Antenna Control: Designing a Closed-Loop Response This chapter has shown that physical subsystems can be modeled mathematically with transfer functions and then interconnected to form a feedback system. The interconnected mathematical models can be reduced to a single transfer function representing

Question

Antenna Control: Designing a Closed-Loop Response This chapter has shown that physical subsystems can be modeled mathematically with transfer functions and then interconnected to form a feedback system. The interconnected mathematical models can be reduced to a single transfer function representing the system from input to output. This transfer function, the closed-loop transfer function, is then used to determine the system response.
The following case study shows how to reduce the subsystems of the antenna azimuth position control system to a single, closed-loop transfer function in order to analyze and design the transient response characteristics.

PROBLEM: Given the antenna azimuth position control system shown on the front endpapers, Configuration 1, do the following:

a. Find the closed-loop transfer function using block diagram reduction.
b. Represent each subsystem with a signal-flow graph and find the state-space representation of the closed-loop system from the signal-flow graph.
c. Use the signal-flow graph found in b along with Mason’s rule to find the closed-loop transfer function.
d. Replace the power amplifier with a transfer function of unity and evaluate the closed-loop peak time, percent overshoot, and settling time for K = 1000.
e. For the system of d, derive the expression for the closed-loop step response of the system.
f. For the simplified model of d, find the value of K that yields a 10% overshoot.

Accepted Answer

Each subsystem’s transfer function was evaluated in the case study in Chapter 2. We first assemble them into the closed-loop, feedback control system block diagram shown in Figure 5.34(a).

a. The steps taken to reduce the block diagram to a single, closed-loop transfer function relating the output angular displacement to the input angular displacement are shown in Figure 5.34(a–d). In Figure 5.34(b), the input potentiometer was pushed to the right past the summing junction, creating a unity feedback system. In Figure 5.34(c), all the blocks of the forward transfer function are multiplied together, forming the equivalent forward transfer function. Finally, the feedback formula is applied, yielding the closed-loop transfer function in Figure 5.34(d).

b. In order to obtain the signal-flow graph of each subsystem, we use the state equations derived in the case study of Chapter 3. The signal-flow graph for the power amplifier is drawn from the state equations of Eqs. (3.87)

[latex] -3x_{1}+x_{2}=-2x_{1} [/latex]

[latex] x_{1}-3x_{2}=-2x_{2} [/latex]

and (3.88),

[latex] x=\left[\begin{matrix} c \ c \end{matrix} ight] [/latex]

and the signal-flow graph of the motor and load is drawn from the state equation of Eq. (3.98).

[latex]x=\left[\begin{matrix} x_{1} \ x_{2} \ e_{a} \end{matrix} ight] [/latex]

Other subsystems are
pure gains. The signal-flow graph for Figure 5.34(a) is shown in Figure 5.35 and consists of the interconnected subsystems.

The state equations are written from Figure 5.35. First define the state variables as the outputs of the integrators. Hence, the state vector is

[latex]x=\left[\begin{matrix} x_{1} \ x_{2} \ e_{a} \end{matrix} ight] [/latex]

Using Figure 5.35, we write the state equations by inspection:

[latex] \dot{x}_{1}= [/latex] [latex] +x_{2} [/latex]

[latex] \dot{x}_{2}= [/latex] [latex] -1.71x_{2}+2.083 e_{a} [/latex]

[latex] \dot{e}_{a}=-3.18Kx_{1} [/latex] [latex] -100 e_{a} +31.8K heta _{i} [/latex]

along with the output equation,

[latex] y = heta _{o}=0.1x_{1} [/latex]

where 1/π = 0.318.
In vector-matrix form,

[latex] \dot{x}=\left[\begin{matrix} 0 & 1 & 0 \ 0 & -1.71 & 2.083 \ -3.18K & 0 & -100 \end{matrix} ight] x+\left[\begin{matrix} 0 \ 0 \ 31.8K \end{matrix} ight] heta _{i} [/latex]

[latex] y=\left[\begin{matrix} 0.1 & 0 & 0\end{matrix} ight] x [/latex]

c. We now apply Mason’s rule to Figure 5.35 to derive the closed-loop transfer function of the antenna azimuth position control system. First find the forward-path gains. From Figure 5.35 there is only one forward-path gain:

[latex] T_{1}=\left(\frac{1}{\pi } ight)(K)(100)\left(\frac{1}{s} ight)(2.083)\left(\frac{1}{s} ight)\left(\frac{1}{s} ight)(0.1)=\frac{6.63K}{s^{3} } [/latex]

Next identify the closed-loop gains. There are three: the power amplifier loop, [latex] G_{L1}(s) [/latex], with [latex]e_{a}[/latex] at the output; the motor loop, [latex] G_{L2}(s) [/latex],with [latex]x_{2}[/latex] at the output; and the entire system loop,[latex] G_{L3}(s) [/latex],with [latex] heta _{0}[/latex] at the output.

