Transfer function of a PMDC Motor Drive
In this tutorial we are going to investigate the response of a DC motor, by means of the circuit model and a block diagram transfer function of the motor. The parameters for the motor are:
- La=20mH
- Ra=2.5
- J=2m (Overshoot) of 20m (Damped)
- B=0.03 or Tload=9.55
- Pnom=3kw
- Nnom=3000 rpm
- Vnom=400
Using an efficiency of \(\eta=95\%\), the parameters \(K\) and \(R_a\) are calculated
- η=95%
- K=0.95*400/314=1.2
- Pe=Pnom/0.95
- Irated=Pe/Vnom=3158/400=7.9
- Ra=0.05*400/7.9=2.5
Transfer function of the Permanent Magnet DC motor
The circuit for a PMDC motor is given below. Scope1 shows the angular speed of the motor, while in scope2 the current through the motor armature is displayed. Clearly visible is the electrical time constant \[ T_a= \frac{L_a}{R_a} \]
The motor itself is only modeled using the motor constant \(K\) \[ T = K_t \cdot I_A \\ U = K_e \cdot \omega_m \] Because electrical equals mechanical power, \(K_t\) and \(K_t\) are replaced by the same value \(K\).
The same circuit can be modeled using block diagram representation
The block diagram can be simplified, if we use the first order transfer functions for the electrical (\(R_a\) and \(L_a\)) and mechanical system (\(B\) and \(J\)).
The block diagram can be studied in time domain, but we can also obtain the bode diagram for the transfer function using the small signal analysis in Caspoc.
The bode diagram shows the frequency response, where the amplitude and phase of the transfer function \(\frac{\omega}{V}\) are calculated.
After rewriting the block diagram into \[ \frac{\omega}{V}= \frac{\frac{K}{L \cdot J}}{s^2 + (\frac{R \cdot J + L \cdot B}{L \cdot J})s + (\frac{R \cdot B}{K}+K)\frac{K}{L \cdot J}} \] we have a second order system with the time constant \(T_a=\frac{L_a}{R_a}\) and the mechanical time constant \(T_m=\frac{J R_a}{K^2}\), if \(B=0\).
In the following we will derive the equations for the second order system as well as the poles/zeros for the second order system
Starting with the equations for the electrical and mechanical system:
$$\begin{aligned}
V_a &= R_a i + L_a s i + K \omega\\
T_e &= K i= J s \omega + B \omega
\end{aligned}\tag{1}$$
using equation (1):
\[
i= \frac{J s +B}{K} \omega \tag{2}
\]
we can derive the final second order equation
$$\begin{aligned}\label{eq:va}
V_a &= (R_a + L_a s) i + K \omega\\
V_a &= (R_a + L_a s) \frac{J s +B}{K} \omega + K \omega
\end{aligned}$$
Divide by \(\omega\)
$$\begin{aligned}
\frac{V_a}{\omega}&= (R_a + L_a s) \frac{J s +B}{K} + K\\
\\
\frac{V_a}{\omega}&= \frac{R_a J s}{K} + \frac{R_a B}{K} + \frac{L_a J s^2}{K} + \frac{L_a s B}{K} + K\\
\\
\frac{V_a}{\omega}&= \frac{L_a J }{K}s^2 + (\frac{R_a J + L_a B}{K})s+ \frac{R_a B}{K} + K\\
\\
\frac{\omega}{V_a}&=\frac{1}{ \frac{L_a J }{K}s^2 + (\frac{R_a J + L_a B}{K})s+ \frac{R_a B}{K} + K}\\
\\
\frac{\omega}{V_a}&=\frac{\frac{K}{L_a J }}{ s^2 + (\frac{R_a J + L_a B}{L_a J})s+ (\frac{R_a B}{K} + K)\frac{K}{L_a J}}
