Baseband Transmission MCQ || Baseband Transmission Questions and Answers - Page 2 of 5

11. Output data ratio of an 8-bit PCM-TDM system sampling 24 voice channels, comparing these using μ-law at the rate of 8 kHz with a 1 frame alignment word is:

1.2 × 10⁶ bit/sec
1.4 × 10⁶ bit/sec
1.6 × 10⁶ bit/sec
1.8 × 10⁶ bit/sec

Answer.3. 1.6 × 10⁶ bits/sec

Explanation:-

Given that

Each frame consists of 24 8-bit words, plus a single synchronizing bit.

Total number of pulses sent over sampling period = 24 × 8 + 1 = 193

Sampling frequency (f_s) = 8 kHz

Sampling Time(T_s) = 1 / f_s = 1 / 8 kHz = 0.125 ms

Time for each bit(T = T_s / 193 = 0.648 μs

Output data ratio(R = 1 / T_b = 1 / 0.648 μs = 1.544 mbps ≈ 1.6 × 10⁶ bits/sec.

Alternative method:

R_b = sampling frequency × total number of pulses over T_s

= 8 × 10³ × 193

= 1.544 mbps ≈ 1.6 × 10⁶ bits/sec

12. Which type is used and preferred in digital logic circuits?

NRZ-L
NRZ-M
NRZ-S
None of the mentioned

Answer.1. NRZ-L

Explanation:-

Non-Return Zero- Level or NRZ-L PCM is the one that is used extensively in digital logic circuits. A binary 1 is represented by one voltage level and a binary 0 is represented by another voltage level. There is a change in level whenever the data change from 1 /to 0 or from 0 to 1.

13. In digital transmission, the modulation technique that requires the minimum bandwidth is:

PCM
PAM
DPCM
Delta modulation

Answer.4. Delta modulation

Explanation:-

In PCM an analog signal is sampled and encoded into different levels before transmission
The bandwidth of PCM depends on the number of levels
If each sample is encoded into n bits, then the bandwidth of PCM is nf_s
The bandwidth of DPCM is almost the same as that of PCM signal, the only difference between PCM and DPCM is that the dynamic range is reduced in the DPCM signal
However, in the case of Delta modulation, each sample is sent using only 1 bit which is +Δ or -Δ
Hence there is bandwidth saving in Delta modulation

14. The number of channels in a TDMA system is given by

$N = \dfrac{m(2B_{Tot} – B_{guard})}{B_c}$
$N = \dfrac{m(B_{Tot} – 2B_{guard})}{B_c}$
$N = \dfrac{m(B_{Tot} – 2B_{guard})}{2B_c}$
$N = \dfrac{2m(B_{Tot} – B_{guard})}{B_c}$

Answer.2. $N = \dfrac{m(B_{Tot} – 2B_{guard})}{B_c}$

Explanation:-

Time-division multiple access (TDM is a channel access method for shared-medium networks.

It allows several users to share the same frequency channel by dividing the signal into different time slots. Number of channels in TDMA system is given by-

$\rm N=\dfrac{m*(B_{tot}-2*B_{gaurd})}{B_c}$

N = number of channels

m = number of TDMA users per ratio channel

B_tot = total spectrum allocation

B_guard = Guard Band

B_c = Channel bandwidth

Note:-

Number of channels in FDMA system is given by-

$\rm N=\dfrac{B_t-B_{guard}}{B_c}$

N = number of channels

B_t = total spectrum allocation

B_guard = guard band

B_c = Channel bandwidth

15. NRZ-S is complement of _______

NRZ-L
NRZ-M
NRZ-L & NRZ-M
None of the mentioned

Answer.2. NRZ-M

Explanation:-

NRZ-S PCM waveform is the complement of NRZ-M PCM waveform. A one is represented by no change in level and zero by a change in level.

16. To transmit N signals each band limited to fm Hz by time-division multiplexing will require a minimum bandwidth of

f_m
2fm
N fm
fm/N

Answer.3. N fm

Explanation:-

For Time division multiplexing (TDM) minimum Band width is given by-

(BW)_min = Rb/2

Where, R_b = Nnf_s

N = No. of signals

n = no. of Bit

f_s = sampling frequency = 2 × f_m

f_m = msg signal Bandwidth

(BW)_min = N_n2fm/2 = N_nfm

Given:

N signal with f_m Bandwidth.

TDM minimum B.W = Nnf_m

Assume n = 1 Bit

BW_min = Nf_m

17. To reduce the distortion, which one of the following can be used?

Multiplexer
Sampler
Companders
Equalizer

Answer.4. Equalizer

Explanation:-

Inter-symbol interference (ISI): In communication ISI is a form of distortion symbols interfere with subsequent symbols and hence all the subsequent symbols bring the noise to the original symbol. This phenomenon can greatly affect the decision of the device at the receiver. Therefore, transmitting and receiving devices are designed in a way to reduce ISI as much as possible.

Equalizer: The equalizer is a device that attempts to reverse the distortion introduced when a signal is transmitted through a channel. The main purpose is to reduce inter-symbol interference.

Zero-forced equalizer: The ZF strategy for the linear equalizer design is to choose the equalizer to be the inverse of the channel. It attenuates the response at frequencies that are amplified by the channel, and it amplifies the response at frequencies that are attenuated by the channel.

18. The return to zero waveform consists of

Unipolar RZ
Bipolar RZ
RZ-AMI
All of the mentioned

Answer.4. All of the mentioned

Explanation:-

The return-to-zero (RZ) waveforms consist of unipolar-RZ, bipolar-RZ, and RZ-AMI (alternate mark inversion). These codes find applications in baseband data transmission and in magnetic recording.

19. An analog voltage is in the range of 0 to 8 V is divided in eight equal intervals for conversion to 3-bit digital output. The maximum quantization error is

0 V
0.5 V
1 V
2 V

Answer.2. 0.5 V

Explanation:-

The maximum quantization is given as:

Q_e(max) = Δ/2

Δ = step size given by:

Δ = (V_max − V_min)/L

L = Number of levels

Calculation:

With n = 3, the number of levels will be:

L = 2³ = 8

With analog input in the range 0 to 8 V and L = 8, the step size will be:

Δ = (8 − 0)/8 =1

Now, the maximum quantization error will be:

Q_e(max) = 1/2 = 0.5

20. An AM Scheme is working with a modulation index of 0.8. Power efficiency of this modulation Scheme is

Answer.1. 24%

Explanation:-

The efficiency of a modulation scheme is defined as:

Efficiency = Sideband Power/Total Power

Mathematically, the power efficiency of an AM signal with single tone modulation is given by:

ηAM = µ²/(2 + µ²)

μ = Modulation index

Calculation:

With μ = 0.8, the efficiency will be:

ηAM = 0.8²/(2 + 0.8²)

η = 24.24 %

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