QUESTION 1 - ENSEEIHTthomas.perso.enseeiht.fr/QUIZ1_anglais.pdf · QUESTION 2 The binary...
Transcript of QUESTION 1 - ENSEEIHTthomas.perso.enseeiht.fr/QUIZ1_anglais.pdf · QUESTION 2 The binary...
Series of bits to transmit: 1001100
Generated signal:
Ts 2Ts 3Ts 4Ts 5Ts 6Ts 7Ts
t
Mapping : 0 -> -1, 1 -> +1
Mapping : 0 -> +1, 1 -> -1
Mapping : 0 -> -1, 1 -> +1
Not enough elements to answer the question
t
h(t)
t
h(t)
Ts
Ts
t
h(t)
Ts
1
-1
1
-1
1
1
-1
QUESTION 1The mapping and the shaping filter impulse response used to generate this signal are:
A
D
B
C
Click on the bullet
correspondingto the good
answer
BAD ANSWERClick here to CHANGE YOUR ANSWER
GOOD ANSWER
Click here to find the SECOND ANSWER
BE CAREFUL TWO GOOD ANSWERS FOR THIS QUESTION
Click here for the FOLLOWING QUESTION
QUESTION 2The binary information transmitted by the generated signal is:
A
B
C
Generated signal:
t
Mapping : 0 -> -1, 1 -> +1
1
-1
10010110100101
1001100
0110011
Not enough elements to answer the questionD
BAD ANSWERClick here to CHANGE YOUR ANSWER
GOOD ANSWERClick here for the FOLLOWING QUESTION
Without the knowledge of the symbol period Ts, It is impossible to answer the question
QUESTION 3
The symbol rate will be:
A
B
C
D
V
-3V
3V
-Vt
Ts 2Ts 3Ts 4Ts
Equal to the bit rate
Higher than the bit rate
Lower than the bit rate
Not enough elements to answer the question
Series of bits to transmit: 00100111
Generated signal:
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GOOD ANSWERClick here for the FOLLOWING QUESTION
One symbol carries here 2 bits => the symbol duration Ts=2Tb, if Tb represents the bit durationThe symbol rate Rs=1/Ts is so lower than the bit rate Rb=1/Tb : Rs=Rb/2
More generally, if one symbol codes n bits => the symbol duration is Ts=nTb, The symbol rate Rs=1/Ts=Rb/log2(M) if M=2n represents the number of possible symbol (modulation order).
Here M=4=22 (4 possible symbols -3V, -V, +V, +3V each one carrying 2 bits : 00,10,01,11)
V
-3V
3V
-Vt
Ts 2Ts 3Ts 4Ts
0 0 1 0 0 1 1 1
Series of bits to transmit: 00100111
Generated signal:
QUESTION 4
To transmit a same bit rate, with a same given transmitted power, the necessary bandwidth to transmit x1(t) will be :
A
B
C
D
V
-3V
3V
-Vt
Ts 2Ts 3Ts 4Ts
V
-Vt
Ts 2Ts 3Ts 4Ts 5Ts 6Ts 7Ts 8Ts
X1(t) X2(t)
Higher than the one needed to transmit x2(t)
Smaller than the one needed to transmit x2(t)
The same as the one needed to transmit x2(t)
Not enough elements to answer the question
Series of bits to transmit: 00100111
Generated signals:
BAD ANSWERClick here to CHANGE YOUR ANSWER
GOOD ANSWER
Click here for the FOLLOWING QUESTION
The bandwidth necessary to transmit a signal is always proportional to the symbol rate Rs, the proportionality coefficient depends on the used shaping filter.
Here:- The used shaping filter is the same for both signals (rectangular impulse response of length), thus
same proportionality coefficient in both cases (for a same kept power(1)). - But the symbol rate is higher for the signal x1(t): Ts = Tb, and thus Rs = Rb, for signal x1(t), while Ts = 2Tb,
and thus Rs = Rb/2, for signal x2(t).
To transmit the same bit rate Rb, the bandwidth necessary to transmit x1(t) will then be higher to the one necessary to transmit x2(t).
(1) To transmit a NRZ signal, theoritically an infinite bandwidth is required. In practice it will be truncated to keep x % of the total signal power.. Typical values are from 98 to 99 %.
