 Hello everyone, I am Prashant S. Malge, Assistant Professor, Department of Electronics Engineering, Walton Institute of Technology, Swalapur. Today, we will discuss symmetric and anti-symmetric FIR filters, learning outcome. At the end of this session, student will be able to explain symmetric and anti-symmetric FIR filters, design of digital filters. For design of frequency selective networks, generally the characteristics are specified in the frequency domain in terms of magnitude characteristics and phase characteristics. Once these characteristics are specified, it is our aim to determine the coefficients of causal FIR or IR filters that closely approximates the specified characteristics. Generally, FIR filters are used whenever there is a requirement of linear phase characteristics in the pass band. IR filters have lower side lobes in the stop band than FIR filters with the same number of parameters. Also, for implementation of IR filters, it involves fewer parameters, requires less memory and has lower computational complexity. So, whenever there is a linear phase requirement, we have to go for FIR filters, otherwise we may use IR filters. Now, consider an ideal low pass filter. If you consider frequency response characteristics of an ideal low pass filter, it is of the form H omega equal to 1 between mod omega less than or equal to omega c and 0 for omega between omega c to pi. So, in case of a digital filter characteristics, the response is specified from minus pi to plus pi. So, for low pass filter, it should be 1 from minus omega c to plus omega c. From this, we can obtain the impulse response of the filter by taking the inverse Fourier transform of this, which we get it as given by this particular equation. If this impulse response is plotted with cutoff frequency equal to pi by 4, you will get the impulse response of this nature. Clearly, as shown in the figure, this particular impulse response is present from minus infinity to plus infinity. Of course, it is not causal and therefore, it is not practically realizable. So, to make it practically realizable, we can convert it into a causal response. The causality of the impulse response implies that the frequency response characteristics will not be 0, except at few possible points in the frequency range. Also, the transition from pass band to stop band will not be sharp. That is, the response characteristics will not abruptly change from unity to 0 when it is going from pass band to stop band. Although the ideal frequency response characteristics is desirable, but in most of the practical applications, it is not required. Usually, some ripple, a non-zero value is tolerable in the pass band. Also, a non-zero value or a ripple is tolerable in the stop band. The transition from pass band to stop band defines the transition band. For example, in case of a low pass filter, this particular frequency is a bandage frequency omega p known as a pass bandage frequency. So, the frequencies from 0 to this frequencies are said to be in the pass band. Another frequency that is a stop bandage frequency is indicated by omega s. Omega s onwards all the frequency will be a stop band and omega s minus omega p in this case defines the transition band. The frequencies which are in the pass band are defined the bandwidth of the filter. For example, in this case, this omega p is the bandwidth of a low pass filter. So, if we consider the ripple which is allowed in the pass band to be a delta 1, in that case the frequency response magnitude will lie between 1 plus delta 1 to 1 minus delta 1. Similarly, if we consider delta 2 as the ripple which is allowed in the stop band, then the frequency response magnitude will be less than delta 2 in the stop band. So, usually for designing the filter these parameters are specified that is a maximum allowable ripple in the pass band delta 1, maximum allowable ripple in the stop band delta 2, the pass bandage frequency omega p and stop bandage frequency omega s. Given these specifications, we need to select the values of the coefficients a k's and b k's for the filter which will meet these specifications as close as possible. Now, consider the metric and anti symmetric FIR filters. FIR filter of length m with input x and input y n is given by the equation y of n is equal to b 0 x of n plus b 1 x of n minus 1 plus dot dot dot b of m minus 1 x of n minus m plus 1. So, the length of the filter is m, the total number of coefficients are 0 to m minus 1, so total m coefficients. So, this equation can be written as k equal to 0 to m minus 1 b k x of n minus k, where b k's are the coefficients of the filter. And as a convolution of unit impulse response h n and the input x n, it can be expressed as y of n is equal to k equal to 0 to m minus 1 h of k x of n minus k. Clearly, the summation range that is 0 to m minus 1 tells us that it is causal and as