Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] Problem on LCR resonator

Status
Not open for further replies.

Debdut

Full Member level 3
Joined
Oct 29, 2013
Messages
174
Helped
7
Reputation
14
Reaction score
7
Trophy points
1,298
Activity points
2,572
Hello :roll:, can you help me solve this problem. Its from An Introduction to Analog and Digital Communications - Simon Haykin

A carrier wave of frequency 1 MHz is modulated 50% by a sinusoidal wave of frequency 5 kHz. The resulting AM wave is transmitted through the resonant circuit of the Fig. below, which is tuned to the carrier frequency and has a Q factor of 175. Determine the modulated wave after transmission through this circuit. What is the percentage modulation of the modulated wave?

Untitled.jpg
 

The problem is wrongly stated: there is no resonator-coupling circuit, nor source impedance.
I would fire your professor for giving you such nonsense!
 

Jiripolivka - I am afraid you didn`t notice that the signal source symbol in the diagram represents a CURRENT source.
In this case, there is of course a filtering effect.
 

The problem is wrongly stated...
Looks OK to me, the input is an AC current source.

3dB bandwidth = 1MHz/175 = 5.7 KHz
i.e. sidebands are attenuated by 3dB at +-2.85KHz from carrier.

Thus the sidebands at +- 5KHz are attenuated by a factor of SQRT((2.85^2) / (2.85^2 + 5^2)) = 0.5,
so the modulation will be 50% * 0.5 = 25%
 
  • Like
Reactions: LvW

    LvW

    Points: 2
    Helpful Answer Positive Rating
godfreyl, could you explain a bit. I didn't understand : SQRT((2.85^2) / (2.85^2 + 5^2)) = 0.5
 

The calculation is the same as if you have a simple first order RC low pass filter with corner frequency = 2.85 KHz, and you want to find the attenuation at 5 KHz.
 
  • Like
Reactions: Debdut

    Debdut

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top