sftrans: Transform filter band edges

Description

Transform band edges of a generic lowpass filter to a filter with different band edges and to other filter types (high pass, band pass, or band stop).

Usage

## S3 method for class 'default':
sftrans(Sz, Sp, Sg, W, stop = FALSE, \ldots)

## S3 method for class 'Arma':
sftrans(Sz, W, stop = FALSE, \ldots)

## S3 method for class 'Zpg':
sftrans(Sz, W, stop = FALSE, \ldots)

Arguments

In the generic case, a model to be transformed. In the default case, a vector containing the zeros in a pole-zero-gain model.

a vector containing the poles in a pole-zero-gain model.

a vector containing the gain in a pole-zero-gain model.

critical frequencies of the target filter specified in radians. W must be a scalar for low-pass and high-pass filters, and W must be a two-element vector c(low, high) specifying the lower and uppe

stop

FALSE for a low-pass or band-pass filter, TRUE for a high-pass or band-stop filter.

...

additional arguments (ignored).

Value

For the default case or for sftrans.Zpg, an object of class "Zpg", containing the list elements:
zerocomplex vector of the zeros of the transformed model
polecomplex vector of the poles of the transformed model
gaingain of the transformed model
For sftrans.Arma, an object of class "Arma", containing the list elements:
bmoving average (MA) polynomial coefficients
aautoregressive (AR) polynomial coefficients

Details

Given a low pass filter represented by poles and zeros in the splane, you can convert it to a low pass, high pass, band pass or band stop by transforming each of the poles and zeros individually. The following summarizes the transformations: Low-Pass Transform $$S -> C S/Fc$$ ll{ Zero at x Pole at x zero: $F_c x/C$ $F_c x/C$ gain: $C/F_c$ $F_c/C$ } High-Pass Transform $$S -> C F_c/S$$ ll{ Zero at x Pole at x zero: $F_c C/x$ $F_c C/x$ pole: $0$ $0$ gain: $-x$ $-1/x$ } Band-Pass Transform $$S -> C \frac{S^2+F_hF_l}{S(F_h-F_l)}$$ ll{ Zero at x Pole at x zero: $b � sqrt(b^2-F_hF_l)$ $b � sqrt(b^2-F_hF_l)$ pole: $0$ $0$ gain: $C/(F_h-F_l)$ $(F_h-F_l)/C$ $b = x/C (F_h-F_l)/2$ $b=x/C (F_h-F_l)/2$ } Band-Stop Transform $$S -> C \frac{S(F_h-F_l)}{S^2+F_hF_l}$$ ll{ Zero at x Pole at x zero: $b � sqrt(b^2-F_hF_l)$ $b � sqrt(b^2-F_hF_l)$ pole: $�sqrt(-F_hF_l)$ $�sqrt(-F_hF_l)$ gain: $-x$ $-1/x$ $b = C/x (F_h-F_l)/2$ $b=C/x (F_h-F_l)/2$ } Bilinear Transform $$S -> \frac{2}{T} \frac{z-1}{z+1}$$ ll{ Zero at x Pole at x zero: $(2+xT)/(2-xT)$ $(2+xT)/(2-xT)$ pole: $-1$ $-1$ gain: $(2-xT)/T$ $(2-xT)/T$ } where $C$ is the cutoff frequency of the initial lowpass filter, $F_c$ is the edge of the target low/high pass filter and $[F_l,F_h]$ are the edges of the target band pass/stop filter. With abundant tedious algebra, you can derive the above formulae yourself by substituting the transform for $S$ into $H(S)=S-x$ for a zero at $x$ or $H(S)=1/(S-x)$ for a pole at $x$, and converting the result into the form: $$H(S) = g \mbox{prod}(S-Xi) / \mbox{prod}(S-Xj)$$ Please note that a pole and a zero at the same place exactly cancel. This is significant for High Pass, Band Pass and Band Stop filters which create numerous extra poles and zeros, most of which cancel. Those which do not cancel have a `fill-in' effect, extending the shorter of the sets to have the same number of as the longer of the sets of poles and zeros (or at least split the difference in the case of the band pass filter). There may be other opportunistic cancellations, but it does not check for them. Also note that any pole on the unit circle or beyond will result in an unstable filter. Because of cancellation, this will only happen if the number of poles is smaller than the number of zeros and the filter is high pass or band pass. The analytic design methods all yield more poles than zeros, so this will not be a problem.

References

Proakis & Manolakis (1992). Digital Signal Processing. New York: Macmillan Publishing Company. Octave Forge http://octave.sf.net