EL220 Comms II 03/04 master solution
1. Fourier Transform
a) The Fourier series applies to periodic signals only. With period T0 , the spectrum of such a
signal will only consist of sinusoidal contributions at the fundamental frequency f0 = 1/T0 ,
and potentially (an infinite number of) harmonics thereof (including DC).
b) x(t) periodic −→ X(jω) must be discrete (Fourier series);
x(t) is a sampled signal −→ X(jω) must be periodic (duality);
sampling period Ts = 1ms −→ spectrum periodic with fs = 1/Ts = 1kHz;
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must have sinusoidal component at fundamental frequency f0 = 20·T
= 50Hz
s
might have sinusoidal components at 0Hz, 100Hz, 150Hz, . . . 500Hz (maximum representable
frequency!);
x(t) is real valued −→ X(jω) = (X(jω))∗.
c)
i) The DFTs in Fig. 1(b) and (c) have different length – in (b) an N = 24 point DFT and
in (c) an N = 20 point DFT has been applied.
ii) In (b), N = 24 is not an integer multiple of the signals fundamental period (20 samples); therefore discontinuities arise, as the DFT periodises the data. This results in
spectral leakage from the main peaks in the speactrum into neighbouring coefficients.
Hence, Figure 1(b) is not a faithful representation of the spectrum of x[n].
In (c), the window of length N = 20 captures exactly one fundemental period of x[n],
the above problems do not arise, and Figure 1(c) is a faithful representation. The frequency bins coincide with the frequency components predicted by the Fourier series
theorem.
iii) Fig. 1(c) cannot be improved. For Fig. 1(b), either the window length could be chosen
differently, such that N = 20k with an integer k, or a tapered window could be applied
to the data in order to reduce discontinuities at the margins of the data window.
d) Tapered windows reduce spectral leakage by disemphasising the interval margins, hence minimising the discontinuity in the DFT’s periodisation of the data. However, this results in the
original spectrum being convolved with the lowpass spectrum of the window function, blurring original spectrum and therefore reducing the spectral resolution.
e) The FFT is an efficient implementation of the DFT, providing the same results at a considerably lower complexity (reduction by log2 N/N).
Both DFT and FFT are based on the computations of multiply-accumulate operations (MACs),
for which a DSP processor is highly optimised.
2. Random Signals
a) Condition:
R
pV (v)dv = 1.
Z∞
−∞
∞
1 −b|v|
2 Z −bv
2
e
dv =
e dv = − e−bv
a
a
ab
0
Hence a = 2/b.
1
∞
0
=
2
ab
(1)
b) CDF:
PV (v) =
Zv
pV (u)du
−∞
v<0:
PV (v) =
Zv
−∞
(2)
b bu
1
1
e du = [ebu ]v−∞ = ebv ;
2
2
2
1
PV (v) = +
2
v≥0:
=
Zv
0
b −bu
1 1
e du = + [−e−bu ]v0 =
2
2 2
1
1 1 −bv
− (e − 1) = 1 − e−bv ;
2 2
2
Sketch: gradient of CDF in v = 0 is pV (0) =
(3)
(4)
(5)
b
2
PV (v)
1
1/2
v
−1/b
1/b
Clearly, for smaller b the PDF is further spread out, the outliers are greater, and therefore the
variance or power larger.
c) Power of v(t):
σv2
=
=
Z∞
v 2 pV (v)dv
(6)
−∞
Z∞
v 2 e−v dv
(7)
0
= [−
v 2 e−v ]∞
0
+2
Z∞
ve−v dv
= 0 + 2 · {[ −
ve−v ]∞
0
+
Z∞
e−v dv}
(9)
0
= 0 + 2 · {0 + [ − e−v ]∞
0 } = 2·1 = 2
d)
(8)
0
(10)
i) Due to the central limit theorem, the PDF of r[n] will tend towards a Gaussian distribution.
2
[n] = 2δ[ncd] (uncorrelated input). The auto-correlation
ii) Auto-correlation of v[n] is rvv
function of the output r[n] is given by
rrr [n] = h[n] ∗ h∗ [−n] ∗ rvv [n]
= h[n] ∗ h[−n]
(11)
(12)
The result can be obtained by graphical evaluation of the correlation/convolution, with
the result shown below:
2
10
6
rr
r [n]
8
4
2
0
−10
−8
−6
−4
−2
0
2
4
6
8
10
lag n
iii) Power / variance of r[n] = rrr [0] = 10 from the above figure.
3. Amplitude Modulation
a) By interpreting the envelope as a cosine signal, γ = 1.5 and γ = 0.5 are found.
b) For γ = 1.5, the signal is undermodulated: (-) a strong carrier component is present, which
wastes transmission power; (+) very simple cricuits can demodulate this signal.
