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Usimov [2.4K]
3 years ago
11

Consider the system whose transfer function is given by: 1()(21)(3)Gsss=++. (a)Use the root-locus design methodology to design a

lead compensator that will provide a closed-loop damping ζ =0.4and a natural frequency ωn=9 rad/sec.The general transfer function for lead compensation is given by(b)Use MATLAB to plot the root locus of the feed-forward transfer function, D(s)*G(s), and verify if the closed loop system pole is located on a root locus for the calculated value of K
Engineering
1 answer:
Hitman42 [59]3 years ago
3 0

Answer:

transfer function from R(s) to E(s) and determine the steady-state error (ess) for a unit-step reference input (c) Select the system parameters (k, kP, kI) such that the closed-loop system has damping coefficient ζ = 0.707 Figure Control system diagram.

Explanation:

hope this helps

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Len [333]

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4 0
3 years ago
What are the rights of people bicycling, under Illinois state law? *
zhuklara [117]

Answer:

○ People biking have the same right to be on the road as motorists, except for roads (such as expressways) where signs prohibit bikes​

Explanation:

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8 0
3 years ago
Consider the following chain-reaction mechanism for the high-temperatureformation of nitric oxide, i.e., the Zeldovich mechanism
vagabundo [1.1K]

Answer: hello attached below is the properly written chain reaction  to your question

answer :

d[NO] / dt = k_{1f} [O] [N_{2}] + K_{2f}  [N][O_{2} ] + K_{3f}[N][OH]

d[N] / dt = k_{1f} [O] [N_{2}] + K_{2f}  [N][O_{2} ] - K_{3f}[N][OH]

Explanation:

<u>write out expressions for  d[NO] / dt and d[N] / dt</u>

Given :

properly written chain reaction ( attached below)

Expression for d[NO] / dt can be written as

k_{1f} [O] [N_{2}] + K_{2f}  [N][O_{2} ] + K_{3f}[N][OH]

Expression for d[N] / dt can be written as

k_{1f} [O] [N_{2}] + K_{2f}  [N][O_{2} ] - K_{3f}[N][OH]

8 0
3 years ago
A capillary tube is immersed vertically in a water container. Knowing that water starts to evaporate when the pressure drops bel
Anuta_ua [19.1K]

Answer:

d=0.414\times 10^{-4}\ m

Explanation:

Given that

P = 4 KPa

Contact angle = 6°

Surface tension = 1 N/m

Lets assume that atmospheric pressure = 100 KPa

Lets take that density of water =1000\ kg/m^3

So the capillarity rise h

h=\dfrac{\Delta P}{\rho g}

h=\dfrac{100\times 1000-4\times 1000}{1000\times 10}

h= 9.61 m

We know that for capillarity rise h

h=\dfrac{2\sigma cos\theta }{r\rho g}

r=\dfrac{2\sigma cos\theta }{h\rho g}

r=\dfrac{2\times 1 cos4^{\circ} }{9.61\times 1000\times 10}

r=0.207\times 10^{-4}\ m

d=0.414\times 10^{-4}\ m

3 0
3 years ago
Consider a solid round elastic bar with constant shear modulus, G, and cross-sectional area, A. The bar is built-in at both ends
Ierofanga [76]

Answer:

\t(x)_{max} =\dfrac{p\times L}{2\times \pi}

Explanation:

Given that

Shear modulus= G

Sectional area = A

Torsional load,

t(x) = p sin( \frac{2\pi}{ L} x)

For the maximum value of internal torque

\dfrac{dt(x)}{dx}=0

Therefore

\dfrac{dt(x)}{dx} = p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}\\ p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}=0\\cos( \frac{2\pi}{ L} x)=0\\ \dfrac{2\pi}{ L} x=\dfrac{\pi}{2}\\\\x=\dfrac{L}{4}

Thus the maximum internal torque will be at x= 0.25 L

t(x)_{max} = \int_{0}^{0.25L}p sin( \frac{2\pi}{ L} x)dx\\t(x)_{max} =\left [p\times \dfrac{-cos( \frac{2\pi}{ L} x)}{\frac{2\pi}{ L}}  \right ]_0^{0.25L}\\t(x)_{max} =\dfrac{p\times L}{2\times \pi}

6 0
3 years ago
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