PLEASE how to implement sigma in matlab
显示 更早的评论
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
2 个评论
Walter Roberson
2018-4-29
Please do not close questions that have an answer.
Rena Berman
2018-5-15
(Answers Dev) Restored edit
采纳的回答
Walter Roberson
2018-4-29
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)
2 个评论
Walter Roberson
2018-4-29
Remember to
syms k
Walter Roberson
2018-4-29
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
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Undirected Graphs 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
