Periodic and Aperiodic Signals MCQ Quiz in తెలుగు - Objective Question with Answer for Periodic and Aperiodic Signals - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

A signal x(t) is a said to be periodic with a period T if x(t ± T) = x(t)

Time period for a signal in the form of A cos (ωt + ϕ) is (2π/ω)

Given signal \(x\left( t \right) = 3\cos \left( {5t + \frac{π }{6}} \right),\)

Time period, \(T = \frac{{2π }}{5}\)

\(\cos \left( {\frac{\pi }{4}t - \frac{\pi }{4}} \right)\)

\(\omega = \frac{\pi }{4}\)

\(T = \frac{{2\pi }}{\omega } = \frac{{2\pi 4}}{\pi }\)

= 8s

A signal x(t) is said to be periodic if for some positive constant T₀;

x(t) = x (t + T₀) for all t

The smallest value of T₀ that satisfies the periodicity condition of the above equation is called the fundamental period of x(t).

This is explained with the help of the below figure:

Concept:

A signal x(t) is a said to be periodic with a period T if x(t ± T) = x(t)

The time period for a signal in the form of A cos (ωt + ϕ) is:

\(T = \frac{{2π }}{\omega}\)

Application:

\(x(t)=\rm cos\ {\left( 2t+\frac{π}{4}\right)}\)

∴ ω = 2 rad/sec

The time period will be:

\(T = \frac{{2π }}{\omega}=\frac{2π }{2}\)

T = π sec

For the sum of the two continuous-time signals with period T 1 and T2 to be periodic, there must exist non-zero integers a, b such that:

a × T1 = b × T2

The fundamental period of a signal = LCM (T1, T2)

A signal x(t) is said to be periodic if for some positive constant T0 it satisfies:

x(t) = x (t + T0) for all t

The smallest value of T0 that satisfies the periodicity condition of the above equation is called the fundamental period of x(t).

This is explained with the help of the below figure:

Concept:

A signal will be called periodic if it is periodic for the entire range of domain, i.e.

If it is periodic for -∞ to +∞, the signal is said to be periodic.

Calculations:

The given signal is defined as:

\(x\left( t \right)=\left\{ \begin{matrix} 2\cos t+\cos \left( 2t \right) & t\le 0 \\ 2\sin t+\sin \left( 2t \right) & t\ge 0 \\ \end{matrix} \right.\)

For t < 0,

x(t) = 2 cos t + cos 2t

2.cos t is period with period = 2π

Cos 2t is also periodic with period \(=\frac{2\pi }{2}=\pi \)

So, 2 cos t + cos (2t) will be periodic with period of 2 π. (LCM of π and 2π)

Similarly,

For, t ≥ 0,

x(t) = 2 sin t + sin (2t)

sin t is periodic with a period of π.

and sin 2t is also periodic with period of π.

2 sin t + sin 2t is therefore periodic as well with period = LCM (π, 2π) = 2 π only

If, x(t) is to be period with period 2π

x (-2π) = x(0) = x(2π)

Checking,

x (-2π) = 2 cos (-2π) +cos (-4π)

= 2 + 1 = 3

x(0) = 2 sin 0 + sin 0 = 0

And x (2π) = 2 sin 2π + sin (4π) = 0

Clearly, x (-2π) ≠ x (0) = x (2π)

Since the signal is not periodic for the complete range of = ∞ to ∞, the given signal is not a periodic signal.

x[n] = 2 sin(π²n)

Let N be the period of the signal, then

x[n] = x[n + N]

⇒ 2 sin(π²n) = sin (π²n + π²N)

⇒ sin(π²n + 2πk) = sin(π²n + π²N)

Where k is an integer.

⇒ 2πk = π²N

\(\Rightarrow N=\frac{2\pi k}{{{\pi }^{2}}}=\frac{2k}{\pi }\) 

For any integer value of k, we cannot have an integer value of N.

So, the given signal is not periodic.

A power signal must have finite power over a time period and infinite energy.

For the given signal x(t) to be a power signal, it must have periodic components.

For 5 sin 2πat to be periodic, the time period will be:

\(T=\frac{{2\pi }}{{2\pi a\;}} = \frac{1}{a}\) must be a real number.

For 7e^jbt to be periodic, the time period (T) must be:

\(T=\frac{{2\pi }}{b}\) must be a rational number.

i.e b is multiple of pi and is irrational

Concept:

A signal x(t) is a said to be periodic with a period T if x(t ± T) = x(t)

The time period for a signal in the form of A cos (ωt + ϕ) is \(T = \frac{{2\pi }}{\omega}\)

For the sum of the two continuous-time signals with period T1 and T2 to be periodic, there must exist non-zero integers a, b such that:

a × T1 = b × T2

The fundamental period of a signal = LCM (T1, T2)

Calculation:

Given signal is x(t) = 2 cos(t/4)

Fundamental period is,

\({T_0} = \frac{{2\pi }}{{\frac{1}{4}}} = 8\pi \)

Concept:

For the sum of the two continuous-time signals with period T1 and T2 to be periodic, there must exist non-zero integers a, b such that:

a × T1 = b × T2

Application:

The period T1 of 9 sin 5t is:

\(T_1=\frac{2\pi}{5}\)

The period T2 of \(7sin\sqrt 5t\) is:

\({T_2} = \frac{{2\pi }}{{\sqrt 5 }}\;s\)

We observe that there does not exist any integer a, b such that:

\(a\times \frac{2\pi}{5}=b\times\frac{2\pi}{\sqrt 5}\)

Hence the sum of the two signals is not periodic.