Julien OSTER, [email protected]/teaching/cdt/A3/4_A3...for all values of the complex...

DEPARTMENT OF ENGINEERING SCIENCE

CDT IN HELTHCARE INNOVATION

Julien OSTER,

[email protected]

Dr. Julien Oster, Postdoctoral Researcher Institute of Biomedical Engineering, University of Oxford

Signals can be characterized by their frequency content:

Music

Light

…

Let’s define tools for characterizing the frequency response of the filter

03 November 2013 More on Digital Filtering, Julien Oster

Let us consider x[n] = eiωn

eiωn are eigenfunctions of LTI systems and is the eigenvalue If x[n]=Asin(ωn+ω0) then y[n]=A sin(ωn+ω0+ϕ)

is the frequency response of the filter


y[n] = h[m]x[n-m]m=-¥

¥

å

y[n] = h[m]e-iwmeiwn

m=-¥

¥

å

y[n] = x[n] h[m]e-iwm = x[n]H (e-iw )m=-¥

¥

å

y[n] = x[n] H (e-iw ) eif

H(e-iw )H(e-iw )

H(e-iw )

Generalization of Frequency response (because Fourier transform doesn’t exist for all time series)

Is equivalent to the Laplace transform in Analogue filtering If x[n]=zn

The output of the system is thus equal to the input, multiplied by a

complex constant H(z). The complex exponential signal zn is said to be an eigenfunction of

the linear, time invariant system h[n]. H(z) is the z-transform of h[n]. z defines a plane, with the real and the imaginary parts (as

orthogonal coordinates).


y[n]=x[n]H[z]

For LTI systems described by an impulse response h[n], H(z)

is referred to as the system function or transfer function.

Consider the class of digital filters described by linear, constant coefficient,

difference equations of the form

Substituting y[n]=H[z]x[n] produces:

Dividing both sides by zn and rearranging terms gives which is the transfer function


The K complex roots of the denominator of H(z) are

called the poles of the filter,

while the M roots of the numerator are called zeros.


1. The Z-transform of simple digital filters can be computed by:

1. Direct application of if the impulse response h[n] is known

2. , if the difference equation is known

2. We will use the notation to denote the relation

between the signal h[n] and its z-transform H(z).

1. Gain: i.e., the z-transform of the unit sample δ[n] is 1. 2. Delay by n0 samples: , so z-1 denotes unit delay.


3. Rectangular filter of length N:

The filter has N-1 zeroes equally spaced on the unit circle (Except for z=1)

4. First-order recursive low-pass filter y[n]=ay[n-1]+x[n]

The filter has a pole at z=a.


The z-transform is a linear operation in the sense that

where c1 and c2 are arbitrary constants. This implies that the z-transform of a parallel combination of

filters is the sum of the transforms for each of the filters.

The most important property of z-transforms is the convolution theorem


Because it replaces the complicated convolution operation by a simpler multiplication,

this theorem is useful for computing the responses of digital filters to signals given

by analytic expressions and for finding the impulse response of cascades of filters.

Proof of convolution theorem: Taking the z-transform of

Interchanging the order of summations over n and

m, and using

Making the change of variable l=n-m, we recognize

the product of H(z) and X(z).


For signals of finite duration, the z-transform is well defined for all values of the complex variable z because it is a finite sum of finite terms. Thus, the z-transforms of FIR filters always converge.

On the other hand, the z-transforms of IIR filters are only defined for certain values of z.

The region of convergence is the portion of the complex z-plane for which the summation

converges. For example, the z-transform of the first-order recursive

low-pass filter h[n]=anu[n] is . The corresponding region of convergence is then |z| > |a|.


We saw earlier that the system is stable is its impulse response is absolutely summable. In this case, the z-transform converges for |z|=1, that is the values of z that are on the unit circle.

Since the region of convergence of a stable filter must include the unit circle and since the region of convergence must be a continuous region and not include any poles, we conclude that the z-transform of a stable filter has all poles either inside or outside the unit circle.

If we restrict our consideration to causal filters which require that the region of convergence be the exterior of a circle, then we conclude that the region of convergence of a causal stable filter is the outside of a circle whose radius is less than unity and the poles of a causal stable filter are all inside the unit circle.

There is no restriction on the location of zeros for causal, stable filters.


Once the poles and zeros have been found for a given z-Transform, they can be plotted onto the z-plane. The z-plane is a complex plane with an imaginary and real axis referring to the complex-valued variable z.


When mapping poles

and zeros onto the

plane, poles are denoted

by an "x" and zeros by

an "o".

When deciding on what filter to design for a given task we need to consider:

IIR or FIR? FIR are always stable but IIR can be computationally efficient...

Lowpass, highpass, bandpass, bandstop? Depends on the task at hand...

Order? Trade-off between computation time and complexity.

The objective is to obtain a reliable transfer function H(z) approximating a desired frequency response. That is best done in the frequency domain.


Low-pass: designed to pass low frequencies from zero to a certain cut-off frequency and to block high frequencies.

High-pass: designed to pass high frequencies from a certain cut-off frequency to π and to block low frequencies.

Band-pass: designed to pass a certain frequency range which does not include zeroand to block other frequencies.

Bandstop: designed to block a certain frequency range which does not include zeroand to pass other frequencies.

The frequency band where the signal is passed is the passband. The frequency band where the signal is removed is the stopband.



All frequencies suffer the same delay In some applications (ECG for example) this is important


What is the delay of FIR filters:

h[n] = δ[n]-δ[n-1]

h[n]=Σ0mδ[n-m]

Ideal low-pass Filter example



Consider and

As M increases, transition band gets narrower, but the ripple remains!


Hamming, Hanning and Blackman windows.

They all provide different trade-offs wrt the width of the main lobe and the ripple effect.


There are much less ripples with the Hanning window but the transition width has increased. That can be improved by increasing the size of the window.


Use filter designed for analogue filtering

Butterworth (flat in the pass-band)

Chebyshev (equi-ripples, faster transition)

In Matlab:

Use functions butter or cheby2.


Julien OSTER, [email protected]/teaching/cdt/A3/4_A3...for all values of the complex...

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