A friend of mine asked me an interesting question: Why do we learn Z-transform and then study designing digital filter? Well, this is interesting. But the clue is simple.

By default, lowercase function name are in time domain and uppercase function names are in Z-domain, a.k.a discrete complex domain. Y or y means the output where as X or x means the input. H or h means the filter.

Abbreviations and notations:

ZT: z-Transform

FIR: Finite Impulse Response

IIR: Infinite Impulse Response

Z(x): performing z-transform on signal x

### The Digital Filter and Transfer Function

A digital filter works on sampled, discrete time signal, rather than continuous signal as analog filters. There are only few operations we can play on time series: summation, scalar multiplication and delay. The flow of data can be backward, the feedback in IIR filter. For example, a smooth filter that averages 3 consecutive points can be built by summarize current input, previous input and the second previous input.

A digital filter has an input and an output, which are the original series and the filtered series respectively. If you consider a digital filter as a blackbox, the relationship between its input and output can be represented by a transfer function H(z) in Z domain:

H(z)= \frac{N(z)}{D(z)} = \frac{ b_0 + b_1 z^{-1} + b_2 z^{-2} + \cdots + b_L z^{-L} }{ 1+ a_1 z^{-1} + a_2 z^{-2} + \cdots + a_M z^{-M} } (1)

such that

Y(z) = X(z) H(z).

The inverse z-transform (iZT) of the transform function is the impulse response, which characterize the system behavior in time domain. Depending on whether the transfer function H(z) has denominators, the impulse response of the system can be finite or infinite and thus lead to two types of filters, FIR filter and IIR filter. Before explaining those two types, I need to introduce a very important property of ZT.

### The Time-shifting Property of z-Transform and the building of Digital Filters

ZT can be considered as a discrete-time version of Laplace transform by discretized the time. ZT has the property known as time-shifting property that

Z[x(n-k)]=z^{-k}Z[x(n)].

The power over z at Z-domain means a delay in time domain. You cannot have positive power over z since in reality the system is causal. You cannot have a signal before it happens. Time domain summation, scalar multiplication and delay can be represented by the summation, coefficients and powers of z on the transfer function. That is why people always relate digital filter design with ZT. For example, in time domain, the I/O equation of a smooth filter is

y(n)=\frac{x(n)+x(n-1)+x(n-2)}{3} (2)

Apply iZT on both side of the equation and the time-shifting property of ZT, we have

Y(z) = X(z) H(z) = X(z) \frac{1+ z^{-1} + z^{-2}}{3}

So, once you see the transfer function, you know how to build such a filter in time domain.

Did we miss something? Oh, the feedback. As you can see, Eq. 2 has no feedback. Why? And what if I want a filter that has back-propagation in time domain?

### FIR and IIR: The Difference on the Denominator

Take a look at Eq. 1, if you multiply the denominator on both sides of the equation, you will notice that the left hand side, the output side, contains the negative power of z, thus, the delay of output. Hence, if one coefficient of the denominators (a_i's) is not zero, the output is not only contributed by the input or its delays but also previous output, the feedback.

A clearer picture can be seen by transforming Eq .1 back to time domain:

y(n) = -a_1 y(n-1) - \cdots - a_M y(n-M) + b_0 x(n) + \cdots + b_L x(n-L)

If one a_i is not zero, the response to any input signal can be infinite since previous output will always contribute current output. For example,

y(n) = 2 y(n-1) + x(n) = 2 (y(n-2)+x(n-1)) + x(n) = 4(y(n-3)+x(n-2)) + 2 x(n-1) + x(n) = \cdots

To any finite input, the output of such system is infinite. The impulse response is also infinite. So, a digital filter that has a transfer function with at least a_i \noteq 0 is called an IIR filter.

The output will continue increase its value. Hence, the system is divergent. If you study the relationship between the transfer function and system stability, the system is stable and causal if and only if all poles lie inside the unit circle of the z-plane. An IIR filter may not be stable if you choose the bad denominator coefficients for its transfer function.

So, let's talk about FIR filter. If all denominator coefficients of the transfer function are 0, the output will purely depend on the input without any feedback from previous output. The response of an FIR filter to a finite input is finite, as in the example in Eq. 2 and 3. You can consider FIR filter as a special case (denominator=1) of IIR filter.

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