Chapter 14, Problem 97P

The crossover circuit in Fig. 14.109 is a high-pass filter that is connected to a tweeter. Determine the transfer function H(ω) = V_o(ω)/V_i(ω).

Figure 14.109

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 14.109.

Answer to Problem 97P

The transfer function $H\left(\omega \right)$ of the crossover circuit shown in Figure 14.109 is $\frac{-j{\omega }^3{R}_{L}L{C}_1{C}_2\left(s{R}_L{C}_2+1\right)}{\left(\begin{array}{l}{\omega }^4\left(R_{L}L{C}_1{C}_2{}^2\left(R_L+R_i\right)\right)-j{\omega }^3\left(R_{L}L{C}_1{C}_2+R_{L}L{C}_1{C}_2+R_{i}L{C}_1{C}_2+R_{L}L{C}_2{}^2+R_i{R}_L{}^2{C}_1{C}_2{}^2\right)\\ -{\omega }^2\left(L{C}_1+L{C}_2+R_i{R}_L{C}_1{C}_2+R_L{}^2{C}_2{}^2+R_i{C}_1{R}_L{C}_2\right)+j\omega \left(R_L{C}_2+R_i{C}_1+R_L{C}_2\right)+1\end{array}\right)}$.

Explanation of Solution

Given data:

Refer to Figure 14.109 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.

Use equation (1) to find $Z_{R_i}$.

$Z_{R_i}=R_i$

Use equation (1) to find $Z_{R_L}$.

$Z_{R_L}=R_L$

Use equation (2) to find $Z_L$.

$Z_L=sL$

Use equation (3) to find $Z_{C_1}$.

$Z_{C_1}=\frac{1}{s{C}_1}$

Use equation (3) to find $Z_{C_2}$.

$Z_{C_2}=\frac{1}{s{C}_2}$

The s-domain circuit of the Figure 1 is drawn as Figure 2.

Refer to Figure 2, the series connected impedances $R_L$ and $\frac{1}{s{C}_2}$ are connected in parallel with the impedance $sL$.

Therefore, the equivalent impedance is calculated as follows.

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

Simplify the above equation to find $Z$.

$Z=\left(\frac{s^2{R}_{L}L{C}_2+sL}{s^{2}L{C}_2+s{R}_L{C}_2+1}\right)$

The reduced circuit of the Figure 2 is drawn as Figure 3.

Apply voltage division rule on Figure 3 to find $V_1$.

$V_1=\frac{Z}{Z+R_i+\frac{1}{s{C}_1}}{V}_i$

Apply voltage division rule on Figure 2 to find $V_o$.

$V_o=\frac{R_L}{R_L+\frac{1}{s{C}_2}}{V}_1$

Substitute $\frac{Z}{Z+R_i+\frac{1}{s{C}_1}}{V}_i$ for $V_1$ in above equation to find $V_o$.

$\begin{array}{c}{V}_o=\frac{R_L}{R_L+\frac{1}{s{C}_2}}\left(\frac{Z}{Z+R_i+\frac{1}{s{C}_1}}{V}_i\right)\\ =\left(\frac{R_L}{\left(\frac{s{R}_L{C}_2+1}{s{C}_2}\right)}\right)\left(\frac{Z}{\left(\frac{Zs{C}_1+s{R}_i{C}_1+1}{s{C}_1}\right)}\right)V_i\\ =\left(\frac{s{R}_L{C}_2}{s{R}_L{C}_2+1}\right)\left(\frac{Zs{C}_1}{Zs{C}_1+s{R}_i{C}_1+1}\right)V_i\end{array}$

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

$\frac{V_o}{V_i}=\left(\frac{s{R}_L{C}_2}{s{R}_L{C}_2+1}\right)\left(\frac{Zs{C}_1}{Zs{C}_1+s{R}_i{C}_1+1}\right)$

Substitute $\left(\frac{s^2{R}_{L}L{C}_2+sL}{s^{2}L{C}_2+s{R}_L{C}_2+1}\right)$ for $Z$ in above equation to find $\frac{V_o}{V_i}$.

