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Choice of Output Signal and Initial Conditions

Recalling that $ {\tilde x}= E\underline{{\tilde x}}$, the output signal from any diagonal state-space model is a linear combination of the modal signals. The two immediate outputs $ x_1(n)$ and $ x_2(n)$ in Fig.G.3 are given in terms of the modal signals $ {\tilde x}_1(n) = \lambda_1^n{\tilde x}_1(0)$ and $ {\tilde x}_2(n)=
\lambda_2^n{\tilde x}_2(0)$ as

y_1(n) &=& [1, 0] {\underline{x}}(n) = [1, 0] \left[\begin{arr...
...\lambda_1^n {\tilde x}_1(0) - \eta \lambda_2^n\,{\tilde x}_2(0).

The output signal from the first state variable $ x_1(n)$ is

y_1(n) &=& \lambda_1^n\,{\tilde x}_1(0) + \lambda_2^n\,{\tilde...
...{j\omega n T} {\tilde x}_1(0) + e^{-j\omega n T}{\tilde x}_2(0).

The initial condition $ {\underline{x}}(0) = [1, 0]^T$ corresponds to modal initial state

$\displaystyle \underline{{\tilde x}}(0) = E^{-1}\left[\begin{array}{c} 1 \\ [2p...
...nd{array}\right] = \left[\begin{array}{c} 1/2 \\ [2pt] 1/2 \end{array}\right].

For this initialization, the output $ y_1$ from the first state variable $ x_1$ is simply

$\displaystyle y_1(n) = \frac{e^{j\omega n T} + e^{-j\omega n T}}{2} = \cos(\omega n T).

A similar derivation can be carried out to show that the output $ y_2(n) = x_2(n)$ is proportional to $ \sin(\omega nT)$, i.e., it is in phase quadrature with respect to $ y_1(n)=x_1(n)$). Phase-quadrature outputs are often useful in practice, e.g., for generating complex sinusoids.

Butterworth Lowpass Poles and Zeros

When the maximally flat optimality criterion is applied to the general (analog) squared amplitude response $ G_a^2(\omega_a)\isdef \left\vert H_a(j\omega_a)\right\vert^2$, a surprisingly simple result is obtained [64]:

$\displaystyle G_a^2(\omega_a) = \frac{1}{1+\omega_a^{2N}} \protect$ (I.1)

where $ N$ is the desired order (number of poles). This simple result is obtained when the response is taken to be maximally flat at $ \omega_a=\infty$ as well as dc (i.e., when both $ G_a^2(\omega_a)$ and $ G_a^2(1/\omega_a)$ are maximally flat at dc).I.1Also, an arbitrary scale factor for $ \omega_a$ has been set such that the cut-off frequency (-3dB frequency) is $ \omega_c = 1$ rad/sec.

The analytic continuationD.2) of $ G_a^2(\omega_a)$ to the whole $ s$-plane may be obtained by substituting $ \omega_a = s/j$ to obtain

$\displaystyle H_a(s)H_a(-s) = \frac{1}{1+\left(\frac{s}{j}\right)^{2N}} =

The $ 2N$ poles of this expression are simply the roots of unity when $ N$ is odd, and the roots of $ -1$ when $ N$ is even. Half of these poles $ s_k$ are in the left-half $ s$-plane ( re$ \left\{s_k\right\}<0$) and thus belong to $ H_a(s)$ (which must be stable). The other half belong to $ H_a(-s)$. In summary, the poles of an $ N$th-order Butterworth lowpass prototype are located in the $ s$-plane at $ s_k = \sigma_k +
j\omega_k = e^{-j\theta_k}$, where [64, p. 168]

\begin{displaymath}\begin{array}{rcrl} \sigma_k &=&-\!&\sin(\theta_k)\\ \omega_k &=&&\cos(\theta_k) \end{array} \protect\end{displaymath} (I.2)


$\displaystyle \theta_k \isdef \frac{(2k+1)\pi}{2N}

for $ k=0,1,2,\dots,N-1$. These poles may be quickly found graphically by placing $ 2N$ poles uniformly distributed around the unit circle (in the $ s$ plane, not the $ z$ plane--this is not a frequency axis) in such a way that each complex pole has a complex-conjugate counterpart.

A Butterworth lowpass filter additionally has $ N$ zeros at $ s=\infty$. Under the bilinear transform $ s = c(z-1)/(z+1)$, these all map to the point $ z = -1$, which determines the numerator of the digital filter as $ (1+z^{-1})^N$.

