Question

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 \(y[n]\)?

• A. Stable and causal response
• B. Stable and non-causal response
• C. Unstable and causal response
• D. Unstable and non-causal response

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.

Answer 1

The Correct Option is A

Step 1: Write the difference equation in standard form.
Rearranging the given relation, \(y[n]-\dfrac{5}{6}y[n-1]+\dfrac{1}{6}y[n-2]=x[n]\).

Step 2: Take the z-transform to get the system function.
Using \(y[n-k]\leftrightarrow z^{-k}Y(z)\),
\[ Y(z)\left(1-\frac{5}{6}z^{-1}+\frac{1}{6}z^{-2}\right)=X(z) \] \[ H(z)=\frac{Y(z)}{X(z)}=\frac{1}{1-\dfrac{5}{6}z^{-1}+\dfrac{1}{6}z^{-2}} \]

Step 3: Factor the denominator to find the poles.
\[ 1-\frac{5}{6}z^{-1}+\frac{1}{6}z^{-2}=\left(1-\frac{1}{2}z^{-1}\right)\left(1-\frac{1}{3}z^{-1}\right) \] Expanding this back gives \(1-\left(\frac12+\frac13\right)z^{-1}+\frac16z^{-2}=1-\frac56z^{-1}+\frac16z^{-2}\), which matches, so the poles of \(H(z)\) are at \(z=\dfrac{1}{2}\) and \(z=\dfrac{1}{3}\).

Step 4: Decide causality.
The difference equation expresses \(y[n]\) using only \(y[n-1]\), \(y[n-2]\) and the present input \(x[n]\), with no dependence on any future sample. Solving this forward in \(n\) starting from rest gives the causal solution, whose region of convergence (ROC) is the exterior of the outermost pole: \(|z|>\dfrac{1}{2}\).

Step 5: Decide stability.
A discrete LTI system is stable exactly when its ROC includes the unit circle \(|z|=1\). Since both poles have magnitude \(1/2\) and \(1/3\), both strictly less than \(1\), the causal ROC \(|z|>1/2\) does include the unit circle.

Step 6: Combine the two conclusions.
The system is both causal (ROC is the exterior of the outermost pole, matching a difference equation that only uses past outputs and the present input) and stable (the unit circle lies inside that ROC). The given input \(x[n]=\delta[n]-\frac{1}{3}\delta[n-1]\) is itself a finite, causal signal, so it does not change this conclusion.

Step 7: Rule out the other options.
Non-causal responses (B, D) would need an anti-causal or two-sided ROC, which does not fit a difference equation using only past outputs and the present input. Unstable responses (C, D) would need the ROC to exclude the unit circle, which does not happen since both poles lie strictly inside the unit circle.

Step 8: Final conclusion.
\[ \boxed{\text{Stable and causal response}} \]