eecs 152b lab report 3

8/18/2019 EECS 152B Lab Report 3

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EECS 152B/CSE 135B

DSP Design & LaboratoryAssignment 3

Implement FIR and IIR Filters using TI c6713 DSK

Milan Patel (15888131)Ali Saad (19141859)

Dacoda Strack (56925997)

Lab Section 3AUniversity of California, Irvine


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Problem 1

The objective of this problem is to implement an FIR lowpass filter with the dsk board.The filter should have a cutoff frequency of 1760 Hz.

itm:questions-fir-lowpassThe following code starts off by declaring an array lowpass[] which has the filter coefficientsfound by the matlab code found in i.. It then begins polling and fills an input vector (of thesame length as the filter) with samples from the dsk board. After filling the input vector,it then enters a loop which accomplishes the following:

1. It sums the convolution of the current input vector with the filter coefficients

2. It outputs this sum to the board

3. It shifts the input vector once to the left and inserts a new sample in its rightmostposition

1 #include "dsk6713_aic23.h"

2 #include "dsk6713_led.h"

3

4 #define DSK6713_AIC23_INPUT_MIC 0x0015

5 #define DSK6713_AIC23_INPUT_LINEIN 0x0011

6

7 #define FILTER_LENGTH 49

8

9 Uint32 fs = DSK6713_AIC23_FREQ_16KHZ; // 1

10 Uint16 inputsource = DSK6713_AIC23_INPUT_LINEIN; // 0x011

11

12

lowpass[FILTER_LENGTH] = {-0.00081696436698336546, -0.00021751423510184715,

13 0.00066640936656965611, 0.00161979322029948370, 0.00214225259628533000,

14 0.00157034732804796000, -0.00047558977253344320, -0.00354590407634487650,

15 -0.00615075099554396580, -0.00624411669947530590, -0.00237847116307618370,

16 0.00500021450790641420, 0.01295156806914578300, 0.01680622510945733700,

17 0.01232212944474056100, -0.00158902124438583140, -0.02095792775291762900,

18 -0.03696902393001178000, -0.03873070207107721900, -0.01779006866361691800,

19 0.02746970212542162300, 0.08966015914615578500, 0.15377710387149290000,

20 0.20195759380955880000, 0.21984511275197313000, 0.20195759380955880000,

21 0.15377710387149290000, 0.08966015914615578500, 0.02746970212542162300,

22 -0.01779006866361691800, -0.03873070207107721900, -0.03696902393001178000,

23 -0.02095792775291762900, -0.00158902124438583140, 0.01232212944474056100,

24 0.01680622510945733700, 0.01295156806914578300, 0.00500021450790641420,

25

-0.00237847116307618370, -0.00624411669947530590, -0.00615075099554396580,26 -0.00354590407634487650, -0.00047558977253344320, 0.00157034732804796000,

27 0.00214225259628533000, 0.00161979322029948370, 0.00066640936656965611,

28 -0.00021751423510184715, -0.00081696436698336546};

29

30 i = 0;

31 j = 0;

32 z = 0;

33

34 main()

1


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35 {

36 comm_poll();

37

38 input[FILTER_LENGTH] = {0};

39 sum = 0;

40 b;

41

42 /* Fill input vector */

43 (z = 0; z < FILTER_LENGTH; z++){

44 input[z] = input_left_sample();

45 }

46

47

(1) {

48 /* Perform multiplication and sum */


50 sum += lowpass[z]*input[FILTER_LENGTH-1-z];

51 }

52 x

53 output_left_sample(sum);

54 sum = 0;55

56 /* Shift input and enter input sample to x[0] */

57

(i = 0;i


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(a) 208 Hz (b) 1.

85 kHz

Figure 1.2: Output waveforms for the FIR lowpass filter resulting fromthe code given in Figure 1.1 for different pure sine frequencies

3


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Problem 2

The objective for this problem is to design a highpass filter with cutoff frequency 4560 Hz.

The code below is exactly the same as for Problem 1 save for the filter coefficient values.



