# concise dsp tutorial by BEERNARD KLAR By BEERNARD KLAR

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The more stringent the filter requirements of stopband attenuation, transition bandwidth and to a lesser extent passband ripple, the more weights that are required. Consider again the design of the above FIR filter (FIR1) which was a low pass filter cutting of at about 1000 Hz. Using SystemView, the above criteria can be varied such that the number of filter weights can be increased and a more stringent filter designed. Consider the design of three low pass filters cutting off at 1000 Hz, with stopband attenuation of 40 dB and transition bandwidths 500 Hz, 200 Hz and 50 Hz: Gain (dB) 0 FIR 1 0 FIR 2 0 -20 -20 -20 -40 -40 -40 -60 -60 -60 -80 -80 -80 0 1000 2000 3000 4000 5000 frequency (Hz) Transition Band: 1000 - 1500Hz No.

The computational savings achieved by the FFT is therefore a factor of K ⁄ log 2K . com  2 Frequency Domain Analysis 31 considerable. 5 x106 There are a number of different FFT algorithms sometimes grouped via the names Cooley-Tukey, prime factor, decimation-in-time, decimation-in-frequency, radix-2 and so on. The bottom line for all FFT algorithms is, however, that they remove redundancy from the direct DFT computational algorithm of Eq. 53. We can highlight the existence of the redundant computation in the DFT by inspecting Eq.

Unlike FIR filters, IIR filters can exhibit instability and must therefore be very carefully designed. The term infinite refers to the fact that the output from a unit pulse input will exhibit nonzero outputs for an arbitrarily long time. If the digital filter is IIR, then two weight vectors can be defined: one for the feedforward weights and one for the feedback weights: x( k – 1) x(k) a0 a1 x( k – 2 ) y(k – 3) y( k – 2 ) y( k – 1 ) b3 b2 b1 a2 Feedforward Zeroes (non-recursive) 2 y(k ) = ∑ y(k) Feedback Poles (recursive) 3 an x( k – n) + n=0 ∑ bn y ( k – n ) n=1 = a 0 x ( k ) + a1 x ( k – 1 ) + a2 x ( k – 2 ) + b1 y ( k – 1 ) + b 2 y ( k – 2 ) + b 3 y ( k – 3 ) x(k ) y( k – 1 ) ⇒ y ( k ) = a T x ( k ) + b T y ( k – 1 ) = a0 a1 a 2 x ( k – 1 ) + b1 b2 b3 y ( k – 2 ) x( k – 2) y( k – 3 ) A signal flow graph and equation for a 3 zero, 4 pole infinite impulse response filter.

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