MATLAB: How to reverse,scale and shift an acquired signal

@signalreversing#signalacquisition#signals processing#signalsandsystems#signalscaling#signalshifting#signalsproject

i want a concise code to perform simple operations on an acquired signal which is in cvv format.

Best Answer

    % code
clc;
clear all;
close all;
s = xlsread('C:\Users\ADMIN\Desktop\DSO00001.csv');
x = s(:,1);
y = s(:,2);
x1 = 0;
y1 = 0;
c = 0;
disp('1)Time Scale:');
disp('2)Time Shift:');
disp('3)Time Reversal:');
disp('4)Scale,Shift and Reversal:');
while((c >=5 || c<=0))
c = input('Enter choice: ');
switch(c)
    case 1
        a = input('Enter Scaling factor: ');
        x1 = x/a;
        y1 = y;
    case 2
        b = input('Enter shifting factor: ');
        x1 = (x-b);
        y1 = y;
    case 3
        x1 = (-1)*x;
        y1 = y;
    case 4
        a = input('Enter Scaling factor: ');
        b = input('Enter shifting factor: ');
        x1 = (-1)*(x-(b*a))/a;
        y1 = y;
     otherwise
        disp('Invalid Input');
end
if(c>0 && c<5)
    subplot(1,2,1)
    plot(x,y)
    xlabel('t')
    ylabel('signal')
    grid on
    title('Original Signal:')
    subplot(1,2,2)
    plot(x1,y1)
    xlabel('t')
    ylabel('signal')
    grid on
    title('Resultant Signal:')
    hold on;
end
end
%given below is the output for reversing%

Related Solutions

MATLAB: Verification of system characteristics

% FOR TIME VARITANT OR INVARIANT
clc
clear all
close all
t=-5:0.001:5
l=length(t)
for i=1:l
    if t(i)>=0
        x1(i)=1;
        
    else x1(i)=0;
    end
end
x2=cos(pi*t)
x3=cos(pi*(t-2))
y1=x1.*x2
y2=x1.*x3
subplot(4,1,1)
plot(t,x1)
xlabel('t')
ylabel('x1')
title('for x1')
grid on
gtext('1841014009')
subplot(4,1,2)
plot(t,y1)
xlabel('t')
ylabel('y1')
title('for y1')
grid on
gtext('1841014009')
subplot(4,1,3)
plot(t,x2)
xlabel('t')
ylabel('x2')
title('for x2')
grid on
gtext('1841014009')
subplot(4,1,4)
plot(t,y2)
xlabel('t')
ylabel('y2')
title('for y2')
grid on
gtext('1841014009')
if round(y1)==round(y2)
    disp('time invariant');text(-2,-0.5, ('System Is Time Invariant'));
    
else disp('time variant');text(-2,-0.5, ('System Is Time Variant'))
end
% FOR LINEARITY TEST
clc
clear all
close all
t=-4:0.001:5
l=length(t)
for i=1:l
    if t(i)>=0
        x1(i)=1;
        
    else x1(i)=0;
    end
end
 
for i=1:l
if t(i)>=1
    x2(i)=1;
    
else x2(i)=0;
end
end
 
a1=2
a2=3
 
y1=a1.*x1.*cos(pi*t)
y2=a2.*x2.*cos(pi*(t-1))
 
y=((a1.*x1)+(a2.*x2)).*cos(pi*t)
 
y3=y1+y2;
 
subplot(4,1,1)
plot(t,y)
title('for y')
xlabel('t')
ylabel('y')
grid on
gtext('1841014009')
subplot(4,1,2)
plot(t,y1)
title('for y1')
xlabel('t')
ylabel('y1')
grid on
gtext('1841014009')
 
subplot(4,1,3)
plot(t,y2)
title('for y2')
xlabel('t')
ylabel('y2')
grid on
gtext('1841014009')
  
subplot(4,1,4)
plot(t,y3)
title('for y3')
xlabel('t')
ylabel('y3')
grid on
gtext('1841014009')
 
if round(y)==round(y3)
    disp('system is linear');
    text(-2,-0.5, ('System Is Linear'));
    
else disp('system is non linear');
    text(-2,-0.5, ('System Is Non-Linear'));
end
% FOR STABILITY TEST
clc
clear all
close all
t=0:0.01:10
l=length(t)
for i=1:l
    if t(i)>=0
        x(i)=1;
        
    else x(i)=0;
    end
end
 
y=exp(-2*t).*x
c=0
for i=1:l-1
    if (y(i)>y(i+1))
        c=c+1;
    end
end
 
subplot(2,1,1)
plot(t,x)
title('for x')
xlabel('t')
ylabel('x')
grid on
gtext('1841014009')
subplot(2,1,2)
plot(t,y)
title('for y')
xlabel('t')
ylabel('y')
grid on
gtext('1841014009')
if(c~=0)
    disp('system is stable');
    text(2,0.5, ('System Is Stable'));
else 
    disp('system is unstable');
    text(2,0.5, ('System Is Untable'));
end

MATLAB: My code says too many output arguments

Give your functions outputs:

newton
function newton
x0 = input('Enter the initial guess value, x: ');
tolerance = input('Enter the tolerance value, tol: ');
max_i = input('Enter the maximum no. of iteration, max_i: ');
r = input('Enter the value of relative roughness, e/D: ');
RN = input('Enter the Reynolds number, RN: ');
f = input('Enter the Friction factor, f: ');
c = colebrook/colebrook_deriv;
x=x0;
for i=1:max_i
    y = x;
    x = y - (c);
    if abs(x-y)<= tolerance
        disp(['root = ', num2str(x)])
        break
    end
end
fprintf('The number of iterations: %d\n', i)
    function CFd = colebrook
        %%Formula for C(fd)
        CFd =(1/sqrt(f)) + 2*log10(r/(3.7)+ 2.51/(RN*sqrt(f)));
    end
    function CFd1 = colebrook_deriv
        CFd1 = 1/(2*f^(3/2)) + 2.51/RN*f^(3/2)*log(10)*(r/3.7) + 2.51/(RN*f^(1/2));
    end
end

Also check the function for colebrook_deriv i had to make a change (missing '*' i guess) tha you should correct if needed. Problems here:

CFd1 = 1/(2*f^(3/2)) + 2.51/RN*f^(3/2)log(10)(r/3.7) + 2.51/(RN*f^(1/2));
%                                     ^      ^
%                                     |      |
%                                      ----------- Check here

Also i added the final end, because i suspect that you wanted to have nested functions.

Best Answer

Related Solutions

MATLAB: Verification of system characteristics

MATLAB: My code says too many output arguments

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