3



6


8



11

12 highpass[FILTER_LENGTH] = {0.00042641152461490183, 0.00112562953175404700,

13 -0.00092337645796045369, -0.00148575066233061150, 0.00208656645683152510,14 0.00193128457176655390, -0.00430111838849391110, -0.00191164049242499900,

15 0.00779811443426204570, 0.00063143422727044536, -0.01257176282544475100,

16 0.00293017932281575520, 0.01834265031540686200, -0.01009072458143068900,

17 -0.02457872732127334600, 0.02287737398573095500, 0.03057290797537401400,

18 -0.04575027393544489600, -0.03556318675588369300, 0.09462042004619715400,

19 0.03887105430451293100, -0.31455060652527644000, 0.46033558445336265000,

20 -0.31455060652527644000, 0.03887105430451293100, 0.09462042004619715400,

21 -0.03556318675588369300, -0.04575027393544489600, 0.03057290797537401400,

22 0.02287737398573095500, -0.02457872732127334600, -0.01009072458143068900,

23 0.01834265031540686200, 0.00293017932281575520, -0.01257176282544475100,

24 0.00063143422727044536, 0.00779811443426204570, -0.00191164049242499900,

25 -0.00430111838849391110, 0.00193128457176655390, 0.00208656645683152510,

26 -0.00148575066233061150, -0.00092337645796045369, 0.00112562953175404700,27 0.00042641152461490183};

28

29 i = 0;

30 j = 0;

31 z = 0;

32

33

main()

34 {

35 comm_poll();

36


38 sum = 0;

39 b;

40


42

(z = 0; z < FILTER_LENGTH; z++){


44 }

45

46 (1) {


4


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50 }

51


53 sum = 0;

54


56 (i = 0;i


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Problem 3

The objective for this problem is to design a highpass filter with cutoff frequencies of 1760 Hzand 4560 Hz.

The code below is exactly the same as for Problem 1 save for the filter coefficient values.



3



6


8



11

12 i = 0;13 j = 0;

14 z = 0;

15

16 bandpass[FILTER_LENGTH]= {0.000031804510880084353, -0.000019789648741247706,

17 -0.001124742421041582600, 0.002285005291701485100, 0.005281289897170116600,

18 0.000264852153473965080, -0.000662268802485741410, 0.004468372773721288400,

19 -0.007818726070629103100, -0.020803397305029305000, -0.002496698628516754200,

20 0.004500191811069565400, -0.013341181860725434000, 0.019690796960738364000,

21 0.062956242466695594000, 0.012430524442952817000, -0.019285086707876285000,

22 0.036841292313531136000, -0.052009560747888622000, -0.247111212040529190000,

23 -0.095827441561315896000, 0.312710665090100040000, 0.312710665090100040000,

24 -0.095827441561315896000, -0.247111212040529190000, -0.052009560747888622000,

25 0.036841292313531136000, -0.019285086707876285000, 0.012430524442952817000,

26 0.062956242466695594000, 0.019690796960738364000, -0.013341181860725434000,

27 0.004500191811069565400, -0.002496698628516754200, -0.020803397305029305000,

28 -0.007818726070629103100, 0.004468372773721288400, -0.000662268802485741410,

29 0.000264852153473965080, 0.005281289897170116600, 0.002285005291701485100,

30 -0.001124742421041582600, -0.000019789648741247706, 0.000031804510880084353};

31

32 main()

33 {

34 comm_poll();

35


37

sum = 0;38

b;

39




43 }

44

45 (1) {


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49 }

50 x


52 sum = 0;

53


55 (i = 0;i


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Problem 4

The objective of this problem is to implement the same filters specified in Problem 1, Problem2, and Problem 3 with IIR filters instead of FIR ones.

While using the following code in lab, we elected to go about having just one set of filtercoefficients uncommented at a time while changing N to fit the size of the filter we wereusing. In the snippet below, all three sets of coefficients are shown uncommented for clarity’ssake.

In any case, after defining N (the length of the given filter) and the filter coefficients properthe code declares input[] and output[] vectors to hold sections of the input and outputsignals of the filter of the same length as the filter.

