Math Editor

Control 2 Problems

Final 21

(2-a) An open loop transfer function as follows

G(s)H(s)=(s+1)(0.1s+1)(0.01s+1)10e−0.7s​

I. Draw the Bode diagram.

II. Determine the gain margin, phase margin, phase crossover frequency, and gain crossover frequency.

III. Determine the value of gain for GM= 10dB.

find the corner frequencies: ω=1,10,100 rad/sec

factor out constant terms: there isn't!

find shift: 20log10=20

starting asymptote slope: 0 dB/decade

ending asymptote slope: -60 dB/decade

here's a quick way to draw the plot: make this table with the help of your calculator

crossover frequencies: phase = 2.44 gain = 7.8

margins: phase: 180-438 = -303 gain = -11.3

shift = -11.3-10 = -21.3 dB

Knew​=Kold​×10shift/20=0.86

Question 3:

Consider a unity feedback control system with the open loop transfer function is

G(s)H(s)=s(0.5s+1)(0.2s+1)10​

Sketch the polar plot.

G(s)H(s)=s(0.5s+1)(0.2s+1)10​

GH(ω)=ω(j0.5ω+1)(j0.2ω+1)−j10​=ω(−0.1ω2+0.7jω+1)−j10​=ω−0.1ω3+j0.7ω2−j10​

=(ω−0.1ω3)2+0.49ω4−j10(ω−0.1ω3−j0.7ω2)​=0.29ω4+ω2+0.01ω6−7ω2​−j×0.29ω4+ω2+0.01ω6w3−10ω​

GH(ω)=1+0.29ω2+0.01ω4−7​−j1+0.29ω2+0.01ω4ω2−10​

find start point: put ω=0⟹(−7,−∞)

find end point: put ω=∞⟹(0,0)

find intersection with real: ω2−10=0⟹ωp​=√10

GH(ωp​)=−1+2.9+17​=−710​

find intersection with imaginary: none can do, bro 😀

find end phase: 3×−90=−270

Final 18

Question 2:

(2-a) Define the following terms:

I- Gain crossover frequency II- Phase crossover frequency
III- Gain margin IV- Phase margin

I- Gain crossover frequency: the frequency at which the magnitude of the open loop transfer function is unity.

II- Phase crossover frequency: the frequency at which the phase angle of the open-loop transfer function equals –180°

III- Gain margin: The amount of additional gain that a system can tolerate before reaching instability.

IV- Phase margin: the amount of additional phase lag at the gain crossover frequency required to bring the system to the verge of instability.

(2-b) Consider the system shown in Fig.2. Draw a Bode diagram of the open loop transfer
function for K=10, and determine the value of the gain K such that the phase
margin is 50ᴼ. What is the gain margin of this system with this gain K?

GH(s)=s(s+3)(s+6)K​

corner frequencies: 3 and 6

shift = 20log3×610​=−5.10

pm=180+∠GH(ωg​)=50

∠GH(ωg​)=−130⟹ωg​=1.5⟹∣GH(ωp​)∣=−9.8

Knew​=Kold​×1020shift​=10×109.8/20=30.9

Question 3:

(3-a) Fig.3 shows a block diagram of a space vehicle control system. Determine the gain K
such that the phase margin is 50ᴼ

easy peasy 😃

put in polar coordinates

ω2∠180K∠0×√4+ω2​∠tan−12ω​​=ω2K√4+ω2​​∠tan−12ω​−180

find gain crossover frequency

PM=180+∠GH(ωg​)

50=tan−12ωg​​⟹ωg​=2.39rad/sec

find K

∣GH(ωg​)∣=1

K=√4+ω2​ω2​=1.82

Question 4

Consider a unity feedback control system with the open loop transfer function

G(s)H(s)=s3+s2+1K(s+0.5)​

Sketch the polar plot and then,

(a) Determine the gain margin with K=5.

(b) Find the K value for the gain margin is 10db, and then find the corresponding phase margin.

(c) Find the K value for critical stability.

(d) Find the K value for the phase margin is 50ᴼ, and then find the corresponding gain margin.

put s=jω

GH(ω)=(1−ω2)−jω3K(0.5+jω)​

GH(ω)=(1−ω2)2+ω6K(0.5+jω)((1−ω2)+jω3)​=(1−ω2)2+ω6K(0.5(1−ω2)−ω4)​+j(1−ω2)2+ω6K(0.5ω3+ω(1−ω2))​

GH(ω)=(1−ω2)2+ω6K(0.5−0.5ω2−ω4)​+j(1−ω2)2+ω6K(ω−0.5ω3)​

put in polar coordinates

GH(ω)=√(1−ω2)2+ω6​∠tan−1−ω3/(1−ω2)K∠0×√0.25+ω2​∠tan−12ω​

GH(ω)=√(1−ω2)2+ω6​K√0.25+ω2​​∠tan−12ω−tan−1(−ω3/(1−ω2))

to plot we need:

start point: (0.5K,0)
end point: (0,0)
intersection with real: ω−0.5ω3=0⟹ωp​=√2⟹(−0.5K,0)
intersection with imaginary: 0.5−0.5ω2−ω4=0⟹ω=√0.5⟹(0,√2K)
end phase: (1−3)×90=−180∘

for gain analysis we need everything in terms of K:

ωp​=√2≈1.4142
GM(K)=−20log0.5K

to find phase margin:

∣GH(ωg​)∣=1⟹K√0.25+ω2​=√(1−ω2)2+ω6​ and solve for ωg​
PM(ωg​)=180+∠(GH(ωg​))=180+tan−12ωg​−tan−1(−ωg3​/(1−ωg2​))

that should do it!

