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OLEGan [10]
3 years ago
7

Let f(t) be an arbitrary signal with bandwidth Ω. Determine the minimum sampling frequencies ωs needed to sample the following a

nalog signals without causing aliasing error. (a) f1(t) = f(t) sin(4000πt) (b) f2(t) = f(t) ∗ sin(4000πt) (c) f3(t) = f(t) ∗ f(sample the following analog signals without causing aliasing error

Engineering
1 answer:
disa [49]3 years ago
7 0

Answer:

See explaination

Explanation:

We can describr Aliasing as a false frequency which one get when ones sampling rate is less than twice the frequency of your measured signal.

please check attachment for the step by step solution of the given problem.

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The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp
Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1

The Lower Surface Cp is given as

Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1

The difference of the Cp over the airfoil is given as

\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32

Now the Lift Coefficient is given as

C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192

Now the coefficient of moment about the leading edge is given as

C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0
3 years ago
Air modeled as an ideal gas enters a well-insulated diffuser operating at steady state at 270 K with a velocity of 180 m/s and e
ZanzabumX [31]

Answer:

exit temperature 285 K

Explanation:

given data

temperature T1 = 270 K

velocity = 180 m/s

exit velocity =  48.4 m/s

solution

we know here diffuser is insulated so here heat energy is negleted

so we write here energy balance equation that is

0 = m (h1-h2) + m ×  (\frac{v1^2}{2}-\frac{v2^2}{2})   .....................1

so it will be

h1 + \frac{v1^2}{2} = h2 + \frac{v2^2}{2}      .....................2

put here value by using ideal gas table

and here for temperature 270K

h1 = 270.11 kJ/kg

270.11 + \frac{180^2\times \frac{1}{1000}}{2} = h2 + \frac{48.4^2\times \frac{1}{1000}}{2}  

solve it we get

h2 = 285.14 kJ/kg

so by the ideal gas table we get

T2 = 285 K

4 0
3 years ago
B)<br>State the essential difference between a plain carbon steel<br>and an alloy steel​
choli [55]

Answer:

Plain carbon steel has no or trace external elements while alloy steel has high amount of other elements.

Explanation:

Plain carbon steel has no or trace amount of other elements while alloy steel has high amount of other elements in their composition.

The presence of other elements in alloy steel improvise several physical properties of the steel while plain carbon steel has the basic properties.

6 0
3 years ago
The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-k
sammy [17]

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

(s_max)² = 0.7046

s_max = 0.8394m

8 0
3 years ago
What is the never repeat rule
Soloha48 [4]
Don't repeat yourself (DRY, or sometimes do not repeat yourself) is a principle of software development aimed at reducing repetition of software patterns,[1] replacing it with abstractions or using data normalization to avoid redundancy.
4 0
3 years ago
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