Chapter 6, Problem 6.40P

To determine

The effect of the damping constant $c$ on the armature-controlled motor’s characteristic roots and on its speed response to a step voltage input. Also, to find the time taken by the motor to attain constant speed and check whether the oscillations will occur in the motor response or not.

Answer to Problem 6.40P

The obtained characteristic roots for the various values of damping constant are as follows:

For $c=0\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$, $s=\frac{25\left(\sqrt{106}-16\right)}{3},s=-\frac{25\left(16+\sqrt{106}\right)}{3}$

For $c=0.01\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$, $s=-\frac{1175}{6}+i\frac{25\sqrt{311}}{6},s=-\frac{1175}{6}-i\frac{25\sqrt{311}}{6}$

For $c=0.1\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$, $s=\frac{25\left(\sqrt{3331}-91\right)}{3},s=-\frac{25\left(91+\sqrt{3331}\right)}{3}$

Thus, we found that for the values of damping constant, $c=0,0.1\text{N}\cdot \text{m}\cdot \text{s/rad}$ the roots lie on left side of the imaginary axis in s-plane. And, these roots are real and distinct in nature. While, for damping constant, $c=0.01\text{N}\cdot \text{m}\cdot \text{s/rad}$ the roots are also on the left of the imaginary axis in s-plane, but these are complex in nature.

The obtained response to step voltage input for the various values of damping constant are as follows:

For $c=0\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$, $\omega \left(t\right)=-20e^{-133.33t}\cosh \left(85.797t\right)-31.08e^{-133.33t}\sinh \left(85.797t\right)+20$

For $c=0.01\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$, $\omega \left(t\right)=-4.762e^{-195.833t}\cos \left(73.479t\right)-12.691e^{-195.833t}\sin \left(73.479t\right)+\frac{100}{21}$

For $c=0.1\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$, $\omega \left(t\right)=-0.606e^{-758.33t}\cosh \left(480.957t\right)-0.956e^{-758.33t}\sinh \left(480.957t\right)+\frac{20}{33}$

Here, for the values of damping constant, $c=0,0.1\text{N}\cdot \text{m}\cdot \text{s/rad}$ the speed responses are stable and free from oscillations. While, for the damping constant, $c=0.01\text{N}\cdot \text{m}\cdot \text{s/rad}$, the response is oscillatory and finally settles for a steady-state in future that is stable.

The time taken for the oscillations to decay for various values of damping constants is as follows:

For $c=0\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$, there won’t be any oscillations as the response includes hyperbolic functions. However, the response reaches the steady state at $t=4\tau =\frac{4}{133.33}=0.03\text{seconds}$.

For $c=0.01\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$, the time taken by the oscillations to decay is $t=4\tau =\frac{4}{195.833}=0.02\text{seconds}$.

For $c=0.1\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$, there won’t be any oscillations as the response includes hyperbolic functions. However, the response reaches the steady state at $t=4\tau =\frac{4}{758.33}=0.005\text{seconds}$.

Explanation of Solution

Given information:

The given parameters for the armature-controlled motor are as follows:

$K_T=K_b=0.05\text{N}\cdot \text{m}/\text{A},L_a=3\times {10}^{-3}\text{H},I=8\times {10}^{-5}\text{kg}\cdot {\text{m}}^2,R_a=0.8\Omega$

Also, the various values of damping constants are as follows:

$c=0\text{N}\cdot \text{m}\cdot \text{s}/\text{rad},0.01\text{N}\cdot \text{m}\cdot \text{s}/\text{rad},0.1\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$.

Concept Used:

The motor speed response with armature voltage as input is as follows:

$\frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{K_T}{L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T}$.

Calculation:

For $c=0\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$

Since,

$\frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{K_T}{L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T}$

Keeping the values of all the known parameters such that

$\begin{array}{l}\frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{K_T}{L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T}\\ \Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{0.05}{3\times {10}^{-3}\times 8\times {10}^{-5}\times s^2+\left(0.8\times 8\times {10}^{-5}+0\right)s+0+0.05\times 0.05}\end{array}$

$\begin{array}{l}\Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{0.05}{24\times {10}^{-8}{s}^2+6.4\times {10}^{-5}s+25\times {10}^{-4}}\\ \Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{1}{4.8\times {10}^{-6}{s}^2+1.28\times {10}^{-3}s+5\times {10}^{-2}}\end{array}$

For the unit step input voltage response, we have

$\begin{array}{l}\frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{1}{4.8\times {10}^{-6}{s}^2+1.28\times {10}^{-3}s+5\times {10}^{-2}}\\ \Rightarrow \Omega \left(s\right)=\frac{1}{s\left(4.8\times {10}^{-6}{s}^2+1.28\times {10}^{-3}s+5\times {10}^{-2}\right)}\\ \Rightarrow \Omega \left(s\right)=\frac{{10}^6}{s\left(4.8s^2+1280s+50000\right)}\end{array}$

