Consider the discrete time system (S) with input \(x[n]\) and output \(y[n]\) as shown in the figure. The two sub-systems, represented by their impulse responses \(h_1[n]\) and \(h_2[n]\), are linear and time invariant. Which of the following statements is necessarily TRUE?
Show Hint
Convolution, addition and a fixed delay are all linear and time invariant, so any interconnection of LTI blocks stays LTI.
Step 1: Represent each branch in the z-domain. Let $X(z)$, $Y(z)$, $H_1(z)$ and $H_2(z)$ denote the z-transforms of $x[n]$, $y[n]$, $h_1[n]$ and $h_2[n]$. Convolution in time becomes multiplication in the z-domain, and a $z^{-1}$ block simply multiplies by $z^{-1}$.
Step 2: Write the upper branch. \[ A(z)=H_1(z)X(z) \]
Step 3: Write the lower branch. \[ C(z)=X(z)-z^{-1}A(z),\qquad D(z)=H_2(z)C(z) \]
Step 4: Combine at the output junction. \[ Y(z)=A(z)-z^{-1}D(z) \]
Step 5: Substitute to get a single relation between Y(z) and X(z). \[ Y(z)=H_1(z)X(z)-z^{-1}H_2(z)\left[X(z)-z^{-1}H_1(z)X(z)\right] \] \[ Y(z)=X(z)\left[H_1(z)-z^{-1}H_2(z)+z^{-2}H_1(z)H_2(z)\right] \]
Step 6: Interpret this relation. The bracketed term is a single fixed function of $z$, call it $H(z)$, so $Y(z)=H(z)X(z)$ for every input, with $H(z)$ not depending on the time index $n$ at all. A relation of the form $Y(z)=H(z)X(z)$ is exactly the z-domain signature of an LTI system, since it corresponds to $y[n]=h[n]*x[n]$ in time.
Step 7: Note what is not guaranteed. $H(z)$ may contain positive powers of $z$ (from the two delay terms), so the equivalent impulse response $h[n]$ need not be zero for $n<0$, meaning causality of $S$ is not guaranteed.
Step 8: Conclude. \[ \boxed{S\text{ is linear and time invariant}} \]
