Signal Processing Under Matlab

7/31/2019 Signal Processing Under Matlab

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Signal Processing under Matlab: IntroductionSignal Processing under Matlab: Introduction 11

Signals and Systems

Lecture-19

Signal Processing under Matlab




Matlab Variables

• Matlab stores variables in the form of matrices having dimensionsM×N:

– M is the number of rows.

– N is the number of columns.

– A 1×1 matrix is a scalar.

– A 1×N matrix is a row vector.

– A M×1 matrix is a column vector.

• Examples:>> x = 5

– Real scalar x=5.

>> x = [1 2 5 7]

– A row vector composed from the 4 elements 1, 2, 5 and 7.

>> x = [1 ; 2 ; 5 ; 7]

– A column vector composed from the 4 elements 1, 2, 5 and 7.




Complex numbers

• If the variable i is not redefined by the user, then complex numbers canbe written under the form z = x+yi.

• Examples:

>> x = 5+2i

– Complex scalar x=5+2i.

>> x = [1+i 2 5-i]

– A row vector composed from the 3 complex numbers 1+i, 2 and 5-i.

>> x = [1 ; (2+i)/(3-i)]

– A column vector composed from the 2 complex numbers 1 and 0.5+0.5i.




Basic operations (1)

• The basic arithmetic operations on matrices (and of course on scalarsthat are special cases of matrices) are:

– addition (+), subtraction (-), multiplication (*), division (/), power (^) and

complex transpose ( ’ ).

• Examples:

>> A = [1 2 ; 3 4]

– Defines the 2×2 matrix

>> B = A’

– Results in the matrix B that is the complex transpose of A:

>> A + B

– Results in the 2×2 matrix

>> A * B


⎥⎦

⎤⎢⎣

⎡=

43

21 A

⎥⎦

⎤⎢⎣

⎡=

42

31 B

⎥⎦

⎤

⎢⎣

⎡

85

51

⎥⎦

⎤⎢⎣

⎡

2511

115





• Examples:

>> A = [1 2+i ; 3 4]

– Defines the 2×2 matrix

>> B = A’ – Results in the matrix B that is the complex transpose of A:

>> B = A .’

– Results in the matrix B that is the transpose of A:

>> B = A^2

– Results in the matrix B=A*A:

⎥⎦

⎤⎢⎣

⎡ +=

43

21 i A

⎥⎦

⎤⎢⎣

⎡

−=

42

31

i B

⎥⎦

⎤

⎢⎣

⎡

+ 42

31

i

⎥⎦

⎤⎢⎣

⎡

+

++

i

ii

32215

51037





• The element-by-element arithmetic operations are defined by: – (.*) , (./) , (.^) and , (.’)

• Examples:

>> A = [1 2 ; 3 4] ; B = [-2 0 ; i 3];

>> A * B


>> A .* B


• Examples:

>> A = [1 2 3] ; B = [-2 0 i];

>> A .* B

– Results in the 1×3 row vector [-2 0 3i].

>> A * B

– Results in an error “inner matrix dimensions must agree”.

⎥⎦

⎤⎢⎣

⎡

+−

+−

1246

622

i

i

⎥⎦

⎤⎢⎣

⎡−

123

02

i




Math functions

• The math functions include: – sin, cos, tan

– asin, acos, atan

– exp, log, log10

– sqrt, sin

– abs (returns the magnitude) and angle (returns the argument in radians).

• Examples:

>> x = 2+3i ; x’

– Results in the scalar 2 - 3i.

>> mag = abs(x)

– Results in the scalar mag = 3.6056.

>> arg_rad = angle(x) ; arg_degree = arg_rad*180/pi;

– Results in the scalars arg_rad = 0.9828 and arg_degree = 56.31.




Plotting (1)

• Let x and y be two vectors of the same length. Plotting y as a functionof x can be done using the function “plot”.

– plot(x,y)

• Examples: Plot the functions te-t and te-tcos(8πt) in the interval [0 8] s.

>> t = [0:0.02:8];

>> x = t.*exp(-t);

>> y = t.*exp(-t).*cos(8*pi*t) ; %(or y = x.*cos(8*pi*t));

>> plot(t,x,b’); >> hold on; >> plot(t,y,’r’);

– Line 1 defines a vector t = [0 0.02 0.04 0.06 0.08 … 8];

– Lines 2 and 3 define the functions x and y.

