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zhenek [66]
3 years ago
12

(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.
Engineering
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

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A heating system must maintain the interior of a building at 20°C during a period when the outside air temperature is 5°C and th
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a. W = 51,194.54 kJ

b. W = 102,390 kJ

c. W = 153,585 kJ

Explanation:

(COP)_{HP} =\frac{Desired-effectx}{Work-done}= \frac{Q_{1} }{W} \\\\(COP)_{HP} =(COP)_{Ideal}\\\\\frac{Q1}{W} =\frac{T_{1} }{T_{1} -T_{2} }

W=Q_{1} \frac{T_{1}-T_{2}  }{T_{1} }

a. the ground at 15°C.

T_{1}=20°C = 273 K + 20 = 293 K

T_{2}=15°C = 273 K + 15 = 288 K

Q_{1}=3x10^{6} kJ

W=3x10^{6} kJ \frac{293 K-288 K}{293 K}=3x10^{6} kJ \frac{5 K}{293 K}=3x10^{6} kJ x 0.017065}

W = 0.051195x10^{6} kJ

W = 51,194.54 kJ

b. a pond at 10°C.

T_{2}=10°C = 273 K + 10 = 283 K

W=3x10^{6} kJ \frac{293 K-283 K}{293 K}=3x10^{6} kJ \frac{10 K}{293 K}=3x10^{6} kJ x 0.034130}

W = 0.102390x10^{6} kJ

W = 102,390 kJ

c. the outside air at 5°C.

T_{2}=5°C = 273 K + 5 = 278 K

W=3x10^{6} kJ \frac{293 K-278 K}{293 K}=3x10^{6} kJ \frac{15 K}{293 K}=3x10^{6} kJ x 0.051195}

W = 0.153585x10^{6} kJ

W = 153,585 kJ

Hope this helps!

3 0
3 years ago
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