Given the system shown in Figure P7.24, find the

For the unity feedback system shown in Figure P7.1, where \(G(s)=\frac{450(s+8)(s+12)(s+15)}{s(s+38)\left(s^2+2 s+28\right)}\) find the steady-state errors for the following test inputs: \(25 u(t), 37 t u(t), 47 t^2 u(t)\). [Section: 7.2]

Figure P7.2 shows the ramp input r(t) and the output c(t) of a system. Assuming the output’s steady state can be approximated by a ramp, find [Section: 7.1] a. the steady-state error; b. the steady-state error if the input becomes r(t) = tu(t).

For the unity feedback system shown in Figure P7.1, where \(G(s)=\frac{60(s+3)(s+4)(s+8)}{s^2(s+6)(s+17)}\) find the steady-state error if the input is \(80 t^2 u(t)\). [Section: 7.2]

For the system shown in Figure P7.3, what steady-state error can be expected for the following test inputs: \(10 u(t), 10 t u(t), 10 t^2 u(t)\). [Section: 7.2]

For the unity feedback system shown in Figure P7.1, where \(G(s)=\frac{500}{(s+28)\left(s^2+8 s+12\right)}\) find the steady-state error for inputs of 20 u(t), 60 tu(t), and \(81 t^2 u(t)\). [Section: 7.3]

An input of \(25 t^3 u(t)\) is applied to the input of a Type 3 unity feedback system, as shown in Figure P7.1, where \(G(s)=\frac{210(s+4)(s+6)(s+11)(s+13)}{s^3(s+7)(s+14)(s+19)}\) Find the steady-state error in position. [Section: 7.3]

The steady-state error in velocity of a system is defined to be \(\left.\left(\frac{d r}{d t}-\frac{d c}{d t}\right)\right|_{t \rightarrow \infty}\) where r is the system input, and c is the system output. Find the steady-state error in velocity for an input of \(t^3 u(t)\) to a unity feedback system with a forward transfer function of [Section: 7.2] \(G(s)=\frac{100(s+1)(s+2)}{s^2(s+3)(s+10)}\)

What is the steady-state error for a step input of 15 units applied to the unity feedback system of Figure P7.1, where [Section: 7.3] \(G(s)=\frac{1020(s+13)(s+26)(s+33)}{(s+65)(s+75)(s+91)}\)

A system has \(K_p=4\). What steady-state error can be expected for inputs of 70 u(t) and 70tu(t)? [Section 7.3]

For the unity feedback system shown in Figure P7.1, where [Section: 7.3] \(G(s)=\frac{5000}{s(s+75)}\) (a) What is the expected percent overshoot for a unit step input? (b) What is the settling time for a unit step input? (c) What is the steady-state error for an input of 5 u(t) ? (d) What is the steady-state error for an input of 5 tu(t) ? (e) What is the steady-state error for an input of \(5 t^2 u(t)\) ?

Given the unity feedback system shown in Figure P7.1, where \(G(s)=\frac{500000(s+7)(s+20)(s+45)}{s(s+30)(s+\alpha)(s+50)}\) find the value of \(\alpha\) to yield a \(K_v=35000\). [Section: 7.4]

For the unity feedback system of Figure P7.1, where \(G(s)=\frac{K(s+2)(s+4)(s+6)}{s^2(s+5)(s+7)}\) find the value of K to yield a static error constant of 10,000. Section: 7.4]

For the system shown in Figure P7.4, [Section: 7.3] (a) Find \(K_p, K_v\), and \(K_a\). (b) Find the steady-state error for an input of 50 u(t), 50 tu(t), and \(50 t^2 u(t)\). (c) State the system type.

A Type 3 unity feedback system has \(r(t)=10 t^3\) applied to its input. Find the steady-state position error for this input if the forward transfer function is [Section: 7.3] \(G(s)=\frac{1030\left(s^2+8 s+23\right)\left(s^2+21 s+18\right)}{s^3(s+6)(s+13)}\)

Find the system type for the system of Figure P7.5. [Section: 7.3]

What are the restrictions on the feedforward transfer function \(G_2(s)\) in the system of Figure P7.6 to obtain zero steady-state error for step inputs if: [Section: 7.3] (a) \(G_1(s)\) is a Type 0 transfer function; (b) \(G_1(s)\) is a Type 1 transfer function; (c) \(G_1(s)\) is a Type 2 transfer function?

