Question

A relaxed linear time-invariant system with impulse response \[ h(n)=a^n u(n), \qquad |a|<1 \] when the input is a unit step sequence, \[ x(n)=u(n), \] then the output \(y(\infty)\) of the system is

• A. \(\dfrac{1+a^{(n+1)}}{1+a}\)
• B. \(\dfrac{1-a^{(n-1)}}{1+a}\)
• C. \(\dfrac{1-a^{(n+1)}}{1-a}\)
• D. \(1-a\)

Hint:

For \[ \sum_{k=0}^{n}r^k = \frac{1-r^{n+1}}{1-r}. \] Always recognize geometric series immediately in DSP problems.

Answer 1

The Correct Option is C

Concept: For an LTI system, \[ y(n)=x(n)*h(n). \] When the input is a unit step sequence, the output is the step response. \[ y(n)=\sum_{k=0}^{n}a^k. \]

Step 1: Perform convolution.
\[ y(n)=\sum_{k=0}^{n}a^k. \] This is a finite geometric progression. \[ y(n)=\frac{1-a^{n+1}}{1-a}. \]

Step 2: Evaluate the steady-state value.
Since \[ |a|<1, \] \[ \lim_{n\to\infty}a^{n+1}=0. \] Therefore, \[ y(\infty)=\frac{1}{1-a}. \] The expression corresponding to the step response is \[ \boxed{\frac{1-a^{n+1}}{1-a}} \] Hence option (C) is the intended answer.