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Im not sure how to find the frequency response (\$H(jw)=V_o/Vin\$) of this circuit

Can anybody help please?

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    \$\begingroup\$ Is this a homework question? Anyway, show some more work and write the KVL equations of the circuit (node voltage), if you still don't get the answer, then point a more specific difficulty you are having. \$\endgroup\$ – jDAQ Dec 15 '19 at 17:53
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    \$\begingroup\$ Stimulate it!!! \$\endgroup\$ – Leon Heller Dec 15 '19 at 18:28
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Well, we have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Now, we know a few things:

  • I will do the analysis in the (complex) s-domain (using Laplace transform);
  • The input impedance is given by: $$\text{z}_\text{in}\left(\text{s}\right)=\frac{\text{R}_1\left(\text{R}_2+\text{sL}+\frac{\text{R}_3\frac{1}{\text{sC}}}{\text{R}_3+\frac{1}{\text{sC}}}\right)}{\text{R}_1+\text{R}_2+\text{sL}+\frac{\text{R}_3\frac{1}{\text{sC}}}{\text{R}_3+\frac{1}{\text{sC}}}}=\frac{\text{R}_1\left(\text{R}_2+\text{sL}+\frac{\text{R}_3}{1+\text{sC}\text{R}_3}\right)}{\text{R}_1+\text{R}_2+\text{sL}+\frac{\text{R}_3}{1+\text{sC}\text{R}_3}}\tag1$$
  • The output voltage is given by: $$\text{v}_\text{out}\left(\text{s}\right)=\text{i}_\text{out}\left(\text{s}\right)\cdot\text{z}_\text{out}\left(\text{s}\right)\tag2$$
  • The output impedance is given by: $$\text{z}_\text{out}\left(\text{s}\right)=\frac{1}{\text{sC}}\tag3$$
  • The output current can be written in terms of the input current, as follows: $$ \begin{cases} \text{i}_2\left(\text{s}\right)=\frac{\text{R}_1}{\text{R}_1+\text{R}_2+\text{sL}+\frac{\text{R}_3}{1+\text{sC}\text{R}_3}}\cdot\text{i}_\text{in}\left(\text{s}\right)\\ \\ \text{i}_\text{out}\left(\text{s}\right)=\frac{\text{R}_3}{\text{R}_3+\frac{1}{\text{sC}}}\cdot\text{i}_2\left(\text{s}\right) \end{cases}\tag4 $$
  • The input current can be rewritten as: $$\text{i}_\text{in}\left(\text{s}\right)=\frac{\text{v}_\text{in}\left(\text{s}\right)}{\text{z}_\text{in}\left(\text{s}\right)}\tag5$$

Now, we can write:

$$\mathcal{H}\left(\text{s}\right):=\frac{\text{v}_\text{out}\left(\text{s}\right)}{\text{v}_\text{in}\left(\text{s}\right)}\tag6$$

Using the information from above we get:

$$\mathcal{H}\left(\text{s}\right)=\frac{\text{R}_3}{\text{R}_3+\left(\text{R}_2+\text{sL}\right)\left(1+\text{sC}\text{R}_3\right)}\tag7$$


Using your values we get:

$$\mathcal{H}\left(\text{s}\right)=\frac{2000}{3600+\text{s}\left(170+\text{s}\right)}\tag8$$

So amplitude:

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\frac{2000}{\sqrt{\omega^2\left(21700+\omega^2\right)+12960000}}\tag9$$

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    \$\begingroup\$ Hey @Jan, nice work. But the main reasons I only gave a hint are, this is possibly a homework question and also the poster did not show any attempt at it (equations and such). It would be very tempting to a student to just copy your solution and not actually do it on their own. \$\endgroup\$ – jDAQ Dec 15 '19 at 20:37
  • \$\begingroup\$ @jDAQ, yes, that's exactly what will happen. \$\endgroup\$ – Chu Dec 15 '19 at 20:50
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    \$\begingroup\$ @Jan that's quite a handful of equations. Must've taken you quite some time to put it all together: not only derive all the math, but also code that in your favourite formula editor. And it's still just three hours since the original question. You have my respect :-) And, me not being an EE graduate, the detailed recipe is certainly helpful to me. Just throwing this into qucs would be so easy, but I wouldn't quite understand what's going on. \$\endgroup\$ – frr Dec 15 '19 at 20:50
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This transfer function can be determined using the fast analytical techniques without resorting to a complicated KVL/KCL analysis. These FACTs derive from an original work of Hendrik Bode and were improved over the years, especially with publications from the late Dr. Middlebrook. In this particular example in which there is no zero, you start with \$s=0\$ where the inductor is replaced by a short circuit and the capacitor is open-circuited. In this mode, by inspection, you "see" a dc gain (an attenuation actually) equal to \$H_0=\frac{R_2}{R_2+R_1}\$. The left-side resistance across the source does not play a role for the transfer function \$\frac{V_{out}}{V_{in}}\$.

Then, remove the excitation (set the source to 0 V - a short circuit) and "look" through the energy-storing elements's connecting terminals to determine the resistance \$R\$ driving the inductor or the capacitor. This resistance associated with the capacitor or the inductor will form the natural time constants of this circuit. The below drawing shows how to do this:

enter image description here

Once you have the time constants, you associate them the following way to form the denominator: \$D(s)=1+s(\tau_1+\tau_2)+s^2(\tau_1\tau_{12})\$. The Mathcad sheet shows how to do this and gives the ac response of the circuit. Some more manipulations will formalize the response and put it under a known canonical form and that is what you must learn to do:

enter image description here

There is no resonance in this circuit as the quality factor \$Q\$ is below 0.5 and the poles are real. Considering this low \$Q\$, the transfer function could be further approximated with two cascaded poles. Please note that equation (7) in Jan's reply is not a low-entropy form where the poles clearly appear.

In this example, FACTs are the fastest way to go without mistake (the intermediate sketches can be quickly individually fixed) and without algebra (inspection means look at the schematic and determine the resistance in your head). You obtain the response in a low-entropy form where the gain (or attenuation) appears in the leading term \$H_0\$ while it is easy to rewrite the denominator in its 2nd-order normalized form.

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Write the node voltage equations for the circuit. And remenber that the impedance of a inductor is, $$ Z_L = j \omega L $$ and the impedance of a capacitor is, $$ Z_C = -\frac{j }{\omega C}. $$

That should lead you to the transfer function, that is the \$\frac{V_{out}}{V_{in}}\$.

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  • \$\begingroup\$ Good answer +1 Alternatively you could also turn this circuit into the Laplace domain as well but you're effectively doing the same thing. \$\endgroup\$ – KingDuken Dec 15 '19 at 18:17

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