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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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What will happen in a wire drawing operation when the cross-sectional area has a reduction of 60% in a single pass?
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Answer:

DRAWING LOAD IS  3.67 A_{O}\sigma

Explanation:

wire drawing is a method of obtaining wire of bigger large diameter from iron rod . it is cold process which need die to obtain wire

drawing load for wire drawing is given as P = A_{F}*\sigma*ln(\frac{A_{O}}{A_{F}})

Where A f is initial area, Ao is original area, σ is yield stress

as given in question sectional area reduce 60%, therefore

A_{f} = A_{O}- 0.6A_{O}

    = 0.4 A_{O}

Due to change in area ,drawing load p is

p = 0.4A_{O}*\sigma*ln(\frac{A_{O}}{0.4A_{O}})

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A 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves throu
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Answer:

The temperature of the strip as it exits the furnace is 819.15 °C

Explanation:

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L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m

The Biot number is given as;

B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952

B_i < 0.1,  thus apply lumped system approximation to determine the constant time for the process;

\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s

The time for the heating process is given as;

t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

T(t) = T_{ \infty} + (T_i -T_{\infty})e^{-t/ \tau}\\\\T(t) = 930 + (20 -930)e^{-300/ 142.5}\\\\T(t) = 930 + (-110.85)\\\\T_{(t)} = 819.15 \ ^0 C

Therefore, the temperature of the strip as it exits the furnace is 819.15 °C

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