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svet-max [94.6K]

3 years ago

15

we wish to send at a rate of 10Mbits/s over a passband channel. Assuming that an excess bandwidth of 50% is used, how much bandw

idth is needed for each of the following schemes: QPSK 64QAM, and 64-ary noncoherent orthogonal modulation useing a Walsh Hadamard code

1 answer:

gayaneshka [121]3 years ago

6 0

Answer:

QPSK: 7.5 MHz

64-QAM:2.5 MHz

64-Walsh-Hadamard: 160 MHz

Explanation:

See attached picture.

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Express the following quantities to the nearest standard prefix using no more than three digits.(a) 20,000,000 Hz(b) 1025 W(c) 0

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Answer:

(a) 20 MHz

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(c) 3.33 ns

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Explanation:

(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = <u>20 MHz</u>

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What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 515 turns, its

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<h2>Answer:</h2>

7532V

<h2>Explanation:</h2>

For a given transformer, the ratio of the number of turns in its primary coil ( $N_{p}$ ) to the number of turns in its secondary coil ( $N_{s}$ ) is equal to the ratio of the input voltage ( $V_{p}$ ) to the output voltage ( $V_{s}$ ) of the transformer. i.e

$\frac{N_p}{N_s}$ = $\frac{V_p}{V_s}$ ----------------(i)

<em>From the question;</em>

$N_{p}$ = number of turns in the primary coil = 8 turns

$N_{s}$ = number of turns in the secondary coil = 515 turns

$V_{p}$ = input voltage = 117V

<em>Substitute these values into equation (i) as follows;</em>

$\frac{8}{515}$ = $\frac{117}{V_s}$

<em>Solve for </em> $V_{s}$ <em>;</em>

$V_{s}$ = 117 x 515 / 8

$V_{s}$ = 7532V

Therefore, the output voltage (in V) of the transformer is 7532

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