FIR Filter Design. FIR Filters and the zdomain. The zdomain model of a general FIR filter is shown in Figure 1. Figure 1


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1 FIR Filters and the Domain FIR Filter Design The domain model of a general FIR filter is shown in Figure. Figure Each  box indicates a further delay of one sampling period. For example, the input to the amplifier a 3 has been delayed by 3 sampling periods and so the amplifier can be represented by a 3 X( 3. The processor output is the sum of the input sequence and the various delayed sequences and is given by: And so: Y( a 0 X( + a X( .+ a X( .+ a 3 X( T( a 0 + a .+ a .+ a IIR Filters and the Domain These are characterised by a feedback loop. Weighted versions of delayed inputs and outputs are added to the input sample to generate the current output sample. Figure shows the domain model of this type of filter which adds only previous outputs to the current input. The transfer function of the IIR filter shown in Figure can be derived, thus: Y( X( + b Y( .+ b Y( .+ b 3 Y( 3 Y( X( + Y((b  + b .+ b 33 Y(( b   b . b 33 X( T ( 3 b b b3
2 Figure A more general form of an IIR filter is shown in Figure 3 Figure 3 By combining the transfer function of the FIR filter with the basic IIR filter the following equation for its transfer function can be derived ( b b b a a a a T Question 4 A filter has a transfer function given by: ( + T
3 (a Draw the domain version of the block diagram for the filter (b Derive an expression for the output sequence y[n], in terms of the input sequence, x[n], and the delayed input and output sequences. Solution 4 Y ( Y ( 0.4 X ( + 0. ( + 0. X ( ( 0.4 Y ( X ( 0.4X (.Y ( Changing to the time domain Y[n] x[n] 0.4[n ] 0.y[n ] Figure 4 Choosing Between FIR and IIR Filters The choice is based on the relative advantages and disadvantages:. FIR can have an exactly linear phase response. This is an important requirement in data transmission, biomedicine, digtal audio and image processing. The phase response of IIR filters are nonlinear.. FIR filters are always stable whereas this stability cannot be guaranteed in IIR filters. 3. The effects of using a limited number of bits to implement filters such as roundoff noise and coefficient quantiation errors are much less severe in FIR filters compared to IIR filters. 4. FIR filters require more coefficients for sharp cutoff filters compared to IIR. Thus more processing time and storage will be required for FIR implementation. 5. For a given number of taps the IIR filter is 5 to 0 times more efficient in controlling its gain response than the FIR filter 3
4 6. Analogue filters can be readily transformed into equivalent IIR digital filters meeting similar specifications. This is not possible with FIR filters as they have no analogue counterpart. 7. In general, the effects of noise introduced is more marked in IIR filters, because of their feedback elements cause the errors to be accumulated over time. FIR filters, however, are simply feed forward circuits so that any errors appear in the output signal only once per sampling period.0 A broad guideline would be to: Use IIR when the only important requirements are sharp cutoff filters, as IIR filters, will give fewer coefficients than FIR. Use FIR if the number of filter coefficients is not too large and, in particular, if little or no phase distortion is desired. The Fourier Transform As with IIR filters there are several methods of designing FIR filters. The method we are going to look at is based around the Fourier Transform. The Fourier series enables us to express a periodic wave in terms of a d.c. value and a series of sinusoidal signals. The sinusoidal components have frequencies equal to the frequency of the original wave and this fundamental frequency. For example, a square wave consists of the fundamental frequency and weighted values of the odd harmonics. However, many signals are not periodic speech is an obvious example of a nonperiodic waveform. The Fourier transform, F(jω, of a signal, f(t, is defined by: F( jω f ( t e jωt. dtkktime frequency It is an extremely useful transform as it enables a nonperiodic signal f(t, to be broken down into its frequency components F(jω. F(jω is normally complex, expressing both magnitude and phase of the frequency components. As the signal is nonperiodic the component frequencies will cover a continuous band, unlike the discrete frequencies composing a periodic wave. The inverse Fourier transform (IFT works the other way round. In other words, if the IFT is used to operate on the frequency spectrum of a signal, the time variation can be recovered, i.e. the signal shape. The IFT is given by: f ( t F( jω e π j ω t. dωll frequency time It is the inverse Fourier transform which is more relevant to the design of FIR filters. 4
5 Phase linearity and FIR filters FIR filters are usually designed to be linearphase (but they don't have to be. A FIR filter is linearphase if (and only if its coefficients are symmetrical around the center coefficient, that is, the first coefficient is the same as the last; the second is the same as the nexttolast, etc. (A linearphase FIR filter having an odd number of coefficients will have a single coefficient in the center which has no mate. Figure 6 Non linear phase response Figure 7 Linear phase response 5
