[dsplab2013][lap2a]tran dinh ta
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báo cáo thí nghiệm dspTRANSCRIPT
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BÁO CÁO LAB 2A
Fourier Series and Gibbs PhenomenonHỌ VÀ TÊN: TRẦN ĐÌNH TÁ
LỚP: 09DT1
NHÓM : 39B
Signal Synthesis :Exercise 1: Synthesizer
Tao file sigsynth.m
load trumpet.matFs=11025; soundsc(ed1,Fs)
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figure, plot(ed1(200:400)); axis tight % looks very periodic
4.Xem phổ tần số của âm thanh:
load trumpet.mat
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Fs=11025; soundsc(ed1,Fs)
figure, plot(ed1(200:400)); axis tight % looks very periodic
wavwrite(ed1,Fs,'trumpet.wav')
Y = fft(ed1, 512); % take the fft of trumpet
Ymag = abs(Y); % take the mag of Y
f = Fs * (0:256)/512; % get a meaningful axis
plot(f, Ymag(1:257)); % plot Ymag (only half the points are needed)
xlabel('Frequency (Hz)')
ylabel('Magnitude')
axis tight
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5. Đoc dư liêu đô thị:chon giá trị tai các tần số có biên độ đỉnh lớn nhất
F=258,4 hz, Ymag=11,8
F= 538,3 hz, Ymag= 37,75
F=796,7 hz, Ymag= 65,11
F=1055 hz, Ymag= 52,24
F=1335 hz, Ymag=52,59
F=1593 hz, Ymag=38,99
F=1852 hz, Ymag=28,3
6. T o ham addcosinesa .m
Funtion [sigsynth]=addcosines(t,freq,mag)
Sigsynth=zero(1,length(t));
for i=1:length(freq)
sigsynth = sigsynth + mag(i)*cos(2*pi*freq(i)*t);
end
if max(abs(sigsynth ))>1
sigsynth = sigsynth/max(abs(sigsynth));
end
end
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7. sư du ng ham addcosines.m va dư liê u thu đươc tư câu 5 .
Fs=11025;
t=0:1/Fs:3;
freq=[258.4 538.3 796.7 1055 1335 1593 1852];
mag=[11.8 37.8 65 52.3 52.6 39 28.3];
sigsynth=zeros(1,length(t));
for i=1:length(freq)
sigsynth = sigsynth + mag(i)*cos(2*pi*freq(i)*t);
end
sigsynth=0.999*sigsynth/max(abs(sigsynth));
soundsc(sigsynth,Fs)
wavwrite(sigsynth,Fs,'sigsynth1.wav');
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8. So sanh hai âm thanh: trumpet va âm thanh tai tạo ở câu 7.Fs=11025;
t=0:1/Fs:3;
freq=[258.4 538.3 796.7 1055 1335 1593 1852];
mag=[11.8 37.8 65 52.3 52.6 39 28.3];
y=addcosines(t,freq,mag);
soundsc(y,Fs);
figure,
subplot(2,1,1);
plot(200:400,ed1(200:400)); axis tight
subplot(2,1,2);
plot(200:400,ed1(200:400)); axis tight
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9. Tổng hơp âm thanh ở một vai tần số.
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9. Tổng hơp âm thanh ở nhiều tần số.
Ex2:
2 . Tạo tin hiêu sig1, sig2.Fs=8000;
t=1:1/Fs:3;
cosine1= 3*cos(2*pi*500*t);
cosine2=5*cos(2*pi*200*t);
sig1= cosine1 + cosine2;
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sig2= cosine1+5*cos(2*pi*200*(t-1/(2*200)));
4. Dùng subplot để vẽ 2 tin hiêu trên 1 hình va so sánh 2 tin hiêu:
Fs=8000;
t=0:1/Fs:3;
cosine1=3*cos(2*pi*500*t);
cosine2=3*cos(2*pi*200*t);
sig1= cosine1 + cosine2;
sig2= cosine1+5*cos(2*pi*200*(t-1/(2*200)));
figure (1),
subplot(2,1,1);
plot(sig1(1:200));
grid on;
title('sig1');
subplot(2,1,2);
plot(sig2(1:200));
grid on;
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Sự chậm trễ theo thời gian làm cho sig2 chính là đối xứng của sig1 qua trục t sau khi bị trể 1 nửa chu kì.
5. Nghe thư hai âm thanh sig 1 va sig 2 ở trên.
soundsc(sig1,Fs); pause(5); soundsc(sig2,Fs); àchung nghe giống nhau
6. Thư delay 1/10 chu ky.
Fs=8000;
t=0:1/Fs:3;
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cosine1=3*cos(2*pi*500*t);
cosine2=3*cos(2*pi*200*t);
sig1= cosine1 + cosine2;
sig2= cosine1+5*cos(2*pi*200*(t-1/(2*200)));
sig2b= cosine1+5*cos(2*pi*200*(t-1/(10*200)));
figure (2),
subplot(3,1,1);
plot(sig1(1:200));
grid on;
title('sig1');
subplot(3,1,2);
plot(sig2(1:200));
grid on;
title('sig2');
subplot(3,1,3);
plot(sig2b(1:200));
grid on;
title('sig2b');
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6. Thư delay 1/100 chu ky.
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Ta nhận thấy thời gian delay càng nhỏ tín hiêu càng giống tín hiêu gốc.
7. T o ra tin hiêa u sig3 va sig4.
sig3=3*cos(2*pi*500*t);
sig4=sig3+3*cos(2*pi*500*(t-1/(500*4)));
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8. Vẽ sig va sig4 trên một đô thi.
Nghe thư hai âm thanh
soundsc(sig3,Fs);
pause(5);
soundsc(sig4,Fs);
2 âm thanh nghe giống nhau
tư thi nghiêm trên ta kêt luân răng tai cua chung ta không nhạy cam vơi sư thay đổi cua âm thanh.
Truncated Fourier Series
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Exercise 3 :Gibbs phenomena.
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-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-0.8
-0.6
-0.4
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0
0.2
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thoi gian t
do lon
N=1
N=3N=5
N=25