J 2 k 4 j4 e k 1 2 k0 bk jk0 jk j4 h

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Unformatted text preview: 1 j�n 1 −j�n 1 j �n/2 1 −j�n/2 = 2ej 0 + e + e − + e + ej�n/4 ej �/4 + e−j�n/4 e−j�/4 e (1) 2 2 2j 2j y [n] = 4 − 2 sin(ωn) + 2 cos(ωn/4) � � � � 1 j�n/4 1 −j�n/4 1 j�n 1 −j�n e−e +2 e +e = 4−2 2j 2j 2 2 1 1 = 4ej 0 − ej�n + e−j�n + ej�n/4 + e−j�n/4 (2) j j Now we can find H (ej� ) using the following two approaches: (I) Consider equation (1) to b e the summation of the eigenfunctions ej�n that comprise the input, x[n]. By superposition, the output y [n] will contain the same eigenfunctions multiplied by the system’s eigenvalues, i.e ej�n H (ej� ) (refer to O & W, Section 3.2, p.183). 1 j�n e+ 2 1 y [n] = 4ej 0 − ej�n + j x[n] = 2ej 0 + 1 −j�n 1 j �n/2 1 −j�n/2 e − e + e + ej�/4 ej �n/4 + e−j�/4 e−j�n/4 2 2j 2j 1 −j�n e + (0)ej �n/2 + (0)e−j�n/2 + ej�n/4 + e−j�n/4 j By matching the eigenfunctions in the input and the output, we can find the values of 11 the frequency response: � ⎧−j 2, ⎧ ⎧ ⎧ j�/4 ⎧e ⎧ , ⎧ ⎧ ⎧ ⎧2, ⎧ � j� � H (e ) = e−j�/4 , ⎧ ⎧ ⎧j 2, ⎧ ⎧ ⎧ ⎧0, ⎧ ⎧ ⎧ ⎧ �?, � = −ω � = −� 4 �=0 �=� 4 �=ω � = ±...
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