page 1 of 4

YTU FACULTY OF ELECTRICAL & ELECTRONICS ENGINEERING DEPARTMENT OF ELECTRICAL ENGINEERING

ELM3101 CONTROL SYSTEMS, MIDTERM EXAM – 2

𝐺(𝑠) =𝐾

𝑠 + 𝑎

Problem 1. The unit step response of a

system is plotted via testing in the lab as

seen on the right hand side. Considering it

as a simple first-order system in the form,

(a) find K and a to model the system,

(b) find also rise and settling times for

the system modelled,

(c) show Ts, Tr and Tc (τ) on the plot.

For 1st order systems: 𝑇𝑟 =

2.2

𝑎, 𝑇𝑠 =

4

𝑎

Name and Surname:

Student number:

Signature: Date:

Marking:

Good luck!

Şeref Naci Engin

𝐺(𝑠) =𝐾

𝑠2 + 𝑎𝑠 + 𝑏

%𝑂𝑆 =𝑐𝑚𝑎𝑥 − 𝑐𝑓𝑖𝑛

𝑐𝑓𝑖𝑛100 =

1.4 − 1

1100

𝜔𝑛 = 0.818 rad/sec

Problem 2. Find the transfer function of

the system in the form of

whose unit step response is plotted on

the right hand side.

Solution 2.

%OS = 40, 𝜁 =−ln(%𝑂𝑆/100)

√𝜋2+ln2(%𝑂𝑆/100)= 0.28

𝑇𝑝 =𝜋

𝜔𝑛√1−𝜁2 𝜔𝑛 =

𝜋

𝑇𝑝√1−𝜁2=

𝜋

4√1−0.282

Solution 1. (a) The time constant, Tc (τ) is the time required for the response to reach the amp-

litude of 63% times the cfinal, which is 0.63 x 2 = 1.26. Then from the curve, Tc = 0.2 sec.= 1/a

a = 5

Now let us find K; cfinal = K / a = 2 K = 10.

Therefore the transfer function is found as 𝐺(𝑠) =10

𝑠+5

(b) Tr= 2.2 / a = 2.2 / 5 = 0.44 sec. and Ts= 4 / a = 4 / 5 = 0.8 sec.

𝐺(𝑠) =𝜔𝑛

2

𝑠2 + 2𝜁𝜔𝑛𝑠 + 𝜔𝑛2 =

0.67

𝑠2 + 0.458𝑠 + 0.67

Formulas for 2nd

order systems:

𝑇𝑝 =𝜋

𝜔𝑛√1−𝜁2 , 𝑇𝑠 ≅

4

𝜁𝜔𝑛,

%OS= 100. 𝑒−𝜁𝜋 √1−𝜁2⁄ , 𝜁 =−ln(%𝑂𝑆/100)

√𝜋2+ln2(%𝑂𝑆/100)

page 2 of 4

Problem 3. For each of the transfer function, (a) find the locations of the poles, (b) find the values of

natural frequency and damping ratio, (c) characterize the nature of the response,

(d) write the general form of the step response.

(i) (𝑠) =42

(𝑠+3)(𝑠+7) , (ii) 𝐺(𝑠) =

29

𝑠2+6𝑠+58 , (iii) 𝐺(𝑠) =

50

𝑠2+100 , (iv) 𝐺(𝑠) =

80

𝑠2+8𝑠+16

Solution 3.

(i) (a) Poles are at -3, -7

(b) Char. Eqn = (𝑠 + 3)(𝑠 + 7) = 𝑠2 + 10𝑠 + 21 = 0 𝜔𝑛 = √21 = 4.583, 𝜁 =10

2𝜔𝑛= 1.1

(c) 𝜁 > 1, hence it is 𝐨𝐯𝐞𝐫𝐝𝐚𝐦𝐩𝐞𝐝.

(d) 𝑐(𝑡) = 𝐴 + 𝐵𝑒−3𝑡 + 𝐶𝑒−7𝑡

(ii) (a) Poles are the roots of the char. eqn., that are −3 ± 7𝑗

(b) Char. Eqn = 𝑠2 + 6𝑠 + 58 = 0 𝜔𝑛 = √58 = 7.616, 𝜁 =6

2𝜔𝑛= 0.4

(c) 𝜁 < 1, hence it is 𝐮𝐧𝐝𝐞𝐫𝐝𝐚𝐦𝐩𝐞𝐝.

(d) 𝑐(𝑡) = 𝐴 + 𝐵𝑒−3𝑡 cos(7𝑡 − 𝜑)

(iii) (a) Poles are at ±10𝑗

(b) Char. Eqn = 𝑠2 + 100 = 0 𝜔𝑛 = 10, 𝜁 = 0

(c) 𝜁 = 0, hence it is 𝐮𝐧𝐝𝐚𝐦𝐩𝐞𝐝.

(d) 𝑐(𝑡) = 𝐴 + 𝐵 cos(10𝑡 − 𝜑)

(iv) (a) Poles are the roots of the char. eqn., that are −4, −4 (repeated)

(b) Char. Eqn = 𝑠2 + 8𝑠 + 16 = 0 𝜔𝑛 = 4, 𝜁 =8

2𝜔𝑛= 1

(c) 𝜁 = 1, hence it is 𝐜𝐫𝐢𝐭𝐢𝐜𝐚𝐥𝐥𝐲 𝐝𝐚𝐦𝐩𝐞𝐝.

(d) 𝑐(𝑡) = 𝐴 + 𝐵𝑒−4𝑡 + 𝐶𝑡𝑒−4𝑡

Problem 4. Considering the transfer function given in Problem 3(ii),

(a) Find the values for the speed (the peak and settling times) of unit step response.

