# RLC Bandpass Filter network

Can anybody tell me what the "s" in this equation is? I am looking for the transfer function of this circuit

• Have you learned the Laplace transform? 's' is the usual variable in Laplace transform space. Possibly you could think of it as a complex frequency. – The Photon Mar 1 at 15:57

The meaning: with $$\omega_0=2\pi f_0$$ write your equation as $$(s^2+2\zeta\omega_0s +\omega_o^2)v_o=(2\zeta\omega_0 s)v_i$$ and then substitute $$s=\frac{d}{dt}$$ that is $$\frac{d^2 v_o}{dt^2} +2\zeta\omega_0 \frac{dv_o}{dt} +\omega_o^2v_o = 2\zeta\omega_0 \frac{dv_i}{dt}.$$ The "transfer function" is by definition the rational function $$H(s) = \frac{2\zeta\omega_0 s}{s^2+2\zeta\omega_0s +\omega_o^2}$$ as a function of the complex variable $$s=\sigma+\mathfrak{j}\omega$$
• $\Re[s]=\sigma, \Im[s]=\omega$; yes, you can do that and it will give you $H(\mathfrak{j} \omega)$ – hyportnex Mar 1 at 16:26