Network Security Questions freshers on “Layers”.
1. If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is its bandwidth?
a) 2500 Hz
b) 900 Hz
c) 800 Hz
d) can’t be determined with the information given
Answer: c
Clarification: The bandwidth is 900-100 = 800 Hz.
2. Assume we need to download text documents at the rate of 100 pages per second. What is the required bit rate of the channel?
a) 1.846 Mbps
b) 1.536 Mbps
c) 2.4 Mbps
d) None of the Mentioned
Answer: b
Clarification: Average No, of lines u=in each page = 24
Each line has 80 char.
Each char required 8 bits
100 x 24 x 80 x 8 = 1,536,000=1536 Mbps.
3. A digitized voice channel is made by digitizing a 4-kHz bandwidth analog voice signal. We need to sample the signal at twice the highest frequency (two samples per hertz). We assume that each sample requires 8 bits. What is the required bit rate?
a) 64 kbps
b) 32 kbps
c) 128 kbps
d) 16 kbps
Answer: a
Clarification: 2 × 4000 × 8 = 64,000 = 64kbps.
4. What is the bit rate for high-definition TV (HDTV)?
a) 1.4 Gbps
b) 2 Gbps
c) 1.5 Gbps
d) 1.8 Gbps
Answer: c
Clarification:
The HDTV screen is normally a ratio of 16: 9 (in contrast to 4: 3 for regular TV)
There are 1920 by 1080 pixels per screen, and
Screen is renewed 30 times per second.
24-bits represent one color pixel.
Therefore Bit Rate = 1920 × 1080 × 30 × 24 = 1,492992,000 = 1.5 Gbps.
5. A device is sending out data at the rate of 1000 bps. How long does it take to send a file of 100,000 characters?
a) 200s
b) 400s
c) 600s
d) 800s
Answer: d
Clarification: ((100,000 × 8) / 1000) s = 800 s.
6. A file contains 2 million bytes. How long does it take to download this file using a 56-Kbps channel?
a) 1-2 mins
b) 5-6 mins
c) 2-4 mins
d) 7-8 mins
Answer: b
Clarification:
With a 56-Kbps channel, it takes 16,000,000/56,000 = 289 s ≈ 5 minutes.
7. We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need?
a) 1024
b) 512
c) 256
d) 128
Answer: d
Clarification:
256000 = 2 × 20000 × log L = 6.625.
L = 98.7 Levels.
Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit rate. If we have 128 levels, the bit rate is 280 kbps.
8. Calculate the theoretical highest bit rate of a regular telephone line. The signal-to-noise ratio is usually 3162.
a) 34.860 kbps
b) 17.40 kbps
c) 11.62 kbps
d) none of the Mentioned
Answer: a
Clarification: C = B log (1+SNR) = 3000 log (1+3162) = 34,860 bps.
Logarithm here is in terms of base 2.
9. Calculate the theoretical channel capacity. If SNR(dB) = 36 and the channel bandwidth is 2 MHz.
a) 12 Mbps
b) 24 Mbps
c) 16 Mbps
d) 32 Mbps
Answer: b
Clarification: None.
10. A channel has a 1-MHz bandwidth. The SNR for this channel is 63. What is the appropriate bit rate?
a) 4 Mbps
b) 6 Mbps
c) 8 Mbps
d) 12 Mbps
Answer: b
Clarification:
C = Blog_2(1 + SNR) = 10^6 log_2(1 + 63) = 6Mbps.
11. A channel has a 1-MHz bandwidth. The SNR for this channel is 63. What is the appropriate signel level?
a) 2
b) 8
c) 4
d) 16
Answer: c
Clarification: As a continuation to the previous question where we get the channel capacity as 6 Mbps. For better performance we choose 4 Mbps. Thus we calculate L from –
4Mbps = 2 x 1 MHz x logL.