Problem 1(O&w 3. 22(a)-only the signal in Figure p3.22(c)) Determine the Fourier series representation for the signal a (t) c(t) 2+t, for -2<t<O c(t for 0<t<1 r(t) periodic with period T=3→==等 a goal of this problem solution is to show different ways to reaching the same answer Finding the Fourier series coefficients of a signal using the analysis equation usually requires the most effort, but can be reverted to if everything else fails. Oftentimes, a signal can be dissected into simpler signals that are easier to analyze or can be derived from a simpler signal by integration, differentiation, time shifting, or any combination of the properties of the Fourier series (see Table 3.1, O&w, p. 206) We will start with finding ao, which is usually straight-forward and doesn't require much effort, and then explore the different methods for finding ak+0 ao T/(t)dt=-(the total area under the curve for one period)I (2+1)=1 The following are four possible methods to calculate ak+0, the Fourier series coefficients of r(t)fork≠0: Method (a): Using the integration property Let g(t)=do -r(t)=g(t)dt+, where p is the value of r(t)at the beginning of the period, and it equals to zero for the period we selected that starts at t=-2. Note that, since we are trying to find ak+, the value of p is not important because it only ffects the DC level of a(t)and we have already calculated it by finding ao g(t) 234 1� � 0 Problem 1 (O&W 3.22 (a) - only the signal in Figure p3.22 (c)) Determine the Fourier series representation for the signal x(t). x(t) 2 · · · · · · x(t) = 2 + t, for − 2 ← ←t 2 − 2t, for 0 ← ←t 1. −5 −4 −3 −2 −1 0 1 2 3 4 t T 2� x(t) periodic with period T = 3 �0 = 2 T � ⇒ = 3 A goal of this problem solution is to show different ways to reaching the same answer. Finding the Fourier series coefficients of a signal using the analysis equation usually requires the most effort, but can be reverted to if everything else fails. Oftentimes, a signal can be dissected into simpler signals that are easier to analyze or can be derived from a simpler signal by integration, differentiation, time shifting, or any combination of the properties of the Fourier series (see Table 3.1, O&W, p.206). We will start with finding a0, which is usually straight-forward and doesn’t require much effort, and then explore the different methods for finding ak=0 √ : ⎪ 1 1 1 a0 = x(t)dt = (the total area under the curve for one period) = (2 + 1) = 1. T T 3 3 The following are four possible methods to calculate ak√=0, the Fourier series coefficients of x(t) for k =√ 0: • Method (a): Using the integration property: Let g(t) = dx(t) x(t) = g(t)dt + p, where p is the value of x(t) at the beginning of dt ⇒ the period, and it equals to zero for the period we selected that starts at t = −2. Note that, since we are trying to find ak=0 √ , the value of p is not important because it only affects the DC level of x(t) and we have already calculated it by finding a0. g(t) −5 −4 −3 2 3 4 −1 t −2 1 · · · · · · 3