Filters and Stability

All filters we consider are linear, time invariant, and causal.

Linear, Time Invariant, Causal Filters

Every signal can be written as a linear combinations of sine functions running at increasing frequencies, as \(\sin(2\pi k t)\), \(k=1, 2, \ldots\), see Fig. 12 for the first sine function in this family. Frequencies are measured in Hertz (Hz): \(n\) Hz = \(n\) cycles per second.

Fig. 12 Sine function running at 1 Hertz, one cycle per second.

Processing a continuous signal happens by taking samples. Sampling at 8 Hertz means taking 8 equidistant samples per second, as illustrated in Fig. 13. Our signal is the represented by \(\{ u_0, u_1, u_2, u_3, u_4, u_5, u_6, u_7, u_0, \ldots \}\), where \(u_k = \sin(2\pi k/8)\), \(k=0,1,2,\ldots,7\).

Fig. 13 Sine function running at 1 Hertz, sampled at 8 Hertz.

We can view the unit delay as a filter, as introduced in Fig. 14.

Fig. 14 The unit delay as a filter.

The unit delay has as input \(u = \{ u_0, u_1, u_2, \ldots \}\) has z-transform \(\displaystyle U(z) = \sum_{k=0}^\infty u_k z^{-k}\). The output \(y = \{ 0, u_0, u_1, u_2, \ldots \}\) has z-transform \(Y(z)\), computed below.

\[\begin{split}\begin{array}{rcl} Y(z) & = & {\displaystyle \sum_{k=1}^\infty u_{k-1} z^{-k}} \\ & = & {\displaystyle z^{-1} \sum_{k=1}^\infty u_{k-1} z^{-(k-1)}} ~=~ {\displaystyle z^{-1} \sum_{k=0}^\infty u_k z^{-k}} \\ & = & z^{-1} U(z). \end{array}\end{split}\]

Therefore, the delay of a signal can be viewed as a filter, schematically represented in Fig. 15.

Fig. 15 Schematic representation of a filter.

Definition of a linear, time invariant, causal filter.

A filter \(F\) turns input \(u = \displaystyle \left\{ u_k \vphantom{u^k} \right\}_{k=0}^\infty\) into output \(y = \displaystyle \left\{ y_k \vphantom{y^k} \right\}_{k=0}^\infty\) and is

1. linear: \(F( \alpha u^{(1)} + \beta u^{(2)}) = \alpha F(u^{(1)}) + \beta F(u^{(2)})\),

2. time invariant: \(\displaystyle F\left( \left\{ u_{k+k_0} \vphantom{u^k} \right\}_{k=0}^\infty \right) = \left\{ y_{k + k_0} \vphantom{y^k} \right\}_{k=0}^\infty\), \(k > k_0\), and

3. causal: \(u_k = 0, k < k_0 \Rightarrow y_k = 0, k < k_0\).

For good understanding of this definition, as an exercise, show that the unit delay is a linear, time invariant, and causal filter.

Unit Impulse and Transfer Function

The unit impulse is \(\delta = \{ 1, 0, 0, \ldots \}\). The impulse response is \(h = F \delta = \{ h_0, h_1, h_2, \ldots \}\). Both unit impulse and impulse response are shown in Fig. 16.

Fig. 16 Schematic representation of the unit impulse and the impulse response.

Theorem charactizing a linear, time invariant, causal filter.

A linear, time invariant, causal filter is entirely determined by the impulse response.

For input \(u\), the \(k\)-th element in the output \(y\) is

\[y_k = u_k h_0 + u_{k-1} h_1 + \cdots + u_1 h_{k-1} + u_0 h_k = \sum_{j=0}^k h_j u_{k-j}.\]

The convolution is denoted by the operator \(\star\), as \(y = h \star u\).

In the frequency domain, the z-transforms of the input and output are

\[U(z) = \sum_{k=0}^\infty u_k z^{-k} \quad \mbox{and} \quad Y(z) = \sum_{k=0}^\infty y_k z^{-k}.\]

The impulse response has as z-transform \(\displaystyle H(z) = \sum_{k=0}^\infty h_k z^{-k}\). The z-transform of \(y = h \star u\) is then

\[Y(z) = H(z) U(z).\]

Definition of the transfer function.

The transfer function of a filter is the z-transform of the impulse response.

