Chapter 6, Problem 6.41P

To determine

(a)

The step response of $i_a\left(t\right)$ and $\omega \left(t\right)$ for the armature voltage $v_a=10\text{V}$.

Answer to Problem 6.41P

The responses are as follows:

$i_a\left(t\right)=-0.125e^{-\frac{201t}{2}}\cos \left(\frac{\sqrt{40399}t}{2}\right)+2487.6875\cdot \frac{2}{\sqrt{40399}}{e}^{-\frac{201t}{2}}\sin \left(\frac{\sqrt{40399}t}{2}\right)+\frac{25}{202}$

$\omega \left(t\right)=-\frac{5000}{101}{e}^{-\frac{201t}{2}}\cos \left(\frac{\sqrt{40399}t}{2}\right)-\frac{1005000}{101\sqrt{40399}}{e}^{-\frac{201t}{2}}\sin \left(\frac{\sqrt{40399}t}{2}\right)+\frac{5000}{101}$.

Explanation of Solution

Given:

The given parameters for an armature-controlled motor are as:

$K_b=K_T=0.2\text{N}\cdot \text{m}/\text{A}$, $c=5\times {10}^{-4}\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$, $R_a=0.8\Omega$, $L_a=4\times {10}^{-3}\text{H}$, $I=5\times {10}^{-4}\text{kg}\cdot {\text{m}}^{\text{2}}$ and $v_a=10\text{V}$.

Concept Used:

For an armature-controlled motor, the output responses armature current $i_a\left(t\right)$ and motor speed $\omega \left(t\right)$ is related to input response armature voltage $v_a$ as follows:

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

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

Calculation:

Since,

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

Keeping the values of all the parameters, we get

$\begin{array}{l}\frac{I_a\left(s\right)}{V_a\left(s\right)}=\frac{Is+c}{L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T}\\ \Rightarrow \frac{I_a\left(s\right)}{V_a\left(s\right)}=\frac{5\times {10}^{-4}s+5\times {10}^{-4}}{4\times {10}^{-3}\times 5\times {10}^{-4}{s}^2+\left(0.8\times 5\times {10}^{-4}+5\times {10}^{-4}\times 4\times {10}^{-3}\right)s+5\times {10}^{-4}\times 0.8+0.2\times 0.2}\\ \Rightarrow \frac{I_a\left(s\right)}{V_a\left(s\right)}=\frac{5\times {10}^{-4}\left(s+1\right)}{2\times {10}^{-6}{s}^2+\left(4\times {10}^{-4}+2\times {10}^{-6}\right)s+4\times {10}^{-4}+0.04}\\ \Rightarrow \frac{I_a\left(s\right)}{V_a\left(s\right)}=\frac{\left(s+1\right)}{0.004s^2+0.804s+80.8}\end{array}$

Now, since input armature voltage is a unit step response of strength $10\text{V}$.

Therefore,

$\begin{array}{l}{I}_a\left(s\right)=\frac{\left(s+1\right)}{\left(0.004s^2+0.804s+80.8\right)}{V}_a\left(s\right)\\ \Rightarrow I_a\left(s\right)=\frac{\left(s+1\right)}{\left(0.004s^2+0.804s+80.8\right)}\cdot \frac{10}{s}\end{array}$

On using the partial fraction expansion in order to simplify above expression, we have

$\begin{array}{l}{I}_a\left(s\right)=\frac{\left(s+1\right)}{s\left(0.004s^2+0.804s+80.8\right)}\cdot \frac{10}{s}=\frac{A}{s}+\frac{Bs+C}{\left(0.004s^2+0.804s+80.8\right)}\\ \Rightarrow \frac{10\left(s+1\right)}{s\left(0.004s^2+0.804s+80.8\right)}=\frac{A\left(0.004s^2+0.804s+80.8\right)+s\left(Bs+C\right)}{s\left(0.004s^2+0.804s+80.8\right)}\end{array}$

On comparing the numerator on both sides, we get

$\begin{array}{l}\Rightarrow 10s+10=\left(0.004A+B\right)s^2+\left(0.804A+C\right)s+80.8A\\ \therefore A=\frac{25}{202},B=-0.0005,C=9.9005\end{array}$

