Level 4 — ApplicationGuidance, Navigation & Control (GNC)

Guidance, Navigation & Control (GNC)

60 marksprintable — key stays hidden on paper

Level: 4 (Application — novel problems, no hints) Time: 60 minutes Total marks: 60

Answer all questions. Show full working. Use $...$ / $...$ for mathematics. Assume standard 3-2-1 Euler convention unless stated.

Question 1 — Attitude kinematics & quaternion propagation [12 marks]

A microsatellite spins about its body $z$ -axis at a constant rate $\omega = (0,\,0,\,\Omega)^\mathsf{T}$ rad/s. At $t=0$ its attitude quaternion (scalar-first, body-to-inertial) is $q(0) = (1,0,0,0)$ .

(a) Write the quaternion kinematics $\dot q = \tfrac12\,\Xi(q)\,\omega$ explicitly and reduce it to a scalar system in $q_0$ and $q_3$ only. [4]

(b) Solve for $q(t)$ in closed form. [4]

(c) A star tracker requires the yaw angle $\psi(t)$ . Using the standard 3-2-1 quaternion-to-Euler relation, find $\psi(t)$ and state the spin period in terms of $\Omega$ . [4]

Question 2 — TRIAD attitude determination [12 marks]

A spacecraft measures two unit vectors in its body frame: $\mathbf{b}_1 = (1,0,0)^\mathsf{T},\qquad \mathbf{b}_2 = \tfrac{1}{\sqrt2}(0,1,1)^\mathsf{T}.$ The corresponding inertial reference directions (Sun, magnetic field) are: $\mathbf{r}_1 = (0,0,1)^\mathsf{T},\qquad \mathbf{r}_2 = \tfrac{1}{\sqrt2}(0,1,1)^\mathsf{T}.$

(a) Build the body triad $\{\mathbf{t}_1,\mathbf{t}_2,\mathbf{t}_3\}$ and reference triad $\{\mathbf{s}_1,\mathbf{s}_2,\mathbf{s}_3\}$ using $\mathbf{b}_1,\mathbf{r}_1$ as the anchor directions. [6]

(b) Form the attitude matrix $C = [\mathbf{t}_1\,\mathbf{t}_2\,\mathbf{t}_3][\mathbf{s}_1\,\mathbf{s}_2\,\mathbf{s}_3]^\mathsf{T}$ (body-from-inertial). [4]

(c) Verify $C$ is a valid rotation ( $\det C = +1$ ) and comment on why TRIAD discards information from the second measurement. [2]

Question 3 — Scalar Kalman filter update [12 marks]

A 1-D altitude estimate has prior mean $\hat x^- = 100$ m with variance $P^- = 9$ m². A barometer gives a measurement $z = 106$ m of the altitude directly ( $H=1$ ) with noise variance $R = 4$ m².

(a) Derive the scalar Kalman gain $K$ that minimizes the posterior variance, and compute its value. [5]

(b) Compute the updated estimate $\hat x^+$ and posterior variance $P^+$ . [4]

(c) A second, independent barometer with the same $R=4$ then reads $z_2 = 103$ m. Update again and give the new $\hat x^+, P^+$ . Comment on the monotonic behaviour of $P$ . [3]

Question 4 — Controllability, pole placement & LQR [14 marks]

A single-axis reaction-wheel attitude plant is $\dot{\mathbf{x}} = \underbrace{\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}}_{A}\mathbf{x} + \underbrace{\begin{bmatrix}0\\1\end{bmatrix}}_{B}u,\qquad \mathbf{x}=(\theta,\dot\theta)^\mathsf{T}.$

(a) Form the controllability matrix and show the system is controllable. [3]

(b) Using state feedback $u=-K\mathbf{x}$ with $K=[k_1\,k_2]$ , place the closed-loop poles at $-2\pm j2$ . Find $k_1,k_2$ . [5]

(c) For the LQR with $Q=\mathrm{diag}(1,0)$ , $R=1$ , write the algebraic Riccati equation $A^\mathsf{T}P+PA-PBR^{-1}B^\mathsf{T}P+Q=0$ for $P=\begin{bmatrix}p_1&p_2\\p_2&p_3\end{bmatrix}$ , solve for the optimal gain $K^\ast = R^{-1}B^\mathsf{T}P$ , and state the resulting closed-loop poles. [6]

Question 5 — Proportional navigation intercept [10 marks]

An interceptor uses pure proportional navigation, commanded lateral acceleration $a_c = N\,V_c\,\dot\lambda$ , with navigation constant $N=4$ . The closing velocity is $V_c = 1000$ m/s. At an instant the line-of-sight rate is measured as $\dot\lambda = 0.02$ rad/s.

