This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Spread Spectrum”.

(Q.1-Q.3) A pseudo-noise (PN) sequence is generated using a feedback shift register of length m = 4. The chip rate is 107 chips per second.

1. The PN sequence length is

a) 10

b) 12

c) 15

d) 18

View Answer

Explanation: The PN sequence length is N = 2

^{m}– 1 = 16 – 1 = 15.

2. The chip duration is

a) 1µs

b) 0.1 µs

c) 0.1 ms

d) 1 ms

View Answer

Explanation: Tc = 1/(10

^{7}) or 0.1 µs.

3. The period of PN sequence is

a) 1.5 µs

b) 15 µs

c) 6.67 ns

d) 0.67 ns

View Answer

Explanation: The period of the PN sequence is T = NTc = 15 x 0.1 = 1.5 s

4. A slow FH/MFSK system has the following parameters.

Number of bits per MFSK symbol = 4

Number of MFSK symbol per hop = 5

The processing gain of the system is

a) 13.4 dB

b) 37.8 dB

c) 6 dB

d) 26 dB

View Answer

Explanation: PG = Wc/Rs = 5 x 4 = 20 or 26 db.

5. A fast FH/MFSK system has the following parameters.

Number of bits per MFSK symbol = 4

Number of pops per MFSK symbol = 4

The processing gain of the system is

a) 0 dB

b) 7 dB

c) 9 dB

d) 12 dB

View Answer

Explanation: PG = 4 x 4 = 16 or 12 db.

(Q.6-Q.7) A rate 1/2 convolution code with dfrec = 10 is used to encode a data sequence occurring at a rate of 1 kbps. The modulation is binary PSK. The DS spread spectrum sequence has a chip rate of 10 MHz.

6. The coding gain is

a) 7 dB

b) 12 dB

c) 14 dB

d) 24 dB

View Answer

Explanation: 0.5 x 10 = 5 or 7 db is the coding gain.

7. The processing gain is

a) 14 dB

b) 37 dB

c) 58 dB

d) 104 dB

View Answer

Explanation: PG = (10

^{7})/(2 x 1000) = 5000 or 37 db.

(Q.8-Q.9) An FH binary orthogonal FSK system employs an m 15 stage liner feedback shift register that generates an ML sequence. Each state of the shift register selects one of L non over lapping frequency bands in the hopping pattern. The bit rate is 100 bits/s. The demodulator employ non coherent detection.[/expand]

8. If the hop rate is one per bit, the hopping bandwidth

for this channel is

a) 6.5534 MHz

b) 9.4369 MHz

c) 2.6943 MHz

d) None of the mentioned

View Answer

Explanation: The length of the shift-register sequence is L = 2

^{m}– 12

^{15}= 32767 bits

For binary FSK modulation, the minimum frequency separation is 2/T, where 1/T is the symbol (bit) rate. The hop rate is 100 hops/sec. Since the shift register has L 32767 states and each state utilizes a bandwidth of 2/T = 200 Hz, then the total bandwidth for the FH signal is 6.5534 MHz.

9. Suppose the hop rate is increased to 2 hops/bit and the receiver uses square law combining the signal over two hops. The hopping bandwidth for this channel is

a) 3.2767 MHz

b) 13.1068 MHz

c) 26.2136 MHz

d) 1.6384 MHz

View Answer

Explanation: If the hopping rate is 2 hops/bit and the bit rate is 100 bits/sec, then, the hop rate is 200 hops/sec. The minimum frequency separation for orthogonality 2/T 400 Hz. Since there are N 32767 states of the shift register and for each state we select one of two frequencies separated by 400 Hz, the hopping bandwidth is 13.1068 MHz.

10. In a fast FH spread spectrum system, the information is transmitted via FSK with non coherent detection. Suppose there are N = 3 hops/bit with hard decision decoding of the signal in each hop. The channel is AWGN with power spectral density 0.5No and an SNR 20 ~13 dB (total SNR over the three hops). The probability of error for this system is

a) 0.013

b) 0.0013

c) 0.049

d) 0.0049

View Answer

Explanation: The total SNR for three hops is 20 ~ 13 dB. Therefore the SNR per hop is 20/3. The probability of a chip error with non-coherent detection is

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