L10

24/08/26 Mon 15:35pm

Given the dynamics convert to state space.

$$\ddot{x}+3\dot{x}+2x=4$$

assign

$$x_1=x$$ $$\dot{x_1}=\dot{x}=x_2$$

on substituting

$$\dot{x_2}+3x_2+2x_1=u$$ $$\left[\begin{array}{c}\dot{x_1}\\ \dot{x_2}\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]+\left[\begin{array}{c}0\\ 1\end{array}\right]u$$ $$Y=C\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]+d.u$$ $$Y=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]+0.u$$

You can determine the stability of the system, by manipulating the matrix A

location of poles: $\det |\lambda I-A|=0$

$$\det \left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]-\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]$$ $$\det \left[\begin{array}{cc}\lambda & -1\\ 2& \lambda +3\end{array}\right]$$ $$\lambda (\Lambda +3)+2=0$$ $${\Lambda }^2+3\lambda +2=0$$ $$\lambda (\lambda +2)+1(\Lambda +2)=0$$ $$(\lambda +1)(\lambda +2)=0$$ $$\lambda$$

---

A digital Twin is developed to monitor the temperature of a reactor, a temperature sensor and its signal conditioning circuit, produce an analog output of 0 → 5 volts corresponding to 0 to $200\text{celsius}$. During operation the useful temperature dynamics are found to vary only up to 8 Hz, however measurements show that the sensor signal is contaminated by significant electrical interference at 50 Hz from nearby power equipment. Further the conditioned signal will be acquired by a 10bit ADC having a 0 → 5 Volt input range and transmit it to the digital twin.

1. Design a suitable signal conditioning circuit such that the useful temperature information is retained while the unwanted interference is suppressed.
2. Determine the magnitude of circuit response at 8Hz and 50Hz.
3. Calculate the output amplitude at 50Hz if the interference entering the circuit has an amplitude of 0.5 Volts.
4. If the actual reactor temperature is $120\text{celsius}$ determine the corresponding sensor voltage.
5. Determine the ADC digital output corresponding to $120\text{celsius}$ assuming the interference is neglected after signal conditioning.
6. Justify why the circuit selected by you is suitable for this application.

Solution:

The question describes a situation with high frequency noise and useful information at lower frequency $\therefore$ we design a low pass filter.

1. Assume C value = 0.1 $\text{micro}F$ (found out after trying with 0.01 because resistance was too high ( in the order of mega ohms))

$$F_c=\frac{1}{2\pi RC}$$

Consider a cutoff frequency of 10Hz

$$10=\frac{1}{2\pi R(0.01\times 10^{-6})}$$ $$R=159.154K\Omega \approx 160K\Omega$$

From the maximum gain formula

$$1+\frac{R_2}{R_1}=\text{max Gain}$$

assuming Gain as 2
and assuming $R_2$ as 10 $K\Omega$

$$1+\frac{10\times 10^{-3}}{2}=2$$ $$R_1=10K\Omega$$

L10-2026-09-15-18.36.16.excalidraw

2. Circuit Response

$$\frac{\frac{V\times 1}{SC}}{R+\frac{1}{SC}}=\frac{V}{RSC+1}$$ $$V_o=2\frac{V}{(RSC+1)}$$ $$\frac{V_o}{V}=\frac{2}{1+RSC}$$ $$G=\frac{2}{1+\frac{j\omega }{\frac{1}{RC}}}$$ $$G=\frac{2}{1+\frac{j.2\pi f}{2\pi f_c}}$$ $$\boxed{|G|=\frac{2}{\sqrt{1+{\left(\frac{f}{f_c}\right)}^2}}}$$

now plugging in the required values i.e. 8Hz and 50Hz.

$$|G{|}_{8Hz}=\frac{2}{\sqrt{1+{\left(\frac{8}{10}\right)}^2}}=\boxed{1.561}$$ $$|G{|}_{50Hz}=\frac{2}{\sqrt{1+{\left(\frac{50}{10}\right)}^2}}=\boxed{0.392}$$

3. The output amplitude at 50Hz given the input is 0.5 Volts
   simply multiplying the gain at 50 Hz with the input i.e.

$$\boxed{0.392\times 0.5=0.196Hz}$$

4. Given reactor temperature $120^{\circ }C$
   the corresponding sensor voltage can be found by first finding the range of the sensor

$$\frac{\text{Voltage range}}{\text{Temperature range}}=\frac{5-0}{200-0}=0.025V{/}^{\circ }C$$

now for $120\text{celsius}$

$$\boxed{120\times 0.025=3V}$$

5. To find the ADC digital output for $120^{\circ }C$ or $3V$
   first we find the resolution of the ADC

$$\text{Resolution}=\frac{5}{2^{10}}=4.88mV/step$$ $$\boxed{\frac{3}{\frac{5}{2^{10}}}=614.4}$$

Note

$$\text{Resolution}=\frac{V_{ref}^{+}-V_{ref}^{-}}{2^n}$$ $$\text{ADC number}=\frac{V_{analog}}{\text{Resolution}}$$