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# Test: Biasing In BJT Amplifier Circuits & Spread Spectrum

## 18 Questions MCQ Test Analog Circuits | Test: Biasing In BJT Amplifier Circuits & Spread Spectrum

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This mock test of Test: Biasing In BJT Amplifier Circuits & Spread Spectrum for Electronics and Communication Engineering (ECE) helps you for every Electronics and Communication Engineering (ECE) entrance exam. This contains 18 Multiple Choice Questions for Electronics and Communication Engineering (ECE) Test: Biasing In BJT Amplifier Circuits & Spread Spectrum (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Biasing In BJT Amplifier Circuits & Spread Spectrum quiz give you a good mix of easy questions and tough questions. Electronics and Communication Engineering (ECE) students definitely take this Test: Biasing In BJT Amplifier Circuits & Spread Spectrum exercise for a better result in the exam. You can find other Test: Biasing In BJT Amplifier Circuits & Spread Spectrum extra questions, long questions & short questions for Electronics and Communication Engineering (ECE) on EduRev as well by searching above.
QUESTION: 1

Solution:

QUESTION: 2

Solution:

QUESTION: 3

### Find R (in 1) for R (sig) = 5 kΩ.

Solution:

It is the parallel combination of the 32 kΩ resistor and 2.6 kΩ resistor respectively.

QUESTION: 4

Find Vb1/Vsig for R(sig) = 5 kΩ.

Solution:

QUESTION: 5

Find R (in 2).

Solution:

It is the parallel combination of the 32 kΩ resistor and 2.6 kΩ resistor respectively.

QUESTION: 6

Find Vb2/Vb1.

Solution:

QUESTION: 7

For Rl = 2 kΩ find Vo/Vb2.

Solution:

QUESTION: 8

Find the overall voltage gain.

Solution:

QUESTION: 9

(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.

Q. The PN sequence length is

Solution:

The PN sequence length is N = 2m – 1 = 16 – 1 = 15.

QUESTION: 10

The chip duration is

Solution:

Tc = 1/(107) or 0.1 µs.

QUESTION: 11

The period of PN sequence is

Solution:

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

QUESTION: 12

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

Solution:

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

QUESTION: 13

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

Solution:

PG = 4 x 4 = 16 or 12 db.

QUESTION: 14

(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.

Q. The coding gain is

Solution:

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

QUESTION: 15

The processing gain is

Solution:

PG = (107)/(2 x 1000) = 5000 or 37 db.

QUESTION: 16

(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]

Q.  If the hop rate is one per bit, the hopping bandwidth for this channel is

Solution:

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.

QUESTION: 17

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

Solution:

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.

QUESTION: 18

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

Solution:

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