Frequency Response and Transfer Functions

Demonstrative Video


Introduction


Transfer Function Representation

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Frequency Response from TF

Obtain the TF \(V_o/V_s\) and its frequency response if \(v_s = V_m \cos\omega t\) image

\[\begin{aligned} &\mathbf{H}(\omega)=\frac{\mathbf{V}_{o}}{\mathbf{V}_{s}}=\frac{1 / j \omega C}{R+1 / j \omega C}=\frac{1}{1+j \omega R C} \\ &H=\frac{1}{\sqrt{1+\left(\omega / \omega_{0}\right)^{2}}}, \quad \phi=-\tan ^{-1} \frac{\omega}{\omega_{0}}, \quad \omega_{0}=1 / R C . \end{aligned}\]

\[\begin{aligned} H&=\frac{1}{\sqrt{1+\left(\omega / \omega_{0}\right)^{2}}} \\ \phi & =-\tan ^{-1} \frac{\omega}{\omega_{0}} \end{aligned}\]

\(\omega/\omega_0\) \(\mathbf{H}\) \(\mathbf{\phi}\) \(\omega/\omega_0\) \(\mathbf{H}\) \(\mathbf{\phi}\)
0 1 0 10 0.1 -84\(^\circ\)
1 0.71 -45\(^\circ\) 20 0.05 -87\(^\circ\)
2 0.45 -63\(^\circ\) 100 0.01 -89\(^\circ\)
3 0.32 -72\(^\circ\) \(\infty\) 0 -90\(^\circ\)
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The Decibel Scale

image \[\begin{aligned} G_{\mathrm{dB}} &=10 \log _{10} \frac{P_{2}}{P_{1}}=10 \log _{10} \frac{V_{2}^{2} / R_{2}}{V_{1}^{2} / R_{1}} \\ &=10 \log _{10}\left(\frac{V_{2}}{V_{1}}\right)^{2}+10 \log _{10} \frac{R_{1}}{R_{2}} \\ G_{\mathrm{dB}} &=20 \log _{10} \frac{V_{2}}{V_{1}}-10 \log _{10} \frac{R_{2}}{R_{1}} \end{aligned}\]

\[\begin{aligned} \text{if}~R_1 & = R_2 \\ &\boxed{G_{\mathrm{dB}} =20 \log _{10} \frac{V_{2}}{V_{1}}} \\ \text{Alternatively}~P_1 & = I_1^2R_1 \quad P_2 = I_2^2R_2 \\ &\boxed{G_{\mathrm{dB}} =20 \log _{10} \frac{I_{2}}{I_{1}}} \end{aligned}\]

\[\boxed{G_{\mathrm{dB}} = 10 \log_{10}\dfrac{P_2}{P_1}} \quad \boxed{G_{\mathrm{dB}} =20 \log _{10} \frac{V_{2}}{V_{1}}} \quad \boxed{G_{\mathrm{dB}} =20 \log _{10} \frac{I_{2}}{I_{1}}}\] Important Points:


Bode Plots