1. To increase the rate of communication by log2M in M-array communication, the power requirement increased by
2M
M2
MM
2M
Answer.2. M2
Explanation
The distance of each signaling point from the origin is given by √Es
Where Es = Eblog2M.
Power in M-array communication is given by Es
As Es = Eblog2M
To increase power by log2M,
Es = Eb(log2M)2
Es = Eblog2M2
Es = 2Eblog2M
Hence to increase the rate of communication by log2M in M-array communication,
the power requirement increased by M2.
2. The limit which represents the threshold Eb/N0 value below which reliable communication cannot be maintained is called as
Probability limit
Error limit
Shannon limit
Communication limit
Answer.3. Shannon limit
Explanation
The Shannon capacity theorem defines the maximum amount of information, or data capacity, which can be sent over any channel or medium (wireless, coax, twisted pair, fiber, etc.). What this says is that the higher the signal-to-noise (SNR) ratio and the more the channel bandwidth, the higher the possible data rate.
Eb/N0 curve has a waterfall shape. Shannon limit gives the threshold value below which reliable communication cannot be maintained.
3. The length of cable required for transmitting data at the rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits and for signal speed 2,00,000 km/s is
2.5 km
2.0 km
1.5 km
1.0 km
Answer.2. 2.0 km
Explanation
Bandwidth = 500 mbps
Frame size = 10,000 bits
Signal speed = 2,00,000 km/sec
Transmission delay = 2 × propagation delay
frame size/bandwidth = 2 × (length of the cable /signal speed)
Length of the cable = (20000 × 10000)/(2 × 500 × 106)
Length of the cable = 2 Km
4. Which modulation scheme will have minimal BER?
QPSK
8-PSK
16-PSK
16-QAM
Answer.1. QPSK
Explanation
Probability of error (Pe) = $Q\sqrt {\frac{{{E_b}}}{{{N_0}}}}$
Pe is minimum when energy (Eb) is maximum.
Now,
Eb ∝ d2 (distance between the symbol)
i.e., Pe is minimum when the distance is maximum.
Analysis:
BER (Bit Error Ratio) Is a Signal Quality Quantitative Measurement of Digital Communication Systems. Out of the given scheme, QPSK has the maximum distance, and hence Pe is minimum in QPSK. For M-ary as M increases, the distance between symbols decreases.
5. M-ary signaling produces _______ error performance with orthogonal signaling and _______ error performance with multiple phase signaling.
Degraded, improved
Improved, degraded
Improved, improved
Degraded, degraded
Answer.2.Improved, degraded
Explanation
In M-ary signaling as k increases, the curve moves towards the degraded error performance. It produces improved error performance in the case of orthogonal signaling and degraded error performance in the case of multiple-phase signaling.
6. A full-duplex binary FSK transmission is made through a channel of bandwidth 10 kHz. In each direction of transmission, the two carriers used for the two states are separated by 2 kHz. The maximum baud rate for this transmission is:
2000 bps
3000 bps
5000 bps
10000 bps
Answer.2. 3000 bps
Explanation
The bandwidth for a full-duplex binary FSK transmission is given by:
B.W. = Baud rate + f2 – f1
Calculation:
Because the transmission is full-duplex only 5000 (5 kHz) is allocated for each direction.
Given f2 – f1 2k – 2000
Band rate = BW – (f2 – f1) – 5000 – 2000
Band rate = 3000 bps
7. What is the relation between bandwidth B of BPSK signal and the bandwidth Bm of the M-ary PSK signal, for a given data rate?
Bm = MB
Bm = B log2 M
B = Bm log2 M
B = M⋅Bm
Answer.3. B = Bm log2 M
Explanation
M-array PSK:
For m-array PSK minimum bandwidth to pass digitally modulated carrier is given as:
An M-ary transmission is a type of digital modulation where instead of transmitting one bit at a time, two or more bits are transmitted simultaneously.
QPSK is a 4-ary system, that is, it has 4 different states each represented by a set of 2 binary digits, i.e., 00, 01, 10, 11. Likewise, an 8-ary system has 8 different states each represented by a set of 3 binary digits, i.e., 000, 001, ….., 111, a 16-ary system has 16 different states represented by a set of 4 binary digits, i.e., 0000, 0001, ….., 1111, and so on.
The minimum energy noise vector for 4-ary system is smaller than 2-ary system. So 4-ary system is more vulnerable to noise.
9. An M-level PSK modulation scheme is used to transmit independent binary digits over a band-pass channel with a bandwidth of 100 kHz. The bit rate is 200 kbps and the system characteristic is a raised-cosine spectrum with 100% excess bandwidth. The minimum value of M is
10
14
16
20
Answer.3. 16
Explanation
For M-PSK bandwidth of signal given by B $ = \frac{{{R_b} \times \left( {1 + \propto } \right)}}{{logM}}$