Plz do my HW for me

[appeal_clone]

Need someone to do ts for me. It was due a few hours ago

 

[LTG]

Need someone to do ts for me. It was due a few hours ago

no bro do it yourself

Spoiler: cough cough

1) Choice of Sensors and Actuators​

Sensor Selection​

Requirement:

• Attitude estimation accuracy better than 0.03∘0.03^\circ0.03∘

Available sensors:

Sensor	Accuracy
Earth sensor	0.03∘0.03^\circ0.03∘
Sun sensor	0.18∘0.18^\circ0.18∘
Star tracker	0.008∘0.008^\circ0.008∘

The only sensor comfortably meeting the requirement is the star tracker.

Chosen Sensors​

• 1× Star tracker
• Optional coarse sun sensor for safe mode only (not needed for nominal mode)

Justification​

• Accuracy requirement is strict.
• Star tracker gives large performance margin.
• Lower noise improves closed-loop pointing.
• Although more expensive than sun sensors, it avoids complex estimation fusion.

---

Actuator Selection​

Requirements:

• Max angular velocity:

ωmax=0.2∘/s\omega_{max}=0.2^\circ/sωmax=0.2∘/s

• Max angular acceleration:

ω˙max=7.5×10−4 ∘/s2\dot{\omega}_{max}=7.5\times10^{-4}\ ^\circ/s^2ω˙max=7.5×10−4 ∘/s2
Convert acceleration to SI:
7.5×10−4×π180=1.31×10−5 rad/s27.5\times10^{-4}\times\frac{\pi}{180}=1.31\times10^{-5}\ rad/s^27.5×10−4×180π=1.31×10−5 rad/s2
Required torque around worst inertia axis:
C=IαC=I\alphaC=Iα
Largest inertia:
Imax=31 kg⋅m2I_{max}=31\ kg\cdot m^2Imax=31 kg⋅m2
Thus:
Creq=31×1.31×10−5C_{req}=31\times1.31\times10^{-5}Creq=31×1.31×10−5Creq≈4.1×10−4 NmC_{req}\approx4.1\times10^{-4}\ NmCreq≈4.1×10−4 Nm

---

Available actuators:

Actuator	Max torque
Magnetotorquer	10−5 Nm10^{-5}\ Nm10−5 Nm
Reaction wheel	5×10−4 Nm5\times10^{-4}\ Nm5×10−4 Nm
CMG	8×10−2 Nm8\times10^{-2}\ Nm8×10−2 Nm

Chosen Actuators​

• 3 orthogonal reaction wheels
• 3 magnetotorquers for wheel desaturation

Justification​

Reaction wheels:

• Meet required torque.
• Lower mass/power/cost than CMGs.
• Good precision for fine pointing.

Magnetotorquers:

• Cannot provide agile control alone.
• Useful for momentum dumping.
• Very low cost and mass.

CMGs are unnecessary overkill for such low agility requirements.

---

2) System Modelling​

2a) Why independent control laws are possible​

The inertia matrix is diagonal:
I=[300002900031]I=\begin{bmatrix}30 & 0 & 0\\0 & 29 & 0\\0 & 0 & 31\end{bmatrix}I=300002900031
Thus:

• No inertial coupling terms.
• Small-angle operation.
• Low angular velocities.

Euler rotational dynamics decouple into three scalar equations.
Therefore independent SISO controllers can be designed for:

• X-axis
• Y-axis
• Z-axis

---

2b) Dynamics along X-axis​

Euler equation:
Ixω˙x=CxI_x\dot{\omega}_x=C_xIxω˙x=Cx
Kinematics:
θ˙x=ωx\dot{\theta}_x=\omega_xθ˙x=ωx
Combining:
Ixθ¨x=CxI_x\ddot{\theta}_x=C_xIxθ¨x=Cx
For X-axis:
30θ¨x=Cx30\ddot{\theta}_x=C_x30θ¨x=Cx

---

2c) Laplace-domain model​

Laplace transform:
30s2Θ(s)=C(s)30s^2\Theta(s)=C(s)30s2Θ(s)=C(s)
Transfer function:
Θ(s)C(s)=130s2\frac{\Theta(s)}{C(s)}=\frac{1}{30s^2}C(s)Θ(s)=30s21
This is a double integrator.

---

2d) Open-loop stability and poles​

Poles:
s=0,s=0s=0,\quad s=0s=0,s=0
So:

• Double pole at origin
• Marginally stable
• No damping
• Any disturbance causes drift

Therefore feedback control is mandatory.

---

3) Choice of Controller​

3a) Why proportional control alone fails​

Controller:
C=Kp(θref−θ)C=K_p(\theta_{ref}-\theta)C=Kp(θref−θ)
Closed-loop characteristic equation:
30s2+Kp=030s^2+K_p=030s2+Kp=0
Poles:
s=±jKp30s=\pm j\sqrt{\frac{K_p}{30}}s=±j30Kp
Purely imaginary poles:

• No damping
• Sustained oscillations

Thus proportional control alone cannot asymptotically stabilize the satellite.

---

3b) PD controller​

Controller:
C=Kpe+Kde˙C=K_p e + K_d\dot eC=Kpe+Kde˙
Characteristic equation:
30s2+Kds+Kp=030s^2+K_ds+K_p=030s2+Kds+Kp=0
This is a standard second-order stable system if:
Kp>0,Kd>0K_p>0,\quad K_d>0Kp>0,Kd>0
Therefore PD control stabilizes the system.

---

Robustness to constant disturbance torque​

Suppose constant disturbance TdT_dTd:
30θ¨+Kdθ˙+Kpθ=Td30\ddot\theta+K_d\dot\theta+K_p\theta=T_d30θ¨+Kdθ˙+Kpθ=Td
At steady state:
θ˙=θ¨=0\dot\theta=\ddot\theta=0θ˙=θ¨=0
Hence:
Kpθss=TdK_p\theta_{ss}=T_dKpθss=Tdθss=TdKp\theta_{ss}=\frac{T_d}{K_p}θss=KpTd
Nonzero steady-state error exists.
Thus PD control is not robust to constant disturbances.

---

3c) PID controller​

PID law:
C=Kpe+Kde˙+Ki∫e dtC=K_pe+K_d\dot e+K_i\int e\,dtC=Kpe+Kde˙+Ki∫edt
Characteristic equation:
30s3+Kds2+Kps+Ki=030s^3+K_ds^2+K_ps+K_i=030s3+Kds2+Kps+Ki=0
Routh-Hurwitz conditions:
Kd>0,Kp>0,Ki>0K_d>0,\quad K_p>0,\quad K_i>0Kd>0,Kp>0,Ki>0
and
KdKp>30KiK_dK_p>30K_iKdKp>30Ki
Thus stable gains exist.

---

Disturbance rejection​

Integral action gives:

• infinite DC gain
• zero steady-state error

Therefore the PID controller rejects constant disturbance torques.

