Question:hard
The response of a discrete time system \(y[n]\) obeys the following relation:
\[ y[n]=\frac{5}{6}y[n-1]-\frac{1}{6}y[n-2]+x[n]. \] The input to the system is \(x[n]=\delta[n]-\frac{1}{3}\delta[n-1]\). Which of the following options is TRUE for \(y[n]\)?
The response of a discrete time system \(y[n]\) obeys the following relation:
\[ y[n]=\frac{5}{6}y[n-1]-\frac{1}{6}y[n-2]+x[n]. \] The input to the system is \(x[n]=\delta[n]-\frac{1}{3}\delta[n-1]\). Which of the following options is TRUE for \(y[n]\)?
\[ y[n]=\frac{5}{6}y[n-1]-\frac{1}{6}y[n-2]+x[n]. \] The input to the system is \(x[n]=\delta[n]-\frac{1}{3}\delta[n-1]\). Which of the following options is TRUE for \(y[n]\)?
Show Hint
Factor the characteristic polynomial to find the pole locations, then check that both poles lie strictly inside the unit circle for the causal ROC.
Updated On: Jul 20, 2026
- Stable and causal response
- Stable and non-causal response
- Unstable and causal response
- Unstable and non-causal response
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Solve the homogeneous recursion first.
The recursion $y[n]-\frac56y[n-1]+\frac16y[n-2]=0$ has a characteristic equation found by trying $y[n]=r^n$:
\[ r^2-\frac56r+\frac16=0 \] Multiplying through by $6$ gives $6r^2-5r+1=0$, which factors as $(2r-1)(3r-1)=0$, giving roots $r=\dfrac12$ and $r=\dfrac13$.
Step 2: Write the general causal solution form.
Because the recursion only uses past values of $y$ and the present $x[n]$, solving it forward from $n=0$ produces a response built from these two roots as base terms:
\[ y[n]=C_1\left(\frac12\right)^n+C_2\left(\frac13\right)^n,\quad n\ge0 \] plus a particular part driven by $x[n]$, which is itself a combination of shifted, scaled versions of these same decaying terms since $x[n]$ is just two impulses.
Step 3: Check boundedness directly from these terms.
Both base terms $(1/2)^n$ and $(1/3)^n$ shrink to zero as $n$ grows, and neither blows up for any finite $n$. A sum of decaying geometric terms stays bounded for a bounded input, which is the direct time-domain meaning of stability, matching the earlier finding from the pole locations.
Step 4: Check causality directly from the recursion structure.
Since $y[n]$ only ever needs $y[n-1]$, $y[n-2]$ and the current $x[n]$, and never a future sample, the response at any time $n$ can be computed purely from the past, which is exactly what causal means.
Step 5: Conclude.
Both the decay of the geometric terms and the forward-only dependency confirm the same result reached earlier from the z-transform poles.
\[ \boxed{\text{Stable and causal response}} \]
The recursion $y[n]-\frac56y[n-1]+\frac16y[n-2]=0$ has a characteristic equation found by trying $y[n]=r^n$:
\[ r^2-\frac56r+\frac16=0 \] Multiplying through by $6$ gives $6r^2-5r+1=0$, which factors as $(2r-1)(3r-1)=0$, giving roots $r=\dfrac12$ and $r=\dfrac13$.
Step 2: Write the general causal solution form.
Because the recursion only uses past values of $y$ and the present $x[n]$, solving it forward from $n=0$ produces a response built from these two roots as base terms:
\[ y[n]=C_1\left(\frac12\right)^n+C_2\left(\frac13\right)^n,\quad n\ge0 \] plus a particular part driven by $x[n]$, which is itself a combination of shifted, scaled versions of these same decaying terms since $x[n]$ is just two impulses.
Step 3: Check boundedness directly from these terms.
Both base terms $(1/2)^n$ and $(1/3)^n$ shrink to zero as $n$ grows, and neither blows up for any finite $n$. A sum of decaying geometric terms stays bounded for a bounded input, which is the direct time-domain meaning of stability, matching the earlier finding from the pole locations.
Step 4: Check causality directly from the recursion structure.
Since $y[n]$ only ever needs $y[n-1]$, $y[n-2]$ and the current $x[n]$, and never a future sample, the response at any time $n$ can be computed purely from the past, which is exactly what causal means.
Step 5: Conclude.
Both the decay of the geometric terms and the forward-only dependency confirm the same result reached earlier from the z-transform poles.
\[ \boxed{\text{Stable and causal response}} \]
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