<span class="text_page_counter">Trang 1</span><div class="page_container" data-page="1">

<b>HO CHI MINH UNIVERSITY OF TECHNOLOGY AND EDUCATIONFACULTY </b>

</div><span class="text_page_counter">Trang 2</span><div class="page_container" data-page="2">

With K=10, frequency range from 0.1 to 100 Using this code to simulate in Matlab:

<b>Fig.2.1.1: Bode diagram of the system when K=10</b>

</div><span class="text_page_counter">Trang 3</span><div class="page_container" data-page="3">

<b>b. Determine cross-over frequency, phase cross-over frequency, phase margin and gain margin:</b>

Phase cross-over frequency is located at the point when the bode plot cut phase= -180. Gain cross-over frequency is located at the point when the bode plot cut gain magnitude = 0

Phase margin is the value at the point when the magnitude reached 0dB (Gain Cross-over frequency). Which means that the magnitude line goes cross the value 0dB, at that point we get the value of the phase margin.

Gain margin is the value at the point when the magnitude reached -180 degree (Phase Cross-over frequency). Which means that the phase line goes cross the value -180dB, at that point we get the value of the gain margin.

From the picture exported from Matlab simulation:

- Cross-over frequency: -0.0065dB (~0dB) at frequency 0.454 rad/s - Phase cross-over frequency: -179 (~-180) degree at frequency 4.59 rad/s - Gain margin: 24.8dB at frequency 4.65 rad/s

- Phase margin: 103 degree at frequency 0.455 rad/s

<b>c. Consider the stability of a closed system and explain</b>

As we can see from the value given above, the closed-system is stable because of the positive of gain margin (24.8dB) and phase margin (103 degree).

<b>+ Compare with Routh Array:</b>

</div><span class="text_page_counter">Trang 4</span><div class="page_container" data-page="4">

From Routh array table, the first column value is all positive, means that the system is stable. So, the system is currently stable.

<b>d.Draw the transient response of the above system with unit step function as input for the time interval t=0÷10s to illustrate the conclusion in c. </b>

Step response with input is unit step function at t = 0 – 10 (s)

According to the step response, the signal goes to an exact value (~0.7) when the time goes to infinite. So that means the system is stable.

<b>e. With K=400, repeat the requirements from question a→d</b>

• <b>Draw the Gain and Phase Bode plot of the open system in the frequency range (0.1, 100)</b>

With K=400, frequency range from 0.1 to 100 Using this code to simulate in Matlab:

</div><span class="text_page_counter">Trang 5</span><div class="page_container" data-page="5">

<b>Fig.2.1.2: Bode diagram of the system when K=10</b>

<b>• Determine cross-over frequency, phase cross-over frequency, phase margin and gain margin:</b>

From the picture exported from Matlab simulation:

- Cross-over frequency: -0.0623dB (~0dB) at frequency 6.74 rad/s - Phase cross-over frequency: -179 (~-180) degree at frequency 4.61 rad/s - Gain margin: -7.27dB at frequency 4.65 rad/s

- Phase margin: -23.4 degree at frequency 6.73 rad/s

As we can see from the value given above, the closed-system is unstable because of the negative of gain margin (-7.27dB) and phase margin (-23.4 degree).

<b>+ Compare with Routh Array:</b>

Characteristic equation:

</div><span class="text_page_counter">Trang 6</span><div class="page_container" data-page="6">

1 21.6 0

From Routh array table, the first column is having a negative value, means that the system is unstable.So the system is currently unstable.

• <b>Draw the transient response of the above system with unit step function as input for the time interval t=0÷10s to illustrate the conclusion in c. </b>

Step response with input is unit step function at t=0 – 10 (s):

According to the step response, the signal oscillates when the time goes to infinite. So that means the system is unstable.

</div><span class="text_page_counter">Trang 8</span><div class="page_container" data-page="8">

<b>b. Find the gain margin, phase margin</b> the amplitude of system’s output at phase crossover frequency (phase crossover frequency is the frequency at which where the Nyquist plot crosses the negative real axis)

From the Nyquist plot,we get the amplitude at phase crossover frequency is about 0.0572.

Compared to the value of gain margin shown by MATLAB, there is a small error (|24.85 - 24.8| = 0.05), which can be accepted.