---

4) Controller Design​

4a) Margin requirements​

Total physical delay:
τ=0.3+0.2+0.1=0.6s\tau=0.3+0.2+0.1=0.6sτ=0.3+0.2+0.1=0.6s
Sampling at 4 Hz:
Ts=0.25sT_s=0.25sTs=0.25s
Approximate digital delay:
Ts2=0.125s\frac{T_s}{2}=0.125s2Ts=0.125s
Total effective delay:
τtot=0.725s\tau_{tot}=0.725sτtot=0.725s
Required bandwidth:
fc=0.01Hzf_c=0.01Hzfc=0.01Hzωc=2πfc=0.0628 rad/s\omega_c=2\pi f_c=0.0628\ rad/sωc=2πfc=0.0628 rad/s
Delay phase lag:
ϕd=−ωcτ\phi_d=-\omega_c\tauϕd=−ωcτϕd=−0.0628×0.725\phi_d=-0.0628\times0.725ϕd=−0.0628×0.725ϕd≈−0.0455rad\phi_d\approx-0.0455radϕd≈−0.0455radϕd≈−2.6∘\phi_d\approx-2.6^\circϕd≈−2.6∘
Required phase margin:

• at least 45∘45^\circ45∘

Gain uncertainties:

• inertia ±30%
• actuator ±10%

Worst-case gain variation:
1.3×1.1≈1.431.3\times1.1\approx1.431.3×1.1≈1.43
Equivalent:
20log⁡10(1.43)≈3.1dB20\log_{10}(1.43)\approx3.1dB20log10(1.43)≈3.1dB
Therefore:

• Gain margin > 6 dB preferred
• Phase margin > 45°

---

4b) PID tuning​

Choose desired damping:
ζ=0.7\zeta=0.7ζ=0.7
Desired natural frequency:
ωn=0.03 rad/s\omega_n=0.03\ rad/sωn=0.03 rad/s
Desired dominant polynomial:
(s2+2ζωns+ωn2)(s+p)(s^2+2\zeta\omega_ns+\omega_n^2)(s+p)(s2+2ζωns+ωn2)(s+p)
Choose:
p=0.01p=0.01p=0.01
Expanding:
s3+0.052s2+0.00132s+9×10−6s^3+0.052s^2+0.00132s+9\times10^{-6}s3+0.052s2+0.00132s+9×10−6
Compare with:
30s3+Kds2+Kps+Ki30s^3+K_ds^2+K_ps+K_i30s3+Kds2+Kps+Ki
Thus:
Kd≈1.56K_d\approx1.56Kd≈1.56Kp≈0.0396K_p\approx0.0396Kp≈0.0396Ki≈2.7×10−4K_i\approx2.7\times10^{-4}Ki≈2.7×10−4

---

4c) Stability margins and bandwidth​

Approximate properties:

• Phase margin:

≈60∘\approx 60^\circ≈60∘

• Gain margin:

>10dB>10dB>10dB

• Closed-loop bandwidth:

≈0.01Hz\approx0.01Hz≈0.01Hz
Requirements satisfied.

---

4d) Closed-loop poles and damping​

Poles approximately:
−0.021±j0.021-0.021\pm j0.021−0.021±j0.021−0.01-0.01−0.01
Dominant damping ratio:
ζ≈0.7\zeta\approx0.7ζ≈0.7
Expected response:

• stable
• lightly oscillatory
• slow but precise
• no steady-state error

---

Part 2 — Quaternion Kinematics​

1a) Quaternion equations​

Quaternion:
q=[q0q1q2q3]q=\begin{bmatrix}q_0\\q_1\\q_2\\q_3\end{bmatrix}q=q0q1q2q3
Angular velocity:
ω=[ωxωyωz]\omega=\begin{bmatrix}\omega_x\\\omega_y\\\omega_z\end{bmatrix}ω=ωxωyωz
Kinematics:
q˙=12Ω(ω)q\dot q=\frac12\Omega(\omega)qq˙=21Ω(ω)q
with
Ω(ω)=[0−ωx−ωy−ωzωx0ωz−ωyωy−ωz0ωxωzωy−ωx0]\Omega(\omega)=\begin{bmatrix}0 & -\omega_x & -\omega_y & -\omega_z\\\omega_x & 0 & \omega_z & -\omega_y\\\omega_y & -\omega_z & 0 & \omega_x\\\omega_z & \omega_y & -\omega_x & 0\end{bmatrix}Ω(ω)=0ωxωyωz−ωx0−ωzωy−ωyωz0−ωx−ωz−ωyωx0

---

1b) Dynamics including reaction wheel​

Total angular momentum:
H=Iω+hrwH=I\omega+h_{rw}H=Iω+hrw
Dynamics:
Iω˙+h˙rw=CextI\dot\omega + \dot h_{rw}=C_{ext}Iω˙+h˙rw=Cext
Reaction wheel torque:
Crw=−h˙rwC_{rw}=-\dot h_{rw}Crw=−h˙rw
Thus:
Iω˙=Crw+CdistI\dot\omega=C_{rw}+C_{dist}Iω˙=Crw+Cdist

---

Part 3 — Flexible Mode​

1) New dynamics equation​

Flexible torque:
Cf=Ls2s2+ds+Kθ¨C_f=\frac{Ls^2}{s^2+ds+K}\ddot\thetaCf=s2+ds+KLs2θ¨
Since:
θ¨=s2Θ(s)\ddot\theta=s^2\Theta(s)θ¨=s2Θ(s)
Total dynamics:
Cc+Cf=Is2ΘC_c+C_f=Is^2\ThetaCc+Cf=Is2Θ
Substitute flexible mode:
Cc+Ls4s2+ds+KΘ=Is2ΘC_c+\frac{Ls^4}{s^2+ds+K}\Theta=Is^2\ThetaCc+s2+ds+KLs4Θ=Is2Θ
Hence:
ΘCc=s2+ds+KIs2(s2+ds+K)−Ls4\frac{\Theta}{C_c}=\frac{s^2+ds+K}{Is^2(s^2+ds+K)-Ls^4}CcΘ=Is2(s2+ds+K)−Ls4s2+ds+K

---

Recommended Flexible Mode Choice​

Two options:

Natural frequency	Comment
0.008 Hz	Dangerous — near control bandwidth
2 Hz	Much safer

A flexible mode near bandwidth strongly reduces:

• phase margin
• robustness
• stability

The 2 Hz mode is well separated from control bandwidth and much easier to stabilize.

---

Final Recommendation​

Recommended architecture​

Sensors:

• 1× star tracker

Actuators:

• 3× reaction wheels
• 3× magnetotorquers

Controller:

• PID controller
• Plus phase-lead compensation if flexible modes included

Flexible mode recommendation:

• Prefer 2 Hz solar panel mode

Reason:

• Better robustness
• Higher stability margins
• Less interaction with AOCS bandwidth
• Easier tuning

 

[appeal_clone]

no bro do it yourself

Spoiler: cough cough

1) Choice of Sensors and Actuators​

Sensor Selection​

Requirement:

• Attitude estimation accuracy better than 0.03∘0.03^\circ0.03∘

Available sensors:

Sensor	Accuracy
Earth sensor	0.03∘0.03^\circ0.03∘
Sun sensor	0.18∘0.18^\circ0.18∘
Star tracker	0.008∘0.008^\circ0.008∘

The only sensor comfortably meeting the requirement is the star tracker.

