https://electricalstudent.com/wp-content/uploads/2022/07/img_0162-1024×856.jpg” alt=”Problem 1. An NMOS common-source amplifier circuit shown as the figure, the transistor parameters are: VTN=0.8V, Kn = 1mA/V2, and the channel modulation effect is ignored. The circuit parameters are VDD=5V, Rs=1KΩ, RD= 4KΩ, R1=225KΩ, and R2=175KΩ.
Find the circuit quiescent values IDQ and VDSQ.
Drawthesmallsignalequivalentcircuit.
FindthesmallsignalvoltagegainforRLasanopencircuit. 4. Find the input impedance Rin as indicated in the figure.
Find the output impedance Rout as indicated in the figure.” class=”wp-image-16958″ width=”840″ height=”702″/>

The above problem is completely solved

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 ECE 476 

Control Systems Laboratory 

Laboratory Experiment 7: Controller Design using Root Locus 

1𝑠(𝑠+2)

+ 

–

r 

y 

𝑘

Objectives: 

1.Design the controller using root locus design method.

2.Learn how changes in the parameters influence the pole locations and how pole locationsinfluence the system performance.

List of Equipment/Software 

Following equipment/software is required: 

MATLAB/Simulink 

Deliverables 

A complete lab report including the following: 

Summarized learning outcomes. 

MATLAB scripts and their results should be reported properly. 

Complete all the assignments. 

Part I: Open-loop Root Locus 

Root locus is a graph displaying the roots of a polynomial equation when one of the parameters in the coefficients of the equation changes from 0 to ∞. 

Example 1: Consider the following unit feedback system. 

The transfer function of the open-loop system is 

𝐻(𝑠)=𝑌(𝑠)𝑈(𝑠)=𝑘𝑠(𝑠+2) 

The root locus with k =1 can be plotted using MatLab. You can use ‘rlocus’ to plot the root locus. 

Exercise 1: Show the root locus of the system in Example 1 and find the breakpoints in the root locus. Exercise 2: Consider the unit feedback system below. Plot the root locus of the system using matlab. 

Part II: Second order system performance in dynamic 

Following transfer function shows the second order system with no finite zeros and two complex poles and with undamped natural frequency ωn , damping ratio ζ, and negative real part σ. 

H(s)=𝜔𝑛2𝑠2+2𝜉𝜔𝑛+𝜔𝑛2=𝜔𝑛2(𝑠+𝜎)2+𝜔𝑑2

In analysis and design, they can be used to estimate rise time, overshoot, and settling time, respectively, for just about any system. In design synthesis we wish to specify rise time tr , overshoot Mp , and settling time ts and to ask where the poles need to be so that the actual responses are less than or equal to these specifications. For specified values of tr , Mp , and ts , the synthesis form of the equation is 

These specifications can be graphed in the s-plane as shown below. Based on the above equations, we can find the regions for possible closed-loop pole locations 

Example 2: Find the allowable regions in the s-plane for the poles of a transfer function of a system if the system response requirements are tr ≤ 0.6 sec, Mp ≤ 10%, and ts ≤ 3 sec. Solution: Regardless of the system, we can obtain a first approximation using the relationships for a second-order system. 

ξ≥√(𝑙𝑛𝑀𝑝)2𝜋2+(𝑙𝑛𝑀𝑝)2=0.6 

The allowable region is anywhere to the left of the solid line in the figure. Note that in this case any pole meeting the ζ and ωn restrictions will automatically meet the σ restriction.

Part III: Determine k from the root locus 

The root locus shows all possible closed-loop pole locations for a pure proportional controller. Obviously not all of those closed-loop poles will satisfy our design criteria, To determine what part of the locus is acceptable, we can use the calculation in Part II to find out the acceptable region in s-plane based on natural frequency, damping ratio, and negative real part. 

Example 3: Design a proportional controller for the following system. The system response requirements are tr ≤ 0.6 sec, Mp ≤ 10%, and ts ≤ 3 sec. 

➢From Example 2 we can get the acceptable regions for pole locations in s-plane is:

ξ≥√(𝑙𝑛𝑀𝑝)2𝜋2+(𝑙𝑛𝑀𝑝)2=0.6 

we can use the command ‘sgrid(zeta,wn)’ to plot constant damping and frequency lines for the damping ratios in the vector zeta and the natural frequencies in the vector wn. Enter the following in the MATLAB command window:

In this example, the pole locations should be in between = 0.6 lines ( > 0.6), and outside the =3 circle ( > 3.). 

➢We can then use the command ‘[k,poles] = rlocfind(sys)’ to find root locus gains for agiven set of roots. Select the poles by clicking on the root locus plot in your acceptableregion. The MATLAB code is:

Then we can get the proportional controller gain k and system poles: 

➢Verify the closed-loop response:

From the figure below, we can see that the step response meets all the repuirements (tr ≤ 0.6sec, Mp ≤ 10%, and ts ≤ 3 sec).

Exercise 3: Design a proportional controller for the system in Example 3. The system response requirements are tr ≤ 0.6 sec, Mp ≤ 10%, and ts ≤ 3 sec.

Exercise 4: Design a proportional controller for the system below. The system response requirements are tr ≤ 0.3 sec, Mp ≤ 15%, and ts ≤ 1.5 sec. Verify the response with your controller. 

Exercise 5: Design a proportional controller for the system below. The system response requirements are tr ≤ 0.9 sec, Mp ≤ 20%, and ts ≤ 4 sec. Verify the response with your controller. 

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Exercise 1: Plot (1) the block diagram of this example, (2) results from two scopes, (3) results from workspace using standard MATLAB commands 

Exercise 2: Use Time block, Transfer Fcn block, a Sum block, Scope, and To Workspace block to create a Simulink model of this closed loop system. Choose proper stop time for the simulation. 

(1) Plot the block diagram of the system. 

(2) Show the step response of the system (Output y) from both scope and workspace 

Exercise 3: Add two more second order closed-loop systems (withand) in the same Simulink file. Put three outputs y, y1, and y2 together in one scope. 0.4  = 1.2  =

(1) Plot the block diagram of the simulation file. 

(2) Plot the outputs from both scope and workspace 

Exercise 4: Consider the feedback system shown in the figure below. 

Exercise 5: Consider the feedback system shown in the figure below. The variables ‘K’, ‘a’ and ‘b’ are k=10.8e8, a=1, b=8 and J=10.8e8 

complete solution solved here along with matlab codes

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