The phase margin is defined by drawing a circle with center O, radius R=1. The intersection point between the circle and the Nyquist plot is the point corresponding to the phase margin, which equals 103.

</div><span class="text_page_counter">Trang 9</span><div class="page_container" data-page="9">

Compared to the value of the gain margin and the phase margin found by using Bode plot

</div><span class="text_page_counter">Trang 10</span><div class="page_container" data-page="10">

<b>c.Consider the stability of a closed system and explain </b>

The Nyquist stability criterion states that: a closed-loop system is defined as being stable if its Nyquist plot encircles the point (-1, 0) ^𝐿

2^{rounds, which L is the number of poles} locating at the right side of the complex plane. (1)

=> The opened-loop transfer function G(s) has 3 poles that all locate at the left of the complex plane, thus this transfer function is stable (2)

According to Fig.2.1., the plot obviously does not encircle the point (-1, 0) (3) From (1), (2) and (3) => The given closed-loop system with K=10 is stable.

</div><span class="text_page_counter">Trang 11</span><div class="page_container" data-page="11">

<b>Fig.2.2.2. Nyquist diagram of the system when K=400• Find gain margin, phase margin:</b>

From the Nyquist plot,we get the amplitude at phase crossover frequency is about 2.31

</div><span class="text_page_counter">Trang 12</span><div class="page_container" data-page="12">

=> = => Gain margin: = = = − (the same as the value from MATLAB)

The phase margin is defined by drawing a circle with center O, radius R=1. The intersection point between the circle and the Nyquist plot is the point corresponding to the phase margin, which equals -23.4.

Compared to the value of the gain margin and the phase margin found by using Bode plot in section 2.1.3:

</div><span class="text_page_counter">Trang 13</span><div class="page_container" data-page="13">

It can be clearly seen that the values by using both methods are the same (GM = 24.8 and PM = 103), so the result is absolutely correct.

• <b>Consider the stability of a closed system and explain </b>

The Nyquist stability criterion states that: a closed-loop system is defined as being stable if its Nyquist plot encircles the point (-1, 0) ^𝐿

2^{rounds, which L is the number of poles} locating at the right side of the complex plane. (1)

The opened-loop transfer function G(s) has 3 poles that all locate at the left of the complex plane. However, according to Fig.2.2.2, the Nyquist plot encircles the point (-1, 0), which breaks the Nyquist stability criterion, Therefore, the given closed-loop system with K=400 is unstable.

</div><span class="text_page_counter">Trang 14</span><div class="page_container" data-page="14">

<b>a. Drawing the system’s root locus using MATLAB:</b>

Enter this code into MATLAB:

<b>Fig.2.3.1. Root locus diagram of the system</b>

The gain margin K<small>gh </small>will be given by the point where the root locus crosses the imaginary axis in the complex plane.

According to Fig.2.3.1, the value of gain of the points that the root locus crosses the imaginary axis is 103, so K is considered to equal 175.

</div><span class="text_page_counter">Trang 15</span><div class="page_container" data-page="15">

<b>b. Find K such that the system has the natural frequency ω = 4:<small>n</small></b>

The two complex conjugate poles of a system are defined as:

It means that the magnitude from the origin to the poles equals 

</div><span class="text_page_counter">Trang 16</span><div class="page_container" data-page="16">

From the root locus, at pole (-3.73; 1.46), we have:  =

(

−

) (

+

)

 At this point, we earn the gain K = 8.

<b>c. Find K so that the system has a damping coefficient = 0.7</b>

</div><span class="text_page_counter">Trang 17</span><div class="page_container" data-page="17">

According to above figure, at pole (-1.64; 1.67) which lies on the line from the origin corresponds to = , we get the gain K = 23.

<b>d. Find K for the system to have overshoot max </b> =

So, K is the point that is cut by a = <small>−</small>  =  line and Root Locus

From the root locus, at pole (-1.15; 2.6) where = , we get the gain K = 43.6

</div><span class="text_page_counter">Trang 18</span><div class="page_container" data-page="18">

<b>e. Find K for the system has settling time (2% standard) txl=4s</b>

</div><span class="text_page_counter">Trang 19</span><div class="page_container" data-page="19">

<b>a. Drawing the system’s root locus using MATLAB:</b>

Enter this code into MATLAB:

<b>Fig.2.4.1. Root locus diagram of the system</b>

The gain margin K<small>gh </small>will be given by the point where the root locus crosses the imaginary axis in the complex plane.

