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Signal Processing - Continuous and Discrete

Problem 1: Prove the linear phase property of FIR filters in which the impulse response { h n } , n = 0 . . . N − 1 (for N odd) is odd and symmetrical about its mid-point, that is when

hn = − h N−1−n

0 ≤ n ≤ (N − 1)/ 2

Determine the phase response, and the equivalent delay.

Note: If N is odd then h ( N − 1 ) / 2 = 0 for the impulse response to be odd and symmetrical about the mid-point.

Problem 2:

Consider the pole-zero plot shown below Im

| z| = 3 /4

z -p I a n |z |

| z| = 4 /3 xx

6 0

R

-6 0

7 poIe

(a) Does the plot represent an FIR filter? (b) Is it a linear phase system? Find the delay imposed by the filter. (c) Use MATLAB to plot the magnitude and phase of the frequency response of the filter.

Problem 3: MATLAB exercise. Use this example to become familiar with the MATLAB functions kaiserord, fi r 1 and fi l t e r . The task is to create a data set that is contaminated with additive random noise and to then examine the effects of a low-pass filter on the noisy signal. You will use fi r 1 to design the filter and fi l t e r to execute the filtering operations. I have deliberately left the problem loosely defined so that you should experiment to find a satisfactory result.

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(a) Create a data array of a 1 Hz. sinusoidal signal sampled at 10 Hz. in MATLAB, for example: t = 0:.1:100; s i g = sin(2*pi*t); (b) Contaminate your signal with a zero-mean noise signal: noise = 2 * ( r a n d ( s i ze ( s i g n a l ) ) - 0 . 5 ) ; n o i sy s i g = s i g + n o i s e ; (c) Use the ka i s e r o rd ( ) and fi r 1 functions to design a Kaiser window based low-pass FIR filter (you choose the parameters) to “clean up” the signal. Use the fi l t e r ( ) function to filter the noisy signal. Make plots of 1) the contaminated data and 2) the filtered output. (d) Experiment with filter design until you are satisfied with the results. Hand in plots from your final choice. (e) Plot out the first 100 or so points of the signal (sinusoid) and the filtered output and estimate the filter delay. (f) Use MATLAB’s f r e q z ( ) to plot the frequency response of you filter. (g) Use MATLAB’s z p l a n e ( ) to make a pole-zero plot of your final filter, and describe how the low-pass action results.

Make sure you hand in your plots and design parameters for your filter.

Problem 4: Repeat Problem 3, but this time use fi r 1 to design a bandpass filter that will attenuate noise components above and below the signal frequency. You can choose any window function here.

Problem 5: Another MATLAB problem using fi r 1 and fi l t e r .

The signal you are interested in consists of the sum of two sinusoids f (t) = 1.5 sin(60πt) + 2 cos(180πt)

You have taken a set of data, 2000 samples, sampled at 300Hz.

Unfortunately your instrumentation suffers from electromagnetic line frequency (60 Hz) interference so that the data you actually recorded was s(t) = 1.5 sin(60πt) + 2 cos(180πt) + 1.0 sin(120πt) Your task is to (h) Create a data record in MATLAB representing the samples of s(t). (i)

Design a “notch” (band-stop) filter, using fi r 1 , that will attenuate the 60 Hz component significantly while leaving the other two components relatively “unscathed”.

(j) Use fi l t e r ( ) to filter you data record, and hand in magnitude spectra plots of the before and after operations. Label your output plots with the cut-off frequencies and filter length that you used. (k) Use f r e q z ( ) and z p l a n e ( ) to plot the frequency response and pole-zero plots.

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P ro b l e m 1: There are a number of ways to prove this – here are two: ( a ) Start with an acausal filter of the same length H a ( z) =

(N−1)/2

Σ

n = − ( N − 1) / 2

h n z −n

with N odd, and with real, odd symmetric coefficients h n = −h −n so that the impulse response is a real, odd function. Since { h n } ⇐⇒ H ( e j Ω ), DT FT

from the properties of the D T F T, namely that a real, odd function has a imaginary, odd D T F T, then H a (e j Ω ) is imaginary and odd. H a ( z) may be made causal by adding a delay of ( N − 1)/2, that is H (z) = z − ( N − 1) / 2 H a (z) so that H ( ej Ω ) = e− j Ω (N − 1)/ 2 H a ( ej Ω )

