I have undamped oscilator which is described with following equation:
$$ \ddot{y}+\omega^2y=u $$
I need to find transfer function $$ H(s) $$ and freguency transfer function for $$ H(j\omega) $$
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I found transfer function, but I don't know how to make Frequency transfer function. $$ H(s)=\frac{1}{s^2+\omega^2} $$
If I use this equation $$ p =j\omega $$ i get this:
$$ H(s)=\frac{1}{(j\omega)^2+\omega^2}= \frac{1}{-\omega^2+\omega^2}$$ Ang I guess this is not right result.
Thank you for advice.
To find:
$$\text{H}\left(\text{s}\right)=\frac{\text{Y}\left(\text{s}\right)}{\text{U}\left(\text{s}\right)}$$
Use Laplace transform:
$$\text{y}''(t)+\omega_\text{k}^2\cdot\text{y}(t)=\text{u}(t)\to\text{s}^2\text{Y}\left(\text{s}\right)-\text{s}\cdot\text{y}(0)-\text{y}'(0)+\omega_\text{k}^2\cdot\text{Y}\left(\text{s}\right)=\text{U}\left(\text{s}\right)$$
When we set all the initial conditions equal to $0$:
$$\text{s}^2\cdot\text{Y}\left(\text{s}\right)+\omega_\text{k}^2\cdot\text{Y}\left(\text{s}\right)=\text{U}\left(\text{s}\right)\Longleftrightarrow\frac{\text{Y}\left(\text{s}\right)}{\text{U}\left(\text{s}\right)}=\frac{1}{\omega_\text{k}^2+\text{s}^2}$$
Where $\omega_\text{k}$ is the constant omega.
When we set $\text{s}=\text{j}\omega$ where $\text{j}^2=-1$:
$$\text{H}\left(\text{s}\right)=\frac{\text{Y}\left(\text{s}\right)}{\text{U}\left(\text{s}\right)}=\frac{1}{\omega_\text{k}^2+\left(\text{j}\omega\right)^2}=\frac{1}{\omega_\text{k}^2-\omega^2}$$