After entering the while loop, it sums the convolution of the current input[] vector andthe b filter coefficients in order to get the first part of the output signal. It then sums theconvolution of the current output[] vector and the a filter coefficients and subtracts this

sum from the sum described above.

This floating-point difference is typecasted to a short and output to the speakers.

The input and output vectors are then shifted to the left and new appropriate values areplaced in their rightmost positions.



3



6



9

10 // Change length of filter and uncomment appropriate filter

11 #define N 15

12

13 i = 0;

14 j = 0;

15 z = 0;

16

17 /* Low pass Filter with length 8 */

18 b[N] = { 0.015495475408834581, -0.044995815898198468,

19

0.063486394663502116, -0.029432173299327589,20 -0.029432173299327756, 0.063486394663502185,

21 -0.044995815898198482, 0.015495475408834577};

22

23 a[N] = { 1.000000000000000000, - 5.36523779557625600,

24 13.245327837339437000, -19.24381272936014400,

25 17.684441942602113000, -10.25448614975332100,

26 3.475008624624797100, - 0.53213396812700375};

27

28 /* High Pass Filter with length 8 */

8


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29 b[N] = {0.049627511046376083, -0.049188451610409481,

30 0.138121314059458600, -0.139309244079199950,

31 0.139309244079199780, -0.138121314059458520,

32 0.049188451610409245, -0.049627511046375986};

33

34 a[N] = {1.000000000000000000, 2.306895654299675700,

35 4.422490037496873000, 5.049223959537021000,36 4.574298241591526100, 2.802757514457913300,

37 1.203775077617120300, 0.289193186820021040};

38

39 /* Band Pass Filter with length 15 */

40 b[N] = { 0.032327592148735554, -0.092434399595495334, 0.139335498243089440,

41 -0.182665720844533200, 0.235427846647907140, -0.202790003150475790,

42 0.089676279279297999, -0.000000000000000000, -0.089676279279297971,

43 0.202790003150476490, -0.235427846647908390, 0.182665720844534420,

44 -0.139335498243090470, 0.092434399595496028, -0.032327592148735811};

45

46 a[N] = { 1.000000000000000000, - 4.027423078991632500, 10.808766498459843000,

47 -21.188830934225550000, 34.523519894086988000, -46.620894233142970000,

48 54.470584082326944000, -54.546702929057446000, 47.782069693527120000,49 -35.774596361036920000, 23.131917894380116000, -12.332159299321756000,

50 5.419161525366078500, - 1.702254530803074800, 0.365819348914442730};

51

52 main()

53 {

54 comm_poll();

55

56 input[N] = {0};

57 output[N] = {0};

58

59 temp =0;

60

temp1 = 0;

61

62 out;

63 outs;

64

65 bs;

66

67 (1)

68 {

69 // Perform multiplication and sum with flipped input and b

70

(z = 0; z < N; z++)

71 {

72 temp += b[z]*input[N-1-z];

73

}74

75 // Perform multiplication and sum with flipped output and a

76 (z = 1; z < N; z++)

77 {

78 temp1 += output[N-z]*a[z];

79 }

80

81 // Subtract away feedback

82 out = temp-temp1;

9


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83

84 // Output to speakers

85 outs = (

)out;

86 output_left_sample(outs);

87

88 temp = 0; temp1 = 0; //reset sum variables for next cycle

89

90 // Shift input vector left (same as reversing input and shifting right)

91 (i = 0;i


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(a) Near 1.4802kHz

(b) Some higher frequency1

Figure 4.3: Output waveforms for the IIR lowpass filter resulting from thecode given in Figure 4.1 for different pure sine frequencies

(a) Some lower frequency2 (b) 4.95 kHz

Figure 4.4: Output waveforms for the IIR highpass filter resulting fromthe code given in Figure 4.1 for different pure sine frequencies

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(a) 2.

