Digital Signal Processing Laboratory, Second Edition by B. Preetham Kumar

By B. Preetham Kumar

Advent TO electronic sign PROCESSING short concept of DSP ConceptsProblem SolvingComputer Laboratory: advent to MATLAB®/SIMULINK®Hardware Laboratory: operating with Oscilloscopes, Spectrum Analyzers, sign SourcesDigital sign Processors (DSPs)ReferencesDISCRETE-TIME LTI indications AND structures short thought of Discrete-Time indications and SystemsProblem SolvingComputer Laboratory: Simulation of continuing Time Read more...

summary: creation TO electronic sign PROCESSING short concept of DSP ConceptsProblem SolvingComputer Laboratory: creation to MATLAB®/SIMULINK®Hardware Laboratory: operating with Oscilloscopes, Spectrum Analyzers, sign SourcesDigital sign Processors (DSPs)ReferencesDISCRETE-TIME LTI signs AND structures short conception of Discrete-Time indications and SystemsProblem SolvingComputer Laboratory: Simulation of continuing Time and Discrete-Time indications and structures ReferencesTIME AND FREQUENCY research OF verbal exchange signs short idea of Discrete-Time Fourier remodel (DTFT), Discrete Fourier remodel

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Similarly, the next term in the table below, h(1 – m), is obtained by shifting h(–m) by one step to the right. The dot product of the vectors x(m) and h(1 – m) gives the convolution output y(1). ( ) () h 1 − m =  0 0 3 2 1 0 0  ; y 1 = 3 • The process is continued until the output y(n) remains at zero. ( ) () ( ) () h 2 − m =  0 0 0 3 2 1 0  ; y 2 = 6 h 3 − m =  0 0 0 0 3 2 1 ; y 3 = 6 23 Discrete-Time LTI Signals and Systems ( ) () ( ) () h 4 − m =  0 0 0 0 0 3 2  ; y 4 = 5 h 5 − m =  0 0 0 0 0 0 3  ; y 5 = 3 Any more shift in the sequence h(–m) will result in a zero output.

The process of creation of the minimum phase system from the original system is clearly explained in the example below. Example Specify the minimum-phase system Hmin(z), for the following system function H(z): (1 + 2z )  1 − 21 z −1 () H z = ( ) −1     1 z −1  1 + z −1  3   ( ) such that H e jω = H min e jω . Solution The solution is given in a series of steps below: • Rewrite the system function H(z) in using z terms instead of z–1 terms. , z > 1. , replace z = c with z = 1/c∗, where the symbol ∗ denotes complex conjugate):  1   z + 2   z − H z =  1  z + 3  () 1 2  ( z + 2)  1  z + 2  () () () which can be separated as H z = H min z × H ap z .

H (e ) , we have to determine the • In order to ensure H e jω jω min magnitude of the all-pass system response as follows: () H ap z = ( z + 2)  1  z + 2  Hence the frequency response of the all-pass system is: H ap e (e ) = ( jω jω  e jω +2 ) 1 +  2 ( ) ( ) (since the system is jω = H e j0 and the magnitude response is H e all-pass, the magnitude response is independent of frequency) H ap e ( e = ( )   e Hence, if we redefine = ( ) j0 jω j0 +2 ) = (1 + 2 ) = 2 1 +  2 H min  1  1 + 2   1  1  z + 2   z − 2  z =2  1  z + 3  () ( ) and Hap(z) ( ) is ensured.

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