Since H a (e j Ω ) is imaginary

/

H (ej Ω ) =

/

j e− j Ω ( N

− 1) / 2

( N − 1)

= −

Ω ±2π/ 2

which is linear with frequency Ω = ω Δ T (apart from possible jump discontinuities where H a ( e j Ω ) changes sign. Note that in practical filters, any jump in the phase response will occur outside the pass-band, and is generally ignored.) The equivalent delay is found from the slope, and represents a delay of (N − 1)/2 samples. ( b ) Assume the causal filter and let H (z)

=

NΣ−1

n=0

h n z −n

=

h0 z 0 + h1 z − 1 + . . . + hN

− 1z

− ( N − 1)

with N odd, and with odd symmetric coefficients about the mid-point, h n = −hN − 1 − n . Group the symmetric components + h2 − 2 together H ( z ) = h0 0 + h1 − 1 − (N − 1) − (N − 2) − (N − 2) − (N − 1)/ 2 z −z

z

−z

z

−z

+ . . . + h(N

− 1) / 2 z

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(N −Σ1)/ 2− 1 n= 0

=

n − (N − 1− n ) h n z −(N − −Σz1)/ 2− 1 h n z − (N − 1)/ 2  n=0

=



+ h ( N − 1) / 2 z − (N − 1)/ 2 z (N − 1)/ 2− n − z − ((N − 1/ 2)− n ) 

since, as noted in the problem statement, for an odd-symmetric function about the mid-point h(N

−1)/2

= 0. T h e frequency response is H ( e j Ω )  = H ( z ) | z = e jΩ

H(e

jΩ

)

=

(N −Σ1)/ 2− 1

= =



(N − 1)/ 2  e− −j Ωj Ω (N − 1)/ 2  je



hn ej Ω ((N − 1)/ 2− n ) − e− j Ω ((N − 1/ 2)− n ) 

(N −Σ1)/ 2− 1 n=0

2h n sin (Ω((N − 1)/ 2 + n)) 

n= 0



j e− j Ω (N − 1)/ 2 H a ( ej Ω ) ,

The function H a ( e j Ω ) is purely real, therefore /

H (ej Ω ) =

/

j e− j Ω ( N

− 1) / 2

( N − 1)

= −

Ω ±2π/ 2

which is linear with frequency Ω = ωΔ T (apart from possible jump discontinuities of π at sign changes in H a (e j Ω ), which as noted above occur outside the pass-band) and represents a delay of ( N − 1)/2 samples.

P ro b l e m 2: (a) The system has 7 zeros and 7 poles at the origin. It is therefore a non-recursive F I R system. (b) The zeros are at: 1,

3 − ,4 — 3 4 , 3 With the help of M AT L A B :

4

± j sin

cosπ 3

π,

3

3 cos π

4

3

± j sin

π

3

z1 = 1 ; z2 = - 4 / 3 ; z3 = - 3 / 4 ; z4 = ( 4 / 3 ) * ( c o s ( p i / 3 ) + i* sin(pi/3)); z5 = (4/3)*(cos(pi/3) i* sin(pi/3)); z6 = (3/4)*(cos(pi/3) + i* sin(pi/3)); z7 = ( 3 / 4 ) * ( c o s ( p i / 3 ) - i * s i n ( p i / 3 ) ) ; z = [ z 1 z2 z3 z4 z5 z6 z 7 ] ; p = [ 0 0 0 0 0 0 0 ] ; K=1; H = tf ( z p k ( z , p , K , - 1 ) ) The transfer function is

H (z)

z 7 − z 6 + 2.79z 4 − 2.79z 3 + z 1 − z 0

= =

z7 − 4 z 0 − z − 1 + 2.79z − 3 − 2.79z + z− 6 − z− 7

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so that this is an odd symmetric filter with N = 8 and { h n } = {1, −1, Then H ( ze j Ω ) has a linear phase shif with frequency /

H(e

jΩ

)=

/

je

0,

2.79,

−2.79,

0,

1,

−1}

= − Ω7 2 + π/2

− j Ω (N − 1)/ 2)

and the delay Δ is the slope, that is Δ = −

( c ) The M AT L A B command

d

dΩ / H ( e j Ω ) = 7/2 steps.

