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?
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?
Show Hint
For a system to be BIBO stable, the output must remain bounded for any bounded input. Summation with decaying weights generally ensures stability.
Updated On: Feb 3, 2026
- \( T_1 \) and \( T_2 \) are BIBO stable
- \( T_1 \) and \( T_2 \) are not BIBO stable
- \( T_1 \) is BIBO stable but \( T_2 \) is not BIBO stable
- \( T_1 \) is not BIBO stable but \( T_2 \) is BIBO stable
Show Solution
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The Correct Option is D
Solution and Explanation
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.
&(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.
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