(a) Compute the commanded acceleration and express it in $g$ ( $g=9.81$ m/s²). [3]

(b) Explain physically why PN drives $\dot\lambda \to 0$ , and what an increasing $\dot\lambda$ implies about the engagement. [3]

(c) The target now manoeuvres with constant lateral acceleration $a_T = 30$ m/s². State the augmented PN (APN) command law and compute the extra acceleration term it adds. State one practical limit on achievable $a_c$ . [4]

Answer keyMark scheme & solutions

Question 1

(a) [4] For scalar-first, body-rate quaternion kinematics: $\dot q = \tfrac12\begin{bmatrix} -q_1 & -q_2 & -q_3\\ q_0 & -q_3 & q_2\\ q_3 & q_0 & -q_1\\ -q_2 & q_1 & q_0\end{bmatrix}\begin{bmatrix}0\\0\\\Omega\end{bmatrix} = \tfrac{\Omega}{2}\begin{bmatrix}-q_3\\ q_2\\ -q_1\\ q_0\end{bmatrix}.$ [2] With $q_1=q_2=0$ initially, $\dot q_1=\dot q_2=0$ , so they stay zero. Remaining system: $\dot q_0 = -\tfrac{\Omega}{2}q_3,\qquad \dot q_3 = \tfrac{\Omega}{2}q_0.$ [2]

(b) [4] This is harmonic with frequency $\Omega/2$ . With $q_0(0)=1,q_3(0)=0$ : $q_0(t)=\cos\!\big(\tfrac{\Omega t}{2}\big),\quad q_3(t)=\sin\!\big(\tfrac{\Omega t}{2}\big).$ So $q(t)=\big(\cos\tfrac{\Omega t}{2},\,0,\,0,\,\sin\tfrac{\Omega t}{2}\big)$ . [2] (This is exactly the axis-angle quaternion for rotation angle $\Omega t$ about $z$ .) [2]

(c) [4] For a pure $z$ -rotation, yaw: $\psi = \operatorname{atan2}\!\big(2(q_0q_3+q_1q_2),\,1-2(q_2^2+q_3^2)\big) = \operatorname{atan2}(2q_0q_3,\,1-2q_3^2).$ Substitute: $2q_0q_3=2\cos\tfrac{\Omega t}{2}\sin\tfrac{\Omega t}{2}=\sin\Omega t$ ; $1-2\sin^2\tfrac{\Omega t}{2}=\cos\Omega t$ . [2] $\psi(t)=\operatorname{atan2}(\sin\Omega t,\cos\Omega t)=\Omega t.$ Spin period $T = 2\pi/\Omega$ . [2]

Question 2

(a) [6] Anchor with vector 1. Body triad:

$\mathbf{t}_1=\mathbf{b}_1=(1,0,0)$ .
$\mathbf{t}_2=\dfrac{\mathbf{b}_1\times\mathbf{b}_2}{|\cdot|}$ . $\mathbf{b}_1\times\mathbf{b}_2=(1,0,0)\times\tfrac1{\sqrt2}(0,1,1)=\tfrac1{\sqrt2}(0\cdot1-0\cdot1,\;0\cdot0-1\cdot1,\;1\cdot1-0\cdot0)=\tfrac1{\sqrt2}(0,-1,1)$ . Magnitude $=1$ , so $\mathbf{t}_2=\tfrac1{\sqrt2}(0,-1,1)$ . [2]
$\mathbf{t}_3=\mathbf{t}_1\times\mathbf{t}_2=(1,0,0)\times\tfrac1{\sqrt2}(0,-1,1)=\tfrac1{\sqrt2}(0,-1,-1)$ . [1]

Reference triad:

$\mathbf{s}_1=\mathbf{r}_1=(0,0,1)$ .
$\mathbf{r}_1\times\mathbf{r}_2=(0,0,1)\times\tfrac1{\sqrt2}(0,1,1)=\tfrac1{\sqrt2}(0\cdot1-1\cdot1,\;1\cdot0-0\cdot1,\;0\cdot1-0\cdot0)=\tfrac1{\sqrt2}(-1,0,0)$ . Magnitude 1 → $\mathbf{s}_2=\tfrac1{\sqrt2}(-1,0,0)=(-1,0,0)$ . [2]
$\mathbf{s}_3=\mathbf{s}_1\times\mathbf{s}_2=(0,0,1)\times(-1,0,0)=(0\cdot0-1\cdot0,\;1\cdot(-1)-0\cdot0,\;0)=(0,-1,0)$ . [1]

(b) [4] $C=[\mathbf t_1\,\mathbf t_2\,\mathbf t_3][\mathbf s_1\,\mathbf s_2\,\mathbf s_3]^\mathsf T=\sum_i \mathbf t_i\mathbf s_i^\mathsf T$ .

