sample question for final exam

7
(a) A new ICU room is equipped with a central station for simultaneous monitoring of electrocardiograms (ECG) of 10 patients. The data for 10 patients are brought to a processing center over wires and are sampled, quantized, binary coded and time division multiplexed. The multiplexed data are then transmitted to the monitoring station as shown in Figure 1. As an engineer, you’re asked to design the ECG monitoring system based on the following specifications: The ECG signal is 200Hz with voltage range of 0-5V The maximum tolerable error in sample amplitudes is 0.25% of peak signal amplitude. The signal must be sampled at least 25% above the Nyquist sampling rate. Framing and synchronization (FS) requires additional 1.25% of total data bits. The FS is added at the beginning of each frame. Figure 1: ECG Monitoring System Source/ patient

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Final Exam

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Page 1: Sample Question for FInal Exam

(a) A new ICU room is equipped with a central station for simultaneous monitoring of

electrocardiograms (ECG) of 10 patients. The data for 10 patients are brought to a

processing center over wires and are sampled, quantized, binary coded and time division

multiplexed. The multiplexed data are then transmitted to the monitoring station as

shown in Figure 1. As an engineer, you’re asked to design the ECG monitoring system

based on the following specifications:

The ECG signal is 200Hz with voltage range of 0-5V

The maximum tolerable error in sample amplitudes is 0.25% of peak signal

amplitude.

The signal must be sampled at least 25% above the Nyquist sampling rate.

Framing and synchronization (FS) requires additional 1.25% of total data bits. The FS

is added at the beginning of each frame.

Figure 1: ECG Monitoring System

Determine the key parameters of the system design:

(i) Quantization level. (Ans: L =200, actual L = 256)

[2 marks]

(ii) Number of bits used. (Ans: 8)

[2 marks]

(iii) Step size. (Ans: = 5/256)

[1 mark]

Source/patient

Page 2: Sample Question for FInal Exam

(iv) Transmission bit rate from each source. (Ans: Rsource = n x fs = 4kbps)

[2 marks]

(v) Number of channels in one frame. (Ans: 10)

[1 mark]

(vi) Minimum transmission cable bandwidth required to transmit data from all

sources. (Bmin = R/2 = 40.5/2 = 20.25kHz)

[3 marks]

(vii) Output SNR from each source. (Ans: 49.92dB)

[1 mark]

(viii) Suggest a strategy to improve the output SNR from each source to become at least

66dB. Determine the new transmission bit rate to support this requirement.

Page 3: Sample Question for FInal Exam

Design a TDM system that can accommodate two analog signals and one synchronous digital

signal. The designed system must have the following specifications:

The analog signals are sampled at 1.5 times the Nyquist rate, quantized and then

binary coded.

They are multiplexed using one commutator producing a total output of 24000

bps.

Each channel has equal information bit rate.

The minimum number of bits per sample is 3.

Present your design using a block diagram and label the output of all stages. Indicate the

choice of (i) baseband bandwidth of the analog signals and (ii) the corresponding

quantization level used in your design. Justify your design with calculations.

Ans: 3 inputs, (2 analog and 1 digital)

Output 24kbps so each user transmitting 24k/3 = 8kbps

Input BW Sampling BitrateAnalog 1 A A x 2 x 1.5 n x A x 2 x 1.5 = 8kbpsAnalog 2 B B x 2 x 1.5 n x B x 2 x 1.5 = 8kbpsDigital C - 8kbps = 8kbps

24kbpsBased on n= 3 bits, A=B = 888.9Hz

If n is chosen to be 4 bits, then A=B = 666.7Hz

Page 4: Sample Question for FInal Exam

A FDM group multiplexer takes 12 voice channels (4 kHz per channel) and puts them on

subcarriers starting at 60 kHz using SSBSC utilizing upper sideband (USB). The groups are then

multiplexer together to produce a supergroup multiplexer (consists of 5 groups) with a

subcarriers starting at 420 kHz utilizing USB as well.

(i) Why does the standard voice channel is set at 4 kHz? (ans: including guard band)

[1 mark](ii) Determine the range of the group spectrum. (Ans: 60-108kHz)

[1 mark](iii) Draw the frequency spectrum of the supergroup multiplexer showing all the

pertinent frequency values (Ans: range from 312kHz – 552kHz)

[4 marks]

(iv) An extra guard band of 1 kHz is now being added between each channel. How

many voice channels can be assigned within the supergroup frequency band?

(Assume the output bandwidth of group/supergroups remain the same)

(Ans : 240/5 = 48 channels)

[2 marks]

(v) If the original FDM system employed DSBSC instead of SSBSC, what happened

to the output of the groups and supergroups multiplexer?

(ans: number of channel will become halves the original if subcarrier frequency

change accordingly

or the output will be overlapping with one another if subcarrier frequency remain

the same)

[2 marks]

Page 5: Sample Question for FInal Exam

Δv

Vvi

(a) μ-law compander uses a compressor which relates output to input by the relations and

characteristics as below.

The range of allowable voltage is V=±10volts and 256 quantization levels are

employed. The parameter μ determines the degree of compression.

(i) What is the voltage interval between levels if there is no compression?

(ii) For μ = 100 what are the minimum and the maximum voltage interval between

levels?

V out=Vmax ln (1+μV in/Vmax )ln (1+μ )

Page 6: Sample Question for FInal Exam

01

-cos ωct

cos ωct

-sin ωct sin ωct(0° reference)

00

1110

(a) Figure below shows a phasor diagram and its truth table for a QPSK modulator. The

carrier frequency is 50 kHz with baud rate = 25 kbps. Given the input stream to the

modulator is 101101.

(i) Determine the input bit rate and bandwidth.

[2 marks]

(ii) Sketch the waveform of the input and modulated signal at the original carrier

frequency.

[3 marks]

(iii) Determine the degree of phase shift of the carrier modulating the Q bit with

respect to the reference oscillator carrier. Justify your answer. Draw a detail block

diagram of the QPSK modulator that will produce the intended output.

Binary Input

QPSK Output PhaseQ I

0 0 135°0 1 45°1 0 -135°1 1 -45°