Question

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?

• A. \(S\) is causal.
• B. \(S\) is linear and time invariant.
• C. \(S\) is linear and time varying.
• D. \(S\) is non-linear.

Hint:

Convolution, addition and a fixed delay are all linear and time invariant, so any interconnection of LTI blocks stays LTI.

Answer 1

The Correct Option is B

Step 1: Trace the upper branch of the diagram.
The input $x[n]$ passes through $h_1[n]$ to produce $a[n]=h_1[n]*x[n]$, where $*$ denotes convolution. This $a[n]$ feeds the final summing junction and is also delayed by one sample through a $z^{-1}$ block.

Step 2: Trace the lower branch.
At the lower summing junction, the delayed value $a[n-1]$ is subtracted from $x[n]$, giving $c[n]=x[n]-a[n-1]$. This passes through $h_2[n]$ to give $d[n]=h_2[n]*c[n]$, which is then delayed by another $z^{-1}$ block.

Step 3: Write the final output.
At the top summing junction, the delayed $d[n-1]$ is subtracted from $a[n]$, so \[ y[n]=a[n]-d[n-1] \]

Step 4: Check linearity.
Every operation used to build $y[n]$ from $x[n]$, convolution with $h_1$ and $h_2$, unit delay, and addition or subtraction, is a linear operation. A network built only from linear building blocks connected in cascade and feedback still obeys superposition at every internal signal, so the overall system $S$ is linear no matter what $h_1[n]$ and $h_2[n]$ actually are.

Step 5: Check time invariance.
Convolution with a fixed impulse response is a time invariant operation, and a unit delay is time invariant by definition, since delaying the input by $n_0$ samples simply delays every internal signal, and hence the output, by the same $n_0$ samples. Since every block in the diagram is time invariant, the interconnection $S$ is time invariant as well.

Step 6: Check causality.
Causality of $S$ depends only on whether $h_1[n]$ and $h_2[n]$ are themselves causal. The question states only that they are LTI, with no restriction that their impulse response is zero for negative $n$, so $S$ is not necessarily causal.

Step 7: Rule out the other options.
Option (A) fails because causality is not guaranteed by the given information. Options (C) and (D) contradict the fact, shown above, that the system is built entirely from linear, time invariant blocks.

Step 8: Final conclusion.
\[ \boxed{S\text{ is linear and time invariant}} \]