Output voltage

Discussion in 'Homework Help' started by Hello, Mar 25, 2009.

  1. Hello

    Thread Starter Active Member

    Dec 18, 2008
    Can you please have a look at Q3 part (ii) in the attached file.
    Not sure if i have the right answer!

    Im assuming its asking vout for the circuit in fig.4 (low pass):

    Vout/Vin = 1/(jωRC + 1)
    |Vout| = |Vin|/((ωRC)^2 +1)^0.5)

    Any help is appreciated.
    Last edited: Mar 25, 2009
  2. jasperthecat


    Mar 26, 2009
    Hello Hello
    Yes you're right - It's a low pass filter (or integrator or hiss remover or voltage smoother) - the higher frequencies are attenuated

    Back to the maths

    Vout = is output of a frequency sensitive voltage divider - the voltage across the capacitor is inversely proportional to the frequency.

    vout = (impedance or reactance of the capacitor) divided by the total impedance of the series circuit made up of R and C

    vout = vin ( 1/jwc) / (R + 1/jwc))


    As this a complex number multiply top and bottom lines by 1/jwc

    vout = vin /(jwcR +1)

    This solution is still a complex number (contains but amplitude and phase information
    You could go one step further as you did and calculate the absolute value of the output voltage but to complete the picture you nedd to do the phase angle

    theta = tan-1 ( / )

    but the complex number tells all.

    You can do a simple diagram with 3 points to show you which way the output is going to go.

    when f=0 (Direct current) no current flows through the capacitor (infinite reactance) and all the input voltage appears at the output (f=0 , output = 100%)

    At the other end of the spectrum when f is very high (zero or almost zero reactance in the capacitor) the voltage across the capacitor is zero or close enough) (f=infinite, output =0%)

    The third point occurs when f = 1/CR and the output is about 70.7% of the input.
    (known as the corner point or 3db point)

    Hope this helps
  3. etuzuner


    Mar 21, 2009
    I think you're wrong about the output.
    If we use voltage divider(as you did);


    I think it is wanted H(jw) in the other part. The question is to calculate

    |H(jw)|=|Vout/Vin|=|((1/jwC)/(R+(1/jwC)))|. In my opinion, when you take the magnitude, you get the solution.