Home | Classical Feedback Control | Previous |

Please refer to the beginning of this document for copyright information and use permissions.

The slope at the crossover frequency 40 Hz is (a) -6 dB

*/*oct (b) -9 dB*/*oct (c) -10 dB*/*oct (d) -12 dB*/*oct. What is the feedback in dB at 1.25 Hz if the slope remains the same down to this frequency? What would be the loop phase lag if the slope is constant at all frequencies?What is the length of the Bode step and

*f*_{b}if

(a) the asymptotic slope is -18 dB*/*oct,*x*= 10 dB,*y*= 1*/*6, the asymptote passes the -10 dB level at 1 kHz, and the n.p. lag at 1 kHz is 0.5 rad;

(b) the asymptotic slope is -12 dB*/*oct, the asymptote passes the 0 dB level at 5 kHz, and the n.p. lag at 5 kHz is 1 rad;

(c) the asymptotic slope is -12 dB*/*oct, the asymptote passes the -10 dB level at 100 Hz, and the n.p. lag at 100 Hz is 0.5 rad;

(d) the asymptotic slope is -24 dB/oct, the asymptote passes the -10 dB level at 50 kHz, and the n.p. lag at 100 kHz is 0.6 rad.Sketch the phase lag response and the Nyquist diagram for the optimal Bode cutoff in which the Bode step is omitted, with low-frequency slope -10 dB

*/*oct and asymptotic slope -18 dB*/*oct. Is the system stable?At the frequency of structural resonance

*f*_{st}= 120 Hz or higher there is a narrow resonance peak in the plant gain response and the plant phase at this frequency is completely uncertain. (The phase uncertainty is the result of using a digital controller with sampling frequency 100 Hz. Digital controllers will be studied in detail in Chapter 5.) The system must be therefore gain-stabilized at the frequency of the structural resonance*f*_{st}.Using the asymptotic slope -18 dB

*/*oct, 30° and 10 dB stability margins, and assuming that*f*_{d}= 2*f*_{b}, and*f*_{c}= 4*f*_{b}(which are typical numbers and should be used for initial estimates), express*f*_{b}as a function of*f*_{st}and*Q*. Make a sketch of the loop gain response similar to that shown in Fig. 4.47, but with numbers on it, for the peak value 20 log*Q*equal to: (a) 20 dB; (b) 25 dB; (c) 30 dB; (d) 40 dB; (e) 50 dB.**Fig. 4.47**Feedback bandwidth limitation due to a structural resonanceThe same as in Problem 4, but the resonance uncertainty range starts at 170 Hz.

The same as in Problem 4, but the resonance uncertainty range starts at 85 Hz.

The crossover frequency

*f*_{b}is 1 kHz,*y*= 1*/*6. The system is phase-stabilized at all frequencies up to*f*_{b}, with Bode optimal loop response. What is the available feedback over the bandwidth (a) [0,50], (b) [0,30], (c) [30,60]?(a) What is the feedback in dB at 1.5 Hz if the crossover frequency is 300 Hz and the main slope is -10 dB/oct?

(b) What is the maximum available feedback in dB at 1.6 Hz when the feedback is kept constant at frequencies below 1.6 Hz, the system is phase-stabilized with 30° stability margins, and the crossover frequency is 300 Hz?The required feedback at 10 Hz is 40 dB, and the feedback should increase at lower frequencies with the slope -10 dB

*/*oct. What is the crossover frequency? What are the frequencies at the beginning and the end of the Bode step if the step's length is 0.8 oct and*x*= 10 dB?In a system with feedback bandwidth 100 Hz, amplitude stability margin 10 dB, phase stability margin 30°, and no n.p. lag, the attenuation in the feedback loop is required to be large over 1.5 kHz where there might be flexible modes in the plant. Calculate what attenuation at 1.5 kHz is available in the loop response with a Bode step if the asymptotic slope is chosen to be (a) -12 dB

*/*oct; (b) -18 dB*/*oct. What is the conclusion? Explain the result by referring to the shape of the weight function in the Bode phase gain relation.In a GaAs microwave feedback amplifier, there are two gain stages, and the length of the feedback loop is 1 mm. The speed of signal propagation is 150,000 km

*/*sec. At what frequency is*B*_{n}= 1 rad? Considering this frequency as*f*_{c }, what is the length of the Bode step? What is the available feedback over the bandwidth (a) 0 to 3 GHz, (b) 1.5 to 3 GHz, (c) 0 to 2 GHz, (d) 2 to 3 GHz?Find the loop transfer function and the prefilter for the system with crossover frequency: (a) 0.2 Hz; (b) 6 Hz; (c) 2 kHz; (d) 6 kHz; (e) 2 kHz; (f) 6 MHz; (g) 2 MHz; (h) 4 rad

*/*sec; (i) 100 rad*/*sec. Use the 1 rad*/*sec crossover prototype described in Section 4.2.3.The actuator and plant transfer function

*AP*is:

(a) 4(10 -*s*)(*s*+ 2)*/*[*s*^{2}(10 +*s*)(*s*+ 7)];

(b) 2(10 -*s*)(*s*+ 2)*/*[*s*^{2}(10 +*s*)(*s*+ 8)];

(c) 3(10 -*s*)(*s*+ 4)*/*[*s*^{2}(10 +*s*)(*s*+ 9)];

(d) 0.6(10 -*s*)(*s*+ 5)*/*[*s*^{2}(10 +*s*)(*s*+ 10)];

(e) 2.72(10 -*s*)(*s*+ 6)*/*[*s*^{2}(10 +*s*)(*s*+ 11)];

Find the compensator that makes the loop transfer function the same as in the example studied in Section 4.2.3, where.

