0.6 Digital filtering and the dft  (Page 10/13)

For the FFT method, the two vectors (input and convolution) must both have length N+M-1 . The raw output has complex values due to numerical roundoff, andthe command real is used to strip away the imaginary parts. Thus, the FFT based method requires more M atlab commands to implement. Observe also that conv(h,x) and conv(x,h) are the same, whereas filter(h,1,x) is not the same as filter(x,1,h) .

To view the frequency response of the filter h , M atlab provides the command freqz , which automatically zero pads By default, the M atlab command freqz creates a length 512 vector containing the specified impulse response followed by zeros. The FFT of this elongated vector is used forthe magnitude and phase plots, giving the plots a smoother appearance than when taking the FFT of the raw impulse response. the impulse response and then plots both the magnitude and the phase. Type     freqz(h) to see that the filter with impulse response h=[1, 1, 1, 1, 1] is a (poor) lowpass filter with two dips at 0.4 and 0.8 of the Nyquist frequency as shown in [link] . The command freqz always normalizes the frequency axis so that “1.0” corresponds to the Nyquist frequency $f_s/2$ . The passband of this filter(all frequencies less than the point where the magnitude drops 3 dB below the maximum) ends just below $0.2$ . The maximum magnitude in the stopband occurs at about $0.6$ , where it is about 12 dB down from the peak at zero.Better (i.e., closer to the ideal) lowpass filters would attenuate more in the stopband,would be flatter across the passband, and would have narrower transition bands.

Implementing iir filters

At first glance it might seem counterintuitive that a useful filter could have an impulse response that isinfinitely long. To see how this infiniteness might arise in a specific case,suppose that the output $y\left[k\right]$ of a LTI system is created from the input $x\left[k\right]$ according to the formula

where $a$ and $b$ are constants. This is shown in the top part of [link] , where the time delay between successive terms(between $k$ and $k+1$ for instance) is represented by $z^{-1}$ . This is an example of the $\mathcal{Z}$ -transform. If the initial value is $y\left[0\right]=0$ , the response of the system [link] to an impulse $x\left[k\right]=\delta \left[k\right]$ (where $\delta \left[k\right]$ is the discrete impulse function [link] ) is

If $\left|a\right|>1$ , this impulse response increases towards infinity and the system is said to be unstable. If $\left|a\right|<1$ , the values in the sequence die away towards (but never quite reaching) zero,and the system is stable. In both cases, the impulse response is infinitely long and so the system [link] is IIR. The special case when $a=1$ is called a summer (or, in analogy with continuous time, an integrator ) because $y\left[k\right]=y\left[0\right]+b{\sum }_{i=0}^{k-1}x\left[i\right]$ sums up all the inputs. The summer is oftenrepresented symbolically as in the lower part of [link] .