\end{aligned}$$
Using:
$$\begin{aligned}
\tau_a &=\frac{L_a}{R_a}\\
\tau_b &=\frac{J}{B}\\
\tau_m &=\frac{J R_a}{K^2}
\end{aligned}$$
rewrite:
$$\begin{aligned}
\frac{\omega}{V_a}&=\frac
{ \frac{1}{K}\frac{R_a}{L_a}\frac{K^2}{R_a J} }
{ s^2 + (\frac{R_a}{L_a}+\frac{B}{J})s + (\frac{R_a}{L_a}\frac{B}{J} + \frac{R_a}{L_a}\frac{K^2}{R_a J})}
\\
\frac{\omega}{V_a}&=\frac
{ \frac{1}{K}\frac{1}{\tau_a}\frac{1}{\tau_m} }
{ s^2 + (\frac{1}{\tau_a}+\frac{1}{\tau_b})s + (\frac{1}{\tau_a}\frac{1}{\tau_b} + \frac{1}{\tau_a}\frac{1}{\tau_m})}
\end{aligned}$$
Natural frequency and damping ratio
To obtain the natural frequency and damping for second order system, we will rewrite the second order transfer function into a more general form: $$\begin{aligned} \frac{\omega}{V_a}&=C \cdot G(s)&= \frac{C \cdot\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2} \\ \\ &=\frac { \frac{1}{K}\frac{1}{\tau_a}\frac{1}{\tau_m} } { s^2 + (\frac{1}{\tau_a}+\frac{1}{\tau_b})s + (\frac{1}{\tau_a}\frac{1}{\tau_b} + \frac{1}{\tau_a}\frac{1}{\tau_m})} &= \frac{C \cdot\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2} \\ \\ &=\frac { \frac{1}{K}\frac{\tau_b}{\tau_a\tau_b\tau_m} } { s^2 + (\frac{1}{\tau_a}+\frac{1}{\tau_b})s + (\frac{\tau_m+\tau_b}{\tau_a\tau_b\tau_m})} &= \frac{C \cdot\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2} \\ \\ &=\frac { \frac{1}{K}\cdot \frac{\tau_b}{\tau_m+\tau_b} \cdot \frac{\tau_m+\tau_b}{\tau_a\tau_b\tau_m} } { s^2 + (\frac{1}{\tau_a}+\frac{1}{\tau_b})s + (\frac{\tau_m+\tau_b}{\tau_a\tau_b\tau_m})} &= \frac{C \cdot\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2} \end{aligned}$$ gives:\[ \omega_n = \sqrt{ \frac{\tau_m + \tau_b}{\tau_a \tau_b \tau_m} } = \sqrt{ \frac{1}{\tau_a}\frac{1}{\tau_b} + \frac{1}{\tau_a}\frac{1}{\tau_m} } \] and:
$$ \begin{aligned} 2 \zeta \omega_n &= \frac{1}{\tau_a} + \frac{1}{\tau_b} \\ 2 \zeta \omega_n &= \frac{\tau_a + \tau_b}{\tau_a \tau_b}\\ \zeta &= \frac{1}{2\omega_n}(\frac{1}{\tau_a} + \frac{1}{\tau_b})\\ \zeta &=\frac{1}{2}\sqrt{ \frac{\tau_a \tau_b \tau_m}{\tau_m + \tau_b} }\frac{\tau_a + \tau_b}{\tau_a \tau_b}\\ \zeta &=\frac{1}{2}\sqrt{ \frac{\tau_a \tau_b \tau_m}{\tau_m + \tau_b} } \sqrt{(\frac{\tau_a + \tau_b}{\tau_a \tau_b})^2}\\ \zeta &=\frac{1}{2}\sqrt{ \frac{\tau_a \tau_b \tau_m}{\tau_m + \tau_b} } \sqrt{\frac{(\tau_a + \tau_b)^2}{(\tau_a \tau_b)^2}}\\ \zeta &=\frac{1}{2}\sqrt{ \frac{\tau_a \tau_b \tau_m}{\tau_m + \tau_b} } \sqrt{\frac{(\tau_a + \tau_b)\tau_m}{(\tau_a \tau_b \tau_m)^2}}\\ \zeta &=\frac{1}{2}\sqrt{ \frac{\tau_a \tau_b \tau_m}{(\tau_m + \tau_b)}\frac{(\tau_a + \tau_b)\tau_m}{(\tau_a \tau_b \tau_m)^2}}\\ \zeta &=\frac{1}{2}\sqrt{ \frac{1}{(\tau_m + \tau_b)}\frac{(\tau_a + \tau_b)\tau_m}{(\tau_a \tau_b \tau_m)}} \end{aligned} $$
To simplify the results we assume that \(B=0\) and the results are easier to understand. The load is still included in the model by the parameter \(T_{Load}\)
if \(B=0\) and therefore \(\frac{1}{\tau_b} =0\):
$$\begin{aligned}
