Question 1 – 7 points

The figure shows one period of the Fourier transform of the sequence obtained by sampling a periodic
signal
[V] 6
5
4
3
2
1
0
0 1 2 3 4 5 6 7 9 10 11 12 13 14 15
8
n
Knowing that the fundamental frequency of the continuous-time signal is f1 = 50 Hz and that the
signal maximum frequency is fM = 350 Hz, the student shall determine the sampling frequency with
which the above sequence has been sampled and shall motivate whether aliasing has occurred or not
and which harmonic component has caused aliasing.

Solution
The maximum harmonic in the signal is:
𝑓𝑓𝑀𝑀
𝑁𝑁𝑚𝑚𝑚𝑚𝑚𝑚 = =7
𝑓𝑓1
Since the period of the Fourier transform of the sequence is 13, this means that the sequence of
samples in the discrete-time domain is composed of M = 13 samples, taken over one period.
Therefore, the sampling frequency is:
𝑓𝑓𝑠𝑠 = 𝑀𝑀 ∙ 𝑓𝑓1 = 650 Hz
When M = 13 samples are taken, the maximum harmonic in the signal, to avoid aliasing, shall be:
𝑀𝑀 − 1
𝑁𝑁 = =6
2
Since the signal has 7 harmonics, aliasing has occurred. The harmonic causing aliasing is the 7th, and
the 6th harmonic seen in the Fourier transform is the replica of the 7th.
Question 2 – 9 points
A triangular signal is given, with zero mean value, peak-to-peak amplitude Vpp = 20 V and period T
= 20 ms. This signal is sampled over one period with a sampling frequency fs = 1.95 kHz. The samples
are converted into digital by an ideal ADC, with full-scale voltage value Vfs = Vpp = 20 V. The student
shall find the minimum ADC resolution, expressed in number of bits, that grants that the aliasing
errors are avoided.
Remind that the peak amplitude of the harmonic components of a triangular waveform is given by:
4Vpp
Vk = , with k the order of the harmonic component. Only odd harmonic components are present
π 2k 2
in a triangular signal.

Solution
Since one period of the signal is sampled, the number of samples acquired is:
𝑀𝑀 = 𝑓𝑓𝑠𝑠 ∙ 𝑇𝑇 = 39
This means that, to avoid aliasing, the maximum harmonic in the signal should be:
𝑀𝑀 − 1
𝑁𝑁 = = 19
2
However, a triangular waveform has an infinite number of harmonics. Therefore, the only way to
avoid aliasing is that the next harmonic of the signal, that is the 21st, has amplitude below the ADC
resolution.
The amplitude of the 21st harmonics is given by:
4∙𝑉𝑉𝑝𝑝𝑝𝑝
𝑉𝑉21 = 𝜋𝜋2 ∙21 2 = 18.38 mV
Therefore, it shall be:
𝑉𝑉𝑓𝑓𝑓𝑓
𝑉𝑉21 < 𝑛𝑛
2
from which we get:
𝑉𝑉𝑓𝑓𝑓𝑓
2𝑛𝑛 <
𝑉𝑉21
and finally:
𝑉𝑉𝑓𝑓𝑓𝑓
𝑛𝑛 < log2
𝑉𝑉21
log 𝑎𝑎
By solving this last equation, remembering that log 2 𝑎𝑎 = 10 2 , we get: n = 10.09, and therefore the
log10
minimum ADC resolution is 10 bits.
Question 3 – 5 points
The student shall explain the theoretical principle on which the design of a maximum asymptotic
decaying window is based and shall show how the coefficients of a third-order maximum asymptotic
window are obtained.

Solution
It is known that the amplitude of the side lobes of the Fourier transform of a cosine window is directly
related to the discontinuity of the window, in the time domain, on his edges (Gibbs phenomena).
Therefore, to design maximum asymptotic decaying windows we must design windows that are
maximally continuous at their edges.
Since the equation of the cosine windows is:
𝐿𝐿−1
2𝜋𝜋𝜋𝜋𝜋𝜋
𝑤𝑤(𝑛𝑛) = � 𝐴𝐴𝑙𝑙 ∙ cos � �
𝑀𝑀
𝑙𝑙=0
where M is the duration of the window, in number of samples, the maximum continuity of the function
at the edges means that the function and all its derivatives must be 0 in n = 0. Due to the symmetry
of the function, this implies that the functions and its derivatives are 0 also in n = M.
For a third-order window (L = 3), we have to assign a value to three coefficients, A0, A1 and A2, so we
need to write three equations. One of these equations is the normality condition.
Since the odd derivatives of a cosine function are sine functions, they’re always 0 in n = 0, and
therefore the three conditions are:

