## Low Pass Filters

Refer to the below diagram to answer the following questions.

- Derive an expression that relates w(actual) to w(d) and F(d).
- Explain why high-gain feedback (i.e. a big proportional gain K) was able to better cancel the effects of friction on the motor shaft and make the motor spin at a speed closer to the desired velocity.
- Derive the transfer function of the closed-loop system shown in the block diagram.
- What type of filter does this transfer function resemble? What is its constant? Recall that transfer functions are found by setting all inputs and disturbances to zero.

##
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__**Hint**

**Hint**

The control law relating K and ω is given by:

$$$F=K(\omega _{d}-\omega)$$$

##
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__**Hint 2**

**Hint 2**

To derive the transfer function, solve for
$$\frac{\omega}{\omega_d}$$

The control law relating K and ω is given by:

$$$F=K(\omega _{d}-\omega)$$$

Adding the disturbance force we then have:

$$$\omega=[K(\omega _{d}-\omega)+F_d]\frac{K_m}{S}\rightarrow \frac{KK_m}{S}\omega_d+\frac{K_m}{s}F_d=\omega + \frac{KK_m}{s}\omega=\omega*\frac{(S+KK_m)}{S}$$$

$$$\omega=\frac{KK_m}{S+KK_m}\omega_d+\frac{K_m}{S+KK_m}F_d$$$

Therefore, if K is large, then the effect from the disturbance force is effectively cancelled and
$$\omega$$
~
$$\omega_d$$

Setting the disturbance to zero, we have:

$$$\omega=\frac{K_m}{S}K(\omega_d-\omega)$$$

$$$\frac{\omega}{\omega_d}=\frac{KK_m}{S+KK_m}=\frac{1}{\frac{1}{KK_m}+1}$$$

The system’s time constant is given by:
$$\tau =1/KK_m$$

The transfer function is similar to that of a low pass filter.

- $$\omega=\frac{KK_m}{S+KK_m}\omega_d+\frac{K_m}{S+KK_m}F_d$$
- If K is large, then the effect from the disturbance force is effectively cancelled and $$\omega$$ ~ $$\omega_d$$
- $$\frac{\omega}{\omega_d}=\frac{KK_m}{S+KK_m}=\frac{1}{\frac{1}{KK_m}+1}$$
- The transfer function is similar to that of a low pass filter. The system’s time constant is given by: $$\tau =1/KK_m$$