Question

Consider the discrete-time systems $ T_1 $ and $ T_2 $ defined as follows: 
$ [T_1x][n] = x[0] + x[1] + \dots + x[n], $ 
$ [T_2x][n] = x[0] + \frac{1}{2}x[1] + \dots + \frac{1}{2^n}x[n]. $ 
Which of the following statements is true?

• A. \( T_1 \) and \( T_2 \) are BIBO stable
• B. \( T_1 \) and \( T_2 \) are not BIBO stable
• C. \( T_1 \) is BIBO stable but \( T_2 \) is not BIBO stable
• D. \( T_1 \) is not BIBO stable but \( T_2 \) is BIBO stable

Hint:

For a system to be BIBO stable, the output must remain bounded for any bounded input. Summation with decaying weights generally ensures stability.

Answer 1

The Correct Option is D

From the question we have given, \[ [T_1x][n] = x[0] + x[1] + \dots + x[n], \] \[ [T_2x][n] = x[0] + \frac{1}{2}x[1] + \dots + \frac{1}{2^n}x[n]. \] Step 1: Analyze \( T_1 \). The system \( T_1 \) sums all inputs without any decay factor. For a bounded input \( x[k] = u[k] \), the output becomes: \begin{align*} &\sum_{k=0}^{\infty} |h[k]|Lt;\infty
&(T_1 x)[n] = \sum_{k=0}^{n} x[k]
x[k] = u[k] = \text{Bounded Input}
&(T_1 x)[n] = \left|\sum_{k=0}^{n} u[k]\right| = \sum_{k=0}^{n} 1 = n \end{align*} which grows unbounded as \( n \to \infty \). Hence, \( T_1 \) is not BIBO stable. Step 2: Analyze \( T_2 \). The system \( T_2 \) includes a decay factor \( \frac{1}{2^k} \). For a bounded input \( x[k] = u[k] \), the output becomes: \[ \{T_{2}x\}[n] = x[0] + \frac{1}{2}x[1] + \cdots + \frac{1}{2^{n}}x[n] \] \begin{align*} \{T_2 x\}[n] &= \sum_{k=0}^{n} \left( \frac{1}{2} \right)^k x[k]
x[k] &= u[k]
\{T_2 x\}[n] &= \sum_{k=0}^{n} \left| \left( \frac{1}{2} \right)^k u[k] \right|
&= \sum_{k=0}^{n} \left( \frac{1}{2} \right)^k
\{T_2 x\}[n] &= \frac{1}{1-\frac{1}{2}} = 2 \end{align*} This output remains bounded for bounded inputs. Hence, \( T_2 \) is BIBO stable.