On taking partial fraction expansion for the above expression, we have

$\begin{array}{l}\Omega \left(s\right)=\frac{1000000}{s\left(4.8s^2+1280s+50000\right)}=\frac{A}{s}+\frac{Bs+C}{4.8s^2+1280s+50000}\\ \Rightarrow \frac{1000000}{s\left(4.8s^2+1280s+50000\right)}=\frac{A\left(4.8s^2+1280s+50000\right)+s\left(Bs+C\right)}{s\left(4.8s^2+1280s+50000\right)}\end{array}$

On comparing the numerator of both sides

$\begin{array}{l}\Rightarrow 1000000=A\left(4.8s^2+1280s+50000\right)+s\left(Bs+C\right)\\ \therefore A=20,B=-96,C=-\text{25600}\end{array}$

Therefore,

$\begin{array}{l}\Omega \left(s\right)=\frac{-96s-25600}{4.8s^2+1280s+50000}+\frac{20}{s}\\ \Rightarrow \Omega \left(s\right)=-20\frac{\left(s+133.33\right)}{\left({\left(s+133.33\right)}^2-7361.11\right)}-2666.67\frac{1}{\left({\left(s+133.33\right)}^2-7361.11\right)}+\frac{20}{s}\end{array}$

On taking the inverse Laplace transform of the above obtained expression, we get

$\omega \left(t\right)=-20e^{-133.33t}\cosh \left(85.797t\right)-31.08e^{-133.33t}\sinh \left(85.797t\right)+20$

That shows the response doesn’t contain any oscillations. However, the steady-state for the response could be obtained at $t=4\tau =\frac{4}{133.33}=0.03\text{seconds}$.

The characteristic polynomial for the system is as follows:

$L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T=0$

On keeping the values, we have

$\begin{array}{l}{L}_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T=0\\ \Rightarrow 3\times {10}^{-3}\times 8\times {10}^{-5}{s}^2+\left(0.8\times 8\times {10}^{-5}+0\right)s+0+0.05\times 0.05=0\\ \Rightarrow 24\times {10}^{-8}{s}^2+6.4\times {10}^{-5}s+25\times {10}^{-4}=0\\ \Rightarrow 24s^2+6400s+250000=0\\ \Rightarrow 3s^2+800s+31250=0\end{array}$

Thus, the roots are as follows:

$s=\frac{25\left(\sqrt{106}-16\right)}{3},s=-\frac{25\left(16+\sqrt{106}\right)}{3}$

For $c=0.01\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$

Since,

$\frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{K_T}{L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T}$

Keeping the values of all the known parameters such that

$\begin{array}{l}\frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{K_T}{L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T}\\ \Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{0.05}{3\times {10}^{-3}\times 8\times {10}^{-5}{s}^2+\left(0.8\times 8\times {10}^{-5}+0.01\times 3\times {10}^{-3}\right)s+0.01\times 0.8+0.05\times 0.05}\\ \Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{0.05}{24\times {10}^{-8}{s}^2+\left(6.4\times {10}^{-5}+3\times {10}^{-5}\right)s+105\times {10}^{-4}}\\ \Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{0.05}{24\times {10}^{-8}{s}^2+9.4\times {10}^{-5}s+105\times {10}^{-4}}\end{array}$

$\Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{1}{4.8\times {10}^{-6}{s}^2+1.88\times {10}^{-3}s+21\times {10}^{-2}}$

For the unit step input voltage response, we have

$\begin{array}{l}\frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{1}{4.8\times {10}^{-6}{s}^2+1.88\times {10}^{-3}s+21\times {10}^{-2}}\\ \Rightarrow \Omega \left(s\right)=\frac{1}{s\left(4.8\times {10}^{-6}{s}^2+1.88\times {10}^{-3}s+21\times {10}^{-2}\right)}\\ \Rightarrow \Omega \left(s\right)=\frac{{10}^6}{s\left(4.8s^2+1880s+210000\right)}\end{array}$

On taking partial fraction expansion for the above expression, we have

$\begin{array}{l}\Omega \left(s\right)=\frac{{10}^6}{s\left(4.8s^2+1880s+210000\right)}=\frac{A}{s}+\frac{Bs+C}{4.8s^2+1880s+210000}\\ \Rightarrow \frac{{10}^6}{s\left(4.8s^2+1880s+210000\right)}=\frac{A\left(4.8s^2+1880s+210000\right)+s\left(Bs+C\right)}{s\left(4.8s^2+1880s+210000\right)}\end{array}$

On comparing the numerator of both sides

$\begin{array}{l}\Rightarrow 1000000=A\left(4.8s^2+1880s+210000\right)+s\left(Bs+C\right)\\ \therefore A=\frac{100}{21},B=-\text{22}\text{.857},C=-\text{8952}\text{.38}1\end{array}$

