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I need to find the transfer function of the second order FIR filter. I searched a lot in the standard textbooks, but I could not get the mathematical expressions and derivations.

I want that the zeros positioned at $z_{1,2} = r(w_{0})e^{jw_0} $ where $ r(w_{0}) = (1-b + b\cos(w_{0})) $ where, $ b \in [0.5,1] $, $ w_{0} \in [0,\pi] $

  • How to find the transfer function if it is this kind of form? This kind of representation confuses me lot.
  • How to determine the direct form and lattice form and their coefficients for such FIR filter?
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1 Answer 1

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Q: How to find the transfer function if it is this kind of form?

Assumption: the zeros are located at $z_{1,2} = r(\omega_0)e^{j\omega_0}$ where $r(\omega_0) = 1-b+b\cos\omega_0$.

We can generically write the z-transform domain representation of a 2nd order FIR filter as: $$h(z) = k(z-z_1)(z-z_2) = k(z^2-(z_2+z_1)z+z_1z_2)$$ where $k$ is an arbitrary scale factor and $z_{1,2}$ are the locations of the zeros.

Converting to the frequency domain by letting $z \to e^{j\omega}$: $$H(\omega) = k(e^{j2\omega} - (z_2+z_1)e^{j\omega} +z_1z_2)$$

Now your choice of $z_{1,2}$ can be substituted in.

How to determine the direct form and lattice form and their coefficients for such FIR filter?

The direct forms pretty much given by the expression for $h(z)$ above.

To convert to a lattice representation, follow the steps below (based on Lattice-Structure for FIR filters). This notation is somewhat different than above, but matches that in the reference more closely.

Let $y(n) = x(n) + \alpha_2(1)x(n-1) + \alpha_2(2)x(n-1)$

2nd order lattice filter

The output of the lattice implementation is: $$f_2(n) = f_1(n) + k_2g_1(n-1)$$

Substituting for $g_1(n-1)$ and $f_1(n)$ $$f_2(n) = x(n) + K_1x(n-1) + K_2K_1x(n-1) + K_2x(n-2)$$ $$f_2(n) = x(n) + (K_1 + K_2K_1)x(n-1) + K_2x(n-2)$$

Comparing that with $$y(n) = x(n) + \alpha_2(1)x(n-1) + \alpha_2(2)x(n-2)$$ gives: $$\alpha_2(1) = K_1(1+K_2)$$ $$K_1 = \frac{\alpha_2(1)}{1+\alpha_2(2)}$$ and: $$\alpha_2(2) = K_2$$ $$K_2 = \alpha_2(2)$$

Mapping this back to the original direct form expression we can make the following relations: $$\alpha_2(1) = z_2 + z_1$$ and $$\alpha_2(2) = z_1z_2$$

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