Question

1. Produce a block diagram for the following differential equation. 5a(t)+0.1a(t)-5U, (t)

Answer 1

Answer:

Given equation is:

Given equation is: 5 d alpha/dt + 0.1 alpha(t) = 5U_s (t) This equation can also be written as: d alpha/dt = U_s (t) - 1/50 alpha (t) We will now integrate this equation from time 0 to t with theta as integration variable: integral_0^t (alpha dot) d theta = alpha(t) - alpha (0) = integral_0^t [U_s (theta) - 1/50 alpha (theta) d theta] that we give: alpha (t) = alpha (0) + integral_0^t [U_s (theta) - 1/50 alpha (theta)] d theta We will use this equation to draw the block diagram as below:

This equation can also be written as:

$\frac{\mathrm{d} \alpha }{\mathrm{d} t} = U_{s}(t) - \frac{1}{50}\alpha \left ( t \right )$

We will now integrate this euqation from time 0 to t with $\theta$ as integration variable:

$\int_{0}^{t}\left \{ \dot{\alpha} \right \}d\theta = \alpha (t) - \alpha (0)$ $= \int_{0}^{t}[ U_{s}(\theta) - \frac{1}{50}\alpha \left ( \theta \right )]d\theta$

that will give:

$\alpha (t)= \alpha (0) + \int_{0}^{t}[ U_{s}(\theta) - \frac{1}{50}\alpha \left ( \theta \right )]d\theta$

We will use this equation to draw the block diagram as below: