I'm trying to understand something which is really confusing me and I made my best to understand it but still confused.

If I have transmitted and receiver. Assume I want to transmit S(t) data and there's two multipath ways L1 , L2 in order to arrive the receiver. Why when I have constructive interference then it's good for me ? in the paths/multipath there's might be destructive interference or constructive interference...my question why constructive interference is good for me..assume I get 2S(t) -constructive interference- in the received signal on the receiver ..why it's good for me ? what does it mean that I have 2S(t) and not S(t) ?

Im confused on the term constructive interference and how it's related to my received signal ..?

My confusion exactly is that- I received 2X although I transmitted X , so 2X isn't X ..and we say that we received a good signal? if we received 2X which isn't equal to transmitted Data X so it's not the same data so it's considered distortion and it's not a good received signal.

thanks alot enter image description here


Constructive interference just increases the gain of the signal at the receiver, which is designed (using automatic gain control) to adjust the signal level over a very wide range to that desired for reception. So the issue the OP refers to is of no real consequence as long as the received signal is in the (relatively wide) range of the receiver between maximum and minimum signal levels allowed.

Multipath distortion can come in different forms based on how fast the multipath is changing (such as in a mobile environment) resulting in "fast" or "slow" fading, and the relative distance between the paths compared the symbol rate, resulting in "flat" or "frequency selective" fading. These are the real issues with multipath distortion which are corrected with equalizers in the receiver.

Please see this post which expands on these different fading models:

Rayleigh fading with frequency selective fading channel

  • $\begingroup$ Appreciated much Dan. $\endgroup$ – Maher Mar 13 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.