The steady-state error is defined to be the difference in position between input and output as time approaches infinity. Let us define a steady-state velocity error, which is the difference in velocity between input and output. Derive an expression for the error in velocity, \(\dot{e}(\infty)=\dot{r}(\infty)-\dot{c}(\infty)\), and complete Table P7.1 for the error in velocity. [Sections: 7.2, 7.3]

For the system shown in Figure P7.7, [Section: 7.4] (a) What value of K will yield a steady-state error in position of 0.01 for an input of (1/10) t? (b) What is the \(K_v\) for the value of K found in Part a? (c) What is the minimum possible steady-state position error for the input given in Part a?

Given the unity feedback system of Figure P7.1, where \(G(s)=\frac{K(s+a)}{s(s+2)(s+15)}\) find the value of Ka so that a ramp input of slope 30 will yield an error of 0.005 in the steady state when compared to the output. [Section: 7.4]

Given the system of Figure P7.8, design the value of K so that for an input of 100tu(t), there will be a 0.01 error in the steady state. [Section: 7.4]

Find the value of K for the unity feedback system shown in Figure P7.1, where \(G(s)=\frac{K(s+3)}{s^2(s+7)}\) if the input is \(10 t^2 u(t)\), and the desired steady-state error is 0.061 for this input. [Section: 7.4]

The unity feedback system of Figure P7.1, where \(G(s)=\frac{K\left(s^2+3 s+30\right)}{s^n(s+5)}\) is to have 1/6000 error between an input of 10 tu(t) and the output in the steady state. [Section: 7.4] (a) Find K and n to meet the specification. (b) What are \(K_p, K_v\), and \(K_a\) ?

For the unity feedback system of Figure P7.1, where [Section: 7.3] G(s) = \(\frac{K\left(s^2+6 s+6\right)}{(s+5)^2(s+3)}\) a. Find the system type. b. What error can be expected for an input of 12u(t)? c. What error can be expected for an input of 12tu(t)?

For the unity feedback system of Figure P7.1, where G(s) = \(\frac{K(s+13)(s+19)}{s(s+6)(s+9)(s+22)}\) find the value of K to yield a steady-state error of 0.4 for a ramp input of 27tu(t). [Section: 7.4]

Given the unity feedback system of Figure P7.1, where G(s) = \(\frac{K(s+6)}{(s+2)\left(s^2+10 s+29\right)}\) find the value of K to yield a steady-state error of 8%. [Section: 7.4]

For the unity feedback system of Figure P7.1, where G(s) = \(\frac{K}{s(s+4)(s+8)(s+10)}\) find the minimum possible steady-state position error if a unit ramp is applied. What places the constraint upon the error?

The unity feedback system of Figure P7.1, where G(s) = \(\frac{K(s+\alpha)}{(s+\beta)^2}\) is to be designed to meet the following specifications: steady-state error for a unit step input = 0.1; damping ratio = 0.5; natural frequency = \(\sqrt{10}\). Find K, \(\alpha\), and p. [Section: 7.4]

A second-order, unity feedback system is to follow a ramp input with the following specifications: the steady-state output position shall differ from the input position by 0.01 of the input velocity; the natural frequency of the closed-loop system shall be 10 rad/s. Find the following: a. The system type b. The exact expression for the forward-path transfer function c. The closed-loop system's damping ratio

The unity feedback system of Figure P7.1 has a transfer function G(s) = \(\frac{C(s)}{E(s)}\) = \(\frac{K}{s(s+\alpha)}\) and is to follow a ramp input, r(t) = tu(t), so that the steady-state output position differs from the input position by 0.01 of the input velocity (e.g., \(e(\infty)\) = \(\frac{1}{K_v}\) = 0.01). The natural frequency of the closed-loop system will be \(\omega_n\) = 5 rad/s. [Section: 7.4] Find the following: a. The system type b. The values of K and \(\alpha\) c. The closed-loop system's damping ratio, \(\zeta\) d. If K is reduced to 4 and \(\alpha\) = 0.4, find the corresponding new values of \(e(\infty)\), \(\omega_n\), and \(\zeta\).

