Lowpass Filter

• Consider the circuit with a resistor R and a capacitor C in series.
• The output is the voltage v_out across the capacitor.
• The following differential equation describes the circuit
  
  $$v_{in}(t) = v_{out}(t) + RC \frac{dv_{out}(t)}{dt}.$$
  
  

• Let the input voltage be a sinusoid of frequency f, i.e., $v_{in}(t) = \sin(2 \pi ft)$.
• We will show that the resulting output is $v_{out}(t) = A \sin(2 \pi ft + \phi)$.
• The amplitude A and the phase $\phi$ are to be determined.
• When we insert v_in(t) and v_out(t) into the differential equation, we obtain
  
  $$\sin(2 \pi ft) = A \sin(2 \pi ft + \phi) + 2 \pi fRC \cos(2 \pi ft + \phi).$$
  
  

• Applying double angle trigonometric identities $\sin(x+y) = \sin x \cos y + \cos x \sin y$ and $\cos(x+y)=\cos x \cos y - \sin x \sin y$, we get
  
  $$\sin(2 \pi ft) = A \sin(2 \pi ft)(\cos \phi - 2 \pi fRC \sin \phi) + A \cos(2 \pi ft)(\sin \phi + 2 \pi fRC \cos \phi).$$
  
  

• Hence, the following two equations must hold
  
  $$A(\cos \phi - 2 \pi fRC \sin \phi) = 1$$
  
  
  and
  
  $$A(\sin \phi + 2 \pi fRC \cos \phi) = 0.$$
  
  

• Therefore,
  
  $$\tan \phi = - 2 \pi fRC$$
  
  
  and
  
  $$A^2 = \frac{1}{1+(2 \pi fRC)^2}.$$