Therefore,

$\Omega \left(s\right)=\frac{-22.857s-8952.381}{4.8s^2+1880s+210000}+\frac{100}{21s}$

$\Rightarrow \Omega \left(s\right)=-4.762\frac{\left(s+195.833\right)}{\left({\left(s+195.833\right)}^2+5399.31\right)}-932.546\frac{1}{\left({\left(s+195.833\right)}^2+5399.31\right)}+\frac{100}{21s}$

On taking the inverse Laplace transform of the above obtained expression, we get

$\omega \left(t\right)=-4.762e^{-195.833t}\cos \left(73.479t\right)-12.691e^{-195.833t}\sin \left(73.479t\right)+\frac{100}{21}$

That shows the response does contain oscillations. Thus, the time taken by the oscillations to decay is $t=4\tau =\frac{4}{195.833}=0.02\text{seconds}$.

The characteristic polynomial for the system is as follows:

$L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T=0$

On keeping the values, we have

$\begin{array}{l}{L}_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T=0t=4\tau =\frac{4}{195.833}=0.02\sec onds\\ \Rightarrow 3\times {10}^{-3}\times 8\times {10}^{-5}{s}^2+\left(0.8\times 8\times {10}^{-5}+0.01\times 3\times {10}^{-3}\right)s+0.01\times 0.8+0.05\times 0.05=0\\ \Rightarrow 24\times {10}^{-8}{s}^2+9.4\times {10}^{-5}s+105\times {10}^{-4}=0\\ \Rightarrow 24s^2+9400s+1050000=0\\ \Rightarrow 3s^2+1175s+131250=0\end{array}$

Thus, the roots are as follows:

$s=-\frac{1175}{6}+i\frac{25\sqrt{311}}{6},s=-\frac{1175}{6}-i\frac{25\sqrt{311}}{6}$

For $c=0.1\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$

Since,

$\frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{K_T}{L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T}$

Keeping the values of all the known parameters such that

$\frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{K_T}{L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T}$

$\begin{array}{l}\Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{0.05}{3\times {10}^{-3}\times 8\times {10}^{-5}{s}^2+\left(0.8\times 8\times {10}^{-5}+0.1\times 3\times {10}^{-3}\right)s+0.1\times 0.8+0.05\times 0.05}\\ \Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{0.05}{24\times {10}^{-8}{s}^2+\left(6.4\times {10}^{-5}+30\times {10}^{-5}\right)s+825\times {10}^{-4}}\\ \Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{0.05}{24\times {10}^{-8}{s}^2+36.4\times {10}^{-5}s+825\times {10}^{-4}}\end{array}$

$\Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{1}{4.8\times {10}^{-6}{s}^2+7.28\times {10}^{-3}s+165\times {10}^{-2}}$

For the unit step input voltage response, we have

$\begin{array}{l}\frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{1}{4.8\times {10}^{-6}{s}^2+7.28\times {10}^{-3}s+165\times {10}^{-2}}\\ \Rightarrow \Omega \left(s\right)=\frac{1}{s\left(4.8\times {10}^{-6}{s}^2+7.28\times {10}^{-3}s+165\times {10}^{-2}\right)}\\ \Rightarrow \Omega \left(s\right)=\frac{{10}^6}{s\left(4.8s^2+7280s+1650000\right)}\end{array}$

On taking partial fraction expansion for the above expression, we have

$\begin{array}{l}\Omega \left(s\right)=\frac{1000000}{s\left(4.8s^2+7280s+1650000\right)}=\frac{A}{s}+\frac{Bs+C}{4.8s^2+7280s+1650000}\\ \Rightarrow \frac{1000000}{s\left(4.8s^2+7280s+1650000\right)}=\frac{A\left(4.8s^2+7280s+1650000\right)+s\left(Bs+C\right)}{s\left(4.8s^2+7280s+1650000\right)}\end{array}$

On comparing the numerator of both sides

$\begin{array}{l}\Rightarrow 1000000=A\left(4.8s^2+7280s+1650000\right)+s\left(Bs+C\right)\\ \therefore A=\frac{20}{33},B=\text{-2}\text{.909}andC=\text{-4412}\text{.121}\end{array}$

Therefore,

$\Omega \left(s\right)=\frac{-2.909s-4412.121}{4.8s^2+7280s+1650000}+\frac{20}{33s}$

$\Rightarrow \Omega \left(s\right)=-0.606\frac{\left(s+758.33\right)}{\left({\left(s+758.33\right)}^2-231319.444\right)}-459.61\frac{1}{\left({\left(s+758.33\right)}^2-231319.444\right)}+\frac{20}{33s}$

On taking the inverse Laplace transform of the above obtained expression, we get

$\omega \left(t\right)=-0.606e^{-758.33t}\cosh \left(480.957t\right)-0.956e^{-758.33t}\sinh \left(480.957t\right)+\frac{20}{33}$

That shows the response doesn’t contain any oscillations. However, the steady-state for the response could be obtained at $t=4\tau =\frac{4}{758.33}=0.005\text{seconds}$.

