y()=x+(t) ase-jku k=-00 Letk=-k→y(t) a bk=a Now to prove Parseval's relation for period signals we can use the results above as follows: r(t)/2=r(t) *(t)=a(t)y(t) 2celkrot, where c =o bk-n, from part( n=-0 ∑ak-nc=∑ an"n-kekw as proven earlier A common mistake is to using the substitution bk-n=a*+n(i.e. just negating the index n) which would be correct if the relationship was shifting in frequency and then taking the conjugate, while the case at hand is taking the conjugate first and then shifting in frequency to perform the convolution Pave=t/=(t)dt= To Jo k==oo n= and dt yot To Notice that the integration has a value of To only when k=0 and zero for all the other values of k, as seen below. This rule, usually referred to as orthogonality holds for any complex exponential signal, integrated over one period Fork≠0 (1)dt= To Therefore we have: P To ana Db=∑|an2 lr(t)dt� � +� � +� +� +� +� +� +� +� +� +� +� +� ⎨ jk�0t ⎨ ke−jk�0t y(t) = x� (t) = ake = a� k=−� k=−� ⎨ jk�0t Let k = −k � y(t) = a� −ke k=−� � bk = a� −k Now to prove Parseval’s relation for period signals we can use the results above as follows: |x(t)| 2 = x(t)x� (t) = x(t)y(t) ⎨ ⎨ = ckejk�0t , where ck = anbk−n, from part(a) k=−� n=−� ⎨ ⎨ ⎨ ⎨ = anbk−nejk�0t = ana� jk�0t , as proven earlier. n−ke k=−� n=−� k=−� n=−� A common mistake is to using the substitution bk−n = a� (i.e. just negating k+n the index n) which would be correct if the relationship was shifting in frequency and then taking the conjugate, while the case at hand is taking the conjugate first and then shifting in frequency to perform the convolution. 1 ⎪ T0 1 ⎪ T0 +� T0 0 | T0 0 k= n−kejk�0t dt ⎨ ⎨ Pave = |x(t) 2 dt = ana� −� n=−� ⎨ ⎨ ⎪ T0 1 ana� = T0 n−k ejk�0t dt 0 n=−� k=−� Notice that the integration has a value of T0 only when k = 0 and zero for all the other values of k, as seen below. This rule, usually referred to as orthogonality, holds for any complex exponential signal, integrated over one period. ⎪ T0 1 � � T0 1 For k jk�0t � √= 0 : 0 ejk�0t dt = jk�0 e � = jk�0 (ejk�0T0 − ej0 ) 0 1 = (ejk(2�) − ej0 ) = 0 jk�0 ⎪ T0 ⎪ T0 ⎪ T0 For k = 0 : ejk�0t dt = ej(0)�0t dt = (1)dt = T0. 0 0 0 ⎨ 1 ⎨ +� ⎪ T0 1 +� Therefore we have: Pave = ana�T0 = 2 = x(t)| 2 dt. T0 n |an| T0 0 | n=−� n=−� 2