nts.uni-duisburg-essen.dents.uni-duisburg-essen.de/downloads/tm/Old_Exams_TM2.pdf · NTS/F5 GMUD...

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Examination date: 11.03.2002
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Exercise 1 (10 Points):
Two signals y(t) and z(t) are given. The spectra Y (ω) and Z(ω) are known.
Z(ω) = rect
1.1 (2 Points)
Sketch Z(ω) and Y (ω). Give all important points on the abscissa and ordinate.
1.2 (2 Points)
Sketch the spectrum X(ω) of the signal x(t) = z(t) · y(t). Give all important points on the abscissa. Denote the maximum value of the magnitude of X(ω) as A and do not determine it.
1.3 (1 Point)
1.4 (1 Point)
Sketch the spectrum V (ω) of the signal v(t) = z(t) · y0(t), where y0(t) is the analytic signal of y(t).
1.5 (1 Point)
Determine the analytic signal x0(t) = F−1{X0(ω)} as a function of v(t).
1.6 (2 Points)
1.7 (1 Point)
Conclude now from above results the relation between the analytic signal x0(t) and the equivalent lowpass signal xELP(t) of x(t). Justify your response.
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Given the signal constellation of a 4-QAM (Quadrature Amplitude Modulation).
f1
f0
x2
x3
x1
x4
d1
d0
The observed process is Y (t) = xi(t)+N(t), i = 1, 2, 3, 4. N(t) is a Gaussian distributed, white noise process with zero mean.
2.1 (1 Point)
Sketch the 4 decision regions Ri, i = 1, 2, 3, 4 pertaining to the signals xi.
2.2 (2 Points)
Calculate the probability of a correct decision Pc|x1 of the symbol x1. Use the function
Pe,binary(dj) = P { Nj < −dj
2
} , j = 0, 1, where dj is the horizontal and vertical distance
respectively.
2.4 (1 Point)
Determine the symbol error probability Pe,4−QAM of 4-QAM if d0 = d1 = d.
The observed process is now Ynew(t) = xi(t) + Λ(t), i = i = 1, 2, 3, 4. Λ(t) is a zero mean independent noise process. Its probability density function (pdf) is given by
pΛ(λ) =
2.5 (1 Point)
Sketch pΛ(λ). Give all important points on the abscissa and ordinate.
2.6 (2 Points)
2
2.7 (1 Point)
Calculate the new symbol error probability Pe,4−QAM of 4-QAM if d0 = d1 = d.
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Given is a DPSK-modulator. The relation between the binary source sequence d(k) and the phase differences φ(k) is given in the following table:
d(k) 00 01 11 10
φ(k) π 4
3.1 (1 Point)
Which code is used for the coding of the binary sequence? Sketch the signal constellation of the sequence. Denote all important points on the abscissa and ordinate.
Now, the binary sequence
k 0 1 2 3 4
d(k) 00 11 01 10 01
is transmitted. The initial phase for k = −1 is admitted to φ(−1) = 0.
3.2 (3 Points)
Determine the phase of the given binary sequence d(k) and sketch it. Denote all important points on the abscissa and ordinate.
3.3 (3 Points)
At k = 2, a phase error occurs. The phase is given by φ(2) = φ(2) + π 2 . Determine the
detected binary sequence d(k).
3.5 (2 Points)
Now, a π 4 -DQPSK-Modulator is used. Explain this modulation scheme by using the se-
quence given in table 1. Give the phase values and sketch the signal constellation of the binary sequence d(k) in table 1.
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Given the following modulator. The used modulation index is mI = 0.5.
∫ φ(t)
2πmI
− sin(ωct)
v(t)
+∞∑ k=−∞
h(t) cos(ωct)
Fig.1: Modulator
The impulse response h(t) of the frequency pulse shaping filter is assumed to be:
h(t) =
{ A for 0 ≤ t ≤ 2Tb
0 else. 4.1 (3 Points)
Determine and sketch the impulse response q(t) of the phase pulse shaping filter. Denote all important points on the abscissa and ordinate. Determine the value of the constant A which fulfill the condition lim
t→+∞ q(t) = 0.5.
