Transfer function and Bode plot

[skrat]

Homework Statement

Find the transfer function for the filter on the attached picture and sketch it in Bode diagram, if you know $Q_0 =\frac{\omega _0L}{R}>>1$ and $\omega _0^2=\frac{1}{LC}$.

Homework Equations

The Attempt at a Solution

If I am not mistaken, we defined the transfer function as $$H(i\omega )=\frac{U_{out}(\omega )}{U_{in}(\omega )}$$
where $U_{out}(\omega )=IZ$. $$\frac 1 Z =\frac{1}{R+i\omega L}+\frac{1}{i\omega C}$$ $$Z=\frac{-\omega ^2 LC + i\omega RC}{R+i\omega (L+C)}$$ So finally $$H(i\omega )=\frac{U_{out}(\omega )}{U_{in}(\omega )}=\frac{Z}{Z+R}=\frac{-\omega ^2LC+iR\omega C}{R^2-\omega ^2LC+i\omega R(L+2C)}$$ But I highly suspect this is wrong from the very start. I have no idea how to solve this kind of problems.

 

[vela]

The calculation is almost right. Your expression for Z can't be right because it involves summing L and C, which you can't do because they don't have the same units. In your first expression for 1/Z, you're adding Z_C instead of 1/Z_C.

 

[skrat]

Thank you!

Well, the end result is not that important (Except that $1/Z_c$ part, that is important). I was just not sure if this: $$H(i\omega )=\frac{U_{out}(\omega )}{U_{in}(\omega )}=\frac{Z}{Z+R}$$ is right. I did the rest of the calculations using Mathematica.

 

[vela]

Yeah, that's fine. That's just a plain-old voltage divider. It holds for impedances, not just resistances.

 

[paranormal]

Your attempt at finding the transfer function is correct. However, in order to sketch it in a Bode plot, we need to plot the magnitude and phase of the transfer function as a function of frequency. This can be done by first converting the transfer function from complex form to polar form, and then plotting the magnitude and phase separately.

To convert the transfer function, we can use the fact that for a complex number in polar form, the magnitude is given by the absolute value of the complex number (i.e. the distance from the origin) and the phase is given by the angle of the complex number with respect to the real axis.

In this case, the magnitude of the transfer function is given by: $$|H(i\omega)|=\frac{\sqrt{(\omega RC)^2+(\omega^2LC-\omega R(L+2C))^2}}{\sqrt{R^2+(\omega LC)^2}}$$

And the phase is given by: $$\arg(H(i\omega))=\arctan\left(\frac{\omega RC}{\omega^2LC-\omega R(L+2C)}\right)-\arctan\left(\frac{R}{\omega LC}\right)$$

Once we have these equations, we can plot the magnitude and phase as a function of frequency on a logarithmic scale using the Bode plot format. This will give us a better understanding of how the filter behaves at different frequencies.

In summary, the transfer function and Bode plot are useful tools in analyzing and understanding the behavior of filters and other systems. By converting the transfer function to polar form and plotting the magnitude and phase, we can gain insights into how the system responds to different frequencies and make informed decisions about its design and performance.