Question 6 of 8: Second-Order Transfer Function — Standard Form and Response Regimes

Problem 6: Second-Order Transfer Function — Standard Form and Response Regimes (20%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A linear second-order ODE with a variable damping coefficient $k$ and unit forcing coefficient on $x$: $\ddot y+k\dot y+2y=x$.

Find. (a) $Y(s)/X(s)$ in standard $K/(\tau^2s^2+2\zeta\tau s+1)$ form; (b) the qualitative response over $-20<k<20$ and its functional form.

Approach. Transform with zero initial conditions to read off the natural time constant and gain, express damping as a function of $k$, then classify the roots of the characteristic polynomial $s^2+ks+2$ by sign and discriminant.

1. (a) Transfer function. With zero initial conditions, $s^2Y+ksY+2Y=X$, so $$\frac{Y(s)}{X(s)}=\frac{1}{s^2+ks+2}.$$
2. (a) Standard form. Divide numerator and denominator by $2$ to make the constant term unity: $$\boxed{\frac{Y}{X}=\frac{K}{\tau^2s^2+2\zeta\tau s+1},\quad K=\tfrac12,\ \tau=\tfrac{1}{\sqrt2}=0.707,\ \zeta=\frac{k}{2\sqrt2}=0.354\,k.}$$ Matching $\tau^2=1/2$ gives $\tau=0.707$; matching $2\zeta\tau=k/2$ gives $\zeta=k/(2\sqrt2)$.
3. (b) Roots and stability. The poles solve $s^2+ks+2=0$: $s=\dfrac{-k\pm\sqrt{k^2-8}}{2}$. Their product is $+2$ (both signs tied) and their sum is $-k$; the real parts are negative only when $k>0$. So the response converges for $0<k<20$ and diverges for $-20<k<0$; $k=0$ gives a sustained (undamped) oscillation. The damping character switches at the discriminant $k^2-8=0$, i.e. $|k|=2\sqrt2=2.83$.
4. (b) Response forms (convergent side). With $\zeta=0.354k$: for $0<k<2.83$ ($\zeta<1$, underdamped) the poles are complex with negative real part, $$y(t)=y_\infty+e^{-\zeta t/\tau}\big(C_1\cos\omega_d t+C_2\sin\omega_d t\big),\quad \omega_d=\tfrac{\sqrt{8-k^2}}{2};$$ at $k=2.83$ ($\zeta=1$, critically damped) $y(t)=y_\infty+(C_1+C_2t)e^{-t/\tau}$; for $2.83<k<20$ ($\zeta>1$, overdamped) $y(t)=y_\infty+C_1e^{r_1t}+C_2e^{r_2t}$ with two negative real roots $r_{1,2}$.
5. (b) Response forms (divergent side). For $-2.83<k<0$ the complex poles have positive real part: $y(t)=y_\infty+e^{+|\sigma|t}(C_1\cos\omega_d t+C_2\sin\omega_d t)$ — a growing oscillation. For $-20<k<-2.83$ the two real roots are both positive: $y(t)=y_\infty+C_1e^{r_1t}+C_2e^{r_2t}$ with $r_{1,2}>0$ — monotone divergence. At $k=0$ the poles are $\pm j\sqrt2$: $y(t)=y_\infty+C_1\cos\sqrt2\,t+C_2\sin\sqrt2\,t$, an undamped oscillation of frequency $\sqrt2$.

In standard form $K=\tfrac12$, $\tau=0.707$, $\zeta=0.354\,k$.