(b) Find the steady-state (final) and peak value of the response.

(c) Propose a new transfer function that produces a unit step response that is 4 times faster while

maintaining the same maximum overshoot and steady-state value

Solution 4. (a) 𝐺(𝑠) =29

𝑠2+6𝑠+58 Poles: −3 ± 7𝑗, 𝜔𝑛 = √58 = 7.616, 𝜁 =

6

2𝜔𝑛= 0.4

𝑇𝑝 =𝜋

𝜔𝑛√1−𝜁2=

𝜋

Im(poles)=

𝜋

7 𝑇𝑝 = 0.4488 sec, 𝑇𝑠 ≅

4

𝜁𝜔𝑛=

4

|Re(poles)|=

4

3 𝑇𝑠 ≅ 1.333 sec

(b) 𝑐𝑠𝑠 = 𝑐𝑓𝑖𝑛 =29

58= 0.5, %OS= 100. 𝑒−𝜁𝜋 √1−𝜁2⁄ %OS = 26.02,

%𝑂𝑆 =𝑐𝑚𝑎𝑥−𝑐𝑓𝑖𝑛

𝑐𝑓𝑖𝑛100 𝑐𝑚𝑎𝑥 =

%OS

100𝑐𝑓𝑖𝑛 + 𝑐𝑓𝑖𝑛 = 0.26 x 0.5 + 0.5 𝑐𝑚𝑎𝑥 = 0.63

(c) Four times faster response with the same %OS (or damping ratio) can be achieved with a new 𝜔𝑛𝑛 that is

four times larger. Since the complex conjugate poles are at −𝜁𝜔𝑛 ± 𝑗𝜔𝑛√1 − 𝜁2, the new poles would

be at 4 × (−3 ± 7𝑗) = −12 ± 28𝑗. So the new char. eqn.: 𝑠2 + 2𝜁𝜔𝑛𝑛𝑠 + 𝜔𝑛𝑛2 = 𝑠2 + 24𝑠 + 928 = 0.

Consequently, the new transfer function would be: 𝐺(𝑠) =928/2

𝑠2+24𝑠+928 𝐺(𝑠) =

464

𝑠2+24𝑠+928

page 3 of 4

Problem 5.

Find 𝑇(𝑠) =𝐶(𝑠)

𝑅(𝑠) for the system given below using block diagram reduction.

Solution 5.

Alternative Solution:

You can also push 𝐺1(𝑠) to the right past the pickoff point. In this case you would get

1

1 + 𝐺1(𝑠)𝐻1(𝑠)× [𝐺1(𝑠)𝐺2(𝑠) + 1] × 𝐺3(𝑠)

The same result could be obtained in an even easier way.

Problem 6. For the system given below find the values of K1 and K2 to yield a peak time of 1.5 second

and a settling time of 3.2 seconds for the closed-loop system’s step response.

Solution 6. The closed loop transfer function can be found as 𝑇(𝑠) =10𝐾1

𝑠2+(2+10𝐾2)𝑠+10𝐾1 .

Using the given values for peak and settling times: 𝑇𝑝 =𝜋

𝜔𝑛√1−𝜁2=

𝜋

Im(𝑝𝑜𝑙𝑒𝑠)= 1.5 Im(poles)=2.09,

𝑇𝑠 ≅4

𝜁𝜔𝑛≅

4

Re(poles)= 3.2 Re(poles)=1.25. Hence the complex conjugate poles are at −1.25 ± 𝑗2.09

The module of the poles gives us 𝜔𝑛 = √1.252 + 2.092 = 2.44 rad/sec

Characteristic equation: 𝑠2 + (2 + 10𝐾2)𝑠 + 10𝐾1 = 𝑠2 + 2𝜁𝜔𝑛𝑠 + 𝜔𝑛2 = 0 𝜔𝑛

2 = 10𝐾1

𝐾1 = 0.595.

Re(poles)=1.25= 𝜁𝜔𝑛. From the equation above, 2 + 10𝐾2 = 2𝜁𝜔𝑛 = 2 × 1.25 = 2.5

𝐾2 = 0.05

page 4 of 4

Problem 7. Find the range of K to keep the system shown in the figure below stable.

Solution 7. The closed-loop transfer function is found first, then the Routh table is setup as below.

From the table, the range of gain for stability is found as −2

3< 𝐾 < 0 using the Routh-Hurwitz

stability criterion

Problem 8. Suppose that a system has the closed-loop transfer function of,

𝑇(𝑠) =𝐾

𝑠4 + 7𝑠3 + 14𝑠2 + 8𝑠 + 𝐾

(a) Find the range of K for stability.

(b) Find the value of K for marginal stability.

(c) Find the frequency of oscillation when the system is marginally stable.

(d) Actual location of closed-loop poles when the system is marginally stable.

Solution 8. The Routh table is setup as,

s4 1 14 K

s3 7 8 0

s2 90 7 = 12.86⁄ K 0

s1 (102.86 − 7𝐾) 12.86⁄ 0

s0 K

(a) The range of K for stability would be found as 0 < 𝐾 < 14.69.

(b) The system is marginally stabile for K for 𝐾 = 14.69.

(c) For 𝐾 = 14.69, the row of zeros (ROZ) would be the row of s2. The auxiliary polynomial is

then found as an even polynomial of 𝑃(𝑠) = 12.86𝑠2 + 14.69 = 0

𝑠1,2 = ±𝑗√14.69 12.86⁄ = ±𝑗1.07 𝜔𝑛 = 1.07rad

sec.

(d) As seen the characteristic polynomial has four (4) roots. Two of them are just found as ±𝑗1.07. The

other two can be found as −3.5 ± 𝑗0.78.