and complex exponentials with a fundamental frequency of wo=2 y(t) 1-e(3)2t1 (-3)t 点3)=∑b be-2,k otherwise b=akH(jk)→H(k)=a, for ak≠0 y(t) has only three non-zero components, therefore H(u) has a non-zero value at those three components, corresponding to A1, A2 and A, as follows H(j(0)o) bo 1 1/2 2=4 H((3)o) b3 (3.)2 4sin(3丌/2 9 j)=7丌2 H(1(-3)o) ÷2 (-3.丌/2) j(9)m2 4sin(-3丌/2) 9 4 HGw) also needs to eliminate the other frequency components of a(t)which do not exist in the output y(t) H(jk0)=0,fork≠0,± To meet the above conditions, the cutoff frequencies 1.2.3 must be chosen to pass the desired components and reject the undesired components. The following inequalities meet that
� � � � � and complex exponentials with a fundamental frequency of �0 = 2 . � � ej 3� t − 1 e−j 3� t 3ω 2 2 y(t) = 1 − cos t = 1 − 2 2 2 1 j(−3) � t = 1 − 2 2 1 ej(3) � t − e 2 2 1 1 j(−3)�0t = bkejk�0t = 1 − ej(3)�0t − e 2 2 k ⎧1, k = 0 − � 1 bk 2 , k = ±3 ⎧0, otherwise � bk = akH(jk�0) � H(jk�0) = bk , for ak ≤= 0. ak y(t) has only three non-zero components, therefore H(j�) has a non-zero value at those three components, corresponding to A1, A2 and A3 as follows: b0 1 H(j(0)�0) = = = 2 = A2. a0 1/2 b3 1 H(j(3)�0) = = − ÷ 2 sin(3 · ω/2) e−j(3)�/2 a3 2 j(3 · ω)2 = − j(9)ω2 ej 3� 2 4 sin(3ω/2) 9 = − j9ω2 (−j) = ω2 = A3. 4(−1) 4 1 sin(−3 · ω/2) e−j(−3)�/2 H(j(−3)�0) = b−3 = − ÷ 2 a−3 2 j(−3 · ω)2 = − 2 j(9)ω2 ej −3� 4 sin(−3ω/2) 9 = −j9ω2 (j) = ω2 = A1. 4(1) 4 H(j�) also needs to eliminate the other frequency components of x(t) which do not exist in the output y(t). � H(jk�0) = 0, for k =≤ 0, ±3 To meet the above conditions, the cutoff frequencies �1,2,3 must be chosen to pass the desired components and reject the undesired components. The following inequalities meet that 6
requirement 0<91<(1)0,(2)0<92<(3)0,(3)0<93<(4)0 (21) Choosing any cutoff frequencies which satisfy (2. 1) would be sufficient. Let's say, for practicality sake, that we want to choose the cutoff frequencies in the midpoints between the desired and undesired frequency components. This gives us the following specific values 2,92,(2+3)u 0+1)u T
requirement: 0 < �1 < (1)�0, (2)�0 < �2 < (3)�0, (3)�0 < �3 < (4)�0 (2.1) Choosing any cutoff frequencies which satisfy (2.1) would be sufficient. Let’s say, for practicality sake, that we want to choose the cutoff frequencies in the midpoints between the desired and undesired frequency components. This gives us the following specific values: �1 = (0 + 1)�0 2 , �2 = (2 + 3)�0 2 , �3 = (3 + 4)�0 2 , where �0 = ω 2 ω 5ω 7ω � �1 = 4 , �2 = 4 , �3 = 2 . 7
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