4.2 (4 Points)
Now, the sequence of the bit stream ak shown in the following figure is considered. ak
k 7630
2 4 51
Sketch the phase φ(t) with φ(0) = φ0 = 0. Give all important points on the abscissa and ordinate.
4.3 (2 Points)
Hint: cos(x) cos(y) sin(x) sin(y) = cos(x ± y)
4.4 (1 Point)
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Four questions are given from the content of the lecture “Transmission and Modulation of Signals 2“ in the following. These questions can be answered independently. The answers shall be restricted to a few short remarks and/or plots or formulas respectively. For each question the correct and complete answer yields 2 points.
Question 1 (2 Points):
Explain the task of clock synchronization and show how an analog PLL has to be modified to become useful for clock synchronization.
Explain what entropy means and give a formula. Under which situation reaches the entropy its maximum ?
What is the principal disadvantage of PSK and how can this disadvantage be circumven- ted ?
Explain the second Nyquist-criterion.
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Solutions for the Examination Paper Transmission and Modulation of Signals 2
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ω
2π Z(ω) ∗ Y (ω).
2π Z(ω) ∗ Y 0(ω).
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Solution 1.6: (2 Points)
= π (δ(ω − ωc) + δ(ω + ωc)) · 2ε(ω)
= 2πδ(ω − ωc).
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f0
f1
R2
R1 = [0,∞]×[0,∞], R2 = [−∞, 0]×[0,∞], R3 = [−∞, 0]×[−∞, 0], R4 = [0,∞]×[−∞, 0].
Y0 = x1,0 + N0 = d0
2 + N0
Pc|x1 = P {Y0 > 0, Y1 > 0} = P {Y0 > 0}P {Y1 > 0} = P
{ d0
= 1 − Pe,binary(d0) − Pe,binary(d1) + Pe,binary(d0)Pe,binary(d1).
Solution 2.3 (2 Points)
= Pe,binary(d0) + Pe,binary(d1) − Pe,binary(d0)Pe,binary(d1).
Solution 2.5: (1 Point)
2d−2d − d 2
2d
pΛ(λ)
λ
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= 2 7
16 − 49
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Solution 3.1 (1 Point
The sequence is Gray-coded. The constellation of the signal is given in the following figure.
0001
k -1 0 1 2 3 4 d(k) 00 11 01 10 01 φ(k) π
4 −3π
3π/4
π/2
π/4
−π/4
−π/2
Solution 3.3 (3 Points) k = 2: φ(2) = φ(2) + π 2 . The detected phase is:
k -1 0 1 2 3 4
φ(k) 0 π 4
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φ(k) π
4 − 0
Solution 3.4: (1 Point)
⇒ 2 Errors.
4 ,−3π
4 ,−π
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2ATb for t > 2Tb.
2πATb for t > 2Tb
4Tb t for 0 ≤ t ≤ 2Tb
π 2
π
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= cos (ωct + φ(t)) .
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Answer 1 (2 Points):
The task of clock synchronization is to minimize the sampling error TS . The analog PLL has to be modified as follows:
1. the multiplicator has to be changed by a sampler
2. the output of the VCO should be a series of Dirac-impulse, switching the sampler on and off, instead of a sinusoidal function.
3. the lowpass has to be substituted by a device that generates an output being pro- portional to the sampling error.
Entropy is the average information content of an alphabet X
H(X ) =
pl · ld(pl).
It takes its maximum if all the probabilities are the same, so
pl = 1
The principal disadvantage are the phase jumps which cause a higher bandwidth occupa- tion. To circumvent this problem, continuous phase modulation techniques can be applied instead.
g0(l TS
for l = ±1 0 else
.
has to be fulfilled. This is called the second Nyquist-criterion.
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Fig. 1 Maximum likelihood detection using correlation receiver
A digital transmission system uses correlation receiver for Maximum likelihood detection as shown in figure 1. The source is binary, discrete and memoryless with symbol set {0, 1}. The waveform generator provides the waveforms x1(t) and x2(t) as shown in figure 2 for the symbols 0 and 1 respectively.
Tb
4
x1(t)
t
−A
+A
Tb
Tb
4
+A
−A
Tb
t
x2(t)
Fig.2 Symbol waveforms x1(t) and x2(t)
In the following we consider transmission of the sequence {0, 1}.