f re q z ( [ 1 - 1 0 2.792 -2.792 0 1 - 1 ] , [ 1 ] ) generates the following plot Magnitude (dB)

20

0

−20 0

0.2

0.4 0.6 Normalized Frequency ( rad/sample)

0.8

1

Phase (degrees)

200 −40 0 −200 −4000

0.2

0.4 Normalized Frequency

0.6

0.8

1

( rad/sample)

−600

which shows a crude “band-pass” characteristic. The phase plot shows a change in phase of −7π/2 over a range of π in frequency, corresponding to a delay of 7/2 steps.

P ro b l e m 3:

The following M AT L A B script was used for this problem:

% Problem Set 6 - - Problem 3 % Enter the s i g n a l and contaminate i t : t = 0 : . 1 : 1 0 0 ; signal = sin(4*pi*t); noise = 2 * ( r a n d ( s i z e ( s i g n a l ) ) - . 5 ) ;

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n o i sy _ s ig n a l = s i g n a l + n o i s e ; % P l o t the waveforms fi g u r e ( 1 ) p l o t ( t , s i g n a l ,’ b ’ ) hold on p l o t ( t , n o i s y _ s i g n a l ,’ r ’ ) ti t l e ( ’ S i g n a l and Ad diti ve N o i s e ’ ) hold o f % % Design the fi l t e r % w_c = 0 . 4 2 ; w_stop = 0 . 4 8 ; passband_ripple = 0 . 1 ; stopband_ripple = 0.01 [N,Wn,BETA,FILTYPE] = kaiserord([w_c w_stop],[1 0 ] , [passband_ripple sto pband_ ripple ],2 ) B = fi r1 (N,Wn ,FILTY P E,kaise r(N+ 1 ,BE TA)); A = 1; % % F i l t e r the noisy data and p l o t the output % fi l t e r e d _ s i g n a l = fi l t e r ( B , A , n o i s y _ s i g n a l ) ; fi g u r e ( 2 ) p l o t ( t , fi l t e r e d _ s i g n a l ,’g ’ ) ti t l e ( ’ F i l t e r e d S i g n a l ’ ) % % P l o t the frequency response and pole-zero p l o t s % fi g u r e ( 3 ) f r e q z ( B , A ) ; fi g u r e ( 4 ) zp la ne(B, A) d f = 1/100.1; f = - 5 :d f : -5 + 1 0 0 0 * d f; fi g u r e ( 5 ) p l o t ( f, fft s h i f ( a b s ( fft ( n o i s y _ s i g n a l ) ) ) ) x labe l( ’Fre que ncy ( H z ) ’ ) ylabel(’Magnitude | F | ’ ) ti t l e (’ M a g n it u d e Spectrum o f Noisy S i g n a l ’ ) fi g u r e ( 6 ) p l o t ( f, fft s h i f ( a b s ( fft ( fi l t e r e d _ s i g n a l ) ) ) ) xla bel(’Fre que ncy ( H z ) ’ ) ylabel(’Magnitude | F | ’ ) ti t l e (’ M a g n it u d e Spectrum o f Low-Pass F i l t e r e d S i g n a l ’ ) This script gave the filter parameters from k a i s e ro rd ( ) as: N = 75, β = 3.395, ωn = 0.45.

The following plots were produced:

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Signal and Additive Noise

2 1.5 1 0.5 02 1.5 −0.5 1 −10

20

40

0.5 −1.5

60

80

100

60

80

100

Filtered Signal

0 −2 −0.5 −1 −1.5 −2 −2.5 0

20

40

The following two spectra have been prepared using fft s h i f ( ) to center them on 0 Hz, Magnitude Spectrum of Noisy Signal

500 450

Magnitud e |F|

400 350 300 250 200 150 100 50 0

−5

−4

−3

−2

−1

0 1 Frequency (Hz)

2

3

4

5

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Magnitude Spectrum of Low−Pass Filtered Signal

500 450

Magnitud e |F|

400 350 300 250 200 150 −5

−4

−3

−2

−1

0 Frequency (Hz)

1

2

3

4

5

100

50 Magnitude (dB)

50

0

0

−50 −1000

0.2

0.4 0.6 Normalized Frequency ( rad/sample)

