When we consider the much more realistic situation when we have
      a channel that introduces attenuation and noise, we can make use
      of the just-described receiver's linear nature to directly
      derive the receiver's output.  The attenuation affects the output
      in the same way as the transmitted signal: It scales the output
      signal by the same amount.  The white noise, on the other hand,
      should be filtered from the received signal before
      demodulation.  We must thus insert a bandpass filter having
      bandwidth
      
	2W
      
	  
	  2
	  W
	 
      and center frequency 
      
	  
	    
	      f
	      c
	    
	  
      
	    
	      f
	      c
	    
	  : This filter has no effect on the received
      signal-related component, but does remove out-of-band noise
      power. As shown in the triangular-shaped signal spectrum, we apply
      coherent receiver to this filtered signal, with the result that
      the demodulated output contains noise that cannot be removed: It
      lies in the same spectral band as the signal.
    
    
      As we derive the signal-to-noise ratio in the demodulated
      signal, let's also calculate the signal-to-noise ratio of the
      bandpass filter's output
      
      
	
	    
	      r
	      ˜
	    
	  t
      
	  
	    
	      r
	      ˜
	    
	  
	  t
	.  The signal component of
      
	
	    
	      r
	      ˜
	    
	  t
      
	  
	    
	      r
	      ˜
	    
	  
	  t
	
      equals
      
	α
	    
	      A
	      c
	    
	  mtcos2π
		
		  f
		  c
		
	      t
      
	  
	  α
	  
	    
	      A
	      c
	    
	  
	  
	    m
	    t
	  
	  
	    
	    
	      
	      2
	      
	      
		
		  f
		  c
		
	      
	      t
	    
	  
	.  This signal's Fourier transform equals
      
	
	  α
		  
		    A
		    c
		  
		2(Mf+
		    
		      f
		      c
		    
		  +Mf−
		    
		      f
		      c
		    
		  )
	
	    
	    
	      
	      
		
		α
		
		  
		    A
		    c
		  
		
	      
	      2
	    
	    
	      
	      
		M
		
		  
		  f
		  
		    
		      f
		      c
		    
		  
		
	      
	      
		M
		
		  
		  f
		  
		    
		      f
		      c
		    
		  
		
	      
	    
	  
      
(1)  
      making the power spectrum, 
      
	
	  α2
		    
		      A
		      c
		    
		  24(|Mf+
			
			  f
			  c
			
		      |2+|Mf−
			
			  f
			  c
			
		      |2)
	
	    
	    
	      
	      
		
		
		  
		  α
		  2
		
		
		  
		  
		    
		      A
		      c
		    
		  
		  2
		
	      
	      4
	    
	    
	      
	      
		
		
		  
		  
		    M
		    
		      
		      f
		      
			
			  f
			  c
			
		      
		    
		  
		
		2
	      
	      
		
		
		  
		  
		    M
		    
		      
		      f
		      
			
			  f
			  c
			
		      
		    
		  
		
		2
	      	   
	    
	  
      
(2)   
    
    
	
	  If you calculate the magnitude-squared of the first
	  equation, you don't obtain the second unless you make an
	  assumption.  What is it?
	
       
	  The key here is that the two spectra  
	  
	    Mf−
		  
		    f
		    c
		  
		
	  
	      M
	      
		
		f
		
		  
		    f
		    c
		  
		
	      
	    ,
	  
	    Mf+
		  
		    f
		    c
		  
		
	  
	      M
	      
		
		f
		
		  
		    f
		    c
		  
		
	      
	    
	  do not overlap because we have assumed that the carrier frequency  
	  
	    
	      
		f 
		c
	      
	    
	  
	      
		f 
		c
	      
	    
	  
	  is much greater than the signal's highest frequency.
	  Consequently, the term
	  
	    Mf−
		    
		      f
		      c
		    
		  Mf+
		    
		      f
		      c
		    
		  
	  
	      
	      
		M
		
		  
		  f
		  
		    
		      f
		      c
		    
		  
		
	      
	      
		M
		
		  
		  f
		  
		    
		      f
		      c
		    
		  
		
	      
	    
	  normally obtained in computing the magnitude-squared equals
	  zero.
	
