ryan88
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Hi,
I am trying to derive the general transfer function for a second order dynamic system, shown below:
\frac{Y(s)}{X(s)}=\frac{K\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}
In order to do this I am considering a mass-spring-damper system, with an input force of f(t) that satisfies the following second-order differential equation:
m\frac{d^2y}{\dt^2}+c\frac{dy}{dt}+ky=f(t)
Using the following two relationships:
c=2\zeta\omega_nm
\frac{k}{m}=\omega_n^2
I get this:
\frac{d^2y}{dt^2}+2\zeta\omega_n\frac{dy}{dt}+\omega_n^2y=\frac{f(t)}{m}
\mathcal{L}\left\{\frac{d^2y}{dt^2}\right\}+2\zeta\omega_n\mathcal{L}\left\{\frac{dy}{dt}\right\}+\omega_n\mathcal{L}\left\{y\right\}=\frac{1}{m}\mathcal{L}\left\{f(t)\right\}
Y(s)\left[s^2+2\zeta\omega_ns+\omega_n^2\right]=\frac{F(s)}{m}
\frac{Y(s)}{F(s)}=\frac{1}{m(s^2+2\zeta\omega_ns+\omega_n^2)}
Wheras my lecturer has the following in his notes:
\frac{d^2y}{dt^2}+2\zeta\omega_n\frac{dy}{dt}+\omega_n^2y=K\omega_n^2x(t)
\mathcal{L}\left\{\frac{d^2y}{dt^2}\right\}+2\zeta\omega_n\mathcal{L}\left\{\frac{dy}{dt}\right\}+\omega_n\mathcal{L}\left\{y\right\}=K\omega_n^2\mathcal{L}\{x(t)\}
Y(s)\left[s^2+2\zeta\omega_ns+\omega_n^2\right]=K\omega_n^2X(s)
\frac{Y(s)}{X(s)}=\frac{K\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}
This obvisously gives the correct transfer function. So, from the two approaches, I have come to the conclusion that:
\frac{f(t)}{m}=K\omega_n^2x(t)
But I do not understand the physical reasoning behind this. Can anyone offer any help with this?
Thanks,
Ryan
I am trying to derive the general transfer function for a second order dynamic system, shown below:
\frac{Y(s)}{X(s)}=\frac{K\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}
In order to do this I am considering a mass-spring-damper system, with an input force of f(t) that satisfies the following second-order differential equation:
m\frac{d^2y}{\dt^2}+c\frac{dy}{dt}+ky=f(t)
Using the following two relationships:
c=2\zeta\omega_nm
\frac{k}{m}=\omega_n^2
I get this:
\frac{d^2y}{dt^2}+2\zeta\omega_n\frac{dy}{dt}+\omega_n^2y=\frac{f(t)}{m}
\mathcal{L}\left\{\frac{d^2y}{dt^2}\right\}+2\zeta\omega_n\mathcal{L}\left\{\frac{dy}{dt}\right\}+\omega_n\mathcal{L}\left\{y\right\}=\frac{1}{m}\mathcal{L}\left\{f(t)\right\}
Y(s)\left[s^2+2\zeta\omega_ns+\omega_n^2\right]=\frac{F(s)}{m}
\frac{Y(s)}{F(s)}=\frac{1}{m(s^2+2\zeta\omega_ns+\omega_n^2)}
Wheras my lecturer has the following in his notes:
\frac{d^2y}{dt^2}+2\zeta\omega_n\frac{dy}{dt}+\omega_n^2y=K\omega_n^2x(t)
\mathcal{L}\left\{\frac{d^2y}{dt^2}\right\}+2\zeta\omega_n\mathcal{L}\left\{\frac{dy}{dt}\right\}+\omega_n\mathcal{L}\left\{y\right\}=K\omega_n^2\mathcal{L}\{x(t)\}
Y(s)\left[s^2+2\zeta\omega_ns+\omega_n^2\right]=K\omega_n^2X(s)
\frac{Y(s)}{X(s)}=\frac{K\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}
This obvisously gives the correct transfer function. So, from the two approaches, I have come to the conclusion that:
\frac{f(t)}{m}=K\omega_n^2x(t)
But I do not understand the physical reasoning behind this. Can anyone offer any help with this?
Thanks,
Ryan