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Step2247 [10]
3 years ago
10

Design a PI controller to improve the steady-state error. The system should operate with a damping ratio of 0.8. Compute the ove

rall value of gain for the PI system and state the transfer function for the controller in the form G_c(s) = K_PI(s + a)/s.
Engineering
1 answer:
blondinia [14]3 years ago
5 0

Answer:

The MATLAB Code for this PI Controller will be:

Kp = 350;

Ki = 300;

Kd = 50;

C = pid(Kp,Ki,Kd)

T = feedback(C*P,1);

t = 0:0.01:2;

step(T,t)

Explanation:

When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response.

Obtain an open-loop response and determine what needs to be improved

Add a proportional control to improve the rise time

Add a derivative control to reduce the overshoot

Add an integral control to reduce the steady-state error

Adjust each of the gains $K_p$, $K_i$, and $K_d$ until you obtain a desired overall response.

The further explanation is attached in the Word File.

Download docx
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Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

pi=3.142

N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

T=1527.689Nm

Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

Where

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d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

d=0.04808m or 48.1mm approx 50mm

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Consider a half-wave rectifier circuit with a triangular-wave input of LaTeX: 6V6 V peak-to-peak amplitude and zero average valu
Varvara68 [4.7K]

The question is not complete. We are supposed to find the average value of v_o.

Answer:

v_o,avg = 0.441V

Explanation:

Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;

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So, 12/T = 0.7/t1

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Also, from symmetry of triangles,

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t2 = 0.4417T

Average of voltage output is;

v_o,avg = (1/T) x Area under small triangle

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T will cancel out to give;

v_o,avg = 0.441V

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Answer:

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