 
    
    
      Thus, the total signal-related power in 
      
      
	
	    
	      r
	      ˜
	    
	  t
      
	  
	    
	      r
	      ˜
	    
	  
	  t
	
      is
      
	α2
		  
		    A
		    c
		  
		22powerm
      
	  
	  
	    
	    
	      
	      
		
		α
		2
	      
	      
		
		
		  
		    A
		    c
		  
		
		2
	      
	    
	    2
	  
	  
	    power
	    m
	  
	.  
      The noise power equals the integral of the noise power spectrum;
      because the power spectrum is constant over the transmission
      band, this integral equals the noise amplitude
      
	
	  
	    N
	    0
	  
	
      
	  
	    N
	    0
	  
	
      times the filter's bandwidth
      
	2W
      
	  
	  2
	  W
	.  The so-called received signal-to-noise
      ratio — the signal-to-noise ratio after the
      de rigeur front-end bandpass filter and
      before demodulation — equals
      
	
	  
	      
		SNR
		r
	      
	    =α2
		    
		      A
		      c
		    
		  2powerm4
		  
		    N
		    0
		  
		W
	
	    
	    
	      
		SNR
		r
	      
	    
	    
	      
	      
		
		
		  
		  α
		  2
		
		
		  
		  
		    
		      A
		      c
		    
		  
		  2
		
		
		  power
		  m
		
	      
	      
		
		4
		
		  
		    N
		    0
		  
		
		W
	      
	    
	  
      
(3) 
     
    The demodulated signal
      
	
	      
		m
		^
	      
	    t=α
		  
		    A
		    c
		  
		mt2+
		
		  n
		  out
		
	      t
      
	  
	  
	    
	      
		m
		^
	      
	    
	    t
	  
	  
	    
	    
	      
	      
		
		α
		
		  
		    A
		    c
		  
		
		
		  m
		  t
		
	      
	      2
	    
	    
	      
		
		  n
		  out
		
	      
	      t
	    
	  
	.  Clearly, the signal power equals 
      
	α2
		
		  A
		  c
		
	      2powerm4
      
	  
	  
	    
	    
	      
	      α
	      2
	    
	    
	      
	      
		
		  A
		  c
		
	      
	      2
	    
	    
	      power
	      m
	    
	  
	  4
	.  To determine the noise power, we must understand how
      the coherent demodulator affects the bandpass noise found in
      
	
	    
	      r
	      ˜
	    
	  t
      
	  
	    
	      r
	      ˜
	    
	  
	  t
	.  Because we are concerned with noise, we must deal
      with the power spectrum since we don't have the Fourier
      transform available to us.  Letting
      
	
	    
	      P
	      ñ
	    
	  f
      
	  
	    
	      P
	      ñ
	    
	  
	  f
	
      denote the power spectrum of
      
	
	    
	      r
	      ˜
	    
	  t
      
	  
	    
	      r
	      ˜
	    
	  
	  t
	's noise component, the power spectrum after
      multiplication by the carrier has the form
      
	
	  
		  
		    P
		    ñ
		  
		f+
		    
		      f
		      c
		    
		  +
		  
		    P
		    ñ
		  
		f−
		    
		      f
		      c
		    
		  4
	
	    
	    
	      
	      
		
		  
		    P
		    ñ
		  
		
		
		  
		  f
		  
		    
		      f
		      c
		    
		  
		
	      
	      
		
		  
		    P
		    ñ
		  
		
		
		  
		  f
		  
		    
		      f
		      c
		    
		  
		
	      
	    
	    4
	  
      
(4)
      The delay and advance in frequency indicated here results in two
      spectral noise bands falling in the low-frequency region of
      lowpass filter's passband.  Thus, the total noise power in this
      filter's output equals
      
	2·
		
		  N
		  0
		
	      2·W·2·14=
		
		  N
		  0
		
	      W2
      
	  
	  ·
	    2
	    
	      
	      
		
		  N
		  0
		
	      
	      2
	    
            W
            2
	    
               1
               4
	    
	  
	  
	    
	    
	      
	      
		
		  N
		  0
		
	      
	      W
	    
	    2
	  
	.
      The signal-to-noise ratio of the receiver's output thus equals
      
	
	  
	      
		SNR
		
		  m
		  ^
		
	      
	    =α2
		    
		      A
		      c
		    
		  2powerm2
		  
		    N
		    0
		  
		W=2
		
		  SNR
		  r
		
	      
	
	    
	    
	      
		SNR
		
		  m
		  ^
		
	      
	    
	    
	      
	      
		
		
		  
		  α
		  2
		
		
		  
		  
		    
		      A
		      c
		    
		  
		  2
		
		
		  power
		  m
		
	      
	      
		
		2
		
		  
		    N
		    0
		  
		
		W
	      
	    
	    
	      
	      2
	      
		
		  SNR
		  r
		
	      
	    
	  
      
(5) 
     
    
      Let's break down the components of this signal-to-noise ratio to
      better appreciate how the channel and the transmitter parameters
      affect communications performance.  Better performance, as
      measured by the SNR, occurs as it
      increases.
      