– plot(t,x,’b’) plots x as a function of t (the color of the curve is blue).

– “hold on” permits to plot more than one curve on the same figure.

– plot(t,y,’r’) plots y as a function of t (the color of the curve is red).




Plotting (2)

• The result is given by:

0 1 2 3 4 5 6 7 8-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4




Plotting (3)

• The x-axis and the y-axis can be labeled using the functions “xlabel”and “ylabel”. A title can be added using the function ‘title’

>> xlabel(‘Time (s)’);

>> ylabel(‘Amplitude’);

>> title(‘x and y’);

0 1 2 3 4 5 6 7 8-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

Time (s)

A m p l i t u d e

x and y




Plotting (4)

• Legends of the different curves can be added using the function“legend”. A grid can be added using the function “grid on”.

>> legend(‘x(t)’, ‘y(t)’)

>> grid on

0 1 2 3 4 5 6 7 8-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

Time (s)

A m p l i t u d e

x and y

x(t)

y(t)




Plotting (5)

• The ranges of the x-axis and the y-axis can be limited using thefunctions “xlim” and “ylim”.

>> xlim([0 4]); % limits the values of the x-axis between 0 and 4.

>> ylim([-0.1 1]); % limits the values of the y-axis between -0.1 and 1.

0 0.5 1 1.5 2 2.5 3 3.5 4-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Time (s)

A m p l i t u d e

x and y

x(t)

y(t)




Polynomials (1)

• Matlab represents polynomials as row vectors of polynomialcoefficients.

• Examples:

>> p = [1 4 -5];

– p corresponds to the representation of the polynomial s2 + 4s -5.

>> poly2str(p,’s’)

– Represents the polynomial as a function of s. The answer will be:

>> s^2 + 4s -5

• Examples: Define the polynomial s5 + 2s2 + s + 5.

>> p = [1 0 0 2 1 5];

– In order to check the result, we type:

>> poly2str(p,’s’)

The result of the previous command is:

>> s^5 + 2s^2 + s + 5




Polynomials (2)

• The function “polyval” is used to evaluate a polynomial for certainvalues of the variable.

• Examples:

>> p = [1 4 -5];

>> a = polyval(p,[0]);

>> b = polyval(p,[1 5 10]);

– In line 1, we defined the polynomial p(s) =s2 + 4s - 5.

– In line 2, we evaluate a = p(0) = -5.

– In line 3, we evaluate p(s) for s = 1, 5 and 10. The answer is the row

vector b = [0 40 135].




Polynomials (3)

• The function “roots” is used to find the poles of a polynomial.

• Examples:

>> p = [1 4 -5];

>> r = roots (p); – The result is the column vector r composed from the roots of the function

s2 + 4s - 5. These roots are given by -5 and 1.

• Example: Determine the roots of the polynomial s4 + s3 + s2 + s – 1.

Check the result by plotting this function.

>> p = [1 1 1 1 -1];

>> r = roots (p);

>> x = -2:0.01:2;

>> y = x.^4 + x.^3 + x.^2 + x -1;

>> plot(x,y);






Polynomials (5)

• Consider the two polynomials p1 and p2. The product of thesepolynomial can be determined using the function “conv”:

– p = conv(p1,p2)

• Example: determine the product of the polynomials x2 + 2x + 3 and

x5 + 3x -7.

>> p1 = [1 2 3];

>> p2 = [1 0 0 0 3 -7];

>> p = conv(p1,p2);

>> X = poly2str(p,’x’);

– The vector p will be given by: p = [1 2 3 0 3 -1 -5 -21].

– The last line will result in the string X given by:X = x^7 + 2 x^6 + 3 x^5 + 3 x^3 - 1 x^2 - 5 x - 21




Convolution (1)

• The function “conv” can be also applied to determine the convolutionbetween 2 DT sequences.

• Example (assignment-7, exercise 4.2): Consider the DT LTI system

whose impulse response is given by: h[n] = 0.2δ[n] – 0.5δ[n-2] + 0.4

δ[n-3]. Determine the response of the system to the input sequencex[n] = {-1 , 1 , 0 , -1}.

