Q6Digital Signal Processing

The system's impulse response is h(n) = 2δ(n-3), i.e., a scaled unit impulse located at n=3 (h(3)=2, h(n)=0 for all other n). Even though the problem describes this as a 'linear time-variant system', for an impulse response of this simple scaled-and-shifted-delta form, the input-output relationship is simply convolution (as would apply for any LTI system with this impulse response), since y(n) = Σ_k x(k)h(n-k) = Σ_k x(k)·2δ(n-k-3) = 2x(n-3), because the delta function picks out exactly the term where n-k=3, i.e., k=n-3.

The input's Z-transform is given as X(z) = z^4 + z^2 - 2z + 2 - 3z^-4. Recalling that X(z) = Σ_n x(n)z^-n, each term z^-n in this expansion corresponds to the coefficient of x(n): the term z^4 = z^-(-4) corresponds to x(-4)=1; the term z^2 = z^-(-2) corresponds to x(-2)=1; the term -2z = -2z^-(-1) corresponds to x(-1)=-2; the constant term 2 = 2z^0 corresponds to x(0)=2; and the term -3z^-4 corresponds to x(4)=-3. All other values of x(n) not listed (including x(1), x(2), x(3), and any n not among {-4,-2,-1,0,4}) are zero, since X(z) is a finite polynomial containing only these five terms.

Using the input-output relation y(n) = 2x(n-3), the output at n=4 is:

Since x(1) is not among the nonzero terms listed in X(z) (the given polynomial has terms only at n = -4, -2, -1, 0, 4), x(1) = 0, and therefore:

This result also makes intuitive sense noting that h(n) simply delays and scales its input by 3 samples, so the output at any time n is entirely determined by the single input sample x(n-3); since x(n) here is a sparse sequence with nonzero values only at the five specific indices listed, and n-3=1 is not one of them, the output at n=4 must be exactly zero.