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(Q.1Q.6) consider the figure shown below and answer the questions that proceed.
For Vcc = 15V, R1 = 100 kΩ, R(E) = 3.9 kΩ, R(C) = 6.8 kΩ and ß = 100, determine the dc collector current for each transistor.
For Vcc = 15V, R1 = 100 kΩ, R(E) = 3.9 kΩ, R(C) = 6.8 kΩ and ß = 100, determine the dc collector voltage for each transistor.
Find R (in 1) for R (sig) = 5 kΩ.
It is the parallel combination of the 32 kΩ resistor and 2.6 kΩ resistor respectively.
Find Vb1/Vsig for R(sig) = 5 kΩ.
It is the parallel combination of the 32 kΩ resistor and 2.6 kΩ resistor respectively.
For Rl = 2 kΩ find Vo/Vb2.
Find the overall voltage gain.
(Q.1Q.3) A pseudonoise (PN) sequence is generated using a feedback shift register of length m = 4. The chip rate is 107 chips per second.
Q. The PN sequence length is
The PN sequence length is N = 2^{m} – 1 = 16 – 1 = 15.
Tc = 1/(10^{7}) or 0.1 µs.
The period of PN sequence is
The period of the PN sequence is T = NTc = 15 x 0.1 = 1.5 s
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
PG = Wc/Rs = 5 x 4 = 20 or 26 db.
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
PG = 4 x 4 = 16 or 12 db.
(Q.6Q.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.
Q. The coding gain is
0.5 x 10 = 5 or 7 db is the coding gain.
PG = (10^{7})/(2 x 1000) = 5000 or 37 db.
(Q.8Q.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]
Q. If the hop rate is one per bit, the hopping bandwidth for this channel is
The length of the shiftregister 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.
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 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.
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
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 noncoherent detection is
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