The unity feedback system of Figure P7.1, where G(s) = \(\frac{K(s+\alpha)}{s(s+\beta)}\) is to be designed to meet the following requirements: The steady-state position error for a unit ramp input equals 1/10; the closed-loop poles will be located at -1 \(\pm\) j 1. Find K, \(\alpha\), and \(\beta\) in order to meet the specifications. [Section: 7.4]

Given the unity feedback control system of Figure P7.1, where G(s) = \(\frac{K}{s^n(s+a)}\) find the values of n, K, and a in order to meet specifications of 12% overshoot and \(K_v\) = 110. [Section: 7.4]

Given the unity feedback control system of Figure P7.1, where G(s) = \(\frac{K}{s(s+a)}\) find the following: [Section: 7.4] a. K and a to yield \(K_v\) = 1000 and a 20% overshoot b. K and a to yield a 1% error in the steady state and a 10% overshoot.

Given the system in Figure P7.9, find the following: [Section: 7.3] a. The closed-loop transfer function b. The system type c. The steady-state error for an input of 5u(t) d. The steady-state error for an input of 5tu(t) e. Discuss the validity of your answers to Parts c and d.

Repeat Problem 33 for the system shown in Figure P7.10. [Section: 7.3]

For the system shown in Figure P7.11, use MATLAB to find the following: [Section: 7.3] a. The system type b. \(K_p\), \(K_v\), and \(K_a\) c. The steady-state error for inputs of 100u(t), 100tu(t), and 100\(t^2\) u(t)

The system of Figure P7.12 is to have the following specifications: \(K_v\) = 20 ; \(\zeta\) = 0.7. Find the values of \(K_1\) and \(K_f\) required for the specifications of the system to be met. [Section: 7.4]

The transfer function from elevator deflection to altitude change in a Tower Trainer 60 Unmanned Aerial Vehicle is P(s) = \(\frac{h(s)}{\delta(s)_e}\) = \(\frac{-34.16 s^3-144.4 s^2+7047 s+557.2}{s^5+13.18 s^4+95.93 s^3+14.61 s^2+31.94 s}\) An autopilot is built around the aircraft as shown in Figure P7.13, with F(s) = H(s) = 1 and G(s) = \(\frac{0.00842(s+7.895)\left(s^2+0.108 s+0.3393\right)}{(s+0.07895)\left(s^2+4 s+8\right)}\) (Barkana, 2005). The steady-state error for a ramp input in this system is \(e_{\mathrm{ss}}\) = 25. Find the slope of the ramp input.

Find the total steady-state error due to a unit step input and a unit step disturbance in the system of Figure P7.14. [Section: 7.5]

Design the values of \(K_1\) and \(K_2\) in the system of Figure P7.15 to meet the following specifications: Steady-state error component due to a unit step disturbance is -0.00001; steady-state error component due to a unit ramp input is 0.002. [Section: 7.5]

In Figure P7.16, let G(s) = 5 and P(s) = \(\frac{7}{s+2}\). a. Calculate the steady-state error due to a command input R(s) = \(\frac{3}{s}\) with D(s) = 0. b. Verify the result of Part a using Simulink. c. Calculate the steady-state error due to a disturbance input D(s) = \(-\frac{1}{s}\) with R(s) = 0. d. Verify the result of Part c Simulink using Simulink. e. Calculate the total steady-state error due to a command input R(s) = \(\frac{3}{s}\) and a disturbance D(s) = \(-\frac{1}{s}\) applied simultaneously.

Derive Eq. (7.72) in the text, which is the final value of the actuating signal for non unity feedback systems. [Section: 7.6]

For each system shown in Figure P7.17, find the following: [Section: 7.6] a. The system type b. The appropriate static error constant c. The input waveform to yield a constant error d. The steady-state error for a unit input of the waveform found in Part c e. The steady-state value of the actuating signal.

For each system shown in Figure P7.18, find the appropriate static error constant as well as the steady-state error, r(\(\infty\))-c(\(\infty\)), for unit step, ramp, and parabolic inputs. [Section: 7.6]

Given the system shown in Figure P7.19, find the following: [Section: 7.6] a. The system type b. The value of K to yield 0.1% error in the steady state.