well as the impulse response is finite in duration. So, if you compare these two, your values of impulse response can be same as the coefficients. For example, h of 0 equal to b 0 h of 1 equal to b 1 and so on. These filters can also be characterized by the system transfer function as h of z equal to k equal to 0 to m minus 1 h of k z raise to minus k. Actually, you can get this particular equation by taking the z transform thus. So, now, these FI filters gives a linear phase if the impulse response satisfies the condition h of n is equal to plus or minus h of m minus 1 minus n. That is either it is symmetric or anti symmetric. So, if this symmetry or anti symmetric condition is incorporated into the equation of the system transfer function. So, there h of z can be written as h of z equal to h of 0 plus h 1 z raise to minus 1 h of 2 z raise to minus 2 and so on. So, now, in case of symmetry or anti symmetry. So, if you can say symmetry h of 0 equal to h of m minus 1 h of 1 equal to h of m minus 2 and so on. So, by taking this z raise to minus m minus 1 by 2 we can write it as h of m minus 1 by 2 plus k equal to 0 to m minus 3 by 2 h of k z raise to m minus 1 minus 2 k by 2 plus or minus z raise to minus m minus 1 minus 2 k by 2 when m is odd because we have the center coefficients h of m minus 1 by 2 and in case when m is even then we can write this as z raise to minus m minus 1 by 2 k equal to 0 to m by 2 minus 1 h of k z raise to m minus 1 minus 2 k by 2 plus or minus z raise to minus m minus 1 minus 2 k by 2 this plus or minus indicates that plus for symmetry and minus for anti symmetry. Now, if we substitute or replace z by z raise to minus 1 in the transfer function we get this as z raise to minus m minus 1 h z raise to minus 1 equal to plus or minus h z. So, here it shows that the roots of this h z must occur in the reciprocal pair that is if z 1 is the root of h z then 1 upon z 1 must also be a root. So, for every root which lies inside the unit circle one root corresponding to reciprocal will lie outside the unit circle and this is the condition basically because of the linear phase requirement. So, this is a symmetry of 0 locations for a linear phase FR filter. Now, think on how to obtain the frequency response characteristics of a linear phase FR filter from transfer function. Pause this video for a minute. So, the frequency response can be obtained by the transfer function by evaluating the equation on the unit circle. With the symmetric condition this h of n equal to this the frequency response can be written as h of omega equal to h r omega e raise to minus j omega m minus 1 by 2. Where h r omega is a real function of frequency written as h r omega equal to h of m minus 1 by 2 plus 2 k equal to 0 to m minus 1 by 2 h k cos of m minus 1 by 2 minus k if m hod and in case m m is even it is equal to 2 k equal to 0 to m by 2 minus 1 h k cos of m minus 1 by 2 minus k. Of course, clearly you can get this equation from our previous equations by substituting here z equal to e raise to minus j omega and simplifying it and the phase characteristics of the filter is phi omega equal to minus omega m minus 1 by 2 if it is positive otherwise it is minus omega m minus 1 by 2 plus pi if it is negative because this e raise to minus j pi is minus 1. So, that negative sign of h r omega is given by this. Same way if in case of anti-symmetry is considered this h r omega can be written as h r omega e raise to j minus omega m minus 1 by 2 plus pi by 2 where h r omega equal to now once again 2 k equal to 0 to m minus 3 by 2 h k sin m minus 1 by 2 minus k and h r omega equal to this when m is even. Of course, in this case h of m minus 1 by 2 will be 0 because of anti-symmetry condition. The phase characteristics of the filter is given by phi omega equal to pi by 2 minus omega m minus 1 by 2 when it is positive otherwise it is 3 pi by 2 minus omega m minus 1 by 2 if it is negative. Now, these frequency characteristic equation can be used for designing this symmetric and anti-symmetric FR filters. Basically the choice of symmetric or anti-symmetric filter depends on the application. For example, if h n is anti-symmetric and m is odd which implies that h r omega or h r 0 as well as h r pi both are 0 therefore, it is not suitable for either as a low pass filter or a high pass filter. Clearly we can get this from this particular equation it is equal to 0 from this therefore, it is not suitable for low pass and high pass filter. Similarly, anti-symmetric with m even also results h r 0 equal to 0. So, it is also not suitable for low pass filter. Symmetric filter yields a linear phase FI filter with nonzero response at omega equal to 0 because of that cost term and hence it is suitable for low pass filter. So, this can be considered and can be used for designing of symmetric and anti-symmetric filter. So, this is a reference. Thank you.