Have I obtained the correct simplified transfer function for this RLC circuit?

Question

Have I obtained the correct simplified transfer function for this RLC circuit?

I am trying to obtain the transfer function for the following RCL circuit. However, the response of the TF that I obtain doesn't correlate to the response of the circuit itself when I simulate it. Please see below:

Therefore, I have obviously done something wrong. I have spent a good couples of days and plenty of scrap papers trying to see where I have missed something, but I keep on arriving at the same answer.

So, in order to save myself the headache of repeatedly doing the same thing only to obtain the same outcome, I've decided to put my dilemma to the community!

Below is my working out from the mid-way point to the end result and if anyone can spot or point out what I have missed, I would really appreciate it.

So the starting expression is:

$$ I_2(s)((\frac{C_2Ls^3+C_2(R_1+R_2)s+1}{C_2Ls^2+C_2R_1s})(\frac{C_1Ls^2+C_1R_1s+1}{C_1s}))-I_2(s)(Ls+R_1)=E_i(s)\tag{1} $$

$$ I_2(s)((\frac{C_2Ls^3+C_2(R_1+R_2)s+1}{C_2Ls^2+C_2R_1s})(\frac{C_1Ls^2+C_1R_1s+1}{C_1s})-(Ls+R_1))=E_i(s) $$

$$ I_2(s)(\frac{(C_2Ls^2+C_2(R_1+R_2)s+1)({C_1Ls^2+C_1R_1s+1})}{(C_2Ls^2+C_2R_1s)(C_1s)})-(Ls+R_1))=E_i(s) $$

$$ I_2(s)(\frac{(C_2Ls^2+C_2(R_1+R_2)s+1)(C_1Ls^2+C_1R_1s+1)-(C_1C_2Ls^3+C_1C_2R_1s^2)(Ls+R_1)}{(C_1C_2Ls^3+C_1C_2R_1s^2)})\tag{2}$$

Expanding the numerator for the positive term

$$ (C_2Ls^2+C_2(R1+R2)s+1)(C_1Ls^2+C_1R_1s+1)$$

$$ (C_2Ls^2)(C_1Ls^2) + (C_2Ls^2)(C_1R_1s) + (C_2Ls^2)(1) + (C_2(R_1+R_2)s)(C_1Ls^2) + (C_2(R_1+R_2)s)(C_1R_1s)+(C_2(R_1+R_2)s)(1)) + (1)(C_1Ls^2) + (1)(C_1R_1s) + (1)(1) $$

$$ (C_1C_2L^2s^4) + (C_1C_2R_1Ls^3) + (C_2Ls^2) + (C_1C_2L(R_1+R_2)s^3) + (C_1C_2R_1(R_1+R_2)s^2) + C_2(R_1+R_2)s + (C_1Ls^2) + (C_1R_1s) + 1 $$

$$ C_1C_2L^2s^4 + C_1C_2R_1Ls^3 + C_2Ls^2 + C_1C_2L(R_1+R_2)s^3 + C_1C_2R_1(R_1+R_2)s^2 + C_2(R_1+R_2)s + C_1Ls^2 + C_1R_1s + 1 $$

$$ C_1C_2L^2s^4 + C_1C_2R_1Ls^3 + C_2Ls^2 + C_1C_2R_1Ls^3 + C_1C_2R_2Ls^3 + C_1C_2R_1^2s^2 + C_1C_2R_1R_2s^2 + C_2(R_1+R_2)s + C_1Ls^2 + C_1R_1s + 1 $$

$$ C_1C_2L^2s^4 + 2C_1C_2R_1Ls^3 + C_2Ls^2 + C_1C_2R_2Ls^3 + C_1C_2R_1^2s^2 + C_1C_2R_1R_2s^2 + C_2(R_1+R_2)s + C_1Ls^2 + C_1R_1s + 1 \tag{3}$$

Expanding the numerator for the negative term

$$ -(C_1C_2Ls^3+C_1C_2R_1s^2)(Ls+R_1)$$

$$ -((C_1C_2Ls^3)(Ls)+(C_1C_2Ls^3)(R_1)+(C_1C_2R_1s^2)(Ls)+(C_1C_2R_1s^2)(R_1))$$

$$ -(C_1C_2L^2s^4 + C_1C_2R_1Ls^3 + C_1C_2R_1Ls^3 + C_1C_2R_1^2s^2)\tag{4}$$

Subtracting \$(4)\$ from \$(3)\$

$$ C_1C_2L^2s^4 + 2C_1C_2R_1Ls^3 + C_2Ls^2 + C_1C_2R_2Ls^3 + C_1C_2R_1^2s^2 + C_1C_2R_1R_2s^2 + C_2(R_1+R_2)s + C_1Ls^2 + C_1R_1s + 1 -C_1C_2L^2s^4 - C_1C_2R_1Ls^3 - C_1C_2R_1Ls^3 - C_1C_2R_1^2s^2 $$

$$ C_1C_2L^2s^4 + 2C_1C_2R_1Ls^3 + C_2Ls^2 + C_1C_2R_2Ls^3 + C_1C_2R_1^2s^2 + C_1C_2R_1R_2s^2 + C_2(R_1+R_2)s + C_1Ls^2 + C_1R_1s + 1 - C_1C_2L^2s^4 - 2C_1C_2R_1Ls^3 - C_1C_2R_1^2s^2 $$

$$ C_2Ls^2 + C_1C_2R_2Ls^3 + C_1C_2R_1R_2s^2 + C_2(R_1+R_2)s + C_1Ls^2 + C_1R_1s + 1 $$

