Question

Consider an LTI system with a system function \( H(z) = \frac{1}{1 - \frac{1}{4}z^{-1}} \). Its difference equation will be

• A. \( y[n] - \frac{1}{2}y[n-1] = x[n] \)
• B. \( y[n] - \frac{1}{4}y[n+1] = x[n] \)
• C. \( y[n] + \frac{1}{2}y[n-1] = x[n] \)
• D. \( y[n] - \frac{1}{4}y[n-1] = x[n] \)

Hint:

To convert a rational transfer function \(H(z) = Y(z)/X(z)\) to a difference equation, cross-multiply and then take the inverse Z-transform term by term, remembering that multiplication by \(z^{-k}\) in the Z-domain corresponds to a time delay of \(k\) samples in the time domain.

Answer 1

The Correct Option is D

Step 1: Start with the definition of the transfer function. The transfer function \(H(z)\) relates the Z-transform of the output \(Y(z)\) to the Z-transform of the input \(X(z)\): \[ H(z) = \frac{Y(z)}{X(z)} \] Given \( H(z) = \frac{1}{1 - \frac{1}{4}z^{-1}} \), we can write: \[ \frac{Y(z)}{X(z)} = \frac{1}{1 - \frac{1}{4}z^{-1}} \]
Step 2: Rearrange the equation to isolate input and output terms. Cross-multiply to get rid of the fraction: \[ Y(z) \left( 1 - \frac{1}{4}z^{-1} \right) = X(z) \] Distribute \(Y(z)\) on the left side: \[ Y(z) - \frac{1}{4}z^{-1}Y(z) = X(z) \]
Step 3: Apply the inverse Z-transform. We use the time-shifting property of the Z-transform, which states that \( \mathcal{Z}^{-1}\{z^{-k}F(z)\} = f[n-k] \). Taking the inverse Z-transform of each term in the equation: \[ \mathcal{Z}^{-1}\{Y(z)\} \rightarrow y[n] \] \[ \mathcal{Z}^{-1}\{\frac{1}{4}z^{-1}Y(z)\} \rightarrow \frac{1}{4}y[n-1] \] \[ \mathcal{Z}^{-1}\{X(z)\} \rightarrow x[n] \] Combining these gives the difference equation: \[ y[n] - \frac{1}{4}y[n-1] = x[n] \]