Lecture 5 BME452 Biomedical Signal Processing 2013 (copyright Ali Işın, 2013)1 BME 452 Biomedical Signal Processing Lecture 5 Digital filtering.

Lecture 5 BME452 Biomedical Signal Processing 2013 (copyright Ali Işın, 2013) 1

BME BME 445252 Bio Biomedimedical Signal cal Signal ProcessingProcessing

Lecture Lecture 55

Digital filteringDigital filtering


Lecture 5 Outline In this lecture, we’ll study digital filtering

methods General considerations and filter specifications

Filtering in frequency domain

Filtering in time domain Sum and difference (SD) filter Finite Impulse Response

(FIR) filter Infinite Impulse Response (IIR) filter using MATLAB


General considerations As mentioned in earlier lectures, filtering is the process of

keeping components of the signal with certain desired frequencies and removing components of the signal with certain undesired frequencies

Very often, we keep the gain of the required frequency components to 1 or close to 1

And the gain of the undesired frequency components will be 0 or close to 0

In general, there are 4 types of filtering: LPF, HPF, BPF, BSF


Filter specifications Passband

the range of frequency components that are allowed to pass Stopband

the range of frequency components that are suppressed Passband ripple

ripples in the passband the maximum amount by which attenuation in the passband may

deviate from gain (which is normally 1) Stopband ripple

Ripples in the stopband The maximum amount by which attenuation in the stopband may

deviate from gain (which is normally 0) Stopband attenuation

the minimum amount by which frequency components in the stopband are attenuated

Transition band The band between the passband and the stopband


Ideal filter and edge frequencies

Frequency response of ideal

filters

The edge frequencies are the end frequencies of passband or stopband

fc= cut-off frequency

LPF: passband: 0 f fc

Stopband: fc f fs/2

HPF: passband: fc f fs/2

Stopband: 0 f fc

BPF: passband: fc1 f fc2

Stopband: 0 f fc1 and fc2 f fs/2

BSF: passband: 0 f fc1 and fc2 f fs/2Stopband: fc1 f fc2

Gain

freqfc

1LPF

Gain

freqfc

1HPF

fs/2Last point in

freq axisGain

freqfc1

1BPF

fc2

Gain

freqfc1

1BSF

fc2

fs/2Last point in

freq axis


Actual LPF

LPF Passes all low-

frequency components below fp and blocks all higher frequency components above fs

In reality, you can’t design ‘square’ type of filters

So, there needs to be transition betweens the bandsLPF: passband: 0f fp

Stopband: fsf fs/2

Transition band: fp<f <fs


LPF example Low-pass filter (LPF)

Eg.: Consider a combination of 3 sinusoidal signals - 2 Hz, 5 Hz and 11 Hz.

The final output signals after 2 LPF are shown

LPF at fp=3 Hz and fs=4 Hz

LPF at fp=8 Hz and fs=9 Hz

0 200 400 600 800 1000

-3

-2

-1

0

1

2

3

Combined signal

LPF, fp=8 Hz, fs=9 Hz

0 200 400 600 800 1000

-3

-2

-1

0

1

2

3

0 200 400 600 800 1000-1.5

-1

-0.5

0

0.5

1

1.5

Only 2 Hz signal remains

Only 2 Hz and 5 Hz signals remain

0 200 400 600 800 1000-1.5

-1

-0.5

0

0.5

1

1.5

0 200 400 600 800 1000-1.5

-1

-0.5

0

0.5

1

1.5

0 200 400 600 800 1000-1.5

-1

-0.5

0

0.5

1

1.5

LPF, fp=3 Hz, fs=4 Hz

2 Hz signal 5 Hz signal 11 Hz signal


HPF Passes all high-frequency

components above fp and blocks all higher frequency components below fs

Eg.: Consider the same combination of 3 sinusoidal signals, 2 Hz, 5 Hz and 11 Hz.

