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QUIZ NO. 6 MEXE 3101

Total questions: 20

Worksheet time: 3mins

Name
Class
Date
1.

Which 802.11 PHY technique spreads each bit of data into 11 chips using a Barker sequence?

a)

FHSS

b)

DSSS

c)

OFDM

d)

HR-DSSS

2.

In FHSS, the maximum number of hops per second in the original 802.11 standard is:

a)

10

b)

50

c)

75

d)

100

3.

Which of the following is a major reason why OFDM performs better than DSSS in multipath environments?

a)

It increases the power spectral density

b)

It uses many narrowband subcarriers with long symbol durations

c)

It relies on chip-level error correction

d)

It spreads each bit across the entire channel bandwidth

4.

In DSSS, the processing gain is primarily determined by:

a)

Signal amplitude and SNR

b)

Chip rate relative to bit rate

c)

The number of subcarriers used

d)

The number of available frequency hops

5.

The original 802.11 FHSS PHY uses hop sets that span which frequency band?

a)

2.400–2.4835 GHz

b)

2.400–2.500 GHz

c)

5.150–5.350 GHz

d)

900–915 MHz

6.

What is the symbol duration of OFDM in 802.11a/g relative to the channel bandwidth?

a)

It becomes shorter as bandwidth increases

b)

It is inversely proportional to the number of chips

c)

It is long because of narrow subcarrier spacing

d)

It is equal to the beacon interval

7.

The OFDM PHY uses how many data subcarriers in its standard 64-subcarrier structure?

a)

26

b)

48

c)

52

d)

64

8.

In OFDM, the cyclic prefix (guard interval) is used to:

a)

Increase throughput by shortening symbols

b)

Protect against ISI and multipath delay spread

c)

Encode preambles for synchronization

d)

Perform spread spectrum coding

9.

The 802.11 HR-DSSS (used in 802.11b) achieves higher data rates by using:

a)

Barker sequence spreading at higher power

b)

Complementary Code Keying (CCK)

c)

Increased FHSS hop sets

d)

More subcarriers per channel

10.

Which channel bandwidth is used by OFDM in 802.11a and 802.11g?

a)

5 MHz

b)

10 MHz

c)

20 MHz

d)

40 MHz

11.

Why do DSSS signals require more bandwidth than the original uncoded data?

a)

They employ multiple antennas

b)

They convert data bits into chip sequences

c)

They use OFDM-like tone spacing

d)

They must compensate for higher transmission power

12.

In 802.11 FHSS, each hop corresponds to:

a)

A change in amplitude

b)

A change in carrier frequency

c)

A change in data modulation scheme

d)

A new SSID identification code

13.

Which modulation is used for the highest data rates in the OFDM PHY?

a)

BPSK

b)

QPSK

c)

16-QAM

d)

64-QAM

14.

Which preamble component is used by OFDM receivers for frequency offset correction and synchronization?

a)

Short training symbols

b)

Long training symbols

c)

PLCP header

d)

CCK preamble

15.

In OFDM, the pilot subcarriers serve to:

a)

Increase channel bandwidth

b)

Maintain phase and frequency reference

c)

Perform encryption key rotation

d)

Support CSMA/CA timing

16.

In DSSS, if the chip rate increases while data rate stays constant, the processing gain:

a)

Decreases

b)

Increases

c)

Remains unchanged

d)

Becomes negative

17.

The PHY header (PLCP header) in 802.11b is transmitted at which rate?

a)

The highest supported rate

b)

The same rate as the payload

c)

The lowest mandatory rate

d)

A variable rate depending on noise

18.

Which 802.11 physical layer supports the highest theoretical data rate among those in Chapter 14?

a)

FHSS

b)

DSSS

c)

HR-DSSS

d)

OFDM

19.

In OFDM, increasing the number of subcarriers while keeping bandwidth constant has what effect?

a)

Decreases subcarrier spacing

b)

Increases symbol rate

c)

Reduces multipath tolerance

d)

Eliminates need for guard intervals

20.

What is the primary reason the original 802.11 FHSS PHY cannot reach the same data rates as OFDM?

a)

FHSS uses only 1 MHz bandwidth per hop

b)

FHSS power levels must be lower

c)

FHSS channels are more prone to fading

d)

OFDM uses stronger encryption