6 Figure 8 Linear phase response The Design of FIR Filters Create a digital filter with a cutoff frequency of 50H. The sampling frequency will be assumed to be 500H, although this is only relevant for interpretation of the frequency and time scales. The ideal low pass filter has a frequency magnitude response that is unity from 0 to 50H and ero elsewhere. The phase response will be set to ero. Figure 9 If this frequency response is transformed using the Inverse Fourier Transform, the result will be the impulse response of the filter in the time domain. Performing an inverse Fourier transform on a square wave will result in a sinc function centered at the origin. Figure 0 6
7 There are a number of problems using these coefficients as the filter: the filter is non causal (there are coefficients to the left of the time origin as it needs to know future values of the function it is filtering. While this is sometimes possible it isn't if the filter is to act in real time. the filter length is infinite, in other words in order to filter with the ideal rectangular frequency response requires an infinite length filter. These two issues can be addressed by truncating the filter coefficients and then shifting the filter to the right so that all coefficients occur at or after the origin of the time axis. For example, Figure shows how Figure 0 has been truncated to 40 filter coefficients and then shifted right by 0 time steps to make it causal. Figure There is a penalty incurred for truncating the filter series to 40 coefficients, that penalty can be seen if the sinc function is transformed back into the frequency domain to view the new frequency response. Instead of the ideal rectangular response there is now "ringing" in the cutoff and the pass band, as well as a more gradual transition between the bands. The ripple is often called the Gibbs phenomenon after Willard Gibbs who documented it in 899. However, there is no way to have a finite length (duration filter and not have an infinite frequency response, the best one can do is to minimise the effect. Figure 7
8 There are two ways to reduce the departures from the ideal filter. The first is to increase the number of coefficients used by the filter. Figure 3 shows the frequency magnitude response for filter lengths of 0, 40, and 80. As expected, the transition becomes sharper as the filter order increases. The height of the ripple decreases but quite slowly. One disadvantage of using increasingly long filter lengths is the compute time required to perform the filtering. Figure 3 The usual way to reduce the ripple is to window the impulse response in the time domain. Strictly speaking the impulse response was windowed by truncating the sinc function resulting in a rectangular window. The effect was a convolution of the ideal frequency response with the appropriate sinc function. There are better windows that have a smoother transition in the time domain and reduced ripples in the frequency domain. The following shows the frequency response for filter lengths (width W of 0, 40, and 80 using a Hanning window. This window is one of many but is a good tradeoff between simplicity and sideband suppression. It is attributed to Julius von Hann and is sometimes called the raised cosine window that follows from its formula: πn w[ n] cos N where N/ < n < N/ 8
9 Figure 4 The Hanning window Figure 5 The result of windowing with the Hanning window results in much reduced ripples, although at the expense of the transition between the cutoff and the pass band as shown in Figure 5. There are many other window types, for example the Bartlett window which is a triangular function and therefore corresponds to a convolution of the ideal frequency response by sinc. A more sophisticated window with optimal characteristics (in a certain sense is the Kaiser window. It is traditional to display the magnitude response on a log 0 scale, as follows for the three different width Hanning windowed example above. 9
10 Figure 6 Before starting the FIR design examples it is imperative that you understand the concept of the impulse response of a filter. The transfer function of a filter is given by H(n Y(n/X(n Now if the input was a single impulse then the formula would become H(n Y(n/ or H(n Y(n In other words the transfer function equals the output. Therefore if we know the impulse response we just have to convolve each of the input samples with the impulse response to obtain the output. Try this in MATLAB [b, a] ellip(4, , 0.4 imp(b, a, 50 The graph gives 50 samples and shows the impulse response of an elliptical filter. Filter Design Example The starting point for this filter design method is the desired frequency response of the filter. For instance a low pass filter with a cutoff frequency of 4kH is required. The required magnitude response is shown in Figure 7. It will be assumed that the desired phase response is ero i.e. that signals of all frequencies experience no phase change on passing through a filter. 0
11 Figure 7 The first step in the design process is to move from the frequency domain to the time domain by using the Inverse Fourier Transform. This transform when calculated mathematically becomes: sin(ω ct f ( t which known as the sinc function and the shape of the waveform is: πt Figure 8 The sinc function Therefore for this filter, with a cutoff frequency of 4kH and as ω πf sin(8000π t h( t ( πt The sinc function generated is the unit impulse response of the filter. Figure 9 shows the plot of h(t [sin(8000πt]/πt, i.e the filter unit impulse response, within the window of .ms to +.ms.