The characteristic polynomial for the system is as follows:

$L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T=0$

On keeping the values, we have

$\begin{array}{l}{L}_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T=0\\ \Rightarrow 3\times {10}^{-3}\times 8\times {10}^{-5}{s}^2+\left(0.8\times 8\times {10}^{-5}+0.1\times 3\times {10}^{-3}\right)s+0.1\times 0.8+0.05\times 0.05=0\\ \Rightarrow 24\times {10}^{-8}{s}^2+36.4\times {10}^{-5}s+825\times {10}^{-4}=0\\ \Rightarrow 24s^2+36400s+8250000=0\\ \Rightarrow 3s^2+4550s+1031250=0\end{array}$

Thus, the roots are as follows:

$s=\frac{25\left(\sqrt{3331}-91\right)}{3},s=-\frac{25\left(91+\sqrt{3331}\right)}{3}$.

Conclusion:

The obtained characteristic roots for the various values of damping constant are as follows:

For $c=0\text{N}\cdot \text{m}\cdot \text{s/rad}$, $s=\frac{25\left(\sqrt{106}-16\right)}{3},s=-\frac{25\left(16+\sqrt{106}\right)}{3}$

For $c=0.01\text{N}\cdot \text{m}\cdot \text{s/rad}$, $s=-\frac{1175}{6}+i\frac{25\sqrt{311}}{6},s=-\frac{1175}{6}-i\frac{25\sqrt{311}}{6}$

For $c=0.1\text{N}\cdot \text{m}\cdot \text{s/rad}$, $s=\frac{25\left(\sqrt{3331}-91\right)}{3},s=-\frac{25\left(91+\sqrt{3331}\right)}{3}$

Thus, we found that for the values of damping constant, $c=0\text{or}0.1\text{N}\cdot \text{m}\cdot \text{s/rad}$ the roots lie on the left side of the imaginary axis in s-plane. And, these roots are real and distinct in nature. While, for damping constant, $c=0.01\text{N}\cdot \text{m}\cdot \text{s/rad}$ the roots are also on the left of the imaginary axis in s-plane, but these are complex in nature.

The obtained response to step voltage input for the various values of damping constant are as follows:

For $c=0\text{N}\cdot \text{m}\cdot \text{s/rad}$, $\omega \left(t\right)=-20e^{-133.33t}\cosh \left(85.797t\right)-31.08e^{-133.33t}\sinh \left(85.797t\right)+20$

For $c=0.01\text{N}\cdot \text{m}\cdot \text{s/rad}$,

$\omega \left(t\right)=-4.762e^{-195.833t}\cos \left(73.479t\right)-12.691e^{-195.833t}\sin \left(73.479t\right)+\frac{100}{21}$

For $c=0.1\text{N}\cdot \text{m}\cdot \text{s/rad}$,

$\omega \left(t\right)=-0.606e^{-758.33t}\cosh \left(480.957t\right)-0.956e^{-758.33t}\sinh \left(480.957t\right)+\frac{20}{33}$

Here, for the values of damping constant, $c=0\text{or}0.1\text{N}\cdot \text{m}\cdot \text{s/rad}$ the speed responses are stable and free from oscillations. While, for the damping constant, $c=0.01\text{N}\cdot \text{m}\cdot \text{s/rad}$ the response is oscillatory and finally settles for a steady-state in future that is stable.

The time taken for the oscillations to decay for various values of damping constants is as follows:

For $c=0\text{N}\cdot \text{m}\cdot \text{s/rad}$, there won’t be any oscillations as the response includes hyperbolic functions. However, the response reaches the steady state at $t=4\tau =\frac{4}{133.33}=0.03\text{seconds}$.

For $c=0.01\text{N}\cdot \text{m}\cdot \text{s/rad}$, the time taken by the oscillations to decay is $t=4\tau =\frac{4}{195.833}=0.02\text{seconds}$.

For $c=0.1\text{N}\cdot \text{m}\cdot \text{s/rad}$, there won’t be any oscillations as the response includes hyperbolic functions. However, the response reaches the steady state at $t=4\tau =\frac{4}{758.33}=0.005\text{seconds}$.

Thus, we see that for values of damping constants $c=0,0.1\left(\text{N}\cdot \text{m}\cdot \text{s/rad}\right)$, the characteristic roots are real and lie on the left-hand side of the s-plane. While for the damping constant $c=0.01\text{N}\cdot \text{m}\cdot \text{s/rad}$, the characteristic roots are complex in nature.