V
-3V
3V
-Vt
Ts 2Ts 3Ts 4Ts
V
-Vt
Ts 2Ts 3Ts 4Ts 5Ts 6Ts 7Ts 8Ts
X1(t)
X2(t)
0 0 1 0 0 1 1 1
0 0 1 0 0 1 1 1
Series of bits to transmit: 00100111
Generated signals:
QUESTION 5
A
B
C
D
V
-3V
3V
-Vt
Ts 2Ts 3Ts 4Ts
V
-Vt
Ts 2Ts 3Ts 4Ts 5Ts 6Ts 7Ts 8Ts
X1(t) X2(t)
The transmission spectral efficiency will be:
Better if we transmit signal x1(t)
Better if we transmit signal x2(t)
The same if we transmit signal x1(t) or signal x2(t)
Not enough elements to answer the question
Series of bits to transmit: 00100111
Generated signals:
Click here to CHANGE YOUR ANSWER
BAD ANSWER
GOOD ANSWER
Click here for the FOLLOWING QUESTION
We saw, in the previous question, that, for the same bit rate Rb, the necessary bandwidth to transmit signal x1(t) was higher than the one necassary to transmit signal x2(t).
To transmit a given bit rate we need a given frequency bandwidth. For a given bit rate to betransmitted, the transmission spectral efficiency is improved when the signal occupied bandwidth is
reduced.
The transmission spectral efficiency is so better here if we transmit the signal x2(t) because we willneed less bandpass to transmit a same bit rate.
V
-3V
3V
-Vt
Ts 2Ts 3Ts 4Ts
V
-Vt
Ts 2Ts 3Ts 4Ts 5Ts 6Ts 7Ts 8Ts
X1(t) X2(t)
Series of bits to transmit: 00100111
Generated signals:
QUESTION 6
A
B
C
D
By using a square root raised cosine shaping filter, the obtained spectral efficiency willbe:
Mapping : –V –V +V –V –V +V +V +V
Higher than the one obtained with a rectangular shaping filter
Lower than the one obtained with a rectangular shaping filter
The same as the one obtained with a rectangular shaping filter
Not enough elements to answer the question
Series of bits to transmit: 00100111
BAD ANSWERClick here to CHANGE YOUR ANSWER
GOOD ANSWERClick here for the FOLLOWING QUESTION
The bandwidth necessary to transmit a signal is always proportional to the symbol rate Rs, the proportionality coefficient depends on the used shaping filter.
Here what will give the occupied bandwidth is the used shaping filter:
- Using a rectangular shaping filter leads theoretically to an infinite occupied bandwidth (truncatedin practice to keep x% of the signal total power).
- Using a square root raised cosine filter leads to generate a signal with a finite bandwidth (includingthe total signal power): B=(1+a)/2Ts for baseband transmissions, with 0<a<1.
Considering a same transmitted power, the signal generated using a square root raised cosine filter willbe more spectrally efficient, because it will require a lower bandpass to transmit the same bit rate.
QUESTION 7
A
B
TRUE
FALSE
Considering a receiver filter with a rectangular impulse response of lengthTs, t0=Ts and a threshold detector with thresholds in –2V, 0, 2V to take the decisions, the obtained bit
error rate will be the minimum one:
Transmissionchanneln(t)
Receivedbinary
information
Receiver filterhr(t)Demapping
Baseband demodulation
Decisions
Sampling
V
-3V
3V
-Vt
Ts 2Ts 3Ts 4Ts
x(t)
Series of bits to transmit: 00100111
Generated signal:
BAD ANSWERClick here to CHANGE YOUR ANSWER
GOOD ANSWERClick here for the FOLLOWING QUESTION
- Without filter in the channel, the channel global impulse response is given by:
Sampling at t0+m Ts with t0= Ts allows to respect the Nyquist criterion. Then we have z(t0+mTs)=amg(t0)+wm at the sampling output, where wm is a sample of filtered noise.
- The rectangular receiver filter of length Ts is the matched filter for a rectangular shaping filter of length Ts (no filter in the channel). The signal to noise ratio at the sampling output is so maximized.
- BUT the thresholds are not correct for the decisions. The received symbols without noise are -3VTs, -VTs, VTs , 3VTs, so we need thresholds in -2VTs , 0, 2VTs.
