Chapter 14, Problem 96P

The crossover circuit in Fig. 14.108 is a low-pass filter that is connected to a woofer. Find the transfer function H(ω) = V_o(ω)/V_i(ω).

To determine

Find the transfer function $H\left(\omega \right)=\frac{V_o\left(\omega \right)}{V_i\left(\omega \right)}$ of the crossover circuit shown in Figure 14.108.

Answer to Problem 96P

The transfer function $H\left(\omega \right)$ of the crossover circuit shown in Figure 14.108 is $\frac{R_L}{\left(R_L+R_i-{\omega }^2\left(R_{L}L{C}_2+R_{i}L{C}_1\right)+j\left(\omega \left(L+R_i{R}_L{C}_2+R_i{R}_L{C}_1\right)-{\omega }^3\left(R_i{R}_{L}L{C}_1{C}_2\right)\right)\right)}$.

Explanation of Solution

Given data:

Refer to Figure 14.108 in the textbook.

Formula used:

Write a general expression to calculate the impedance of resistor in s-domain.

$Z_R=R$ (1)

Here,

$R$ is the value of the resistor.

Write a general expression to calculate the impedance of an inductor in s-domain.

$Z_L=sL$ (2)

Here,

$L$ is the value of the inductor.

Write a general expression to calculate the impedance of a capacitor in s-domain.

$Z_C=\frac{1}{sC}$ (3)

Here,

$C$ is the value of the capacitor.

Write the general expression to calculate the transfer function of the system

$H\left(\omega \right)=\frac{V_o\left(\omega \right)}{V_i\left(\omega \right)}$ (4)

Here,

$V_o\left(\omega \right)$ is the output response of the system, and

$V_i\left(\omega \right)$ is the input response of the system.

Calculation:

The given circuit is redrawn as Figure 1.

The s-domain circuit of the Figure 1 is drawn as Figure 2 using the equations (1), (2), and (3).

Refer to Figure 2, the resistor $R_L$ and the capacitor $C_2$ is connected in parallel form. Therefore, the impedance $Z_1$ is expressed as,

$\begin{array}{c}{Z}_1=R_L\parallel \frac{1}{s{C}_2}\\ =\frac{\left(R_L\times \frac{1}{s{C}_2}\right)}{\left(R_L+\frac{1}{s{C}_2}\right)}\\ =\frac{\left(\frac{R_L}{s{C}_2}\right)}{\left(\frac{1+s{R}_L{C}_2}{s{C}_2}\right)}\\ =\frac{R_L}{1+s{R}_L{C}_2}\end{array}$

The Figure 2 is redrawn as Figure 3.

Refer to Figure 3, the inductor $L$ and impedance $Z_1$ is connected in series form and the combination is connected in parallel with the capacitor $C_1$. Therefore, the impedance $Z_2$ is expressed as,

$Z_2=\frac{1}{s{C}_1}\parallel \left(sL+Z_1\right)$

Substitute $\frac{R_L}{1+s{R}_L{C}_2}$ for $Z_1$ in above equation to find $Z_2$.

$\begin{array}{c}{Z}_2=\frac{1}{s{C}_1}\parallel \left(sL+\frac{R_L}{1+s{R}_L{C}_2}\right)\\ =\frac{1}{s{C}_1}\parallel \left(\frac{sL+s^2{R}_{L}L{C}_2+R_L}{1+s{R}_L{C}_2}\right)\\ =\frac{\frac{1}{s{C}_1}\times \left(\frac{sL+s^2{R}_{L}L{C}_2+R_L}{1+s{R}_L{C}_2}\right)}{\frac{1}{s{C}_1}+\left(\frac{sL+s^2{R}_{L}L{C}_2+R_L}{1+s{R}_L{C}_2}\right)}\\ =\frac{\left(\frac{sL+s^2{R}_{L}L{C}_2+R_L}{s{C}_1\left(1+s{R}_L{C}_2\right)}\right)}{\left(\frac{1+s{R}_L{C}_2+s^{2}L{C}_1+s^3{R}_{L}L{C}_1{C}_2+s{R}_L{C}_1}{s{C}_1\left(1+s{R}_L{C}_2\right)}\right)}\end{array}$

Simplify the above equation to find $Z_2$.