Thus,

$\begin{array}{l}{I}_a\left(s\right)=\frac{A}{s}+\frac{Bs+C}{\left(0.004s^2+0.804s+80.8\right)}\\ \Rightarrow I_a\left(s\right)=\frac{25}{202}\cdot \frac{1}{s}+\frac{9.9005-0.0005s}{\left(0.004s^2+0.804s+80.8\right)}\\ \Rightarrow I_a\left(s\right)=\frac{25}{202}\cdot \frac{1}{s}-0.125\cdot \frac{s+\frac{201}{2}}{{\left(s+\frac{201}{2}\right)}^2+\frac{40399}{4}}+2487.6875\cdot \frac{1}{{\left(s+\frac{201}{2}\right)}^2+\frac{40399}{4}}\end{array}$

On taking the inverse Laplace of above obtained expression, the response $i_a\left(t\right)$ is,

$i_a\left(t\right)=-0.125e^{-\frac{201t}{2}}\cos \left(\frac{\sqrt{40399}t}{2}\right)+2487.6875\cdot \frac{2}{\sqrt{40399}}{e}^{-\frac{201t}{2}}\sin \left(\frac{\sqrt{40399}t}{2}\right)+\frac{25}{202}$

Similarly, for the response $\omega \left(t\right)$,

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 parameters in this transfer function,

$\begin{array}{l}\frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{K_T}{L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T}\\ \Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{0.2}{4\times {10}^{-3}\times 5\times {10}^{-4}{s}^2+\left(0.8\times 5\times {10}^{-4}+5\times {10}^{-4}\times 4\times {10}^{-3}\right)s+5\times {10}^{-4}\times 0.8+0.2\times 0.2}\\ \Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{0.2}{2\times {10}^{-6}{s}^2+\left(4\times {10}^{-4}+2\times {10}^{-6}\right)s+4\times {10}^{-4}+0.04}\end{array}$

$\begin{array}{l}\Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{1}{{10}^{-5}{s}^2+\left(2\times {10}^{-3}+{10}^{-5}\right)s+2\times {10}^{-3}+0.2}\\ \Rightarrow \frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{{10}^5}{s^2+201s+20200}\end{array}$

Now, since input armature voltage is a unit step response of strength $10\text{V}$.

Therefore,

$\begin{array}{l}\frac{\Omega \left(s\right)}{V_a\left(s\right)}=\frac{{10}^5}{s^2+201s+20200}\\ \Rightarrow \Omega \left(s\right)=\frac{{10}^5}{\left(s^2+201s+20200\right)}{V}_a\left(s\right)\\ \Rightarrow \Omega \left(s\right)=\frac{{10}^6}{\left(s^2+201s+20200\right)}\cdot \frac{1}{s}\end{array}$

On using the partial fraction expansion in order to simplify the above expression, we have

$\begin{array}{l}\Omega \left(s\right)=\frac{{10}^6}{\left(s^2+201s+20200\right)}\cdot \frac{1}{s}=\frac{A}{s}+\frac{Bs+C}{\left(s^2+201s+20200\right)}\\ \Rightarrow \frac{{10}^6}{s\left(s^2+201s+20200\right)}=\frac{A\left(s^2+201s+20200\right)+\left(Bs+C\right)s}{s\left(s^2+201s+20200\right)}\end{array}$

Compare the denominator of both sides

Thus,

$\begin{array}{l}\Rightarrow {10}^6=A\left(s^2+201s+20200\right)+\left(Bs+C\right)s\\ \therefore A=\frac{5000}{101},B=-\frac{5000}{101}andC=-\frac{1005000}{101}\end{array}$

$\Omega \left(s\right)=\frac{{10}^6}{\left(s^2+201s+20200\right)}\cdot \frac{1}{s}=\frac{-5000s-1005000}{101\left(s^2+201s+20200\right)}+\frac{5000}{101s}$

