Test: Spread Spectrum
20 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Test: Spread Spectrum
| 1 Crore+ students have signed up on EduRev. Have you? |
A rate 1/2 convolution code with dfrec = 10 is used to encode a data requeence occurring at a rate of 1 kbps. The modulation is binary PSK. The DS spread spectrum sequence has a chip rate of 10 MHz
Que: The coding gain is
A rate 1/2 convolution code with dfrec = 10 is used to encode a data requeence occurring at a rate of 1 kbps. The modulation is binary PSK. The DS spread spectrum sequence has a chip rate of 10 MHz
Que: The processing gain is
The processing gain
A CDMA system is designed based on DS spread spectrum with a processing gain of 1000 and BPSK modulation scheme. If user has equal power and the desired level of performance of an error probability of 10-6, the number of user will be
To achieve an error probability of 10-6 ,we required 
Then, the number of users of the CDMA system is
In previous question if processing gain is changed to 500, then number of users will be
If the processing gain is reduced to W/R = 500, then
A DS spread spectrum system transmit at a rate of 1 kbps in the presets of a tone jammer. The jammer power is 20 dB greater then the desired signal, and the required εb/J0 to achieve satisfactory performance is 10 dB.
Que: The spreading bandwidth required to meet the specifications is
We have a system where (Jav/Pav) = 20 dB, R =1000, and (εb/J0)db = 10 dB
A DS spread spectrum system transmit at a rate of 1 kbps in the presets of a tone jammer. The jammer power is 20 dB greater then the desired signal, and the required εb/J0 to achieve satisfactory performance is 10 dB.
Que: If the jammer is a pulse jammer, then pulse duty cycle that results in worst case jamming is
The duty cycle of a pulse jammer of worst-case jamming is
A DS spread spectrum system transmit at a rate of 1 kbps in the presets of a tone jammer. The jammer power is 20 dB greater then the desired signal, and the required εb/J0 to achieve satisfactory performance is 10 dB.
Que: The correspond probability of error is
The corresponding probability of error for this worst-case jamming is
A CDMA system consist of 15 equal power user that transmit information at a rate of 10 kbps, each using a DS spread spectrum signal operating at chip rate of 1 MHz. The modulation scheme is BPSK
Que: The Processing gain is
A CDMA system consist of 15 equal power user that transmit information at a rate of 10 kbps, each using a DS spread spectrum signal operating at chip rate of 1 MHz. The modulation scheme is BPSK
Que: The value of εb/Jb is
We have Nu = 15 users transmitting at a rate of 10,000 bps each, in a bandwidth of W =1MHz.
A CDMA system consist of 15 equal power user that transmit information at a rate of 10 kbps, each using a DS spread spectrum signal operating at chip rate of 1 MHz. The modulation scheme is BPSK
Que: How much should the processing gain be increased to allow for doubling the number of users with affecting the autopad SNR
An m =10 ML shift register is used to generate the pre hdarandlm sequence in a DS spread spectrum system. The chip duration is Tc = ℓμs and the bit duration is Tb, = NTc, where N is the length (period of the m sequence).
Que: The processing gain of the system is
The period of the maximum length shift register sequence is
An m =10 ML shift register is used to generate the pre hdarandlm sequence in a DS spread spectrum system. The chip duration is Tc = ℓμs and the bit duration is Tb, = NTc, where N is the length (period of the m sequence).
Que: If the required εb /J0 is 10 and the jammer is a tone jammer with an average power Jav, then jamming margin is.
A Jamming margin is
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.
Que: If the hop rate is one per bit, the hopping band width for this channel is
The length of the shift-register sequence is
L = 2m - 1215 - 1 = 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.
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
Que: 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
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 = 2767 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.
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 1/2 N0 and an SNR 20-13 dB (total SNR over the three hops)
Que: The probability of error for this system is
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
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 1/2 N0 and an SNR 20-13 dB (total SNR over the three hops)
Que: In case of one hop per bit the probability of error is
In the case of one hop per bit, the SNR per bit is 20,
A slow FH binary FSK system with non coherent detection operates at ε0/J0 10, with hopping bandwidth of 2 GHz, and a bit rate of 10 kbps.
Que: The processing gain of this system is
We are given a hopping bandwidth of 2 GHz and a bit rate of 10 kbs.
A slow FH binary FSK system with non coherent detection operates at ε0/J0 10, with hopping bandwidth of 2 GHz, and a bit rate of 10 kbps.
Que: If the jammer operates as a partial band jammar, the bandwidth occupancy for worst case jamming is
The bandwidth of the worst partial-band jammer is α * W , where
A slow FH binary FSK system with non coherent detection operates at ε0/J0 10, with hopping bandwidth of 2 GHz, and a bit rate of 10 kbps.
Que: The probability of error for the worst-case partial band jammer is
The probability of error with worst-case partial-band jamming is
A slow FH binary FSK system with non coherent detection operates at ε0/J0 10, with hopping bandwidth of 2 GHz, and a bit rate of 10 kbps.
Que: The minimum hop rate for a FH spread spectrum system that will prevent a jammer from operating five onives away from the receiver is
d = 5 miles = 8050 meters
|
20 docs|261 tests
|
|
Use Code STAYHOME200 and get INR 200 additional OFF
|
Use Coupon Code |
|
20 docs|261 tests
|
Download free EduRev App
Related content with Test: Spread Spectrum
|
|
|
|
Related tests with Test: Spread Spectrum
|
|
|
|
|
|
© EduRev
|
Education Revolution
|
Follow Us
|
