific vehicle traveling at a speed of U = -10 m/sec, the parameters are \(a_{11}\) = -11.6842, \(a_{12}\) = 6.7632, \(b_1\) = -61.5789, \(a_{21}\) = -3.5143, \(a_{22}\) = 24.0257, and \(b_2\) = 66.8571 . d = 5 m is the look-ahead distance (Ünyelio?lu, 1997). Assuming the vehicle will be controlled in closed loop: a. Find the system's characteristic equation as a function of K. b. Find the system's root locus as K is varied. c. Using the root locus found in Part b, show that the system will be unstable for all values K.

It is known that mammals have hormonal regulation mechanisms that help maintain almost constant calcium plasma levels (0.08-0.1 g/L in dairy cows). This control is necessary to maintain healthy functions, as calcium is responsible for diverse physiological functions, such as bone formation, intracellular communications, and blood clotting. It has been postulated that the mechanism of calcium control resembles that of a PI (proportional-plus integral) controller. PI controllers (discussed in detail in Chapter 9) are placed in cascade with the plant and used to improve steady-state error. Assume that the PI controller has the form \(G_c(s)\) = \(\left[K_P+\frac{K_I}{S}\right]\) where \(K_P\) and \(K_I\) are constants. Also assume that the mammal's system accumulates calcium in an integrator-like fashion, namely P(s) = \(\frac{1}{V_s}\), where V is the plasma volume. The closed-loop model is similar to that of Figure P8.3, where G(s) = \(G_c\)(s)P(s) (Khammash, 2004). a. Sketch the system’s root locus as a function of \(K_P\), assuming \(K_I\) > 0 is constant. b. Sketch the system’s root locus as a function of \(K_I\), assuming \(K_P\) > 0 is constant.

Problem 64 in Chapter 7 introduced the model of a TCP/IP router whose packet-drop probability is controlled by using a random early detection (RED) algorithm (Hollot, 2001). Using Figure P8.3 as a model, a specific router queue's open-loop transfer function is G(s) = \(\frac{7031250 L e^{-0.2 s}}{(s+0.667)(s+5)(s+50)}\) The function \(e^{-0.2 s}\) represents delay. To apply the root locus method, the delay function must be replaced with a rational function approximation. A first-order Padé approximation can be used for this purpose. Let (e^{-s D} \approx 1-D s\). Using this approximation, plot the root locus of the system as a function of L.

For the dynamic voltage restorer (DVR) discussed in Problem 47, Chapter 7, do the following: a. When \(Z_L\) = \(\frac{1}{s C_L}\), a pure capacitance, the system is more inclined toward instability. Find the system's characteristic equation for this case. b. Using the characteristic equation found in Part a, sketch the root locus of the system as a function of \(C_L\). Let L = 7.6 mH, C = 11 \(\mu \mathrm{F}\), \(\alpha\) = 26.4, \(\beta\) = 1, \(K_m\) = 25, \(K_v\) = 15, \(K_T\) = 0.09565, and \(\tau\) = 2 ms(Lam, 2004).

The closed-loop vehicle response in stopping a train depends on the train's dynamics and the driver, who is an integral part of the feedback loop. In Figure P8.3, let the input be R(s) = \(v_r\) the reference velocity, and the output C(s) = v, the actual vehicle velocity. (Yamazaki, 2008) shows that such dynamics can be modeled by G(s) = \(G_d(s) G_t(s)\) where \(G_d(s)\) = \(h\left(1+\frac{K}{s}\right) \frac{s-\frac{L}{2}}{s+\frac{L}{2}}\) represents the driver dynamics with h, K, and L parameters particular to each individual driver. We assume here that h = 0.003 and L = 1. The train dynamics are given by \(G_t(s)\) = \(\frac{k_b f K_p}{M\left(1+k_e\right) s(\tau s+1)}\) where M = 8000 kg, the vehicle mass; \(k_e\) = 0.1 the inertial coefficient; \(k_b\) = 142.5, the brake gain; \(K_p\) = 47.5, the pressure gain; \(\tau\) = 1.2 sec, a time constant; and f = 0.24, the normal friction coefficient. a. Make a root locus plot of the system as a function of the driver parameter K. b. Discuss why this model may not be an accurate description of a real driver-train situation.

