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svet-max [94.6K]

2 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]2 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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How may a Professional Engineer provide notice of licensure to clients?

azamat

Find full question attached

Answer:

(b) By including a statement that he or she is licensed by the Board for Professional Engineers and Land Surveyors immediately above the signature line in at least 12 point type on all contracts for services

Explanation:

A PE(professional engineer) licensee must show that he is licensed in order to show and ensure public safety as he is qualified for the job he is handling. The California regulations on professional engineers holds that all professional engineers must be licensed by the board of professional engineers and Land surveyors in order to operate legally as an engineer. The engineer may show licensure through the following options:

The engineer might provide statement to each client to show he is licensed which would then be signed by the client

The engineer may choose to post a wall certificate in his work premises to show he is licensed

The engineer may choose to include a statement of license in a letterhead or contract document which must be above the client's signature line and not less than 12 point type

4 0

2 years ago

Look at the data set below.

Goshia [24]

The answer would be 7

3 0

3 years ago

Read 2 more answers

Consider that a system has two entities, Students, Instructors and Course. The Student has the following properties: student nam

tekilochka [14]

Answer:

There's no answer ?

Explanation:

5 0

2 years ago

A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i

Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter $d_{o}$ = 12.8 mm

Final diameter $d_{f}$ = 10.7

Engineering stress $\alpha _{E}$ = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4( $d_{o}$ )² ; Ag = π/4( $d_{f}$ )²

% = π/4 [ ( ( $d_{f}$ )² - ( $d_{o}$ )²) / ( π/4 ( $d_{o}$ )²) ]

= ( ( $d_{f}$ )² - ( $d_{o}$ )²) / ( $d_{o}$ )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress $\alpha _{T}$ = $\alpha _{E}$ ( 1 + $E_{E}$ )

$E_{E}$ is engineering strain

$E_{E}$ = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49

$E_{E}$ = 0.431

so we substitute the value of $E_{E}$ into our initial equation;

True stress $\alpha _{T}$ = 460 ( 1 + 0.431)

True stress $\alpha _{T}$ = 460 (1.431)

True stress $\alpha _{T}$ = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

6 0

1 year ago

8. Describe and correct the error in stating the domain. Xf * (x) = 4x ^ (1/2) + 2 and g(x) = - 4x ^ (1/2) The domain of (f + g)

konstantin123 [22]

Answer:

hi

Explanation:

4 0

2 years ago