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 TimeRead 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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**Additional resources for Digital Signal Processing Laboratory, Second Edition**

**Sample text**

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.