$\mathbf t_1\mathbf s_1^\mathsf T=(1,0,0)^\mathsf T(0,0,1)=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}$

$\mathbf t_2\mathbf s_2^\mathsf T=\tfrac1{\sqrt2}(0,-1,1)^\mathsf T(-1,0,0)=\begin{bmatrix}0&0&0\\ \tfrac1{\sqrt2}&0&0\\ -\tfrac1{\sqrt2}&0&0\end{bmatrix}$

$\mathbf t_3\mathbf s_3^\mathsf T=\tfrac1{\sqrt2}(0,-1,-1)^\mathsf T(0,-1,0)=\begin{bmatrix}0&0&0\\0&\tfrac1{\sqrt2}&0\\0&\tfrac1{\sqrt2}&0\end{bmatrix}$

Sum: $C=\begin{bmatrix}0&0&1\\[2pt] \tfrac1{\sqrt2}&\tfrac1{\sqrt2}&0\\[2pt] -\tfrac1{\sqrt2}&\tfrac1{\sqrt2}&0\end{bmatrix}.$ [4]

(c) [2] $\det C$ : expand along column 3 → $1\cdot\big(\tfrac1{\sqrt2}\cdot\tfrac1{\sqrt2}-\tfrac1{\sqrt2}\cdot(-\tfrac1{\sqrt2})\big)=1\cdot(\tfrac12+\tfrac12)=1$ . Valid proper rotation. [1] TRIAD uses vector 2 only to define the plane (via cross products); its component along $\mathbf{t}_1/\mathbf{s}_1$ is discarded, so measurement-2 noise is only partially used — hence it is not optimal (unlike QUEST/Davenport). [1]

Question 3

(a) [5] Posterior variance for gain $K$ : $P^+=(1-KH)^2P^-+K^2R$ . Minimize $\frac{dP^+}{dK}=0$ : $-2H(1-KH)P^-+2KR=0\Rightarrow K=\frac{P^-H}{H^2P^-+R}.$ [3] With $H=1$ : $K=\dfrac{9}{9+4}=\dfrac{9}{13}\approx0.6923.$ [2]

(b) [4] $\hat x^+=\hat x^-+K(z-H\hat x^-)=100+\tfrac{9}{13}(106-100)=100+\tfrac{54}{13}=104.1538$ m. [2] $P^+=(1-K)P^-=(1-\tfrac9{13})\cdot9=\tfrac{4}{13}\cdot9=\tfrac{36}{13}\approx2.769$ m². [2]

(c) [3] Now prior $P^-=36/13\approx2.769$ , $R=4$ . $K_2=\dfrac{36/13}{36/13+4}=\dfrac{36}{36+52}=\dfrac{36}{88}=0.4091$ . [1] $\hat x^+=104.1538+0.4091(103-104.1538)=104.1538-0.4720=103.682$ m. [1] $P^+=(1-0.4091)\cdot2.769=1.636$ m². Variance decreases monotonically with each independent measurement — information is additive, never lost. [1]

Question 4

(a) [3] $\mathcal C=[B\;AB]$ . $AB=\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}$ . $\mathcal C=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ , $\det=-1\neq0$ → rank 2 → controllable. [3]

(b) [5] $A-BK=\begin{bmatrix}0&1\\-k_1&-k_2\end{bmatrix}$ . Characteristic: $s^2+k_2 s+k_1=0$ . [2] Desired poles $-2\pm j2$ : $(s+2)^2+4=s^2+4s+8$ . Match: $k_2=4,\;k_1=8$ . So $K=[8\;4]$ . [3]

(c) [6] ARE terms with $R^{-1}B^\mathsf T P = [p_2\;p_3]$ (since $B^\mathsf T P=[p_2\,p_3]$ ). $A^\mathsf T P+PA=\begin{bmatrix}0&0\\1&0\end{bmatrix}\begin{bmatrix}p_1&p_2\\p_2&p_3\end{bmatrix}+\begin{bmatrix}p_1&p_2\\p_2&p_3\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}0&p_1\\p_1&2p_2\end{bmatrix}.$ $PBR^{-1}B^\mathsf TP=\begin{bmatrix}p_2\\p_3\end{bmatrix}[p_2\;p_3]=\begin{bmatrix}p_2^2&p_2p_3\\p_2p_3&p_3^2\end{bmatrix}.$ Equations (add $Q$ ):

(1,1): $-p_2^2+1=0\Rightarrow p_2=1$ (take $p_2>0$ ).
(1,2): $p_1-p_2p_3=0\Rightarrow p_1=p_3$ .
(2,2): $2p_2-p_3^2=0\Rightarrow p_3^2=2\Rightarrow p_3=\sqrt2$ (positive for PD). [3]

$K^\ast=[p_2\;p_3]=[1\;\sqrt2]\approx[1\;1.414]$ . [2] Closed loop: $A-BK^\ast=\begin{bmatrix}0&1\\-1&-\sqrt2\end{bmatrix}$ ; char $