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

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

$\begin{array}{c}\frac{V_o}{V_i}=\left(\frac{s{C}_1\left(s{R}_L{C}_2\right)\left(s^2{R}_{L}L{C}_2+sL\right)}{\left(s{R}_L{C}_2+1\right)\left(\begin{array}{l}s{C}_1\left(s^2{R}_{L}L{C}_2+sL\right)+s^3{R}_{i}L{C}_1{C}_2+s^{2}L{C}_2+s^2{R}_i{R}_L{C}_1{C}_2\\ +s{R}_L{C}_2+s{R}_i{C}_1+1\end{array}\right)}\right)\\ =\left(\frac{\left(s^2{R}_L{C}_1{C}_2\right)\left(s^2{R}_{L}L{C}_2+sL\right)}{\left(s{R}_L{C}_2+1\right)\left(\begin{array}{l}{s}^3{R}_{L}L{C}_1{C}_2+s^{2}L{C}_1+s^3{R}_{i}L{C}_1{C}_2+s^{2}L{C}_2+s^2{R}_i{R}_L{C}_1{C}_2\\ +s{R}_L{C}_2+s{R}_i{C}_1+1\end{array}\right)}\right)\\ =\left(\frac{\left(s^4{R}_L{}^{2}L{C}_1{C}_2{}^2+s^3{R}_{L}L{C}_1{C}_2\right)}{\left(\begin{array}{l}{s}^4{R}_L{}^{2}L{C}_1{C}_2{}^2+s^3{R}_{L}L{C}_1{C}_2+s^3{R}_{L}L{C}_1{C}_2+s^{2}L{C}_1+s^4{R}_i{R}_{L}L{C}_1{C}_2{}^2+s^3{R}_{i}L{C}_1{C}_2\\ +s^3{R}_{L}L{C}_2{}^2+s^{2}L{C}_2+s^3{R}_i{R}_L{}^2{C}_1{C}_2{}^2+s^2{R}_i{R}_L{C}_1{C}_2+s^2{R}_L{}^2{C}_2{}^2+s{R}_L{C}_2\\ +s^2{R}_i{C}_1{R}_L{C}_2+s{R}_i{C}_1+s{R}_L{C}_2+1\end{array}\right)}\right)\\ =\left(\frac{\left(s^4{R}_L{}^{2}L{C}_1{C}_2{}^2+s^3{R}_{L}L{C}_1{C}_2\right)}{\left(\begin{array}{l}{s}^4{R}_L{}^{2}L{C}_1{C}_2{}^2+s^4{R}_i{R}_{L}L{C}_1{C}_2{}^2+s^3{R}_{L}L{C}_1{C}_2+s^3{R}_{L}L{C}_1{C}_2+s^3{R}_{i}L{C}_1{C}_2+s^3{R}_{L}L{C}_2{}^2\\ +s^3{R}_i{R}_L{}^2{C}_1{C}_2{}^2+s^{2}L{C}_1+s^{2}L{C}_2+s^2{R}_i{R}_L{C}_1{C}_2+s^2{R}_L{}^2{C}_2{}^2+s^2{R}_i{C}_1{R}_L{C}_2+s{R}_L{C}_2\\ +s{R}_i{C}_1+s{R}_L{C}_2+1\end{array}\right)}\right)\end{array}$

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

$\frac{V_o}{V_i}=\left(\frac{s^3{R}_{L}L{C}_1{C}_2\left(s{R}_L{C}_2+1\right)}{\left(\begin{array}{l}{s}^4\left(R_{L}L{C}_1{C}_2{}^2\left(R_L+R_i\right)\right)\\ +s^3\left(R_{L}L{C}_1{C}_2+R_{L}L{C}_1{C}_2+R_{i}L{C}_1{C}_2+R_{L}L{C}_2{}^2+R_i{R}_L{}^2{C}_1{C}_2{}^2\right)\\ +s^2\left(L{C}_1+L{C}_2+R_i{R}_L{C}_1{C}_2+R_L{}^2{C}_2{}^2+R_i{C}_1{R}_L{C}_2\right)+s\left(R_L{C}_2+R_i{C}_1+R_L{C}_2\right)+1\end{array}\right)}\right)$

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)}=\left(\frac{{\left(j\omega \right)}^3{R}_{L}L{C}_1{C}_2\left(s{R}_L{C}_2+1\right)}{\left(\begin{array}{l}{\left(j\omega \right)}^4\left(R_{L}L{C}_1{C}_2{}^2\left(R_L+R_i\right)\right)\\ +{\left(j\omega \right)}^3\left(R_{L}L{C}_1{C}_2+R_{L}L{C}_1{C}_2+R_{i}L{C}_1{C}_2+R_{L}L{C}_2{}^2+R_i{R}_L{}^2{C}_1{C}_2{}^2\right)\\ +{\left(j\omega \right)}^2\left(L{C}_1+L{C}_2+R_i{R}_L{C}_1{C}_2+R_L{}^2{C}_2{}^2+R_i{C}_1{R}_L{C}_2\right)\\ +\left(j\omega \right)\left(R_L{C}_2+R_i{C}_1+R_L{C}_2\right)+1\end{array}\right)}\right)\\ \\ =\left(\frac{j^3{\omega }^3{R}_{L}L{C}_1{C}_2\left(s{R}_L{C}_2+1\right)}{\left(\begin{array}{l}{j}^4{\omega }^4\left(R_{L}L{C}_1{C}_2{}^2\left(R_L+R_i\right)\right)\\ +j^3{\omega }^3\left(R_{L}L{C}_1{C}_2+R_{L}L{C}_1{C}_2+R_{i}L{C}_1{C}_2+R_{L}L{C}_2{}^2+R_i{R}_L{}^2{C}_1{C}_2{}^2\right)\\ +j^2{\omega }^2\left(L{C}_1+L{C}_2+R_i{R}_L{C}_1{C}_2+R_L{}^2{C}_2{}^2+R_i{C}_1{R}_L{C}_2\right)\\ +j\omega \left(R_L{C}_2+R_i{C}_1+R_L{C}_2\right)+1\end{array}\right)}\right)\\ \\ =\left(\frac{j^{2}j{\omega }^3{R}_{L}L{C}_1{C}_2\left(s{R}_L{C}_2+1\right)}{\left(\begin{array}{l}{\left(j^2\right)}^2{\omega }^4\left(R_{L}L{C}_1{C}_2{}^2\left(R_L+R_i\right)\right)\\ +j^{2}j{\omega }^3\left(R_{L}L{C}_1{C}_2+R_{L}L{C}_1{C}_2+R_{i}L{C}_1{C}_2+R_{L}L{C}_2{}^2+R_i{R}_L{}^2{C}_1{C}_2{}^2\right)\\ +j^2{\omega }^2\left(L{C}_1+L{C}_2+R_i{R}_L{C}_1{C}_2+R_L{}^2{C}_2{}^2+R_i{C}_1{R}_L{C}_2\right)\\ +j\omega \left(R_L{C}_2+R_i{C}_1+R_L{C}_2\right)+1\end{array}\right)}\right)\\ =\left(\frac{-j{\omega }^3{R}_{L}L{C}_1{C}_2\left(s{R}_L{C}_2+1\right)}{\left(\begin{array}{l}{\left(-1\right)}^2{\omega }^4\left(R_{L}L{C}_1{C}_2{}^2\left(R_L+R_i\right)\right)\\ -j{\omega }^3\left(R_{L}L{C}_1{C}_2+R_{L}L{C}_1{C}_2+R_{i}L{C}_1{C}_2+R_{L}L{C}_2{}^2+R_i{R}_L{}^2{C}_1{C}_2{}^2\right)\\ -{\omega }^2\left(L{C}_1+L{C}_2+R_i{R}_L{C}_1{C}_2+R_L{}^2{C}_2{}^2+R_i{C}_1{R}_L{C}_2\right)\\ +j\omega \left(R_L{C}_2+R_i{C}_1+R_L{C}_2\right)+1\end{array}\right)}\right)\left\{\because j^2=-1\right\}\end{array}$