1.1 (1 point)
Plot xm(t) for 0 ≤ t ≤ 2Tb. Denote all important points on abscissa and ordinate.
1.2 (1 point)
The additive noise n(t) is shown in figure 3. Assume that it is Gaussian distributed. Plot the received signal y(t). Denote all important points on abscissa and ordinates.
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1.3 (2 points)
Plot the instantaneous amplitude eA(t). Denote all important points on abscissa and ordinates.
1.4 (2 points)
Plot the instantaneous amplitude eB(t). Denote all important points on abscissa and ordinates.
1.5 (1 point)
Plot the sampled values of eA(t) and eB(t) at the sampling instants t = nTb, n = 1, 2. Denote all important points on abscissa and ordinates.
1.6 (2 points)
Determine the likelihood values λ1 and λ2. Does error occur in decision process?
1.7 (1 point)
The maximum likelihood receiver of figure 1 can be also implemented by using matched filters. Plot the impulse responses h1(t) and h2(t) of the matched filters for the detection of symbol waveforms x1(t) and x2(t) respectively.
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π 4
Fig.1 Constellation diagram of QPSK
Given is a uniform and symmetric signal constellation diagram of a QPSK signal. The signal exhibits four possible phases φm = (2m−1)π
4 , where m = 1, 2, 3, 4. The basis vectors
f0 and f1 are
sin(2πfct) are the orthonormal basis
functions. Ts is the symbol duration and fc is the carrier frequency.
2.1 (1 point)
In general the transmitted QPSK signal x1(t) corresponding to φ1 can be expressed as
x1(t) =
2.2 (2 points)
Express x1(t) in terms of Ac, fc and φ1. Determine the general expression for xm(t).
2.3 (1 point)
The transmitted QPSK signal xm(t) can also be expressed as
xm(t) = amAc cos(2πfct) cos((2m − 1) π
4 ) + bmAc sin(2πfct) sin((2m − 1)
π
4 )
where m = 1, 2, 3, 4. Plot am and bm as a function of m for the constellation of figure 1.
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2.4(3 points)
Give the constellation diagram of a signal for which the variation of am and bm is as shown in figure 2. Does this constellation represent a QPSK signal? Justify your answer.
m
1
2
3
1
2
3
am bm
Fig. 2 Variation of am and bm as a function of m
2.5 (3 points)
Sketch the decision regions for the signal points of the constellation of 2.4 under the as- sumption that signal is corrupted by additive Gaussian noise and the symbols are equally likely.
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Problem 3 (10 points):
Given is a π 4 -DQPSK-modulator. The relation between the binary source sequence d(k)
and the phase differences φ(k) is given in the following table:
d(0) 00 01 11 10
φ(0) π 4
Table 1: Coded binary source sequence
The initial phase for k = −1 is admitted to φ(−1) = 0.
3.1 (1 point)
Which code is used for the coding of the binary sequence?
Now, the binary sequence
is transmitted.
3.2 (2 points)
Determine the phase φ(k) of the given binary sequence d(k) and sketch it. Denote all important points on the abscissa and ordinate.
3.3 (3 points)
At k = 1, a phase error occurs. The phase is given by φ(1) = φ(1) − π 2 . Determine the
detected binary sequence d(k).
3.5 (2 points)
k 0 1 2 3 4 5
d(k) 11 00 11 01 10 11
Determine those k, where no phase error occurs.
3.6 (1 point)
Explain the π 4 -DQPSK modulation scheme by using the sequence given in table 1. Give
the phase values and sketch the signal constellation of the binary sequence d(k) in table 1.
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Given is a circuit for the generation of ASK signals.
Upsampler
4
xM−ASK(t)
sM−ASK(t)
The mode of operation of the upsampler is shown in the figure below. It makes from 1 input sample N samples, where the first output sample is identical with the input sample and the remaining (N − 1) output samples are zeros.
Input Output
N
We assume that the DAC (Digital/Analog Converter) is ideal. Its output signal is given by
sM−ASK(t) = ∞∑
k=−∞ s(k)rect
1
−1
6
5
432
and the impulse response of the impulse shaping filter as
h(k) = γ0(k) + γ0(k − 1) + γ0(k − 2) + γ0(k − 3) with γ0(k) =
{ 1 for k = 0 0 else.
