Amplitude modulation Problem

Replies

• 

  RRITESH

  Read this #-Link-Snipped-#
  
  

  A 1000-kHz carrier is simultaneously modulated with 300-Hz

  As i remember the max side band freq will be Fmax= 1000+300 and Fmin = 1000-300
• 

  Sagar07

  i think, if its amplitude modulation problem(as you say in the title), frequency of the output will be frequency of carrier itself which is 1000 Hz.
• 

  silverscorpion

  ^^ Yes.
  
  If amplitude is modulated, then there will be different amplitudes at various times in the output, but the frequency of the output will be the same as the carrier frequency.
• 

  Arnav Joshi

  Sagar07

  i think, if its amplitude modulation problem(as you say in the title), frequency of the output will be frequency of carrier itself which is 1000 Hz.

  silverscorpion

  ^^ Yes.
  
  If amplitude is modulated, then there will be different amplitudes at various times in the output, but the frequency of the output will be the same as the carrier frequency.

  
  I am afraid this is not correct. Amplitude modulation does not mean that the output frequency will be same as carrier frequency. Since carrier frequency is usually very large than signal frequency , the sidebands produced have frequency very near to that of carrier. What you say about the amplitude at different times is of right though.
  
  Each sine wave applied at input of Amplitude Modulator will produce two side bands around the carrier. For instance the sine wave has frequency f[sub]1[/sub] and carrier has frequency f[sub]c[/sub]. Output will have the frequency f[sub]c[/sub] - f[sub]1[/sub] as well as f[sub]c[/sub] + f[sub]1[/sub] and if carrier is not suppressed f[sub]c[/sub] will also be present.
  
  In this problem three sine waves are applied, each will produce two side bands.
  
  Of course I am assuming standard AM. There are forms of AM in which carrier and/or one of the side bands may be suppressed.(Modulation Analog Modulation Methods)
  
  As to why two side bands are produced ? Expand the AM signal cos(2pi f[sub]c[/sub]t )cos(2pi f[sub]1[/sub]t ) by the rule
  cos(A)cos(B) = 1/2( cos(A+B) + cos(A-B) )
• 

  Eng.Micah

  frequency will be same for both output and input. the lower and upper side bands will bring the diffferent in amplitude.