Embarrassing Laplace question using ilaplace

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Robert
Robert 2017 年 1 月 16 日
回答済み: Star Strider 2017 年 1 月 20 日
I always get stuck using ilaplace And then 6 months goes by and i get stuck again because i don't use it often enough
Im just trying to plot the step response of a first order system using the step command which works fine as always. Just for kicks i want to take the same function take the inverse laplace function and plot it against t manually They should yield the EXACT same answer but yet it doesn't.
Can anyone see my mistake?
clear all
clc
syms R C S
R = linspace(1,10000,5);
C = .0001;
for r = 1:5
Vout_in_t = ilaplace(5/(R(r)*C*S+1))
ezplot(Vout_in_t,[0,2])
hold on
my = tf([1], [1 1/(R(r)*C)])
figure
opt = stepDataOptions
opt.StepAmplitude = 5
step(my,opt)
hold on
end

回答 (2 件)

Jyotish Robin
Jyotish Robin 2017 年 1 月 20 日
編集済み: Jyotish Robin 2017 年 1 月 20 日
Hi Robert,
From the given code , what I understood is that the transfer function you are trying to use is the one below:
S/(1+RCS)
In that case , you have to rewrite the line which defines your transfer function as:
>>my = tf([1 0], [(R(r)*C) 1])
Hope this helps!

Star Strider
Star Strider 2017 年 1 月 20 日
If you want to see the response to the step input, you need to convolve it with the Laplace transform of the unit step, (1/s).
See if this does what you want:
syms R C s
R = linspace(1,10000,5);
C = .0001;
for r = 1:5
Vout_in_t = ilaplace(5/(R(r)*C*s+1) * 1/s)
ezplot(Vout_in_t,[0,2])
hold on
my = tf([1], [1 1/(R(r)*C)])
figure
opt = stepDataOptions
opt.StepAmplitude = 5
step(my,opt)
end
hold off

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