A digital filter takes a series of numbers x_{0},x_{1},\ldots as input and produces the series of numbers y_{0},y_{1},\ldots as output. The type of filter we are going to talk about is called a linear time invariant filter. In its most general form the output is related to the input as follows.

\displaystyle y_{n}= \displaystyle\;c_{0}x_{n}+c_{1}x_{n-1}+c_{2}x_{n-2}+\ldots+c_{N}x_{n-N} (1)

The equation shows that the current output y_{n} is a weighted sum of the current input x_{n}, the N previous inputs and the M previous outputs. The weight of the k^{th} previous input, x_{n-k}, is the constant c_{k}, and the weight of the k^{th} previous output, y_{n-k}, is the constant d_{k}. Equation 1 can be written more succinctly as

y_{n}=\sum_{k=0}^{N}c_{k}x_{n-k}+\sum_{k=1}^{M}d_{k}y_{n-k} (2)

To find out how the filter behaves you have to take the z-transform of this equation. To take the z-transform of any sequence of numbers x_{0},x_{1},\ldots you multiply each x_{k} by z^{-k} and sum up all the products. Assume for now that z is just some complex number and let X(z) be the z-transform of the sequence, then the equation is

X(z)=\sum_{k=0}^{\infty}x_{k}z^{-k} (3)

This definition of the z-transform is usually called the one-sided z-transform since the summation goes from 0 to \infty. The full z-transform takes the summation from -\infty to \infty but we will only deal with sequences for which x_{k}=0 for k<0 so it becomes equivalent to the one sided transform in this case.

The z-transform is a way to compactly represent a possibly infinite sequence of numbers. The following are some examples of z-transforms (in all cases x_{n}=0 for n<0).

0,\;n>0\end{cases}\qquad\qquad X(z)=1 (4)
x_{n}=1,\;n\geq 0\qquad\qquad X(z)=\frac{1}{1-z^{-1}}=\frac{z}{z-1} (5)
x_{n}=n,\;n\geq 0\qquad X(z)=\frac{z^{-1}}{(1-z^{-1})^{2}}=\frac{z}{(z-1)^{2}} (6)

In general X(z) may not have a simple form as in these examples. If you are familiar with generating functions then the z-transform looks like a generating function for the x_{n} sequence in the variable z^{-1} and this is essentially what it is. If you are not familiar with generating functions, don’t worry, they won’t come up again.

The system function for a digital filter is a z-transform that can be used to analyze how the filter behaves with different inputs. The system function will always have the form of the ratio of two polynomials. To find the system function, multiply both sides of equation 2 by z^{-n} and sum over n from 0 to \infty. On the left side of the equation, you have

\sum_{n=0}^{\infty}y_{n}z^{-n}=Y(z) (7)

which is the z-transform of the output sequence. On the right side you have terms of the form


For the purpose of describing the operation of a digital filter we can assume zero initial conditions which simply means that both the x_{n} and y_{n} sequence is zero for n<0. In this case the above equations are equivalent to

z^{-k}\sum_{n=0}^{\infty}x_{n}z^{-n}\quad\text{and}\quad z^{-k}\sum_{n=0}^{%

The summations are the z-transforms of x_{n} and y_{n} so the two equations are just

z^{-k}X(z)\quad\text{and}\quad z^{-k}Y(z)

Using these results, the z-transform of equation 2 becomes

Y(z)=\sum_{k=0}^{N}c_{k}z^{-k}X(z)+\sum_{k=1}^{M}d_{k}z^{-k}Y(z) (8)

Rearranging the terms in this equation gives you the filter’s system function

^{-k}} (9)

The system function is the ratio of the z-transform of the output to the input. By definition H(z) must also be the z-transform of some sequence which we will call h_{n}. In terms of h_{n}, we can write H(z) as

H(z)=\sum_{n=0}^{\infty}h_{n}z^{-n} (10)

The sequence h_{0},h_{1},\ldots is called the impulse response of the filter. The name comes from the fact that it is the response of the filter to the input given by eq. 4 which is called an impulse. Equation 9 says that Y(z)=H(z)X(z) but for an impulse X(z)=1 so we have Y(z)=H(z) or

\sum_{n=0}^{\infty}y_{n}z^{-n}=\sum_{n=0}^{\infty}h_{n}z^{-n} (11)

Equating coefficients of z^{-n} gives y_{n}=h_{n} as the output when the input is an impulse. For a general sequence of inputs x_{0},x_{1},x_{2},\ldots the output can be found by convolving the inputs with the impulse response. To see what this means, write Y(z)=H(z)X(z) as follows

