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e-lub [12.9K]
3 years ago
14

For the unity negative feedback system G(s) = K(s+6)/ (s + 1)(s + 2)(s + 5) It's known that the system is operating with a domin

ant-pole damping ratio of 0.5. Design a PD controller so that the settling time is reduced by a factor of 3, compared with the uncompensated unity negative feedback system. Compare the percentage overshoot and resonant frequencies of the uncompensated and compensated systems.
Engineering
1 answer:
Ad libitum [116K]3 years ago
4 0

Answer:The awnser is 5

Explanation:Just divide all of it

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If a wooden handle hammer becomes loose, what should you do?
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Like the person above said buy a new handle or buy a new hammer. Safety precaution
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3 years ago
Carbon dioxide used as a natural refrigerant flows through a cooler at 10 MPa, which is supercritical, so no condensation occurs
kupik [55]

Answer:

The answer which is a calculation can be found as an attached document

Explanation:

5 0
3 years ago
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Steam at 500 bar and 500°C undergoes a throttling expansion to 1 bar. What will be the temperature of the steam after the expans
aleksandr82 [10.1K]

T1=T2=500°C

<u>Explanation:</u>

Given-

Pressure, P1 = 500 bar

Temperature, T1 = 500°C

P2 = 1 bar

T2 after expansion, = ?

We know,

P1/T1 = P2/T2

500/ 500 = 1/T2

T2 = 1°C

If the steam were replaced by an ideal gas, since enthalpy of ideal gas is a function of temperature only, we easily obtain T2 = T1 = 500°C

8 0
3 years ago
(a) The reverse-saturation current of a pn junction diode is IS = 10−11 A. Determine the diode voltage to produce currents of (i
kirill115 [55]

Answer:

The equation used to solve a diode is

i_d = I_se^\frac{V_d}{V_T}-1

  • i_d is the current going through the diode
  • I_s is your saturation current
  • V_D is the voltage across your diode
  • V_T is the voltage of the diode at a certain room temperature. by default, you always use V_T=25.9mV for room temperature.

If you look at the equation, i_d = I_se^\frac{V_d}{V_T}-1, you'd notice that the e^\frac{V_d}{V_T} grow exponentially fast, so we can ignore the -1 in the equation because it's so small compared to the exponential.

i_d = I_se^\frac{V_d}{V_T}-1

i_d\approx I_se^\frac{V_d}{V_T}

Therefore, use i_d= I_se^\frac{V_d}{V_T} to solve your equation.

Rearrange your equation to solve for V_D.

V_D=V_Tln(\frac{i_D}{I_s})

a.)

i.)

You're given I_s=10^{-11}A

at i_d=10\mu A,     V_D=V_Tln(\frac{i_D}{I_s})=(25.9\cdot10^{-3})ln(\frac{10\cdot10^{-6}}{10\cdot10^{-11}})=.298V

at i_d=100\mu A,   V_D=V_Tln(\frac{i_D}{I_s})=(25.9\cdot10^{-3})ln(\frac{100\cdot10^{-6}}{10\cdot10^{-11}})=.358V

at i_d=1mA,      V_D=V_Tln(\frac{i_D}{I_s})=(25.9\cdot10^{-3})ln(\frac{1\cdot10^{-3}}{10\cdot10^{-11}})=.417V

<em>note: always use</em>  V_T=25.9mV

ii.)

Just repeat part (i) but change to I_s=-5\cdot10^{-12}A

b.)

same process as part A. You do the rest of the problem by yourself.

4 0
4 years ago
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