Exercises — Transfer function — Laplace domain, poles and zeros

Zeros = roots of the numerator $2(s+4)=0\Rightarrow s=-4$. One finite zero. Poles = roots of the denominator $(s+1)(s+5)=0\Rightarrow s=-1,s=-5$. Two poles. Large-$s$ trend. Why we compare only the highest powers: when $s$ is enormous, every term inside a polynomial is dwarfed by its top-degree term (e.g. in $s^2+6s+5$, the $s^2$ swamps $6s$ and $5$). So the ratio of the two polynomials collapses to the ratio of their leading terms. Here $G\approx \frac{2s}{s^2}=\frac{2}{s}\to 0$. So the output shrinks at high frequency — this machine is a low-pass shape.

(a) Both real parts negative → stable. (b) Real part is $-1<0$ for both (complex pair) → stable (it rings at $\omega =3$ but decays). (c) One pole at $+0.5$ has $\sigma >0$ → unstable. (d) A pole exactly at $s=0$ has $\sigma =0$ — the term is $e^{0\cdot t}=1$, a constant that never decays. This is marginally stable (not asymptotically stable). Since our stability test demands strictly $\sigma <0$, it is not stable.

Step 1 — Laplace each term, zero initial conditions. Why: the parent note's derivative rule turns $\dot{y}\to sY$, $\ddot{y}\to s^{2}Y$, provided $y(0)=\dot{y}(0)=0$. $s^{2}Y+5sY+6Y=2U.$ Step 2 — factor out $Y$, because we want the ratio $Y/U$: $(s^2+5s+6)Y=2U.$ Step 3 — form the ratio: $G(s)=\frac{Y}{U}=\frac{2}{s^2+5s+6}=\frac{2}{(s+2)(s+3)}.$ Poles: $s=-2,-3$. Both in the LHP → stable, non-oscillatory. What the time response looks like. Why exponentials appear: each pole $p$ inverse-Laplace-transforms to a term $e^{pt}$ (the parent note's ${\mathcal{L}}^{-1}\{1/(s-p)\}=e^{pt}$). So the response to any bounded input has the shape $y(t)\sim c_1{e}^{-2t}+c_2{e}^{-3t}$, where $c_1$ and $c_2$ are constants (residues) — fixed numbers that the input and the partial-fraction split assign to each pole. They set the size of each decaying piece; the poles $-2,-3$ set the rate. Their exact values depend on the particular input, but whatever they are, both terms decay → the system settles.

Compare coefficients term-by-term with the standard denominator $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$.

• Constant term: Why this one first: it holds only ${\omega }_n$, so it isolates the natural frequency. ${\omega }_n^2=9\Rightarrow {\omega }_n=3\text{rad/s}$.
• Coefficient of $s$: Why $2\zeta {\omega }_n$ appears here: in the standard form the whole middle term is $2\zeta {\omega }_{n}s$, so its numeric coefficient (the number multiplying $s$) must equal $2\zeta {\omega }_n$. Setting them equal, $2\zeta {\omega }_n=2\Rightarrow \zeta =\frac{2}{2\cdot 3}=\frac{1}{3}\approx 0.333$.

Since $0<\zeta <1$ the system is underdamped (it rings). Poles: $s=-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}=-1\pm j3\sqrt{1-\frac{1}{9}}=-1\pm j3\cdot \frac{2\sqrt{2}}{3}=-1\pm j2\sqrt{2}.$ So $s=-1\pm j2.828$. Real part $-1$ (decays), imaginary part $2.828$ (rings) — stable, oscillatory.

Read the figure below: the two purple × marks are these poles. The coral arrow is the real part $\sigma =-1$ (how fast the envelope decays); the slate arrow is the imaginary part $\omega =2.828$ (how fast it rings). Both sit inside the mint LHP band, confirming stability at a glance.

Forecast (the 80/20 shortcut) — and its precondition. In the standard form $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$ the coefficient of $s$ is $+2\zeta {\omega }_n$. For a monic second-order polynomial, the shortcut "sign of the $s$-coefficient tells you stability" is only valid when the constant term is positive, i.e. ${\omega }_n^2>0$. Why the precondition matters: if the constant term were negative, the roots would already be real with opposite signs (one guaranteed in the RHP) no matter what the middle coefficient does — so you must check the constant term first. Here the constant term is $+5>0$ ✓, so the shortcut applies. The $s$-coefficient is $-2$ — negative — so we smell instability. Confirm with the quadratic formula: $s=\frac{2\pm \sqrt{(-2{)}^2-4\cdot 5}}{2}=\frac{2\pm \sqrt{4-20}}{2}=\frac{2\pm \sqrt{-16}}{2}=1\pm 2j.$ Real part $=+1>0$ → poles in the RHP → term $e^{+t}(\dots )$ grows → UNSTABLE (it oscillates at $\omega =2$ while blowing up).

