This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: UCLA Dept. of Electrical Engineering EE 114, Winter 2009 Problem Set 2 Solutions 1. During spectral analysis of speech, a preemphasis filter is generally used to boost higher frequencies, since they tend to contain discriminative information. A common form of the preemphasis filter is: H ( z ) = 1 G 1 az 1 (1) (a) If we want to design the filter to have 5 times the gain at the Nyquist frequency relative to DC, i.e:  H ( )  = 5  H (0)  , (2) what should the parameter a be set to? (Hint: a should lie in the range (0 , 1)). (b) If it is desired that the total energy of the filter be unity, i.e.: X n =  h ( n )  2 = 1 , (3) what should the parameter G be set to? Answer: (a) The frequency response of the preemphasis filter is determined as:  H ( )  2 = H ( ) H * ( ) = 1 G 2 1 ae jn 1 ae jn = 1 G 2 1 + a 2 2 a cos( )  H ( )  = 1 G q 1 + a 2 2 a cos( ) Using this expression:  H ( )   H (0)  = s 1 + a 2 + 2...
View
Full
Document
This note was uploaded on 05/10/2010 for the course EE EE114 taught by Professor Vanderschaar during the Spring '09 term at UCLA.
 Spring '09
 VANDERSCHAAR
 Electrical Engineering

Click to edit the document details