Electronic Devices and Circuits Questions and Answers – Spread Spectrum

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

1. A pseudo-noise (PN) sequence is generated using a feedback shift register of length m = 4. The chip rate is 107 chips per second. What is the PN sequence length?
a) 10
b) 12
c) 15
d) 18
View Answer

Answer: c
Explanation: The PN sequence length is N = 2m – 1 = 16 – 1 = 15.

2. A pseudo-noise (PN) sequence is generated using a feedback shift register of length m = 4. The chip rate is 107 chips per second. The chip duration will be?
a) 1µs
b) 0.1 µs
c) 0.1 ms
d) 1 ms
View Answer

Answer: b
Explanation: Tc = 1/(107) or 0.1 µs.

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. The period of PN sequence is?
a) 1.5 µs
b) 15 µs
c) 6.67 ns
d) 0.67 ns
View Answer

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

Answer: d
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

Answer: d
Explanation: PG = 4 x 4 = 16 or 12 db.
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6. 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.The coding gain is?
a) 7 dB
b) 12 dB
c) 14 dB
d) 24 dB
View Answer

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

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. The processing gain is?
a) 14 dB
b) 37 dB
c) 58 dB
d) 104 dB
View Answer

Answer: b
Explanation: PG = (107)/(2 x 1000) = 5000 or 37 db.
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8. 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. 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

Answer: a
Explanation: The length of the shift-register sequence is L = 2m – 1215 = 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. 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. 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

Answer: b
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.
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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

Answer: b
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
The total SNR for three hops is 20 13 dB for the SNR per hop 20/3

Sanfoundry Global Education & Learning Series – Electronic Devices and Circuits.

To practice all areas of Electronic Devices and Circuits, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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