Each pole $p$ gives a term $e^{pt}$ with time constant $\tau =1/|p|$ (the time to decay to $1/e$ of its value).

• Pole $-1$: ${\tau }_1=1\text{s}$.
• Pole $-10$: ${\tau }_2=0.1\text{s}$. The $-10$ pole's term ($e^{-10t}$) has essentially vanished by $0.3\text{s}$, while the $-1$ pole's term ($e^{-t}$) lingers for seconds. So the slow pole $s=-1$ (nearest the imaginary axis) dominates the long-term response. Ratio of time constants ${\tau }_1/{\tau }_2=10$: the slow mode lasts $10\times$ longer.

Read the figure below: the purple curve is the slow mode $e^{-t}$, the coral curve the fast mode $e^{-10t}$. Notice how the coral one has crashed near zero (its shaded area gone) before the purple one has barely bent — that visual gap is domination by the nearest pole.

Stability is identical — both have poles at $-1,-2$ in the LHP. Zeros never enter the exponents, so they cannot break stability.

The mechanism, made quantitative. Why partial fractions: they split $G$ into a sum of one-pole pieces $\frac{A_1}{s+1}+\frac{A_2}{s+2}$, each of which inverts to $A_k{e}^{p_{k}t}$. The residue $A_k$ is the weight of that decaying mode.

For $G_1(s)=\frac{2}{(s+1)(s+2)}$, cover-up gives $A_1=\frac{2}{(s+2)}{|}_{s=-1}=\frac{2}{1}=2,A_2=\frac{2}{(s+1)}{|}_{s=-2}=\frac{2}{-1}=-2,$ so the impulse response is $2e^{-t}-2e^{-2t}$. At $t=0$ this is $2-2=0$, and it rises smoothly to a single hump — no overshoot beyond its own peak, the two same-signed exponentials just add and subtract cleanly.

For $G_2(s)=\frac{4(s+0.5)}{(s+1)(s+2)}$, the zero at $-0.5$ reweights the residues: $A_1=\frac{4(s+0.5)}{(s+2)}{|}_{s=-1}=\frac{4(-0.5)}{1}=-2,A_2=\frac{4(s+0.5)}{(s+1)}{|}_{s=-2}=\frac{4(-1.5)}{-1}=6.$ Now the response is $-2e^{-t}+6e^{-2t}$. Notice the numbers changed while the poles did not: the fast mode's weight leapt from $-2$ to $+6$ because the zero sits close to the origin and amplifies the near-pole residues. A large fast-mode weight makes the output shoot up early and then relax down toward its final value — that early spike is the overshoot. So the zero location literally sets the residue arithmetic, and the residues set the overshoot.

RHP zero ($s=+0.5$): redo the cover-up with numerator $\propto (s-0.5)$. The residue signs flip in a way that makes the initial slope point the wrong way: the response first moves opposite to its final direction. This is non-minimum-phase "wrong-way" motion — pull an aircraft's stick up and it first dips before climbing. Still stable (poles unchanged), but the transient starts backwards.

Step 1 — start from the standard form (chosen because it directly parametrises ${\omega }_n,\zeta$): $G(s)=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}.$ The numerator ${\omega }_n^2$ is exactly what makes $G(0)={\omega }_n^2/{\omega }_n^2=1$, giving unity DC gain for free. Step 2 — plug numbers: ${\omega }_n^2=16$, $2\zeta {\omega }_n=2(0.5)(4)=4$: $G(s)=\frac{16}{s^2+4s+16}.$ Step 3 — poles: $s=-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}=-2\pm j4\sqrt{1-0.25}=-2\pm j4\cdot \frac{\sqrt{3}}{2}=-2\pm j2\sqrt{3}.$ So $s=-2\pm j3.464$. LHP → stable. Verify DC gain: $G(0)=\frac{16}{0+0+16}=1.$ ✓