>> h = [0.2 0 -0.5 0.4];

>> x = [-1 1 0 -1];

>> y = conv(h,x);

– The output sequence y will be given by:

y = [-0.2 0.2 0.5 -1.1 0.4 0.5 -0.4].




Convolution (2)

• To plot y, we use the function “stem”.>> L = length(y); % returns the length of vector y.

>> n = 0:1:L-1; % n = 0, 1, 2, … L-1

>> stem(n,y)

-1 0 1 2 3 4 5 6 7-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6




Laplace Transform (1)

• Consider a LTI system whose transfer function is given by:

• Factorizing the numerator and the denominator of H(s) results in:

Where – z1 … zm are the zeros of H(s).

– p1 … pn are the poles of H(s).

– k is the gain of the transfer function.

0

1

1

0

1

1)(bsbsb

asasas H

n

n

n

n

m

m

m

m

L

L

++

++=

−−

−−

( ) ( )( ) ( )n

m

ps ps

zs zsk s H

−−

−−=

L

L

1

1)(





• To determine the zeros, poles and gain of a transfer function, we usethe function “tf2zp” (transfer function to zero-poles):

[z,p,k] = tf2zp(num,den)

Where

– num and den correspond to the numerator and denominator of the

transfer function respectively.

– The zeros will be stored in the vector z, the poles in vector p and the gain

in the scalar k.

• Example: Determine the zeros and poles of:

>> num = [2 3];

>> den = [1 4 0 5];

>> [z,p,k] = tf2zp(num,den);

– The result will be z = -1.5, k = 2 and

54

32)(23 ++

+=ss

ss H

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

1.07i-0.13

1.07i+0.13

4.27-

p





• To visualize the location of the zeros and poles of a transfer function,we use the function “pzmap”

pzmap(num,den)

where:



• Example: In the previous example:

>> num = [2 3] ;

>> den = [1 4 0 5];

>> pzmap(num,den);

– From this plot, we can determine if the system is stable or not.






-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5-1.5

-1

-0.5

0

0.5

1

1.5Pole-Zero Map

Real Axis

I m a

g i n a r y A x i s





• The inverse of the function “tf2zp” is given by the function “zp2tf”:[num,den] = zp2tf(z,p,k)

Where:

– z is a vector containing the zeros, p is a vector containing the poles and k

is a scalar that stands for the gain.

– The numerator and denominator of the transfer function will be stored in

the vectors num and den respectively.

• Example: Determine transfer function of a system whose gain is equalto 1, whose poles are given by 1+i, 1-i and 2 and that has no zeros.

>> z = [ ]; % the system has no zeros (the numerator is a constant).

>> p = [1+i 1-i 2] ; k = 1;

>> [num,den] = zp2tf(z,p,k);

– The answer is num = [0 0 0 1] and den = [1 -4 6 -4].

– In other words, the transfer function is equal to:464

1)(

23 −+−=

ssss H







– The result is given by:

– Note that the first system (in blue) is not stable. The second system (in

red) is stable since all of its poles lie in the open LHP.

0 1 2 3 4 5 60

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Time (sec)

A m p l i t u d e

Step response





• To plot the impulse response of CT systems, we use the function“impulse”:

y = impulse(num,den,t)



– t corresponds to the time interval over which the impulse response y

must be plotted.

• Example: Plot the impulse response of a system whose gain is equal

to 1, whose poles are given by 1+i, 1-i and 2 and that has no zeros.Plot the impulse response of a system whose gain is equal to 1, whose

poles are given by -1+i, -1-i and -2 and that has no zeros.

>> z = [ ] ; k = 1; p = [1+i 1-i 2] ; p1 = [-1+i -1-i -2];

>> [num,den] = zp2tf(z,p,k) ; [num1,den1] = zp2tf(z,p1,k);

>> t = 0:0.01:4; y = impulse(num, den, t) ; y1 = impulse(num1,den1,t);

>> plot(t,y); hold on; plot(t,y1, ‘r’); ylim([0 0.5])





– The result is given by:

– Note that the first system (in blue) is not stable. The second system (in

red) is stable since all of its poles lie in the open LHP.





• The partial fraction expansion (PFE) of a transfer function can bedetermined using the function “residue”

[r,p,k] = residue(num,den)

• In this case, the transfer function is written as:

where k(s) is different from zero when m ≥ n.

