help me to write a code for processing an audio signal using taylor series

clc;

clear all;

[x,Fs] = audioread('audioe.wav');

sound(x,Fs);

whos x

disp(Fs)

x = x(:,1);

dt = 1/Fs;

t = 0:dt:(length(x)*dt)-dt;

figure(1)

plot(t,x); xlabel('Seconds'); ylabel('Amplitude');

h = spectrum.periodogram; % create a periodogram spectral estimator.

figure(2)

psd(h,x,'Fs',Fs); % Calculates and plots the two-sided PSD.

%plot(psd(spectrum.periodogram,x,'Fs',Fs,'NFFT',length(x)));

y = interp(x,4);%resample data at a higher rate using lowpass interpolation

%[y, ty] = resample(x,t,Fs);

%y =resample(x,3,2);

figure(3)

subplot(211);

stem(x);

title('Original Signal');

subplot(212);

stem(y);

title('Interpolated Signal');

SNR = snr(x);

SNR1 = snr(y);

a=2;

N =2;

z = taylor(x,a,N);

plot(z);

[Adam]

You didn't actually ask a question.

[Elavarasi E]

i need to sum up three different dimensions, which is in taylor series. this is in matrix and the equation is :

(x+h) = x + (h.*diff(x)) + (h^2.*diff(x,2))./factorial(2);

(x+h) is of same length of 'x' while first derivative is one lessthan 'x' and second derivative is of two lessthan of 'x'.

[Elavarasi E]

In an audio signal, I need to find the intermediate samples. To read the intermediate I make use of taylor series. Audio signal is read thro’ [x,Fs] = audioread('audioe.wav');

The taylor series am using is f(x+h) = f(x) + hf’(x) + h2f’’(x)/2! .

Were f(x) is ‘x’ . Accordingly f(x+h) should be x(t)+h but this doesn’t work. ‘h’ is some intermediate value.

clc;

close all;

clear all;

[x,Fs] = audioread('audioe.wav');

sound(x,Fs);

whos x

disp(Fs)

x= x(:,1);

dt = 1/Fs;

t = 0:dt:(length(x)*dt)-dt;

figure(1)

plot(t,x);

xlabel('Seconds');

ylabel('Amplitude');

dx = gradient(x);

dx2 = gradient(dx);

for t = 0:dt:(length(x)/4*dt)-dt

for x = 0:dt:(length(x)/4*dt)-dt

h = 0.347;

y(t+h) = x + (h.*dx) + (h^2.*dx2)./factorial(2);

end

end

figure(2)

% hold on

plot(t,x,'blue',t,y, 'red')

% hold off

legend('actual','taylorseries')

for loop doesnt work

Answer 1

[darova]

Try gradient instead of diff

dx = gradient(x);
dx2 = gradient(dx);
result = x + (h.*dx) + (h^2.*dx2)./factorial(2);

[Elavarasi E]

this is my code to read an audio signal

clc;

close all;

clear all;

[x,Fs] = audioread('audioe.wav');

sound(x,Fs);

whos x

disp(Fs)

x= x(:,1);

dt = 1/Fs;

t = 0:dt:(length(x)*dt)-dt;

figure(1)

plot(t,x);

xlabel('Seconds');

ylabel('Amplitude');

h = 0.347;

dx = gradient(x);

dx2 = gradient(dx);

for t1 = 0:dt:(length(x)*dt)-dt

(x+h) == x + (h.*dx) + (h^2.*dx2)./factorial(2);

end

figure(2)

hold on

plot(t,x,'blue',t1(:,1:length(x+h)),(x+h), 'red')

hold off

legend('actual','taylorseries')

the lengths same now. but, the for loop does not excecute nor do it shows me any error except doble equal sign. the second figure does appear at all.

[darova]

What does it mean?

for  t1 = 0:dt:(length(x)*dt)-dt
    (x+h) == x + (h.*dx) + (h^2.*dx2)./factorial(2);
end

[Elavarasi E]

am equating the time t1 to t which nothing but dt . in this for loop i get an error wrt equal sign

[Elavarasi E]

i need to process the read audio signal through taylor series. the length of the fine processed audio signal will be same as that of the actual signal. so i use a 'for' loop as the length of time 't' and plot the fine processed signal wrt 't'. but the for loop is not exceuting nor does it show me any error.

[Elavarasi E]

the only error it shows is about equal sign n it guides me to use double equal sign.

[Elavarasi E]

f(x+h) = f(x) + hf'(x) + h^2f''(x)/2!

i need to find the intermediate value of f(x) at instance z = x+dt where both time n signal are variable.

[darova]

Please show the error occurs and why it guides you to use double sign

Why it doesn't work?

result = x + (h.*dx) + (h^2.*dx2)./factorial(2);

[Elavarasi E]

In an audio signal, I need to find the intermediate samples. To read the intermediate I make use of taylor series. Audio signal is read thro’ [x,Fs] = audioread('audioe.wav');

The taylor series am using is f(x+h) = f(x) + hf’(x) + h2f’’(x)/2! .