The final output signals after 2 HPF are shown

HPF at fs=3 Hz and fp=4 Hz

HPF at fs=8 Hz and fp=9 Hz

0 200 400 600 800 1000

-3

-2

-1

0

1

2

3

0 200 400 600 800 1000

-3

-2

-1

0

1

2

3

Combined signal0 200 400 600 800 1000

-1.5

-1

-0.5

0

0.5

1

1.5

Only 5 Hz and 11 Hz signals remain


HPF, fs=3 Hz, fp=4 Hz

HPF, fs=8 Hz, fp= 9 Hz

Gain

freq

Stopband ripple

Stopband

Stopbandattenuation

Passband ripple

Passband

Transitionband

1

Fs/2fpfs

From this point onwards, we will use Fs for sampling frequency


BPF Passes all frequency components between

edge passband frequencies, fp1<freq(allow)<fp2 and blocks all frequencies below and above edge stopband frequencies, freq(block)<fs1; freq(block)>fs2


The final output signal after BPF at fp1=4 Hz, fp2=6 Hz, fs1=3 Hz, fs2=7 Hz is shown

0 200 400 600 800 1000

-3

-2

-1

0

1

2

3

Combined signal

BPF

0 200 400 600 800 1000-1.5

-1

-0.5

0

0.5

1

1.5


Gain

freq

Stopband ripple

Stopband

Passband ripple

Passband

Transitionband

Stopband ripple

Stopband

Transitionband

1

freqFs/2fs1 fs2fp1 fp2


BSF Band-stop filter (BSF)

Passes all frequency components lower and higher than edge passband frequencies, freq(allow)<fp1; freq(allow)>fp2

and blocks all frequencies between fs1<freq(block)<fs2


The final output signal after BSF at fp1=4 Hz, fp2=6 Hz, fs1=3 Hz, fs2=7 Hz is shown

0 200 400 600 800 1000

-3

-2

-1

0

1

2

3

Combined signal

BSF

0 200 400 600 800 1000-1.5

-1

-0.5

0

0.5

1

1.5

5 Hz signal is filtered out, only 2 Hz and 11 Hz signals remain

Gain

Passband

Passband ripple

Stopband

Transitionband

Stopband ripple

Passband

Transitionband

1

freq

Passband ripple

Fs/2fp1 fp2fs1 fs2


Direct filtering (in frequency domain) A simple method of doing this

Obtain the DFT of the signal (from 0 to fs) Set to zero the values that are not in the required frequency range i.e. apply a

RECTANGULAR window Compute the Inverse Discrete Fourier Transform (IDFT)

Example: f1=8 Hz and f2=25 Hz with N=100, fs=200 Hz Say, we wish to design a LPF with fp=10 Hz and fs=12 Hz

Compute y=fft(x) in MATLAB Set the values y(7:95)=0 WHY? This is important! Compute yf=ifft(y,’symmetric’) And you get the low pass filtered signal! What is the main disadvantage of this technique?

High computation and time Another disadvantage??

0 20 40 60 80 100-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 20 40 60 80 100-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 20 40 60 80 100-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

+ =

0 20 40 60 80 100-1.5

-1

-0.5

0

0.5

1

1.5

In MATLAB, you have to force conjugate symmetry, else you will get complex values due round-off errors in doing FFT and IFFT

Low pass filtered signal

Gain

Fs/2

198 Hz

LPF

1 2 3 4 5 67 … …….. .. ...… …….. ..95

96 97 98 99 100

51 Hzfs


Direct filtering (in frequency domain) –cont.

How it works:

We can also use a different window say Hanning window, instead of rectangular window to obtain a smoother filtered output

H=hanning(12);Hf(1:6) =H(7:12);H(7:95)=0;Hf(96:100)=H(2:6);

y=fft(x);yy=y.*Hf;yf=ifft(yy,'symmetric');plot(yf);

0 20 40 60 80 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Hanning window

0 20 40 60 80 100-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Time domain

Original signal X with length=100

DFT (or FFT)0 20 40 60 80 100

-40

-20

0

20

40

0 20 40 60 80 100-50

0

50

Real values

Imaginary values

0 20 40 60 80 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Multiply with rectangular window on both sides/ends

0 20 40 60 80 1000

5

10

15

20

0 20 40 60 80 100-50

0

50

0 20 40 60 80 100-1.5

-1

-0.5

0

0.5

1

1.5


IDFT (or IFFT) Real values

Imaginary values


Filtering in time domain Two disadvantages of direct filtering

computational time and complexity is high there is distortion

To solve, we should filter in time domain

There are many types of time domain filtering methods

We will look at Simple FIR filters IIR filters using MATLAB codes


Filtering in time domain (in equation form) The output from a IIR digital filter in made

up of previous inputs and previous outputs as well

where B and A are the filter coefficients and the operation * is convolution

Convolution in time domain is equivalent to multiplying in frequency domain –do you remember that we did some window multiplication for direct filtering