Voltage droop control is a technique in which loads are driven at lower voltages than those provided by the source. In general, the voltage is decreased as current demand increases in the load. The advantage of voltage droop is that it results in lower sensitivity to load current variations. Voltage droop can be applied to the power distribution of several generators and loads linked through a dc bus. In (Karlsson, 2003) generators and loads are driven by 3-phase ac power, so they are interfaced to the bus through ac/dc converters. Since each one of the loads works independently, a feedback system shown in Figure P8.14 is used in each to respond equally to bus voltage variations. Given that \(C_s\) = \(C_r\) = 8,000 \(\mu \mathrm{F}\), \(L_{\text {cable }}\) = 50 \(\mu \mathrm{H}\), \(R_{\text {cable }}\) = 0.06 \(\Omega\), \(Z_r\) = \(R_r\) = 5 \(\Omega\), \(\omega_{l p}\) = 200 rad/s, \(G_{\text {conv }}(s)\) = 1, \(V_{d c-r e f}\) = 750 V, and \(P_{\text {ref-ext }}\) = 0, do the following: a. If \(Z_{r_{e q}}\) is the parallel combination of \(R_r\) and \(C_r\), and \(G_{\text {conv }}(s)\) = 1, find G(s) = \(\frac{V_s(s)}{I_s(s)}=\frac{V_s(s)}{I_{s-r e f}(s)}\) b. Write a MATLAB M-file to plot MATLAB and copy the full root locus ML for that system, then zoom-in the locus by setting the x-axis (realaxis) limits to -150 to 0 and the y-axis (imaginary-axis) limits to -150 to 150. Copy that plot, too, and find and record the following: (1) The gain, K, at which the system would have complex-conjugate closed-loop dominant poles with a damping ratio \(\zeta\) = 0.707 (2) The coordinates of the corresponding point selected on the root-locus (3) The values of all closed-loop poles at that gain (4) The output voltage \(v_s\)(t) for a step input voltage \(v_{dc-ref}\) (t)= 750 u(t) volts c. Plot that step response and use the MATLAB Characteristics tool (in the graph window) to note on the curve the following parameters: (1) The actual percent overshoot and the corresponding peak time, \(T_p\) (2) The rise time, \(T_r\), and the settling time, \(T_s\) (3) The final steady-state value in volts

It has been suggested that the use of closed-loop feedback in ventilators can highly reduce mortality and health problems in patients in need of respiratory treatments (Hahn, 2012). A good knowledge of the transfer functions involved is necessary for the design of an appropriate controller. In a study with 18 patients it was found that the open-loop transfer function from minute ventilation (MV) to end-tidal carbon dioxide partial pressure \(\left(\mathrm{P}_{\mathrm{ET}} \mathrm{CO}_2\right)\) can be nominally modeled as: G(s) = \(\frac{0.415 k_c(s+0.092)(s+0.25)}{s(s+0.007)(s+0.207)}\) a. Make a sketch of the root locus of the system indicating the breakaway points and the value \(k_c\) takes in each of them. b. In the design of ventilators it is very important to have negligible overshoot with the fastest possible settling time. It has been suggested that a value of \(k_c\) = 5.35 will achieve these specifications. Mark the position of the closed-loop poles for this value of \(k_c\) and explain why this is a reasonable gain choice.

Figure P8.15 shows a simplified drawing of a feedback system that includes the drive system G(s) = \(\frac{25\left(s^2+1.2 s+12500\right)}{s\left(s^2+5.6 s+62000\right)}\) presented in Problem 67, Chapter 5 (Thomsen, 2011). Referring to Figures P5.43 and P8.15, \(G_p(s)\) in Figure P8.15 is given by: \(G_p(s)\) = \(K_M \frac{G(s)}{1+0.1 G(s)}\) Given that the controller is proportional, that is, \(G_C(s)\) = \(K_P\), use MATLAB and a procedure similar to that developed in Problem 40 in this chapter plot the root locus \({ }^4\) and obtain the output response, c(t) = \(\omega_L(t)\), when a step input, r(t) = \(\omega_r(t)\) = 260 u(t) rad/ sec, is applied at t = 0. Mark on the time response graph, c(t), all relevant characteristics, such as the percent over- shoot (which should not exceed 168 ), peak time, rise time, settling time, and final steady-state value.