4.1 (2 points)
Sketch the output sequence dup(k) of the upsampler. Give all important points on abscissa and ordinate.
4.2 (1 point)
Sketch the output sequence s(k) of the impulse shaping filter. Give all important points on abscissa and ordinate.
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4.4 (2 points)
Sketch the analog output signal sM−ASK(t) of the DAC. Give all important points on abscissa and ordinate.
4.5 (1 point)
The same signal sM−ASK(t) can be obtained without upsampling and without impulse shaping. Write down sM−ASK(t) as a function of d(k).
Now, a new impulse shaping filter is used with the impulse response
1
4.6 (2 points)
Sketch for this case the output signal s′M−ASK(t) of the DAC.
4.7 (1 point)
Which Fourier-transform SM−ASK(ω) or S ′ M−ASK(ω) does have a lower decay? Explain
your answer.
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Four questions are given from the content of the lecture “Transmission and Modulation of Signals 2“ in the following. These questions can be answered independently. The answers shall be restricted to a few short remarks and/or plots or formulas respectively. For each question the correct and complete answer yields 2 points.
Explain how a signal transmitted over a slowly time-variant linear channel can be equalized. Give an application for such a channel.
What maximum bit rate (bits/s) would you expect for
1. a high quality analog telephone channel (signal-to-noise power ratio 50dB)
2. a low quality analog telephone channel (signal-to-noise power ratio 13, 806dB)
3. an ISDN channel
Hint: Assume that a speech signal is concentrated on the frequency region [300Hz, 3400Hz] and the channel noise is Gaussian distributed.
The available bandwidth for GSM is 200kHz and the bit rate is 270833 bit/s. What signal-to-noise power ratio (in dB) would you expect for Gaussian channel noise ?
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Tb
−A
+A
t
xm(t)
2Tb
Tb
4
Solution 1.2:(1 point)
Solution 1.3:(2 points)
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Solution 1.5: (1 point)
Fig.8 eA(t = nTb) and eB(t = nTb) for duration of 2Tb
A2Tb
−5A2Tb
xm(t) 0 1
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xm(t) = √
am = 1 and bm = 1 for m = 1, 2, 3, 4, which is shown in figure 3.
bm
1
2
3
1
2
3
am
m m Fig.3 Discrete am and bm as a function of m
The magnitude of envelope remains constant although the phase changes.
xm =
( am
)
Substituting m = 1, 2, 3, 4 and corresponding values of am and bm, we get the constellation as shown in figure 4.
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Fig.4 Constellation diagram for given am and bm
Yes, the constellation represents a QPSK modulated signal. The amplitude of the envelope is still constant.
Line L1 is the perpendicular bisector of the lines joining the signal points corresponding to m = 1&m = 2 and m = 3&m = 4. L2 is the perpendicular bisector of the line joining signal points corresponding to m = 1&m = 4 and m = 2&m = 3. The decision regions are shown in figure 5.
R1
f0
L1
L2
R2
f1
R3
R4
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The sequence is Gray-coded. Solution 3.2: (2 points)
The phase follows from φ(k) = φ(k) + φ(k − 1).
k -1 0 1 2 3 4 5 d(k) 01 00 10 00 10 01 φ(k) 3π
4 π 4
π
−3π/4
−π
φ(k)
k
k = 1: φ(1) = φ(1) − π 2 . The detected phase is:
k -1 0 1 2 3 4 5
φ(k) 0 3π 4
k 0 1 2 3 4 5
φ(k) 3π 4
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⇒ 2 Errors.
Solution 3.5: (2 points)
From d(k) we derive φ(k). Using now the relation φ(k) = φ(k) + φ(k − 1), we get the phase φ(k) of the detected binary sequence d(k). The comparison of φ(k) with the phase of the transmitted sequence d(k) delivers the result: At k = 2 (k2) and k = 7 (k7) occurs no phase error.