[latex] G_{L1}(s)=\frac{-100}{s} [/latex]

[latex] G_{L2}(s)=\frac{-1.71}{s} [/latex]

[latex] G_{L3}(s)=(K)(100)\left(\frac{1}{s} ight)(2.083)\left(\frac{1}{s} ight)\left(\frac{1}{s} ight)(0.1)\left(\frac{-1}{\pi } ight) =\frac{-6.63K}{s^{3} } [/latex]

Only [latex] G_{L1}(s) [/latex] and[latex] G_{L2}(s) [/latex] are nontouching loops. Thus, the nontouching-loop gain is

[latex] G_{L1}(s)G_{L2}(s)=\frac{171}{s^{2}} [/latex]

Forming Δ and [latex] \Delta _{k}[/latex] in Eq. (5.28),

[latex]G(s)=\frac{C(s)}{R(s)} =\frac{\Sigma _{k}T_{k}\Delta _{k}}{\Delta } [/latex]

we have

[latex] \Delta =1-\left[G_{L1}(s)+G_{L2}(s)+G_{L3}(s) ight] +\left[G_{L1}(s)G_{L2}(s) ight] [/latex]

[latex] =1+\frac{100}{s}+\frac{1.71}{s}+\frac{6.63K}{s^{3} } +\frac{171}{s^{2} } [/latex]

and

[latex]\Delta _{1} =1[/latex]

Substituting Eqs. (5.102),

[latex] T_{1}=\left(\frac{1}{\pi } ight)(K)(100)\left(\frac{1}{s} ight)(2.083)\left(\frac{1}{s} ight)\left(\frac{1}{s} ight)(0.1)=\frac{6.63K}{s^{3} } [/latex]

(5.105),

[latex] \Lambda =1-\left[G_{L1}(s)+G_{L2}(s)+G_{L3}(s) ight] +\left[G_{L1}(s)G_{L2}(s) ight] [/latex]

[latex] =1+\frac{100}{s}+\frac{1.71}{s}+\frac{6.63K}{s^{3} } +\frac{171}{s^{2} } [/latex]

and (5.106) [latex]\Delta _{1} =1[/latex]

into Eq. (5.28),we obtain the closedloop transfer function as

[latex]T(s)=\frac{C(s)}{R(s)}=\frac{T_{1}\Delta _{1} }{\Delta }=\frac{6.63K}{s^{3}+101.71s^{2}+171s+6.63K } [/latex]

d. Replacing the power amplifier with unity gain and letting the preamplifier gain, K, in Figure 5.34(b) equal 1,000 yield a forward transfer function, G(s), of

[latex]G(s)=\frac{66.3}{s(s+1.71)}[/latex]

Using the feedback formula to evaluate the closed-loop transfer function, we obtain

[latex]T(s)=\frac{66.3}{s^{2}+1.71s+66.3}[/latex]

From the denominator,[latex]\omega _{n} =8.14[/latex], ζ = 0.105. Using Eqs. (4.34),

[latex]T_{p} =\frac{\pi }{\omega _{n}\sqrt{1-\zeta ^{2} } } [/latex]

(4.38),

[latex]\%OS=e^{-(\zeta \pi /\sqrt{1-\zeta ^{2}} )} imes 100[/latex]

and (4.42),

[latex]T_{s}=\frac{4}{\zeta \omega _{n} } [/latex]

the peak time = 0.388 second, the percent overshoot = 71.77%, and the settling time = 4.68 seconds.

e. The Laplace transform of the step response is found by first multiplying Eq. (5.109)

[latex]T(s)=\frac{66.3}{s^{2}+1.71s+66.3}[/latex]

by 1/s, a unit-step input, and expanding into partial fractions:

[latex]C(s)=\frac{66.3}{s(s^{2}+1.71s+66.3)} =\frac{1}{s}-\frac{s+1.71}{s^{2}+1.71s+66.3} [/latex]

[latex]=\frac{1}{s}-\frac{(s+0.855)+0.106(8.097)}{(s+0.855)^{2}+(8.097)^{2} } [/latex]

Taking the inverse Laplace transform, we find

[latex] c(t)=1-e^{-0.855r}(\cos 8.097t+0.106\sin 8.097t) [/latex]

f. For the simplified model we have

[latex] G(s)=\frac{0.0663K}{s(s+1.71)} [/latex]

from which the closed-loop transfer function is calculated to be

[latex] T(s)=\frac{0.0663K}{s^{2}+1.71s+0.0663K} [/latex]

From Eq.(4.39)

[latex] \zeta =\frac{-\ln (\%OS/100)}{\sqrt{\pi ^{2}+\ln ^{2}(\%OS/100) } } [/latex]

a 10% overshoot yields ζ = 0.591 .Using the denominator of Eq. (5.113),

[latex] T(s)=\frac{0.0663K}{s^{2}+1.71s+0.0663K} [/latex]

[latex]\omega _{n} =\sqrt{0.0663K}[/latex] and [latex]2\zeta \omega _{n} =1.71[/latex]. Thus,

[latex]\zeta =\frac{1.71}{2\sqrt{0.0663K} } =0.591[/latex]

from which K = 31.6.

CHALLENGE: You are now given a problem to test your knowledge of this chapter’s objectives: Referring to the antenna azimuth position control system shown on the front endpapers, Configuration 2, do the following:

a. Find the closed-loop transfer function using block diagram reduction.
b. Represent each subsystem with a signal-flow graph and find the state-space representation of the closed-loop system from the signal-flow graph.
c. Use the signal-flow graph found in (b) along with Mason’s rule to find the closed-loop transfer function.
d. Replace the power amplifier with a transfer function of unity and evaluate the closed-loop percent overshoot, settling time, and peak time for K = 5.
e. For the system used for (d), derive the expression for the closed-loop step response.
f. For the simplified model in (d), find the value of preamplifier gain, K, to yield 15% overshoot.

Question 5.CS.1: Antenna Control: Designing a Closed-Loop Response This chapt...

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