For the system shown in Figure P7.20, [Section: 7.6] a. What is the system type? b. What is the appropriate static error constant? c. What is the value of the appropriate static error constant? d. What is the steady-state error for a unit step input?

For the system shown in Figure P7.21, use MATLAB to find the following for K = 10, and K = \(10^6\): [Section: 7.6] a. The system type b. \(K_p\), \(K_v\), and \(K_a\) c. The steady-state error for inputs of 30u (t), 30tu (t), and 30\(t^2\)u (t)

A dynamic voltage restorer (DVR) is a device that is connected in series to a power supply. It continuously monitors the voltage delivered to the load, and compensates voltage sags by applying the necessary extra voltage to maintain the load voltage constant. In the model shown in Figure P7.22, \(u_r\) represents the desired reference voltage, \(u_o\) is the output voltage, and \(Z_L\) is the load impedance. All other parameters are internal to the DVR (Lam, 2004). a. Assuming \(Z_L\) = \(\frac{1}{s C_L}\), and \(\beta \neq 1\), find the system's type. b. Find the steady-state error to a unit step input as a function of \(\beta\).

Derive Eq. (7.69) in the text. [Section: 7.6]

Given the system shown in Figure P7.23, do the following: [Section: 7.6] a. Derive the expression for the error, E(s) = R(s) - C(s), in terms of R(s) and D(s). b. Derive the steady-state error, \(e(\infty)\), if R(s) and D(s) are unit step functions. c. Determine the attributes of \(G_1(s)\), \(G_2(s)\), and H(s) necessary for the steady-state error to become zero.

Given the system shown in Figure P7.24, find the sensitivity of the steady-state error to parameter a. Assume a step input. Plot the sensitivity as a function of parameter a. [Section: 7.7]

a. Show that the sensitivity to plant changes in the system of Figure P7.13 is \(S_{T: P}\) = \(\frac{P}{T} \frac{\delta T}{\delta P}=\frac{1}{1+L(s)}\) where L(s) = G(s) P(s) H(s) T(s) = \(\frac{C(s)}{R(s)}=\frac{F(s)}{H(s)} \cdot \frac{L(s)}{1+L(s)}\). b. Show that \(S_{T: P}(s)+\frac{T(s) H(s)}{F(s)}\) = 1 for all values of s.

In Figure P7.13, P(s) = \(\frac{5}{s}\), H(s) = 1 T(s) = \(\frac{C(s)}{R(s)}=\frac{200 K}{(s+1)(s+3)\left(s^2+2 s+20\right)}\) and \(S_{T: P}\) = \(\frac{P}{T} \frac{\delta T}{\delta P}=\frac{s^2+2 s}{s^2+2 s+20}\) a. Find F(s) and G(s). b. Find the value of K that will result in zero steady-state error for a unit step input.

For the system shown in Figure P7.25, find the sensitivity of the steady-state error for changes in \(K_1\) and in \(K_2\), when \(K_1\) = 100 and \(K_2\) = 0.1. Assume step inputs for both the input and the disturbance. [Section: 7.7]

Given the block diagram of the active suspension system shown in Figure P5.36, (Lin, 1997) a. Find the transfer function from a road disturbance r to the error signal e. b. Use the transfer function in Part a to find the steady state value of e for a unit step road disturbance. c. Use the transfer function in Part a to find the steady-state value of e for a unit ramp road disturbance. d. From your results in Parts b and c, what is the system’s type for e?

For each of the following closed-loop systems, find the steady-state error for unit step and unit ramp inputs. Use both the final value theorem and input substitution methods. [Section: 7.8] a. \(\dot{\mathbf{x}}\) = \(\left[\begin{array}{rrr}-5 & -4 & -2 \\ -3 & -10 & 0 \\ -1 & 1 & -5\end{array}\right] \mathbf{x}+\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right] r ; y=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right] \mathbf{x}\) b. \(\dot{\mathbf{x}}\) = \(\left[\begin{array}{rrr}0 & 1 & 0 \\ -5 & -9 & 7 \\ -1 & 0 & 0\end{array}\right] \mathbf{x}+\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right] r ; y=\left[\begin{array}{lll}1 & 0 & 0\end{array}\right] \mathbf{x}\) c. \(\dot{\mathbf{x}}\) = \(\left[\begin{array}{rrr}-9 & -5 & -1 \\ 1 & 0 & -2 \\ -3 & -2 & -5\end{array}\right] \mathbf{x}+\left[\begin{array}{l}2 \\ 3 \\ 5\end{array}\right] r ; y=\left[\begin{array}{lll}1 & -2 & 4\end{array}\right] \mathbf{x}\)