According to Fig.2.4.1, the value of gain of the points that the root locus crosses the imaginary axis is 103, so K <small>gh</small>is considered to equal 103.

+ Theoretically calculation: Characteristic equation:

</div><span class="text_page_counter">Trang 20</span><div class="page_container" data-page="20">

From the above results, it can be concluded that the gain margin K <small>gh</small>is exactly 103.

<b>b. Find K such that the system has a natural oscillation frequency ω = 4<small>n</small></b>

The two complex conjugate poles of a system are defined as: = −   −

Performing them on the complex plane:

It means that the magnitude from the origin to the poles equals 

</div><span class="text_page_counter">Trang 21</span><div class="page_container" data-page="21">

From the root locus, at pole (-0.205; 3.99), we have:

The characteristic equation is quadratic equation, so it is supposed to have two real poles and two complex conjugate poles

</div><span class="text_page_counter">Trang 22</span><div class="page_container" data-page="22">

+ Comment: The error between the gain K found by using MATLAB and the theoretically calculated one is very small (|78.4 - 78.47| = 0.07), which can be accepted.

<b>c.Find K such that the system has a damping coefficient = 0.7</b>

</div><span class="text_page_counter">Trang 23</span><div class="page_container" data-page="23">

According to above figure, the red line from the origin corresponds to = does not across the root locus, which means that there is no value of K such that the system has the damping coefficient = .

+ Theoretical calculation:

</div><span class="text_page_counter">Trang 24</span><div class="page_container" data-page="24">

+ Comment: Both methods show that there is no value of K such that the system has the damping coefficient = , so this conclusion is correct.

</div><span class="text_page_counter">Trang 25</span><div class="page_container" data-page="25">

<b>d. Find K such that the system has an overshoot σmax% = 25%</b>

</div><span class="text_page_counter">Trang 26</span><div class="page_container" data-page="26">

+ Comment: The error between the gain K found by using MATLAB and the theoretically calculated one is very small (|9.11 – 9.09| = 0.02), which can be accepted.

</div><span class="text_page_counter">Trang 27</span><div class="page_container" data-page="27">

<b>e.Find K such that the system has a settling time (2% standard) = 4s</b>

</div><span class="text_page_counter">Trang 28</span><div class="page_container" data-page="28">

+ Comment: The error between the gain K found by using MATLAB and the theoretically calculated one is very small (|9.11 – 9.09| = 0.02), which can be accepted.

<b>Investigate the system using Bode plot and Nyquist plot when </b> =====

</div><span class="text_page_counter">Trang 30</span><div class="page_container" data-page="30">

<b>Fig.2.4.3: The Nyquist diagram of the system</b>

<b>2.5 </b>

<b>1. Comparing control system surveying methods:</b>

<b>Algebraic stability criteria: Including Routh-Hurwitz criteria. These methods only </b>

using algebraic, which means they determine if the control system stable or not. Not illustrating any chart or geometry form so it hard to see to whole thing.

<b>Root Locus: Root locus use the characteristic equation to plot the chart of roots. The </b>

method contains information as damping value, overshoot value, gain and frequency and the stability of the system.

<b>Frequency stability criteria: Including Mikhailov, Bode plot and Nyquist frequency </b>

stability. Mikhailov based on rotate angle fundamental. Nyquist using closed system formula to sketch the Nyquist curves to check the stability. Bode plot use closed system to sketch the chart of magnitude of gain and phase to check the stability.

<b>2. When to use control system surveying methods?</b>

</div><span class="text_page_counter">Trang 31</span><div class="page_container" data-page="31">

Control system surveying methods are used in various scenarios and stages of a control system's lifecycle to assess and improve its performance, stability, and overall functionality.

Using the Algebraic stability criteria when just checking if the system is stable or not. Using the Root Locus when to setting the damping value, overshoot, gain and frequency to control design, stability analysis, tunning the controllers, transient response analysis. Using the Frequency stability criteria when analyzing frequency domain, filter design, delay time, …

<b>3. Bode plot and Nyquist relationship: </b>

Frequency Response Information: Both Bode and Nyquist plots provide information about a system's frequency response. They show how the system behaves at different frequencies.

While Bode plots provide insights into the magnitude and phase response of a system, Nyquist plots are primarily used for stability analysis. In a Nyquist plot, the stability of a system is determined by the number of encirclements of the critical point (-1, 0) in the complex plane.

