A definite advantage of the FFT is that it reduces  considerably the computation in the convolution sum. Thus if  x[n], 0 ‰¤ n ‰¤ N ˆ’ 1, is the input of an FIR filter with impulse response  h[n], 0 ‰¤ n ‰¤ M ˆ’ 1, their convolution sum y[n] = (x^ˆ—h) [n] will be of length M + N ˆ’ 1. Now if X[k] and H[k] are the DFTs (computed by the  FFT) of x[n] and h[n], then Y[k] = X[k] H[k] is the DFT of the convolution  sum, of length M + N ˆ’ 1. To multiply the FFTs X[k] and H[k] they both  should be of the same length as Y[k] , i.e., M + N ˆ’ 1. Consider what  happen when the filter is IIR which has possibly an impulse response of  very large length. Let

y[n] ˆ’ 1.755y[n ˆ’ 1] + 0.81y[n ˆ’ 2] = x[n] + 0.5x[n]

be the difference equation representing an IIR filter with input x[n] and  output y[n]. Assume the initial conditions are zero, and the input is  x[n] = u[n] ˆ’ u[n ˆ’ 50]. Use MATLAB to obtain the filter output.

(a) Compute using filterthe first 40 values of the impulse response h[n], and call them  hÌ‚[n], an approximate of h[n]. Compute the filter output yÌ‚[n] using the FFT as indicated above. In this case, we are  approximating the IIR filter by and FIR filter of length 40. Plot the  input and the output. Use FFTs of length 128.

(b) Suppose now that we do not want to approximate h[n], so consider the following procedure. Find the transfer function of the IIR  filter, say H(z) = B(z)/A(z), and if X(z) is the Z-transform of the  input then

Compute as before the FFT for x[n], of length 128, call it X[k],  and compute the 128-length of the coefficients of B(z) and A(z) to  obtain DFTs B[k] and A[k]. Multiplying X[k] by B[k] and dividing  by A[k], all of length 128, results in a sequence of length 128 that  should correspond to Y[k], the DFT of y[n]. Compute the inverse FFT to get y[n] and plot it.

(c) Use filterto solve the difference equation and obtain y[n] for  x[n] = u[n] ˆ’ u[n ˆ’ 50]. Considering this the exact solution, calculate the error with respect to the other responses in (a) and (b).  Comment on your results.

Transcribed Image Text:

B(z)X(z) Y (z) : A(z)