Discrete Deconvolution in DSP Systems

Discrete Deconvolution: To find h(n) if both x(n) and y(n) are given To know the type of system if it is FIR or IIR Here, we usually use the z-transform method since it is valid for both the FIR and IIR DSP systems. h(n)= Z-1[H(z)]= Z-1[Y(z)/X(z)]

Ex: Find h(n) if x(n)=u(n) and y(n)=2u(n)-(0.5)n u(n). Solution: X(z)=z/(z-1) and Y(z)= [2z/(z-1)] [z/(z-0.5)] Hence h(n)=(0.5)n u(n).

DSP system Implementations Here h(n) has finite no of elements: Y (Z)= h(0) X(Z) + h(1) X(Z) + h(2) X(Z) + ..h(m) X(Z)

b)-IIR system: For IIR system having m-zeros and r-poles, y(n)=ao x(n)+a1 x(n-1)+a2 x(n-2)+ .+am x(n-m)- b1 y(n-1)- b2 y(n-2)- .- bry(n-r)

Spectral Analysis of Discrete Signals: a-periodic discrete signals: If x(n) is a periodic signal having a period of N samples, i.e. x(n) is repeated every N samples, then the frequency content of x(n) (spectral analysis) is obtained using discrete Fourier series (DFS), = 0 n N 2 n k 1 1 N j = ( ) ( ) X k x n e N N- 1 k 0 Where: n is the time index k is the frequency index 2 n k 1 N j = k = ( ) ( ) x n X k e N The inverse DFS is given by: N- 1 n 0 0

Ex: Find the DFS of the signals x(n) if x(n) is periodic with period N=4 samples and is defined for full period by: x(n)={2,-1,3,-3}. 2 n k 1 N 1 N j = n = ( ) ( ) X k x n e N 0 2 ) 0 ( n + + + + 3 1 ) 0 ( x ) 1 ( x ) 2 ( x ) 3 ( x 2 1 3 3 1 j = = = = ) 0 ( X ( ) x n e 4 =dc value 4 4 4 4 = 0 n 2 ) 1 ( 3 n 3 1 1 j j j = j = + + + ) 1 ( X ( ) [ ) 0 ( x ) 1 ( x ) 2 ( x ) 3 ( x ] x n e e e e 4 2 2 4 4 = 0 n X(1)=0.25[2-1(-j)+3(-1)-3(j)] = 0.25(-1-2j) = -0.25-0.5j X(2)=9/4 and X(3)=-0.25+0.5j Note that X(3)=X*(1) (* conjugate)

b- aperiodic or random discrete signals: If x(n)={x(0), x(1),x(2), .x(N-1)} is the sampled signal from x(t), then the DFT of x(n) is given by:

Ex: Speech signal is sampled at 8000Hz. Find the min number of samples to the DFT analysis such that frequency analysis is done with a resolution of 20Hz. for 20 Hz resolution, then 20=fs/N=8000/N, then N=400 samples, i.e.: x(n)={x(0),x(1),x(2), .x(399)} and: 1 ) ( k X 2 n k 399 j = n = ( ) x n e 400 400 0 Number of complex multiplications = NXN = N2

Ex: Find the DFT of the sampled sequence representing a sampled pulse: x(n)={1,1,1,0,0,0}. E 2 n k 1 N 1 N j = n = ( ) ( ) X k x n e N 0

Fast Fourier Transform (FFT) This is a fast method to find the DFT 1) For FFT base-2 (radix-2), then N must be a power of 2, i.e. N=2r (4,8,16,32,64,128,256,512, ). If N 2r, then 0's are added to complete the sequence to the nearest 2r value. 2 n k 1 N j = n = ( ) ( ) X k x n e N 0 3) Separate the samples into N/2 even samples and N/2 odd samples 4) X(k)=G(k)+WNk * H(k) .(1) where G(k)=DFT of the even numbered samples (N/2 samples) H(k)=DFT of the odd numbered samples (N/2 samples). WNk is the twiddle factor = e-j2 k/N

DFT of 2-samples: Where the unity path gain is not marked. The 2-point (samples) DFT is called "Butterfly'' Number of complex multiplications in FFT= N log2N

Ex: Draw the signal flow graph of the 4-point FFT, then use to it to find the spectrum of the sequence x(n)={1,-1,2,3}. The even numbered samples are x(0) and x(2) The odd numbered samples are x(1) and x(3) c X(0)=G(0)+W40 H(0) X(1)=G(1)+W41 H(1) X(2)=G(2)+W42 H(2) since G(2)=G(0)and H(2)=H(0) where 2-point DFT has a period of 2 samples, then X(2)=G(0)+W42 H(0) X(3)=G(3)+W43 H(3) And again G(3)=G(1)and H(3)=H(1), then: X(3)=G(1)+W43 H(1) E=