How can I shift and add a discrete-time signal simultaneously

%Its good that you have both the signals of same length otherwise things would have been a bit different and we would have to %make signals of equal length first. But here i will try my best to cover both the aspects and please give my answer a vote if you %find it worthy.

%Let us first define new shifted range of x1.

n1=n-2;

%Now the approach I am going to explain can be used for both the cases i.e. where signals are of equal length or not.

%Define a signal 's1' of length same as the required output filled with zeros.

%Since the current signals are ranging from -8 to 8 (17 samples) but after shifting of one signal (n-2), its range will be from -10 %to 6,

%Thus our addition of signals will be from -10 to 6 for x1 after shifting 2 units and from -8 to 8 for x2. Therefore we must keep in %mind that our addition result y wil range from -10 to 8 overall.

%so we must define signals of this range.

n2= -10:8; %(simply speaking n1=min(min(n1),min(n)) : max(max(n1),max(n));)

s1=zeros(1,length(n2));

s2=zeros(1,length(n2));

%Now we shall be using find function to fill in the values at the indexes. For s1 the values will be of x1 with 2 units of left shift while %for s2 the values will be of x2 as it is since it is not being shifted.

s1(find(n2>=min(n1)&n2<=max(n1)==1))=x1;

%Here n2>=min(n1)&n2<=max(n1) is a logical condition which returns an array of 1 or 0 if the condition is true and the find %function finds those indices and simply assigns the value of x1 over there. This way we have successfully shifted the signal x1 %by 2 units leftwards using the logical condition of where to put values.

%Similarly

s2(find(n2>=min(n)&n2<=max(n)==1))=x2;

%Now since the dimensions of 2 signals are equal we can add our shifted and non shifted signal simultaneously.

y=zeros(1,length(n2));

y=s1+s2;

%To check the result:

figure;

subplot(4,1,1)

stem(n,x1)

title('x1[n]')

grid on

axis([-12 10 -5 5])

subplot(4,1,2)

stem(n,x2)

title('x2[n]')

grid on

axis([-12 10 -5 5])

subplot(4,1,3)

stem(n2,s1)

title('x1[n-2]')

grid on

axis([-12 10 -5 5])

subplot(4,1,4)

stem(n2,y)

title('y[n]')

grid on

axis([-12 10 -5 5])

% I hope I have answered your concern.

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Answer 2

Default 2020-6-23

1
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https://ww2.mathworks.cn/matlabcentral/answers/355460-how-can-i-shift-and-add-a-discrete-time-signal-simultaneously#answer_456058

编辑：Default 2020-6-23

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This is easier than you might think. The two signals have the same length and are already zero-padded. Just use circshift to shift x1 to the right by 2 (the sign is negative). Then add the two signals.

n  = -8:8;
x1 = [0 0 0 0 0 0 0 1 2 3 2 1 0 0 0 0 0];
x2 = [0 0 0 0 0 3 -3 2 -2 1 -1 0.5 -0.5 0 0 0 0];
yn = circshift(x1,2) + x2; % y[n] = x1[n-2]+x2[n]
stem(n,yn)
grid on

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Answer 3

Vishwas 2017-9-14

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/355460-how-can-i-shift-and-add-a-discrete-time-signal-simultaneously#answer_281614

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In order to simultaneously shift and add two discrete time signals, we can only make use of indices of x1 and x2 which can only be positive integers. So, for y[n] = x1[n-2]+x2[n], n should always be greater than 3.

Based on your example, we can consider the scenarios with positive values of n

n = 3:17;
y(n) = x1(n-2) + 2*x2(n)

or a we can write a function and build a resultant signal values

function result = cal(n)
      if n<3 || n>17
      result = 0;
      else
      result = x1(n-2)+2*x2(n);    
      end
  end
for i=1:max(size(x1))
      y(i) = cal(i);
end