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2 months ago

Communication Systems physics ncert exemplar solutions class 12th chapter two

A sinusoidal wave y(t) = 40 sin (10 * 10⁶pt) is amplitude modulated by another sinusoidal wave x(t) = 20sin(1000πt). The amplitude of minimum frequency component of modulated signal is:

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Vishal Baghel

Contributor-Level 10

$C_{m} (t) = A_{C} s i n ω_{C} t + \frac{μ A_{C}}{2} [c o s (ω_{C} - ω_{m}) t - c o s (ω_{C} + ω_{m}) t]$

$μ = \frac{A m}{A_{C}}$

$A_{C} = 40$

$μ = \frac{1}{2}$

So amplitude of minimum frequency = $\frac{μ A_{C}}{2}$

= 10

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2 months ago

Communication Systems Class 12th

Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an signal requires a bandwith of 8 kHz, how many channels can be accommodated for transmission:

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alok kumar singh

Contributor-Level 10

$v = f λ$

$f = \frac{v}{λ} = \frac{3 * 1 0^{8}}{1 0^{3} * 1 0^{- 9}} = 3 * 1 0^{14} H_{z}$

Channels = $\frac{2 % o f 3 * 1 0^{14} H_{z}}{8 * 1 0^{3}}$

$\frac{2 % o f 3 * 1 0^{14} H_{z}}{8 * 1 0^{3}}$

$\frac{= \frac{2 x}{100} 3 * 1 0^{14}}{8 * 1 0^{3}} = 75 * 1 0^{7}$

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Communication Systems Waves

A longitudinal wave is represented by x = 10 sin 2p $(n t - \frac{x}{λ}) c m .$ The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to:

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alok kumar singh

Contributor-Level 10

x = 10 $2 ? (n t ? \frac{x}{?}) c m$

V (wave velocity) = $\frac{? ?}{2 ?}$

v_max = 10 * 2p n

10 * 2pn = 4 * $\frac{? ?}{2 ?}$

10 * 2pn = $4 \frac{2 ? ? n ?}{2 ?}$

x = 5?

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2 months ago

Communication Systems Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

If the highest frequency modulating a carrier is 5 kHz, then the number of AM broadcast stations accommodated in a 90 kHz bandwidth are

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Payal Gupta

Contributor-Level 10

Band Width = 2 * n * the highest modulation frequency

$\Rightarrow n = \frac{90 k H_{z}}{2 * 5 k H z} = 9$

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2 months ago

Communication Systems Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

The maximum and minimum voltage of an amplitude modulated signal are 60V and 20V respectively. The percentage modulation index will be:

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alok kumar singh

Contributor-Level 10

Percentage modulation = $\frac{(V_{m a x} - V_{m i n})}{(V_{m a x} + V_{m i n})} * 100 %$

Percentage modulation = $(\frac{60 - 20}{80 + 20}) * 100$

= 50%

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2 months ago

Communication Systems Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

Draw the output signal Y in the given combination of gates.

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Payal Gupta

Contributor-Level 10

$y = \bar{\bar{A} + B} = A . \bar{B}$

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Communication Systems Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

A radio can tune to any station in 6 MHz to 10 MHz band. The value of corresponding of wavelength bandwidth will be:

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Vishal Baghel

Contributor-Level 10

$λ_{1} = 6 M H z = 6 * 1 0^{6} H z$

$λ_{1} = \frac{C}{υ_{1}} = \frac{3 * 1 0^{8}}{6 * 1 0^{6}} = 50 m$

$λ_{2} = \frac{C}{υ_{2}} = 10 * 1 0^{6} = \frac{3 * 1 0^{8}}{10 * 1 0^{6}} = 30 m$

Wavelength bandwidth

$= | λ_{1} - λ_{2} | = 20 m$

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2 months ago

Communication Systems Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

A FM Broad cast transmitter, using modulating signal of frequency 20 kHz has a deviation ratio of 10. The Bandwidth required for transmission is:

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Vishal Baghel

Contributor-Level 10

f = 20 kHz

Deviation ratio = 10

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2 months ago

Communication Systems Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

Match List – I with List – II

	List – I		List – II
A.	Television signal	I.	03 KHz
B.	Radio signal	II.	20 KHz
C.	High Quality Music	III.	02 MHz
D.	Human speech	IV.	06 MHz

Choose the correct answer from the options given below:

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Payal Gupta

Contributor-Level 10

Using factual data.

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2 months ago

Communication Systems Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

The current I flowing through the given circuit will be_________A.

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Vishal Baghel

Contributor-Level 10

$i = \frac{v}{R_{N e t}} = \frac{6}{3} = 2 A$

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