Filters are often used to modify the spectrum of given data.
With input *X*(*Z*), filters *B*(*Z*), and output *Y*(*Z*) we have
*Y*(*Z*) = *B*(*Z*)*X*(*Z*) and the Fourier conjugate
.
Multiplying these two relations together we get

(25) |

Figure 13

We will have frequent occasion to deal with sinusoidal time
functions. A simple way to represent a sinusoid by *Z*
transforms is

(26) |

The time function associated with this *Z* transform is
, but it is ``turned on'' at *t* = 0. Actually,
the left-hand side of (26) contains a pole exactly
on the unit circle, so that the series sits on the borderline
between convergence and divergence. This can cause paradoxical
situations
[you could expand (26) so that the sinusoid turns off at *t* = 0]
which we will avoid by pushing the pole
from the unit circle to a small distance outside
the unit circle. Let Then define

(27) |

The time function corresponding to *B*(*Z*) is zero before
*t* = 0 and is after *t* = 0.
It is a sinusoidal function which decreases gradually with time
according to .The coefficients are shown in Figure 14.

2-14
The time function associated with
a simple pole just outside the unit circle at .Figure 14 |

It is intuitively obvious, although we will prove it later,
that convolution with the coefficients of (27),
which are sketched in Figure 14, is a narrow-banded
filtering operation. If the pole is chosen very close
to the unit circle, the filter bandpass becomes narrower
and the coefficients of *B*(*Z*) drop off more and more slowly.
To actually perform the convolution it is necessary to
truncate, that is, to drop powers of *Z* beyond a certain
practical limit. It turns out that there is a very much
cheaper method of narrow-band filtering than convolution
with the coefficients of *B*(*Z*). This method is polynomial
division by *A*(*Z*). We have for the output *Y*(*Z*)

(28) | ||

(29) |

(30) |

For definiteness, let us suppose the *x*_{t} and *y*_{t}
vanish before *t* = 0. Now identify coefficients of successive
powers of *Z*. We get

(31) | ||

(32) |

(33) |

so

(34) |

To a good approximation this function may be thought of as A plot of (34) is shown in Figure 15.

2-15
Spectrum associated with
a single pole at .Figure 15 |

Now it should be apparant why this is called
a narrowband filter. It amplifies a vary narrow
band of frequencies and
attenuates all others. The frequency window
of this filter is said to be
in width.
The time window is
,the damping time constant of the dampend sinusoid *b*_{t}.

One practical disadvantage of the filter under discussion is that although its input may be a real time series its output will be a complex time series. For many applications a filter with real coefficients may be preferred.

One approach is to follow the filter
by the time-domain, complex conjugate filter
.The composit time-domain operator is now
which is real.
[Note that the complex conjugate
in the frequency domain is
but in the time domain it is
.The composite filter may be denoted by
.The spectrum of this filter is
.One may quickly verify that the spectrum of
is like that of *B*(*Z*),
but the peak is at instead of .Thus, the composite spectrum is the product of
Figure 15 with itself reversed along the
frequency axis. This is shown in Figure 16.

2-16
Spectrum of a two-pole filter
where one pole is like Figure 15 and
the other is at the conjugate position.
Figure 16 |

- A simple feedback operation is
.This operation is called leaky integration.
Give a closed form expression for the output
*y*_{t}if*x*_{t}is an impulse. What is the decay time of your solution (the time it takes for*y*_{t}to drop to*e*)? For small ,say = 0.1, .001, or 0.0001, what is ?^{-1}y_{0} - How far from the unit circle are the poles of
1/(1 - .1
*Z*+ .9*Z*)? What is the decay time of the filter and its resonant frequency?^{2} - Find a three-term real feedback filter to pass 59-61 Hz on data which are sampled at 500 points/sec. Where are the poles? What is the decay time of the filter?

10/30/1997