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masha68 [24]
2 years ago
15

The velocity of a particle which moves along the s-axis is given by = 40 − 3 2/ , ℎ t is in seconds. Calculate the displacement

∆ of the particle during the interval from t = 2sec to 4sec.
Engineering
1 answer:
scoundrel [369]2 years ago
8 0

The displacement ∆S of the particle during the interval from t = 2sec to 4sec is; 210 sec

<h3>How to find the displacement?</h3>

We are given the velocity equation as;

s' = 40 - 3t²

Thus, the speed equation will be gotten by integration of the velocity equation to get;

s = ∫40 - 3t²

s = 40t - ¹/₂t³

Thus, the displacement between times of t = 2 sec and t = 4 sec is;

∆S = [40(4) - ¹/₂(4)³] - [40(2) - ¹/₂(2)³]

∆S = 210 m

Read more about Displacement at; brainly.com/question/4931057

#SPJ1

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Using the characteristics equation, determine the dynamic behavior of a PI controller with τI = 4 applied to a second order proc
Sladkaya [172]

Answer:

The values of Kc that render this closed-loop process unstable are in the interval

(Kc < 0)

Explanation:

The transfer function of a PI controller is given as

Gc = Kc {1 + (1/sτI)}

τI = 4

Gc = Kc {1 + (1/4s)}

Gc = Kc {(4s+1)/(4s)}

Divide numerator and denominator by 4

Gc = Kc {(s+0.25)/(s)}

For a second order process, the general transfer function is given by

Gp = Kp {1/(τn²s² + 2ζτns + 1)}

Kp = 2, τn = 5 and ζ = 1.5

Gp = 2/(25s² + 15s + 1)

Divide numerator and denominator by 25

Gp = 0.08/(s² + 0.6s + 0.04)

Ga = 1

Gs = 1

We need to find the value(s) of Kc that makes the closed loop transfer function unstable. Gp*Ga*Gc*Gs + 1 = 0

The closed loop transfer function is unstable when the solution(s) of the characteristic equation obtained is positive.

Gp*Ga*Gc*Gs + 1 = 0

Becomes

[0.08/(s² + 0.6s + 0.04)] × [Kc (s+0.25)/(s)] + 1 = 0

[0.08Kc (s + 0.25)/(s³ + 0.6s² + 0.04s)] = - 1

0.08Kc (s + 0.25) = -s³ - 0.6s² - 0.04s

0.08Kc s + 0.02Kc = -s³ - 0.6s² - 0.04s

s³ + 0.6s² + 0.04s + 0.08Kc s + 0.02Kc = 0

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

We will use the direct substitution method to evaluate the values of Kc that matter. The values of Kc at the turning points of the closed loop transfer function.

For the substitution,

We put s = jw into the equation. (frequency analysis)

Note that j = √(-1)

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

(jw)³ + 0.6(jw)² + (0.08Kc + 0.04)(jw) + 0.02Kc = 0

-jw³ - 0.6w² + (0.08Kc + 0.04)(jw) + 0.02Kc = 0

we then collect terms with j and terms without.

(0.08Kcw + 0.04w - w³)j + (0.02Kc - 0.6w²) = 0

Meaning,

0.08Kcw + 0.04w - w³ = 0 (eqn 1)

0.02Kc - 0.6w² = 0 (eqn 2)

0.02 Kc = 0.6 w²

Kc = 15w²

Substituting this into eqn 1

0.08Kcw + 0.04w - w³ = 0

Kc = 15w²

0.08(15w²)w + 0.04w - w³ = 0

1.2w³ + 0.04w - w³ = 0

0.2w³ + 0.04w = 0

w = 0 or 0.2w² + 0.04 = 0

0.2w² = -0.04

w² = -0.2

w = ± √(-0.2)

w = ± 0.4472j or w = 0

Recall, Kc = 15w² = 15(-0.2) = -3 or Kc = 0

The turning points for the curve of the closed loop transfer function occur when

Kc = 0 or Kc = -3

To investigate, we pick values around these turning points to investigate the behaviour of the closed loop transfer function at those points.

Kc < -3, Kc = -3, (-3 < Kc < 0), Kc = 0 and Kc > 0

Note that, one positive characteristic root or pole is enough to make the system unstable.

We pick a value for Kc in that interval and evaluate the closed loop transfer function.

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

- First of, let Kc = - 4 (Kc < -3)

s³ + 0.6s² - 0.28s - 0.08 = 0

Solving the polynomial

s = (-0.22002), 0.44223, (-0.82221)

One positive pole means the closed loop transfer function is unstable in this region

Let Kc = -3

s³ + 0.6s² - 0.20s - 0.06 = 0

s = 0.37183, (-0.21251) or (-0.75933)

One positive pole still means that the closed loop transfer function is still unstable.

Then the next interval

Let Kc = -1

s³ + 0.6s² - 0.04s - 0.02 = 0

Solving this polynomial,

s = 0.18686, (-0.1749) or (-0.61196)

The function is unstable in the region being investigated.

Let Kc = 0

s³ + 0.6s² + 0.04s = 0

s = 0, -0.0769, -0.5236

One zero, all negative roots, indicate that the closed loop transfer function is marginally stable at this point.

Let Kc = 1, Kc > 0

s³ + 0.6s² + 0.12s + 0.02 = 0

s = (-0.42894), (-0.08553 + 0.1983j) or (-0.08553 - 0.1983j)

All the real negative parts of the poles are all negative, this indicates stability.

Hence, after examining the turning points of the closed loop transfer function, it is evident that, the region's of Kc where the closed loop transfer function is unstable is (Kc < 0)

Hope this Helps!!!

8 0
3 years ago
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