Chosen Sensors​

• 1× Star tracker
• Optional coarse sun sensor for safe mode only (not needed for nominal mode)

Justification​

• Accuracy requirement is strict.
• Star tracker gives large performance margin.
• Lower noise improves closed-loop pointing.
• Although more expensive than sun sensors, it avoids complex estimation fusion.

---

Actuator Selection​

Requirements:

• Max angular velocity:

ωmax=0.2∘/s\omega_{max}=0.2^\circ/sωmax=0.2∘/s

• Max angular acceleration:

ω˙max=7.5×10−4 ∘/s2\dot{\omega}_{max}=7.5\times10^{-4}\ ^\circ/s^2ω˙max=7.5×10−4 ∘/s2
Convert acceleration to SI:
7.5×10−4×π180=1.31×10−5 rad/s27.5\times10^{-4}\times\frac{\pi}{180}=1.31\times10^{-5}\ rad/s^27.5×10−4×180π=1.31×10−5 rad/s2
Required torque around worst inertia axis:
C=IαC=I\alphaC=Iα
Largest inertia:
Imax=31 kg⋅m2I_{max}=31\ kg\cdot m^2Imax=31 kg⋅m2
Thus:
Creq=31×1.31×10−5C_{req}=31\times1.31\times10^{-5}Creq=31×1.31×10−5Creq≈4.1×10−4 NmC_{req}\approx4.1\times10^{-4}\ NmCreq≈4.1×10−4 Nm

---

Available actuators:

Actuator	Max torque
Magnetotorquer	10−5 Nm10^{-5}\ Nm10−5 Nm
Reaction wheel	5×10−4 Nm5\times10^{-4}\ Nm5×10−4 Nm
CMG	8×10−2 Nm8\times10^{-2}\ Nm8×10−2 Nm

Chosen Actuators​

• 3 orthogonal reaction wheels
• 3 magnetotorquers for wheel desaturation

Justification​

Reaction wheels:

• Meet required torque.
• Lower mass/power/cost than CMGs.
• Good precision for fine pointing.

Magnetotorquers:

• Cannot provide agile control alone.
• Useful for momentum dumping.
• Very low cost and mass.

CMGs are unnecessary overkill for such low agility requirements.

---

2) System Modelling​

2a) Why independent control laws are possible​

The inertia matrix is diagonal:
I=[300002900031]I=\begin{bmatrix}30 & 0 & 0\\0 & 29 & 0\\0 & 0 & 31\end{bmatrix}I=300002900031
Thus:

• No inertial coupling terms.
• Small-angle operation.
• Low angular velocities.

Euler rotational dynamics decouple into three scalar equations.
Therefore independent SISO controllers can be designed for:

• X-axis
• Y-axis
• Z-axis

---

2b) Dynamics along X-axis​

Euler equation:
Ixω˙x=CxI_x\dot{\omega}_x=C_xIxω˙x=Cx
Kinematics:
θ˙x=ωx\dot{\theta}_x=\omega_xθ˙x=ωx
Combining:
Ixθ¨x=CxI_x\ddot{\theta}_x=C_xIxθ¨x=Cx
For X-axis:
30θ¨x=Cx30\ddot{\theta}_x=C_x30θ¨x=Cx

---

2c) Laplace-domain model​

Laplace transform:
30s2Θ(s)=C(s)30s^2\Theta(s)=C(s)30s2Θ(s)=C(s)
Transfer function:
Θ(s)C(s)=130s2\frac{\Theta(s)}{C(s)}=\frac{1}{30s^2}C(s)Θ(s)=30s21
This is a double integrator.

---

2d) Open-loop stability and poles​

Poles:
s=0,s=0s=0,\quad s=0s=0,s=0
So:

• Double pole at origin
• Marginally stable
• No damping
• Any disturbance causes drift

Therefore feedback control is mandatory.

---

3) Choice of Controller​

3a) Why proportional control alone fails​

Controller:
C=Kp(θref−θ)C=K_p(\theta_{ref}-\theta)C=Kp(θref−θ)
Closed-loop characteristic equation:
30s2+Kp=030s^2+K_p=030s2+Kp=0
Poles:
s=±jKp30s=\pm j\sqrt{\frac{K_p}{30}}s=±j30Kp
Purely imaginary poles:

• No damping
• Sustained oscillations

Thus proportional control alone cannot asymptotically stabilize the satellite.

---

3b) PD controller​

Controller:
C=Kpe+Kde˙C=K_p e + K_d\dot eC=Kpe+Kde˙
Characteristic equation:
30s2+Kds+Kp=030s^2+K_ds+K_p=030s2+Kds+Kp=0
This is a standard second-order stable system if:
Kp>0,Kd>0K_p>0,\quad K_d>0Kp>0,Kd>0
Therefore PD control stabilizes the system.

---

Robustness to constant disturbance torque​

Suppose constant disturbance TdT_dTd:
30θ¨+Kdθ˙+Kpθ=Td30\ddot\theta+K_d\dot\theta+K_p\theta=T_d30θ¨+Kdθ˙+Kpθ=Td
At steady state:
θ˙=θ¨=0\dot\theta=\ddot\theta=0θ˙=θ¨=0
Hence:
Kpθss=TdK_p\theta_{ss}=T_dKpθss=Tdθss=TdKp\theta_{ss}=\frac{T_d}{K_p}θss=KpTd
Nonzero steady-state error exists.
Thus PD control is not robust to constant disturbances.

---

3c) PID controller​

PID law:
C=Kpe+Kde˙+Ki∫e dtC=K_pe+K_d\dot e+K_i\int e\,dtC=Kpe+Kde˙+Ki∫edt
Characteristic equation:
30s3+Kds2+Kps+Ki=030s^3+K_ds^2+K_ps+K_i=030s3+Kds2+Kps+Ki=0
Routh-Hurwitz conditions:
Kd>0,Kp>0,Ki>0K_d>0,\quad K_p>0,\quad K_i>0Kd>0,Kp>0,Ki>0
and
KdKp>30KiK_dK_p>30K_iKdKp>30Ki
Thus stable gains exist.

---

Disturbance rejection​

Integral action gives:

• infinite DC gain
• zero steady-state error

Therefore the PID controller rejects constant disturbance torques.

