$T_1$ System BIBO Stability Analysis

A discrete-time system is BIBO (Bounded Input, Bounded Output) stable if every bounded input sequence produces a bounded output sequence.

The system $T_1$ is defined as: $\{T_1x\}[n] = x[0] + x[1] + \dots + x[n] = \sum_{k=0}^{n} x[k]$

To check for BIBO stability, consider a bounded input $x[n] = 1$ for all $n \ge 0$. This input satisfies $|x[n]| \le 1$ for all $n$.

The corresponding output is:

As $n$ increases, the output $n+1$ grows without bound ($n+1 \to \infty$ as $n \to \infty$). Since a bounded input leads to an unbounded output, the system $T_1$ is not BIBO stable.

$T_2$ System BIBO Stability Analysis

The system $T_2$ is defined as: $\{T_2x\}[n] = x[0] + \frac{1}{2}x[1] + \dots + \frac{1}{2^n}x[n] = \sum_{k=0}^{n} \frac{1}{2^k} x[k]$

Assume a bounded input sequence $x[n]$ such that $|x[n]| \le M$ for all $n$, where $M$ is a finite constant.

Examine the magnitude of the output:

Using the triangle inequality:

Substitute $|x[k]| \le M$:

The sum $\sum_{k=0}^{n} (\frac{1}{2})^k$ is a partial sum of a convergent geometric series. The infinite sum converges to:

Thus, for any finite $n$, the sum $\sum_{k=0}^{n} (\frac{1}{2})^k$ is bounded and strictly less than 2.

The output $\{T_2x\}[n]$ is therefore bounded (specifically, $|\{T_2x\}[n]| < 2M$). Since any bounded input produces a bounded output, the system $T_2$ is BIBO stable.

Conclusion

System $T_1$ is not BIBO stable, while system $T_2$ is BIBO stable.