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# 电子代写|数字信号处理代写Digital Signal Processing代考|Recursive Structures for Causal IIR Filters

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## 电子代写|数字信号处理代写Digital Signal Processing代考|Recursive Structures for Causal IIR Filters

Having designed a DT filter $H(z)$ that we wish to implement on a computer, we need to decide between several methods for doing so. If $H(z)$ has at least one nonzero pole, then the impulse response $h[n]$ does not have finite-duration. Such systems are called infinite impulse response (IIR) filters. In order to implement an IIR filter, we must encode the difference equations corresponding to
$$H(z)=H_i(z)+H_o(z)$$
with the appropriate initial conditions. This can be done using a variety of computational schemes.
Several issues must be considered when deciding which scheme to use. For example, the method chosen can affect computational speed, memory requirements, and numerical stability. A detailed study of these issues is beyond the scope of this course. A “quick and dirty” approach is to design several different computational structures and decide through simulation which works best. The best choice turns out to be quite application-dependent.
Suppose we are given a difference equation
$$y[n+N]+a_{N-1} y[n+N-1]+\ldots+a_K y[n+K]=b_M x[n+M]+\ldots+b_0 x[n]$$
or rational function
$$H(z)=\frac{b_M z^{M-N}+\ldots+b_0 z^{-N}}{1+a_{N-1} z^{-1}+\ldots+a_K z^{K-N}}$$
corresponding to a BIBO stable system. Recall that, if the system is neither causal nor anti-causal, then we may decompose $H(z)$ into inner and outer parts, resulting in the sum of a causal system and an anti-causal system. Hence, it suffices to consider the implementation of systems that are either causal or anti-causal. We begin with causal systems.

## 电子代写|数字信号处理代写Digital Signal Processing代考|The Anti-Causal Case

The same computational structures may be used for anti-causal systems. The only change is that the recursion is backward, rather than forward. Suppose (10.9) corresponds to an anti-causal system. Backward recursion is based on the form
\begin{aligned} y[n] & =-\frac{1}{a_K} y[n+N-K]-\frac{a_{N-1}}{a_K} y[n+N-K-1]-\ldots-\frac{a_{K+1}}{a_K} y[n+1] \ & +\frac{b_M}{a_K} x[n+M-K]+\ldots+\frac{b_0}{a_K} x[n-K] \end{aligned}
and
$$H(z)=\frac{\frac{b_M}{a_K} z^{M-K}+\ldots+\frac{b_0}{a_K} z^{-K}}{\frac{1}{a_K} z^{N-K}+\frac{a_{N-1}}{a_K} z^{N-K-1}+\ldots+\frac{a_{K-1}}{a_K} z+1}$$
For Direct Form I, we factor
$$H(z)=\left(\frac{1}{\frac{1}{a_K} z^{N-K}+\frac{a_{N-1}}{a_K} z^{N-K-1}+\ldots+\frac{a_{K-1}}{a_K} z+1}\right)\left(\frac{b_M}{a_K} z^{M-K}+\ldots+\frac{b_0}{a_K} z^{-K}\right),$$
which yields the equations
$$\begin{gathered} v[n]=\frac{b_M}{a_K} x[n+M-K]+\ldots+\frac{b_0}{a_K} x[n-K] \ y[n]=-\frac{1}{a_K} y[n+N-K]-\frac{a_{N-1}}{a_K} y[n+N-K-1]-\ldots-\frac{a_{K+1}}{a_K} y[n+1]+v[n] \end{gathered}$$
For Direct Form II,
$$\begin{gathered} H(z)=\left(\frac{b_M}{a_K} z^{M-K}+\ldots+\frac{b_0}{a_K} z^{-K}\right)\left(\frac{1}{\frac{1}{a_K} z^{N-K}+\frac{a_{N-1}}{a_K} z^{N-K-1}+\ldots+\frac{a_{K-1}}{a_K} z+1}\right) \ v[n]=-\frac{1}{a_K} v[n+N-K]-\frac{a_{N-1}}{a_K} v[n+N-K-1]-\ldots-\frac{a_{K+1}}{a_K} v[n+1]+x[n], \ y[n]=\frac{b_M}{a_K} v[n+M-K]+\ldots+\frac{b_0}{a_K} v[n-K] . \end{gathered}$$

## 电子代写|数字信号处理代写Digital Signal Processing代考|Recursive Structures for Causal IIR Filters


H (z) = H_i (z) + H_o (z)



y (n + n) +现代{n} y (n + n – 1) + \ ldots + a_K y (n + K) = b_M x (n + M) + \ ldots + b_0 x [n]



H (z) = \压裂{b_M z ^ {m n} + \ ldots + b_0 z ^ {n}}{1 +现代z ^ {n} {1} + \ ldots + a_K z ^ {k – n}}


## 电子代写|数字信号处理代写Digital Signal Processing代考|The Anti-Causal Case

$$y[n]=-\frac{1}{a_K} y[n+N-K]-\frac{a_{N-1}}{a_K} y[n+N-K-1]-\ldots-\frac{a_{K+1}}{a_K} y[n+1] \quad+\frac{b_M}{a_K} x[n+M-K]+\ldots+\frac{b_0}{a_K} x[n-K]$$

$$H(z)=\frac{\frac{b_M}{a_K} z^{M-K}+\ldots+\frac{b_0}{a_K} z^{-K}}{\frac{1}{a_K} z^{N-K}+\frac{a_{N-1}}{a_K} z^{N-K-1}+\ldots+\frac{a_{K-1}}{a_K} z+1}$$

$$H(z)=\left(\frac{1}{\frac{1}{a K} z^{N-K}+\frac{a_{N-1}}{a_K} z^{N-K-1}+\ldots+\frac{a_{K-1}}{a_K} z+1}\right)\left(\frac{b_M}{a_K} z^{M-K}+\ldots+\frac{b_0}{a_K} z^{-K}\right),$$

$$v[n]=\frac{b_M}{a_K} x[n+M-K]+\ldots+\frac{b_0}{a_K} x[n-K] y[n]=-\frac{1}{a_K} y[n+N-K]-\frac{a_{N-1}}{a_K} y[n+N-K-1]-\ldots-\frac{a_{K+1}}{a_K} y[n+1]+v[n]$$

$$\begin{gathered} H(z)=\left(\frac{b_M}{a_K} z^{M-K}+\ldots+\frac{b_0}{a_K} z^{-K}\right)\left(\frac{1}{\frac{1}{a_K} z^{N-K}+\frac{a_{N-1}}{a_K} z^{N-K-1}+\ldots+\frac{a_{K-1}}{a_K} z+1}\right) \ v[n]=-\frac{1}{a_K} v[n+N-K]-\frac{a_{N-1}}{a_K} v[n+N-K-1]-\ldots-\frac{a_{K+1}}{a_K} v[n+1]+x[n], \ y[n]=\frac{b_M}{a_K} v[n+M-K]+\ldots+\frac{b_0}{a_K} v[n-K] . \end{gathered}$$

## MATLAB代写

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