How to find transfer function to a second order ODE having a constant term?

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PONNADA
PONNADA 2024-2-29
编辑： PONNADA 2024-3-2
and are constants.
Transfer function
[edited by @Sam Chak]
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Paul
Paul 2024-3-2
As far as I can tell, there is no transfer function of the said problem, so there also is not characteristic of that transfer function.
If you set c1 = 0, then the the system is LTI and you can fnd the transfer function of that system and the characteristic equation of that transfer function.
Are you sure that the problem statement is correct as written?

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Sam Chak
Sam Chak 2024-3-2
Ran in:
Hi @PONNADA
This type of differential equation is commonly encountered in examples involving an ideal undamped mass-spring system subjected to an input force and constant gravitational acceleration. The equation can be rearranged and expressed as a 2-input, 1-output state-space system.
When you transform the state-space system into transfer function form, you'll obtain two transfer functions because the system's response is influenced by two external inputs: one from the manipulatable force and the other from the constant effect of gravity.
You can observe somewhat similar dynamics in a free-falling object:
https://wethestudy.com/mathematics/free-falling-bodies-differential-equations/
c2 = 2;
A = [0 1;
-c2 0];
B = [0 0;
1 -1];
C = [1, 0];
D = 0*C*B;
%% State-space system
sys = ss(A,B,C,D, 'StateName', {'Position' 'Velocity'}, ...
'InputName', {'Force', 'Constant'}, 'OutputName', 'MyOutput')
sys = A = Position Velocity Position 0 1 Velocity -2 0 B = Force Constant Position 0 0 Velocity 1 -1 C = Position Velocity MyOutput 1 0 D = Force Constant MyOutput 0 0 Continuous-time state-space model.
%% Transfer functions
G = tf(sys)
G = From input "Force" to output "MyOutput": 1 ------- s^2 + 2 From input "Constant" to output "MyOutput": -1 ------- s^2 + 2 Continuous-time transfer function.
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Sam Chak
Sam Chak 2024-3-2
@PONNADA: ... by removing the constant term from the given equation, and if I find the transfer function, will it affect the characteristic equation?
No, it won't affect the characteristic equation. As you can see above, both transfer functions have the same characteristic polynomial, , due to there being only 1 degree of freedom in this single mass-spring system.
By taking the Laplace transform of the differential equation with zero initial conditions, we establish the relationship between the inputs and output.
To determine how the output behaves in response to a given input, we must determine the transfer functions: one from force input to plant output, and the other from constant gravity to output.
PONNADA
PONNADA 2024-3-2
编辑：PONNADA 2024-3-2
What a wonderful explanation you provided, sir. Your patience and commitment are evident in the response you provided. Thank you so much. It is really helpful to me.

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Alexander
Alexander 2024-2-29
编辑：Alexander 2024-2-29
Are you looking for something like this:
AnregeAmpl = 1; % mm
fmax = 20; % Hz
f=0.1:0.1:fmax; % Hz
w=2*pi*f; % 1/s
s=ones(size(f))*AnregeAmpl; % mm
c=200000; % N/m
m=20; % kg
d=100; % Ns/m
x=(c+d*j*w)./(c+d*j*w+j*w.*j.*w.*m);
plot(f,abs(x),f,abs(s));grid minor
Replace jw by S and you have your transfere function in S. But maybe your professor is not amused about this. ;=)

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