\displaystyle\sum_{n=0}^{\infty}y_{n}z^{-n}= \displaystyle\left(\sum_{n=0}^{\infty}h_{n}z^{-n}\right)\left(\sum_{n=0}^{%
\infty}x_{n}z^{-n}\right) (12)
\displaystyle= \displaystyle\left(h_{0}+h_{1}z^{-1}+h_{2}z^{-2}+\ldots\right)

When you perform the multiplication on the right and equate coefficients of z^{-n} on the two sides of the equation, you find that

y_{n}=\sum_{k=0}^{n}h_{k}x_{n-k} (13)

The summation on the right is called the convolution of the h_{n} and x_{n} sequence. This equation shows why the system function and the impulse response are so important. Suppose the input is x_{n}=z^{n} so that the n^{th} input is the n^{th} power of the complex number z. According to eq. 13, the output is then

y_{n}=\sum_{k=0}^{n}h_{k}z^{n-k}=z^{n}\sum_{k=0}^{n}h_{k}z^{-k}=z^{n}H_{n}(z) (14)

The output is the same as the input multiplied by the function H_{n}(z) which looks like the system function H(z). It is not quite the same since the summation only goes to n whereas the system function summation goes to infinity, as defined in equation 10.

But we are only interested in stable filters for which |z|\leq 1 and the terms in the impulse response, h_{n}, decrease with increasing n so that H_{n}(z) becomes closer and closer to H(z) as n increases, and in the limit H_{\infty}(z)=H(z). This means that after the filter has been running for awhile, its output for the input x_{n}=z^{n} will, to a good approximation, be y_{n}=z^{n}H(z). The filter simply multiplies the input by the factor H(z) to get the output.

For inputs of the form x_{n}=z^{n} the system function tells you all you need to know about what the output will be. One important class of inputs of this form occurs when z=e^{j\theta} and x_{n}=e^{jn\theta}=\cos n\theta+j\sin n\theta. The x_{n} are points on the unit circle in the complex plane (we are using j=\sqrt{-1} which is the more common convention in engineering work). An example of such a sequence is shown in figure 1.

Figure 1: Sequence of points on the unit circle.

As the index n increases, the points x_{n} move around the unit circle in angular steps of size \theta. The angle \theta acts as a dimensionless frequency. To see how this can be related to a real frequency, recall that a periodic function of time, u(t+T)=u(t), can be expressed as a Fourier series which is a weighted sum of the complex exponentials e^{j2\pi kt/T}. When the function is sampled at intervals \Delta t then the complex exponentials become e^{j2\pi kn\Delta t/T}=e^{jn\theta} where \theta=2\pi k\Delta t/T=2\pi f_{k}/s, f_{k}=k/T, k=\ldots-1,0,1,\ldots, and s=1/\Delta t is the sampling rate.

The value of H(e^{j\theta}) as \theta ranges from 0 to 2\pi or -\pi to \pi is the frequency response of the filter. In polar form it can be written as follows

H(e^{j\theta})=|H(e^{j\theta})|e^{j\Phi} (15)

The magnitude |H(e^{j\theta})| measures how much the filter amplifies or attenuates the input x_{n}=e^{jn\theta} and the phase \Phi measures how much the filter shifts its phase.

Since the value of H(e^{j\theta}) will generally be complex, we need to represent H(z) in the complex plane. The simplest way to do that is with a pole-zero plot. H(z) will be a rational function of two polynomials as shown in eq. 9. The poles of H(z) are those values of z where H(z) goes to infinity. These values are the roots of the denominator polynomial, and z=\infty if the numerator degree is greater than the denominator degree. The zeros of H(z) are those values of z where H(z) is zero. These values are the roots of the numerator polynomial, and z=\infty if the denominator degree is greater than the numerator degree. A pole-zero plot of H(z) simply shows the location of the poles and zeros in the complex plane along with the unit circle. Poles are represented by a filled circle “\bullet”, and zeros by an unfilled circle “\circ”.

Let the the zeros of H(z) be a_{i}e^{j\theta_{i}}, i=1,2,\ldots,M, and the poles be b_{i}e^{j\phi_{i}}, i=1,2,\ldots,N, then H(z) can be written as

{j\phi_{i}})} (16)

In eq. 9 the coefficients of the numerator and denominator polynomial are real. For the filters we will consider, this will always be true. This means that complex poles or zeros must come in conjugate pairs. If ae^{j\theta} is a complex zero then there must be another zero equal to ae^{-j\theta} and likewise for poles.

Substituting z=e^{j\theta} into equation 16 and taking the magnitude and phase gives the following equations

{2})}{\prod_{i=1}^{N}(1-2b_{i}\cos(\theta-\phi_{i})+b_{i}^{2})} (17)
\cos\theta-b_{i}\cos\phi_{i}} (18)

The following sections will show how these equations are used.

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