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

analogcircuit analysistransfer function

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:

enter image description here

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 🙂

Best Answer

Your schematic is as follows:

schematic

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:

enter image description here

And got this result:

enter image description here

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.