4 ,−3π
4 ,−π
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Solution Problem 4:
In the following we have: a = 1 and b = 0.5 Solution 4.1 (2 points)
a b
−b −a
0 1
2 3
4 5
Solution 4.3 (1 point) M = 5 Solution 4.4 (2 points)
5
Solution 4.5 (1 point)
Yes. We can write
Solution 4.7 (1 point)
S ′ 5−ASK(ω) has a lower decay, because each bit of s′5−ASK(t) has a narrower width.
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In the data stream a training sequence, which is known to the receiver, is periodically inserted. The receiver can estimate the channel impulse repsonse, since for a certain time slot it knows the transmitted signal – the training sequence – as well as the received signal. Hence, by cross-correlation, which leads to a linear system of equations, the channel impulse response can be estimated and similiarly by use of this estimated impulse response the user data can be equalized. Due to the periodically inserting of a training sequence the distortions caused through a slowly time-variant channel can be compensated by periodically updating the impulse response estimate. A typical practical scenario is given by a wireless communication channel where at least one user is moving in time.
Cs|Gauss = fco ld
For the signal-to-noise power ratio we obtain 1.9215dB.
• The Viterbi-algorithm is of recursive nature. For incorporating a new sample only two additions, one convolution and one multiplication (squaring) is required.
• Only a limited number of state transitions must be calculated by using the Viterbi- algorithm, those transitions which are impossible are not calculated. This additio- nally saves numerically costs.
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A digital transmission system with additive white Gaussian noise is given.
Source generator Binary
Log- likelihood Computer
A digital signal transmission chain
A binary, discrete and memoryless source emits equally likely bits 1 and 0. The output of a waveform generator is given by
x0(t) = A sin( πt
) for bit 1
The zero mean noise N(t) has a variance of σ2 N . The basis function f0(t) = kx0(t), k ≥ 0.
1.1 (2 points)
Calculate constant k in terms of A and Tb such that f0(t) is a unit energy signal.
Hint: 2 sin2 x = 1 − cos 2x, ∫ π
0 cos(2x) dx = 0.
1.2 (3 points)
Calculate the output y0,m (m = 0, 1) of the correlation receiver in terms of A, Tb and N0,
where N0 = ∫ Tb
0 N(t)f0(t) dt.
4 .
y0,m can also be represented as Y0|Hm, where hypothesis H0 is true if bit 0 is transmitted and H1 is true if bit 1 is transmitted.
1.3 (2 points)
Determine the conditional probability density functions pY0|H0(y|H0) and pY0|H1(y|H1).
Hint: Assume that N0 has the same probability density function as that of N(t).
1.4 (3 points)
Calculate the log-likelihood ratio function λ(y0). Determine the decision value yth that seperates the decision regions corresponding to hypotheses H0 and H1.
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Given is the constellation diagram of a modulated signal, where −1 ≤ k ≤ 1. x0 and x1
represent the signal points corresponding to equally likely transmitted bits 0 and 1. The basis vectors f0 and f1 are
f1
[ √ Eb
0
] f0
x0
tions. Tb is the bit duration and fc = n Tb
is the carrier frequency, where n is an integer.
2.1 (2 points)
Calculate d′ ≥ 0 in terms of Eb and k, if the average energy transmitted per bit is Eb.
In the following use d′ of 2.1
2.2 (2 points)
Express the transmitted signals x0(t) and x1(t) corresponding to signal points x0 and x1, respectively in terms of f0(t), f1(t), k and Eb.
2.3 (1 point)
Calculate the normalized correlation coefficient ρx0,x1 of the signals x0(t) and x1(t).
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Hint: ∫ Tb
0 f0(t)f1(t) dt = 0
In the presence of zero mean additive white Gaussian noise with variance σ2 N , the average
bit error probability can be expressed as
Pe,b = 1
2 erfc
Calculate the value of k for which Pe,b is minimum.
Hint: Computation of derivative of error function is not necessary.
2.5 (2 points)
Sketch the constellation diagrams with the decision regions for k = 0 and k = 1.
2.6 (2 points)
Plot the trajectory of x1 as k varies from -1 to 1. Which modulation method does this constellation represent?
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The binary source sequence D = 0 0 1 0 0 1
is given.
a) a 2-ASK signal x2−ASK(t) (Binary Modulation),
b) a 4-ASK signal x4−ASK(t) (Quaternary Modulation), and
c) an 8-ASK signal x8−ASK(t) (Octonary Modulation).