Convolution operation will not be discussed in this course

The output from a FIR digital filter in made up of previous inputs only, so no feedback

N

j

M

k

jnyjAknxkBny11

][*][][*][][

M

k

knxkBny1

][*][][

0 20 40 60 80 100-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Time domain

Original signal X with length=100

DFT (or FFT)0 20 40 60 80 100

-40

-20

0

20

40

0 20 40 60 80 100-50

0

50

Real values

Imaginary values

0 20 40 60 80 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Multiply with rectangular window on both sides/ends

0 20 40 60 80 1000

5

10

15

20

0 20 40 60 80 100-50

0

50

0 20 40 60 80 100-1.5

-1

-0.5

0

0.5

1

1.5


IDFT (or IFFT) Real values

Imaginary values

Convolution with coefficients A and B that represents the suitable ‘filter’(IIR filter)

Convolution with coefficients B that represents the suitable ‘filter’(FIR filter)


Simple low pass FIR filter –sum filter Consider, y[n]=x[n]+x[n-1] for every data in the signal

This filter is also known as the sum filter This simple addition acts as a LPF! This can be proved by using z=transform but is not needed for the purpose of this

course

For hardware design, the block diagram would look like

Advantages: You just need one adder and one delay circuit – simple and cost effective (i.e.

cheap) The filter coefficients are integer values (in this case they are 1), so no round-off

errors It is an FIR filter so it is stable – why are FIR filters stable? No feedback

Disadvantage: Not very good LPF (see next slide)

+z-1x(n) x(n-1) y(n)


Sum filter (cont.) The frequency response of the sum filter: not very good as it is far from the

ideal ‘square’ filter

The gain at freq=0 is 1 The stopband frequency (when gain=0) is at rad/sample or at Fs/2 Hz So, there is no stopband So how to define the passband frequency and transition bands? For these cases, we use the 3 dB cut-off as the passband frequency

Figure from S.K.Mitra, DSP 3e

Magnitude is also known as gain


3 dB cut-off frequency

The 3 dB cut-off frequency is defined as the frequency when the gain drops 3 dB from maximum gain of 1, which is assumed to be 0 dB

Gain=1, 20log10(1)=0 dB

When energy is half, i.e. gain=(1/2)0.5=0.7071, we have 20log10(0.7071)=-3 dB

From the figure We can see that the 3 dB cut-off frequency (when gain=0.7071) is

approximately 0.5 rad/sample or Fs/4 Hz This is the passband frequency So, the passband is from 0 to Fs/4 Hz And transition band is from Fs/4 to Fs/2


Increasing the order of sum filter

For order 1, we had y[n]=x[n]+x[n-1] Now assume, this is fed to another same filter in cascade connection

Now, we have z[n]=y[n]+y[n-1]

Solving, we have z[n]=y[n]+y[n-1]=x[n]+x[n-1]+x[n-1]+x[n-2]=x[n]+2x[n-1]+x[n-2]

+z-1y(n) y(n-1) z(n)

+z-1x(n) x(n-1) y(n)

+z-1x(n) x(n-1) z(n)z-1 x(n-2)

2

Verify this using x[1]=3, x[2]=2, x[3]=5

Hint:

Compute y[2] and y[3]

Compute z[3] using single cascaded filter and two filters and compare

You will notice that z[n] will be defined only for n=3 onwards if x[1] is the starting point, i.e. for every order M, you lose M initial data points in filteringLikewise y[n] is defined only from y[2] onwards.


Increasing the order of sum filter (cont.)

So for order M, we have

The frequency response for M=3 is given =>

The passband is about 0.302 rad or Fs/6 We can see that with increasing order, the passband is becoming smaller without any

change in stopband Also, the curve is becoming closer to the ideal ‘square’ So, we can increase/decrease M depending on our needs The 3-dB cut-off frequency is given by fp= cos-1(2-1/(2*M)) *2/pi

][0

)( rnxM

rrC

Mny

where )!(!

