Chapter 6, Problem 6.47P

To determine

(a)

The transfer function $\frac{X\left(s\right)}{V\left(s\right)}$ for the speaker model for the parameter values L and c equal to zero. Consequently, also compute the expressions for its characteristic roots.

Answer to Problem 6.47P

The transfer function $\frac{X\left(s\right)}{V\left(s\right)}$ is $\frac{X\left(s\right)}{V\left(s\right)}=\frac{K_f}{mR{s}^2+K_f{K}_{b}s+kR}$ and the characteristic roots are $s=-\frac{K_f{K}_b}{2mR}\pm \frac{1}{2}\sqrt{\frac{K_f{}^2{K}_b{}^2}{m^2{R}^2}-\frac{4k}{m}}$.

Explanation of Solution

Given:

1. The electrical and mechanical sub-system of the speaker system is shown in figure below:



Also, the inductance (L) and damping constant (c) are zero.

2. Characteristic equation is the denominator of the transfer function of the corresponding system. Thus, a transfer function for a system could be represented as follows:

$T\left(s\right)=\frac{K}{s^2+As+B}$ for the characteristic equation $s^2+As+B=0$.

Concept Used:

The model equations for the speaker system as shown in the figure are:

For mechanical sub-system:

$m\frac{d^{2}x}{d{t}^2}=-c\frac{dx}{dt}-kx+K_{f}i$

For electrical sub-system:

$v=L\frac{di}{dt}+Ri+k_b\frac{dx}{dt}$.

Calculation:

Since for the speaker system, we have

$m\frac{d^{2}x}{d{t}^2}=-c\frac{dx}{dt}-kx+K_{f}i\left(1\right)$

$v=L\frac{di}{dt}+Ri+k_b\frac{dx}{dt}\left(2\right)$

On using equations (1) and (2), while keeping initial conditions equal to zero such that

$\begin{array}{l}m\frac{d^{2}x}{d{t}^2}=-c\frac{dx}{dt}-kx+K_{f}i\\ \Rightarrow m{s}^{2}X\left(s\right)=-csX\left(s\right)-kX\left(s\right)+K_{f}I\left(s\right)\\ \Rightarrow \left(m{s}^2+cs+k\right)X\left(s\right)=K_{f}I\left(s\right)\\ \Rightarrow \frac{X\left(s\right)}{I\left(s\right)}=\frac{K_f}{\left(m{s}^2+cs+k\right)}\left(3\right)\end{array}$

Similarly,

$\begin{array}{l}v=L\frac{di}{dt}+Ri+K_b\frac{dx}{dt}\\ \Rightarrow V\left(s\right)=LsI\left(s\right)+RI\left(s\right)+K_{b}sX\left(s\right)\\ \Rightarrow V\left(s\right)=\left(Ls+R\right)I\left(s\right)+K_{b}sX\left(s\right)\\ \Rightarrow V\left(s\right)=\frac{\left(Ls+R\right)\left(m{s}^2+cs+k\right)}{K_f}X\left(s\right)+K_{b}sX\left(s\right)\because \frac{X\left(s\right)}{I\left(s\right)}=\frac{K_f}{\left(m{s}^2+cs+k\right)}\end{array}$

$\begin{array}{l}\Rightarrow \frac{X\left(s\right)}{V\left(s\right)}=\frac{K_f}{\left(Ls+R\right)\left(m{s}^2+cs+k\right)+K_f{K}_{b}s}\\ \Rightarrow \frac{X\left(s\right)}{V\left(s\right)}=\frac{K_f}{mL{s}^3+s^2\left(cL+mR\right)+s\left(Lk+cR+K_f{K}_b\right)+kR}\left(4\right)\end{array}$

Thus, at $L=c=0$, we get

$\begin{array}{l}\frac{X\left(s\right)}{V\left(s\right)}=\frac{K_f}{mL{s}^3+s^2\left(cL+mR\right)+s\left(Lk+cR+K_f{K}_b\right)+kR}\\ \Rightarrow \frac{X\left(s\right)}{V\left(s\right)}=\frac{K_f}{mR{s}^2+K_f{K}_{b}s+kR}\end{array}$

