Problem 2 O w323(a) Given ak, the Fourier series coefficients of a periodic continuous time signal with period 4 determine the signal a(t) The Fourier series coefficients ak are given as follows in k/4 T, otherwise k Here are some of the facts we know about a(t) a0=0- no DC component in a(t) ·T=4→o=2x/4=丌/2 Gr* sin(-k/ 4) sin(k /4) (-)smkx/4= Thus r(t) is a real signal (O&W, Section 3.5.6, p. 204) Noting that j=ej/2-G)=(ei/2)=ekm/2= ejkwo =e-jkuwo( -1), we can consider r(t) to be a time-shifted version of another signal y(t)such that x()=y/(t +1), where v(0)-, bo =0, ba 0- sin k/4 and ak =bge/ko() By backtracking the derivation equation of bk, we can find the signal i(t)which has the same bk but can have a different DC level (i.e. bo) Sin b≠0=b≠0 k丌/41/c1kx/4-c- k 2 (1)e?kott The integration above suggests that <t< 0(t lo, elsewhere in the same period T=4� � � � � � Problem 2 O & W 3.23 (a) Given ak , the Fourier series coefficients of a periodic continuous time signal with period 4, determine the signal x(t). The Fourier series coefficients ak are given as follows: a �0, k = 0 k = �(j)k sin k�/4 , otherwise. k� Here are some of the facts we know about x(t): • a0 = 0 ⇒ no DC component in x(t) • ⇒ T = 4 �0 = 2�/4 = �/2 • � �k (j) −k sin(−k�/4) 1 − sin(k�/4) a−k = = −k� j −k� (−j) k sin(k�/4) = = a� k. k� Thus x(t) is a real signal (O&W, Section 3.5.6, p.204). e−jk�0(−1) Noting that j = ej�/2 ⇒ (j)k = ej�/2 �k = ejk�/2 = ejk�0 = , we can consider x(t) to be a time-shifted version of another signal y(t) such that: F x(t) = y(t + 1), where y(t) �⇒ b0 = 0, bk=0 = sin k�/4 and ak = bkejk�0(1) √ k� By backtracking the derivation equation of bk, we can find the signal ˆ b y(t) which has the same k but can have a different DC level (i.e. b0) : ˆbk=0 √ = bk=0 = sin k�/4 = 1 ejk�/4 − e−jk�/4 √ k� k� 2j 1 1 � = (4) jk(� ejk�/4 − e−jk�/4 � 2 ) ⎪ 1 2 ) 1 1 1 2 jk�0( 1 − ejk�0(− 2 ) � 1 (1)ejk�0t = e = dt. 1 2 T jk�0 T − The integration above suggests that 1 yˆ(t) = 1, −2 < t < 1 2 0, elsewhere in the same period T=4. 9