$(i) x(t) =2\cos \dfrac {2\pi t}3+3\cos \dfrac {2\pi t}7 \ (ii) x[n]=\cos^2 [\dfrac \pi4 n] $

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : DEC 2014

(i)

$x(t) = 2 \cos \dfrac {2πt}3 + 3 \cos \dfrac {2πt}7 $

To determine the fundamental period of x(t)

First, check whether the signal is periodic or not

Compare the given equation with $x(t) = A \cos 2πf_1t + B \cos 2πf_2t \\ ∴2πf_1t = \dfrac {2πt}3 \\ ∴f_1 = \dfrac 13 \\ \text {Hence}, T_1 = 3 \\ \text {And } 2πf_2t = \dfrac {2πt}7 \\ ∴f_2 = \dfrac 17 \\ \text {Hence,} T_2 = 7$

Therefore, the ratio of two periods is

$∴\dfrac {T_1}{T_2} = \dfrac 37 $

Since, the ratio of two periods is a rational number

Therefore, this signal is periodic.

To determine the fundamental period (T)

The fundamental period is the least common multiple of $T_1$ and $T_2$.

Now least common multiple of $T_1 = 3$ and $T_2 = 7$ is 21

Therefore, the fundamental period (T) = 21

(ii)

$x[n] = \cos^2 [\dfrac π4 n]$

To determine the fundamental period of x[n]

First, check whether the signal is periodic or not

$∴x[n] = \dfrac{1 + \cos 2 \dfrac π4 n} 2 \hspace {1cm} \because \cos^2A = \dfrac {1 + cos2A}2 \\ ∴x[n] = \dfrac 12 + \dfrac 12 \cos \dfrac π2 n $

Here, $\dfrac 12$ is the DC shift added to the signal $x_1[n] = \dfrac 12 \cos \dfrac π2 n$

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 π2 n \\ ∴f = \dfrac kN = \dfrac 14 $

Here f is expressed as ratio of two integers with k = 1 and N = 4.

Hence, the signal is periodic with N = 4

Therefore, the fundamental period (N) = 4