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.

Inverting a modulated signal (OOK).

Status
Not open for further replies.

imrankhanPNU

Member level 1
Member level 1
Joined
Feb 21, 2018
Messages
34
Helped
0
Reputation
0
Reaction score
0
Trophy points
6
Activity points
461
Hello,
I'm modulating the message signal with OOK (on-off keying) modulation. Multisim software is used to model the schematic [attached]. At the resistor R1, the modulated signal is obtained. I wanted to invert the modulated single, for that I used a simple NOT gate (U3), however at the NOT's gate output, I got the same signal as input means without inverting [attached]. So my question is, why isn't the NOT gate inverting? Am I doing something wrong?

As hit and trail, I added two more NOT gates (U4 and U5) but this time I got an inverted signal, although further improvement is needed to achieve a 180-degree phase shift. So my second question is: Why is it an inverter this time? and how to get exactly a 180-degree phase shift between two signals?

Thanks for your time.
 

Attachments

  • 3.JPG
    3.JPG
    97.9 KB · Views: 152
  • 1.JPG
    1.JPG
    114.1 KB · Views: 157
  • 2.JPG
    2.JPG
    101.1 KB · Views: 175

Consider gate propagation delay.

Suggest to make a setup with a more reasonable operating frequencyFrom previous
Thanks, @FvM for your response. Due to the signal delay of the inverter, a shift is observed in the output signal, which is not applicable in my application.
Moreover, I am keen to achieve, an oscillating frequency of 100Mhz for a 1Mhz input signal. For that, i tried another schematic [attached]. however I could not get the desired result although everything seems correct to me, I do not know where I am making mistake? could you please review it? Thanks.

Fc=Oscillating freq, R=resistor, and C= capacitor
Fc=1/2.2RC

 

Attachments

  • 4.JPG
    4.JPG
    88.6 KB · Views: 139
  • 5.JPG
    5.JPG
    51.9 KB · Views: 133
  • 6.JPG
    6.JPG
    39 KB · Views: 134

The reactance of a 1nF capacitor at 100MHz is 1.592 Ohms and the series resistor is 4.55 Ohms so even if the 4093 could switch enough, you would be trying to draw I=V/R 5/6.146 = 814mA from its output. That is rather more than it is rated to deliver!

You need faster logic gates, a much smaller capacitor and higher feedback resistor value.

Brian.
 
Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top