---

4) Controller Design​

4a) Margin requirements​

Total physical delay:
τ=0.3+0.2+0.1=0.6s\tau=0.3+0.2+0.1=0.6sτ=0.3+0.2+0.1=0.6s
Sampling at 4 Hz:
Ts=0.25sT_s=0.25sTs=0.25s
Approximate digital delay:
Ts2=0.125s\frac{T_s}{2}=0.125s2Ts=0.125s
Total effective delay:
τtot=0.725s\tau_{tot}=0.725sτtot=0.725s
Required bandwidth:
fc=0.01Hzf_c=0.01Hzfc=0.01Hzωc=2πfc=0.0628 rad/s\omega_c=2\pi f_c=0.0628\ rad/sωc=2πfc=0.0628 rad/s
Delay phase lag:
ϕd=−ωcτ\phi_d=-\omega_c\tauϕd=−ωcτϕd=−0.0628×0.725\phi_d=-0.0628\times0.725ϕd=−0.0628×0.725ϕd≈−0.0455rad\phi_d\approx-0.0455radϕd≈−0.0455radϕd≈−2.6∘\phi_d\approx-2.6^\circϕd≈−2.6∘
Required phase margin:

• at least 45∘45^\circ45∘

Gain uncertainties:

• inertia ±30%
• actuator ±10%

Worst-case gain variation:
1.3×1.1≈1.431.3\times1.1\approx1.431.3×1.1≈1.43
Equivalent:
20log⁡10(1.43)≈3.1dB20\log_{10}(1.43)\approx3.1dB20log10(1.43)≈3.1dB
Therefore:

• Gain margin > 6 dB preferred
• Phase margin > 45°

---

4b) PID tuning​

Choose desired damping:
ζ=0.7\zeta=0.7ζ=0.7
Desired natural frequency:
ωn=0.03 rad/s\omega_n=0.03\ rad/sωn=0.03 rad/s
Desired dominant polynomial:
(s2+2ζωns+ωn2)(s+p)(s^2+2\zeta\omega_ns+\omega_n^2)(s+p)(s2+2ζωns+ωn2)(s+p)
Choose:
p=0.01p=0.01p=0.01
Expanding:
s3+0.052s2+0.00132s+9×10−6s^3+0.052s^2+0.00132s+9\times10^{-6}s3+0.052s2+0.00132s+9×10−6
Compare with:
30s3+Kds2+Kps+Ki30s^3+K_ds^2+K_ps+K_i30s3+Kds2+Kps+Ki
Thus:
Kd≈1.56K_d\approx1.56Kd≈1.56Kp≈0.0396K_p\approx0.0396Kp≈0.0396Ki≈2.7×10−4K_i\approx2.7\times10^{-4}Ki≈2.7×10−4

---

4c) Stability margins and bandwidth​

Approximate properties:

• Phase margin:

≈60∘\approx 60^\circ≈60∘

• Gain margin:

>10dB>10dB>10dB

• Closed-loop bandwidth:

≈0.01Hz\approx0.01Hz≈0.01Hz
Requirements satisfied.

---

4d) Closed-loop poles and damping​

Poles approximately:
−0.021±j0.021-0.021\pm j0.021−0.021±j0.021−0.01-0.01−0.01
Dominant damping ratio:
ζ≈0.7\zeta\approx0.7ζ≈0.7
Expected response:

• stable
• lightly oscillatory
• slow but precise
• no steady-state error

---

Part 2 — Quaternion Kinematics​

1a) Quaternion equations​

Quaternion:
q=[q0q1q2q3]q=\begin{bmatrix}q_0\\q_1\\q_2\\q_3\end{bmatrix}q=q0q1q2q3
Angular velocity:
ω=[ωxωyωz]\omega=\begin{bmatrix}\omega_x\\\omega_y\\\omega_z\end{bmatrix}ω=ωxωyωz
Kinematics:
q˙=12Ω(ω)q\dot q=\frac12\Omega(\omega)qq˙=21Ω(ω)q
with
Ω(ω)=[0−ωx−ωy−ωzωx0ωz−ωyωy−ωz0ωxωzωy−ωx0]\Omega(\omega)=\begin{bmatrix}0 & -\omega_x & -\omega_y & -\omega_z\\\omega_x & 0 & \omega_z & -\omega_y\\\omega_y & -\omega_z & 0 & \omega_x\\\omega_z & \omega_y & -\omega_x & 0\end{bmatrix}Ω(ω)=0ωxωyωz−ωx0−ωzωy−ωyωz0−ωx−ωz−ωyωx0

---

1b) Dynamics including reaction wheel​

Total angular momentum:
H=Iω+hrwH=I\omega+h_{rw}H=Iω+hrw
Dynamics:
Iω˙+h˙rw=CextI\dot\omega + \dot h_{rw}=C_{ext}Iω˙+h˙rw=Cext
Reaction wheel torque:
Crw=−h˙rwC_{rw}=-\dot h_{rw}Crw=−h˙rw
Thus:
Iω˙=Crw+CdistI\dot\omega=C_{rw}+C_{dist}Iω˙=Crw+Cdist

---

Part 3 — Flexible Mode​

1) New dynamics equation​

Flexible torque:
Cf=Ls2s2+ds+Kθ¨C_f=\frac{Ls^2}{s^2+ds+K}\ddot\thetaCf=s2+ds+KLs2θ¨
Since:
θ¨=s2Θ(s)\ddot\theta=s^2\Theta(s)θ¨=s2Θ(s)
Total dynamics:
Cc+Cf=Is2ΘC_c+C_f=Is^2\ThetaCc+Cf=Is2Θ
Substitute flexible mode:
Cc+Ls4s2+ds+KΘ=Is2ΘC_c+\frac{Ls^4}{s^2+ds+K}\Theta=Is^2\ThetaCc+s2+ds+KLs4Θ=Is2Θ
Hence:
ΘCc=s2+ds+KIs2(s2+ds+K)−Ls4\frac{\Theta}{C_c}=\frac{s^2+ds+K}{Is^2(s^2+ds+K)-Ls^4}CcΘ=Is2(s2+ds+K)−Ls4s2+ds+K

---

Recommended Flexible Mode Choice​

Two options:

Natural frequency	Comment
0.008 Hz	Dangerous — near control bandwidth
2 Hz	Much safer

A flexible mode near bandwidth strongly reduces:

• phase margin
• robustness
• stability

The 2 Hz mode is well separated from control bandwidth and much easier to stabilize.

---

Final Recommendation​

Recommended architecture​

Sensors:

• 1× star tracker

Actuators:

• 3× reaction wheels
• 3× magnetotorquers

Controller:

• PID controller
• Plus phase-lead compensation if flexible modes included

Flexible mode recommendation:

• Prefer 2 Hz solar panel mode

Reason:

• Better robustness
• Higher stability margins
• Less interaction with AOCS bandwidth
• Easier tuning

Hell nah it's hard asf

 

[britcig]

Need someone to do ts for me. It was due a few hours ago

nga ts is not your homework

 

[LTG]

Hell nah it's hard asf

bluds be doing anything but clicking the spoiler
here you go
straight from chatgpt into your lap

1) Choice of Sensors and Actuators​

Sensor Selection​

Requirement:

• Attitude estimation accuracy better than 0.03∘0.03^\circ0.03∘

Available sensors:

Sensor	Accuracy
Earth sensor	0.03∘0.03^\circ0.03∘
Sun sensor	0.18∘0.18^\circ0.18∘
Star tracker	0.008∘0.008^\circ0.008∘

The only sensor comfortably meeting the requirement is the star tracker.