A simplified block diagram of a human pupil servomechanism is shown in Figure P8.16. The term \(e^{-0.18 s}\) represents a time delay. This function can be approximated by what is known as a Padé approximation. This approximation can take on many increasingly complicated forms, depending upon the degree of accuracy required. If we use the Padé approximation \(e^{-x}\) = \(\frac{1}{1+x+\frac{x^2}{2 !}}\) then \(e^{-0.18 s}\) = \(\frac{61.73}{s^2+11.11 s+61.73}\) Since the retinal light flux is a function of the opening of the iris, oscillations in the amount of retinal light flux imply oscillations of the iris (Guy, 1976). Find the following: a. The value of K that will yield oscillations b. The frequency of these oscillations c. The settling time for the iris if K is such that the eye is operating with 20% overshoot

A hard disk drive (HDD) arm has an open-loop unstable transfer function, P(s) = \(\frac{X(s)}{F(s)}=\frac{1}{I_b s^2}\) where X(s) is arm displacement and F(s) is the applied force (Yan, 2003). Assume the arm has an inertia of \(I_b\) = \(3 \times 10^{-5} \mathrm{~kg}-\mathrm{m}^2\) and that a lead controller, \(G_c(s)\) (used to improve transient response and discussed in Chapter 9), is placed in cascade to yield \(P(s) G_c(s)\) = G(s) = \(\frac{K}{I_b} \frac{(s+1)}{s^2(s+10)}\) as in Figure P8.3. a. Plot the root locus of the system as a function of K. b. Find the value of K that will result in dominant complex conjugate poles with a \(\zeta\) = 0.7 damping factor.

A robotic manipulator together with a cascade PI controller (used to improve steady-state response and discussed in Chapter 9) has a transfer function (Low, 2005) G(s) = \(\left(K_p+\frac{K_1}{s}\right) \frac{48,500}{s^2+2.89 s}\) Assume the robot's joint will be controlled in the configuration shown in Figure P8.3. a. Find the value of \(K_I\) that will result in \(e_{s s}\) = 2% for a parabolic input. b. Using the value of \(K_I\) found in Part a, plot the root locus of the system as a function of \(K_P\). c. Find the value of \(K_P\) that will result in a real pole at -1. Find the location of the other two poles.

Wind turbines, such as the one shown in Figure P8.17 (a), are becoming popular as a way of generating electricity. Feedback control loops are designed to control the output power of the turbine, given an input power demand. Blade-pitch control may be used as part of the control loop for a constant-speed, pitch-controlled wind turbine, as shown in Figure P8.17(b). The drivetrain, consisting of the windmill rotor, gearbox, and electric generator (see Figure P8.17(c)), is part of the control loop. The torque created by the wind drives the rotor. The windmill rotor is connected to the generator through a gearbox. The transfer function of the drivetrain is \(\begin{aligned} & \frac{P_o(s)}{T_R(s)}=G_{d t}(s) \\ & =\frac{3.92 K_{L S S} K_{H S S} K_G N^2 s}{\left\{N ^ { 2 } K _ { H S S } ( J _ { R } S ^ { 2 } + K _ { L S S } ) \left(J_G s^2\left[\tau_{e l} s+1\right]\right.\right.} \\ & \left.+K_G s\right)+J_R s^2 K_{L S S}\left[\left(J_G s^2+K_{H S S}\right)\right. \\ & \left.\left.\left(\tau_{e l} s+1\right)+K_G s\right]\right\} \\ & \end{aligned}\) where \(P_o(s)\) is the Laplace transform of the output power from the generator and \(T_R(s)\) is the Laplace transform of the input torque on the rotor. Substituting typical numerical values into the transfer function yields \(\begin{aligned} \frac{P_o(s)}{T_R(s)}= & G_{d t}(s) \\ = & \frac{(3.92)\left(12.6 \times 10^6\right)\left(301 \times 10^3\right)(688) N^2 s}{\left\{N^2\left(301 \times 10^3\right)\left(190,120 s^2+12.6 \times 10^6\right)\right.} \\ & \times\left(3.8 s^2\left[20 \times 10^{-3} s+1\right]+668 s\right) \\ & +190,120 s^2\left(12.6 \times 10^6\right) \\ & \times\left[\left(3.8 s^2+301 \times 10^3\right)\right. \\ & \left.\left.\times\left(20 \times 10^{-3} s+1\right)+668 s\right]\right\} \end{aligned}\) (Anderson, 1998). Do the following for the drivetrain dynamics, making use of any computational aids at your disposal: a. Sketch a root locus that shows the pole locations of \(G_{dt}\)(s) for different values of gear ratio, N. b. Find the value of N that yields a pair of complex poles of \(G_{dt}\)(s) with a damping ratio of 0.5.

