notes/Worked Example RC Low Pass.md

For the case where a signal is required to settle before being sampled by an ADC. For this example the aim will be the signal is 99% settled at time of ADC~sampling.

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:

\tau \frac{dv_0}{dt} +v_0 = vi \; .

Take Laplace transforms (zero initial condition, i.e. capacitor not charged):

\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}

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)}

Partial Fractions

Use Partial fractions to break this equation up into parts that can be individually transformed back form Laplace to real~time.

Convert to partial fractions with A,B unknown:

\frac{1}{s(1+s\tau)}= \frac{A}{s} + \frac{B}{(1+s\tau)}

rearranging to solve quickly; set s = 0 to tease out $A$ $$1 = A(1+s\tau)+ Bs$$ Set s = 0 and this shows A=1 : so knowing A is 1:

$ 1 = 1(1+s\tau)+B

$$ 1 = 1 + s\tau + Bs $$
$$0 - s\tau = Bs $$
$$ B= -\tau $$
 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.

\frac{1}{s(1+s\tau)}= \frac{1}{s} - \frac{\tau}{(1+s\tau)}

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)}

Inverse Laplace -- Back to reality!

Taking inverse Laplace transforms:

$\mathcal{L}^{-1} \Big(\frac{1}{s}\Big) = 1

 \mathcal{L}^{-1} \Big(\tau . \frac{1}{1/\tau +s}\Big) = \tau . \frac{1}{\tau}e^{-t/\tau}

Now the real~time response to the step function can be applied

 f(t) = 1 -   e^{-t/\tau} 

This is the RC filter step response.

Percentage step function settled

Now the percentages for settling can be determined.

where say e^{-t/\tau} = 0.9 for instance would determine the 90% settled point.

 0.9 = 1 - e^{-t/\tau} 

0.1 = e^{-t/\tau}

Take logs $$ln(0.1) = {-t}{\tau}$$ $$\tau = ln(0.1)/-t$$

Plug in some component values as a `real' example

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:-

 \tau = RC = \frac{-2.3}{250E-6}= 9.21E-3

So RC must be 9.21E-3: Taking R as 22k RC=9.21\times 10^{-3}

 C = \frac{9.21\times10^{-3}}{22 \times10^{3}}

So C = 419E-9 or 418pF.

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.

Some handy setting \tau figures

t_{63\%} = 1τ 

t_{95\%} = 3τ 

t_{99\%} = 4.6τ 

t_{99.9\%} = 6.9τ 

---