Question

Suppose we are given the following information about a continuous-time periodic signal with period 3 and Fourier coefficients

aka_k

: 1.

ak=ak+2.a_k=a_{k+2}.

ak=ak.a_k=a_{-k}.

0.50.5x(t)dt=1.∫_{-0.5}^{0.5}x(t)dt=1.

4.(∫_{0.5}^{1.5}x(t)dt=2.

Determinex(t).Determine x(t).

Solution

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The Fourier series of continuous-time periodic signal x(t)x(t) can be expressed as follows:

x(t)=k=akejkω0t\textcolor{#4257b2}{x(t) = \sum_{k=-\infty}^{\infty} a_k e^{j k \omega_0 t}}

From problme description we have:

T=3 ω0=2πT=2π3T = 3 \ \rightarrow \omega_0 = \frac{2 \pi}{T} = \frac{2\pi}{3}

Taking into account information ak=ak+2a_k = a_{k+2} we should apply frequency shifting property:

ak+M ejM2πTtx(t)\textcolor{#4257b2}{a_{k+M} \ \rightarrow e^{-j M \frac{2\pi}{T}t} x(t)}

After substitution we have:

ak+2 ej4π3tx(t)=ej2π23tx(t)\begin{align*} a_{k+2} \ \rightarrow e^{-j \frac{4\pi}{3}t} x(t) = e^{-j 2\pi \frac{2}{3}t} x(t) \end{align*}

Furthermore, information ak=aka_k = a_{-k} indicates conjugate symmetry property: x(t)=x(t)x(t) = x(-t). Notice that x(t)x(t) achieves nonzero values for t=±32nt= \pm \dfrac{3}{2}n.

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