# Unsure about RLC low pass transfer function

I'm unsure about the RLC low-pass filter transfer and frequency response functions I've been trying to calculate. I need $$H(s)=\frac{Y(s)}{X(s)}$$ where x(t) is Vin and y(t) is Vr. I calculated $$\frac{d^2v_c(t)}{dt^2}+\frac{1}{LC}v_c(t)=\frac{1}{LC}x(t)$$or$$H(s)=\frac{1}{s^2LC+1}$$

This transfer/frequency response has been giving me the correct magnitudes but obviously doesn't give a phase angle. Have I calculated the transfer function correctly in this instance? Any assistance would be appreciated.

• Your TF is incorrect. It should have R in the formula. Oct 26 '19 at 14:29
• Why do I need the R? Doesn't V_in = V_L + V_c? Oct 26 '19 at 15:02

Have I calculated the transfer function correctly in this instance?

The transfer function starts from the potential divider equation for the components: -

$$\dfrac{\dfrac{\frac{R}{sC}}{R + \frac{1}{sC}}}{\dfrac{\frac{R}{sC}}{R + \frac{1}{sC}}+sL} \Longrightarrow \dfrac{R}{R+s^2RLC+sL}$$

And when you drill down it becomes this: -

$$\dfrac{\dfrac{1}{LC}}{s^2+s\dfrac{1}{CR}+\dfrac{1}{LC}}$$

So immediately you can see that your first error is in not including "R" in the formula.

This transfer/frequency response has been giving me the correct magnitudes but obviously doesn't give a phase angle.

Only the formula above will give you the correct magnitudes at all frequencies. And, it does give you the phase angle. You need to substitute s with jw: -

$$H(j\omega) = \dfrac{\dfrac{1}{LC}}{\dfrac{1}{LC}-\omega^2 +j\omega\dfrac{1}{CR}}$$

So, you have a complex number that can give you amplitude and phase angles.

It might be easier if you convert it into "the standard form" of a 2nd order low pass filter namely: -

$$\dfrac{1}{1-\dfrac{\omega^2}{\omega_n^2 } +j\dfrac{\omega}{\omega_n}\cdot 2\zeta}$$

Where $$\\omega_n^2 = \dfrac{1}{LC}\$$

and, $$\2\zeta = \dfrac{1}{CR\cdot \omega_n}\$$

I recommend this route because it's intuitively easier to examine the phase angles. For instance, at DC, $$\\omega\$$ is zero and the formula reduces to unity hence gain is 1 and phase angle is 0 degrees. At very high frequencies, the formula reduces to a very small negative number thus impling that the phase is 180 degrees and the amplitude is very small.

At resonance ($$\\omega = \omega_n\$$), the formula reduces to $$\\dfrac{1}{j2\zeta}\$$ hence the phase angle is -90 degrees and the amplitude response equals the Q of the circuit (Q = $$\\dfrac{1}{2\zeta}\$$}.

See this web page for an interactive RLC filter tool/calculator: - • Thanks. That makes a lot more sense! Oct 26 '19 at 15:25
• Great post Andy, but your original potential divider is incorrect and needs editing. Parallel combination is (R/SC)/(R+(1/SC)). Oct 26 '19 at 20:32
• @James fixed now. Oct 27 '19 at 10:04

You are correct:-

Vin = VL + VC

BUT VL = L.d(iC + iR)/dt

where iR = VC/R

That eventually gives a transfer function of:-

1/(LCS^2 + SL/R + 1)

Well, we get the following transfer function:

$$\mathcal{H}\left(\text{s}\right)=\frac{\text{R}\text{||}\frac{1}{\text{sC}}}{\left(\text{R}\text{||}\frac{1}{\text{sC}}\right)+\text{sL}}=\frac{1}{\text{CL}\cdot\text{s}^2+\frac{\text{L}}{\text{R}}\cdot\text{s}+1}\tag1$$

If we want to plot the bode-plot we need to look at the value of the function when:

$$\text{s}=\text{j}\omega\tag2$$

Where $$\\text{j}^2=-1\$$.

So, we get:

$$\underline{\mathcal{H}}\left(\text{j}\omega\right)=\frac{1}{\text{CL}\cdot\left(\text{j}\omega\right)^2+\frac{\text{L}}{\text{R}}\cdot\text{j}\omega+1}=\frac{1}{1-\text{CL}\omega^2+\frac{\text{L}}{\text{R}}\cdot\omega\text{j}}\tag3$$

So, the amplitude function is given by:

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\left|\frac{1}{1-\text{CL}\omega^2+\frac{\text{L}}{\text{R}}\cdot\omega\text{j}}\right|=\frac{\left|1\right|}{\left|1-\text{CL}\omega^2+\frac{\text{L}}{\text{R}}\cdot\omega\text{j}\right|}=$$ $$\frac{1}{\sqrt{\left(1-\text{CL}\omega^2\right)^2+\left(\frac{\text{L}}{\text{R}}\cdot\omega\right)^2}}\tag4$$

Some interesting points:

1. $$\\omega\to0\$$: $$\lim_{\omega\to0}\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=1\tag5$$
2. $$\\omega\to\infty\$$: $$\lim_{\omega\to\infty}\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=0\tag6$$
3. The maximum/resonsancefrequency: $$\frac{\partial\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|}{\partial\omega}=0\space\Longrightarrow\space\omega=\frac{\sqrt{2\text{CR}^2-\text{L}}}{\text{CR}\sqrt{2\text{L}}}\tag7$$
4. Corner frequencies can be found solving: $$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\frac{2\text{CR}^2}{\sqrt{\text{L}\left(4\text{CR}^2-\text{L}\right)}}\cdot\frac{1}{\sqrt{2}}\space\Longleftrightarrow\space$$ $$\frac{1}{\sqrt{\left(1-\text{CL}\omega^2\right)^2+\left(\frac{\text{L}}{\text{R}}\cdot\omega\right)^2}}=\frac{2\text{CR}^2}{\sqrt{\text{L}\left(4\text{CR}^2-\text{L}\right)}}\cdot\frac{1}{\sqrt{2}}\tag8$$

Now, the phase information is given by:

$$\arg\left(\underline{\mathcal{H}}\left(\text{j}\omega\right)\right)=- \begin{cases} \frac{\pi}{2},\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space1-\text{CL}\omega^2=0\\ \\ \arctan\left(\frac{\text{L}}{\text{R}}\cdot\frac{\omega}{1-\text{CL}\omega^2}\right),\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space1-\text{CL}\omega^2>0\\ \\\frac{\pi}{2}+\arctan\left(\frac{\text{R}}{\text{L}\omega}\cdot\left|1-\text{CL}\omega^2\right|\right),\space\space\space\space\space\space\space\space\space\text{when}\space1-\text{CL}\omega^2<0 \end{cases} \tag9$$