Combining like terms

$$ C_1C_2R_2Ls^3 + (C_1(C_2R_1R_2 + L)+ C_2L)s^2 + (C_2(R_1+R_2) + C_1R_1)s + 1 $$

Therefore, I get

$$ I_2(s)\frac{C_1C_2R_2Ls^3 + (C_1(C_2R_1R_2 + L)+ C_2L)s^2 + (C_2(R_1+R_2) + C_1R_1)s + 1}{C_1C_2Ls^3+C_1C_2R_1s^2} = E_i(s)$$

And I arrive at this final transfer function each and every time:

$$ \frac{I_2(s)}{E_i(s)}=\frac{C_1C_2Ls^3+C_1C_2R_1s^2}{C_1C_2R_2Ls^3 + (C_1(C_2R_1R_2 + L)+ C_2L)s^2 + (C_2(R_1+R_2) + C_1R_1)s + 1} \tag{5}$$

Therefore, if anyone has noted any step(s) that I might have missed, I would really appreciate it, if you could please point it out :)

Answer 1

Your schematic is as follows:

^{simulate this circuit – Schematic created using CircuitLab}

Above, you can see that \$R_2\$ and \$Z_2\$ form a voltage divider that divides \$V_Y\$ into \$V_\text{OUT}\$. It follows that:

$$\begin{align*} \text{Each Stage} &\left\{ \begin{array}{rl} V_\text{OUT} &= V_Y\,\frac{1}{1+\frac{R_2}{Z_2}}&\text{where}&Z_2 &= \left(Z_{C_2}\mid\mid \infty\right)=Z_{C_2}\\\\ V_Y &= V_\text{IN}\,\frac{1}{1+\frac{Z_{C_1}}{Z_1}}&\text{where}&Z_1 &= \left(Z_{L_1}+R_1\right)\mid\mid \left(Z_2 + R_2\right) \end{array} \right.\\\\ &\therefore \frac{V_\text{OUT}}{V_\text{IN}}=\left[\frac{1}{1+\frac{Z_{C_1}}{Z_1}}\right]\cdot\left[\frac{1}{1+\frac{R_2}{Z_2}}\right]\\\\ &\therefore \frac{I_{C_2}}{V_\text{IN}}=\left[\frac{1}{1+\frac{Z_{C_1}}{Z_1}}\right]\cdot\left[\frac{1}{1+\frac{R_2}{Z_2}}\right]\cdot\bigg[C_2\,s\bigg] \end{align*}$$

Let's do this in sympy:

var('l1 c2 r1 r2 c1 vy vout vin')
zc1=1/s/c1                               # C1 impedance
zc2=1/s/c2                               # C2 impedance
zl1=s*l1                                 # L1 impedance
z2=zc2                                   # Z2 is just C2's impedance
z1=(zl1 + r1)*(z2+r2)/(zl1+r1+z2+r2)     # Z1 = parallel combination, as shown
f1=1/(1+zc1/z1)                          # first fraction
f2=1/(1+r2/z2)                           # second fraction
simplify(f1*f2*c2*s)                     # transfer function of I(C2)/V(IN)
    c1*c2*s**2*(l1*s + r1)/(c1*s*(l1*s + r1)*(c2*r2*s + 1) + c2*s*(l1*s + r1 + r2) + 1)
simplify(f1*f2)                          # transfer function of V(x)/V(IN)
    c1*s*(l1*s + r1)/(c1*s*(l1*s + r1)*(c2*r2*s + 1) + c2*s*(l1*s + r1 + r2) + 1)

Then I just built this schematic:

And got this result:

So what's the problem?? I think you need to look more closely either at your algebra process or else for typos in your laplace function. I used a direct "copy/paste" operation directly from sympy to LTspice, to minimize the chances for error.

Answer 2

When I see the amount of lines thrown on the blank page by the OP as an attempt to derive this 3rd-order transfer function, I think it is about time time that fast analytical circuits techniques or FACTs are taught in universities. The principle is simple and, very often, works with inspection: just look at the circuit to infer the time constants without writing a single line of algebra. The exercise consists of determining the resistance \$R\$ offered by each energy-storing element when the circuit is observed with a zeroed excitation or with a nulled response. Once you have the \$R\$, you can form a time constant expressed as \$\tau=RC\$ or \$\tau=\frac{L}{R}\$.

So we start with a bunch of small sketches to determine the natural time constants of this linear circuit. We reduce the excitation to zero volt (the input source is replaced by a short circuit) and you "look" through the energy-storing element temporarily disconnected from the circuit to find the resistance \$R\$ driving that element:

As you can see, you just inspect the drawing to see, in your head, what resistance the terminals in question offer (where the arrow points). And if you make a mistake, it is easy and fast to come back to the circuit and correct the resistance \$R\$ you mistakenly determined.

For the zeroes, you can do what is a called a null double injection (NDI) but I often prefer for the passive case, to determine high-frequency gains: the source is back in place and you inspect the circuit to determine what the gains are when energy-storing elements are alternately placed in their high-frequency state:

You can see how easy it is! When you have all these elements on hand, you assemble them in a Mathcad sheet which a) will check homogeneity of the formulas and b) will let you compare the response obtained by the FACTs and that delivered by the brute-force approach (Thévenin in this case):

Then you can plot the magnitude and phase while comparing the two responses in similar plots. They perfectly match as shown below:

You can see how easy the FACTs are compared to the classical KVL and KCL methods. You apply the divide-and-conquer strategy promoted by Dr. Middlebrook and solve your circuit step-by-step with the ability to come back and solve an intermediate wrong result. There is no way you can do that with the brute-force approach.

Acquiring the FACTs skill is simple and you can start with 1st-order circuits as shown in the APEC seminar I taught in 2016. Then, for an in-depth description of the method up to the order \$N\$, you can have a look at the book I published on the subject.