!

rQr

QrC

Q


What would be y[n] for N=3?


Simple high pass FIR filter

Similarly, a HPF can be designed using y[n]=x[n]-x[n-1] For hardware design, the block diagram would look like

The stopband frequency (when gain=0) is at 0 Hz, i.e. there is no stopband The gain at freq=Fs/2 Hz or rad is 1 The passband frequency is at 0.5 rad or Fs/4 Hz (using 3 dB cut-off) Passband width is from Fs/4 to Fs/2 Hz The orders, N can be increased to obtain a smaller passband width and to obtain a

frequency response closer to the ideal ‘square’ filter The 3-dB cut-off frequency is given by fp= sin-1(2-1/(2*N)) *2/pi For order N, we have

+z-1x(n)

x(n-1)

y(n)-

N

rrnx

rCNrny

0][)1()(

As homework, try y[n] for N=3?


This is also known as a difference filter


Simple HPF FIR filter example

HPF can be used to remove mean values and low frequency polynomial trends, i.e. to detrend the data

As an example, we saw the passenger data plot in Lecture 4

The detrending can be simply done by using, y[n]=x[n]-x[n-1], where x[n] is the data

Figures from R.Shiavi, Introduction to applied statistical signal analysis


Simple band pass FIR filter

Magnitude response

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.00 0.25 0.50 0.75 1.00

Normalised Frequency (pi rad/sample)

Gai

n

M=7, N=2

M=28 N=8

Similarly, a BPF can be designed using a combination of LPF and HPF

This is known as sum and difference (SD) filter

Different orders, M and N can be chosen to obtain the required frequency response

Where Gaincf is the gain at centre frequency given by

cfGain

NfTMfTfNMG /sin2)cos2()(,

NM

NMsf

fcentre 1cos2

Example (As homework, verify these later on your own) For filter orders of M=28 and N=8 gives

The centre frequency is 40 Hz when fs=256 Hz Approximate 3 dB bandwidth from 32 to 48 Hz (rounded to the nearest integer) The gain amplification at 40 Hz is approximately 47.211


Simple band pass FIR filter (cont.) Let us compute a band pass FIR filter equation:

For orders, LPF, M=4 and HPF, N=1, obtain the band pass FIR equation that expresses z[n] in terms of x[n] and delays of x[n]

Solution (can you get the answer?)

z[n]=x[n]+3x[n-1]+2x[n-2]-2x[n-3]-3x[n-4]-x[n-5]

LPF HPFx(n) y(n) z(n)

N

rrny

rCNrnz

0][)1()(

][0

)( rnxM

rrC

Mny


IIR filter The problem with FIR filter is that you do not get ‘square’

passband like in ideal filters unless you use a very high order

To solve this problem, we can use IIR filter

The disadvantages of IIR filters are The are not stable (due to feedback) The filter coefficients are not normally integer, so can have

round-off errors Hardware design is more complicated Most importantly, their phase response is not linear


Phase response? In you refresh your memory, you’ll know

that frequency responses have two parts: magnitude and phase responses

In the diagrams, the magnitude and phase responses of Butterworth and Elliptic IIR filters are shown

You can see that the gains are relatively stable at 1 for the bandpass range of 0.2 to 0.6

But the phase response is not linear, i.e. not a straight line

In MATLAB, this problem is solved by filtering twice, once forward and once reverse

By doing so, the magnitude response is squared while the phase response becomes zero

0 0.2 0.4 0.6 0.8 1-2000

-1000

0

1000

Normalized Frequency ( rad/sample)

Pha

se (

de

gre

es)

0 0.2 0.4 0.6 0.8 1-400

-200

0

200


Mag

nitu

de

(dB

)

Butterworth

Elliptic

0 0.2 0.4 0.6 0.8 1-500

0

500


Pha

se (

de

gre

es)

0 0.2 0.4 0.6 0.8 1-80

-60

-40

-20

0


Mag

nitu

de

(dB

)


IIR filter design It is not possible to design IIR filters digitally So, the approach is to design analogue IIR filters, then use a bilinear method to

transform the filter to digital This ‘bilinear’ method will not be discussed in this course as it requires z –transofrm

knowledge So, we will use MATLAB functions directly

There many types of IIR filters Butterworth Elliptic (Cauer) Chebyshev I Chebyshev II Bessel