$\Rightarrow \Omega \left(s\right)=-\frac{5000}{101}\cdot \frac{s+\frac{201}{2}}{{\left(s+\frac{201}{2}\right)}^2+\frac{40399}{4}}-\frac{502500}{101}\cdot \frac{1}{{\left(s+\frac{201}{2}\right)}^2+\frac{40399}{4}}+\frac{5000}{101s}$

On taking the inverse Laplace of the above obtained expression, the response $\omega \left(t\right)$ is,

$\omega \left(t\right)=-\frac{5000}{101}{e}^{-\frac{201t}{2}}\cos \left(\frac{\sqrt{40399}t}{2}\right)-\frac{1005000}{101\sqrt{40399}}{e}^{-\frac{201t}{2}}\sin \left(\frac{\sqrt{40399}t}{2}\right)+\frac{5000}{101}$.

Conclusion:

The responses $i_a\left(t\right)$ and $\omega \left(t\right)$ for the armature-controlled motor is,

$i_a\left(t\right)=-0.125e^{-\frac{201t}{2}}\cos \left(\frac{\sqrt{40399}t}{2}\right)+2487.6875\cdot \frac{2}{\sqrt{40399}}{e}^{-\frac{201t}{2}}\sin \left(\frac{\sqrt{40399}t}{2}\right)+\frac{25}{202}$

$\omega \left(t\right)=-\frac{5000}{101}{e}^{-\frac{201t}{2}}\cos \left(\frac{\sqrt{40399}t}{2}\right)-\frac{1005000}{101\sqrt{40399}}{e}^{-\frac{201t}{2}}\sin \left(\frac{\sqrt{40399}t}{2}\right)+\frac{5000}{101}$.

To determine

(b)

The step response of $i_a\left(t\right)$ and $\omega \left(t\right)$ for the load torque $T_L=0.2\text{N}\cdot \text{m}$.

Answer to Problem 6.41P

The responses are follows:

$i_a\left(t\right)=-\frac{100}{101}{e}^{-\frac{201t}{2}}\cos \left(\frac{\sqrt{40399}t}{2}\right)-\frac{20100}{101\sqrt{40399}}{e}^{-\frac{201t}{2}}\sin \left(\frac{\sqrt{40399}t}{2}\right)+\frac{100}{101}$

$\omega \left(t\right)=-3.9604e^{-\frac{201t}{2}}\cos \left(\frac{\sqrt{40399}t}{2}\right)+\frac{3.9604}{\sqrt{40399}}{e}^{-\frac{201t}{2}}\sin \left(\frac{\sqrt{40399}t}{2}\right)+\frac{400}{101}$.

Explanation of Solution

Given:

The given parameters for an armature-controlled motor are as:

$K_b=K_T=0.2\text{N}\cdot \text{m}/\text{A}$, $c=5\times {10}^{-4}\text{N}\cdot \text{m}\cdot \text{s}/\text{rad}$, $R_a=0.8\Omega$, $L_a=4\times {10}^{-3}\text{H}$, $I=5\times {10}^{-4}\text{kg}\cdot {\text{m}}^{\text{2}}$ and $v_a=10\text{V}$.

Concept Used:

For an armature-controlled motor, the output responses armature current $i_a\left(t\right)$ and motor speed $\omega \left(t\right)$ is related to input load torque $T_L$ as follows:

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

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

Calculation:

Since,

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

Keeping the values of all the parameters, we get

$\begin{array}{l}\frac{I_a\left(s\right)}{T_L\left(s\right)}=\frac{K_b}{L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T}\\ \Rightarrow \frac{I_a\left(s\right)}{T_L\left(s\right)}=\frac{0.2}{4\times {10}^{-3}\times 5\times {10}^{-4}{s}^2+\left(0.8\times 5\times {10}^{-4}+5\times {10}^{-4}\times 4\times {10}^{-3}\right)s+5\times {10}^{-4}\times 0.8+0.2\times 0.2}\end{array}$