Given the poles and zeros of the analog prototype, it is straightforward to convert to digital form by means of the bilinear transformation.

Example: Second-Order Butterworth Lowpass

In the second-order case, we have, for the analog prototype,

$\displaystyle H_a(s) = \frac{1}{(s + a)(s + \overline{a})}

where, from Eq.$ \,$(I.2), $ a = e^{j\pi/4}$, so that

$\displaystyle H_a(s) = \frac{1}{(s + e^{j\pi/4})(s + e^{-j\pi/4})} = \frac{1}{s^2 + \sqrt{2}s + 1} \protect$ (I.3)

To convert this to digital form, we apply the bilinear transform

$\displaystyle s = c\frac{1-z^{-1}}{1+z^{-1}}

(from Eq.$ \,$(I.9)), where, as discussed in §I.3, we set

$\displaystyle c = \cot(\omega_cT/2) \isdef \frac{\cos(\omega_cT/2)}{\sin(\omega_cT/2)}

to obtain a digital cut-off frequency at $ \omega_c$ radians per second. For example, choosing $ \omega_c T = \pi/2$ (a cut off at one-fourth the sampling rate), we get

$\displaystyle c = \frac{\cos(\pi/4)}{\sin(\pi/4)} = 1

and the digital filter transfer function is
$\displaystyle H_d(z)$ $\displaystyle =$ $\displaystyle H_a\left(\frac{1-z^{-1}}{1+z^{-1}}\right) =
\frac{1}{\left(\frac{1-z^{-1}}{1+z^{-1}}\right)^2 + \sqrt{2}\left(\frac{1-z^{-1}}{1+z^{-1}}\right) + 1}$ (I.4)
  $\displaystyle =$ $\displaystyle \frac{(1+z^{-1})^2}{(1-2z^{-1}+z^{-2}) + (\sqrt{2} - \sqrt{2}z^{-2}) + (1+2z^{-1}+z^{-2})}$ (I.5)
  $\displaystyle =$ $\displaystyle \frac{(1+z^{-1})^2}{(2+\sqrt{2}) + (2-\sqrt{2})z^{-2}}$ (I.6)
  $\displaystyle =$ $\displaystyle \frac{1}{2+\sqrt{2}}\frac{(1+z^{-1})^2}{1 + \frac{2-\sqrt{2}}{2+\sqrt{2}}z^{-2}}$ (I.7)

Note that the numerator is $ (1+z^{-1})^2$, as predicted earlier. As a check, we can verify that the dc gain is 1:

$\displaystyle H_d(1) = \frac{2^2}{2+\sqrt{2} + 2-\sqrt{2}} = 1

It is also immediately verified that $ H_d(-1) = 0$, i.e., that there is a (double) notch at half the sampling rate.

In the analog prototype, the cut-off frequency is $ \omega_a=1$ rad/sec, where, from Eq.$ \,$(I.1), the amplitude response is $ G_a(j)=1/\sqrt{2}$. Since we mapped the cut-off frequency precisely under the bilinear transform, we expect the digital filter to have precisely this gain. The digital frequency response at one-fourth the sampling rate is

$\displaystyle H_d(j) = \frac{(1-j)^2}{2+\sqrt{2} - (2-\sqrt{2})} = -\frac{j}{\sqrt{2}}, \protect$ (I.8)

and $ 20\log_{10}(\left\vert H_d(j)\right\vert)=-3$ dB as expected.

Note from Eq.$ \,$(I.8) that the phase at cut-off is exactly -90 degrees in the digital filter. This can be verified against the pole-zero diagram in the $ z$ plane, which has two zeros at $ z = -1$, each contributing +45 degrees, and two poles at $ z=\pm
j\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}$, each contributing -90 degrees. Thus, the calculated phase-response at the cut-off frequency agrees with what we expect from the digital pole-zero diagram.

In the $ s$ plane, it is not as easy to use the pole-zero diagram to calculate the phase at $ \omega_a=1$, but using Eq.$ \,$(I.3), we quickly obtain

$\displaystyle H_a(j\cdot 1) = \frac{1}{j^2 + \sqrt{2}j + 1} = -\frac{j}{\sqrt{2}},

and exact agreement with $ H_d(e^{j\pi/2})$ [Eq.$ \,$(I.8)] is verified.

A related example appears in §9.2.4.

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