For a causal LTI system described by the difference equation y[n] + a1 y[n−1] = b0 x[n], which of the following is its transfer function H(z)?

• A. b0 / (1 + a1 z^{-1})
• B. b0 / (1 - a1 z^{-1})
• C. b0 z^{-1} / (1 + a1 z^{-1})
• D. b0 (1 + a1 z^{-1})

Answer: (A)

When you form the transfer function, you relate Y(z) to X(z) for a causal LTI system. Take the Z-transform of the difference equation: Y(z) + a1 z^{-1} Y(z) = b0 X(z). Factor Y(z) to get Y(z)[1 + a1 z^{-1}] = b0 X(z), so the transfer function is H(z) = Y(z)/X(z) = b0 / (1 + a1 z^{-1}). This matches the given form. You can also rewrite it as H(z) = b0 z / (z + a1) if you multiply top and bottom by z, but the essential relationship is the same. The pole is at z = -a1, and the input uses x[n] directly (no delay), so there’s no extra delay factor in the numerator.