$\begin{array}{l}\Rightarrow \frac{I_a\left(s\right)}{T_L\left(s\right)}=\frac{0.2}{2\times {10}^{-6}{s}^2+\left(4\times {10}^{-4}+2\times {10}^{-6}\right)s+4\times {10}^{-4}+0.04}\\ \Rightarrow \frac{I_a\left(s\right)}{T_L\left(s\right)}=\frac{1}{{10}^{-5}{s}^2+\left(2\times {10}^{-3}+{10}^{-5}\right)s+2\times {10}^{-3}+0.2}\\ \Rightarrow \frac{I_a\left(s\right)}{T_L\left(s\right)}=\frac{{10}^5}{s^2+201s+20200}\end{array}$

Now, since input load torque is a unit step response of strength $0.2\text{N}\cdot \text{m}$.

Therefore,

$\begin{array}{l}\frac{I_a\left(s\right)}{T_L\left(s\right)}=\frac{{10}^5}{s^2+201s+20200}\\ \Rightarrow I_a\left(s\right)=\frac{{10}^5}{s^2+201s+20200}{T}_L\left(s\right)\\ \Rightarrow I_a\left(s\right)=\frac{2\times {10}^4}{s\left(s^2+201s+20200\right)}\end{array}$

On using the partial fraction expansion in order to simplify above expression, we have

$\begin{array}{l}{I}_a\left(s\right)=\frac{2\times {10}^4}{s\left(s^2+201s+20200\right)}\\ I_a\left(s\right)=\frac{2\times {10}^4}{s\left(s^2+201s+20200\right)}=\frac{A}{s}+\frac{Bs+C}{\left(s^2+201s+20200\right)}\\ \Rightarrow \frac{2\times {10}^4}{s\left(s^2+201s+20200\right)}=\frac{A\left(s^2+201s+20200\right)+\left(Bs+C\right)s}{s\left(s^2+201s+20200\right)}\end{array}$

On comparing the numerator on both sides, we get

$\begin{array}{l}\Rightarrow 2\times {10}^4=A\left(s^2+201s+20200\right)+\left(Bs+C\right)s\\ \therefore A=\frac{100}{101},B=-\frac{100}{101},C=-\frac{20100}{101}\end{array}$

Thus,

$\begin{array}{l}{I}_a\left(s\right)=\frac{{10}^6}{\left(s^2+201s+20200\right)}\cdot \frac{1}{s}=\frac{-100s-20100}{101\left(s^2+201s+20200\right)}+\frac{100}{101s}\\ \Rightarrow I_a\left(s\right)=-\frac{100}{101}\cdot \frac{s+\frac{201}{2}}{{\left(s+\frac{201}{2}\right)}^2+\frac{40399}{4}}-\frac{10050}{101}\cdot \frac{1}{{\left(s+\frac{201}{2}\right)}^2+\frac{40399}{4}}+\frac{100}{101s}\end{array}$

On taking the inverse Laplace of the above obtained expression, the response $i_a\left(t\right)$ is as shown

$i_a\left(t\right)=-\frac{100}{101}{e}^{-\frac{201t}{2}}\cos \left(\frac{\sqrt{40399}t}{2}\right)-\frac{20100}{101\sqrt{40399}}{e}^{-\frac{201t}{2}}\sin \left(\frac{\sqrt{40399}t}{2}\right)+\frac{100}{101}$

Similarly, for the response $\omega \left(t\right)$,

Since,

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

Keeping the parameters in this transfer function,

$\begin{array}{l}\frac{\Omega \left(s\right)}{T_L\left(s\right)}=\frac{L_{a}s+R_a}{L_{a}I{s}^2+\left(R_{a}I+c{L}_a\right)s+c{R}_a+K_b{K}_T}\\ \Rightarrow \frac{\Omega \left(s\right)}{T_L\left(s\right)}=\frac{4\times {10}^{-3}s+0.8}{4\times {10}^{-3}\times 5\times {10}^{-4}{s}^2+\left(0.8\times 5\times {10}^{-4}+5\times {10}^{-4}\times 4\times {10}^{-3}\right)s+5\times {10}^{-4}\times 0.8+0.2\times 0.2}\\ \Rightarrow \frac{\Omega \left(s\right)}{T_L\left(s\right)}=\frac{4\times {10}^{-3}s+0.8}{2\times {10}^{-6}{s}^2+\left(4\times {10}^{-4}+2\times {10}^{-6}\right)s+4\times {10}^{-4}+0.04}\\ \Rightarrow \frac{\Omega \left(s\right)}{T_L\left(s\right)}=\frac{\left(s+200\right)}{0.0005s^2+0.1005s+10.1}\end{array}$

Now, since input load torque is a unit step response of strength $0.2\text{N}\cdot \text{m}$.