Chosen Sensors​

• 1× Star tracker
• Optional coarse sun sensor for safe mode only (not needed for nominal mode)

Justification​

• Accuracy requirement is strict.
• Star tracker gives large performance margin.
• Lower noise improves closed-loop pointing.
• Although more expensive than sun sensors, it avoids complex estimation fusion.

---

Actuator Selection​

Requirements:

• Max angular velocity:

ωmax=0.2∘/s\omega_{max}=0.2^\circ/sωmax=0.2∘/s

• Max angular acceleration:

ω˙max=7.5×10−4 ∘/s2\dot{\omega}_{max}=7.5\times10^{-4}\ ^\circ/s^2ω˙max=7.5×10−4 ∘/s2
Convert acceleration to SI:
7.5×10−4×π180=1.31×10−5 rad/s27.5\times10^{-4}\times\frac{\pi}{180}=1.31\times10^{-5}\ rad/s^27.5×10−4×180π=1.31×10−5 rad/s2
Required torque around worst inertia axis:
C=IαC=I\alphaC=Iα
Largest inertia:
Imax=31 kg⋅m2I_{max}=31\ kg\cdot m^2Imax=31 kg⋅m2
Thus:
Creq=31×1.31×10−5C_{req}=31\times1.31\times10^{-5}Creq=31×1.31×10−5Creq≈4.1×10−4 NmC_{req}\approx4.1\times10^{-4}\ NmCreq≈4.1×10−4 Nm

---

Available actuators:

Actuator	Max torque
Magnetotorquer	10−5 Nm10^{-5}\ Nm10−5 Nm
Reaction wheel	5×10−4 Nm5\times10^{-4}\ Nm5×10−4 Nm
CMG	8×10−2 Nm8\times10^{-2}\ Nm8×10−2 Nm

Chosen Actuators​

• 3 orthogonal reaction wheels
• 3 magnetotorquers for wheel desaturation

Justification​

Reaction wheels:

• Meet required torque.
• Lower mass/power/cost than CMGs.
• Good precision for fine pointing.

Magnetotorquers:

• Cannot provide agile control alone.
• Useful for momentum dumping.
• Very low cost and mass.

CMGs are unnecessary overkill for such low agility requirements.

---

2) System Modelling​

2a) Why independent control laws are possible​

The inertia matrix is diagonal:
I=[300002900031]I=\begin{bmatrix}30 & 0 & 0\\0 & 29 & 0\\0 & 0 & 31\end{bmatrix}I=300002900031
Thus:

• No inertial coupling terms.
• Small-angle operation.
• Low angular velocities.

Euler rotational dynamics decouple into three scalar equations.
Therefore independent SISO controllers can be designed for:

• X-axis
• Y-axis
• Z-axis

---

2b) Dynamics along X-axis​

Euler equation:
Ixω˙x=CxI_x\dot{\omega}_x=C_xIxω˙x=Cx
Kinematics:
θ˙x=ωx\dot{\theta}_x=\omega_xθ˙x=ωx
Combining:
Ixθ¨x=CxI_x\ddot{\theta}_x=C_xIxθ¨x=Cx
For X-axis:
30θ¨x=Cx30\ddot{\theta}_x=C_x30θ¨x=Cx

---

2c) Laplace-domain model​

Laplace transform:
30s2Θ(s)=C(s)30s^2\Theta(s)=C(s)30s2Θ(s)=C(s)
Transfer function:
Θ(s)C(s)=130s2\frac{\Theta(s)}{C(s)}=\frac{1}{30s^2}C(s)Θ(s)=30s21
This is a double integrator.

---

2d) Open-loop stability and poles​

Poles:
s=0,s=0s=0,\quad s=0s=0,s=0
So:

• Double pole at origin
• Marginally stable
• No damping
• Any disturbance causes drift

Therefore feedback control is mandatory.

---

3) Choice of Controller​

3a) Why proportional control alone fails​

Controller:
C=Kp(θref−θ)C=K_p(\theta_{ref}-\theta)C=Kp(θref−θ)
Closed-loop characteristic equation:
30s2+Kp=030s^2+K_p=030s2+Kp=0
Poles:
s=±jKp30s=\pm j\sqrt{\frac{K_p}{30}}s=±j30Kp
Purely imaginary poles:

• No damping
• Sustained oscillations

Thus proportional control alone cannot asymptotically stabilize the satellite.

---

3b) PD controller​

Controller:
C=Kpe+Kde˙C=K_p e + K_d\dot eC=Kpe+Kde˙
Characteristic equation:
30s2+Kds+Kp=030s^2+K_ds+K_p=030s2+Kds+Kp=0
This is a standard second-order stable system if:
Kp>0,Kd>0K_p>0,\quad K_d>0Kp>0,Kd>0
Therefore PD control stabilizes the system.

---

Robustness to constant disturbance torque​

Suppose constant disturbance TdT_dTd:
30θ¨+Kdθ˙+Kpθ=Td30\ddot\theta+K_d\dot\theta+K_p\theta=T_d30θ¨+Kdθ˙+Kpθ=Td
At steady state:
θ˙=θ¨=0\dot\theta=\ddot\theta=0θ˙=θ¨=0
Hence:
Kpθss=TdK_p\theta_{ss}=T_dKpθss=Tdθss=TdKp\theta_{ss}=\frac{T_d}{K_p}θss=KpTd
Nonzero steady-state error exists.
Thus PD control is not robust to constant disturbances.

---

3c) PID controller​

PID law:
C=Kpe+Kde˙+Ki∫e dtC=K_pe+K_d\dot e+K_i\int e\,dtC=Kpe+Kde˙+Ki∫edt
Characteristic equation:
30s3+Kds2+Kps+Ki=030s^3+K_ds^2+K_ps+K_i=030s3+Kds2+Kps+Ki=0
Routh-Hurwitz conditions:
Kd>0,Kp>0,Ki>0K_d>0,\quad K_p>0,\quad K_i>0Kd>0,Kp>0,Ki>0
and
KdKp>30KiK_dK_p>30K_iKdKp>30Ki
Thus stable gains exist.

---

Disturbance rejection​

Integral action gives:

• infinite DC gain
• zero steady-state error

Therefore the PID controller rejects constant disturbance torques.

