Signal Constellation MCQ Quiz in मल्याळम - Objective Question with Answer for Signal Constellation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Apr 11, 2025

നേടുക Signal Constellation ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Signal Constellation MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Signal Constellation MCQ Objective Questions

Top Signal Constellation MCQ Objective Questions

Signal Constellation Question 1:

For the two signals to be orthonormal, the value of a + b = _________.

Gate EC 2016 Communication ChapterTest 4 Images-Q17

Answer (Detailed Solution Below) 1.4 - 1.414

Signal Constellation Question 1 Detailed Solution

The signals are orthogonal requires that 02s1(t)s2(t)dt=0

Thus, we have,

ab2ab2ab2+ab2=0

So, we see that the signals are orthogonal for all values of a and b.

Now, for the signals to be normal, the signal of each should be unity, i.e,

02s12(t)dt=1 and 02s22(t)dt=1

Checking for f1(t) we have,

02s12(t)dt=2a2=1a=12

Similarly,

02s22(t)dt=2b2=1b=12

Now, a+b=22=2=1.414

Signal Constellation Question 2:

A four-array signal constellation is shown below. All the four symbols are equiprobable

Gate EC 2016 Communication ChapterTest 4 Images-Q5

The average energy of the signal for a=5 is __________.

Answer (Detailed Solution Below) 100

Signal Constellation Question 2 Detailed Solution

Energy of a symbol can be calculated from constellation diagram by simply taking square of the distance of the symbol from origin.

Thus,

E(s1)=0

E(s2)=(2a)2=4a2

E(s3)=(22a)2=8a2

E(s4)=(2a)2=4a2

Now, all symbols are equiprobable, so, average energy is

Eavg=14s1+14s2+14s3+14s4=14×0+14×4a2+14×8a2+14×4a2

=a2+2a2+a2

Eavg=4a2=100

Signal Constellation Question 3:

A BPSK schemes operating over an AWGN channel with noise spectral density No/2 uses equiprobable signals s1(t)=2ETsin(ωct) and s2(t)=2ETsin(ωct) over the symbol interval, (0,T). If the local oscillator in a coherent receiver is ahead in phase by 45° with respect to the received signal, the probability of error in the resulting system is:

  1. Q(2ENo)
  2. Q(E4No)
  3. Q(E2No)
  4. Q(ENo)

Answer (Detailed Solution Below)

Option 4 : Q(ENo)

Signal Constellation Question 3 Detailed Solution

Based on the information given for the two symbols waveforms, we see that the basis function is 2Tsin(ωct). Now the oscillator frequency is 2Tsin(ωct+45)

ECasd6

Using this, we calculate the vectors of demodulated symbols of  s1(t) and s2(t) along the basis function and get the following result.

ECasd7

S1=E2and S2=E2

Now, distance between vectors d=2E2

Now, using the relation that Pe=Q(d22No)=Q(4×E22No)=Q(ENo)

Alternate Solution:

For BPSK

For coherent detection when there is phase synchronization between the carrier and received signal:

Pe=Q(2EbNo)

For coherent detection when there is phase shift of ϕ  between the carrier and received signal:

Pe=Q(2EbCos2ϕNo)

Substituting the values Cos2 ϕ = 1/2

Pe=Q(ENo)

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