0.8

1

0.2

0.4 0.6 Normalized Frequency ( rad/sample)

0.8

1

Phase (degrees)

0 −150 −1000 −2000 −30000 −4000 1.5

Imaginary Part

1

75

0.5 0 −0.5

−1

−1.5

−2.5

−2

−1.5

−1 −0.5 Real Part

0

0.5

1

Note that the pole-zero plot is composed with zeros on the unit-circle at angles Ω corresponding to frequencies above the cut-of frequency – thus forcing the magnitude response to zero at those

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frequencies - this is therefore a low-pass filter. It is interesting to note that below the cut-of frequency, the zeros appear to be in pairs where their magnitudes are reciprocals - one inside and one outside the unit circle (in addition to be in complex conjugate pairs as well).

Think about what this means (it also is evident in the pole-zero plot given in Problem 2, and we will see the same thing in the results of Problems 4 and 5).

P ro b l e m 4: The M AT L A B script from Problem 3 was modified slightly for this problem to design a band-pass filter: % Problem Set 6 - - Problem 4 % Enter the s i g n a l and contaminate i t : t = 0:.1:100; signal = sin(4*pi*t); noise = 2 * ( r a n d ( s i z e ( s i g n a l ) ) - . 5 ) ; n o isy _ si g n a l = s i g n a l + n o i s e ; % P l o t the waveforms fi g u r e ( 1 ) p l o t ( t , s i g n a l ,’ b ’ ) hold on p l o t ( t , n o i s y _ s i g n a l ,’ r ’ ) ti t l e ( ’ S i g n a l and Ad diti ve N o i s e ’ ) hold of % % Design the fi l t e r % % Frequency specs - normalized to the Nyquist frequency w_sl=0.35; w_cl=0.39; w_cu = 0 . 4 1 ; w_su = 0 . 4 5 ; % Pass and stop band r i p p l e p_rip = 0 . 1 ; s_rip = 0.01; [N,Wn,BETA,TYPE] = ka ise ro rd ( [w _ s l w_cl w_cu w_su],[0 1 0 ] , [ s _ r i p p_rip s _ r i p ] , 2 ) B = fi r1(N,Wn,TYPE, kaiser(N+ 1 ,B ETA)); A = 1; % % F i l t e r the no isy data and p l o t the output % fi l t e r e d _ s i g n a l = fi l t e r ( B , A , n o i s y _ s i g n a l ) ; fi g u r e ( 2 ) p l o t ( t , fi l t e r e d _ s i g n a l ,’g ’ ) ti t l e ( ’ F i l t e r e d S i g n a l ’ ) % % P l o t the frequency response and pole-zero p l o t s % fi g u r e ( 3 ) f r e q z ( B , A ) ; fi g u r e ( 4 ) zplane (B,A) d f = 1/100.1;

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f = - 5 : d f: - 5 + 1 0 0 0 * d f; fi g u r e ( 5 ) p l o t ( f, fft s h i f ( a b s ( fft ( n o i s y _ s i g n a l ) ) ) ) xla bel(’Freque ncy ( H z ) ’ ) ylabel(’Magnitude | F | ’ ) ti t l e ( ’ M a g n i t u d e Spectrum o f Noisy S i g n a l ’ ) fi g u r e ( 6 ) p l o t ( f, fft s h i f ( a b s ( fft ( fi l t e r e d _ s i g n a l ) ) ) ) xlabe l( ’Fre que ncy (Hz)’) ylabel(’Magnitude | F | ’ ) ti t l e ( ’ M a g n i t u d e Spectrum of Band-Pass F i l t e r e d S i g n a l ’ ) This script gave the filter parameters from k a i s e ro rd ( ) as: N = 112, β = 3.395, ωn = [0.37 The following plots were produced:

0.43].

