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3 years ago

Write down a transfer function of a stable system for which pure proportional feedback could drive the system unstable.

Engineering

1 answer:

gtnhenbr [62]3 years ago

6 0

Answer:

$Gs = 1 / (s + 1)^3$

Explanation:

A system is known to be a stable system if the poles of the system lie on the negative real axis of the s plane.

A system that could be driven unstable by a P-controller would be:

$Gs = 1 / (s + 1)^3$

As the value of K increases, the closed loop poles move into the right half s plane

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If a shear stress acts in one plane of an element, there must be an equal and opposite shear stress acting on a plane that is

xxMikexx [17]

Answer:

90 degrees

Explanation:

In the case when the sheer stress acts in the one plane of an element so it should be equal and opposite also the shear stress acted on a plan i.e. 90 degrees from the plane

Therefore as per the given situation it should be 90 degrees from the plane

hence, the same is to be considered and relevant too

5 0

2 years ago

Read 2 more answers

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Alex777 [14]

Answer:

Use a resume header

Explanation:

Create a Summary

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3 years ago

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2. The block is released from rest at the position shown, figure 1. The coefficient of

denis23 [38]

Answer:

Velocity = 4.73 m/s.

Explanation:

Work done by friction is;

W_f = frictional force × displacement

So; W_f = Ff * Δs = (μF_n)*Δs

where; magnitude of the normal force F_n is equal to the component of the weight perpendicular to the ramp i.e; F_n = mg*cos 24

Over the distance ab, Potential Energy change mgΔh transforms into a change in Kinetic energy and the work of friction, so;

mg(3 sin 24) = ΔKE1 + (0.22)*(mg cos 24) *(3).

Similarly, Over the distance bc, potential energy mg(2 sin 24) transforms to;

ΔKE2 + (0.16)(mg cos 24)(2).

Plugging in the relevant values, we have;

1.22mg = ΔKE1 + 0.603mg

ΔKE1 = 1.22mg - 0.603mg

ΔKE1 = 0.617mg

Also,

0.813mg = ΔKE2 + 0.292mg

ΔKE2 = 0.813mg - 0.292mg

ΔKE2 = 0.521mg

Now total increase in Kinetic Energy is ΔKE1 + ΔKE2

Thus,

Total increase in kinetic energy = 0.617mg + 0.521m = 1.138mg

Putting 9.81 for g to give;

Total increase in kinetic energy = 11.164m

Finally, if v = 0 m/s at point a, then at point c, KE = ½mv² = 11.164m

m cancels out to give; ½v² = 11.164

v² = 2 × 11.164

v² = 22.328

v = √22.328

v = 4.73 m/s.

5 0

3 years ago

What are some paradigms from history that have been proven inaccurate or incomplete

frozen [14]

That the world is flat. Brainliest?

4 0

2 years ago

Consider a Carnot cycle executed in a closed system with 0.0058 kg of air. The temperature limits of the cycle are300 K and 940

cricket20 [7]

Answer:0.646 KJ

Explanation:

Using First law for cycle

$\sum Q=\sum W$

$\sum Q=Q_{1-2}+Q_{3-4}$

For adiabatic process heat transfer is zero and for isothermal process

d(Q)=d(W)

$Q_{1-2}=mRT_1\ln {\frac{P_1}{P_2}}$

Given $P_1=2000KPa$

$P_3=20KPa$

$\left (\frac{T_2}{T_3}\right )^{\frac{\gamma }{\gamma -1}$ = $\left (\frac{P_2}{P_3}\right )}$

$P_2=1089.06K$

$Q_{1-2}=0.0058\dot 0.287\dot 940\ln \frac{2000}{1089.06}$ = $0.95KJ$

$Q_{3-4}=mRT_2\ln {\frac{P_3}{P_4}}$

$\left (\frac{T_1}{T_4}\right )^{\frac{\gamma }{\gamma -1}$ = $\left (\frac{P_1}{P_4}\right )}$

Now we have to find $P_4=36.72KPa$

$Q_{3-4}=0.0058\dot 0.287\dot 300\ln \frac{20}{36.72}$ = $-0.30341KJ$

$Q_{net}=Q_{1-2}+Q_{3-4}$

$Q_{net}=0.95-0.303=0.646KJ$

$Q_{net}=W_{net}=0.646KJ$

7 0

2 years ago