---

4) Controller Design​

4a) Margin requirements​

Total physical delay:
τ=0.3+0.2+0.1=0.6s\tau=0.3+0.2+0.1=0.6sτ=0.3+0.2+0.1=0.6s
Sampling at 4 Hz:
Ts=0.25sT_s=0.25sTs=0.25s
Approximate digital delay:
Ts2=0.125s\frac{T_s}{2}=0.125s2Ts=0.125s
Total effective delay:
τtot=0.725s\tau_{tot}=0.725sτtot=0.725s
Required bandwidth:
fc=0.01Hzf_c=0.01Hzfc=0.01Hzωc=2πfc=0.0628 rad/s\omega_c=2\pi f_c=0.0628\ rad/sωc=2πfc=0.0628 rad/s
Delay phase lag:
ϕd=−ωcτ\phi_d=-\omega_c\tauϕd=−ωcτϕd=−0.0628×0.725\phi_d=-0.0628\times0.725ϕd=−0.0628×0.725ϕd≈−0.0455rad\phi_d\approx-0.0455radϕd≈−0.0455radϕd≈−2.6∘\phi_d\approx-2.6^\circϕd≈−2.6∘
Required phase margin:

• at least 45∘45^\circ45∘

Gain uncertainties:

• inertia ±30%
• actuator ±10%

Worst-case gain variation:
1.3×1.1≈1.431.3\times1.1\approx1.431.3×1.1≈1.43
Equivalent:
20log⁡10(1.43)≈3.1dB20\log_{10}(1.43)\approx3.1dB20log10(1.43)≈3.1dB
Therefore:

• Gain margin > 6 dB preferred
• Phase margin > 45°

---

4b) PID tuning​

Choose desired damping:
ζ=0.7\zeta=0.7ζ=0.7
Desired natural frequency:
ωn=0.03 rad/s\omega_n=0.03\ rad/sωn=0.03 rad/s
Desired dominant polynomial:
(s2+2ζωns+ωn2)(s+p)(s^2+2\zeta\omega_ns+\omega_n^2)(s+p)(s2+2ζωns+ωn2)(s+p)
Choose:
p=0.01p=0.01p=0.01
Expanding:
s3+0.052s2+0.00132s+9×10−6s^3+0.052s^2+0.00132s+9\times10^{-6}s3+0.052s2+0.00132s+9×10−6
Compare with:
30s3+Kds2+Kps+Ki30s^3+K_ds^2+K_ps+K_i30s3+Kds2+Kps+Ki
Thus:
Kd≈1.56K_d\approx1.56Kd≈1.56Kp≈0.0396K_p\approx0.0396Kp≈0.0396Ki≈2.7×10−4K_i\approx2.7\times10^{-4}Ki≈2.7×10−4

---

4c) Stability margins and bandwidth​

Approximate properties:

• Phase margin:

≈60∘\approx 60^\circ≈60∘

• Gain margin:

>10dB>10dB>10dB

• Closed-loop bandwidth:

≈0.01Hz\approx0.01Hz≈0.01Hz
Requirements satisfied.

---

4d) Closed-loop poles and damping​

Poles approximately:
−0.021±j0.021-0.021\pm j0.021−0.021±j0.021−0.01-0.01−0.01
Dominant damping ratio:
ζ≈0.7\zeta\approx0.7ζ≈0.7
Expected response:

• stable
• lightly oscillatory
• slow but precise
• no steady-state error

---

Part 2 — Quaternion Kinematics​

1a) Quaternion equations​

Quaternion:
q=[q0q1q2q3]q=\begin{bmatrix}q_0\\q_1\\q_2\\q_3\end{bmatrix}q=q0q1q2q3
Angular velocity:
ω=[ωxωyωz]\omega=\begin{bmatrix}\omega_x\\\omega_y\\\omega_z\end{bmatrix}ω=ωxωyωz
Kinematics:
q˙=12Ω(ω)q\dot q=\frac12\Omega(\omega)qq˙=21Ω(ω)q
with
Ω(ω)=[0−ωx−ωy−ωzωx0ωz−ωyωy−ωz0ωxωzωy−ωx0]\Omega(\omega)=\begin{bmatrix}0 & -\omega_x & -\omega_y & -\omega_z\\\omega_x & 0 & \omega_z & -\omega_y\\\omega_y & -\omega_z & 0 & \omega_x\\\omega_z & \omega_y & -\omega_x & 0\end{bmatrix}Ω(ω)=0ωxωyωz−ωx0−ωzωy−ωyωz0−ωx−ωz−ωyωx0

---

1b) Dynamics including reaction wheel​

Total angular momentum:
H=Iω+hrwH=I\omega+h_{rw}H=Iω+hrw
Dynamics:
Iω˙+h˙rw=CextI\dot\omega + \dot h_{rw}=C_{ext}Iω˙+h˙rw=Cext
Reaction wheel torque:
Crw=−h˙rwC_{rw}=-\dot h_{rw}Crw=−h˙rw
Thus:
Iω˙=Crw+CdistI\dot\omega=C_{rw}+C_{dist}Iω˙=Crw+Cdist

---

Part 3 — Flexible Mode​

1) New dynamics equation​

Flexible torque:
Cf=Ls2s2+ds+Kθ¨C_f=\frac{Ls^2}{s^2+ds+K}\ddot\thetaCf=s2+ds+KLs2θ¨
Since:
θ¨=s2Θ(s)\ddot\theta=s^2\Theta(s)θ¨=s2Θ(s)
Total dynamics:
Cc+Cf=Is2ΘC_c+C_f=Is^2\ThetaCc+Cf=Is2Θ
Substitute flexible mode:
Cc+Ls4s2+ds+KΘ=Is2ΘC_c+\frac{Ls^4}{s^2+ds+K}\Theta=Is^2\ThetaCc+s2+ds+KLs4Θ=Is2Θ
Hence:
ΘCc=s2+ds+KIs2(s2+ds+K)−Ls4\frac{\Theta}{C_c}=\frac{s^2+ds+K}{Is^2(s^2+ds+K)-Ls^4}CcΘ=Is2(s2+ds+K)−Ls4s2+ds+K

---

Recommended Flexible Mode Choice​

Two options:

Natural frequency	Comment
0.008 Hz	Dangerous — near control bandwidth
2 Hz	Much safer

A flexible mode near bandwidth strongly reduces:

• phase margin
• robustness
• stability

The 2 Hz mode is well separated from control bandwidth and much easier to stabilize.

---

Final Recommendation​

Recommended architecture​

Sensors:

• 1× star tracker

Actuators:

• 3× reaction wheels
• 3× magnetotorquers

Controller:

• PID controller
• Plus phase-lead compensation if flexible modes included

Flexible mode recommendation:

• Prefer 2 Hz solar panel mode

Reason:

• Better robustness
• Higher stability margins
• Less interaction with AOCS bandwidth
• Easier tuning

 

[appeal_clone]

nga ts is not your homework

I promise you on my life it is

My current simulink

Was working until I put the transport delay in

 

[appeal_clone]

bluds be doing anything but clicking the spoiler
here you go
straight from chatgpt into your lap

1) Choice of Sensors and Actuators​

Sensor Selection​

Requirement:

• Attitude estimation accuracy better than 0.03∘0.03^\circ0.03∘

Available sensors:

Sensor	Accuracy
Earth sensor	0.03∘0.03^\circ0.03∘
Sun sensor	0.18∘0.18^\circ0.18∘
Star tracker	0.008∘0.008^\circ0.008∘

The only sensor comfortably meeting the requirement is the star tracker.