-  More transmitter power — increasing 
	  
	    
	      
		A
		c
	      
	    
	  
	      
		A
		c
	      
	     — increases the signal-to-noise
	  ratio proportionally.
	 
 -  The carrier frequency 
	  
	    
	      
		f
		c
	      
	    
	  
	      
		f
		c
	      
	    
	  has no effect on SNR, but we have assumed that
	  
	    
		
		f
		  c
		
	      ≫W
	  
	    ≫
	      
		
		f
		  c
		
	      
	      W
	    .
	 
 -  The signal bandwidth WW enters the
	  signal-to-noise expression in two places: implicitly through
	  the signal power and explicitly in the expression's
	  denominator.  If the signal spectrum had a
	  constant amplitude as we increased the bandwidth,
	  signal power would increase proportionally. On the other
	  hand, our transmitter enforced the criterion that  signal amplitude was
	  constant.  Signal amplitude essentially equals the
	  integral of the magnitude of the signal's spectrum.
	  
	    This result isn't exact, but we do know that
	    
	      m0=∫−∞∞Mfdf
	    
		
		
		  m
		  0
		
		
		  
		  f
		  
		    
		       
		      
		    
		  
		  
		  
		    M
		    f
		  
		
	      
	    .
 
	  Enforcing the signal amplitude specification means that as
	  the signal's bandwidth increases we must decrease the
	  spectral amplitude, with the result that the signal power
	  remains constant.  Thus, increasing signal bandwidth does
	  indeed decrease the signal-to-noise ratio of the receiver's
	  output.
	  -  Increasing channel attenuation — moving the
	  receiver farther from the transmitter — decreases the
	  signal-to-noise ratio as the square.  Thus, signal-to-noise
	  ratio decreases as distance-squared between transmitter and
	  receiver.
	 
 -  Noise added by the channel adversely affects the
	  signal-to-noise ratio.
	
 
   
    
    
      In summary, amplitude modulation provides an effective means for
      sending a bandlimited signal from one place to another.  For
      wireline channels, using baseband or amplitude modulation makes
      little difference in terms of signal-to-noise ratio.  For
      wireless channels, amplitude modulation is the only
      alternative.  The one AM parameter that does not affect
      signal-to-noise ratio is the carrier frequency
      
	
	  
	    f
	    c
	  
	
      
	  
	    f
	    c
	  
	: We can choose any value we want so long as the
      transmitter and receiver use the same value.  However, suppose
      someone else wants to use AM and chooses the same carrier
      frequency.  The two resulting transmissions will add, and
      both receivers will produce the sum of the
      two signals. What we clearly need to do is talk to the other
      party, and agree to use separate carrier frequencies. As more
      and more users wish to use radio, we need a forum for agreeing
      on carrier frequencies and on signal bandwidth.  On earth, this
      forum is the government.  In the United States, the Federal
      Communications Commission (FCC) strictly controls the use of the
      electromagnetic spectrum for communications.  Separate frequency
      bands are allocated for commercial AM, FM, cellular telephone
      (the analog version of which is AM), short wave (also AM), and
      satellite communications.
    
    
	
	  Suppose all users agree to use the same signal bandwidth.
	  How closely can the carrier frequencies be while avoiding
	  communications crosstalk?  What is the signal bandwidth for
	  commercial AM?  How does this bandwidth compare to the
	  speech bandwidth?
	
       
	  Separation is   
	  
	    2W
	  
	      
	      2
	      W
	    .  Commercial AM signal bandwidth is   
	  
	    5kHz
	  
	      
	      5
	      kHz
	    .  Speech is well contained in this bandwidth, much
	  better than in the telephone!
	
 
    
   
        
"Electrical Engineering Digital Processing Systems in Braille."