But we will look at two only: Butterworth and Elliptic filters Because Butterworth filter gives flat magnitude responses in the passband and

stopband (i.e. no ripples or very little ripple) while Elliptic filter requires the lowest order among all the IIR filters


Butterworth IIR filter Remember the IIR filter equation

We have to determine the orders M and Nand then determine the coefficients A and B

Very often, we use M=N

N

j

M

k

jnyjAknxkBny11

][*][][*][][

In MATLAB, we can find the required minimum order for our specification using

buttord (Wp, Ws, Rp, Rs)

where Wp is the passband edge frequency, Ws is the stopband edge frequency, Rp is maximum ripple in passband and Rs is the minimum attenuation in stopband

Rp and Rs will be in dB, while Ws and Wp will be in normalised radian/sample

Low pass Butterworth filter with different orders Figure from

S.K.Mitra, DSP 3e


Butterworth IIR filter (cont.) After finding the order using buttord function, we have to find out the filter

coefficients B and A using

[B,A]=butter (N,Wp)

where N is the filter order chosen earlier

Finally to actually do the filtering, we use filtfilt (B,A,x) where x is the signal to be filtered

Filtfilt function filter the signal twice and uses convolution operations

Example: Design a lowpass filter with Fs=200 Hz Passband = 0 to 40 Hz Passband ripple = less than 3 dB Stopband = 50 Hz to the Fs/2 Stopband attenuation = at least 30 dB

Plot the filter's frequency response – use freqz (B,A) function for this purpose.


Butterworth IIR filter example - solution Fs=200 Hz Rp=3 and Rs=30 Wp=40/100 Ws=50/100 Use buttord (Wp, Ws, Rp, Rs)

N=buttord(40/100, 50/100, 3, 30) which gives N=11

Next obtain the coefficients, B and A using [B,A]=butter (11, 40/100) which gives

B = 0.0002 0.0026 0.0129 0.0386 0.0772 0.1081 0.1081 0.0772 0.0386 0.0129 0.0026 0.0002

A =1.0000 -2.1931 3.5467 -3.6414 2.9012 -1.7020 0.7749 -0.2625 0.0658 -0.0114 0.0012 -0.0001

Using freqz(B,A), we get


Butterworth IIR filter example applied to reduce 50 Hz noise from ECG

Assume we the use the designed Butterworth filter to reduce 50 Hz noise from the ECG below

plot(ecg);ecgf=filtfilt(B,A,ecg);plot(ecgf);


Elliptic filter The same approach could be used for designing an Elliptic filter.

The functions are

N=ellipord(Wp, Ws, Rp, Rs)

[B,A] = ellip(N,Rp,Rs,Wp)

The advantage of Elliptic filter as compared to Butterworth filter is that for the same specification, we require a lower order but there are ripples in the passband that is not so evident for Butterworth filter

For the bandpass filter example, we need only order 3 for Elliptic filter but have to use order 8 for Butterworth filter

As a homework, try the other IIR filter functions – chebyshev I, chebyshev II, etc. using MATLAB help, if necessary


Advantages and disadvantages of FIR/IIR

Let’s sum up the advantages and disadvantages of FIR/IIR filters

FIR advantages over IIR Stability (as there is no feedback) Linear phase response Simpler hardware design Desirable numerical properties (less round off/finite precision

problem) Can be designed using fractional arithmetic (coefficients are either

integers or less than 1.0)

FIR disadvantages over IIR: Requires higher order (hence more memory, computation time)

Though there are more advantages in using FIR, very often we use IIR as we have powerful computers and software like MATLAB, which solve most of the disadvantages


Study guide (Lecture 5) From this week’s lecture, you should know how to perform

filtering using

Direct filtering in frequency domain

Sum and difference (SD) FIR filtering

IIR filtering using MATLAB

End of lecture 5

Lecture 5 BME452 Biomedical Signal Processing 2013 (copyright Ali Işın, 2013)1 BME 452 Biomedical Signal Processing Lecture 5 Digital filtering.

Documents

copyright ali

range of frequency components

undesired frequency

f fcbpf

f fpstopband

f fcstopband

higher frequency components

lowfrequency components