The obtained BER won’t be the minimum one.
0 2TSTS
Ts
QUESTION 8
A
B
Considering a receiver filter with a rectangular impulse response of lengthTs, t0= Ts/2and a threshold detector with thresholds in –2VTs, 0, 2VTs to take the decisions, the
obtained bit error rate will be the minimum one:
TRUE
FALSE
Transmissionchanneln(t)
Receivedbinary
information
Receiver filterhr(t)Demapping
Baseband demodulation
Decisions
Sampling
V
-3V
3V
-Vt
Ts 2Ts 3Ts 4Ts
x(t)
Series of bits to transmit: 00100111
Generated signal:
BAD ANSWERClick here to CHANGE YOUR ANSWER
GOOD ANSWERClick here for the FOLLOWING QUESTION
Without filter in the channel, the channel global impulse response is given by:
Sampling at t0+m Ts with t0= Ts/2 does not allow to respect the Nyquist criterion:
The obtained BER won’t be the minimum one.
0 2TSTS
Ts
0 2TSTS
Ts
TS/2
TS/2
3TS/2
For t0= Ts/2 g(t0) = 0 but g(t0+pTs) is not equal to 0 for all p non zero integers: g(t0+Ts) = Ts/2
QUESTION 9
A
B
Considering a receiver filter with a rectangular impulse response of lengthTs, t0= Ts and a threshold detector with thresholds in –2VTs, 0, 2VTs to take the decisions, the obtained
bit error rate will be the minimum one:
TRUE
FALSE
Transmissionchanneln(t)
Receivedbinary
information
Receiver filterhr(t)Demapping
Baseband demodulation
Decisions
Sampling
V
-3V
3V
-Vt
Ts 2Ts 3Ts 4Ts
x(t)
Series of bits to transmit: 00100111
Generated signal:
BAD ANSWERClick here to CHANGE YOUR ANSWER
GOOD ANSWERClick here for the FOLLOWING QUESTION
- Without filter in the channel, the channel global impulse response is given by:
Sampling at t0+m Ts with t0= Ts allows to respect the Nyquist criterion. Then we have z(t0+mTs)=amg(t0)+wm at the sampling output, where wm is a sample of filtered noise.
- The rectangular receiver filter of length Ts is the matched filter for a rectangular shaping filter of length Ts (no filter in the channel). The signal to noise ratio at the sampling output is so maximized.
- The thresholds are optimal for the decisions.
- BUT the mapping is not a Gray mapping. Between two symbols at minimum distance more than one
bit is changing: 2 bits are modified between -3V (00) and –V (11), as well as between +V(01) and +3V (10).
The symbol error rate (SER) is the minimum one but it is not true for the BER.
0 2TSTS
Ts
QUESTION 10
A
B
Considering a receiver filter with a rectangular impulse response of lengthTs, t0= Ts and a threshold detector with thresholds in –2VTs, 0, 2VTs to take the decisions, the obtained
bit error rate will be the minimum one:
V
-3V
3V
-Vt
Ts 2Ts 3Ts 4Ts
x(t)
TRUE
FALSE
Transmissionchanneln(t)
Receivedbinary
information
Receiver filterhr(t)Demapping
Baseband demodulation
Decisions
Sampling
Series of bits to transmit: 00100111
Generated signal:
BAD ANSWERClick here to CHANGE YOUR ANSWER
GOOD ANSWERClick here for the FOLLOWING QUESTION
0 2TSTS
Ts
- Without filter in the channel, the channel global impulse response is given by:
Sampling at t0+m Ts with t0= Ts allows to respect the Nyquist criterion. Then we have z(t0+mTs)=amg(t0)+wm at the sampling output, where wm is a sample of filtered noise.
- The rectangular receiver filter of length Ts is the matched filter for a rectangular shaping filter of length Ts (no filter in the channel). The signal to noise ratio at the sampling output is so maximized.
- The thresholds are optimal for the decisions.
- The mapping is a Gray mapping. Between two symbols at minimum distance only one bit ischanging: -3V (00), –V (01), +V (11), +3V (10). Then, we obtain here a BER SER/2, neglecting the probability to make an error between symbols at a distance which is not the minimum one.
The bit error rate will be the minimum one.
THE QUIZ IS FINISHED