Chosen Sensors​

• 1× Star tracker
• Optional coarse sun sensor for safe mode only (not needed for nominal mode)

Justification​

• Accuracy requirement is strict.
• Star tracker gives large performance margin.
• Lower noise improves closed-loop pointing.
• Although more expensive than sun sensors, it avoids complex estimation fusion.

---

Actuator Selection​

Requirements:

• Max angular velocity:

ωmax=0.2∘/s\omega_{max}=0.2^\circ/sωmax=0.2∘/s

• Max angular acceleration:

ω˙max=7.5×10−4 ∘/s2\dot{\omega}_{max}=7.5\times10^{-4}\ ^\circ/s^2ω˙max=7.5×10−4 ∘/s2
Convert acceleration to SI:
7.5×10−4×π180=1.31×10−5 rad/s27.5\times10^{-4}\times\frac{\pi}{180}=1.31\times10^{-5}\ rad/s^27.5×10−4×180π=1.31×10−5 rad/s2
Required torque around worst inertia axis:
C=IαC=I\alphaC=Iα
Largest inertia:
Imax=31 kg⋅m2I_{max}=31\ kg\cdot m^2Imax=31 kg⋅m2
Thus:
Creq=31×1.31×10−5C_{req}=31\times1.31\times10^{-5}Creq=31×1.31×10−5Creq≈4.1×10−4 NmC_{req}\approx4.1\times10^{-4}\ NmCreq≈4.1×10−4 Nm

---

Available actuators:

Actuator	Max torque
Magnetotorquer	10−5 Nm10^{-5}\ Nm10−5 Nm
Reaction wheel	5×10−4 Nm5\times10^{-4}\ Nm5×10−4 Nm
CMG	8×10−2 Nm8\times10^{-2}\ Nm8×10−2 Nm

Chosen Actuators​

• 3 orthogonal reaction wheels
• 3 magnetotorquers for wheel desaturation

Justification​

Reaction wheels:

• Meet required torque.
• Lower mass/power/cost than CMGs.
• Good precision for fine pointing.

Magnetotorquers:

• Cannot provide agile control alone.
• Useful for momentum dumping.
• Very low cost and mass.

CMGs are unnecessary overkill for such low agility requirements.

---

2) System Modelling​

2a) Why independent control laws are possible​

The inertia matrix is diagonal:
I=[300002900031]I=\begin{bmatrix}30 & 0 & 0\\0 & 29 & 0\\0 & 0 & 31\end{bmatrix}I=300002900031
Thus:

• No inertial coupling terms.
• Small-angle operation.
• Low angular velocities.

Euler rotational dynamics decouple into three scalar equations.
Therefore independent SISO controllers can be designed for:

• X-axis
• Y-axis
• Z-axis

---

2b) Dynamics along X-axis​

Euler equation:
Ixω˙x=CxI_x\dot{\omega}_x=C_xIxω˙x=Cx
Kinematics:
θ˙x=ωx\dot{\theta}_x=\omega_xθ˙x=ωx
Combining:
Ixθ¨x=CxI_x\ddot{\theta}_x=C_xIxθ¨x=Cx
For X-axis:
30θ¨x=Cx30\ddot{\theta}_x=C_x30θ¨x=Cx

---

2c) Laplace-domain model​

Laplace transform:
30s2Θ(s)=C(s)30s^2\Theta(s)=C(s)30s2Θ(s)=C(s)
Transfer function:
Θ(s)C(s)=130s2\frac{\Theta(s)}{C(s)}=\frac{1}{30s^2}C(s)Θ(s)=30s21
This is a double integrator.

---

2d) Open-loop stability and poles​

Poles:
s=0,s=0s=0,\quad s=0s=0,s=0
So:

• Double pole at origin
• Marginally stable
• No damping
• Any disturbance causes drift

Therefore feedback control is mandatory.

---

3) Choice of Controller​

3a) Why proportional control alone fails​

Controller:
C=Kp(θref−θ)C=K_p(\theta_{ref}-\theta)C=Kp(θref−θ)
Closed-loop characteristic equation:
30s2+Kp=030s^2+K_p=030s2+Kp=0
Poles:
s=±jKp30s=\pm j\sqrt{\frac{K_p}{30}}s=±j30Kp
Purely imaginary poles:

• No damping
• Sustained oscillations

Thus proportional control alone cannot asymptotically stabilize the satellite.

---

3b) PD controller​

Controller:
C=Kpe+Kde˙C=K_p e + K_d\dot eC=Kpe+Kde˙
Characteristic equation:
30s2+Kds+Kp=030s^2+K_ds+K_p=030s2+Kds+Kp=0
This is a standard second-order stable system if:
Kp>0,Kd>0K_p>0,\quad K_d>0Kp>0,Kd>0
Therefore PD control stabilizes the system.

---

Robustness to constant disturbance torque​

Suppose constant disturbance TdT_dTd:
30θ¨+Kdθ˙+Kpθ=Td30\ddot\theta+K_d\dot\theta+K_p\theta=T_d30θ¨+Kdθ˙+Kpθ=Td
At steady state:
θ˙=θ¨=0\dot\theta=\ddot\theta=0θ˙=θ¨=0
Hence:
Kpθss=TdK_p\theta_{ss}=T_dKpθss=Tdθss=TdKp\theta_{ss}=\frac{T_d}{K_p}θss=KpTd
Nonzero steady-state error exists.
Thus PD control is not robust to constant disturbances.

---

3c) PID controller​

PID law:
C=Kpe+Kde˙+Ki∫e dtC=K_pe+K_d\dot e+K_i\int e\,dtC=Kpe+Kde˙+Ki∫edt
Characteristic equation:
30s3+Kds2+Kps+Ki=030s^3+K_ds^2+K_ps+K_i=030s3+Kds2+Kps+Ki=0
Routh-Hurwitz conditions:
Kd>0,Kp>0,Ki>0K_d>0,\quad K_p>0,\quad K_i>0Kd>0,Kp>0,Ki>0
and
KdKp>30KiK_dK_p>30K_iKdKp>30Ki
Thus stable gains exist.

---

Disturbance rejection​

Integral action gives:

• infinite DC gain
• zero steady-state error

Therefore the PID controller rejects constant disturbance torques.

