78 lines
2.8 KiB
Markdown
78 lines
2.8 KiB
Markdown
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For the case where a signal is required to settle before being sampled by an ADC.
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For this example the aim will be the signal is 99% settled at time of ADC~sampling.
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Starting first with with an RC low pass filter: signal to R, R to C, C to ground. The voltage into the capacitor is $dv_o/dt . RC$. Where $\tau = RC$:
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$$\tau \frac{dv_0}{dt} +v_0 = vi \; .$$
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Take Laplace transforms (zero initial condition, i.e. capacitor not charged):
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$$\tau s V_o(1+s\tau)=V_i(s) \rightarrow H(s)=\frac{V_o(s)}{V_i(s)} = \frac{1}{1+s\tau}$$
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Apply the step input; consider a switch going on, applying a constant voltage, the way an ADC signal could stabalise; so $$\frac{1}{s}$$$$H(s)V_i(s) = \frac{1}{1+s\tau} . \frac{1}{s} = \frac{1}{s(1+s\tau)}$$
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## Partial Fractions
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Use Partial fractions to break this equation up into parts that can be individually transformed back form Laplace to real~time.
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Convert to partial fractions with A,B unknown:
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$$\frac{1}{s(1+s\tau)}= \frac{A}{s} + \frac{B}{(1+s\tau)}$$
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rearranging to solve quickly; set s = 0 to tease out $A$
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$$1 = A(1+s\tau)+ Bs$$
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Set s = 0 and this shows A=1 :
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so knowing A is 1:
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$$ 1 = 1(1+s\tau)+Bs$$
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$$ 1 = 1 + s\tau + Bs $$
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$$0 - s\tau = Bs $$
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$$ B= -\tau $$
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s=0 and A=1 B=$-\tau$ : the equation is now in a format where the parts can be individually converted back from the $s$ domain to real time.
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$$\frac{1}{s(1+s\tau)}= \frac{1}{s} - \frac{\tau}{(1+s\tau)}$$
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divide the last fraction by $\tau$ on both sides, i.e. $$ \frac{\tau}{1+s\tau} \equiv \frac{1}{1/\tau +s}$$$$\frac{1}{s(1+s\tau)}= \frac{1}{s} - \tau . \frac{1}{(s+1/\tau)}$$
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## Inverse Laplace -- Back to reality!
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Taking inverse Laplace transforms:
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$$\mathcal{L}^{-1} \Big(\frac{1}{s}\Big) = 1 $$
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$$ \mathcal{L}^{-1} \Big(\tau . \frac{1}{1/\tau +s}\Big) = \tau . \frac{1}{\tau}e^{-t/\tau}$$
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Now the real~time response to the step function can be applied
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$$ f(t) = 1 - e^{-t/\tau} $$
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This is the RC filter step response.
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## Percentage step function settled
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Now the percentages for settling can be determined.
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where say $e^{-t/\tau} = 0.9$ for instance would determine the 90% settled point.
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$$ 0.9 = 1 - e^{-t/\tau} $$
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$$0.1 = e^{-t/\tau}$$
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Take logs
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$$ln(0.1) = {-t}{\tau}$$
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$$\tau = ln(0.1)/-t$$
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## Plug in some component values as a `real' example
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So $\tau= -ln(0.1) = 2.3$ for 90% settled. Putting some numbers in this were the settling time to be $250\mu s $ for 90% settling of the step function:-
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$$ \tau = RC = \frac{-2.3}{250E-6}= 9.21E-3$$
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So RC must be 9.21E-3: Taking R as 22k $RC=9.21\times 10^{-3}$
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$$ C = \frac{9.21\times10^{-3}}{22 \times10^{3}}$$
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So C = 419E-9 or $418pF$.
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In other words, a simple low pass filter; 22k to 418pF to ground; will settle a step function to 90% from 0V within $250 \mu s$.
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## Some handy setting $\tau$ figures
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$$t_{63\%} = 1τ $$
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$$t_{95\%} = 3τ $$
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$$t_{99\%} = 4.6τ $$
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$$t_{99.9\%} = 6.9τ $$
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---
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