# Problem multiplying exponential function with negative value

조회 수: 54 (최근 30일)
Hasib Ryan Rahman . 2022년 2월 11일
편집: Voss . 2022년 2월 11일
I want to plot x(t) = t*(e−0.15t), − 20 ≤ t ≤ 20
I've tried:
t = -20:1:20;
x(t) = t.*(exp(-0.15*t));
Array indices must be positive integers or logical values.
plot(t,x(t))

댓글을 달려면 로그인하십시오.

### 채택된 답변

Voss 2022년 2월 11일
This statement
x(t) = t.*(exp(-0.15*t));
says to set the value of x at indexes given by t to the expression on the right-hand side. Indexes must be positive integers (1, 2, 3, ...), but t has some negative values and a zero value in it, so that's why the error happens.
In fact the syntax for what you want to do is slightly simpler. You can just say:
t = -20:1:20;
% no need to specify indexes into x, just set the whole thing:
x = t.*(exp(-0.15*t));
Then to plot:
plot(t,x) ##### 댓글 수: 1없음 표시없음 숨기기
Voss 2022년 2월 11일
편집: Voss 님. 2022년 2월 11일
That makes t and x each vectors. Alternatively, you can make x a function of t if you want, and do the same thing:
x = @(t)t.*(exp(-0.15*t));
In this case the syntax x(t) evaluates the function x at the values of t given in t. Then your original syntax in the plot statement would be correct:
t = -20:1:20;
plot(t,x(t)); 댓글을 달려면 로그인하십시오.

### 추가 답변 (1개)

Yongjian Feng 2022년 2월 11일
Try this:
t = -20:1:20;
x_t = t.*(exp(-0.15*t));
plot(t, x_t)

댓글을 달려면 로그인하십시오.

### 카테고리

Help CenterFile Exchange에서 Logical에 대해 자세히 알아보기

R2015a

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!