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love history [14]
3 years ago
13

A signal is sampled at 10000 Hz and a total of 1024 data samples are taken, what are the highest and lowest frequencies that can

be resolved by discrete Fourier analysis. What is the time period T?
Physics
1 answer:
tia_tia [17]3 years ago
7 0

Since the discrete Fourier series, the Sampling rate, would be the equivalent of the inverse of the passage of time, that is, to the frequency, mathematically this can be written as,

\frac{1}{\Delta t} = 10000Hz

In turn, the time can be described depending on the period and the amount of data samples taken. This would be,

\Delta t = \frac{T}{m}

Here,

m = Data Samples

T = Period

Replacing,

\Delta t = \frac{T}{1024}

Replacing the value of the time from the first equation,

\frac{1024}{T} = 10000

T = 102.4ms

At the same time, the range then will be given between the basic frequency to the half of the sample, that is,

f_{min} = \frac{1}{T} = 9.165Hz

f_{max} = \frac{1024}{2} (9.165Hz)

f_{max} = 5000Hz

Therefore the lowest frequency is 5000Hz and highest 9.165Hz

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Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The f
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Answer:

The Entropy generated by the steam = 2.821 kJ/K

Explanation:

Total volume of container = 5m³

Heat transfer does not exist between system and surrounding, dQ = 0

At the first chamber, temperature of water at saturated liquid is 300°C

From the steam table:

Specific enthalpy of saturated liquid at 300°C , h_{f} = 1344.8 kJ/kg

Specific internal energy of saturated liquid at 300°C, U_{f1} =  1332.7 kJ/kg

For closed system, the first law of thermodynamics state that:

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work done for free expansion, dw =0

0 = 0 + dU

dU = 0 , i.e. U₁ = U₂

At the second chamber,

The final pressure, P₂ = 50 kPa

From the steam table, at P₂ = 50 kPa,  U_{f2} = 340.49 kJ/kg

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Let the dryness fraction at the second chamber = x

U_{2} = U_{f2} + U_{fg2}

U_{2} = 340.49 + x2140.7Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is, u_{f2} = 0.00103 m^{3} /kg\\

u_{2} = u_{f2} + xu_{fg2}

u_{2} = 0.00103 + 0.463(3.2393)\\u_{2} = 1.5 m^{3} /kg\\

u_{2} = \frac{v_{2} }{m_{2} }

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C S_{1} = S_{f} = 3.2548 kJ/kg-k\\

S_{2} = S_{f2} + xS_{fg2}

From the steam table,

S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

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Entropy = 2.821 kJ/K

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Explanation:

Comment

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Same with this question.

You should just develop a rule.  The rule will look like this

The greater the heat capacity the (higher or lower) the change in temperature.

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That's not your question. You want to know which substance will have the greatest temperature change given their heat capacities.

Answer

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