The phase information in the Bode plot corresponds to the phase angle of G(jω) at different frequencies. In the Nyquist plot, the phase information is represented by the angle of the complex numbers on the plot.

Bode plots can be used to directly read the gain and phase margins of a system, which indicate how close the system is to instability. Nyquist plots can indirectly provide information about these margins but are primarily used to assess stability by analyzing encirclement.

</div><span class="text_page_counter">Trang 32</span><div class="page_container" data-page="32">

<b>APPLICATION MATLAB IN SYSTEM QUALITY SURVEY3.3.</b>

<b>a. With the K limit value found above, draw the transient response with the input as a unit step function. Verify whether the output fluctuates or not?</b>

Using the result of Kgh from task 2.3.3, we have: Kgh=176

Applied this code in Matlab, we have the simulation:

The output waveform from step respond is oscillating because the gain K<small>gh</small> is at the unspecify zone on Root Locus that we surveyed before.

</div><span class="text_page_counter">Trang 33</span><div class="page_container" data-page="33">

<b>b. With the K value found in question 2.3 d of experiment no.2, draw the transient response of the closed system with the input to the unit step function in the time range from 0 ÷ 5s. Find the overshoot and steady-state error of the system. Verify whether the system has σmax% = 25% or not?</b>

With K found at 3.3 d at experiment 2, we have: K=43.8

Using this code to simulate on Matlab, we have:

</div><span class="text_page_counter">Trang 34</span><div class="page_container" data-page="34">

<b>c. With the K value found in question 2.3 e of experiment no.2, draw the transient response of the closed system with the input to the unit step function in the time range from 0 ÷ 5s. Find the overshoot and steady-state error of the system. Verify that the system has txl = 4s?</b>

With K found at experiment 2.3.3 e, we have: K=52.7

Apply this code in Matlab simulation, we have:

Overshoot: 29%

Settling time: 3.81s. This value is less than the experiment 2.3.3 e (4s) about 0.19s

</div><span class="text_page_counter">Trang 35</span><div class="page_container" data-page="35">

<b>d. Draw the two transient responses of questions b and c on the same figure. </b>

Apply this code to Matlab simulation, we have:

The K=52.7 (1) is higher than K=43.8 (2), leads to:

The overshoot of (1) is higher. This makes the system takes more time to set to stable state: settling time = 3.81s, higher than settling time of (2) (3.12s).

The steady state error of (1) is lower than (2): 7.1% < 8.4%. This means that higher gain makes the output more accurate.

</div><span class="text_page_counter">Trang 36</span><div class="page_container" data-page="36">

<b> 1) Why we have to survey to system quality?</b>

Stability is the condition for a control system, though it is not enough to use in real life. Many requirements need to be satisfied by the system at the same time as accuracy, stability, overshoot, sensitivity, noise reduction…

<b>2) Systems have steady stat equals zero with step unit input signal: </b>

Type 0 systems can achieve zero steady-state error when subjected to a step input because they do not accumulate error over time.

Type 1 with Integral control systems can also achieve zero steady-state error with a step input, provided they have an integral component in their control action. The integral term acts to eliminate any steady-state error by continuously accumulating and correcting the error over time.

<b>3) Systems have steady state equals zero with ramp input signal:</b>

Type 1 with Integral control can achieve zero steady-state error when subjected to a ramp input. The integral component accumulates and eliminates any steady-state error caused by a constant slope input (ramp).

Type 2 with Double integration systems can eliminate steady-state error not only for a step input but also for a ramp input.

<b>4) Describe a system response when the system has a pair of conjugated poles on real axis of the complex plane</b>

The presence of a pair of conjugated poles with real parts on the real axis means that the system is inherently unstable or marginally stable, it exhibits a particular type of response characterized by oscillatory behavior. This situation is commonly referred to as having "underdamped" or "oscillatory" poles.

<b>5) Describe a system response when the system has a pair of conjugated poles on imaginary axis of the complex plane</b>

The response of a system with conjugated poles on the imaginary axis is purely oscillatory. There is no damping to reduce the oscillations over time, it exhibits a special type of response known as "purely oscillatory" or "undamped" response. This configuration is characterized by the absence of any damping in the system, resulting in continuous, undamped oscillations.

</div>