Use the Gray-code and the amplitude values ±d 2 , ±d
2 , ±3d
2 , ±5d
important points on abscissa and ordinate.
Compute the symbol duration TS for each base-band signal as a function of the bit duration TD.
3.2 (3 points)
Now, a QPSK-(4-PSK)-modulator with a symbol rate Ts = 2TD is used. The initial phase φ0 is assumed to be φ0 = π
4 . The phase is Gray-coded according to the scheme
Φ(k) =
3π 4
5π 4
7π 4
for D(n) = 0 and D(n + 1) = 1.
Sketch the phase Φ(k), the in-phase component xip,4−PSK(t), and the quadrature component xqu,4−PSK(t). Assume that the distance between two symbols ist d. Denote all important points in abscissa and ordinate.
3.3 (3 points)
Sketch the phases ΦMSK(t) and ΦGMSK(t) of the MSK and GMSK signals. Denote all important points in abscissa and ordinate.
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Three questions are given from the content of the lecture “Transmission and Modulation of Signals 2“ in the following. These questions can be answered independently. The answers shall be restricted to a few short remarks and/or plots or formulas respectively. For each question the correct and complete answer yields 2 points.
Explain the band-spreading factor β and state its value for DSBAMwC, QAM, SSBAM- wC, VSBAMsC.
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∫ Tb
0
0
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√ E0 = A
)
P{H0} = P{H1} = 0.5. The decision value can be obtained as,
λ(y0) = ln P{H0} P{H1}
1
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Eb = 1
∫ Tb
0
Pe,b = 1
2 erfc
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f0
f1
√ Eb
R0R1
Trajectory of x1 as k varies from -1 to 1
Phase shift keying.
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Solution 3.2 (3 points)
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2BT -Theorem: For large bandwidth(B)-duration (T ) product the required dimension N for the vector representation of a signal with approximate duration T and approximate bandwidth B has the approximate dimension N = 2BT .
Shannon-bandwidth:
2T .
In contrast to more concrete definitions than this abstract definition, the Shannon bandwidth is the minimum amount of bandwidth that the signal needs. Any of the more practical definitions of bandwidth expresses the amount of bandwidth that the signal actually uses.
sum of bandwidth of all source signals sn(t) ,
which is a measure for a more or less efficient use of the available bandwidth. A band spreading factor near, but not lower than one indicates a very effective modulation method, whereas β < 1 automatically leads to a loss of information. Obviously, the higher the band spreading factor is,the worse is the modulation method concering the valuable bandwidth. It should be pointed out that the notation bandwidth only concerns positive frequencies.
DSBAMwC: β = 2, QAM: β = 1, SSBAMwC: β = 1, VSBAMsC: 2 > β > 1.
(' $ &
- $ &
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$ ! # # !
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Problem 1 (10 Points):
Two signals y(t) and z(t) are given. The spectra Y (ω) and Z(ω) are known.
Z(ω) = rect
1.1 (2 Points)
Sketch Z(ω) and Y (ω). Give all important points on the abscissa and ordinate.
1.2 (2 Points)
Sketch the spectrum X(ω) of the signal x(t) = z(t) · y(t). Give all important points on the abscissa. Denote the maximum value of the magnitude of X(ω) as A and do not determine it.
1.3 (1 Point)
1.4 (1 Point)
Sketch the spectrum V (ω) of the signal v(t) = z(t) · y0(t), where y0(t) is the analytic signal of y(t).
1.5 (1 Point)
Determine the analytic signal x0(t) = F−1{X0(ω)} as a function of v(t).
1.6 (2 Points)
1.7 (1 Point)
Conclude now from above results the relation between the analytic signal x0(t) and the equivalent lowpass signal xELP(t) of x(t). Justify your response.
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Problem 2 (10 Points)
Given the signal constellation (see figure below) of an 8-QAM (Quadrature Amplitude Modulation).
f0
f1
d1
d0
x2 x6 x5 x1
x3 x7 x8 x4
The observed process is Y (t) = xi(t) + N(t), i = 1, 2, . . . , 8 where N(t) is a Gaussian distributed, white noise process with zero mean. The signals xi(t) are received with equal probability.