Signal and Additive Noise

2 1.5 1 0.5 0 −0.5 −10

20

40

60

80

100

60

80

100

Filtered Signal

−1.5 1.5 1 −2 0.5 0 −0.5 −1 −1.5

0

20

40

matlabassignmentexperts.com

The following two spectra have been prepared using fft s h i f ( ) to center them on 0 Hz, Magnitude Spectrum of Noisy Signal

500 450

Magnitud e |F|

400 350 300 250 200 150 −5

−4

−3

−2

−1

100

0 Frequency (Hz)

1

2

3

4

5

2

3

4

5

Magnitude Spectrum of Band−Pass Filtered Signal

450 50 400 0

Magnitud e |F|

350 300 250 200 150 −5 100

−4

−3

−2

−1

0 Frequency (Hz)

1

0.1

0.2

0.3 0.4 0.5 0.6 0.7 Normalized Frequency ( rad/sample)

0.8

0.9

1

0.1

0.2

0.3 0.4 0.5 0.6 0.7 Normalized Frequency ( rad/sample)

0.8

0.9

1

Magnitude (dB)

50 0 0 −20 −40 −600

Phase (degrees)

−80 600 400 200 0 −200 0 −400 −600

matlabassignmentexperts.com

1.5

Imaginary Part

1 112

0.5 0 −0.5

−1

−1.5

−3

−2.5

−2

−1.5

−1 Real Part

−0.5

0

0.5

1

Note that the pole-zero plot is composed with zeros on the unit-circle at angles Ω corresponding to frequencies above and below the cut-of frequency – thus forcing the magnitude response to zero at those frequencies - this is therefore a band-pass filter. As in Problem 3, it is interesting to note that within the pass-band the four zeros appear to be in pairs where their magnitudes are reciprocals one inside and one outside the unit circle (in addition to be in complex conjugate pairs as well).

P ro b l e m 5:

The following M AT L A B script was used to design and test a band-stop filter:

% Problem Set 6 - - Problem 5 % Enter the s i g n a l and contaminate i t : Dt = 1/300; t = 0:Dt:1999*Dt; f_sample = 1 / D t ; f_ nyquist = f_sample/2; % s i g n a l = 1 .5 * sin(2 * pi* 30 * t ) + 2*cos(2*pi*90*t); noise = s i n ( 2 * p i * 6 0 * t ) ; n o isy _ s ig n a l = s i g n a l + n o i s e ; % P l o t the magnitude spectrum of the % contaminated s i g n a l f = -150:1/(2000*Dt):-150+1999/(2000*Dt); fi g u r e ( 1 ) p l o t ( f, fft s h i f ( a b s ( fft ( n o i s y _ s i g n a l ) ) ) ) ti t le ( ’ M a g n i t u d e Spectrum o f S i g n a l with 60 Hz Contaminati on’) xlab el(’Fre que ncy ( H z ) ’ ) ylabel(’Magnitude |F_m|’) f_norm = 6 0 / f _ nyq ui st ; % % Design the fi l t e r % w_sl = f_norm*0.9; w_cl = f_norm*0.7; w_cu = f_norm*1.3; w_su = f_norm*1.1;

