Test: Signal Classification - Electrical Engineering (EE) MCQ

Explanation: A discrete time signal can be obtained from a continuous time signal by replacing t by nT, where T is the reciprocal of the sampling rate or time interval between the adjacent values. This procedure is known as sampling.

Explanation: The behavior of the signal is known and can be represented by a saw tooth wave form. So, the signal is deterministic.

Explanation: Let x(t)=x_e(t)+x_o(t)
=>x(-t)=x_e(-t)-x_o(-t)
By adding the above two equations, we get
x_e(t)=(1/2)*(x(t)+x(-t)).

Explanation: Let x(t)=e^(jt)
Now, x_o(t)=(1/2)*(x(t)-x(-t))
=(1/2)*(e^(jt) – e^(-jt))
=(1/2)*(cost+jsint-cost+jsint)
=(1/2)*(2jsint)
=j*sint.

Explanation: If a signal x(t) is said to be periodic with period T, then x(t+mT)=x(t) for all t and any integer m.

Explanation: Let T be the period of the signal x(t)
=>x(t+T)=x₁(t+mT₁)+x₂(t+nT₂)
Thus, we must have
mT₁=nT₂=T
=>(T₁/T₂)=(k/m)= a rational number.

Explanation: For the sum of x₁(t) and x₂(t) to be periodic the ratio of their periods should be a rational number, then the fundamental period is the LCM of T₁ and T₂.

Explanation: For any energy signal, the average power should be equal to 0 i.e., P=0.

Explanation: The energy signal should have total energy value that lies between 0 and infinity.

Explanation: Period of cos2t=(2*pi)/2=pi
Period of sin3t=(2*pi)/3
LCM of pi and (2*pi)/3 is 2*pi.