PLEASE how to implement sigma in matlab

there are two input signals x1[nT] and x2[nT] given, and sample spacing is given as T=0.1. i'm struggling to code x2[nT].

    t = -0.7:0.7;                
    T = 0.1;                            
    n = -0.7/T:0.7/T;                  % discrete-time values (from t=nT)
    x1=tan(pi*n/3)+2*exp(-0.8*abs(n));               
    k=-7:1:n;
    s=0.2.^abs(k);
    S=sum(s);
    x2=0.6.^abs(n)+S;

when i implement stem(n*T, x2), the graph shows wrong as the value k stops at -7 and doesn't move on. how can i fix this code? Thanks

[Walter Roberson]

Please do not close questions that have an answer.

[Rena Berman]

(Answers Dev) Restored edit

Answer 1

[Walter Roberson]

Your n is a vector. When you use k=-7:1:n then you have a vector on the right hand side. MATLAB says that if you have a vector on the left or right side of a ":" then the result is the same as if you used only the first element. So your statement is effectively k=-7:1:n(1) which is k=-7:1:(-0.7/.1) which is k=-7:1:-7 which is just -7 .

Note: 0.1 cannot be exactly represented in binary floating point, -0.7/.1 could come out slightly more negative than -7, leading to k=-7:1:(-7-something) which would be empty. You should avoid division in specifying a colon range. You would be better off with

   n = -7 : 1 : 7;
   T = 0.1;
   t = n * T;

As for your difficulty with k=-7:n : you are assuming that you have a vector of n, whereas the formula is for scalar n. But if you want to vectorize then,

n = -7:7;
syms N
subs(symsum(sym(0.2)^abs(k),k,-7,N), N, n)

[Walter Roberson]

Remember to

syms k

[Walter Roberson]

0.2^2 is less than 0.2^1, so sum(0.2.^[7 6 5 4 3 2 1 0]) must be strictly less than 8 * 0.2, which is 1.6 . So 1+(0.2^7 + 0.2^6 + ...+0.2^0) must be strictly less than 2.6, and nowhere near 11.

For 1+sum(x.^[7 6 5 4 3 2 1 0]) to be equal to 11, your x must be slightly greater than 1 -- about 1.06286984041761