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.

 

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))

Simplify

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

 

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

Vout=Vin*((1/jwC)/(R+(1/jwC)))

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.

Regards

Emre