Question 4.3: Modeling a continuous-time feedback system For the system il......

(a) From the diagram we can write the differential equation for this system by realizing that the output signal from the summing junction is  y^{\prime\prime}(t) and it must equal the sum of its input signals

\mathbf{y}^{\prime\prime}(t)=\mathbf{x}(t)-[b\mathbf{y}^{\prime}(t)+c\mathbf{y}(t)]

With b = 0 and c = 4, the response is described by the differential equation \mathbf{y}^{\prime\prime}(t)+4\mathbf{y}(t)=\mathbf{x}(t).  The eigenfunction is the complex exponential e^{st} and the eigenvalues are the solutions of the characteristic equation s^{2}+4=0\Rightarrow s_{1,2}=\pm j2. The homogeneous solution is  \mathbf{y}(t)=K_{h1}e^{j2t}+K_{h2}e^{-j2t}.  Since there is no excitation, this is also the total solution. Applying the initial conditions \mathbf{y}(0)=K_{h\mathbf{1}}+K_{h\mathbf{2}}=1 and y^{\prime}(t)\vert_{t=0}=j2K_{h1}-j2K_{h2}=0 and solving,  K_{h1}=K_{h2}=0.5. The total solution is y(t)=0.5\left(e^{j2t}+e^{-j2t}\right)=\cos(2t){\mathrm{~,~}}t\geq0. So, with b = 0, the zero-input response is a sinusoid .

(b) Now b = 5. The differential equation is \ y^{\prime\prime}(t)+5y^{\prime}(t)+4y(t)=x(t), the eigenvalues are s_{1,2}=-1,-4 and the solution is y(t) = K_{h1}e^{-t} + K_{h2} e^{-4t}  Applying initial conditions, y(0) = K_{h1} + K_{h2} =1 and {y^\prime}(t)|_{t=0}=-K_{h1}-4K_{h2}=0. Solving for the constants , K_{h1}=4/3,K_{h2}=-1/3 and y({t}){=}(4/3)e^{-t}-(1/3)e^{-4t},t\geq0. This zero-input response approaches zero for times, t > 0.

(c) In this case x(t) is not zero and the total solution of the differential equation includes the particular solution. After t = 0 the input signal is a constant, so the particular solution

is also a constant K_{p}. The differential equation is y^{\prime \prime}(t)+b y^{\prime}(t)+4y(t)=x(t).  Solving for K_{p} we get K_{p} = 0 25 . and the total solution is \mathrm{y}(t)=K_{h1}e^{s_{1}t}+K_{h2}e^{s_{2}t}+0.25 where s_{1,2}=\left(-b\pm{\sqrt{b^{2}-16}}\right)/2. The response and its first derivative are both zero at t = 0. Applying initial conditions and solving for the remaining two constants,

(table (a))

The solutions are (table(b))

These zero-state responses are graphed in Figure 4.13.

Obviously, when b = −1 the zero-state response grows without bound and this feedback
system is unstable. System dynamics are strongly infl uenced by feedback.