Question

The output \(y[n]\) of a digital moving average filter for an input \(x[n]\) is given by
\[ y[n] = \frac{1}{4}\big(x[n] + 2x[n-1] + x[n-2]\big) \]
Which one of the following options is a correct statement about the transfer function of this filter?

• A. It has two poles and two zeros
• B. It has two poles and no zeros
• C. It has no poles and two zeros
• D. It has one pole and one zero

Hint:

Write \(H(z)\) over positive powers of \(z\): the largest delay sets the pole count at the origin, and the numerator \((z+1)^2\) gives the zeros.

Answer 1

The Correct Option is A

Step 1: Take the Z-transform of the difference equation.
Using the time-shift property of the Z-transform, \(x[n-1]\) transforms to \(z^{-1}X(z)\) and \(x[n-2]\) transforms to \(z^{-2}X(z)\). Applying this to \(y[n] = \frac{1}{4}(x[n]+2x[n-1]+x[n-2])\):
\[ Y(z) = \frac{1}{4}\big(X(z) + 2z^{-1}X(z) + z^{-2}X(z)\big) \]

Step 2: Write the transfer function.
The transfer function is \(H(z) = Y(z)/X(z)\), so
\[ H(z) = \frac{1}{4}\big(1 + 2z^{-1} + z^{-2}\big) \]

Step 3: Convert to positive powers of \(z\) to see the poles and zeros clearly.
Multiply numerator and denominator by \(z^2\):
\[ H(z) = \frac{1}{4}\cdot\frac{z^2 + 2z + 1}{z^2} \]
The numerator \(z^2+2z+1\) is a perfect square, \((z+1)^2\), so
\[ H(z) = \frac{(z+1)^2}{4z^2} \]

Step 4: Read off the poles and zeros.
The zeros are the roots of the numerator: \((z+1)^2=0\) gives a double zero at \(z=-1\), which counts as two zeros.
The poles are the roots of the denominator: \(z^2=0\) gives a double pole at \(z=0\) (the origin), which counts as two poles.
So this filter, even though it is a simple FIR (finite impulse response) filter, has two poles, both located at the origin from the \(z^{-1}\) and \(z^{-2}\) delay terms, and two zeros at \(z=-1\).

Final Answer:
The filter has two poles (at the origin) and two zeros (at \(z=-1\)), matching option (A) and ruling out (B), (C) and (D).
\[ \boxed{\text{Two poles and two zeros}} \]