Therefore, the characteristic equation of this transfer function is as:

$\begin{array}{l}mR{s}^2+K_f{K}_{b}s+kR=0\\ \Rightarrow s^2+\frac{K_f{K}_b}{mR}s+\frac{k}{m}=0\end{array}$

On taking out characteristic roots that are,

$\begin{array}{l}{s}^2+\frac{K_f{K}_b}{mR}s+\frac{k}{m}=0\\ \Rightarrow s=-\frac{K_f{K}_b}{2mR}\pm \frac{1}{2}\sqrt{\frac{K_f{}^2{K}_b{}^2}{m^2{R}^2}-\frac{4k}{m}}\end{array}$.

Conclusion:

The transfer function $\frac{X\left(s\right)}{V\left(s\right)}$ is,

$\frac{X\left(s\right)}{V\left(s\right)}=\frac{K_f}{mR{s}^2+K_f{K}_{b}s+kR}$.

And the characteristic roots are as follows:

$s=-\frac{K_f{K}_b}{2mR}\pm \frac{1}{2}\sqrt{\frac{K_f{}^2{K}_b{}^2}{m^2{R}^2}-\frac{4k}{m}}$.

To determine

(b)

The third order transfer function and the corresponding characteristic roots of the speaker system for the given parameter values. Also, to compare this result with the one obtained in sub-part (a).

Answer to Problem 6.47P

The transfer function and characteristic roots for third order case of speaker system are:

$\begin{array}{l}\frac{X\left(s\right)}{V\left(s\right)}=\frac{16}{2\times {\text{10}}^{-6}{s}^3+0.024s^2+608s+\text{48}\times {\text{10}}^5}\\ \Rightarrow s=-8715.433,-1642.28\pm j16512.92\end{array}$

Similarly, for the second order case of the speaker system, the transfer function and characteristic roots are:

$\begin{array}{l}\frac{X\left(s\right)}{V\left(s\right)}=\frac{16}{0.024s^2+208s+\text{48}\times {\text{10}}^5}\\ \Rightarrow s=-4333.33\pm j13461.88\end{array}$

On comparing these cases, it is concluded that system is stable in both the cases as roots are lying left to the imaginary axis in s-plane. However, the complex roots in the second order case are far into imaginary axis in comparison to the complex roots in third order case of the speaker system.

Explanation of Solution

Given:

Here, the parameter values for the speaker system are as follows:

$m=0.002\text{kg,}\text{k=4}\times {\text{10}}^5\text{N/m,}{K}_f=16\text{N/A, }{K}_b=13\text{V}\cdot \text{s/m,}R\text{=12}\text{ohm,}L{\text{=10}}^{-3}\text{H,}c\text{=0}$.

Concept Used:

1. Here, the transfer function $\frac{X\left(s\right)}{V\left(s\right)}$ for the speaker system obtained in sub-part (a) is as follows:

$\begin{array}{l}\frac{X\left(s\right)}{V\left(s\right)}=\frac{K_f}{mL{s}^3+s^2\left(cL+mR\right)+s\left(Lk+cR+K_f{K}_b\right)+kR}\\ \because \text{from}\text{equation}\left(\text{4}\right)\text{in}\text{sub-part}\left(\text{a}\right)\end{array}$

Also, at $L=c=0$, we get

$\frac{X\left(s\right)}{V\left(s\right)}=\frac{K_f}{mR{s}^2+K_f{K}_{b}s+kR}$

2. Characteristic equation is the denominator of the transfer function of the corresponding system. Thus, a transfer function for a system could be represented as follows:

$T\left(s\right)=\frac{K}{s^2+As+B}$ for the characteristic equation $s^2+As+B=0$.