65 kHz (b) Some frequency3

Figure 4.5: Output waveforms for the IIR bandpass filter resulting fromthe code given in Figure 4.1 for different pure sine frequencies

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Questions

100mA 4 × 10−2 µΩ

x =

∫ ∞

−∞

ω(y) dy

1. Document the resulting parameters of the filters you design (i.e. filterorder, transition band frequencies, function used) & their associated code. Include the filter frequency plots using for the coefficientsused in all of the above problems. Plot the detailed passband and stopbandto demonstrate that the ripple requirements are met. Calculate the gainvalues at the test frequencies used and verify that these correspond to thaton an oscilloscope.

For the gain tables below, the theoretical gain is found in matlab and compared tothe experimental gain. The experimental gain was assumed to be unitary when the

input signal frequency was in the passband, since we never actually probed the inputsignal itself. For the signal not in the passband, the gain is calculated by

gain = 20 log

signal amplitude

passband signal amplitude

dB

As noted in Footnote 1, we didn’t record the frequencies of any of the input signals notin the IIR filters’ passbands. Because of this, there are no gain tabulations in thosefilters’ sections.

All calculated experimental gains seem to be within ±10 dB of their theoretical values,though this margin is quite high.

i. FIR Lowpass Filter

Command used fir1()Filter order 48Transition bands ω/π ∈ (0.1992, 0.2637)

Table 5.i.1: Parameters used for the FIR lowpass filter

bl = fir1(48,[0.23]);

Figure 5.i.1: Matlab code used to generate FIR lowpass filter coefficients

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Frequency (kHz) ω (×π rad/sample) Theo. Gain (dB) Exp. Gain (dB)0.208 0.026 0.033 0.0001.850 0.231 −14.066 −18.196

Table 5.i.2: Theoretical and experimental gain values for the FIR lowpassfilter

Normalized Frequency (#: rad/sample)0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

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

-300

-200

-100

0

100

Lowpass Filter Response


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

-25

-20

-15

-10

-5

0

Figure 5.i.2: Magnitude and phase responses for the FIR lowpass filter

Normalized Frequency (#: rad/sample)0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2


-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

Lowpass Filter Response

Figure 5.i.3: Close-up of magnitude response in the passband of the FIRlowpass filter

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ii. FIR Highpass Filter

Command used fir1()Filter order 48Transition bands ω/π ∈ (0.5059, 0.5762)

Table 5.ii.1: Parameters used for the FIR highpass filter

bh = fir1(44,0.54,’high’)

Figure 5.ii.1: Matlab code used to generate FIR highpass filter coeffi-cients

Frequency (kHz) ω (×π rad/sample) Theo. Gain (dB) Exp. Gain (dB)4.090 0.511 −32.615 −39.043

5.260 0.657 0.028 0.000

Table 5.ii.2: Theoretical and experimental gain values for the FIR high-pass filter

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M a g n i t u d e

( d B )

-300

-200

-100

0

100

Highpass Filter Response



-40

-30

-20

-10

0

10

Figure 5.ii.2: Magnitude and phase responses for the FIR highpass filter

Normalized Frequency (#: rad/sample)0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1


-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

Highpass Filter Response

Figure 5.ii.3: Close-up of magnitude response in the passband of the FIRhighpass filter

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iii. FIR Bandpass Filter

Command used fir1()Filter order 48Transition bands ω/π ∈ (0.1641, 0.2363) ∪ (0.5566, 0.6509)

Table 5.iii.1: Parameters used for the FIR bandpass filter

bb = fir1(43,[0.20 0.59],’band’);

Figure 5.iii.1: Matlab code used to generate FIR bandpass filter coeffi-cients

Frequency (kHz) ω (×π rad/sample) Theo. Gain (dB) Exp. Gain (dB)1.289 0.161 −43.016 −33.172

4.120 0.515 −0.026 0.000

Table 5.iii.2: Theoretical and experimental gain values for the FIR band-pass filter

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M a g n i t u d e

( d B )

-300

-200

-100

0

100

Bandpass Filter Response



-40

-30

-20

-10

0

10

Figure 5.iii.2: Magnitude and phase responses for the FIR bandpass filter

Normalized Frequency (#: rad/sample)0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6


-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5


Figure 5.iii.3: Close-up of magnitude response in the passband of the FIRbandpass filter

