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3 years ago

5

Consider a C.T. system in s-plane below. Draw DF1 (Direct Form 1) realization. (by hand) Perform system realization using MATLAB

Simulink. (Use same parameters as [Q 11] for "Step Function" and "Delay Block".) Use the "step function", "delay", and "summer" to build the input, X(s). Use a "scope" as the output, Y(s). Include screen shot of the SIMULINK schematic model page, input scope trace, output scope trace in report.Discuss the results in your own words. s +5 H(S) = 52 + 11s + 10

1 answer:

spin [16.1K]3 years ago

5 0

Answer:

See the attached file for the answer.

Explanation:

See the attached file for the explanation

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Mining is an example of this type of business

luda_lava [24]

Answer:

Mining would go under Industry organization.

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4 years ago

4. When the ESC system senses oversteer,

bagirrra123 [75]

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2 years ago

Please help me with this question

Lera25 [3.4K]

Answer:

4.2

Explanation:

The "allowance" is the difference between the maximum shaft size and the minimum hole size.

maximum shaft = C +D = 8.5 +0.1 = 8.6

minimum hole = A -B = 4.8 -0.4 = 4.4

Allowance = 8.6 -4.4 = 4.2

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For an interference fit, the allowance is negative.

7 0

3 years ago

A piston-cylinder device contains 1.329 kg of nitrogen gas at 120 kPa and 27 degree C. The gas is now compressed slowly in a pol

Shtirlitz [24]

Answer:-0.4199 J/k

Explanation:

Given data

mass of nitrogen(m)=1.329 Kg

Initial pressure $\left ( P_1\right )$ =120KPa

Initial temperature $\left ( T_1\right )=27\degree \approx$ 300k

Final volume is half of initial

R=particular gas constant

Therefore initial volume of gas is given by

PV=mRT

V=0.986\times 10^{-3}

Using $PV^{1.49}$ =constant

$P_{1}V^{1.49}$ = $P_2\left (\frac{V}{2}\right )$

$P_2$ =337.066KPa

$V_2$ = $0.493\times 10^{-3} m^{3}$

and entropy is given by

$\Delta s$ = $C_v \ln \left (\frac{P_2}{P_1}\right )$ + $C_p \ln \left (\frac{V_2}{V_1}\right )$

Where, $C_v$ = $\frac{R}{\gamma-1}$ =0.6059

$C_p$ = $\frac{\gamma R}{\gamma -1}$ =0.9027

Substituting values we get

$\Delta s$ = $0.6059\times\ln \left (\frac{337.066}{120}\right )$ + $0.9027 \ln \left (\frac{1}{2}\right )$

$\Delta s$ =-0.4199 J/k

4 0

3 years ago

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude

Fofino [41]

Answer:

$u = 260.22m/s$

$S_{max} = 1141.07ft$

Explanation:

Given

$S_0 = 89.6ft$ --- Initial altitude

$S_{16.5} = 0ft$ -- Altitude after 16.5 seconds

$a = -g = -32.2ft/s^2$ --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

$S = ut + \frac{1}{2}at^2$

The final altitude after 16.5 seconds is represented as:

$S_{16.5} = S_0 + ut + \frac{1}{2}at^2$

Substitute the following values:

$S_0 = 89.6ft$ $S_{16.5} = 0ft$ $a = -g = -32.2ft/s^2$ and $t = 16.5$

So, we have:

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2$

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45$

$0 = 89.6 + 16.5u- 4383.225$

Collect Like Terms

$16.5u = -89.6 +4383.225$

$16.5u = 4293.625$

Make u the subject

$u = \frac{4293.625}{16.5}$

$u = 260.21969697$

$u = 260.22m/s$

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

$v=u + at$

At the maximum height:

$v =0$ --- The final velocity

$u = 260.22m/s$

$a = -g = -32.2ft/s^2$

So, we have:

$0 = 260.22 - 32.2t$

Collect Like Terms

$32.2t = 260.22$

Make t the subject

$t = \frac{260.22}{ 32.2}$

$t = 8.08s$

The maximum height is then calculated as:

$S_{max} = S_0 + ut + \frac{1}{2}at^2$

This gives:

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2$

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22$

$S_{max} = 89.6 + 260.22 * 8.08 - 1051.11$

$S_{max} = 1141.0676$

$S_{max} = 1141.07ft$

Hence, the maximum height is 1141.07ft

8 0

3 years ago