$Z_2=\frac{sL+R_L+s^2{R}_{L}L{C}_2}{1+s{R}_L{C}_2+s^{2}L{C}_1+s{R}_L{C}_1+s^3{R}_{L}L{C}_1{C}_2}$

The Figure 3 is redrawn as Figure 4.

Apply voltage division rule on Figure 4 to find $V_1$.

$V_1=\frac{Z_2}{Z_2+R_i}{V}_i$

Apply voltage division rule on Figure 3 to find $V_o$.

$V_o=\frac{Z_1}{Z_1+sL}{V}_1$

Substitute $\frac{Z_2}{Z_2+R_i}{V}_i$ for $V_1$ in above equation to find $V_o$.

$V_o=\frac{Z_1}{Z_1+sL}\left(\frac{Z_2}{Z_2+R_i}{V}_i\right)$

Rearrange the above equation to find $\frac{V_o}{V_i}$.

$\frac{V_o}{V_i}=\left(\frac{Z_1}{Z_1+sL}\right)\left(\frac{Z_2}{Z_2+R_i}\right)$

Substitute $\frac{R_L}{1+s{R}_L{C}_2}$ for $Z_1$ and $\frac{sL+R_L+s^2{R}_{L}L{C}_2}{1+s{R}_L{C}_2+s^{2}L{C}_1+s{R}_L{C}_1+s^3{R}_{L}L{C}_1{C}_2}$ for $Z_2$ in above equation to find $\frac{V_o}{V_i}$.

$\begin{array}{c}\frac{V_o}{V_i}=\left(\frac{\left(\frac{R_L}{1+s{R}_L{C}_2}\right)}{\left(\frac{R_L}{1+s{R}_L{C}_2}\right)+sL}\right)\left(\frac{\left(\frac{sL+R_L+s^2{R}_{L}L{C}_2}{1+s{R}_L{C}_2+s^{2}L{C}_1+s{R}_L{C}_1+s^3{R}_{L}L{C}_1{C}_2}\right)}{\left(\frac{sL+R_L+s^2{R}_{L}L{C}_2}{1+s{R}_L{C}_2+s^{2}L{C}_1+s{R}_L{C}_1+s^3{R}_{L}L{C}_1{C}_2}\right)+R_i}\right)\\ =\left(\begin{array}{c}\left(\frac{\left(\frac{R_L}{1+s{R}_L{C}_2}\right)}{\left(\frac{R_L+sL+s^2{R}_{L}L{C}_2}{1+s{R}_L{C}_2}\right)}\right)\times \\ \left(\frac{\left(\frac{sL+R_L+s^2{R}_{L}L{C}_2}{1+s{R}_L{C}_2+s^{2}L{C}_1+s{R}_L{C}_1+s^3{R}_{L}L{C}_1{C}_2}\right)}{\left(\frac{sL+R_L+s^2{R}_{L}L{C}_2+R_i+s{R}_i{R}_L{C}_2+s^2{R}_{i}L{C}_1+s{R}_i{R}_L{C}_1+s^3{R}_i{R}_{L}L{C}_1{C}_2}{1+s{R}_L{C}_2+s^{2}L{C}_1+s{R}_L{C}_1+s^3{R}_{L}L{C}_1{C}_2}\right)}\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(\frac{R_L}{R_L+sL+s^2{R}_{L}L{C}_2}\right)\times \\ \left(\frac{sL+R_L+s^2{R}_{L}L{C}_2}{s^3\left(R_i{R}_{L}L{C}_1{C}_2\right)+s^2\left(R_{L}L{C}_2+R_{i}L{C}_1\right)+s\left(L+R_i{R}_L{C}_2+R_i{R}_L{C}_1\right)+R_L+R_i}\right)\end{array}\right)\\ =\frac{R_L\left(s^2{R}_{L}L{C}_2+sL+R_L\right)}{\left(\left(s^2{R}_{L}L{C}_2+sL+R_L\right)\left(\begin{array}{l}{s}^3\left(R_i{R}_{L}L{C}_1{C}_2\right)+s^2\left(R_{L}L{C}_2+R_{i}L{C}_1\right)\\ +s\left(L+R_i{R}_L{C}_2+R_i{R}_L{C}_1\right)+R_L+R_i\end{array}\right)\right)}\end{array}$

Simplify the above equation to find $\frac{V_o}{V_i}$.