• Example: Determine the PFE of

>> num = 1;

>> den = [1 7 14 8];

>> [r , p ,k] = residue(num,den);

The answer is k = [ ], ,

)()(1

1

0

0 sk ps

r

ps

r

bsb

asas H

n

n

n

n

mm +

−++

−=

++++= L

L

L

81471)(

23 +++=

ssss H

⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

=

0.33

0.5-

0.16

r

⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

−

−

−

=

1

2

4

p





• Consequently:

• If p(j) = … = p(j+m-1) is a pole of multiplicity m, then the expansionincludes terms of the form:

• Example: Determine the PFE of:

>> z = -1 ; k = 1 ; p = [0 0 0 -2 -2];

>> [num,den] = zp2tf(z,p,k) ;

>> [r , p ,k] = residue(num,den);

1

1

3

1

2

1

2

1

4

1

6

1)(

++

+−

+=

ssss H

( ) ( )m j

m j

j

j

j

j

ps

r

ps

r

ps

r

−++

−+

−−++ 1

2

1L

23 )2(

1)(++=

ss

ss H





• The result is:

• Consequently:

⎥⎥⎥

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢

⎣

⎡

=

⎥⎥⎥

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢

⎣

⎡

−

−

==

0.25

0

0.0625-

0.1250

0.0625

;

0

0

0

2

2

;][ r pk

32

25.00625.0

)2(

1250.0

2

0625.0)(

sssss H +−

++

+=




z-transform (1)

• To plot the zeros and poles of a transfer function H(z), we use thefunction “zplane”

zplane(num,den)

• Example: Plot the zeros and poles of

H(z) can be written as:

>> num = [0 0 1 1];

>> den = [1 2 3 -2];

>> zplane(num,den)

232

1)(

23−++

+=

z z z

z z H

321

32

2321)(

−−−

−−

−++

+=

z z z

z z z H

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

-1.5

-1

-0.5

0

0.5

1

1.5

Real Part

I m a g i n a r y P a r t




z-transform (2)

• The step response can be obtained using the function “dstep”:y = dstep(num,den,n)

Where:

– num and den stand for the numerator and denominator of H(z)

respectively.

– The output is the sequence y[n].

• Example: Determine the step response of

>> k = 1; % the gain

>> z = [ ]; %the zeros

>> p = [1/2 1/4]; %the poles


>> n = 0:1:20;

>> y = dstep(num,den,n);

>> stem(n,y);

⎟ ⎠

⎞

⎜⎝

⎛ −

⎟ ⎠

⎞

⎜⎝

⎛ −

=

4

1

2

1

1)(

z z

z H




z-transform (3)

• The result is given by

– we can directly conclude that the system is stable.

0 2 4 6 8 10 12 14 16 18 200

0.5

1

1.5

2

2.5

3

f (4)




z-transform (4)

• The impulse response can be obtained using the function “dimpulse”:y = dimpulse(num,den,n)

Where:

– num and den stand for the numerator and denominator of H(z)

respectively.

– The output is the sequence y[n].

• Example: Determine the impulse response of

>> k = 1; % the gain

>> z = [ ]; %the zeros

>> p = [1/2 1/4]; %the poles


>> n = 0:1:20;

>> y = dimpulse(num,den,n);

>> stem(n,y);

⎟ ⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ −

=

4

1

2

1

1)(

z z

z H

t f (5)




z-transform (5)

• The result is given by

– we can directly conclude that the system is stable.

0 2 4 6 8 10 12 14 16 18 200

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

F R (1)




Frequency Responses (1)

• The function “freqs” is used to determine the frequency response of analog filters.

[h,w] = freqs(num,den)

Where:

– num and den are the numerator and denominator of the transfer function

H(s) respectively.

– H = H(jw) is the frequency response.

– w is a row vector corresponding to the angular frequency.

• Example: Plot the amplitude and phase spectra of

>> num = [0.2 0.3 1];

>> den = [1 0.4 1];

>> [h,w] = freqs(num,den);

14.0

13.02.0)(

2

2

++++=

ss

sss H

F R (2)





• To plot the amplitude and phase spectra as a function of the logarithmof the angular frequency w:

>> subplot(2,1,1)

>> semilogx(w,20*log10(abs(h)));

>> xlabel(‘Frequency’) ; ylabel(‘Magnitude (dB)’) ; grid on

>> subplot(2,1,2)

>> semilogx(w,angle(h)*180/pi);

>> xlabel(‘Frequency’) ; ylabel(‘Phase (degrees)’) ; grid on

– where the function semilogx(x,y) plots y as a function of x (in the same

way as the function “plot”). The difference is that a logarithmic scale is

used on the x-axis.