An active system for the elimination of floor vibrations due to human presence is presented in (Nyawako, 2009). The system consists of a sensor that measures the floor’s vertical acceleration and an actuator that changes the floor characteristics. The open-loop transmission of the particular setup used can be described by G(s) = \(KG_a\)(s)F(s)\(G_m\)(s), where the actuator’s transfer function is \(G_a(s)\) = \(\frac{10.26}{s^2+11.31 s+127.9}\) The floor's dynamic characteristics can be modeled by F(s) = \(\frac{6.667 \times 10^{-5} s^2}{s^2+0.2287 s+817.3}\) The sensor's transfer function is \(G_m(s)\) = \(\frac{s}{s^2+5.181 s+22.18}\) and K is the gain of the controller. The system operations can be described by the unity-gain feedback loop of Figure P8.3. a. Use MATLAB's SISO Design Tool to obtain the root locus of the system in terms of K. b. Find the range of K for closed-1oop stability. c. Find, if possible, a value of K that will yield a closed-loop overdamped response.

Many implantable medical devices such as pacemakers, retinal implants, deep brain stimulators, and spinal cord stimulators are powered by an in-body battery that can be charged through a transcutaneous inductive device. Optimal battery charge can be obtained when the out-of-body charging circuit is in resonance with the implanted charging circuit (Baker, 2007). Under certain conditions, the coupling of both resonant circuits can be modeled by the feedback system in Figure P8.3 where G(s) = \(\frac{K s^4}{\left(s^2+2 \zeta \omega_n s+\omega_n^2\right)^2}\) The gain K is related to the magnetic coupling between the external and in-body circuits. K may vary due to positioning, skin conditions, and other variations. For this problem let \(\zeta\) = 0.5 and \(\omega_n\) = 1. a. Find the range of K for closed-loop stability. b. Draw the corresponding root locus.

It is important to precisely control the amount of organic fertilizer applied to a specific crop area in order to provide specific nutrient quantities and to avoid unnecessary environmental pollution. A precise delivery liquid manure machine has been developed for this purpose (Saeys, 2008). The system consists of a pressurized tank, a valve, and a rheological flow sensor. After simplification, the system can be modeled as a closed-loop negative feedback system with a forward-path transfer function G(s) = \(\frac{2057.38 \mathrm{~K}\left(s^2-120 s+4800\right)}{s(s+13.17)\left(s^2+120 s+4800\right)}\) consisting of an electrohydraulic system in cascade with the gain of the manure flow valve and a variable gain,K. The feedback path is comprised of H(s) = \(\frac{10\left(s^2-4 s+5.333\right)}{(s+10)\left(s^2+4 s+5.333\right)}\) a. Use the SISO Design Tool in MATLAB to obtain the root locus of the system. b. Use the SISO Design Tool to find the range of K for closed-loop stability. c. Find the value of K that will result in the smallest settling time for this system. d. Calculate the expected settling time for a step input with the value of $K$ obtained in Part c. e. Check your result through a step response simulation.

Harmonic drives are very popular for use in robotic ML manipulators due to their low backlash, high torque transmission, and compact size (Spong, 2006). The problem of joint flexibility is sometimes a limiting factor in achieving good performance. Consider that the idealized model representing joint flexibility is shown in Figure P8.18. The input to the drive is from an actuator and is applied at \(\theta_m\). The output is connected to a load at \(\theta_l\). The spring represents the joint flexibility and \(B_m\) and \(B_l\) represent the viscous damping of the actuator and load, respectively. Now we insert the device into the feedback loop shown in Figure P8.19. The first block in the forward path is a PD controller, which we will study in the next chapter. The PD controller is used to improve transient response performance. Use MATLAB to find the gain \(K_D\) to yield an approximate 58 overshoot in the step response given the following parameters: \(J_1\) = 10 ; \(B_I\) = 1 ; k = 100 ; \(J_m\) = 2 ; \(B_m\) = 0.5 ; \(\frac{K_P}{K_D}\) = 0.25 ; \(p_1(s)\) = \(J_1 s^2+B_1 s+k\); and \(p_m(s)\) = \(J_m s^2\) + \(B_m s\) + k

Using LabVIEW, the Control Design Labview and Simulation Module, and the LV MathScript RT Module, open and customize the Interactive Root Locus VI from the Examples to implement the system of Problem 64. Select the parameter \(K_D\) to meet the requirement of Problem 64 by varying the location of the closed-loop poles on the root locus. Be sure your front panel shows the following: (1) open-loop transfer function, (2) closed-loop transfer function, (3) root locus, (4) list of closed-loop poles, and (5) step response.