Therefore,

$\begin{array}{l}\frac{\Omega \left(s\right)}{T_L\left(s\right)}=\frac{\left(s+200\right)}{0.0005s^2+0.1005s+10.1}\\ \Rightarrow \Omega \left(s\right)=\frac{\left(s+200\right)}{0.0005s^2+0.1005s+10.1}{T}_L\left(s\right)\\ \Rightarrow \Omega \left(s\right)=\frac{0.2\left(s+200\right)}{s\left(0.0005s^2+0.1005s+10.1\right)}\end{array}$

On using the partial fraction expansion in order to simplify above expression, we have

$\begin{array}{l}\Omega \left(s\right)=\frac{0.2\left(s+200\right)}{s\left(0.0005s^2+0.1005s+10.1\right)}=\frac{A}{s}+\frac{Bs+C}{\left(0.0005s^2+0.1005s+10.1\right)}\\ \Rightarrow \frac{0.2\left(s+200\right)}{s\left(0.0005s^2+0.1005s+10.1\right)}=\frac{A\left(0.0005s^2+0.1005s+10.1\right)+s\left(Bs+C\right)}{s\left(0.0005s^2+0.1005s+10.1\right)}\end{array}$

Compare the denominator of both sides

Thus,

$\begin{array}{l}\Rightarrow 0.2s+40=\left(0.004A+B\right)s^2+\left(0.804A+C\right)s+80.8A\\ \therefore A=\frac{400}{101},B=-\text{0}\text{.0019},C=-\text{0}\text{.1980}\end{array}$

$\begin{array}{l}\Omega \left(s\right)=\frac{A}{s}+\frac{Bs+C}{\left(0.0005s^2+0.1005s+10.1\right)}\\ \Rightarrow \Omega \left(s\right)=\frac{-0.0019802s-0.19802}{0.0005s^2+0.1005s+10.1}+\frac{400}{101s}\\ \Rightarrow \Omega \left(s\right)=-3.9604\cdot \frac{s+\frac{201}{2}}{{\left(s+\frac{201}{2}\right)}^2+\frac{40399}{4}}+1.9802\cdot \frac{1}{{\left(s+\frac{201}{2}\right)}^2+\frac{40399}{4}}+\frac{400}{101s}\end{array}$

On taking the inverse Laplace of the above obtained expression, the response $\omega \left(t\right)$ is as shown

$\omega \left(t\right)=-3.9604e^{-\frac{201t}{2}}\cos \left(\frac{\sqrt{40399}t}{2}\right)+\frac{3.9604}{\sqrt{40399}}{e}^{-\frac{201t}{2}}\sin \left(\frac{\sqrt{40399}t}{2}\right)+\frac{400}{101}$.

Conclusion:

The responses $i_a\left(t\right)$ and $\omega \left(t\right)$ for the armature-controlled motor is as follows:

$i_a\left(t\right)=-\frac{100}{101}{e}^{-\frac{201t}{2}}\cos \left(\frac{\sqrt{40399}t}{2}\right)-\frac{20100}{101\sqrt{40399}}{e}^{-\frac{201t}{2}}\sin \left(\frac{\sqrt{40399}t}{2}\right)+\frac{100}{101}$

$\omega \left(t\right)=-3.9604e^{-\frac{201t}{2}}\cos \left(\frac{\sqrt{40399}t}{2}\right)+\frac{3.9604}{\sqrt{40399}}{e}^{-\frac{201t}{2}}\sin \left(\frac{\sqrt{40399}t}{2}\right)+\frac{400}{101}$.