– The function subplot divides a figure into sub-figures.

– subplot(m,1,i) corresponds to the i-th subfigure of a figures composed

from m vertical subfigures.







10-1

100

101

-20

-15

-10

-5

0

5

10

Frequency

M a g n i t u d e

( d B )

10-1

100

101

-120

-100

-80

-60

-40

-20

0

Frequency

P h a s e ( d

e g r e e s )






• The function “freqz” is used to determine the frequency response of DTfilters.

[h,w] = freqz(num,den)

Where:

– H(z) = num(z)/den(z).

– h = H(e jw) is the frequency response.

– w is always included between 0 and π.

• Example (from lecture 17, example 2): Plot the amplitude and phasespectra of H(z) = 1+z-2.

>> num = [1 0 1];

>> den = [1 0 0];

>> [h,w] = freqz(num,den);






• To plot the amplitude and phase spectra:

>> subplot(2,1,1)

>> plot(w,abs(h));

>> xlabel(‘Frequency’) ;>> ylabel(‘Magnitude’) ;

>> grid on

>> subplot(2,1,2)>> plot(w,angle(h)*180/pi);

>> xlabel(‘Frequency’) ;

>> ylabel(‘Phase (degrees)’) ;>> grid on

0 0.5 1 1.5 2 2.5 3 3.50

0.5

1

1.5

2

Frequency

M a g n i t u d e

0 0.5 1 1.5 2 2.5 3 3.5-100

-50

0

50

100

Frequency

P h a s e ( d e g r e e s )

Analog filters (1)




Analog filters (1)

• The function “buttap” is used for designing an analog lowpass

Butterworth filter with a cutoff angular frequency of 1 rad/s:

[z,p,k] = buttap(n)

Where:

– n is the order of the filter.

– The zeros, poles and gain of the filter will be stored in z, p and k

respectively.

– By construction, the filter will be a lowpass filter with cutoff frequency 1

rad/s (angular frequency w=2πf).

• The cutoff frequency is defined as the frequency at which the gain of

the filter is equal to -3dB.

Analog filters (2)




Analog filters (2)

• The function “bode” is used to plot the Bode diagram of analog filters:

bode(num,den)

where:



• Example: Plot the Bode diagrams of the LP Butterworth filters of orders

2 and 9 having cutoff frequencies of 1 rad/s.

>> [z,p,k] = buttap(2);>> [num,den] = zp2tf(z,p,k);

>> [z1,p1,k1] = buttap(9);

>> [num1,den1] = zp2tf(z1,p1,k1);>> bode(num,den,‘b’);

>> hold on; grid on;

>> bode(num1,den1,‘r’);

Analog filters (3)




Analog filters (3)


-400

-350

-300

-250

-200

-150

-100

-50

0

M a g n i t u d

e ( d B )

10-2

10-1

100

101

102

-900

-720

-540

-360

-180

0

P h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)

Analog filters (4)




Analog filters (4)

• Limiting the range of the magnitude spectrum between -30 and 10:

>> ylim([-30 10]);

– Butterworth filters are characterized by a magnitude response that is flat

in the passband. Note how the rolloff steepness enhances with the order

of the filter.

-3 0

-2 5

-2 0

-1 5

-1 0

-5

0

5

10

M a g n i t u d e ( d B )

10-2

10-1

100

101

102

-900

-720

-540

-360

-180

0

P h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)

-3 dB at w = 1 rad/s



Analog filters (6)




Analog filters (6)

>> [z,p,k] = buttap(5);


>> [num1,den1] = lp2lp(num,den,2*pi*100);

>> bode(num1,den1); grid on;

-200

-150

-100

-5 0

0

50

M a

g n i t u d e ( d B )

101

102

103

104

105

-450

-360

-270

-180

-9 0

0

P h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)

Analog filters (7)




g ( )

• To design a highpass Butterworth filter with cutoff angular frequency w

rad/s, the function “lp2hp” is used:

[num1,den1] = lp2hp(num,den,w)

Where:

– num and den are the numerator and the denominator of the transfer function of a LP Butterworth filter with cutoff frequency 1.

