Question

2). Solve the following first order system response to the ramp function (2t) as described in the equation below. (The initial condition is zero). Derive an expression for the steady state error. At what time does it reach 98% of its value? Plot a graph containing the input and system response. 4x + 2x = 21

Answer 1

Solution: 2). Given equation first order equation is

$4\dot{x}+2x=2t~~~~~~~~~~~~~~~(i)$

or
dt

or
dr 1 1 dt 22

which is linear equation.

So integrating factor(IF) is $e^{\int \frac{1}{2}dt}=e^{ \frac{1}{2}t}$

Therefore solution is

$x\times e^{ \frac{1}{2}t}=\int e^{ \frac{1}{2}t}{ \frac{1}{2}t}dt+C~~~~~~(ii)$

where C is arbitrary constant of integration.

Now integrating, RHS of (ii), partially, we have

$\int e^{ \frac{1}{2}t}{ \frac{1}{2}t}dt=\frac{1}{2}\left [ 2e^{ \frac{1}{2}t} t-\int2e^{ \frac{1}{2}t}.1.dt \right ]=\frac{1}{2}\left [ 2e^{ \frac{1}{2}t} t-4e^{ \frac{1}{2}t} \right ]$

$=\left [ e^{ \frac{1}{2}t} t-2e^{ \frac{1}{2}t} \right ]$

Therefore

$x\times e^{ \frac{1}{2}t}=\left [ e^{ \frac{1}{2}t} t-2e^{ \frac{1}{2}t} \right ]+C$

$\Rightarrow x =\left [ t-2 \right ]+Ce^{- \frac{1}{2}t}$

When t=0, x=0 then C=2

Therefore$x =\left [ t-2 \right ]+2e^{- \frac{1}{2}t}$

Thus required solution is

$x =\left [ t-2 \right ]+2e^{- \frac{1}{2}t}$