Now we can explain the \(z^{-1}\) in the schematic Fig. 14 representing the unit delay as a filter by asking What is the transfer function of the unit delay filter? The transfer function is the z-transform of the impulse response:

• The response of \(\{1, 0, 0, \ldots \}\) is \(\{0, 1, 0, 0, \ldots \}\).

• The z-transform of \(\{0, 1, 0, 0, \ldots \}\) is \(z^{-1}\).

Thus \(z^{-1}\) is the transfer function of the unit delay filter.

Consider computing the output of a filter. Denote

\[\begin{split}\begin{array}{rcl} \delta^{(0)} & = & \{ 1, 0, 0, \ldots \}, \\ \delta^{(1)} & = & \{ 0, 1, 0, \ldots \}, \\ \delta^{(2)} & = & \{ 0, 0, 1, \ldots \}, \quad \mbox{etc.} \end{array}\end{split}\]

then we can write any input \(u\) as \(\displaystyle u = \sum_{k=0}^\infty u_k \delta^{(k)}\).

The output \(y = F u\) can then be written as

\[y = F\left( \sum_{k=0}^\infty u_k \delta^{(k)} \right) = \sum_{k=0}^\infty u_k F \delta^{(k)},\]

because \(F\) is linear.

After decomposing any input \(u\) as linear combinations of properly delayed unit impulses, we construct a formula for the output:

\[\begin{split}\begin{array}{rcl} y & = & u_0 F \delta^{(0)} + u_1 F \delta^{(1)} + u_2 F \delta^{(2)} + \cdots \\ & = & u_0 \{ h_0, h_1, h_2, \ldots \} + u_1 \{ 0, h_0, h_1, h_2, \ldots \} \\ & & +~ u_2 \{ 0, 0, h_0, h_1, h_2, \ldots \} \\ & = & \{ u_0 h_0, u_0 h_1 + u_1 h_0, u_0 h_2 + u_1 h_1 + u_2 h_0, \ldots \} \\ & = & {\displaystyle \sum_{k=0}^\infty ~ \underbrace{\sum_{j=0}^k u_j h_{k-j}}_{y_k}} \end{array}\end{split}\]

The formula for \(y_k\) is a convolution of the input with the impulse response. We proved that the impulse response entirely determines a filter, as stated earlier in the theorem characterizing a linear, time invariant, and causal filter.

Let us now turn to the original function of a filter and removing a signal shown earlier in Fig. 13. Our signal is \(\{ u_0, u_1, u_2, u_3, u_4, u_5, u_6, u_7, u_0, \ldots \}\), where \(u_k = \sin(2\pi k/8)\), \(k=0,1,2,\ldots,7\). Observe, looking at Fig. 13, the symmetry of the samples:

\[u_5 = -u_1, \quad u_6 = -u_2, \quad u_7 = -u_3.\]

Adding a delay will remove the signal. Let the output be defined as follows

\[y_0 = u_0 + u_4, \quad y_1 = u_1 + u_5, \quad y_2 = u_2 + u_6, \quad y_3 = u_3 + u_7.\]

This results in the removal of the signal, as \(y_0 = 0\), \(y_1 = 0\), \(y_2 = 0\), \(y_3 = 0\). The z-transform of \(y\) is \(Y(z)\):

\[Y(z) = U(z) + z^{-4} U(z) = \left( 1 + z^{-4} \right) U(z).\]

The transfer function of the filter is \(H(z) = 1 + z^{-4}\).

Stability

In addition to working with linear, time invariant, and causal filters, we want filters that are stable.

Definition of a stable filter.

A filter is stable if bounded inputs produce bounded outputs.

What are bounded inputs and outputs? Some examples follow.

• \(x = \{ 1, 1, 1, 1, 1, 1, 1, \ldots \}\) is bounded, and so is any constant signal.

• \(x = \{ 1, 2, 4, 8, 16, 32, 64, \ldots \}\) is not bounded.

• \(\displaystyle x = \left\{ 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \frac{1}{64}, \ldots \right\}\) is bounded.

The impulse response determines a linear, time invariant, causal filter.

The transfer function of a stable filter.

Let \(\displaystyle H(z) = \sum_{k=0}^\infty h_k z^{-k}\) be the transfer function of a filter \(F\).

\[F \mbox{ is stable} \quad \Leftrightarrow \quad \sum_{k=0}^\infty |h_k| < \infty.\]

As a corollary, the transfer function \(H(z)\) of a stable filter converges absolutely and uniformly for all \(|z| \geq 1\) to a continuous function.

Amplitude Gain and Phase Shift

The theorem below provides another characterization of a filter.