– w is the cutoff frequency of the filter (in rad/s).

– num1 and den1 are the numerator and the denominator of the transfer

function of a HP Butterworth filter with cutoff frequency w.

• Example: plot the Bode diagram of a HP Butterworth filter of order 5

having a cutoff frequency of 100 Hz.

Analog filters (8)




g ( )

>> [z,p,k] = buttap(5);


>> [num1,den1] = lp2hp(num,den,2*pi*100);

>> bode(num1,den1); grid on;

-250

-200

-150

-100

-50

0

M a g

n i t u d e ( d B )

101

102

103

104

105

0

90

180

270

360

450

P h a s e ( d e g

)

Bode Diagram

Fr e q u e n c y ( r a d / s e c )

Analog filters (9)




g ( )

• To design bandpass and bandstop Butterworth filters with center

frequency w and bandwidth B, the functions “lp2bp” and “lp2bs” are

used:

[num1,den1] = lp2bp(num,den,w,B)

[num1,den1] = lp2bs(num,den,w,B)

Where:

– num and den are the numerator and the denominator of the transfer

function of a LP Butterworth filter with cutoff frequency 1.

– w is the center frequency of the filter (in rad/s).

– B is the bandwidth of the filter.

– num1 and den1 are the numerator and the denominator of the transfer

function of a BP (resp. BS) Butterworth filter with center frequency w andbandwidth B.

• Example: plot the Bode diagrams of 2 BP Butterworth filters of order 5

having a center frequency of 200 Hz and whose bandwidths are equal

to 20 Hz and 100 Hz respectively.

Analog filters (8)




g ( )

>> [z,p,k] = buttap(5);


>> [num1,den1] = lp2bp(num,den,2*pi*200,2*pi*20);

>> [num2,den2] = lp2bp(num,den,2*pi*200,2*pi*100);

>> bode(num1,den1); grid on; hold on;

>> bode(num2,den2,’r’); ylim([-100 5])

- 1 0 0

- 8 0

- 6 0

- 4 0

- 2 0

0

M a g n i t u d e ( d B )

1 0- 1

1 00

1 01

1 02

1 03

1 04

1 05

- 7 2 0

- 3 6 0

0

3 6 0

7 2 0

P h a s e ( d e g )

B o d e D i a g r a m

Fre q u e n c y ( r a d / s e c )

Analog filters (9)




• Another type of filters is the Chebyshev filter. The function “cheb1ap” is

used for designing an analog lowpass Chebyshev filter (type 1) with

cutoff frequency 1 rad/s:

[z,p,k] = cheb1ap(n,R)

Where:


– R is the ripple in the passband (in dB) .

– The zeros, poles and gain of the filter will be stored in z, p and k

respectively.

– By construction, the filter will be a lowpass filter with cutoff frequency 1

rad/s (angular frequency w=2πf).

• To design lowpass and highpass Chebyshev filters at other cutoff frequencies, the functions “lp2lp” and ‘lp2hp” are used.

• To design bandpass and bandstop Chebyshev filters, the functions

“lp2bp” and ‘lp2bs” are used.

Analog filters (10)




• Example: Compare 2 LP Chebyshev filters of order 4 having a cutoff

frequency of 10 Hz. The first filter introduces 1 dB of ripple at low

frequencies while the second one introduces 3 dB of ripple.

>> [z,p,k] = cheb1ap(4,1);


>> [NUM,DEN] = lp2lp(num,den,2*pi*10);

>> [z1,p1,k1] = cheb1ap(4,3);

>> [num1,den1] = zp2tf(z1,p1,k1);

>> [NUM1,DEN1] = lp2lp(num1,den1,2*pi*10);

>> bode(NUM,DEN);


>> bode(NUM1,DEN1,’r’);

>> ylim([-30 5]);

Analog filters (11)





– The filter with more ripple has a steeper rolloff. This enhanced filtering is

obtained at the expense of introducing more distortions to the signal.