Simplify the above equation to find $\frac{V_o\left(\omega \right)}{V_i\left(\omega \right)}$.

$\frac{V_o\left(\omega \right)}{V_i\left(\omega \right)}=\left(\frac{-j{\omega }^3{R}_{L}L{C}_1{C}_2\left(s{R}_L{C}_2+1\right)}{\left(\begin{array}{l}{\omega }^4\left(R_{L}L{C}_1{C}_2{}^2\left(R_L+R_i\right)\right)\\ -j{\omega }^3\left(R_{L}L{C}_1{C}_2+R_{L}L{C}_1{C}_2+R_{i}L{C}_1{C}_2+R_{L}L{C}_2{}^2+R_i{R}_L{}^2{C}_1{C}_2{}^2\right)\\ -{\omega }^2\left(L{C}_1+L{C}_2+R_i{R}_L{C}_1{C}_2+R_L{}^2{C}_2{}^2+R_i{C}_1{R}_L{C}_2\right)\\ +j\omega \left(R_L{C}_2+R_i{C}_1+R_L{C}_2\right)+1\end{array}\right)}\right)$        (5)

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

$H\left(\omega \right)=\frac{V_o\left(\omega \right)}{V_i\left(\omega \right)}=\frac{-j{\omega }^3{R}_{L}L{C}_1{C}_2\left(s{R}_L{C}_2+1\right)}{\left(\begin{array}{l}{\omega }^4\left(R_{L}L{C}_1{C}_2{}^2\left(R_L+R_i\right)\right)\\ -j{\omega }^3\left(R_{L}L{C}_1{C}_2+R_{L}L{C}_1{C}_2+R_{i}L{C}_1{C}_2+R_{L}L{C}_2{}^2+R_i{R}_L{}^2{C}_1{C}_2{}^2\right)\\ -{\omega }^2\left(L{C}_1+L{C}_2+R_i{R}_L{C}_1{C}_2+R_L{}^2{C}_2{}^2+R_i{C}_1{R}_L{C}_2\right)\\ +j\omega \left(R_L{C}_2+R_i{C}_1+R_L{C}_2\right)+1\end{array}\right)}$

Conclusion:

Thus, the transfer function $H\left(\omega \right)$ of the crossover circuit shown in Figure 14.109 is $\frac{-j{\omega }^3{R}_{L}L{C}_1{C}_2\left(s{R}_L{C}_2+1\right)}{\left(\begin{array}{l}{\omega }^4\left(R_{L}L{C}_1{C}_2{}^2\left(R_L+R_i\right)\right)-j{\omega }^3\left(R_{L}L{C}_1{C}_2+R_{L}L{C}_1{C}_2+R_{i}L{C}_1{C}_2+R_{L}L{C}_2{}^2+R_i{R}_L{}^2{C}_1{C}_2{}^2\right)\\ -{\omega }^2\left(L{C}_1+L{C}_2+R_i{R}_L{C}_1{C}_2+R_L{}^2{C}_2{}^2+R_i{C}_1{R}_L{C}_2\right)+j\omega \left(R_L{C}_2+R_i{C}_1+R_L{C}_2\right)+1\end{array}\right)}$.