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p_ ripple = 0 . 1 ; s_ripple = 0.01; [N,Wn,BETA,TYPE] = ka ise ro rd( [ w _ c l w_sl w_su w_cu],[1 0 1 ] , [ p _ r i p s _ r i p p _ r i p ] , 2 ) B = fi r1(N,Wn,TYPE, ka ise r(N+ 1,B E TA )); A = 1; % F i l t e r the no isy data and p l o t the output magnitude spectrum % fi g u r e ( 2 ) fi l t e r e d _ s i g n a l = fi l t e r ( B , A , n o i s y _ s i g n a l ) ; p l o t ( f , fft s h i f ( a b s ( fft ( fi l t e r e d _ s i g n a l ) ) ) ) ti t le ( ’ M a g n i t u d e Spectrum o f the F i l t e r e d S i g n a l ’ ) xlabe l(’Fre que ncy ( H z ) ’ ) ylabel(’Magnitude |F_m|’) % % P l o t the frequency response and pole-zero p l o t s % fi g u r e ( 3 ) f r eqz ( B, A) ; fi g u r e ( 4 ) z pl ane( B, A) % E x t ra p l o t s - not asked f o r ! % P l o t segments o f the input and fi l t e r e d signals % F i r s t i nte rp o l ate the s i g n a l s ant ti me by a fa c t o r o f e ig ht to make the % p l o t s more re adable. s_8 = re s a m p l e ( s i g n a l , 8 , 1 ) ; ns_8 = re s a m p l e ( n o i sy _ s i g n a l , 8 , 1 ) ; fs_8 = re s a m p l e ( fi l t e re d _ s i g n a l , 8 , 1 ) ; t_8 = re s a m p l e ( t , 8 , 1 ) ; % P l o t the o r i g i n a l s i g n a l and the contaminated s i g n a l fi g u r e ( 5 ) p l o t (t _ 8 ( 3 0 0 : 5 0 0 ) , s _ 8 ( 3 0 0 : 5 0 0 ) ,’ b - - ’,t _ 8 ( 3 0 0 : 5 0 0 ) , n s _ 8 ( 3 0 0 : 5 0 0 ) ,’r ’ ) x la be l( ’ T im e ( s e c ) ’ ) y l a b e l ( ’ S i g n a l ’ ) ti t le ( ’ Se g m e nt of S i g n a l and Contaminated S i g n a l ’ ) fi g u r e ( 6 ) p l o t (t _ 8 ( 3 0 0 : 5 0 0 ) , s _ 8 ( 3 0 0 : 5 0 0 ) ,’ b - - ’,t _ 8 ( 3 0 0 : 5 0 0 ) , f s _ 8 ( 3 0 0 : 5 0 0 ) ,’r ’ ) x la be l( ’ T im e ( s e c ) ’ ) y l a b e l ( ’ S i g n a l ’ ) ti t le ( ’ Se g m e nt o f Uncontaminated S i g n a l and F i l t e r Output ’) fi g u r e ( 7 ) p l o t ( t _ 8 ( 1 : 5 0 0 ) , s _ 8 ( 1 : 5 0 0 ) ,’ b - - ’,t _ 8 ( 1 : 5 0 0 ) , f s _ 8 ( 1 : 5 0 0 ) ,’ r ’ ) x la be l( ’ T im e ( s e c ) ’ ) y l a b e l ( ’ S i g n a l ’ ) ti t l e ( ’ I n i ti a l Segment o f Uncontaminated S i g n a l and F i l t e r Output ’) This script gave the filter parameters from k a i s e ro rd ( ) as: N = 56, β = 3.395, ωcl = 360 rad/s, ωcu = 480 rad/s. The following plots were produced

matlabassignmentexperts.com

Magnitude Spectrum of Signal with 60 Hz Contamination 2500

m

Magnitude |F |

2000

1500

1000

500

0 −150

−100

−50

0 Frequency (Hz)

50

100

150

50

100

150

Magnitude Spectrum of the Filtered Signal

2000 1800

m

Magnitude |F |

1600 1400 1200 1000 800 600 −150

−100

−50

400

0 Frequency (Hz)

Magnitude (dB)

20 200 0 0 −20 −40 −60 0

0.2

0.4 0.6 Normalized Frequency ( rad/sample)

0.8

1

0.2

0.4 0.6 Normalized Frequency ( rad/sample)

0.8

1

Phase (degrees)

−80 0 −1000 −2000 −3000 −40000 −5000

matlabassignmentexperts.com

1

Imaginary Part

0.5 56

0

−0.5 −1 −1.5

−1

−0.5

0 Real Part

0.5

1

1.5

The following plots were not asked for in the problem. To look at the waveforms it was necessary to interpolate the data records to show details of the waveforms, using resample(). See the script for details.

Segment of Signal and Contaminated Signal

4 3

Sign al

2 1 0 −1 −4 −2

Desired signal Signal contaminated with 60 Hz component

−5 0.12

0.13

0.14

0.15

−3

0.16

0.17

0.18

0.19

0.2

0.21

Time (sec) Segment of Uncontaminated Signal and Filter Output 4

3

Sign al

2

1

0

−1 −3 −4 −2 0.12

Uncontaminated input signal Filter output 0.13

0.14

0.15

0.16 0.17 Time (sec)

0.18

0.19

0.2

0.21

matlabassignmentexperts.com

Initial Segment of Uncontaminated Signal and Filter Output

4 3

Sign al

2 1 0 Uncontaminated input signal Filter output

−3 −1 −4

−2

0

0.02

0.04

0.06

0.08

0.1 Time (sec)

0.12

0.14

0.16

0.18

0.2

Note the delay in the filter output.

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