(a) Determine the gain margin with K=5.
GM(K)=−20log0.5×5=−7.9588

(b) Find the K value for the gain margin is 10db, and then find the corresponding phase margin.
GM(K)=−20log0.5K=10⟹K=√102​≈0.6325
√102​=√0.25+ω2(1−ω2)2+ω6​​⟹104​=0.25+x(1−x)2+x3​:x=ω2

10+10x2−20x+10x3−1−4x=0

all roots have imaginary component

ωg​=none

PM=∞ 🙃

(c) Find the K value for critical stability.
0.5K=1⟹K=2

(d) Find the K value for the phase margin is 50ᴼ, and then find the corresponding gain margin.

PM=50=180+tan−12ωg​−tan−1(−ωg3​/(1−ωg2​)) => negative frequency

let's try the rectangular coordinates

∠=−130=tan−10.5−0.5ω2−ω4ω−0.5ω3​

1.191753593=0.5−0.5ω2−ω4ω−0.5ω3​⟹ωg​=0.463rad/sec

∣GH(ωg​)∣=1⟹K=√0.25+ω2(1−ω2)2+ω6​​=1.162051648

GM=−20log0.5×1.162051648=4.716091295dB

That doesn't look right!

Ah I get it tan−310=tan−130

but that would be a false positive!

conclusion: phase margin would never be 50∘

Final 17

(1-a) Define the following terms:

I- Gain crossover frequency II- Gain margin III- Phase margin

I. Gain crossover frequency is the frequency at which the magnitude of the transfer function is unity. I.E the output signal amplitude follows the input.

II. Gain margin is the additional gain that would make the system critically stable.

III. Phase margin is the additional phase shift at unity gain that would make the system critically stable (gain and phase would cross over).

Question 2:

(2-a) Draw a Bode diagram of the open-loop transfer function of the closed-loop system shown in Fig.2. Determine the gain margin, phase margin, phase crossover frequency and gain crossover frequency. Then,

I. Determine the value of gain for GM= 30dB.

II. Determine the value of gain for ϕm=45⁰.

find corner frequencies: 1,√10,5

factor out constants: K=5×1020​=0.4

find shift: 20log0.4=−7.96dB

GM=10dB,PM=180−76=104∘ωp​=4rad/sec,ωg​=0.44rad/sec

shift=GMold−GMnew=10−30=−20dB

Knew=Kold×10shift/20=20×10−20/20=2

PM=45⟹∠=−135⟹ω=3.1⟹Lm=−4.8⟹shift=4.8dB

Knew=Kold×10shift/20=20×104.8/20=34.75

(2-b) Consider a unity feedback control system with the open loop transfer function

G(s)H(s)=s(s2+s+4)K​

Determine the value of the gain k such that the phase margin is 50⁰. What is the gain margin with this gain k?

you know the drill by now 😀

put in polar coordinates:

ω∠90×√(4−ω2)2+ω2​∠tan−1ω/4−ω2K∠0​

=ω√(4−ω2)2+ω2​K​∠−90−tan−14−ω2ω​

PM=50⟹∠=−130⟹4−ω2ω​=tan40⟹ω=1.491003449

K=ω√(4−ω2)2+ω2​=3.458516079

Question 4

Consider a unity feedback control system with the open loop transfer function

G(s)H(s)=s3(1+Ts)2​

Sketch the polar plot and then,

(a) Determine the gain margin with T=5.

(b) Find the T value for the gain margin is 20db, and then find the corresponding phase margin.

(c) Find the T value for critical stability.

(d) Find the T value for the phase margin is 37ᴼ, and then find the corresponding gain margin.

here we go again!

find everything in terms of T

put in rectangular coordinates

GH=ω3j(1−T2ω2+j2Tω)​=−ω22T​+jω31−T2ω2​

find start point: (−∞,∞)

find end point: (0,0)

find intersection with real axis:

1−T2ω2=0⟹ωp​=T1​

−ω22T​=−2T3

find intersection with imaginary axis:

T=0orω=∞⟹ that's the end point #Rejected

find end phase: (2−3)×90=−90∘

GM(T)=−20log2T3⟹GM(5)=−47.95880017

20=−20log2T3⟹T=0.3684031499

To find phase margin:

put in polar coordinates:

GH=ω31+T2ω2​∠2tan−1ωT−270

find ωg​

ω31+T2ω2​=1⟹ω3=1+T2ω2⟹ solve for ωg​ with calculator

PM=180+2tan−1ωT−270=2tan−1ωg​T−90

for T=0.3684031499⟹ωg​=1.047348573⟹PM=−47.80213173

for critical stability put GM=0

2T3=1⟹T=3√0.5

for PM=37=2tan−1ωg​T−90⟹ωg​=T2.005689708​

substitute in magnitude and solve for T

ω31+T2ω2​=1⟹ω3−T2ω2−1=0⟹T=1.171157681