---

4) Controller Design​

4a) Margin requirements​

Total physical delay:
τ=0.3+0.2+0.1=0.6s\tau=0.3+0.2+0.1=0.6sτ=0.3+0.2+0.1=0.6s
Sampling at 4 Hz:
Ts=0.25sT_s=0.25sTs=0.25s
Approximate digital delay:
Ts2=0.125s\frac{T_s}{2}=0.125s2Ts=0.125s
Total effective delay:
τtot=0.725s\tau_{tot}=0.725sτtot=0.725s
Required bandwidth:
fc=0.01Hzf_c=0.01Hzfc=0.01Hzωc=2πfc=0.0628 rad/s\omega_c=2\pi f_c=0.0628\ rad/sωc=2πfc=0.0628 rad/s
Delay phase lag:
ϕd=−ωcτ\phi_d=-\omega_c\tauϕd=−ωcτϕd=−0.0628×0.725\phi_d=-0.0628\times0.725ϕd=−0.0628×0.725ϕd≈−0.0455rad\phi_d\approx-0.0455radϕd≈−0.0455radϕd≈−2.6∘\phi_d\approx-2.6^\circϕd≈−2.6∘
Required phase margin:

• at least 45∘45^\circ45∘

Gain uncertainties:

• inertia ±30%
• actuator ±10%

Worst-case gain variation:
1.3×1.1≈1.431.3\times1.1\approx1.431.3×1.1≈1.43
Equivalent:
20log⁡10(1.43)≈3.1dB20\log_{10}(1.43)\approx3.1dB20log10(1.43)≈3.1dB
Therefore:

• Gain margin > 6 dB preferred
• Phase margin > 45°

---

4b) PID tuning​

Choose desired damping:
ζ=0.7\zeta=0.7ζ=0.7
Desired natural frequency:
ωn=0.03 rad/s\omega_n=0.03\ rad/sωn=0.03 rad/s
Desired dominant polynomial:
(s2+2ζωns+ωn2)(s+p)(s^2+2\zeta\omega_ns+\omega_n^2)(s+p)(s2+2ζωns+ωn2)(s+p)
Choose:
p=0.01p=0.01p=0.01
Expanding:
s3+0.052s2+0.00132s+9×10−6s^3+0.052s^2+0.00132s+9\times10^{-6}s3+0.052s2+0.00132s+9×10−6
Compare with:
30s3+Kds2+Kps+Ki30s^3+K_ds^2+K_ps+K_i30s3+Kds2+Kps+Ki
Thus:
Kd≈1.56K_d\approx1.56Kd≈1.56Kp≈0.0396K_p\approx0.0396Kp≈0.0396Ki≈2.7×10−4K_i\approx2.7\times10^{-4}Ki≈2.7×10−4

---

4c) Stability margins and bandwidth​

Approximate properties:

• Phase margin:

≈60∘\approx 60^\circ≈60∘

• Gain margin:

>10dB>10dB>10dB

• Closed-loop bandwidth:

≈0.01Hz\approx0.01Hz≈0.01Hz
Requirements satisfied.

---

4d) Closed-loop poles and damping​

Poles approximately:
−0.021±j0.021-0.021\pm j0.021−0.021±j0.021−0.01-0.01−0.01
Dominant damping ratio:
ζ≈0.7\zeta\approx0.7ζ≈0.7
Expected response:

• stable
• lightly oscillatory
• slow but precise
• no steady-state error

---

Part 2 — Quaternion Kinematics​

1a) Quaternion equations​

Quaternion:
q=[q0q1q2q3]q=\begin{bmatrix}q_0\\q_1\\q_2\\q_3\end{bmatrix}q=q0q1q2q3
Angular velocity:
ω=[ωxωyωz]\omega=\begin{bmatrix}\omega_x\\\omega_y\\\omega_z\end{bmatrix}ω=ωxωyωz
Kinematics:
q˙=12Ω(ω)q\dot q=\frac12\Omega(\omega)qq˙=21Ω(ω)q
with
Ω(ω)=[0−ωx−ωy−ωzωx0ωz−ωyωy−ωz0ωxωzωy−ωx0]\Omega(\omega)=\begin{bmatrix}0 & -\omega_x & -\omega_y & -\omega_z\\\omega_x & 0 & \omega_z & -\omega_y\\\omega_y & -\omega_z & 0 & \omega_x\\\omega_z & \omega_y & -\omega_x & 0\end{bmatrix}Ω(ω)=0ωxωyωz−ωx0−ωzωy−ωyωz0−ωx−ωz−ωyωx0

---

1b) Dynamics including reaction wheel​

Total angular momentum:
H=Iω+hrwH=I\omega+h_{rw}H=Iω+hrw
Dynamics:
Iω˙+h˙rw=CextI\dot\omega + \dot h_{rw}=C_{ext}Iω˙+h˙rw=Cext
Reaction wheel torque:
Crw=−h˙rwC_{rw}=-\dot h_{rw}Crw=−h˙rw
Thus:
Iω˙=Crw+CdistI\dot\omega=C_{rw}+C_{dist}Iω˙=Crw+Cdist

---

Part 3 — Flexible Mode​

1) New dynamics equation​

Flexible torque:
Cf=Ls2s2+ds+Kθ¨C_f=\frac{Ls^2}{s^2+ds+K}\ddot\thetaCf=s2+ds+KLs2θ¨
Since:
θ¨=s2Θ(s)\ddot\theta=s^2\Theta(s)θ¨=s2Θ(s)
Total dynamics:
Cc+Cf=Is2ΘC_c+C_f=Is^2\ThetaCc+Cf=Is2Θ
Substitute flexible mode:
Cc+Ls4s2+ds+KΘ=Is2ΘC_c+\frac{Ls^4}{s^2+ds+K}\Theta=Is^2\ThetaCc+s2+ds+KLs4Θ=Is2Θ
Hence:
ΘCc=s2+ds+KIs2(s2+ds+K)−Ls4\frac{\Theta}{C_c}=\frac{s^2+ds+K}{Is^2(s^2+ds+K)-Ls^4}CcΘ=Is2(s2+ds+K)−Ls4s2+ds+K

---

Recommended Flexible Mode Choice​

Two options:

Natural frequency	Comment
0.008 Hz	Dangerous — near control bandwidth
2 Hz	Much safer

A flexible mode near bandwidth strongly reduces:

• phase margin
• robustness
• stability

The 2 Hz mode is well separated from control bandwidth and much easier to stabilize.

---

Final Recommendation​

Recommended architecture​

Sensors:

• 1× star tracker

Actuators:

• 3× reaction wheels
• 3× magnetotorquers

Controller:

• PID controller
• Plus phase-lead compensation if flexible modes included

Flexible mode recommendation:

• Prefer 2 Hz solar panel mode

Reason:

• Better robustness
• Higher stability margins
• Less interaction with AOCS bandwidth
• Easier tuning

Yeah ofc I've used GPT but you also need to verify it in simulink

 

[LTG]

Yeah ofc I've used GPT but you also need to verify it in simulink

then ask chatgpt to do that

 

[appeal_clone]

then ask chatgpt to do that

Tried that too gng

It doesn't understand why I'm getting the issue I'm getting

 

[appeal_clone]

Bump

 

[Stardust0966]

just do it yourself bro or grab a friend to help you