Question 10.8.6: The block diagram of a speed control system is shown in Figu......

(a) After some algebra we obtain the equation for the motor torque from the block diagram.

K_T=K_b=0.04  N·m / A \quad N=1.5 \quad K_P=0.63

I_e=1.802 \times 10^{-3}  kg·m ^2 \quad c_e=4.444 \times 10^{-4}  N·m·s / rad

R_a=0.6  \Omega \quad L_a=2 \times 10^{-3}  H

T_m(s)=N K_P K_T \frac{L_a I_e s^2  +  \left(R_a I_e  +  c_e L_a\right) s  +  R_a c_e}{\left(L_a s  +  R_a\right) D(s)} \Omega_r(s)+\frac{K_T\left(K_P  +  N K_b\right) / N}{D(s)} T_L(s)            (1)

where

D(s)=N L_a I_e s^2+N\left(R_a I_e+c_e L_a\right) s+N R_a c_e+N K_T K_b+K_P K_T

The current is found from

I_a(s)=\frac{1}{K_T} T_m(s)

Equation (1) should raise our suspicions because the denominators of the two transfer functions T_m(s)/Ω_r (s)  and  T_m(s)/T_L (s) differ by the factor L_as + R_a. Always keeping in mind the physics of the problem, we observe that there are only two ways energy can be stored in this system (as electromagnetic energy in the inductor L_a and as kinetic energy in the inertia I_e), so the system model should be second order. However, the denominator of the transfer function T_m(s)/Ω_r (s) appears to be third order. The only way the denominators could be the same is if the term L_as + R_a is canceled by an identical term in the numerator. If this is the case we could express the numerator as

(a s+b)\left(L_a s+R_a\right)=a L_a s^2+\left(b L_a+a R_a\right) s+b R_a

Comparing the coefficients with those of the numerator of equation (1), we see that a = I_e  and  b = c_e. Thus, our suspicion is confirmed, and we may write equation (1) as

T_m(s)=\frac{N K_P K_T\left(I_e s  +  c_e\right)}{D(s)} \Omega_r(s)+\frac{K_T\left(K_P  +  N K_b\right) / N}{D(s)} T_L(s)

Another way to detect the cancellation of the factor L_as + R_a is to obtain the partial-fraction expansion for either the free or forced response. The expansion coefficient (the residue) corresponding to the factor s + R_a/L_a would be zero and thus this factor has no influence on the response. A zero residue indicates that a denominator factor has been canceled by a numerator factor.

(b) Use the transfer function I_a(s)/Ω_r (s) with Ω_r (s) = 104.7/s to compute the required armature current as a function of time. This can be done analytically, using the methods of Chapter 2, or numerically, using the methods of Chapter 5. The result is shown in Figure 10.8.9. The plot shows that the motor will require 96 A! This is a large value, as most motors with similar parameter values have a demagnetization current of less than 50 A. We can reduce the maximum current by accepting a steady-state speed error greater than 10%, by using a smaller value for K_P . For example, a speed error of 20% with K_P = 0.28 results in a maximum current of 45 A.

One way to reduce the maximum required current is to use a speed command that increases slowly, such as a ramp function. The step and impulse functions are the most severe inputs that can be applied to a system, because they change their value instantaneously. In many systems it is physically difficult to apply such an input, but these functions are often used in analysis because they result in simpler mathematics.

Suppose K_P =0.63, which corresponds to a 10% speed error. Consider the command input ω_r that is a ramp for 0 ≤ t ≤ t_1, and for t > t_1 is a constant value of 104.7 rad/s (1000 rpm), the desired speed. By experimenting with the value of the time t_1 while examining the resulting maximum current, we can arrive at a value for t_1 that results in a maximum current of less than a desired value. The result for t_1 = 0.5 s is shown in Figure 10.8.10, which can be obtained with MATLAB. The top graph shows the current and the bottom graph shows the modified-ramp command input and the resulting load speed (assuming that T_L = 0). The speed reaches its steady-state value in about 0.6 s, as compared to 0.054 s for the step input. Now, however, the maximum current is 14 A, which is much less than the 96 A required for the step command input.