-3 0

-2 5

-2 0

-1 5

-1 0

-5

0

5

M a g n

i t u d e ( d B )

100

101

102

103

-360

-270

-180

-9 0

0

P h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)

Analog filters (12)




• Example: Compare a BP Butterworth filter and a BP Chebyshev filter.

Both filters have an order of 4, center frequency of 100 rad/s and a

bandwidth of 50 rad/s. The Chebyshev filter introduces 1 dB of ripple.

>> w = 100 ; B = 50;

>> [z,p,k] = buttap(4);


>> [NUM,DEN] = lp2bp(num,den,w,B);

>> [z1,p1,k1] = cheb1ap(4,1);

>> [num1,den1] = zp2tf(z1,p1,k1);

>> [NUM1,DEN1] = lp2bp(num1,den1,w,B);

>> bode(NUM,DEN);


>> bode(NUM1,DEN1,’r’);

>> ylim([-30 5]) ; xlim([10 1e3]);

Analog filters (13)





• In the passband, the Chebyshev filters are not as flat as the Butterworth filters.

However, the Chebyshev filters have a better rolloff.

-20

-15

-10

-5

0

M a g n

i t u d e ( d B )

101 102 103

-360

-180

0

180

360

P h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)

Digital filters (1)




• The function “butter” is used to design digital Butterworth filters:

[num,den] = butter(n,w,type)

Where:


– type = ‘low’, ‘high’, ‘stop’ and ‘bandpass’ stands for the type of the filter.

– num and den are the numerator and the denominator of the transfer

function H(z) of the digital Butterworth filter.

– For lowpass and highpass filters, w is a scalar:w = 2f c / f s

where f c is the cutoff frequency and f s is the frequency at which the data is

sampled.

– For bandpass and bandstop filters, w is a 2-dimensional row vector:

w = 2 [f 1 f 2] / f s

where f 1 and f 2 correspond to the lower and higher frequencies respectively.

Digital filters (2)




• Example: For data sampled at 1000 samples/s, design a bandstop

Butterworth filter of order 8 in the band [300 400] Hz.

>> [num,den] = butter(8, [300 400]*2/1000,‘stop’);

>> [h,w] = freqz(num,den);

>> f = w*(1000/2)/pi; % given that w is between 0 and π

>> plot(f,abs(h)) ; grid on % f is the frequency range in Hz. Based

%% on the sampling theorem, f can not exceed 1000/2 Hz = 500 Hz.

0 5 0 1 0 0 1 5 0 2 0 0 2 5 0 3 0 0 3 5 0 4 0 0 4 5 0 5 0 0

0

0 . 2

0 . 4

0 . 6

0 . 8

1

1 . 2

1 . 4

Digital filters (3)




• To filter discrete-time signals, the function “filtfilt” is used:

y = filtfilt(num,den,x)

Where:

– x is the input signal (before filtering).

– y is the output signal (after filtering).

– num and den are the numerator and denominator of the DT filter

respectively.

• Example: Filter the high frequency component of the signal x(t) =cos(2π50t) + cos(2π300t) sampled at 1000 Hz using a Butterworth filter

of order 4.

>> [num,den] = butter(4,50*2/1000, ‘ low’);

>> t = 0:0.001:0.2;

>> x = cos(2*pi*50*t) + cos(2*pi*300*t);

>> y = filtfilt(num,den,x);

Digital filters (4)




>> plot(t,x,’b’);

>> hold on;

>> plot(t,y,’r’);

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Fourier transform (1)




• The Fourier transform of discrete signals can be determined using the

“fft” function:

X = fft(x,N)

Where:

– x is the DT signal.

– X is the N-point DFT of x.

• Note that:

– The k-th point of X corresponds to the frequencyk f s / N

where f s is the sampling rate.





• Example: plot the 1024-point DFT of the signal x(t) = cos(2π50t)

sampled at the rate of 1000 samples/s.

>> N = 1024;

>> t = [0:1:N-1]*0.001;

>> x = cos(2*pi*50*t);

>> X = fft(x,N);

>> Y = X/N; % It is more significant to normalize the magnitude by N

>> f = [0:1:N-1]*1000/N;

>> plot(f,abs(Y));

>> xlabel(‘Frequency (Hz)’);






0 100 200 300 400 500 600 700 800 900 10000

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Frequency (Hz)





• Note that since the points in the interval [N/2 N-1] are redundant, it is

more convenient to plot the first N/2 points of the DFT.

