How do you add a multiplication in the numerator of a transfer function.
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The equation I need to represent as a transfer function is 5*e-ts/s+5. I coded as follows but my numerator returns as + 5 i need it to be * 5. Is there a way to achieve this?
clc
clear all
num=[exp(-1) 5];
den=[1 5];
g=tf(num,den)
g =
0.3679 s + 5
------------
s + 5
0 件のコメント
回答 (2 件)
Star Strider
2023 年 4 月 16 日
編集済み: Star Strider
2023 年 4 月 16 日
Perhaps this —
num=5*exp(-1);
den=[1 5];
g=tf(num,den)
The exponential function here is a constant. If you intend to introduce a delay, use InputDelay or one of the other options.
EDIT — (16 Apr 2023 at 10:44)
The transfer function would be —
num = 5;
den = [1 5];
g = tf(num, den, 'InputDelay',1)
.
6 件のコメント
Walter Roberson
2023 年 4 月 17 日
heaviside(1) is 1.
num = 5;
den = [1 5];
g = tf(num, den, 'InputDelay',1)
Star Strider
2023 年 4 月 17 日
Of course, however it is a function of time, with the time argument not otherwise defined. This seems to be getting a bit less certain.
Walter Roberson
2023 年 4 月 16 日
s = tf('s');
g = 5 * exp(-t*s)/s + 5
You cannot use an undefined or symbolic variable in a tf().
Are you sure that you want to add 5 to the delayed signal? Rather than using (s+5) as the denominator ?
4 件のコメント
Walter Roberson
2023 年 4 月 17 日
If you apply a unity step to s, the result would be the same as putting in the restriction that s >= 0
Paul
2023 年 4 月 18 日
What does "Tau is a unity step" mean? Tau is place holder for a number, isn't it?
How can a unity step be applied to s? s is the independent variable of the Laplace transform. In that same context, what does "restriction that s>=0" mean?
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