$\frac{V_o}{V_i}=\frac{R_L}{s^3\left(R_i{R}_{L}L{C}_1{C}_2\right)+s^2\left(R_{L}L{C}_2+R_{i}L{C}_1\right)+s\left(L+R_i{R}_L{C}_2+R_i{R}_L{C}_1\right)+R_L+R_i}$

Substitute $j\omega$ for $s$ in above equation to find $\frac{V_o\left(\omega \right)}{V_i\left(\omega \right)}$.

$\begin{array}{c}\frac{V_o\left(\omega \right)}{V_i\left(\omega \right)}=\frac{R_L}{\left(\begin{array}{l}{\left(j\omega \right)}^3\left(R_i{R}_{L}L{C}_1{C}_2\right)+{\left(j\omega \right)}^2\left(R_{L}L{C}_2+R_{i}L{C}_1\right)\\ +\left(j\omega \right)\left(L+R_i{R}_L{C}_2+R_i{R}_L{C}_1\right)+R_L+R_i\end{array}\right)}\\ =\frac{R_L}{\left(\begin{array}{l}{j}^{2}j{\omega }^3\left(R_i{R}_{L}L{C}_1{C}_2\right)+j^2{\omega }^2\left(R_{L}L{C}_2+R_{i}L{C}_1\right)\\ +\left(j\omega \right)\left(L+R_i{R}_L{C}_2+R_i{R}_L{C}_1\right)+R_L+R_i\end{array}\right)}\\ =\frac{R_L}{\left(\begin{array}{l}-j{\omega }^3\left(R_i{R}_{L}L{C}_1{C}_2\right)-{\omega }^2\left(R_{L}L{C}_2+R_{i}L{C}_1\right)\\ +j\omega \left(L+R_i{R}_L{C}_2+R_i{R}_L{C}_1\right)+R_L+R_i\end{array}\right)}\left\{\because j^2=-1\right\}\end{array}$

$\frac{V_o\left(\omega \right)}{V_i\left(\omega \right)}=\frac{R_L}{\left(\begin{array}{l}{R}_L+R_i-{\omega }^2\left(R_{L}L{C}_2+R_{i}L{C}_1\right)\\ +j\left(\omega \left(L+R_i{R}_L{C}_2+R_i{R}_L{C}_1\right)-{\omega }^3\left(R_i{R}_{L}L{C}_1{C}_2\right)\right)\end{array}\right)}$ (5)

From the equation (5), the equation (4) becomes,

$H\left(\omega \right)=\frac{V_o\left(\omega \right)}{V_i\left(\omega \right)}=\frac{R_L}{\left(R_L+R_i-{\omega }^2\left(R_{L}L{C}_2+R_{i}L{C}_1\right)+j\left(\omega \left(L+R_i{R}_L{C}_2+R_i{R}_L{C}_1\right)-{\omega }^3\left(R_i{R}_{L}L{C}_1{C}_2\right)\right)\right)}$

Conclusion:

Thus, the transfer function $H\left(\omega \right)$ of the crossover circuit shown in Figure 14.108 is $\frac{R_L}{\left(R_L+R_i-{\omega }^2\left(R_{L}L{C}_2+R_{i}L{C}_1\right)+j\left(\omega \left(L+R_i{R}_L{C}_2+R_i{R}_L{C}_1\right)-{\omega }^3\left(R_i{R}_{L}L{C}_1{C}_2\right)\right)\right)}$.