Problem 3 Consider a causal discrete-time LTI system whose input an] and output g/nl are related by the following difference equation yn-=yn-1=n+ 2 rn-4 Find the Fourier series representation of the output yIn] when the input is a[n=2+sin(rn/4)-2 cos(Tn/2) First, let's find the frequency response of the system from the difference equation by injecting an input, a[n, that is an eigenfunction of the LTI system r]=en→yn]=H(e)e H(eu) is the frequency response characterizing the system or the eigenvalue of the system By substituting a[n] and yn] in the difference equation y-n-1=m+2nn-4 H(e)e H(e)e-4. H(e)emn e= e+2.eune H(e)el eum 1+2e-Ju4] 1+2e-1u4 Then, we find the Fourier series coefficients, ak, of the given input, possibly by dissecting the input expression into a summation of complex exponentials ]=2+sin(丌n/4)-2cos(m/2)=2+sin(0n)-2cos(2u0m), where wo=- is the greatest Common Factor for the sinusoids frequencies Won won 2e(0×0n+ 2e0k0n⊥1 +c(1)0n_。(-1)on-a(2)nc(-2)un 23 N=8- ak has only eight distinct values and is periodic with a period of N=8� � Problem 3 Consider a causal discrete-time LTI system whose input x[n] and output y[n] are related by the following difference equation: 1 y[n] − y[n − 1] = x[n] + 2x[n − 4] 4 Find the Fourier series representation of the output y[n] when the input is x[n] = 2 + sin(ωn/4) − 2 cos(ωn/2). First, let’s find the frequency response of the system from the difference equation by injecting an input, x[n], that is an eigenfunction of the LTI system: j�n � y j�) ej�n x[n] = e [n] = H(e H(ej�) is the frequency response characterizing the system or the eigenvalue of the system. By substituting x[n] and y[n] in the difference equation: 1 y[n] − y[n − 1] = x[n] + 2x[n − 4] 4 j�) ej�(n−1) j�(n−4) H(ej�) ej�n − 1 · H(e = ej�n + 2 · e 4 1 j�) ej�n − j�) ej�n ej�(−1) j�n ej�(−4) H(e · H(e = ej�n + 2 · e 4 H(ej�) ej�n 1 j�n � 1 + 2 e−j�4 � 1 − e−j� = e 4 � H(ej�) = 1 + 2 e−j�4 . 1 − 1 e−j� 4 Then, we find the Fourier series coefficients, ak, of the given input, possibly by dissecting the input expression into a summation of complex exponentials: x[n] = 2 + sin(ωn/4) − 2 cos(ωn/2) = 2 + sin(�0n) − 2 cos(2�0n), ω where �0 = is the Greatest Common Factor for the sinusoids frequencies 4 ej�0n − e−j�0n ej2�0n + e−j2�0n = 2 ej(0)�0n + − 2 · 2j 2 1 1 = 2 ej(0)�0n + ej(1)�0n − j(−1)�0n − ej(2)�0n − ej(−2)�0n e 2j 2j � N = 8 � ak has only eight distinct values and is periodic with a period of N = 8. 8