I keep getting A(I) = B, the number of elements in B and I must be the same. matlab error

Question

John 2013-11-29

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评论： Image Analyst 2013-11-29

在 MATLAB Online 中打开

last=0;

upflag=0;

p=zeros(length(detsq),1);

for i=1:length(detsq)

if(detsq(i)>1000)

 if(upflag ==0)
 if(last>0)
 t= i-last;
 p=1000/t*60
 end
 last=i;
 end
 upflag=100;
 else
     if(upflag>0)
         upflag=upflag-1;
     end
 end
 pulse(i)=p;
end
figure(3),plot(pulse)

In an assignment A(I) = B, the number of elements in B and I must be the same

This is the error keep popping. Any experts who can help me identify where I went wrong?

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Answer 1

Image Analyst 2013-11-29

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Is it your intention that p should be a vector, like you said here: p=zeros(length(detsq),1);

or is it your intention that p should be a scalar like you say here: p=1000/t*60

If you do this: pulse(i)=p; then p must be a scalar. Though I'm just guessing at the line of code producing the error because, for some strange reason, you left out that crucial part of the error message that identifies the line of code that caused the error.

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John 2013-11-29

在 MATLAB Online 中打开

%Step 8

det =filter (b2, 1, firs6);

figure(17),plot (det)

detsq=det .^2;

figure(18),plot(detsq)

last=0;

upflag=0;

p=zeros(length(detsq),1);

for i=1:length(detsq)
 if(detsq(i)>0.1)      %threshold
 if(upflag ==0)
 if(last>0)
 t= i-last;
 p=1000/t*60           %pulse rate
 end
 last=i;
 end
 upflag=100;
 else
     if(upflag>0)
         upflag=upflag-1;
     end
 end
 pulse(i)=p;
end
figure(19),plot(pulse)
I run again and line 25 which is pulse(i)=p the cause of the error. I show my full code for the whole program.

Image Analyst 2013-11-29

在 MATLAB Online 中打开

Undefined function or variable 'b2'.
Error in test (line 3)
det =filter (b2, 1, firs6);

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Answer 2

John 2013-11-29

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在 MATLAB Online 中打开

%step 3

function output = myFIR1(input,hn)

global in

in = [in input];

output =0;

r =(length(hn))*(length(in)>=length(hn))+(length(in))*(length(in)<length(hn));

    for C = 0:r-1 
       output = output+(in(length (in)-C)*hn(C+1));
    end

end

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Answer 3

John 2013-11-29

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编辑：John 2013-11-29

load 'ecg_8.dat'

y=ecg_8(:,2);

%Step 2

A=ecg_8(:,2)-mean(ecg_8(:,2));

fs=1000; %Sampling Frequency

t=0:1/fs:(length(ecg_8)-1)/fs;

B=((4.096+4.096)/4096)/1000; %converts data to millivolts

plot(t*1000,A*B*1000)

y2=fft(A);

L=length(y2);

freq=(0:fs/(L-1):fs);

plot(freq,abs(y2));

axis([0 500 0 3000000])

%step 4

n=(-100:100); %Set number of taps

w1=2*pi*0.03; %Cut-off frequency 1

w2=2*pi*0.08; %Cut-off frequency 2

h4=(1./(n*pi)).*(sin(w1*n)-sin(w2*n)); %Impulse Response

h4(101)=1+(w1-w2)/pi; %Sets middle value

h4=h4 .*blackman(201)'; %Mutilpied by Blackman function

figure(1);subplot(2,1,1),plot(20*log10(abs(fft(h4)))); %Plot in decibel against frequency

axis([-50 1050 -30 10]);

subplot(2,1,2),plot(h4);

for row=1:length(A)

firs4(row)=myFIR1(A(row),h4);

end

clearvars in row

figure(2),plot(firs4),title('Step 4 Impulse Response');

%step 5

n1=(-200:200);

ws=2*pi*0.008;

h5=-(sin(ws.*n1)./(n1*pi));

h5(201)=1-(ws/pi);

h5=h5 .*blackman(401)';

figure(3),subplot(2,1,1),plot(20*log10(abs(fft(h5))));

axis([-50 1050 -0.1 1.1]);

subplot(2,1,2),plot(h5);

for row=1:length(firs4)

firs5(row)=myFIR1(firs4(row),h5);

end

clearvars in row

figure(4),plot(firs5),title('Step 5')

xlabel('time (ms)'),ylabel('voltage(mv)')

%step 6

h6(1:501)=1;

h6(1:2)=0;

h6(500:501)=0;

h6(36:66)=0;

h6(501-65+1:501-35+1)=0;

plot(h6)

h=real(ifft(h6));

plot(h)

h2(1:251)=h(251:501);

h2(251:501)=h(1:251);

plot(h2);

h3=h2.*blackman(501)';

figure(5),plot(h3),title('Impulse Response');

xlabel('Frequency (Hz)'),ylabel('Ampltitude');

for row=1:length(firs5)

firs6(row)=myFIR1(firs5(row),h3);

end

clearvars in row

figure(6),plot(firs6),title('Step 6')

xlabel('Time(ms)'),ylabel('Voltage(mv)')

%Step 7

H=(A)*(B)*(1000);

hb1=H(4000:4600);

figure(7),plot(hb1)

b=flipud(hb1);

figure(8),plot(b) %Original Graph

xlabel('Frequency (Hz)'),ylabel('Amplitude'),title('Original Hb')

det=filter(b,1,H); %Original det vs Orignal Hb

figure(9),plot(det)

det=filter(b,1,firs5); %Original vs Step 5

figure(10),plot(det)

det=filter(b,1,firs6); %Original vs Step6

figure(11),plot(det)

w=firs6;

plot(w)

hb2=w(2200:2800);

figure(12),plot(hb2)

b2=fliplr(hb2);

figure(13),plot(b2) %Perfect Graph

det=filter(b2,1,H);

figure(14),plot(det) %Perfect vs original

det=filter(b2,1,firs5);

figure(15),plot(det) %Perfect vs step 5

det=filter(b2,1,firs6);

figure(16),plot(det) %Perfect vs step 6

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John 2013-11-29

编辑：John 2013-11-29

Sir, what is the exact problem that causes the program not able to run the last step? Can you let me know which one needs to change? Thank you!

Image Analyst 2013-11-29

You can't have negative or fractional indexes, OR stuff a bunch of elements into a single element. Step through your code until it throws an error, then examine all the variables and see what violates what I said in the first sentence of this comment.

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