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Question 36: The network’s Subnet Mask is 255.255.255.224, determine
its broadcast address if that a computer which has the address
192.168.1.1?
A. 192.168.1.255 B. 192.168.1.15 C. 192.168.1.31 D. 192.168.1.96
Đáp án: C
Question 37: Consider building a CSMA/CD network running at 1 Gbps
over a 1–km cable with no repeaters. The signal speed in the cable is
200000 km/sec. What is the minimum frame size?
A. 500 bits B. 5000 bits C. 1000 bits D. 10000 bits
Đáp án: D
H ng d n: ướ ẫ
Th i gian c n truy n đi cho 1 km là: 1/200000 = 5 x 10^-6 = 5 µsec ờ ầ ề
Th i gian truy n c đi và v 1 km là: 5 x 2 = 10 µsec ờ ề ả ề
S bit truy n đ c là: 10^9 bps x 10 x 10^-6 sec = 10^4 bits = 10000 ố ề ượ
bits
Question 38: What is EIRP of a dipole antenna fed with a power of 100
mW?
A. 100mW B. 0.1W
C. 0.16W D. A, B and C are wrong
Đáp án: D
H ng d n: (d a vào slide 29 – Ch ng 3) ướ ẫ ự ươ
100 mW = 0,1 W
Gain = 0 dBd (do chuy n t anten l ng c c sang đ ng h ng) ể ừ ưỡ ự ẳ ướ
Do đó công su t nh n là 0,1W và công su t phát là ERP = 0,1W. ấ ậ ấ
Ta có: 2,14 = 10log10 (EIRP / ERP)
EIRP / ERP = 1,639.
EIRP = 16,39 W
Question 39: We need to send 56 kbps over a noiseless channel with a
bandwidth of 4 kHz. How many signal levels do we need?
A. 1 B. 64 C. 2 D. 128
Đáp án: D
H ng d n: ướ ẫ
Áp d ng công th c Nyquist, ta có: ụ ứ
• 56000 = 2 x 4000 x log2L
log 2 L = 7, suy ra L = 2^7 = 128.
Question 40: Which modulation does a constellation diagram correspond
to?
A. QAM-4 B. PSK-4 C. PSK-2 D. QAM-2
Đáp án: B (câu này có cái hình)
- M y cái ghi logarit past t word qua đây l i. Ví d log2L hi u là ấ ừ ỗ ụ ể
log c s 2 c a L. ơ ố ủ
- Có ph i d ch câu h i ra ko v y, ch c kh i nha. ả ị ỏ ậ ắ ỏ
Question 1: How many frequencies does a full-duplex QAM-64 modem use?
A. 1 B. 12 C. 6 D. 2
Đáp án: D
Question 2: An analog cellular system has 20 MHz of band-width and
uses two channels simplex of 25 kHz to provide a transfer service of
full-duplex. What is the number of full-duplex channels available by
cell for a reuse factor K = 4?
A. 400 B. 200
C. 100 D. A, B and C are wrong
Đáp án: C
H ng d n: ướ ẫ
Băng thông: 20 000 000 Hz
Băng thông c a kênh là: 25 000 x 2 (simplex channel) = 50 000 Hz ủ
(duplex channel)
S kênh: 20 000 000/50 000 = 400 (channels) ố
K = 4: S kênh có trên t bào (cell) là: 400/4 = 100 (channels) ố ế
Question 3: A modem constellation diagram similar to figure. How many
bps can a modem with these parameters achieve at 1200 baud?
A. 4800bps B. 2400bps
C. 1200bps D. A, B and C are wrong
Đáp án: B
2 x 1200 = 4800 bps
Question 4: Ten signals, each requiring 4000Hz, are multiplexed on to
a single channel using FDM. How much minimum bandwidth is required for
the multiplexed channel? Assume that the guard bands are 400Hz wide.
A. 40800Hz B. 40400Hz
C. 44000Hz D. A, B and C are wrong
Đáp án: D
H ng d n: ướ ẫ
Có 10 tín hi u, m i tín hi u 4000 Hz. Dó đó, c n 9 d i t n (guard ệ ỗ ệ ầ ả ầ
bands)
V y, băng thông t i thi u c n đ ghép kênh là: 4000x10+400x9 = 43600 ậ ố ể ầ ể
Hz.
Question 31: One class A address borrowed 21 bits to divide subnet,
the Subnet Mask is?
A. 255.255.255.248 B. 255.255.224.0 C. 255.255.192.0 D. 255.255.248.0
M n 21 bit =>2 octet gi a m n t ng c ng 16 bit(255.255), octet ượ ư ượ ô ô
cu i m n 5 bit:128+64+32+16+8=248 ố ượ
=>Câu A
Question 32: The radius of a cell in cellular network is 1600m. What
is the distance between centers of adjacent cells?