2.1 (1 Point)
Sketch the 8 decision regions Ri, i = 1, 2, . . . , 8 pertaining to the signals xi.
2.2 (2 Points)
Calculate the probability of a correct decision Pc|x1 of the symbol x1. Use the function
Pe,binary(dj) = P{Nj < −dj
2 }, j = 0, 1, where dj is the horizontal and vertical distance
respectively.
2.3 (3 Points)
Calculate the probability of a correct decision Pc|x5 of the symbol x5. Use again the
function Pe,binary(dj) = P{Nj < −dj
2 }, j = 0, 1, where dj is the horizontal and vertical
distance respectively.
2.4 (2 Points)
Calculate the average probability of a correct symbol decision Pc,8−QAM.
2.5 (1 Point)
2.6 (1 Point)
Determine the symbol error probability Pe,8−QAM if d0 = d1 = d and Pe,binary(d) 1.
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The binary source sequence D = 0 0 1 0 0 1
is given.
a) a 2-ASK signal x2−ASK(t) (Binary Modulation),
b) a 4-ASK signal x4−ASK(t) (Quaternary Modulation), and
c) an 8-ASK signal x8−ASK(t) (Octonary Modulation).
Use the Gray-code and the amplitude values ±d 2 , ±d
2 , ±3d
2 , ±5d
important points on abscissa and ordinate.
Compute the symbol duration TS for each base-band signal as a function of the bit duration TD.
3.2 (3 points)
Now, a QPSK-(4-PSK)-modulator with a symbol rate Ts = 2TD is used. The initial phase φ0 is assumed to be φ0 = π
4 . The phase is Gray-coded according to the scheme
Φ(k) =
3π 4
5π 4
7π 4
for D(n) = 0 and D(n + 1) = 1.
Sketch the phase Φ(k), the in-phase component xip,4−PSK(t), and the quadrature component xqu,4−PSK(t). Assume that the distance between two symbols ist d. Denote all important points in abscissa and ordinate.
3.3 (3 points)
Sketch the phases ΦMSK(t) and ΦGMSK(t) of the MSK and GMSK signals. Denote all important points in abscissa and ordinate.
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Three questions are given from the content of the lecture “Transmission and Modulation of Signals 2“ in the following. These questions can be answered independently. The answers shall be restricted to a few short remarks and/or plots or formulas respectively. For each question the correct and complete answer yields 2 points.
Explain what entropy means and give a formula. Under which situation reaches the entropy its maximum ?
NTS/F5 GMUD
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ω
2π Z(ω) ∗ Y (ω).
2π Z(ω) ∗ Y 0(ω).
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= π (δ(ω − ωc) + δ(ω + ωc)) · 2ε(ω)
= 2πδ(ω − ωc).
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Solution 2.1 (1 Point)
d1
2
−d1
2
f1
f0
R1 = [d0, +∞] × [0, +∞], R2 = [−∞,−d0] × [0, +∞], R3 = [−∞,−d0] × [−∞, 0], R4 = [d0, +∞] × [−∞, 0], R5 = [0, d0] × [0, +∞], R6 = [−d0, 0] × [0, +∞], R7 = [−d0, 0] × [−∞, 0], R8 = [0, d0] × [−∞, 0]
2 + N0
2 + N1
Pc|x1 = P{Y0 > d0, Y1 > 0} = P{Y0 > d0} P{Y1 > 0}
Pc|x1 = P{3 d0
2 + N0 > d0} P{d1
2 + N1 > 0}
= P{N0 > −d0
2 + N1
Pc|x5 = P{0 < Y0 < d0, Y1 > 0} = P{0 < Y0 < d0} P{Y1 > 0}
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2 + N1 > 0}
2 } ]
= 2
[ 1
]
= − Pe,binary(d1) + 2 Pe,binary(d0) Pe,binary(d1)
] = 1 − 3
= 3
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Entropy is the average information content of an alphabet X
H(X ) = L∑
l=1
pl · ld
pl · ld(pl).
It takes its maximum if all the probabilities are the same, so
pl = 1

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