Can anybody tell me what the "s" in this equation is? I am looking for the transfer function of this circuit
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1$\begingroup$ 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. $\endgroup$– The PhotonCommented Mar 1, 2020 at 15:57
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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 $s=\sigma+\mathfrak{j}\omega$
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$\begingroup$ Thank you for the answer! What is σ in this case? Given a real frequency, can we just set it to 0 and insert the frequency as ω=2πf? $\endgroup$ Commented Mar 1, 2020 at 16:22
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$\begingroup$ $\Re[s]=\sigma, \Im[s]=\omega$; yes, you can do that and it will give you $H(\mathfrak{j} \omega)$ $\endgroup$ Commented Mar 1, 2020 at 16:26