• Normality condition: ∑2𝑙𝑙=0|𝐴𝐴𝑙𝑙 | = 1

• Continuity of the function: ∑2𝑙𝑙=0 𝐴𝐴𝑙𝑙 = 0
• Continuity of the second derivative: ∑2𝑙𝑙=0 𝑙𝑙 2 𝐴𝐴𝑙𝑙 = 0
Therefore, it is possible to write the following system of equations:
|𝐴𝐴0 | + |𝐴𝐴1 | + |𝐴𝐴2 | = 1
� 𝐴𝐴0 + 𝐴𝐴1 + 𝐴𝐴2 = 0
𝐴𝐴1 + 4𝐴𝐴2 = 0
That provides the following results: A0 = 0.375, A1 = -0.5 and A2 = 0.125.
Question 4 – 5 points
The student shall describe the equations that allow to estimate the position of a GPS receiver and
correct the receiver clock, knowing the positions of the GPS satellites. The student shall explain why
the signals from at least 4 satellites are needed to solve those equations.

Solution
Geolocalization is based on the measurement of the time-of-flight (TOF) of the signals broadcasted
by the GPS satellites and received by the GPS receiver.
Each satellite i, i = 1, 2, 3, 4, …, transmits a message that contains the spatial coordinates (xi, yi, zi)
of the actual satellite’s position and the time si at which the message is sent, measured by the high-
precision clock on board the satellite.
Let’s now denote with 𝑡𝑡𝑖𝑖̃ the time, measured with the receiver’s clock, at which the GPS receiver,
positioned on the Earth surface at the location of unknown coordinates (x, y, z), receives the messages
sent by satellite i. Under the initial assumption that the receiver’s clock is synchronized with the
satellite clocks, the time-of-flight is given by 𝑡𝑡𝑖𝑖̃ − 𝑠𝑠𝑖𝑖 , and since the message travels at light speed c,
we might conclude that the distance between the receiver position and satellite i is given by:
𝑑𝑑̃𝑖𝑖 = (𝑡𝑡̃𝑖𝑖 − 𝑠𝑠𝑖𝑖 ) ∙ 𝑐𝑐
The actual distance between the receiver and the satellite is given by:
𝑑𝑑𝑖𝑖 = �(𝑥𝑥 − 𝑥𝑥𝑖𝑖 )2 + (𝑦𝑦 − 𝑦𝑦𝑖𝑖 )2 + (𝑧𝑧 − 𝑧𝑧𝑖𝑖 )2
If the assumption of synchronization of the receiver’s and satellites’ clocks were correct, 𝑑𝑑̃𝑖𝑖 = 𝑑𝑑𝑖𝑖 , so
we have only three unknown – x, y, and z – to determine and we would need the messages from only
three satellites to determine the position.
However, the receiver’s clock is generally not synchronized with the satellites’ clocks (that, on the
contrary, are all synchronized. Therefore, the true reception time is given by 𝑡𝑡𝑖𝑖 = 𝑡𝑡̃𝑖𝑖 − 𝑏𝑏, where b is
the deviation between the receiver’s clock and the satellites clock. It can be assumed that b is the
same for all received messages, since we can assume that the receiver’s clock bias doesn’t vary in the
period of time needed to receive the messages from all satellites that are in sight of the receiver’s
position.
Therefore, the distance seen by the receiver is not the actual distance di, but, instead, we have:
𝑑𝑑̃𝑖𝑖 = 𝑑𝑑𝑖𝑖 + 𝑏𝑏 ∙ 𝑐𝑐
To evaluate the correct position, we must hence determine a fourth unknown: b. It is then necessary
to have a fourth equation and, therefore, the messages from at least four satellites are needed.
If the messages from more that four satellites are available, it is possible to write a system of equations
with more equations than unknown (overdetermined system of equation) that can be solved with a
fitting method, thus reducing the effect of noise and measurement uncertainty of the final result and
providing a more accurate localization.