Question

please show detailed work/proof

3. The input and output of a causal LTI system satisfy the following difference equation (d.e.) y[n] = ayln-1] + x[n]-a"x[n-N], N > 0 a. Determine the impulse response h[n]. Hint: solve it iteratively starting from n=0, 1, , n=N+1; x[n] = δ[n] then think what is y[n] ? b. Sketch the impulse response h[n] c. Is this an FIR or an IIR system? d. For what values of the parameter a is the system stable?

Answer 1

Y(n) = ay(n - 1) + x(n) - a^2 x(n - N), N > 0 take z-transforms on both sides z[y(n)] = az[y(n - 1)] + z[x(n)] - a^N z[x(n - N)] y(z) = az^- 1 y(z) + x(z) - a^N z^- N x(z) y(z) [1 - az^- 1] = x(z)[1 - a^N z^- N] y(z)/x(z) = 1 - a^N z^- N/1 - az^- 1 y(z)/x(z) = 1/1 - az^- 1 = a^N z^- N/1 - az^- 1 = H(z) formulae used: z(n) x(z) (time shifting) impulse response = z transforms output/z transforms input = y(n)/x(n) l(n) H(z) = a^n (n) - a^N a(n - N) Union (n - N) impulse response consider N = 2 rightarrow L(n) = a^n (n) - a^n a^(n - 2) Union (n - 2) L(n) = a^n (n) - a^n Union(n - 2) given system is casual that means it exist