Question 2.5.2: A System of Equations a. Obtain the transfer functions X(s)/...

a. Here two outputs are specified, x and y, with one input, v. Thus, there are two transfer functions. To obtain them, transform both sides of each equation, assuming zero initial conditions.

sX(s)=-3X(s)+2Y(s) \\ sY(s)=-9Y(s)-4X(s)+3V(s)

These are two algebraic equations in the two unknowns, X(s) and Y(s). Solve the first equation for Y(s):

Y(s)=\frac{s+3}{2}X(s) \quad (1)

Substitute this into the second equation.

s\frac{s+3}{2}X(s)=-9\frac{s+3}{2}X(s)-4X(s)+3V(s)

Then solve for X(s)/V(s) to obtain

\frac{X(s)}{V(s)}=\frac{6}{s^2+12s+35} \quad (2)

Now substitute this into equation (1) to obtain

\frac{Y(s)}{V(s)}=\frac{s+3}{2} \frac{X(s)}{V(s)}=\frac{s+3}{2} \frac{6}{s^2+12s+35}=\frac{3(s+3)}{s^2+12s+35} \quad (3)

The desired transfer functions are given by equations (2) and (3). Note that denominators of both transfer functions have the same factors, s = −5 and s = −7, which are the roots of the characteristic equation: s² + 12s + 35.

b. From equation (2),

X(s)=\frac{6}{s^2+12s+35}V(s)=\frac{6}{s^2+12s+35}\frac{5}{s}=\frac{30}{s(s^2+12s+35)}

The denominator factors a re s = 0, s= −5, and s = −7, and thus the partial-fraction expansion is

X(s)=\frac{C_1}{s}+\frac{C_2}{s+5}+\frac{C_3}{s+7}

where C_1 = 6/7, C_2 = −3, and C_3 = 15/7. The forced response is

x(t)=\frac{6}{7} – 3e^{-5t}+\frac{15}{7}e^{-7t} \quad (4)

From (1) we have y = (ẋ + 3x)/2. From (4) we obtain

y(t)=\frac{9}{7}+3e^{-5t}-\frac{30}{7}e^{-7t}