For γ = 0.5, the signal is overmodulated: (-) more sophisticated modulation is required; (+)
transmission energy is better utilised.
c) Use Euler and that ejω0 t ◦—• 2πδ(ω − ω0 ):
1
1
s(t) = (γ + (ejωi t + e−jωi t )) · (ejωc t + e−jωc t )
2
2
1 j(ωc −ωi )t γ jωc t 1 j(ωc +ωi )t
= { e
+ e
+ e
} + (u(t))∗
4
2
4
{z
}
|
(13)
(14)
u(t)
1
γ
1
S(jω) = 2π
δ(ω − ωc + ωi ) +
δ(ω − ωc ) + δ(ω − ωc − ωi ) + (15)
4
2
4
γ
1
1
2π
δ(ω + ωc − ωi ) +
δ(ω + ωc ) + δ(ω + ωc + ωi )
(16)
4
2
4
Sketch (only components on the positive frequency axis are shown; |S(jω)| is symmetric to
ω = 0):
γ/2
S(jω) /2π
1/4
1/4
ω
ω c− ω i
ωc
ω c+ ω i
d) Due to Parseval, the power components are independent:
2
carrier power: 2 γ22 =
γ2
2
sideband power: 4 412 =
=
1
2
1
4
e) Demodulation (given γ = 0):
Figure 3.
- ×j
s(t)
signal before lowpass filtering with LPF:
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- LPF
-
x̂(t)
2 cos(ωc t − φ)
x′ (t) = s(t) · 2 cos(ωc t − φ) = x(t) · cos(ωc t) · 2 cos(ωc t − φ)
= x(t) · {cos(φ) + cos(2ωc t − φ)}
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(17)
(18)
The lowpass filter has to remove the component oscillating at twice the carrier frequency,
resulting in
x̂(t) = x(t) · cos(φ)
(19)
which is the originally transmitted signla apart from a carrier frequency offset φ.
4. Baseband Channel Model
a) CIR: transmit- and receive filters, modulation / demodulation, and the physical channels (any
system in between the baseband Tx and Rx). AWGN: additive white Gaussian noise caused
by e.g. athmospheric, other users transmitting in the same band & time slot etc.
b) A dispersive CIR causes intersymbol interference; AWGN acffects the performance of the
detector in the receiver.
c) h[n] is a two-path model. The transport delay between Tx and Rx is equivalent tp one symbol period; a second echoic components arives 3 sampling periods later (it has the same
amplitude, hence no significant loss has been incurred in reflections.
d) Block diagram consisting of TDL, multipliers and adders:
x [n]
0
x [n −1]
x [n−2]
1
0
x [n −3]
adder
1
multiplier
x [n]
delay element
e) We exploit δ[n] ◦—• 1, and the time-shift property of the DFT:
H(ejΩ ) = e−jΩ + e−j3Ω
= e−j2Ω (ejΩ + e−jΩ ) = 2 cos Ωe−j2Ω
(20)
(21)
Magnitude: |H(ejΩ )| = 2 cos Ω. This is a bandstop filter, passing DC and frequencies at 1/2
sampling rate, but freuqnecy components at 1/4 of the sampling rate.
f) Time variation can be due to changes in the physical channel i.e. due to mobility of one
handset, or due to a carrier frequency mismatch between transmitter and receiver.
5. Detection
a) A bit stream is de-multiplexed into a two-bit word. This two-bit word is mapped onto a complex valued symbol by means of a constellation diagram. This symbol can then be quadrature
amplitude modulated and transmitted.
b) DQPSK encodes the change between the present input symbol and the previously transmitted
one. This circumvents the ambiguity of the constellation with respect to phase shifts on π/2,
π, or 3π/2 in the receiver, as encountered with standard QPSK. However, a bit error affects
two successive symbols.
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c) Note: there was an error in the exam question, it should be written as : “Consider the transmission of symbols A and B over a noisy channel utilising the signal levels ∓d to encode A
and B, respectively”.
By inspection of the Gaussian PDF, the noise variance is σn2 = 1 , the signal power d2 = 4.
Hence
d2
SNR = 10 log10 2 = 6dB
σn
d) bit error probability:
P (error) = P (A) · P (r > θ|A) + P (B) · P (r < θ|B)
=
1
4
Z∞
θ
1
2
√ e−(r+d) /2 dr
2π
+
3
4
√
√
substitutions: y = −(r + d)/ 2 and z = (r − d)/ 2;
1
P (error) = − √
4 π
=
1 2
√
8 π
−∞
Z
−∞
2
e−y dy +
−∞
3 2
√
8 π
3 θ−d
1
θ+d
=
Q(− √ ) + Q( √ )
8
8
2
2
(22)
1
2
√ e−(r−d) /2 dr
2π
3
√
4 π
2
e−y dy +
√
−(θ+d)/ 2
√
2
−(θ+d)/
Z
Zθ
√
2
(θ−d)/
Z
−∞
√
2
(θ−d)/
Z
2
e−z dz
2
e−z dz
(23)
(24)
(25)
−∞
(26)
e) Increasing d effectively raises the signal’s power level and therefore improves the SNR at the
receiver. This will lead ot a reduced processing cost.
Furthermore, the detector θ better optimally selected, since the transmitted symbols A and B
are not equiprobably and transmitted symbols B stand a higher chance of being corrupted.
The ideal threshold will balance the number of errors made when transmitting a or B.
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