>> N = 1024;

>> t = [0:1:N-1]*0.001;

>> x = cos(2*pi*50*t);

>> X = fft(x,N);

>> Y = X/N;

>> f1 = [0:1:N/2-1]*1000/N;

>> Y1 = Y(1:1:N/2);

>> plot(f1,abs(Y1));

>> xlabel(‘Frequency (Hz)’); 0 50 100 150 200 250 300 350 400 450 5000

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Frequency (Hz)





• Example: Filter the high frequency component of the signal x(t) =

cos(2π50t) + cos(2π200t) sampled at 1000 Hz using a Butterworth filter of order 4. Plot the 1024-point DFT of x(t) and its filtered version.

>> N =1024;

>> t = [0:1:N-1]*0.001;

>> x = cos(2*pi*50*t) + cos(2*pi*200*t);

>> [num,den] = butter(4,50*2/1000, ‘ low’);

>> y = filtfilt(num,den,x);

>> X = fft(x,N)/N ; X1=X(1:1:N/2);

>> Y = fft(y,N)/N ; Y1=Y(1:1:N/2);

>> f = [0:1:N/2-1]*1000/N;

>> plot(f,abs(X1)) ; hold on;

>> plot(f,abs(Y1),‘r’);






0 50 100 150 200 250 300 350 400 450 5000

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5


Th I F i t f b l l t d i th f ti




• The Inverse Fourier transform can be calculated using the function

“ifft”:

x = ifft(X,N)

Where:

– x is the DT signal.

– X is the N-point DFT of x.

Example: filtering the noise (1)

E l C id th i l




• Example: Consider the signal

x(t) = 1 + cos(2π50t) + 0.5sin(2π70t)

sampled at the rate of 2000 samples/s. Let

y(t) = x(t) + n(t)

where n(t) stands for the noise term. Apply a convenient LP Butterworth

filter of order 3 that permits to reduce the impact of the noise.

%%% Design the Butterworth filter

>> [num,den] = butter(3,70*2/2000, ‘low’); % since the highest

% frequency component of the signal must not be filtered out.

%%% Generate x(t)

>> t = 0:0.0005:0.25;

>> x = 1 + cos(2*pi*50*t) + 0.5*sin(2*pi*70*t);


%%% G t th i t (t)




%%% Generate the noise term n(t)

>> n = 2 * randn(size(t)); %% The noise here is considered to be

% Gaussian with a standard deviation of 2.

%%% Generate y(t)

>> y = x + n;

%%% filter the noise and plot the result

>> z = filtfilt(num,den,y);

>> plot(t,y);

>> hold on;

>> plot(t,z, ‘r’);

>> legend(‘signal corrupted by noise’, ‘ filtered signal’);








• Analyzing the result in the frequency domain




• Analyzing the result in the frequency domain

>> N = 1024;

>> f = [0:1:N/2-1]*2000/N;

>> Y = fft(y,N)/N ; Y1 = Y(1:1:N/2);

>> Z = fft(z,N)/N ; Z1 = Z(1:1:N/2);

>> plot(f,abs(Y1));

>> hold on;

>> plot(f,abs(Z1), ‘r’);

>> ylim([0 0.5]);

>> xlim([0 500]);







Summary of functions (1)

• Basic functions:




Basic functions:

– (+) , (-) , (*) , (/) , (^) , (‘).

– (.*) , (./) , (.^) , (.‘).

• Math functions:

– sin, cos, tan, asin, acos, atan.

– exp, log, log10, sqrt, sin, abs, angle.

– length.

• Plotting functions:

– plot, hold on, xlabel, ylabel, title, legend.

– grid on, xlim, ylim.

– stem, figure. – semilogx, semilogy.

• Polynomials:

– poly2str, polyval, roots, conv


• Laplace Transform:




Laplace Transform:

– tf2zp, zp2tf, pzmap

– step, impulse

– residue

• z-transform:

– zplane

– dstep, dimpulse

• Frequency responses:

– freqs, freqz

• Analog filters:

– buttap, cheb1ap – lp2lp, lp2hp, lp2bp, lp2bs

– bode


• Digital filters:




Digital filters:

– butter

– filtfilt

• Fourier transform:

– fft, ifft

Documents

Signal Processing Under Matlab