Determine the band-pass transform from the low-pass optimal cutoff with frequency range

(a) [0, 1] rad/sec to the bandwidth [30, 70] Hz;

(b) [0, 2] rad/sec to the bandwidth [50, 70] Hz;

(c) [0, 3] rad/sec to the bandwidth [60, 70] Hz;

(d) [0, 4] rad/sec to the bandwidth [40, 70] Hz;

(e) [0, 5] rad/sec to the bandwidth [30, 120] Hz;

(f) [0, 1] rad/sec to the bandwidth [30, 100] Hz.

Draw a Nyquist diagram on the

*T*-plane (not necessarily to scale, only to show the shape) for the Nyquist-stable system in Fig. 4.19.Initial analysis with a low-order compensator has shown that in the plant hardware configuration A, larger feedback is available than in configuration B. The system engineer assumed (wrongly) that feedback will be larger in configuration A even when, later, a better controller will be developed. Therefore, he decided that configuration A should be chosen. Devise a counterexample to prove that optimal shaping for the Bode diagram must be used for initial analysis as well. (Hint: Use a plant with a flexible mode.)

Using the phase-gain chart in Fig. 3.42 (or the program from Appendix 5), calculate the phase response for a Nyquist-stable system whose loop gain response is: from 0 to 10 Hz, 60 dB; from 10 to 20 Hz, -50 dB

*/*oct; from 20 to 80 Hz, 10 dB; from 80 to 320 Hz, -10 dB*/*oct; from 320 to 640 Hz, -10 dB; from 640 Hz, -18 dB*/*oct.An extra management level was added to a four-level management system. How will it affect the speed of accessing the market and adjusting the product quantities and features (make a rough estimate)?

In a Nyquist-stable system with a response like that shown in Fig. 4.19,

*f*_{b}= 100 Hz, the phase stability margin is 30°, and the upper and lower gain stability margins are 10 dB. Calculate the frequencies at the ends of the upper and lower Bode steps if the slope at lower frequencies is (a) -12 dB*/*oct, (b) -18 dB*/*oct, and the asymptotic slope is (a) -18 dB*/*oct, (b) -24 dB*/*oct.The loop gain plot crosses the 0 dB level at 200 kHz, and the rest is as in Problem 19, versions (a) and (b).

The unstable plant can be equivalently represented as a stable plant with a feedback path

*B*_{1}. The path from the plant output to the plant input via the regular feedback loop links is 40 dB larger than via path*B*_{1}at most frequencies, but only by 20 dB larger over some narrow frequency range. How to approach the design of the feedback system?Two parallel links have the following transfer functions respectively:

(a) and ;

(b) and ;

(c) and ;

(d) and .

Is the composite link

*W*_{1}(*s*) +*W*_{2}(*s*) m.p.?Two minimum phase links in parallel made a nonminimum phase link. A third link with constant gain coefficient

*k*has been added in parallel to the two links. How would you find the minimum*k*for the total transfer function to be m.p.?Find transfer function

*G*(*s*) of a high-pass in parallel to low-pass 12*/*[(*s*+ 3)(*s*+ 4)] so that the total transfer function is 1. Is this*G*(*s*) realizable?Prove that if a transfer function of a linear passive two-port is m.p. with some passive impedances of the signal source and the load, the transfer function is m.p. with any other passive source and load impedances.

(a) 1.25 Hz is 5 octaves below 40 Hz. Therefore, the feedback is 6 × 5 = 30 dB.

(a) The frequency

*f*_{c}= 1 kHz; then, from (4.2)*f*_{c}*/f*_{d}= 0.6 × 3 + 0.5 = 2.3 (i.e., the step length is 3.32 log 2.3 = 1.2 oct),*f*_{d}= 0.435, and*f*_{b}=*f*_{d}*/*2 = 0.22 kHz.(a) Since y = 1

*/*6, the slope is -10 dB*/*oct. There are 3.32 log(1000*/*50) = 4.3 oct down to 50 Hz from the crossover. Then, according to (4.5), the available feedback is (4.3 + 1) × 10 = 53 dB.(a) 18 log

_{2}(*f*_{c}*/f*_{st}) = 20 log_{10}*Q*, and*f*_{b}=*f*_{c}*/*4, then*f*_{b}=*f*_{st}*/*2^{(20 log Q)/18+2}, i.e.,fst = 120; Q = 10; fb = fst/2^[20*log10(Q)/18 + 2] fb = 13.8881

The diagram is shown in Fig. 4.48.

**Fig. 4.48**Loop response Bode diagramWith the slope of the Bode diagram -10 dB/oct, disturbance rejection in dB at frequency

*f*<*f*_{b}*/*2 depends on the feedback bandwidth*f*_{b}approximately as 10 log_{2}(*f*_{b}*/f*). At lower frequencies*f*<*f*_{b}*/*6, the slope can be increased to -12 dB*/*oct as shown by the dashed line. (In this case, a nonlinear dynamic compensator must be employed to assure global stability as described in Chapters 10 and 11.)When discussing the tradeoff between the resonance frequency and the resonance quality (

*Q*) of the object of control and the available disturbance rejection at a meeting with mechanical designers, it is helpful for control engineers to have prepared plots like those shown in Fig. 4.49, exemplifying the available disturbance rejection for two structural resonance frequencies,*f*_{st}= 50 Hz and 100 Hz.(a) (b) **Fig. 4.49**Dependence of disturbance rejection

on the structural mode frequency and damping

Home | Classical Feedback Control | Previous |