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iv. IIR Lowpass Filter

Command used ellip()Filter order 7Transition bands ω/π ∈ (0.2207, 0.2266)

Table 5.iv.1: Parameters used for the IIR lowpass filter

[bl,al] = ellip(7,1,40,0.22);

Figure 5.iv.1: Matlab code used to generate IIR lowpass filter coeffi-cients


M a g n i t u d e

( d B )

-300

-200

-100

0

100

IIR Lowpass Filter Response



-10

-8

-6

-4

-2

0

Figure 5.iv.2: Magnitude and phase responses for the IIR lowpass filter

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v. IIR Highpass Filter

Command used ellip()Filter order 7Transition bands ω/π ∈ (0.5645, 0.5703)

Table 5.v.1: Parameters used for the IIR highpass filter

[bh,ah]=ellip(7,1,40,0.57,’high’)

Figure 5.v.1: Matlab code used to generate IIR highpass filter coeffi-cients


M a g n i t u d e

( d B )

-200

-150

-100

-50

0

50

IIR Highpass Filter Response



0

2

4

6

8

10

Figure 5.v.2: Magnitude and phase responses for the IIR highpass filter

21


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Normalized Frequency (#: rad/sample)0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1


-2.5

-2

-1.5

-1

-0.5

0

IIR Highpass Filter Response

Figure 5.v.3: Close-up of magnitude response in the passband of the IIRhighpass filter

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vi. IIR Bandpass Filter

Command used ellip()Filter order 7Transition bands ω/π ∈ (0.2188, 0.2207) ∪ (0.5723, 0.5762)

Table 5.vi.1: Parameters used for the IIR bandpass filter

[bb,ab] = ellip(7,1,40,[0.22 0.57]);

Figure 5.vi.1: Matlab code used to generate IIR bandpass filter coeffi-cients


M a g n i t u d e

( d B )

-200

-150

-100

-50

0

50




-10

-5

0

5

10

Figure 5.vi.2: Magnitude and phase responses for the IIR bandpass filter

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Normalized Frequency (#: rad/sample)0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6


-25

-20

-15

-10

-5

0


Figure 5.vi.3: Close-up of magnitude response in the passband of the IIRbandpass filter

24


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Conclusion

Through the course of this lab, we learned how filters can be implemented and convolutionsperformed using c code, along with how to generate the filter coefficients for FIR and IIRlow-, high- and bandpass filters using standard matlab functions.

While our experimental values seem to differ from our theoretical ones a bit, this likely hasto do with our failing to record the original input signal amplitude.

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Code

1 %% Preliminary shit

2 ytexlabel = @(str) ylabel(str, ’interpreter’, ’latex’, ’FontUnits’, ’points’, ’FontSize’, 16);

3 xtexlabel = @(str) xlabel(str, ’interpreter’, ’latex’, ’FontUnits’, ’points’, ’FontSize’, 16);

4

textitle = @(str) title(str, ’interpreter’, ’latex’, ’FontUnits’, ’points’, ’FontSize’, 20);5 redline = @(x,y) plot(x, y, ’:r’);

6

7 magnitude = @(h) 20*log(abs(h));

8

9 %% Low Pass Filter (cutoff w = 0.22)

10 bl = fir1(48,[0.23]);

11

12 [h,w] = freqz(bl, 1, 512);

13 phi = phasez(bl, 1, 512);

14

15 % Plot

16 figure(1);

17

18 % Magnitude

19 subplot(2, 1, 1);

20 plot(w/pi, magnitude(h))

21 grid on;

22 xtexlabel(’Normalized Frequency ($\times\pi$ rad/sample)’)

23 ytexlabel(’Magnitude (dB)’);

24 textitle(’Lowpass Filter Response’);

25

26 hold on;

27 transitionBandStart = w( find(magnitude(h) < -3, 1))/pi;

28 transitionBandEnd = w( find(magnitude(h) < -40, 1))/pi;

29 redline([transitionBandStart transitionBandStart], [-300, 100]);

30 redline([transitionBandEnd transitionBandEnd],[-300, 100]);31 hold off;

32

33 % Phase

34 subplot(2, 1, 2);

