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x ( α ) = - s ( λ ) h ( α - λ ) d λ

of the matched filter with the impulse response h ( t ) . Given the pulse shape p ( t ) and the assumption that the noise has flat power spectral density, it follows that

h ( t ) = p ( α - t ) , 0 t T 0 , otherwise ,

where α is the delay used in the matched filter. Because h ( t ) is zero when t is negative and when t > T , h ( α - λ ) is zero for λ > α and λ < α - T . Accordingly, the limits on the integration can be converted to

x ( α ) = λ = - α - T α s ( λ ) p ( α - ( α - λ ) ) d λ = λ = - α - T α s ( λ ) p ( λ ) d λ .

This is the cross-correlation of p with s as defined in [link] .

When P n ( f ) is not a constant, [link] becomes

v 2 ( τ ) P w = | - H ( f ) G ( f ) e j 2 π f τ d f | 2 - P n ( f ) | H ( f ) | 2 d f .

To use the Schwarz inequality [link] , associate a with H P n and b with G e j 2 π f τ / P n . Then [link] can be replaced by

v 2 ( τ ) P w - | H ( f ) | 2 P n ( f ) d f - | G ( f ) e j 2 π f τ | 2 P n ( f ) d f - | H ( f ) | 2 P n ( f ) d f ,

and equality occurs when a ( · ) = k b * ( · ) ; that is,

H ( f ) = k G * ( f ) e - j 2 π f τ P n ( f ) .

When the noise power spectral density P n ( f ) is not flat, it shapes the matched filter.Recall that the power spectral density of the noise can be computed from its autocorrelation.

Let the pulse shape be a T -wide Hamming blip. Use the code in matchfilt.m to find the ratio of the power in the downsampled v to that in the downsampled w when:

  1. the receive filter is a SRRC with beta= 0, 0.1, 0.5,
  2. the receive filter is a rectangular pulse, and
  3. the receive filter is a 3 T -wide Hamming pulse.

When is the ratio largest?

Let the pulse shape be a SRRC with beta=0.25 . Use the code in matchfilt.m to find the ratio of the power in the downsampled v to that in the downsampled w when

  1. the receive filter is a SRRC with beta= 0, 0.1, 0.25, 0.5,
  2. the receive filter is a rectangular pulse, and
  3. the receive filter is a T -wide Hamming pulse.

When is the ratio largest?

Let the symbol alphabet be 4-PAM.

  1. Repeat [link] .
  2. Repeat [link] .

Create a noise sequence that is uniformly distributed (using rand ) with zero mean.

  1. Repeat [link] .
  2. Repeat [link] .
Baseband communication system considered in Exercise 11-29.
Baseband communication system considered in [link] .

Consider the baseband communication system in [link] . The symbol period T is an integer multiple of the sample period T s , i.e.  T = N T s . The message sequence is nonzero only each N th k , i.e.  k = N T s + δ T s where the integer δ is within 0 δ < N and δ T s is the “on-sample” baud timing offset at the transmitter.

  1. Suppose that f [ k ] = { 0 . 2 , 0 , 0 , 1 . 0 , 1 . 0 , 0 , 0 , - 0 . 2 } for k = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 and zero otherwise. Determine the causal g [ k ] that is the matched filter to f [ k ] . Arrange g [ k ] so that the final nonzero element is as small as possible.
  2. For the g [ k ] specified in part (a), determine the smallest possible N so the path from m [ k ] to y [ k ] forms an impulse response that would qualify as a Nyquist pulse for some well-chosenbaud-timing offset σ when y is downsampled by N .
  3. For the g [ k ] chosen in part (a) and the N chosen in part (b), determine the downsampler offset σ for s [ i ] to be the nonzero entries of m [ k ] when w [ k ] = 0 for all k .

Matched transmit and receive filters

While focusing separately on the pulse shaping and the receive filtering makes sense pedagogically,the two are intimately tied together in the communication system. This section notes thatit is not really the pulse shape that should be Nyquist, but rather the convolution ofthe pulse shape with the receive filter.

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Source:  OpenStax, Software receiver design. OpenStax CNX. Aug 13, 2013 Download for free at http://cnx.org/content/col11510/1.3
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