\omega_n &= \frac{1}{\sqrt{\tau_a \tau_m}}\\
\zeta &= \frac{1}{2}\sqrt{\frac{\tau_m}{\tau_a}}
\end{aligned}$$
poles
To calculate the poles form the transfer function we use the following abc formula: $$ax^2+bx+c$$ $$ x= \frac{-b \pm \sqrt{b^2 - 4 a c}}{2a}$$
setting:
$$\begin{aligned}
a&=1\\
b&=\frac{1}{\tau_a} + \frac{1}{\tau_b}\\
c&=\frac{1}{\tau_a \tau_b} + \frac{1}{\tau_a \tau_m}
\end{aligned}$$
gives the poles:
$$\begin{aligned}
x_{1,2}&= \frac{-(\frac{1}{\tau_a} + \frac{1}{\tau_b}) \pm \sqrt{(\frac{1}{\tau_a} + \frac{1}{\tau_b})^2 - 4 ( \frac{1}{\tau_a \tau_b} + \frac{1}{\tau_a \tau_m})}}{2}\\
\\
x_{1,2}&= -\frac{1}{2\tau_a} - \frac{1}{2\tau_b} \pm \sqrt{\frac{1}{4}(\frac{1}{\tau_a} + \frac{1}{\tau_b})^2 - \frac{4}{4} ( \frac{1}{\tau_a \tau_b} + \frac{1}{\tau_a \tau_m})}
\\
x_{1,2}&= -\frac{1}{2\tau_a} - \frac{1}{2\tau_b} \pm \sqrt{
\frac{1}{4\tau_a^2}+\frac{2}{4\tau_a\tau_b}+\frac{1}{4\tau_b^2}-\frac{1}{\tau_a\tau_b}-\frac{1}{\tau_a\tau_m}}
\\
x_{1,2}&= -\frac{1}{2\tau_a} - \frac{1}{2\tau_b} \pm \sqrt{
\frac{1}{4\tau_a^2}+\frac{1}{4\tau_b^2}-\frac{0.5}{\tau_a\tau_b}-\frac{1}{\tau_a\tau_m}}
\end{aligned}$$
In Caspoc the poles are calculated using the following block-diagram
The poles are visualized in a scope by two arrows, pointing at the poles.
The complete calculation of the second order transfer functions together with the evaluation of the poles is given in the example BlockDiagramParameters which can be downloaded here:
Download Simulation BlockDiagramParameters
Tasks
Task-1
In this simulation we use the circuit model. Change the Machine parameters and observe the speed current and angular speed response in the scopes.
Run the simulation and observe in the scopes:
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Task-2
The same model as above is now given as a block-diagram model.
Change the Machine parameters and observe the speed current and angular speed response in the scopes.
Run the simulation and observe in the scopes:
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Task-3
The same model as above is now simplified using two first order model.
Change the Machine parameters and observe the speed current and angular speed response in the scopes.
Run the simulation and observe in the scopes:
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Task-4
The same model as above is now extended by adding a Load torque. For this the angular speed is multiplied with \(K_{Load}\)
Change the Machine parameters and observe the speed current and angular speed response in the scopes.
Run the simulation and observe in the scopes:
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Task-5
The times constants \(\tau_a\) and \(\tau_m\) are calculated as well as the parameters for the second order functions.
Change the Machine parameters and observe the speed current and angular speed response in the scopes.
Run the simulation and observe in the scopes:
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