35 plot(w/pi, phi)

36 grid on;


38 ytexlabel(’Phase (degrees)’);

39

40 set(gcf, ’PaperPositionMode’, ’auto’);

41 % print -dpdf /home/dacoda/Documents/school/dsp-design-152b/lab/lab-3/plots/fir-lowpass-plot.pdf

42

43 % In close magnitude44 figure(2)

45 plot(w(1:103)/pi, magnitude(h(1:103)))

46 grid on;




50


26


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52 % print -dpdf /home/dacoda/Documents/school/dsp-design-152b/lab/lab-3/plots/fir-lowpass-close-up

53

54 %% High Pass Filter (cutoff w =0.57)

55 bh = fir1(44,0.54,’high’)

56

57 [h, w] = freqz(bh, 1, 512);

58 phi = phasez(bh, 1, 512);59

60 % Plot

61 figure(1);

62

63 % Magnitude

64 subplot(2, 1, 1);


66 grid on;



69 textitle(’Highpass Filter Response’);

70

71 hold on;72 transitionBandStart = w( find( magnitude(h) > -40, 1) )/pi;

73 transitionBandEnd = w( find( magnitude(h) > -3, 1) )/pi;


75 redline([transitionBandEnd transitionBandEnd],[-300, 100]);

76 hold off;

77

78 % Phase

79 subplot(2, 1, 2);

80 plot(w/pi, phi)

81 grid on;



84


86 % print -dpdf /home/dacoda/Documents/school/dsp-design-152b/lab/lab-3/plots/fir-highpass-plot.pd

87

88 % In close magnitude

89 figure(2)

90 plot(w( find(magnitude(h) > -3, 1) : )/pi, magnitude(h( find(magnitude(h) > -3, 1) : )))

91 grid on;




95

96

set(gcf, ’PaperPositionMode’, ’auto’);97 % print -dpdf /home/dacoda/Documents/school/dsp-design-152b/lab/lab-3/plots/fir-highpass-close-u

98

99 %% Band Pass Filter (between wc =0.22 and 0.57 )

100 bb = fir1(43,[0.20 0.59],’band’);

101

102 [h,w] = freqz(bb, 1, 512);

103 phi = phasez(bb, 1, 512);

104

105 % Plot

27


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106 figure(1);

107

108 % Magnitude

109 subplot(2, 1, 1);


111 grid on;

112 xtexlabel(’Normalized Frequency ($\times\pi$ rad/sample)’)113 ytexlabel(’Magnitude (dB)’);

114 textitle(’Bandpass Filter Response’);

115

116 hold on;

117 firstTransitionBandStartIndex = find(magnitude(h) > -40, 1);

118 firstTransitionBandEndIndex = find(magnitude(h) > -3, 1);

119 secondTransitionBandStartIndex = firstTransitionBandEndIndex + find( magnitude(h(firstTransition

120 secondTransitionBandEndIndex = firstTransitionBandEndIndex + find( magnitude(h(firstTransitionBa

121

122 firstTransitionBandStart = w(firstTransitionBandStartIndex)/pi;

123 firstTransitionBandEnd = w(firstTransitionBandEndIndex)/pi;

124 secondTransitionBandStart = w(secondTransitionBandStartIndex)/pi;

125 secondTransitionBandEnd = w(secondTransitionBandEndIndex)/pi;126

127 redline([firstTransitionBandStart firstTransitionBandStart], [-300, 100]);

128 redline([firstTransitionBandEnd firstTransitionBandEnd],[-300, 100]);

129 redline([secondTransitionBandStart secondTransitionBandStart], [-300, 100]);

130 redline([secondTransitionBandEnd secondTransitionBandEnd], [-300, 100]);

131 hold off;

132

133 % Phase

134 subplot(2, 1, 2);

135 plot(w/pi, phi)

136 grid on;



139


141 % print -dpdf /home/dacoda/Documents/school/dsp-design-152b/lab/lab-3/plots/fir-bandpass-plot.pd

142


144 figure(2)

145 plot(w(firstTransitionBandEndIndex:secondTransitionBandStartIndex)/pi, magnitude(h(firstTransiti

