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can anyone help in finding the expression for the voltage at the output of the circuit if the input is a step function with 1V amplitude.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ What have you tried already? In the question you mention a step function (unit step?), and in the schematic a square wave? \$\endgroup\$ – Bart Nov 18 '19 at 13:26
  • \$\begingroup\$ T = (jw) = (R + (1/jwc)) / 5R + (1/jwc) = jwRC + 1 / 5jwRC + 1 therefore T(s) = 1 + 5(tau) / 1 + 5s(tau) and then applying the step function: Vout (s) = 1/s * T(s) = 1/s [(1/1+5s(tau)) + (s(tau)/1+5s(tau)..... this is as far as ive but struggling to simplifiy down \$\endgroup\$ – mwilliams25 Nov 18 '19 at 13:33
  • \$\begingroup\$ Because this is a simple first order network excited by a Heaviside step, the time dependent part of the output is an exponential. All you need to do is find the time constant, and apply the boundary conditions to find the static parts of the output function. \$\endgroup\$ – Bart Nov 18 '19 at 13:36
  • \$\begingroup\$ Could you explain how i would do that? \$\endgroup\$ – mwilliams25 Nov 18 '19 at 13:38
  • \$\begingroup\$ @mwilliams25 Are you aware of partial fractions? Memorising 1/(s+a)(s+b) transforms to {e^(-at) - e^(-bt)} /(b-a) is quite useful and time saving. \$\endgroup\$ – Deep Nov 18 '19 at 13:41
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You can find the output voltage in the s domain by simply using a voltage divider formula. $$V_{out}=\frac{V_{in}(R+\frac{1}{sC})}{R+4R+\frac{1}{sC}} => V_{in}=1/s => Vout(s)=\frac{RCs+1}{s(5RCs+1)}$$ and now just using the inv. laplace transform you have Vout in the time domain. $$ V_{out}(t)=1 - \frac{4e^{\frac{-t}{5CR}}}{5} =1 - \frac{4\exp\left(\dfrac{-t}{5CR}\right)}{5} $$

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    \$\begingroup\$ Please use the equations which are more readable than plain text :) \$\endgroup\$ – Jakob Halskov Nov 18 '19 at 13:50
  • \$\begingroup\$ I don't know how to make them. \$\endgroup\$ – user229923 Nov 18 '19 at 13:52
  • \$\begingroup\$ In addition to proper equation formatting, please also use lowercase s for the complex frequency. \$\endgroup\$ – Bart Nov 18 '19 at 14:00
  • \$\begingroup\$ See here: math.meta.stackexchange.com/questions/5020/… \$\endgroup\$ – Jakob Halskov Nov 18 '19 at 14:01
  • \$\begingroup\$ I don't have time right now to do it but I will edit it tomorrow. \$\endgroup\$ – user229923 Nov 18 '19 at 14:04

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