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RCR Circuit Response
RCR Circuit Response  
I have a difficulty in solving some questions on my EE course.
One of them is about designing an RCR circuit that would have a frequency response as on the attached image.

I have to design a network using one 100nF capacitor and resistors with the appropriate values.
The corner frequencies are meant to be at 1krad/s and at 10krad/s.
Its pretty simple to figure out what circuit is meant to be used there however, I am asked to show the steps in working it out from the plot. Could you show me how I could work out the circuit from the information given on the attached bode plot? Im not too sure, but I think it might have to do with working out the network function first and working backwards from there. Thank you for your response.

I cannot read the divisions on the x or y axis.  

There are many different solutions to the circuit and every college or university uses texbook that uses the methods employed by its author.  Since my textbook was never published, I will give you my ideas on how to solve the problem.

FIRST: The high frequencies are attenuated by so many db below the lower frequencies.  This can be accomplished by using a resistor divider circuit.   You can calculate the ratio of attenuation needed by simple ohms law or percent reduction in level needed.  From this simple calculation you can determine the resistor ratios.

SECONDLY:  Now, there needs to be a capacitor in series with the resistor dividers. Here is what I would do:  Take the output line from between the two dividing resistors.  Place the capacitor in series with the bottom resistor but after the series capacitor.  This way at very low frequencies the output voltage is not reduced because the the series resistor only adds to the output impedance.  As the frequency increases the cap lowers the voltage at the dividing point due to the lowering reactance.  At the -3db point the cap and series resistor rc network gives the first knee of the curve.  The freqnency increases until the point of no further attenuation takes place. This is the -3db knee of the flatening of the high frequency response.

See how that works for you.  


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Electronics questions about AC, DC and digital theory.


Graduate electrical engineer with over 40 years in electronic design, manufacturing, project organization and patent review. Experience in fields of industrial and consumer electronics (audio, video, acoustics, etc.)

IEEE (Institute of Electrical and Electronics Engineers); Senior Life member AES (Audio Engineering Society), Fellow Life member

BSEE University of North Dakota

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