My course text only says that if we have a conductive coupling then the digital current will affect the analog signal because it causes certain voltage drop in the common wire. But it doesn't say anything about analog signals affecting the digital.

Can someone give me a simple explanation?

  • \$\begingroup\$ Digital signals are sensitive to positive edges, while an analog signal is sensitive to... basically any noise that is picking up! In other words, adding a small analog signal to a digital device won't cause much problem since the edges are preserved. But digital signal creates lot of undesired frequency content in an analog signal. \$\endgroup\$ – lucas92 Apr 25 '16 at 19:03

[The term "digital signal" is only a convenient shorthand. There is digital information imposed on an analog signal.]

The analog current will generate a voltage drop in the common wire too, and it will affect the digital line. However, the drop in the common wire would have to be larger than a logic threshold in order to affect a digital signal. That's why digital signal often is less susceptible to such interference.

  • 3
    \$\begingroup\$ My favourite sig line from usenet was this: "There is no such thing as a digital signal; EMC testing proves this daily". \$\endgroup\$ – Peter Smith Apr 26 '16 at 6:49

In most cases, a digital signal is much faster than an analog signal, meaning that it contains a higher frequency content. As you probably know by now, traces that are close to each other can have parasitic capacitance. As frequency increases, the capacitance between two traces begins to look like a short circuit and can result in something commonly known as cross-talk.

An analog signal can just as easily jump to an analog or digital signal via crosstalk; however, in reality, analog signals are most frequently composed of much lower frequencies than the digital signals.

Additionally, digital signals are simply more immune to noise. A 100mV offset on a digital signal is no big deal past the transition point. In a 10-bit A/D referenced at 3.3V, a 100mV can be 31 codes off!

  • \$\begingroup\$ Ok, i was just curious that, do the parasitic capacitance effect also happens between digital traces? \$\endgroup\$ – Hilton Khadka Apr 26 '16 at 7:49
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    \$\begingroup\$ Yes, it does and it can be a nightmare to debug! This is why your professor probably told you to use the slowest logic chips that will get the job done.... well, mine did... we trained on 7400 series logic chips :) \$\endgroup\$ – slightlynybbled Apr 26 '16 at 12:28
  1. Signals are affected by the magnetic field a current produces in a wire. The bigger the Jump of the Amplitude of the disturbing Signal, the bigger the indicadet Voltage to your Signal. Where digital Signals have sharp edges,analoge Signals often don't have this sharp pitch.

  2. Digital Signals are less affected because of its hysteresis behaviour. Where a ADC detect disturbing Signals in mV, a Digital I/O-Port hase bigger tolerances. Read TTL Logic for more info.



  • \$\begingroup\$ What do you mean by "Read TTL Logic"? \$\endgroup\$ – pipe Apr 25 '16 at 19:31
  • \$\begingroup\$ en.wikipedia.org/wiki/Transistor–transistor_logic chapter Interfacing \$\endgroup\$ – classMate Apr 25 '16 at 19:36
  • \$\begingroup\$ en.wikipedia.org/wiki/Hysteresis \$\endgroup\$ – classMate Apr 25 '16 at 19:38
  • \$\begingroup\$ It's allowed to update your answer with more information, please add the links to the answer if that's what you refer to. \$\endgroup\$ – pipe Apr 25 '16 at 19:39

It's not that anolog stuff can't affect digital stuff, it's just that digital circuits are (usually) more noise tollerant than analog circuits, and digital circuits tend to create more intense transients and intense transients couple better to adjacent circuits. An analog system can still create havoc with a digital one under the right conditions though, e.g. a cell phone in a microwave


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