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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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A 650-kN column load is supported on a 1.5 m square, 0.5 m deep spread footing. The soil below is a well-graded, normally consol
insens350 [35]

<u>Explanation:</u>

Determine the weight of footing

W_{f}=\gamma(L)(B)(D)

Where W_{f} is the weight of footing, γ is the unit weight of concrete,  L is the length of footing is the width of footing, and D is the depth of footing

Substitute 2 m \text { for } L, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 23.6 kN / m ^{3} for γ in the equation

\begin{aligned}W_{f} &=\left(23.6 kN / m ^{3}\right)(2 m )(1.5 m )(0.5 m ) \\&=35.4 kN\end{aligned}

Therefore, the weight of the footing is 35.4 kN

Determine the initial vertical effective stress.

\sigma_{z p}^{\prime}=\gamma(D+B)-u

Here,   \sigma_{z^{p}}^{\prime} is initial vertical stress at a depth below ground surface  γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute 18 kN / m ^{3} \text { for } \gamma, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 0 for u in the equation

\begin{aligned}\sigma_{z p}^{\prime} &=\left(18 kN / m ^{3}\right)(1.5+0.5) m -0 \\&=36 kPa\end{aligned}

Therefore, the initial vertical stress is 36 kPa

Determine the vertical effective stress.

\sigma_{z D}^{\prime}=\gamma D

Here,   \sigma_{z^{p}}^{\prime} is initial vertical stress at a depth below ground surface  γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute \(18 kN / m ^{3}\) for \(\gamma, 0.5 m\) for \(D\) and 0 for \(u\) in the equation.

\begin{aligned}\sigma_{z b}^{\prime} &=\left(18 kN / m ^{3}\right)(0.5 m )-0 \\&=9 kPa\end{aligned}

Therefore, the vertical stress at a depth below the ground surface is

9 kPa

Determine the influence factor at the midpoint of soil layer,

I_{e p}=0.5+0.1 \sqrt{\frac{q-\sigma_{s 0}^{\prime}}{\sigma_{z p}^{\prime}}}

Here I_{e p} is the influence factor at the midpoint of soil layer  \sigma_{z^{p}}^{\prime} is initial vertical stress, \sigma_{z^{p}}^{\prime} is vertical effective stress, and Q is bearing pressure

Substitute 36 kPa for \(\sigma_{z p}^{\prime}, 228.47\) kPa for \(q,\) and 9 kPa for \(\sigma_{z D}^{\prime}\) in the equation\begin{aligned}I_{\epsilon P} &=0.5+0.1 \sqrt{\frac{228.47 kPa -9 kPa }{36 kPa }} \\&=0.75\end{aligned}

Therefore the influence factor at the midpoint of the soil layer is 0.693

6 0
3 years ago
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Answer:

118_{10}= 0110111 2) -49_{10}=110001_{2} 3) 0_{10}=0:16 \Rightarrow 0_{10}=0_{16}

Explanation:

1) Expressing the Division as the summation of the quotient and the remainder

for

118, knowing it is originally a decimal form:

118:2=59  +(0), 59/2 =29 + 1, 29/2=14+1, 14/2=7+0, 7/2=3+1, 3/2=1+1, 1/2=0+1

118_{10}= 0110111

2) -49_{10}

Similarly, we'll start the process with the absolute value of -49 since we want the positive value of it. Then let's start the successive divisions till zero.

|-49|=49

49:2=24+1, 24:2=12+0,12:2=6+0,6:2=3+0,3:2=1+1,1:2=0+1

100011

-49_{10}=110001_{2}

3) (-0)_{10}

The first step on that is dividing by 16, and then dividing their quotient again by 16, so on and adding their remainders. Simply put:

(-0)_{10}=0:16=0 \Rightarrow (0)_{10}=0_{16} \:or\\(0)_{16}=0000000000000000

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