Were f(x) is ‘x’ . Accordingly f(x+h) should be x(t)+h but this doesn’t work. ‘h’ is some intermediate value.

this my modified code again does not work for the "for" loop

clc;

close all;

clear all;

[x,Fs] = audioread('audioe.wav');

sound(x,Fs);

whos x

disp(Fs)

x= x(:,1);

dt = 1/Fs;

t = 0:dt:(length(x)*dt)-dt;

figure(1)

plot(t,x);

xlabel('Seconds');

ylabel('Amplitude');

dx = gradient(x);

dx2 = gradient(dx);

for t = 0:dt:(length(x)/4*dt)-dt

for x = 0:dt:(length(x)/4*dt)-dt

h = 0.347;

y(t+h) = x + (h.*dx) + (h^2.*dx2)./factorial(2);

end

end

figure(2)

% hold on

plot(t,x,'blue',t,y, 'red')

% hold off

legend('actual','taylorseries')

[darova]

You used one loop previously. Now you use two loops

I don't understand?

[Elavarasi E]

ya . the signal is variable in both time n amplitude. hence two loops

[Elavarasi E]

length of x is large. so i made it equal to t.

[Elavarasi E]

clc;

close all;

clear all;

[x,Fs] = audioread('audioe.wav');

sound(x,Fs);

whos x

disp(Fs)

x= x(:,1);

dt = 1/Fs;

t = 0:dt:(length(x)*dt)-dt;

figure(1)

plot(t,x);

xlabel('Seconds');

ylabel('Amplitude');

for t = 1:dt:(length(x)/4*dt)-dt

for k = 1:length(x)/32

dx(k) = gradient(x(k));

dx2(k) = gradient(dx(k));

h = 0.347;

y(t+h) = x(k) + (h.*dx(k)) + (h^2.*dx2(k))./factorial(2);

end

end

figure(2)

% hold on

plot(t,x,'blue',t,y, 'red')

% hold off

legend('actual','taylorseries')

Results :

Name Size Bytes Class Attributes

x 2395137x2 38322192 double 44100

Array indices must be positive integers or logical values.

Error in eaudiodiff (line 24)

y(t+h) = x(k) + (h.*dx(k)) + (h^2.*dx2(k))./factorial(2);

1) guide me to solve theabove error.

2) when i execute these statements individually on comment window

k = 1:length(x)/32

dx(k) = gradient(x(k))

dx2(k) = gradient(dx(k))

h = 0.347

y(t+h) = x(k) + (h.*dx(k)) + (h^2.*dx2(k))./factorial(2)

I get an error as memoryspace exceeded.

[darova]

I don't have your data so i generated some random

n = 100;
Fs = 2000;
dt = 1/Fs;
t = 0:dt:(n*dt)-dt;
x = 0.05*sin(2000*t) + cos(100*t);        % random data

dx = gradient(x);
dx2 = gradient(dx);
h = 0.347;
x1 = x + h.*dx + h^2.*dx2./factorial(2);

plot(t,x,'blue',t,x1, '.r')
legend('actual','taylorseries')
xlabel('Seconds');
ylabel('Amplitude');

And here are the results

Slightly differs is it?

[Elavarasi E]

Appreciate your kindness. Ya it differs. Actually I have to process it on any audio signal.

when I execute the entire code I get these errors

Array indices must be positive integers or logical values.

Error in eaudiodiff (line 24)

y(t+h) = x(k) + (h.*dx(k)) + (h^2.*dx2(k))./factorial(2);

1) guide me to solve the above error.

2) when i execute these statements individually on comment window

k = 1:length(x)/32

dx(k) = gradient(x(k))

dx2(k) = gradient(dx(k))

h = 0.347

I get the correct values but when the below equation is executed

y(t+h) = x(k) + (h.*dx(k)) + (h^2.*dx2(k))./factorial(2)

I get an error as memoryspace exceeded.

need to rectify these two errors.

[Elavarasi E]

need help to reduce the memory size of the program.

[darova]

What will happen if you use my code?

[Elavarasi E]

ur code is correct but its only for one instance at x1, which should actually be x+h, i.e x1 = (x+h). i need to evaluate it for all the values of x displaced by the instance 'h' wrt to 't'.

i.e., y(t+h) = x(k) + (h.*dx(k)) + (h^2.*dx2(k))./factorial(2)

y(0+h) is the output at instance x(0). were x(0) is calculated using taylor series. this has to repeat for all 't'.

[darova]

Maybe try this

result = x + (h.*dx) + (h^2.*dx2)./factorial(2);
result = result + h;            
% result = result - h;

[Elavarasi E]

Very much appreciate your prompt response.

here I need to find the value at instance (t+h) but result = result + h will give me output + h and not output at time (t+h).

i.e if output is 20 , t= 5 and h = 0.112

as per result = result + h; i'll get the result as 20.112 at time t = 5 but at the instance t = 5.112 it will be some 'xx' value.

I need this 'xx' value. this can be done only by varying both time and amplitude.

[darova]

So add it to t

x1 = x + (h.*dx) + (h^2.*dx2)./factorial(2);
t1 = t + h;
plot(t,x,t1,x1)

[Elavarasi E]

hi !

let 'n' be a sequence. it should be broken down into samples of length 1024.

i.e n/1024 = y spectrum.

i should plot this 'y' spectrum as overlapped spectrums with an overlap of 48 samples on each spectrum. the last spectrum can be padded with zeroes at the last.