A. 2770m B. 1600m
C. 3200m D. A, B and C are wrong
Kho ng cách = r x căn b c 2 c a 3 = 2770 (đánh d u căn nó l i ả ậ ủ ấ a
ko hi n lên :D) ệ
=>Câu A
Question 33: Consider a noiseless channel with a bandwidth of 3000 Hz
transmitting a signal with 16 signal levels. The maximum bit rate can
be?
A. 48000 bps B. 3000 bps C. 12000 bps D. 6000 bps
Công th c Nyquist:BitRate = 2 x 3000 x log2(16)=24000 ứ
=>câu E(ch c dáp án sai ^^) ắ
Question 34: A group of N stations share a Slotted ALOHA channel of 1
Mb/s. Each station emits in average a frame of 12000 bits/second. By
taking again the results obtained for Smax (maximum efficiency)
determine the maximum value of N that one can have?
A. 300 B. 700
C. 600 D. A, B and C are wrong
1 byte = 8 bit
1Mb = 1024 x 1024 byte
=>N = 1024 x 1024 x 8 / 12000 = 700
=>Câu B
Question 35: One takes the antenna gains into account; microwaves
transmitter has an output power of 0.1W at 2 GHz. The EIRP of the
transmitted signal is 10W. What is the wavelength of receiving
antennas parabolas?
A. 1.2m B. 0.64m
C. 0.32m D. A, B and C are wrong
lamda = c/f = 3x10^8 / 2x10^9 = 0,15(m)(đánh d u lamda nó l i ko ấ a
hi n lên :D) ệ
=>Câu D
Question 11: A typical Bluetooth data frame describe in figure. Which
field determines the active devices?
A. Header B. Access code C. Address D. Checksum
Ans: C
Question 12: If the diameter of receiving antennas parabolas is 30 cm,
what is the frequency concerned?
A. 1GHz B. 1MHz
C. 1kHz D. A, B and C are wrong
Ans: D
Question 13: The signal can be written .
What is the modulation type?
A. ASK-2 B. ASK-4 C. PSK-2 D. QAM-2
Ans: A
Question 14: The output power at transmitter is 1W, power received is
100mW. The gain is?
A. 10 dB B. 100 C. 10 dB D. 1
Ans: D
Question 15: A telephone line has a bandwidth of 4kHz. The
signal-to-noise ratio is 3162. For this channel the capacity is?
A. 46480 bps B. 40000 bps
C. 37944 bps D. A, B and C are wrong
Ans: A
C = B log2 (1 + SNR)
B is the bandwidth .SNR is the signal to noise ratio, capacity is the
capacity of the channel in bits per second
*Question 6:* Suppose a transmitter produces 50W of power. The transmitter’s
power is applied to a *unity gain antenna* with a 900MHz carrier frequency.
The close-in distance is d0 = 100m. The value Pr (d0) may be predicted from
equation, where Gt, Gr are antenna gain at transmitter, receiver, and L ³ 1
is the system loss factor. If Pr is in unit of dBm, the received power is
given by Pr(d) = 10log10[Pr(d0)/0.001] + 20log10(d0/d), where d³d0. What is
the received power at a free space distance of 10km?
*A. *» -24.5dBm *B. *»
-64.5dBm
*C. *» +64.5dBm *D. A,
B and C are wrong*
*Answer: **unity gain antenna: Gain = 0; *
* *
= c / f = 3 * 10^8 / 900000000 = 1/3
L = (4d / ) ^ 2
=> Pr (d0) = Pt * ^4 / (4d) ^ 4 = 2.475 *10^-10 (mW)
*=> Pr(d = 10km) = -106.0635447 dBm.***
* *
* *
*Question 7:* A channel has a bit rate of 5 kbps and a propagation delay of
20 msec. For what range of frame sizes does stop-and-wait give an efficiency
of at least 50%?
*A. *frame size < 100 bits *B. 100
bits < frame size < 200 bits*
*C. *frame size > 200 bits *D. *frame
size < 200 bits
*Answer**: * Tf / Tt >= 50% (1)
Tf = Fs / Bw (2) ; Tt
= 20*10^-3 + Tf (3*)*
From (1) and (3) => Tf >= 0.02 (4);
From (4) and (2) => *Fs >= 102.4 (bits);*
With:
Tf: Time to transmit the frame; Tt: Total time include propagation
delay time;
Fs: frame size; Bw: bandwidth
*Question 8:* Need to share with each subnet have 510 hosts maximum in a
class B address, should use the Subnet Mask?
*A. *255. 255. 255.0 *B. *255. 255. 255.192 *C. *255. 255.252.0
*D. 255.255.254.0*
* *
*Answer:** hosts = 510 => 9bits host => 23 bits net, IP class B => borrow
7bits => 255.255.254.0*
*Question 9:* Suppose that 1 MHz is dedicated to the control channels and
that only a control channel is necessary by cluster; determine a reasonable
distribution of the *voice traffic* channels for each cell, for the 9 reuse
factors given previously.