An automatic regulator is used to control the field current of a three-phase synchronous machine with identical symmetrical armature windings (Stapleton, 1964). The purpose of the regulator is to maintain the system voltage constant within certain limits. The transfer function of the synchronous machine is \(G_{s m}(s)\) = \(\frac{\Delta \delta(s)}{\Delta P_{m I}(s)}\) = \(\frac{M\left(s-z_1\right)\left(s-z_2\right)}{\left(s-p_1\right)\left(s-p_2\right)\left(s-p_3\right)}\) which relates the variation of rotor angle, \(\Delta \delta(s)\), to the change in the synchronous machine's shaft power, \(\Delta P_m(s)\). The closed-loop system is shown in Figure P8.3, where G(s) = \(K G_C(s) G_{s m}(s)\) and K is a gain to be adjusted. The regulator's transfer function, \(G_C(s)\), is given by \(G_C(s)\) = \(\frac{\mu / T_e}{s+\frac{1}{T_e}}\) Assume the following parameter values: \(\mu\) = 4, M = 0.117, \(T_e\) = 0.5, \(z_{1,2}\) = \(-0.071 \pm j 6.25\), \(p_1\) = -0.047, and \(p_{2,3}\) = \(-0.262 \pm j 5.1\), and do the following: Write a MATLAB M-file to plot the root locus for the system and to find the following: a. The gain K at which the system becomes marginally stable b. The closed-loop poles, p, and transfer function, T(S), corresponding to a 16% overshoot c. The coordinates of the point selected on the root-locus corresponding to 16% overshoot d. A simulation of the unit-step response of the closed-loop system corresponding to your 16% overshoot design. Note in your simulation the following values: (1) actual percent overshoot, (2) corresponding peak time, \(T_p\), (3) rise time, \(T_r\), (4) settling time, \(T_s\), and (5) final steady-state value.

It is well known that when a person ingests a significant amount of water, the blood volume increases, causing an increase in arterial blood pressure until the kidneys are able to excrete the excess volume and the pressurè returns to normal (Shahin, 2010). In order to describe mathematically this process, water-loading experiments are performed in various subjects while their mean arterial pressure is monitored. It was found that the open-loop transfer function of this process is G(s) = \(\frac{b_p\left(1.759 s^3+2.318 s^2+2.173 \times 10^{-4}\right)}{3.362 s^3+11.34 s^2+7.803 s+0.00293}\) where \(b_p\) is an autonomous nervous activity parameter. a. Make a sketch of the root locus of the system, indicating the breakaway points and the value of \(b_p\) for each point. b. Indicate the range of \(b_p\) for which the system is overdamped. c. Indicate the values of \(b_p\) for which the system is critically damped. d. Indicate the range of \(b_p\) for which the system is underdamped. e. Explain why the system will have a larger settling time for larger values of \(b_p\)

One of the treatments for Parkinson's disease in some patients is Deep Brain Stimulation (DBS) (Davidson, 2012). In DBS a set of electrodes is surgically implanted and a vibrating current is applied to the subthalamic nucleus, also known as a brain pacemaker. Root locus has been used on a linearized model of the system to help explain the dynamics of DBS. The DBS model can be obtained by substituting G(s) = \(\frac{k s}{(s-b)^2}(b>0)\) in the unity-feedback diagram of Figure P8.3. a. Make a sketch of the resulting root locus as a function of k and find the break-in point and its corresponding value of gain. b. Find the range of k for closed-loop stability in terms of b. c. Find the frequency of oscillation when the system has closed-loop poles on the \(j \omega\) axis.