Requirement (i): "block DC" means $G(0)=0$, i.e. a zero at $s=0$ → numerator carries a factor $s$. Requirement (ii): poles at $-3,-8$ → denominator $(s+3)(s+8)$, and we must keep both, so nothing in the numerator may cancel $(s+3)$ or $(s+8)$. Requirement (iii): high-frequency gain $\to 1$. First try the simplest numerator that satisfies (i): $G(s)=\frac{ks}{(s+3)(s+8)}?$ Check large $s$: $G\approx \frac{ks}{s^2}=\frac{k}{s}\to 0$, not $1$. The numerator degree ($1$) is one short of the denominator degree ($2$). Why this matters: high-frequency gain is the ratio of the two leading (top-degree) terms, which is nonzero only when the degrees match. So we need a second zero, placed somewhere that does not sit on a pole (that would cancel it). Choose the second zero at $s=-1$ (any value $\ne -3,-8$ works): $G(s)=\frac{ks(s+1)}{(s+3)(s+8)}.$ Now the leading terms are $k{s}^2/s^2=k$, so set $k=1$ for unit high-frequency gain. Check all three requirements:

• $G(0)=\frac{1\cdot 0\cdot (0+1)}{(3)(8)}=0$ ✓ (blocks DC),
• poles still at $s=-3$ and $s=-8$ ✓ (nothing cancelled — $-1\ne -3,-8$),
• $\underset{s\to \infty }{\lim }G(s)=1$ ✓. $\boxed{G(s)=\frac{s(s+1)}{(s+3)(s+8)}}.$ This is a washout / high-pass shape (zero at DC, unity gain up high) — exactly what a rate-damping GNC sensor uses, and it keeps both required poles.

(a) Transform, zero initial conditions. Why we may drop initial-condition terms: the derivative rule is $\mathcal{L}\{f^{\prime }\}=sF(s)-f(0)$, but a transfer function assumes zero ICs — here $y(0)=\dot{y}(0)=\ddot{y}(0)=0$ and $u(0)=0$. That last one matters on the right-hand side: $\mathcal{L}\{\dot{u}\}=sU-u(0)=sU$ only because $u(0)=0$. So:

• Left side: $s^{3}Y+4s^{2}Y+6sY+3Y$.
• Right side: $\mathcal{L}\{\dot{u}+2u\}=sU+2U$ (using $u(0)=0$). $(s^3+4s^2+6s+3)Y=(s+2)U\Rightarrow G(s)=\frac{s+2}{s^3+4s^2+6s+3}.$ (b) Zeros: numerator $s+2=0\Rightarrow s=-2$. One zero, at $s=-2$. (c) Stability. Denominator $D(s)=s^3+4s^2+6s+3$. Factoring a cubic is ugly, so use Routh–Hurwitz, which tells us pole signs without finding the poles. Coefficients $a_3=1,a_2=4,a_1=6,a_0=3$ are all positive (necessary condition passed). Build the array:

where $b=\frac{4\cdot 6-1\cdot 3}{4}=\frac{24-3}{4}=\frac{21}{4}=5.25$, and the last row is $c=\frac{b\cdot 3-4\cdot 0}{b}=3.$ First column: $1,4,5.25,3$ — all positive, so there are no sign changes → zero poles in the RHP → the system is stable. Final answer. $\boxed{G(s)=\frac{s+2}{s^3+4s^2+6s+3}}$ — one zero at $s=-2$, all three poles in the LHP, therefore stable. (The zero shapes the transient but, as always, does not touch stability.)

Read the time signature. A term $e^{-2t}$ comes from a pole at $s=-2$. A term $t{e}^{-2t}$ is the fingerprint of a repeated pole at $s=-2$ (a double root). Why: ${\mathcal{L}}^{-1}\{1/(s+2{)}^2\}=t{e}^{-2t}$. So the denominator is $(s+2{)}^2$. Structure: $G(s)=\frac{k}{(s+2{)}^2}$ (no evidence of a zero from the response shape). Fix $k$ from the final value. The final value theorem says the step's steady state is ${\lim }_{s\to 0}s\cdot \frac{1}{s}\cdot G(s)=G(0)$ (the $\frac{1}{s}$ is the Laplace of a unit step). Set $G(0)=3$: $G(0)=\frac{k}{(0+2{)}^2}=\frac{k}{4}=3\Rightarrow k=12.$ Answer: $G(s)=\frac{12}{(s+2{)}^2}$. Poles: double at $s=-2$ (LHP → stable); DC gain $G(0)=3$ ✓; response contains $e^{-2t}$ and $t{e}^{-2t}$ ✓.

Stability is decided by ::: the sign of the real part of the poles (all $\sigma <0$ = stable); zeros never affect it.