*A. *8 cells with 53 voice channels and 4 cells with 54 voice channels
*B. *8 cells with 71 voice channels and 1 cell with 72 voice channels
*C. *4 cells with 91 voice channels and 3 cells with 92 voice channels
*D. *A, B and C are wrong
*Question 10:* The signal can be written.
What is the modulation type?
*A. *QAM-4 *B. PSK-4* *C. *PSK-2
*D. *QAM-2
[image: s_n(t) = \sqrt{\frac{2E_s}{T}} \cos \left ( 2 \pi f_c t + (2n -1)
\frac{\pi}{4}\right ),\quad n = 1, 2, 3, 4.]
This yields the four phases /4, 3 /4, 5 /4 and 7 /4 as needed. π π π π
Còn câu 9 n lát n a post nhe, tranh th m y câu này tr c! ^^ ợ ư ủ ấ ướ
Question 27: In the cellular network, Q = D/R called co-channel reuse
ratio. Suppose i = 3 and j = 1. What is the Q?
A. 6 B. 6.24
C. 4.58 D. A, B and C are wrong
Đáp án : C. Theo slide 13 ch ng 5 ươ
Question 28: For the network address 172.16.0.0/16. How many subnets
this network can be split up?
A. 16 B. 214
C. 216 D. A, B and C are wrong
Đáp án : C
Question 29: The telephone line is now to have a loss of 20dB. The
input signal power is measured as 0.5W, and the output noise level is
measured as 4.5W. Calculate the output SNR in dB?
A. 30.5dB B. +30.5dB
C. 89dB D. A, B and C are wrong
Đáp án : B
Gi i : ả
Ta có SNRdB = 10log10(Psignal /Pnoise) = 10 log10 [(0,5)/(4,5*10-6)] ~
50,5dB.
Do loss = 20dB => output SNRdB = 50,5 – 20 = 30,5 dB
2 câu ch a làm đ c : ư ượ
Question 26: The relationship between bit rate and modulation rate?
A. R ≤ 2B B. C = B log2 (1 + ) C. R = D. D = R n
Question 30: Suppose a transmitter produces 50W of power. If the
transmitter’s power is applied to a unity gain antenna with a 900MHz
carrier frequency, what is the received power in dBm at a free space
distance of 100m? The value Pr(d0) may be predicted from equation ,
where Gt, Gr are antenna gain at transmitter, receiver, and L 1 is
the system loss factor.
A. +47dBm B. 24.5dBm
C. +24.5dBm D. A, B and C are wrong
Câu 9:
Suppose that 1 MHz is dedicated to the control channels and that only a
control channel is necessary by cluster; determine a reasonable distribution
of the *voice traffic* channels for each cell, for the 9 reuse factors given
previously.
*A. *8 cells with 53 voice channels and 4 cells with 54 voice channels
*B. 8** **cells with 71 voice channels and 1 cell with 72 voice channels*
*C. *4 cells with 91 voice channels and 3 cells with 92 voice channels
*D. *A, B and C are wrong
*Answer: B*
Đ này thi u: ề ế
Total bandwidth: 33MHz
Channel bandwidth: 25 kHz simplex channels => 50 kHz duplex channels.
=> Total available channels = 33000 / 50 = 660 channels.
1MHz control channels => 1000 / 50 = 20 control channels
=> 660 – 20 = 640 voice channels.
=> N = 9 => Number of available voice channels per cell = 640 / 9 = 71.1111
=> Ta có 8 * 71 + 1 * 72 = 640 => câu B là chính xác.
*Question 21:* Four transmitter having key A: (-1 -1 -1 +1 +1 -1 +1 +1),
B: (-1 -1 +1 -1 +1 +1 +1 -1), C: (-1 +1 -1 +1 +1 -1 -1 -1) and D: (-1
+1 -1 -1 -1 -1 +1 -1). The received chip sequence is (-2 -2 0 -2 0 -2 +4
0). What happened?
*A. *A, B send bit 1 and C, D send bit 0 *B. *A, B, C and D
send bit 1
*C. *A, B, C and D send bit 0 *D. *A, B and
D send bit 1, C send bit 0
Lý thuy t : Đ bi t user g i bit gì thì ta l n l t l y key c a user tích ế ể ế ở ầ ượ ấ ủ
trong v i chu i tín hi u nh n đ c ớ ô ệ ậ ượ
- +1 : user phát bit 1
- 0 : user không phát gì
- -1 : user phát bit 0
CT tích trong : S x T = (1/m). ∑ Si.Ti
đây mình tính ra A,B,D nhân v i T =1 còn C x T = ½ => câu này có v đ Ở ớ ẻ ề
sai, m y b n ki m tra l i giùm ấ a ể a