A simplified model of the steering of a four-wheel drive vehicle is shown in Figure P7.26. In this block diagram, the output r is the vehicle's yaw rate, while \(\delta_f\) and \(\delta_r\) are the steering angles of the front and rear tires, respectively. In this model, \(r^*(s)\) = \(\frac{\frac{s}{300}+0.8}{\frac{s}{10}+1}\), \(G_f(s)\) = \(\frac{h_1 s+h_2}{s^2+a_1 s+a_2}\) \(G_r(s)\) = \(\frac{h_3 s+b_1}{s^2+a_1 s+a_2}\) and K(s) is a controller to be designed. (Yin, 2007). a. Assuming a step input for \(\delta_f\), find the minimum system type of the controller K(s) necessary so that in steady-state the error, as defined by the signal e in Figure P7.26, is zero if at all possible. b. Assuming a step input for \(\delta_f\), find the system type of the controller K(s) necessary so that in steady state the error as defined by \(\delta_f(\infty)-r(\infty)\) is zero if at all possible.

Glycolysis is a feedback process through which living cells use glucose to generate adenosine triphosphate (ATP), necessary for cell operations. A linearized glycolysis model (Chandra, 2011) is given by \(\left[\begin{array}{c} \dot{\Delta} x \\ \dot{\Delta} y \end{array}\right]=\left[\begin{array}{cc} -k & a+g+h \\ (q+1) k & -q a-g(q+1)+q h \end{array}\right]\left[\begin{array}{l} \Delta x \\ \Delta y \end{array}\right]+\left[\begin{array}{l} 0 \\ 1 \end{array}\right] \delta\) where \(\delta\) is the perturbation (disturbance input) on ATP production, \(\Delta y\) is the change in ATP level (output). \(\alpha\) > 0 is the cooperativity of ATP binding to PFK, g > 0 is the feedback strength of ATP on PK (PFK and PK are two different types of glycolytic enzymes), k > 0 is the intermediate reaction rate, q > 0 is the autocatalytic stoichiometry, and h > 0 is the feedback strength of ATP on the PFK enzyme. a. Since in this system \(\delta\) is a disturbance input, zero steady-state error is achieved when \(\frac{\Delta y}{\delta}\) = 0. Show that in steady state \((\dot{\Delta} x\) = 0, \(\dot{\Delta} y\) = 0), \(\frac{\Delta y}{\delta}\) = \(\frac{1}{a-h}\). b. Use the Routh-Hurwitz stability criterion to show that the system will be closed-loop stable as long as \(0<h-a<\frac{k+g(q+1)}{q}\) c. Assuming that h is the only parameter of choice for steady-state error adjustments, show that zero steady-state error is not achievable.

As part of the development of a textile cross-lapper machine (??o, 2010), a torque input, u(t) = \(\left\{\begin{array}{rc}1 & 0 \leq t<50 \\ -1 & 50 \leq t<100\end{array}\right\}\), is applied to the motor of one of the movable racks embedded in a feedback loop. The corresponding velocity output response is shown in Figure P7.27. a. What is the open-loop system's type? b. What is the steady-state error? c. What would be the steady-state error for a ramp input?

The block diagram in Figure P7.28 represents a motor driven by an amplifier with double-nested tachometer feedback loops (Mitchell, 2010). a. Find the steady-state error of this system to a step input. b. What is the system type?

PID control, which is discussed in Chapter 9, may be recommended for Type 3 systems when the output in a feedback system is required to perfectly track a parabolic as well as step and ramp reference signals (Papadopoulos, 2013). In the system of Figure P7.29, the transfer functions of the plant, GP(s), and the recommended controller, GC(s), are given by: \(G_P(s)\) = \(\frac{127 e^{-0.2 s}}{s(s+1)(s+2)(s+5)^2(s+10)}\) \(G_C(s)\) = \(\frac{\left(92.9 s^2+13.63 s+1\right)}{97.6 s^2(0.1 s+1)}\) Use Simulink to model this system and plot its response (from 0 to 300 seconds) to a unit-step reference input, r(t), applied at t = 0, and (on the same graph) to a disturbance, d(t) = 0.25 r (t), applied at t = 150 seconds. What are the values of the steady-state error due to the reference input and due to the disturbance? What about the relative stability of this Type 3 system as evidenced by the percent overshoot in response to the unit-step reference input?

