LAPLACE TIME DELAY

$$\mathcal{L}{f(t - a)u(t - a)} = e^{-as} F(s)$$

Time Delay Example:

After energizing relay, it takes time before power reaching other end.

Essentially, delay from signal sent to signal received to output motion carried out.

Transfer Function is in polynomial form, traditionally to see poles in the CE, denominator.

When it isn't given in polynomial form, it becomes hard to find poles.

Make Bode Diagram and look at -3 dB

Read Breakpoint Frequency, w_b $$s = a \pm jb$$

In transfer function G(s), substitute $$s=jw_b=\frac{j}{\tau}$$

Solve the Transfer Function

Convert to Polar Form $$z = a + jb , $$ $$z = r \angle \theta , \quad r = \sqrt{a^2 + b^2}$$ $$ \quad \theta = \tan^{-1}!\left(\frac{b}{a}\right)$$

$$r = k_dc$$

No transfer function for returning pass (unity)

Input=Output for returning pass.

ex: cruise control. r=desired speed, y=actual speed

• $$r-y = e$$

• \(e\) represents the tracking error, the difference between the desired output \(r\) and the actual output \(y\).

$$ u(t) = K_p e(t) + K_i \int e(t)\,dt + K_d \frac{de}{dt} $$

• The control signal \(u\) to the plant is equal to the proportional gain \(K_p\) times the magnitude of the error plus the integral gain \(K_i\) times the integral of the error plus the derivative gain \(K_d\) times the derivative of the error.

When you route error into the control system.

Example gravel depositing on weighted belt * Weight too heavy, system controls flow rate to reduce weight * Weight is sufficient, system does no work to correct error = neutral zone