$$y[n] = \frac{1}{4}(x[n] + 2x[n - 1] + x[n - 2])$$
Which one of the following options is a correct statement about the transfer function of this filter?
Digital Filter Transfer Function Analysis
The given digital moving average filter has the input-output relationship:
$y[n] = \frac{1}{4}(x[n] + 2x[n - 1] + x[n - 2])$To determine the properties of the transfer function $H(z)$, we first find $H(z)$ by taking the Z-transform of the difference equation.
Transfer Function Derivation
Using the time-shifting property of the Z-transform ($Z\{x[n-k]\} = z^{-k}X(z)$):
$Y(z) = \frac{1}{4}(X(z) + 2z^{-1}X(z) + z^{-2}X(z))$Factor out $X(z)$:
$Y(z) = X(z) \frac{1}{4}(1 + 2z^{-1} + z^{-2})$The transfer function is $H(z) = \frac{Y(z)}{X(z)}$:
$H(z) = \frac{1}{4}(1 + 2z^{-1} + z^{-2})$To find the poles and zeros, it is helpful to express $H(z)$ as a ratio of polynomials in $z$. Multiply the numerator and denominator by $z^2$:
$H(z) = \frac{1}{4} \frac{z^2(1 + 2z^{-1} + z^{-2})}{z^2} = \frac{1}{4} \frac{z^2 + 2z + 1}{z^2}$Factor the numerator polynomial:
$H(z) = \frac{1}{4} \frac{(z + 1)^2}{z^2}$Poles Identification
The poles are the roots of the denominator polynomial, $z^2$.
Setting $z^2 = 0$ gives $z=0$ as a root with multiplicity 2.
Thus, the filter has two poles at $z=0$.
Zeros Identification
The zeros are the roots of the numerator polynomial, $\frac{1}{4}(z+1)^2$.
Setting $(z + 1)^2 = 0$ gives $z=-1$ as a root with multiplicity 2.
Thus, the filter has two zeros at $z=-1$.
Conclusion on Transfer Function Properties
The transfer function $H(z) = \frac{1}{4} \frac{(z + 1)^2}{z^2}$ has two poles (at $z=0$) and two zeros (at $z=-1$).
This matches the description: "It has two poles and two zeros".
Important Questions from Z Transform
The z transform of e −t sampled at 10 Hz will be:
What is the set of all values of z for which X(z) attains a finite value?
The z transform of the following real exponential sequence
x(n) = {a n ;n >= 0} , {= 0 ; n < 0} and a > 0 is given by
What will be the z-transform of a Unit step function ?
The z-transform of a causal periodic signal can be determined from the knowledge of the z-transform of its:
