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2 years ago

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Harmonics problem. A square wave of frequency f contains harmonics (sine waves) at f, 3f, 5f, 7f, ... . Suppose a system respond

s to frequencies in the range 20-40 kHz but is insensitive outside of this range.
EXAMPLE:
Imagine applying a square wave with f = 30 kHz to this system. What is the shape of the response?
[Answer: a 30 kHz sine wave. ]
Imagine applying a square wave with f = 10 kHz to this system.

What is the shape of the response?
A. A 10 kHz sine wave
B. A combination of 20, 30 and 40 kHz sine waves
C. A 30 kHz sine wave
D. No response

1 answer:

ad-work [718]2 years ago

4 0

Answer:

C.

Explanation:

If the fundamental frequency of the square wave is 10 Khz, and contains harmonics at f, 3f, 5f, etc., it will have a component at 10 Khz, other at 30 Khz, and another one at 50 Khz.
If the system responds to frequencies in the range between 20 Khz and 40 KHz only, the shape of the response will contain only a 30 Khz sine wave (3rd harmonic of 10 Khz).

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Answer:

Second object is located at 42.03 cm in front of mirror

Explanation:

In this question we have given,

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put values of u and v in equation (1)

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$\frac{1}{f} =\frac{1}{7.05}+ \frac{1}{-13.5}$

$\frac{1}{f}=\frac{13.5-7.05}{13.5\times 7.05}$

$\frac{1}{f}=\frac{6.45}{13.5\times 7.05}\\f=13.5\times 1.09\\f=14.75$

we have given that

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it means

$o_{2}=2o_{1}\\i_{1}=i_{2}$ .............(2)

we know that

$\frac{v}{u}=\frac{i}{o}\\i=\frac{o\times v}{u}$

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$i_{1}=\frac{o_{1}\times v}{u}$ .................(3)

put values of v and u in equation 3

$i_{1}=-\frac{o_{1}\times 7.05}{13.5}$

$i_{1}=-0.52o_{1}$

therefore from equation 2

$i_{2}=-0.52o_{1}$

we know that

$i_{2}=\frac{o_{2}\times V}{U}$ .................(4)

put value of $i_{2}$ and $o_{2}$ in equation 4

$-.52o_{1}=\frac{2o_{1}\times V}{U}$

$U=\frac{2o_{1}\times V}{-.52o_{1}} \\U=-3.85V$

we know that U,V and f are related by following formula

$\frac{1}{f} =\frac{1}{V}+ \frac{1}{U}$ .............(5)

put values of f and U in equation 5

we got

$\frac{1}{14.75} =\frac{1}{V}- \frac{1}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}\\V=\frac{2.85\times 14.75}{3.85}\\V=10.91 cm$

Therefore,

U=-10.91\times 3.85

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