A Type 3 feedback control system (Papadopoulos, 2013) was presented in Problem 60. Modify the Simulink model you developed in that problem to plot its response (from 0 to 100 seconds) to a unit-ramp reference input, r(t)=tu(t), applied at t = 0, and (on the same graph) to a disturbance, d(t) = 0.25tu(t), applied at t = 50 seconds. What are the values of the steady-state position error due to the reference-input and disturbance ramps? Copy this model and paste it in the same file. Then, in that copy, change the reference input to a unit parabola, r(t) = 0.5 \(t^2\) u(t), applied at t = 0, and the disturbance to d(t) = 0.125\(t^2\) u(t), applied at t = 50 seconds, and plot, on a new graph (Scope 1), the system’s response to these parabolic signals.

Motion control, which includes position or force control, is used in robotics and machining. Force control requires the designer to consider two phases: contact and noncontact motions. Figure P7.30(a) is a diagram of a mechanical system for force control under contact motion. A force command, \(F_{cmd}(s), is the input to the system, while the output, F(s), is the controlled contact force. In the figure a motor is used as the force actuator. The force output from the actuator is applied to the object through a force sensor. A block diagram representation of the system is shown in Figure P7.30(b). \(K_2\) is velocity feedback used to improve the transient response. The loop is actually implemented by an electrical loop (not shown) that controls the armature current of the motor to yield the desired torque at the output. Recall that \(T_m\) = \(K_t i_a\) (Ohnishi, 1996). Find an expression for the range of \(K_2\) to keep the steady-state force error below 10% for ramp inputs of commanded force.

An open-loop swivel controller and plant for an industrial robot has the transfer function \(G_e(s)\) = \(\frac{\omega_o(s)}{V_i(s)}=\frac{K}{(s+10)\left(s^2+4 s+10\right)}\) where \(\omega_o(s)\) is the Laplace transform of the robot's angular swivel velocity and \(V_i(s)\) is the input voltage to the controller. Assume \(G_e(s)\) is the forward transfer function of a velocity control loop with an input transducer and sensor, each represented by a constant gain of 3 (Schneider, 1992). a. Find the value of gain, K, to minimize the steady state error between the input commanded angular swivel velocity and the output actual angular swivel velocity. b. What is the steady-state error for the value of K found in Part a? c. For what kind of input does the design in Part a apply?

Packet information flow in a router working under TCP/IP can be modeled using the linearized transfer function P(s) = \(\frac{Q(s)}{f(s)}=\frac{\frac{C^2}{2 N} e^{-s R}}{\left(s+\frac{2 N}{R^2 C}\right)\left(s+\frac{1}{R}\right)}\) where C = link capacity (packets/second) N = load factor (number of TCP sessions) Q = expected queue length R = round trip time (second) p = probability of a packet drop The objective of an active queue management (AQM) algorithm is to automatically choose a packet-drop probability, p, so that the queue length is maintained at a desired level. This system can be represented by the block diagram of Figure P7.13 with the plant model in the P(s) block, the AQM algorithm in the G(s) block, and F(s) = H(s) = 1. Several AQM algorithms are available, but one that has received special attention in the literature is the random early detection (RED) algorithm. This algorithm can be approximated with G(s) = \(\frac{L K}{s+K}\), where L and K are constants (Hollot, 2001). Find the value of L required to obtain a 10% steady-state error for a unit step input when C = 3750 packets/s, N = 50 TCP sessions, R = 0.1 s, and K = 0.005.