12 Figure 9 The next step in the design is to sample the unit impulse response. This is the crux of the design as these sampled values will be the filter coefficients. For example if just five samples (a, b, c, b, a were taken and used as the coefficients of the transfer function, then this gives T( a + b  + c  + b 3 + a 4 ( Therefore if the values of a, b, c, b, a can be found by sampling, a digital filter that has the desired unit impulse response which corresponds to the required frequency response, will have been designed. One of the problems with the design so far is that the sinc function will go on forever (theoretically, although the values will be very small. As they are so small a compromise can be found if we limit the sample to ±5 x 04 secs. If one of our samples is to be at the central peak, then the coefficients must be symmetric about the centre to ensure phase linearity, which means that an odd number of samples will be required. If sampling occurs every ± 0.5 x 04 secs then exactly samples can be fitted into the 03 time slot. Sampling will therefore be at 0kH. This gives a Nyquist frequency of 0kH, which is well above the cutoff frequency of 4kH and will therefore be satisfactory. The response shown in Figure 8 starts at a negative time, which means that any filter would be noncausal. However, to solve this problem all that is required is to shift the required time response to the right by 0.5 x 03 secs. To reflect this time shifting by, the equation ( needs to be changed to: f sin t 4 ( 8000π ( t π ( t 5 0 ( The signal can now be sampled to obtain the filter coefficients. To do this t has to be replaced with nt in equation (, where n is the sample number, 0 to 0, and T is the sampling period of 0.5 x 04 secs f sin nt 4 ( 8000π ( nt π ( nt 5 0 ( The timewindowed, shifted and sampled signal is shown in Figure 0 below, and the sampled values shown in Table. (3 (4
13 Figure 0 n f(nt 0/0 0 /9673 / / / /5 0 6/454 7/347 8/ 87 9/ Table From Table 5., the filter must have the transfer function of: T(s When this filter is plotted in MATLAB it can be seen that the phase response is linear as expected and the cutoff frequency is correct, however, the magnitude response is far too big. Normally a filter has a gain less than unless an amplifier has been added. The magnitude response in the passband is approximately 86dB rater than the 0dB expected, which is approximately 0,000 times to much, so we need to divide the coefficients by 0,000. This gives T(s You will notice that we are sampling at 0kH and that we needed to scale by 0,000. It is possible to show that, we need to divide the sampled values by the sampling frequency. To speed up the design of FIR filters we can use set formula s. For example 3
14 Highpass Lowpass fc n sin π fn fc n sin π fn Bandpass sin fu fn fl sin fn Bandstop sin fl fn fu sin fn Example Design a low pass FIR filter with a cutoff frequency of 3kH. The filter is to be used with a sampling frequency of kh and is to have coefficients. The formula for calculating the coefficients for a low pass filter is: f c sin f N C( n 8(b To reduce the ripple redesign the filter using a Hanning window. The Hanning window function is given by: w( n cos N 4
15 Answer f c 3kH, f N 6kH C n f sin f c N sin(0.5 As there are coefficients then substituting n 5 to +5 into this equation gives coefficients of: , 0, , 0, 0.383, 0.5, 0.383, 0, , 0, T ( N.B. The central value involves dividing 0 by 0, which is a problem. However, if the sine wave is expanded using the power series, we get: 3 x sin x x 3! + 5 x 5! 7 x 7! + As sin x approaches x, then x approaches ero, and so (sin x/x must approach. It follows that: sin( 0.5 f ( t must become 0.5 when n 0. For example when x starts to get small e,g. 0.0 then: sin x ! 5! 7! e sin x sin b The Hanning Window function is given by: w( n cos N π w ( cos 0 [ cos( ]
16 etc. giving: 0, , , , ,, , , , , 0 Multiplying the filter coefficients by part (a by the corresponding Hanning coefficient gives a modified filter transfer function of: T( or T( The Dicrete Fourier Transform Another common design method for FIR filters is based on a variation of the Inverse Fourier Transform called the Discrete Fourier Transform (DFT. The difference between the Fourier Transform and the Discrete Fourier Transform is that the DFT converts the sampled time signal into it s sampled frequency spectrum X k N n 0 x[ n] e πnk j N Where X k is the kth value of the frequency samples. The inverse discrete Fourier transform (sampled frequency to sampled time is defined by: x[ n] N N n 0 X k e πnk j N Where x[n] in the nth sample in the sampled signal. 6
17 Fast Fourier Transform Although the DFT is a very efficient method of determining the freqency spectrum of any signal, it s drawback lies in the amount of time required to compute the output. This is because both indices k and n must progress through N values to produce the full range of output phasors and therefore N computations must be performed. Looking at the two equations above, it can be seen that each part of the equation will be a multiply and add function, so a DFT with N 000 will require,000,000 multiply and add operations. This would take a fast DSP processor about 0.05seconds which would limit the sample rate to 0H (/0.05! Two individuals, Cooley and Tukey, noticed that if you perform a DFT there are many redundancies in the points which you calculate. First of all, the sine and cosine waves are the same waves at different positions. One could get away with calculating just the sine wave, and shifting the phase to get the cosine wave. Also, one can use the symmetry of the waves to even further reduce calculation time. Plus, the waves, as the frequency being checked increases, overlap more and more. They devised the FFT which reduces the computation, and runs approximately 00 times faster than a DFT on the same sample wave. 7
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