Calculation:

Since for the speaker system, the transfer function is as follows:

$\frac{X\left(s\right)}{V\left(s\right)}=\frac{K_f}{mL{s}^3+s^2\left(cL+mR\right)+s\left(Lk+cR+K_f{K}_b\right)+kR}$

On substituting the parameter values in the above expression, we get

$\frac{X\left(s\right)}{V\left(s\right)}=\frac{K_f}{mL{s}^3+s^2\left(cL+mR\right)+s\left(Lk+cR+K_f{K}_b\right)+kR}$

$\begin{array}{l}m=0.002\text{kg,}\text{k=4}\times {\text{10}}^5\text{N/m,}{K}_f=16\text{N/A, }{K}_b=13\text{V}\cdot \text{s/m,}R\text{=12}{\text{ohm, L=10}}^{-3}\text{H,}\text{c=0}\\ \Rightarrow \frac{X\left(s\right)}{V\left(s\right)}=\frac{16}{0.002\times {\text{10}}^{-3}{s}^3+s^2\left(0+0.002\times \text{12}\right)+s\left(\text{4}\times {\text{10}}^5\times {\text{10}}^{-3}+0+16\times 13\right)+\text{4}\times {\text{10}}^5\times 12}\\ \Rightarrow \frac{X\left(s\right)}{V\left(s\right)}=\frac{16}{2\times {\text{10}}^{-6}{s}^3+0.024s^2+608s+\text{48}\times {\text{10}}^5}\end{array}$

Therefore, the characteristic equation would be:

$\begin{array}{l}2\times {\text{10}}^{-6}{s}^3+0.024s^2+608s+\text{48}\times {\text{10}}^5=0\\ \Rightarrow s^3+12\times {\text{10}}^3{s}^2+304\times {\text{10}}^{6}s+24\times {\text{10}}^{11}=0\end{array}$

On taking out characteristic roots that are,

$\begin{array}{l}{s}^3+12\times {\text{10}}^3{s}^2+304\times {\text{10}}^{6}s+24\times {\text{10}}^{11}=0\\ \Rightarrow s=-8715.433,-1642.28\pm j16512.92\end{array}$

Similarly, for the transfer function at $L=c=0$, we have

$\frac{X\left(s\right)}{V\left(s\right)}=\frac{K_f}{mR{s}^2+K_f{K}_{b}s+kR}$

On keeping the parameter values in this expression, we have

$\begin{array}{l}\frac{X\left(s\right)}{V\left(s\right)}=\frac{K_f}{mR{s}^2+K_f{K}_{b}s+kR}\\ \because m=0.002\text{kg,}\text{k=4}\times {\text{10}}^5\text{N/m,}{K}_f=16\text{N/A, }{K}_b=13\text{V}\cdot \text{s/m,}R\text{=12}\text{ohm}\\ \Rightarrow \frac{X\left(s\right)}{V\left(s\right)}=\frac{16}{0.002\times 12s^2+16\times 13s+\text{4}\times {\text{10}}^5\times 12}\\ \Rightarrow \frac{X\left(s\right)}{V\left(s\right)}=\frac{16}{0.024s^2+208s+\text{48}\times {\text{10}}^5}\end{array}$

Therefore, the characteristic equation would be:

$\begin{array}{l}0.024s^2+208s+\text{48}\times {\text{10}}^5=0\\ \Rightarrow s^2+8666.67s+2\times {\text{10}}^8=0\end{array}$

On taking out characteristic roots that are,

$\begin{array}{l}{s}^2+8666.67s+2\times {\text{10}}^8=0\\ \Rightarrow s=-4333.33\pm j13461.88\end{array}$.

Conclusion:

The transfer function and characteristic roots for third order case of speaker system are:

$\begin{array}{l}\frac{X\left(s\right)}{V\left(s\right)}=\frac{16}{2\times {\text{10}}^{-6}{s}^3+0.024s^2+608s+\text{48}\times {\text{10}}^5}\\ \Rightarrow s=-8715.433,-1642.28\pm j16512.92\end{array}$

Similarly, for the second order case of the speaker system, the transfer function and characteristic roots are:

$\begin{array}{l}\frac{X\left(s\right)}{V\left(s\right)}=\frac{16}{0.024s^2+208s+\text{48}\times {\text{10}}^5}\\ \Rightarrow s=-4333.33\pm j13461.88\end{array}$

On comparing these cases, it is concluded that a system is stable in both the cases as roots are lying left to the imaginary axis in s-plane. However, the complex roots in the second order case are far into imaginary axis in comparison to the complex roots in third order case of the speaker system.