Question

A continuous-time periodic signal x(t) is real valued and has a fundamental period T = 8. The nonzero Fourier series coefficients for x(t) are specified as
a_1 = a*_-1 = j, a_5 = a_-5 = 2.
Express x(t) in the form
x(t) = sigma^infinity_k=0 A_k cos(W_k t + phi_k).

Answer 1

Answer:

x(t) = −2 cos ( $\frac{\pi }{4}t$  − $\frac{\pi }{2}$ ) + 4 cos ( $\frac{5\pi }{4}t$ )

Explanation:

Given:

Fundamental period of real valued continuous-time periodic signal x(t) = T = 8

Non-zero Fourier series coefficients for x(t) :

a₁ = $a^{*}_{-1}$ = j

a₅ = $a_{-5}$ = 2  

To find:

Express x(t) in the form

               ∞

x(t) =        ∑    A$_{k}$ cos ( w$_{k}$ t + φ$_{k}$ )

              $_{k=0}$  

Solution:

Compute fundamental frequency of the signal:

w₀ = 2 π / T

     = 2 π / 8               Since T = 8

w₀  = π / 4

                   ∞

                    ∑ $a_{k}e^{jw_{0}t }$

x(t) =          k=⁻∞

   

     =  $a_{1}e^{jw_{0} t} + a_{-1}e^{-jw_{0} t} + a_{5}e^{5jw_{0} t}+a_{-5}e^{-5jw_{0} t}$  

     =  $je^{j(\pi/4)t} - je^{-j(\pi/4)t} +2e^{(5\pi/4)t}+2e^{-(5\pi/4) t}$

     = −2 sin ( $\frac{\pi }{4}t$ ) + 4 cos ( $\frac{5\pi }{4}t$ )

     = −2 cos ( $\frac{\pi }{4}t$  − $\frac{\pi }{2}$ ) + 4 cos ( $\frac{5\pi }{4}t$ )