146 grid on;




150


152 % print -dpdf /home/dacoda/Documents/school/dsp-design-152b/lab/lab-3/plots/fir-bandpass-close-u

153

154 %% Print coefficients

155 fprintf(’lowpass = ’)

156 fprintf(’[’);

157 fprintf(’%0.20f ,’, low);

158 fprintf(’\n’);

159

28


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160 fprintf(’highpass = ’)


162 fprintf(’%0.20f, ’, high);


164

165

166 fprintf(’bandpass = ’)167 fprintf(’[’);

168 fprintf(’%0.20f ,’, bpass);


Figure 7.1: Matlab code used to generate FIR filter coefficients and plotsin its entirety.

1 %% Preliminary shit

2 ytexlabel = @(str) ylabel(str, ’interpreter’, ’latex’, ’FontUnits’, ’points’, ’FontSize’, 16);

3 xtexlabel = @(str) xlabel(str, ’interpreter’, ’latex’, ’FontUnits’, ’points’, ’FontSize’, 16);

4 textitle = @(str) title(str, ’interpreter’, ’latex’, ’FontUnits’, ’points’, ’FontSize’, 20);

5 redline = @(x,y) plot(x, y, ’:r’);6

7 magnitude = @(h) 20*log(abs(h));

8

9 %% Low Pass Filter (cutoff w = 0.22)

10 bl = fir1(48,[0.23]);

11

12 [h,w] = freqz(bl, 1, 512);

13 phi = phasez(bl, 1, 512);

14

15 % Plot

16 figure(1);

17

18 % Magnitude19 subplot(2, 1, 1);


21 grid on;




25

26 hold on;

27 transitionBandStart = w( find(magnitude(h) < -3, 1))/pi;

28 transitionBandEnd = w( find(magnitude(h) < -40, 1))/pi;



31 hold off;32

33 % Phase

34 subplot(2, 1, 2);

35 plot(w/pi, phi)

36 grid on;



39

29


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41 % print -dpdf /home/dacoda/Documents/school/dsp-design-152b/lab/lab-3/plots/fir-lowpass-plot.pdf

42


44 figure(2)

45 plot(w(1:103)/pi, magnitude(h(1:103)))

46 grid on;47 xtexlabel(’Normalized Frequency ($\times\pi$ rad/sample)’)



50


52 % print -dpdf /home/dacoda/Documents/school/dsp-design-152b/lab/lab-3/plots/fir-lowpass-close-up

53

54 %% High Pass Filter (cutoff w =0.57)

55 bh = fir1(44,0.54,’high’)

56

57 [h, w] = freqz(bh, 1, 512);

58 phi = phasez(bh, 1, 512);

59

60 % Plot

61 figure(1);

62

63 % Magnitude

64 subplot(2, 1, 1);


66 grid on;




70

71 hold on;

72 transitionBandStart = w( find( magnitude(h) > -40, 1) )/pi;

73 transitionBandEnd = w( find( magnitude(h) > -3, 1) )/pi;



76 hold off;

77

78 % Phase

79 subplot(2, 1, 2);

80 plot(w/pi, phi)

81 grid on;



84


86 % print -dpdf /home/dacoda/Documents/school/dsp-design-152b/lab/lab-3/plots/fir-highpass-plot.pd

87


89 figure(2)

90 plot(w( find(magnitude(h) > -3, 1) : )/pi, magnitude(h( find(magnitude(h) > -3, 1) : )))

91 grid on;



30


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150


152 % print -dpdf /home/dacoda/Documents/school/dsp-design-152b/lab/lab-3/plots/fir-bandpass-close-u

153

154 %% Print coefficients155 fprintf(’lowpass = ’)


157 fprintf(’%0.20f ,’, low);


159

160 fprintf(’highpass = ’)


162 fprintf(’%0.20f, ’, high);


164

165

166 fprintf(’bandpass = ’)

167 fprintf(’[’);168 fprintf(’%0.20f ,’, bpass);


Figure 7.2: Matlab code used to generate IIR filter coefficients and plotsin its entirety.