A linear dynamic model of the \(\alpha\)-subsystem of a grid connected voltage-source converter (VSC) using a Y-Y transformer is shown in Figure P8.20(a) (Mahmood, 2012). Here, C = \(135 \mu \mathrm{F}\), \(R_1\) = 0.016 \(\Omega\), \(L_1\) = 0.14 mH, \(R_2\) = 0.014 \(\Omega\), \(L_2\) = 10 \(\mu \mathrm{H}\), \(R_g\) = 1.1 \(\Omega\), and \(L_g\) = 0.5 mH. a. Find the transfer function \(G_P(s)\) = \(\frac{V_a(s)}{M_a(s)}\). b. If \(G_P(s)\) is the plant in MATLAB Figure P8.20 (b) and \(G_C(s)\) = K, use MATLAB to plot the root locus. On a closeup of the locus (from -300 to 0 on the real axis and from -50 to 5000 on the imaginary axis), find K and the coordinates of the dominant poles, which correspond to \(\zeta\) = 0.012. Plot the output response, c(t) = \(v_a(t)\), at that value of the gain when a step input, r(t) = \(v_r(t)\) = 208 u(t) volts, is applied at t = 0. Mark on the time response graph, c(t), all relevant characteristics, such as the percent overshoot, peak time, rise time, settling time, and final steady-state value.

Control of HIV/AIDS. In the linearized model of Chapter 6 , Problem 68 , where virus levels are controlled by means of RTIs, the open-loop plant transfer function was shown to be P(s) = \(\frac{Y(s)}{U_1(s)}\) = \(\frac{-520 s-10.3844}{s^3+2.6817 s^2+0.11 s+0.0126}\) The amount of RTIs delivered to the patient will automatically be calculated by embedding the patient in the control loop as G(s) shown in Figure P6.17 (Craig, 2004). a. In the simplest case, G(s) = K, with K > 0. Note that this effectively creates a positive-feedback loop because the negative sign in the numerator of P(s) cancels out with the negative-feedback sign in the summing junction. Use positive-feedback rules to plot the root locus of the system. b. Now assume G(s) = -K with K > 0. The system is now a negative-feedback system. Use negative-feedback rules to draw the root locus. Show that in this case the system will be closed-loop stable for all K > 0.

Hybrid vehicle. In Chapter 7, Figure P7.31 shows the block diagram of the speed control of an HEV rearranged as a unity feedback system (Preit1, 2007). Let the transfer function of the speed controller be \(G_{S C}(s)\) = \(K_{P_{s e}}+\frac{K_{I_{s e}}}{s}=\frac{K_{P_{s c}}\left(s+\frac{K_{P_{s s}}}{K_{P_{s c}}}\right)}{s}\) a. Assume first that the speed controller is configured as a proportional controller \((K_{I_{s C}}\) = 0 and \(G_{S C}(s)\) = \(K_{P_{S C}}\). Calculate the forward path open-loop poles. Now use MATLAB to plot the system's root locus and find the gain, \(K_{P_{s c}}\) that yields a critically damped closed-loop response. Finally, plot the time-domain response, c(t), for a unit-step input using MATLAB. Note on the curve the rise time, \(T_r\), and settling time, \(T_s\). b. Now add an integral gain, \(K_{I_{S C}}\), to the controller, such that \(K_{I_{S C}} / K_{P_{S C}}\) = 0.4. Use MATLAB to plot the root locus and find the proportional gain, \(K_{P_{s c}}\), that could lead to a closed loop unit-step response with 108 overshoot. Plot c(t) using MATLAB and note on the curve the peak time, \(T_p\), and settling time, \(T_S\). Does the response obtained resemble a second-order underdamped response?

Parabolic trough collector. Consider the fluid temperature control of a parabolic trough collector (Camocho, 2012) embedded in the unity feedback structure as shown in Figure P8.3, where the open-loop plant transfer function is given by G(s) = \(\frac{137.2 \times 10^{-6} K}{s^2+0.0224 s+196 \times 10^{-6}} e^{-39 s}\) Approximating the time-delay term with \(e^{-s T} \approx \frac{1-\frac{T}{2} s}{1+\frac{T}{2} s}\), make a sketch of the resulting root locus (Note: After substituting the approximation, \(G(\infty)\) < 0, the positive feedback rules of Section 8.9 must be used). Mark where appropriate in the plot and find: a. The asymptotes and their intersection with the real axis; b. The break-in and breakaway points. (The procedures presented in Section 8.5 are also valid for positive feedback systems); c. The range of K for closed-loop stability; d. The value of K that will make the system oscillate and the oscillation frequency.