Spring 2017
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GUC
Tutorial 3
Modulation
Q1) Suppose a CATV system uses coaxial cable to carry 100 channels,
each of 6 MHz bandwidth. Suppose that QAM modulation is used.
m
QAM modulation with 2 point ⇒ m bits/pulse
T = pulse duration = 1/W ⇒ pulse rate = baud rate = W = 6 x 10
6
T = pulse duration = 1/W ⇒ pulse rate = baud rate = W = 6 x 106
a. What is the symbol rate/channel if a four-point constellation is used?
Eight-point? Given a 8 level quantizer.
R(bit rate) = 8W, if m = 8 ⇒ R =Sampling rate*bits= 2*W*8=96 Mbps.
Moving From base-band to pass band channel doubles the Bandwidth
W=2*96
Symbol rate= Bit rate/ number of bits per symbol
For a 4 point constellation 2 bits per symbol therefore symbol rate =
96*2/2=96 symbols/second
.
Q3) Suppose that the receiver in a QAM system is not perfectly
synchronized to the carrier of the received signal; that is, it multiplies the
received signal by 2cos(2πfc t + φ) and by 2sin(2πfc t + φ) where φ is a
small phase error. What is the output of the demodulator?
Demodulation occurs at the receiver:
Spring 2017
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GUC
However due to the imperfect synchronization the Received signal is
instead multiplied by 2cos(2πfc t + φ) and by 2sin(2πfc t + φ).
Solution:
The transmitted signal in QAM is y(t) = Ak cos(2πfc t) + Bk sin(2πfc t )
y(t)* 2cos(2πfc t + φ) = 2cos(2πfc t + φ) Ak cos(2πfc t) + 2cos(2πfc t + φ)
Bk sin(2πfc t )
= Ak {cos(φ) + cos(4πfc t + φ) } + Bk {-sin(φ) + sin(4πfc t + φ) }
The low pass filter removes the double-frequency component, so the
output of the upper demodulator circuit is: Ak cos(φ) - Bk sin(φ) .
When the phase error φ is small, then cos φ is approximately 1, and
sin φ is approximately 0, so the phase error causes a small error in the
demodulator output. As the phase error increases however, the desired
signal Ak becomes harder to discern because the cosine term decreases
and the sine term increases.
It can be similarly shown that the output of the lower demodulator is: B k
cos(φ) + Ak sin(φ) .
Spring 2017
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GUC
Q4) A phase modulation system transmits the modulated signal Acos(2πf c
t + φ) where the phase φ is determined by the 2 information bits that are
accepted every T-second interval:
for 00, φ = 0 ; for 01, φ = π/2 ; for 10, φ = π; for 11, φ = 3π/2.
Solutions follow questions:
a. Plot the signal constellation for this modulation scheme.
The transmitted signals corresponding to the phase values are as
follows:
for 00 φ = 0, so x(t) = Acos(2πfc t)
for 01 φ = π/2, so x(t) = Acos(2πfc t + π/2) = -Asin(2πfc t)
for 10 φ = π, so x(t) = Acos(2πfc t + π) = Acos(2πfc t + φ)
for 11 φ = 3π/2, so x(t) = Acos(2πfc t - π/2) =- Asin(2πfc t)
The signal constellation is shown below:
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b. Explain how an eight-point phase modulation scheme would
operate.
The generalization to an eight-point constellation is straightforward. In
the above figure we can see that the four constellation points are
placed at equidistant points in a circle about the origin. The figure
below shows how eight points can be placed in a circle with angle π/4
between them.
Midterm Question- In class Assignment
Use Figure 2 to draw the signal constellation for the modulation system
described in Figure 1
Assuming that Ak takes the values -1 and 1 and Bk takes the values -2 , 1 ,1 and 2 ( please define the axis you use on Figure 2) [ 3 marks]
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GUC
b. How many bits per symbol can the modulator in part _a (of this
question) support? [ 2 marks]
Assume a digital broadcast system that supports 50 stereo music
channels. The stereo music from each
Channel is digitized using a 12 bit per sample quantizer. The stereo music
signal has a bandwidth of 22Khz. The system is operating in an
environment that has noise level equal to -50 dB
a. What is the sampling rate [ 2 marks]
b. Calculate the bandwidth required to support the system assuming that
the system uses a 16 QAM Modulator