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3 years ago

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(TCO 4) A system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without

further modification. Determine the frequency that the sampling system will generate in its output.

1 answer:

steposvetlana [31]3 years ago

5 0

Answer:

The frequency that the sampling system will generate in its output is 70 Hz

Explanation:

Given;

F = 190 Hz

Fs = 120 Hz

Output Frequency = F - nFs

When n = 1

Output Frequency = 190 - 120 = 70 Hz

Therefore, if a system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification, the frequency that the sampling system will generate in its output is 70 Hz

You might be interested in

What impact does modulus elasticity have on the structural behavior of a mechanical design?

devlian [24]

Answer with Explanation:

The modulus of elasticity has an profound effect on the mechanical design of any machine part as explained below:

1) Effect on the stiffness of the member: The ability of any member of a machine to resist any force depends on the stiffness of the member. For a member with large modulus of elasticity the stiffness is more and hence in cases when the member has to resist a direct load the member with more modulus of elasticity resists the force better.

2)Effect on the deflection of the member: The deflection caused by a force in a member is inversely proportional to the modulus of elasticity of the member thus in machine parts in which we need to resist the deflections caused by the load we can use materials with greater modulus of elasticity.

3) Effect to resistance of shear and torque: Modulus of rigidity of a material is found to be larger if the modulus of elasticity of the material is more hence for a material with larger modulus of elasticity the resistance it offer's to shear forces and the torques is more.

While designing a machine element since the above factors are important to consider thus we conclude that modulus of elasticity has a profound impact on machine design.

8 0

3 years ago

Find the power and the rms value of the following signal square: x(t) = 10 sin(10t) sin(15t)

ArbitrLikvidat [17]

Answer:

$\mathbf{P_x =25 \ watts}$

$\mathbf{x_{rmx} = 5 \ unit}$

Explanation:

Given that:

x(t) = 10 sin(10t) . sin (15t)

the objective is to find the power and the rms value of the following signal square.

Recall that:

sin (A + B) + sin(A - B) = 2 sin A.cos B

x(t) = 10 sin(15t) . cos (10t)

x(t) = 5(2 sin (15t). cos (10t))

x(t) = 5 × ( sin (15t + 10t) + sin (15t-10t)

x(t) = 5sin(25 t) + 5 sin (5t)

From the knowledge of sinusoidial signal Asin (ωt), Power can be expressed as:

$P= \dfrac{A^2}{2}$

For the number of sinosoidial signals;

Power can be expressed as:

$P = \dfrac{A_1^2}{2}+ \dfrac{A_2^2}{2}+ \dfrac{A_3^2}{2}+ ...$

As such,

For x(t), Power $P_x = \dfrac{5^2}{2}+ \dfrac{5^2}{2}$

$P_x = \dfrac{25}{2}+ \dfrac{25}{2}$

$P_x = \dfrac{50}{2}$

$\mathbf{P_x =25 \ watts}$

For the number of sinosoidial signals;

$RMS = \sqrt{(\dfrac{A_1}{\sqrt{2}})^2+(\dfrac{A_2}{\sqrt{2}})^2+(\dfrac{A_3}{\sqrt{2}})^2+...$

For x(t), the RMS value is as follows:

$x_{rmx} =\sqrt{(\dfrac{5}{\sqrt{2}} )^2 +(\dfrac{5}{\sqrt{2}} )^2 }$

$x_{rmx }=\sqrt{(\dfrac{25}{2} ) +(\dfrac{25}{2} ) }$

$x_{rmx }=\sqrt{(\dfrac{50}{2} )}$

$x_{rmx} =\sqrt{25}$

$\mathbf{x_{rmx} = 5 \ unit}$

8 0

3 years ago

Estimate the rotor inertia assuming that the rotor is a cylinder of radius 8.98 mm, and length 25 mm, with a material of 100% co

Softa [21]

Answer:

The moment of inertia of the rotor is approximately $1.105\times 10^{-6}$ kilogram-square meters.

The rotor inertia may differ from these assumption due to differences in the shape of cross section.

Explanation:

We assume that rotor can be represented as a solid cylinder of radius $r$ , length $l$ , made of cooper ( $\rho = 8960\,\frac{kg}{m^{3}}$ ) and whose axis of rotation passes through its center of mass and is parallel to its cross section. By definition of Moment of Inertia and Theorem of Parallel Axes, the moment of inertia of the rotot is:

$I = \frac{1}{4}\cdot \rho \cdot \left(\frac{\pi}{4} \right) \cdot R^{3}\cdot (3\cdot R^{2}+L^{2})$

$I = \frac{\pi}{16}\cdot \rho \cdot R^{3}\cdot (3\cdot R^{2}+L^{2})$ (Eq. 1)

Where:

$\rho$ - Density of copper, measured in kilograms per cubic meter.

$R$ - Radius of the rotor, measured in meters.

$L$ - Length of the rotor, measured in meters.

$I$ - Moment of inertia, measured in kilogram-square meters.

If we know that $\rho = 8960\,\frac{kg}{m^{3}}$ , $L = 25\times 10^{-3}\,m$ and $R = 8.98\times 10^{-3}\,m$ , the estimated moment of inertia of the rotor is:

$I = \frac{\pi}{16}\cdot \left(8960\,\frac{kg}{m^{3}} \right)\cdot (8.98\times 10^{-3}\,m)^{3}\cdot [3\cdot (8.98\times 10^{-3}\,m)^{2}+(25\times 10^{-3}\,m)^{2}]$

$I \approx 1.105\times 10^{-6}\,kg\cdot m^{2}$

The moment of inertia of the rotor is approximately $1.105\times 10^{-6}$ kilogram-square meters.

From D'Alembert's Formula we know that net force of rigid bodies experimenting rotation equals the product of moment of inertia and angular acceleration. In this case, the purpose is minimizing moment of inertia and it is done by modifying the shape of the cross section so that rotor could be aerodynamically more efficient.

5 0

3 years ago

Based on bonding theory, explain why coefficient of expansion increases when you consider ceramics, metals and polymers

Mrrafil [7]

Answer:

The ratio that a material expands in accordance with changes in temperature is called the coefficient of thermal expansion. Because Fine Ceramics possess low coefficients of thermal expansion, their distortion values, with respect to changes in temperature, are low. The coefficients of thermal expansion depend on the bond strength between the atoms that make up the materials.

Explanation:

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3 0

3 years ago

Consider the same piping system. this time, the same pipe is buried underground. assuming that there is a constant heat flux of

charle [14.2K]

Complete Question

Complete Question is attached below

Answer:

Option A

Explanation:

From the question we are told that:

inner Diameter of pipe $d_i=100^c$

Thickness $t=50mm$

Outer diameter of pipe $d_o=1.1m$

Length $l=5m$

Temperature $T_i=130^oC$

Generally the equation for Heat Balance is mathematically given by

$q*\pi d_oL=mC_p(T_i-T_o)$

Therefore

$T_o=T_i+\frac{q*\pi d_oL}{mC_p}$

$T_o=130+\frac{100*3.142 *1.1*5}{0.5*4000}$

$T_o=129.136^oC$

Therefore the exit temperature of the water.is $T_o=129.136^oC$

Option A

7 0

2 years ago