Theorem on the amplitude and phase shift of a filter.

Let \(\displaystyle H(z) = \sum_{k=0}^\infty h_k z^{-k}\) be the transfer function of a filter \(F\).

For input \(\displaystyle u = \left\{ u_k = \sin(\omega k T) \vphantom{u^k} \right\}_{k=0}^\infty\), \(\displaystyle y = \left\{ y_k = r \sin(\omega k T + \phi) \vphantom{y^k} \right\}_{k=0}^\infty\) is the output, where

• \(r = |H(e^{i \omega T})|\) is the amplitude gain, and

• \(\phi = \mbox{arg} H(e^{i \omega T})\) is the phase shift.

We interpret the input \(u\) as

• \(u = \sin(\omega t)\), \(\omega = 2 \pi n\), at \(n\) Hertz,

• sampled at \(t = kT\), \(T\) is the sampling rate.

Complex numbers appear, the imaginary unit is \(i = \sqrt{-1}\).

The proof of the theorem uses complex numbers. We look at the response of the signal \(s = e^{i \omega T}\).

The input \(\displaystyle u = \left\{ s^k \vphantom{u^k} \right\}_{k=0}^\infty\). Let us compute \(y = h \star u\).

\[y_k = \sum_{j=0}^k h_j u_{k-j} = \sum_{j=0}^k h_j s^{k-j} = s^k \sum_{j=0}^k h_j s^{-j}\]

By the definition of \(H\): \(\displaystyle H(s) = \sum_{j=0}^\infty h_j s^{-j}\). So, we have:

\[y_k = s^k H(s) - s^k \sum_{j=k+1}^\infty h_j s^{-j}.\]

Consider \(k \rightarrow \infty\) and apply the complex exponentiation to the sine function. Our input signal is \(s = e^{i \omega T}\).

\[\begin{split}\begin{array}{cccccr} & s & = & e^{i \omega T} & = & \cos(\omega T) + i \sin(\omega T) \hphantom{\left. \right)} \\ - \left( \right. & s^{-1} & = & e^{-i \omega T} & = & \cos(\omega T) - i \sin(\omega T) \left. \right) \\ \hline & s - s^{-1} & & & = & 2 i \sin(\omega T) \hphantom{\left. \right)} \end{array}\end{split}\]

It then follows:

\[\sin(\omega T) = \frac{1}{2i} \left( s - s^{-1} \vphantom{\frac{1}{2}} \right)\]

and

\[\sin(k \omega T) = \frac{1}{2i} \left( s^k - s^{-k} \vphantom{\frac{1}{2}} \right).\]

We derived \(y_k = s^k H(s)\). Suppose \(H(e^{i \omega T}) = r e^{i \phi}\). What are \(r\) and \(\phi\)?

\[\begin{split}\begin{array}{rcl} { \displaystyle \frac{s^k H(s) - s^{-k} H(s^{-1})}{2 i} } & = & { \displaystyle \frac{1}{2i} \left( \vphantom{\frac{1}{2}} s^k r e^{i \phi} - s^{-k} r e^{-i \phi} \right), \quad s = e^{i \omega T} } \\ & = & { \displaystyle \frac{1}{2i} \left( \vphantom{\frac{1}{2}} e^{k i \omega T + \phi} - e^{-(k i \omega T + \phi)} \right) } \\ & = & r \sin(k \omega T + \phi) \end{array}\end{split}\]

Then indeed, \(r = |H(e^{i \omega T})|\) and \(\phi = \mbox{arg} H(e^{i \omega T})\). The theorem is proven.

Exercises

1. What is the z-transform of a one Hertz sine function sampled at eight Hertz?

2. Show that the unit delay \(F = z^{-1}\) is linear, time invariant, and causal.

3. Noise often happens at high frequency. Give the transfer function of a filter to remove a signal at 60 Hz, sampled at 720 Hz.

   1. Justify the design of your filter.

   2. Which signals are also removed by your filter?

4. Consider the following transfer function

   \[H(z) = \frac{2 z - 1}{z^2 - (5/2)z + 1}.\]

   Is the filter with this transfer function \(H\) stable? Justify your answer.

Bibliography

1. Richard C. Dorf and Robert H. Bishop: Modern Control Systems. Nineth Edition, Prentice Hall, 2001.

2. Charles R. MacCluer, middle of Chapter 3 of Industrial Mathematics. Modeling in Industry, Science, and Government, Prentice Hall, 2000.