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stiv31 [10]
4 years ago
10

Complex poles cmd zeros. Sketch the asymptotes of the Bode plot magnitude and phase for each of the listed open-loop mmsfer fuoc

tions. and approximate the transitillll at the stX'ond-order break point. based on the value of the ... ,. ,, 410. The Frequency-Response Design Method damping ratio. After completing the hand sketches, verify your result using Mallab. Tum in your hand sketches and the Matlab results on the same scales.
(a) L(s) = s2 + 3~r + 10
(b) L(s) = s(s2 + ~s + 10)
(c) L(s) = (s2 + 2s + 8) s(s2 + 2s + 10)
(d) L(s) = (s2 +I) s(s• +4)
(e) L(s) = (s:? +4) s(.r~ + I) 6
Engineering
1 answer:
Roman55 [17]4 years ago
8 0

Answer: e is the best answer

Explanation:

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about edubrainly. it seems it takes the questions from this website, C&P them, and also try's to get you to have viruses. ca
Y_Kistochka [10]

Answer:

yes

Explanation:

6 0
3 years ago
A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls
Inessa05 [86]

Answer:

the percent increase in the velocity of air is 25.65%

Explanation:

Hello!

The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.

m1=m2

Now remember that mass flow is given by the product of density, cross-sectional area and velocity

(α1)(V1)(A1)=(α2)(V2)(A2)

where

α=density

V=velocity

A=area

Now we can assume that the input and output areas are equal

(α1)(V1)=(α2)(V2)

\frac{V2}{V1} =\frac{\alpha1 }{\alpha 2}

Now we can use the equation that defines the percentage of increase, in this case for speed

i=(\frac{V2}{V1} -1) 100

Now we use the equation obtained in the previous step, and replace values

i=(\frac{\alpha1 }{\alpha 2} -1) 100\\i=(\frac{1.2}{0.955} -1) 100=25.65

the percent increase in the velocity of air is 25.65%

6 0
3 years ago
What are the steps to execute an instruction by cpu?What is the function of DMA controller.
JulijaS [17]

Explanation:

1. A sequence of instructions is stored in memory.

2. The memory address wherever the first instruction is found is copied to the instruction pointer.

3. The CPU sends the address within the instruction pointer to memory on the address bus.

4. The CPU sends a “read” signal to the control bus.

5. Memory responds by sending a copy of the state of the bits at that memory location on the

data bus, that the CPU then copies into its instruction register.

6. The instruction pointer is automatically incremented to contain the address of the next

instruction in memory.

7. The CPU executes the instruction within the instruction register.

8. Go to step 3

Steps 3, 4, and 5 are called an instruction fetch. Notice that steps 3 – 8 constitute a cycle, the instruction execution cycle. It is shown graphically below.

A DMA controller can generate memory addresses and initiate memory read or write cycles. It contains several hardware registers that can be written and read by the CPU. These include a memory address register, a byte count register, and one or more control registers.

4 0
3 years ago
Determine the enthalpy, volume and density of 1.0 kg of steam at a pressure of 0.5 MN/m2 and with a dryness fraction of 0.96
Viktor [21]

Answer:

Enthalpy, hsteam = 2663.7 kJ/kg

Volume, Vsteam = 0.3598613 m^3 / kg

Density = 2.67 kg/ m^3

Explanation:

Mass of steam, m = 1 kg

Pressure of the steam, P = 0.5 MN/m^2

Dryness fraction, x = 0.96

At P = 0.5 MPa:

Tsat = 151.831°C

Vf = 0.00109255 m^3 / kg

Vg = 0.37481 m^3 / kg

hf = 640.09 kJ/kg

hg = 2748.1 kJ/kg

hfg = 2108 kJ/kg

The enthalpy can be given by the formula:

hsteam = hf + x * hfg

hsteam = 640.09 + ( 0.96 * 2108)

hsteam = 2663.7 kJ/kg

The volume of the steam can be given as:

Vsteam = Vf + x(Vg - Vf)

Vsteam = 0.00109255 + 0.96(0.37481 - 640.09)

Vsteam = 0.3598613 m^3 / kg

From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg/ m^3

5 0
3 years ago
This operating mode assists steering return after completing a turn.
castortr0y [4]

Answer:

In return mode, the system assists steering return after completing a turn. Information from the steering position sensor prevents the system from overshooting the center position. In dampener mode, the system acts like a hydraulic dampener to prevent kick back and bump steer.

3 0
3 years ago
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