Hey, based on the picture, what does it means by creating a baseband signal ?

I dont get it. I understand that baseband filter is a filter that will not filter a specific range of frequency.

Thanks

How about this ?

What does this means ?

Thanks

 

The function of a low pass filter is to pass low frequencies and attenuate frequencies above the filter's corner frequency.

In this case the LP filter is used to attenuate the RF frequency but pass the modulation frequency.

Note that a standard op amp will not work very well as a filter here because of the op amps limited frequency response. An active filter requires the op amp to have significant gain at the RF frequency and a standard op amp doesn't. Better to use just a passive RC LP filter.

Actually, what is the effect of the RC value to the demodulated signal waveform in terms of charging and discharging ?

Below is what I attached in my report. I am not so sure why is the graph like that. My theory is that if R increases, the discharge rates get slower whereas if C increases, the charge rate will be slower too. I am not very sure how can I observe this from my waveform.

Thanks for help !

p/s The simulation result is under 7.0 Simulation Result

 

Dear all, I am confuse for the envelope detector circuit. Can anyone clarify to me for the picture below, is the thing in the box a Low pass filter or what is it ?

Thanks and appreciate

 

The box acts as a low-pass filter to by-pass the high frequency RF from the rectified modulated signal. It's operation is as follows:

The capacitor is charged to near the peak of the RF sine-wave through the diode. For the rest of the RF cycle (when it's below its positive peak) the capacitor slightly discharges through R1 until it's recharged at the next cycle peak. The RC time-constant is chosen so that the decay time is fast enough to follow the fastest change in the RF peaks due to the highest modulation frequency but slow enough so the RF ripple is suppressed. Typically the RC time-constant would be chosen so that the corner frequency of the filter [1 / (2\(\pi\)RC)] is at or slightly above the highest modulation frequency. (For example the circuit you show has a corner frequency of 159Hz so any modulation frequency above that would be filtered.)

 

My audio frequency is 100Hz. So I guese this circuit is good enough to recover back my modulated signal ? No audio wave form will be filtered out right ?

(The RC time-constant is chosen so that the decay time is fast enough to follow the fastest change in the RF peaks due to the highest modulation frequency but slow enough so the RF ripple is suppressed) <-- I dont understand this though.

Thanks

 

you really need to connect the circuit and check the result on a scope (or simulate it). i suggest to use two channels as in post 23 and start with no capacitor. you will see signal before and after diode. then add capacitor but use smaller value such as 1nF. see what happens. then put 10nF and see what changes (eep increasing capacitor value and monitor changes on the scope).

time constant of simple RC element is tau or T=R1*C1
f=1/T = 1kHz

so 100Hz would go through (very little attenuation) but 10kHz would not (a lot more attenuation).

 

I do simulate it...refer to my previous post ya..

Thanks )

 

The box acts as a low-pass filter to by-pass the high frequency RF from the rectified modulated signal. It's operation is as follows:

The capacitor is charged to near the peak of the RF sine-wave through the diode. For the rest of the RF cycle (when it's below its positive peak) the capacitor slightly discharges through R1 until it's recharged at the next cycle peak. The RC time-constant is chosen so that the decay time is fast enough to follow the fastest change in the RF peaks due to the highest modulation frequency but slow enough so the RF ripple is suppressed. Typically the RC time-constant would be chosen so that the corner frequency of the filter [1 / (2\(\pi\)RC)] is at or slightly above the highest modulation frequency. (For example the circuit you show has a corner frequency of 159Hz so any modulation frequency above that would be filtered.)

Hey yea, so do you means that if I choose a RC value whic give me a cut off frequency below 100 Hz, then I wont get the demodulated signal ?

Thanks

 

Hello,

Have a look at page 5 and 6 of the following PDF:
http://educypedia.karadimov.info/library/tut331_04am_dem.pdf

Bertus

What is fc and W ? fc = cut off frequency ?

How about W ?

Thanks

 

I'm not wildly impressed by the linked PDF--It leaves a fair bit of important stuff out.

The "W" is because the original printer could not produce the symbol "ω".which in turn means "2∏f".
Unfortunately,that is probably not something you have learned at this point.

 

I'm not wildly impressed by the linked PDF--It leaves a fair bit of important stuff out.

The "W" is because the original printer could not produce the symbol "ω".which in turn means "2∏f".
Unfortunately,that is probably not something you have learned at this point.

2∏f, so what is the f ?

Thanks

 

2∏f, so what is the f ?

Frequency in Hertz.

 

Frequency of which thing ?

Thanks

 

Frequency of which thing ?

Thanks

The signal.

 

2∏f, so what is the f ?

Thanks

Well,actually,after looking at the PDF again,I was wrong!(see P4)
The "W" wasn't an attempt at ω.
For some unknown reason the author was using "W" to symbolise the bandwidth of the modulated signal.
Sorry about that!

 

I do simulate it...refer to my previous post ya..

Thanks )

prove it, post some screenshots and interpretation.

 

prove it, post some screenshots and interpretation.

It is in my previous post.

Have a look

Thanks !

 

ok,

your RC values for filter are poor, signal is distorted and all of your chosen value move the filter characteristic in one and same direction (only large time constant).

to prove understanding of the circuit operation i would expect to see ripple too, not just envelope.

 

The only resistor-capacitor values that give a proper output with the given modulation is the first, shown in 7.0, where R1 = 100kΩ and C1 - .01μF. All the rest give a distorted output because the time constant is too long (filter corner frequency is below the 100Hz modulation frequency).