$(i) x(t) = 14 + 40\cos (60πt)\ (ii) x[n] = \cos^2[\dfrac \pi an]$

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 04

Year : MAY 2015

$$(i) x(t) = 14 + 40 \cos (60πt)$$

To determine the fundamental period of x(t)

First, check whether the signal is periodic or not

Here, 14 is the DC shift added to the signal $x_1(t) = 40 \cos 60πt$

The given signal is a cosine signal and therefore it is periodic.

It will remain periodic after adding the DC shift.

To determine the fundamental period (T)

Compare the equation with $x[t] = A + B \cos (2πft) \\ ∴2πft = 60πt \\ ∴f = 30 \\ ∴T = \dfrac 1{30} $

Therefore, the fundamental period $(T) = \dfrac 1{30} $

$(ii) x[n] = \cos^2\dfrac πa n$

To determine the fundamental period of x[n]

First, check whether the signal is periodic or not

$∴x[n] = \dfrac {1 + \cos 2 \dfrac πa n}2 \hspace{1cm} \because \cos^2A =\dfrac { 1 + \cos2A }2 \\ ∴x[n] = \dfrac 12 + \dfrac12 \cos \dfrac {2π}a n$

Here, $\dfrac12$ is the DC shift added to the signal $x_1[n] = \dfrac12 \cos\dfrac { 2π}a n$

The given signal is a cosine signal and therefore it is periodic.

It will remain periodic after adding the DC shift.

To determine the fundamental period (N)

Compare the equation with $x[n] = A + A \cos (2πfn)\\ ∴2πfn = \dfrac{2π}a n \\ ∴f = \dfrac kN = \dfrac 1a $

Here f is expressed as ratio of two integers with k = 1 and N = a.

Hence, the signal is periodic with N = a

Therefore, the fundamental period (N) = a