In Figure P7.16, the plant, P(s) = \(\frac{48,500}{s^2+2.89 s}\), represents the dynamics of a robotic manipulator joint. The system's output, C(s), is the joint's angular position (Low, 2005). The system is controlled in a closed loop configuration as shown with G(s) = \(K_P+\frac{K_I}{s}\), a proportional-plus-integral (PI) controller to be discussed in Chapter 9. R(s) is the joint's desired angular position. D(s) is an external disturbance, possibly caused by improper dynamics modeling, Coulomb friction, or other external forces acting on the joint. a. Find the system's type. b. Show that for a step disturbance input, \(e_{\mathrm{ss}}\) = 0 when \(K_I \neq 0\). c. Find the value of \(K_I\) that will result in \(e_{\mathrm{ss}}\) = 5% for a parabolic input. d. Using the value of \(K_I\) found in Part c, find the range of \(K_P\) for closed-loop stability.

Control of HIV/AIDS. Consider the HIV infection model of Problem 68 in Chapter 6 and its block diagram in Figure P6.17 (Craig, 2004). a. Find the system's type if G(s) is a constant. b. It was shown in Problem 68, Chapter 6, that when G(s) = K the system will be stable when \(K<2.04 \times 10^{-4}\). What value of K will result in a unit step input steady-state error of 10%? c. It is suggested that to reduce the steady-state error the system's type should be augmented by making G(s) = \(\frac{K}{s}\). Is this a wise choice? What is the resulting stability range for K?

Hybrid vehicle. Figure P7.31 shows the block diagram of the speed control of an HEV taken from Figure P5.53, and rearranged as a unity feedback system (Preitl, 2007). Here the system output is C(s) = \(K_{S S}\)V(s), the output voltage of the speed sensor/transducer. a. Assume the speed controller is given as \(G_{S C}(\mathrm{~s})\) = \(K_{P_{S C}}\). Find the gain, \(K_{P_{S C}}\), that yields a steady-state error, \(e_{\text {step }}(\infty)\) = 1%. b. Now assume that in order to reduce the steady-state error for step inputs, integration is added to the controller yielding \(G_{S C}(s)\) = \(K_{P_{S C}}+\left(K_{I_{S C}} / s\right)\) = \(\left.100+\left(K_{I_{s c}} / s\right)\right)\). Find the value of the integral gain, \(K_{I_{s c}}\), that results in a steady-state error, \(e_{\text {ramp }}(\infty)\) = 2.5%. c. In Parts a and b, the HEV was assumed to be driven on level ground. Consider the case when, after reaching a steady-state speed with a controller given by \(G_{SC}(s)\) = 100 + \(\frac{40}{s}\), the car starts climbing up a hill with a gradient angle, \(\alpha\) = \(5^{\circ}\). For small angles sin \(\alpha\) = \(\alpha\) (in radians) and, hence, when reflected to the motor shaft the climbing torque is \(T_{s t}\) = \(\frac{F_{s t} r}{i_{\text {tot }}}=\frac{m g r}{i_{\text {tot }}} \sin \alpha=\frac{m g r \alpha}{i_{\text {tot }}}\) = \(\frac{1590 \times 9.8 \times 0.3 \times 5}{4.875 \times 57.3}=83.7 \mathrm{Nm}\). The block diagram in Figure P7.32 represents the control system of the HEV rearranged for Part c. In this diagram, the input is \(T_{s t}(t)\) = 83.7 u(t), corresponding to \(\alpha\) = \(5^{\circ}\), and the output is the negative error, -e(t) = -c(t) = \(-K_{S S} v(t)\), proportional to the change in car speed, v(t). Find the steady-state error \(e(\infty)\) due to a step change in the disturbance; e.g., the climbing torque, \(T_{s t}(t)\) = 83.7 u(t).

Parabolic trough collector. The parabolic trough collector (Camacho, 2012) is embedded in a unit feedback configuration as shown in Figure P7.1, where G(s) = \(G_C(s) P(s)\) and P(s) = \(\frac{137.2 \times 10^{-6}}{s^2+0.0224 s+196 \times 10^{-6}} e^{-39 s}\) a. Assuming \(G_C(s)\) = K, find the value of K required for a unit-step input steady-state error of 3%. Use the result you obtained in Problem 70, Chapter 6, to verify that the system is closed-loop stable when that value of K is used. b. What is the minimum unit-step input steady-state error achievable with \(G_C(s)\) = K? c. What is the simplest compensator, \(G_C(s)\), that can be used to achieve a steady-state error of 0%?