QUESTION 1 - ENSEEIHTthomas.perso.enseeiht.fr/QUIZ1_anglais.pdf · QUESTION 2 The binary information transmitted by the generated signal is: A B C Generated signal: t Mapping : 0

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


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



Generated signal:



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


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)



Generated signals:


GOOD ANSWER


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


X1(t)

X2(t)

0 0 1 0 0 1 1 1

0 0 1 0 0 1 1 1


Generated signals:

QUESTION 5

A

B

C

D

V

-3V

3V

-Vt

Ts 2Ts 3Ts 4Ts

V

-Vt


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)



Generated signals:

Click here to CHANGE YOUR ANSWER

BAD ANSWER

GOOD ANSWER


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


X1(t) X2(t)


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





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)


Generated signal:



- 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


Receivedbinary

information



Decisions

Sampling

V

-3V

3V

-Vt

Ts 2Ts 3Ts 4Ts

x(t)


Generated signal:



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


Receivedbinary

information



Decisions

Sampling

V

-3V

3V

-Vt

Ts 2Ts 3Ts 4Ts

x(t)


Generated signal:






- 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


Receivedbinary

information



Decisions

Sampling


Generated signal:



0 2TSTS

Ts




- 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

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QUESTION 1 - ENSEEIHTthomas.perso.enseeiht.fr/QUIZ1_anglais.pdf · QUESTION 2 The binary information transmitted by the generated signal is: A B C Generated signal: t Mapping : 0