Problem 1 on Bilinear Transformation Method of IIR Filter Design – Discrete Time Signal Processing


Hi friends let us see how can we utilize a bilinear transformation method in solving a problem let us take up a problem given is HA of S as S plus 0.1 divided by S plus 0.1 the whole square plus 16 and they have said that are the resonant frequency requirement Omega R of the digital filter is Pie by 4 Pie by 2 so so let us change it to Pie by 2 rather than Pie by 4 let us take it as Phi by 2 now whenever you solve a bilinear transformation method the solution is we will replace or we know that H of s is equal to H of Z if I substitute s is equal to 2 by T 1 minus Z inverse upon 1 plus Z inverse so if I replace all SS in the given transfer function into H of s I am going to get H of Z now we need to obtain this T so how to operate this T so to obtain this T we will use this and something from this in general a transfer function of analog filter is written as s plus a upon s plus a the whole square plus Omega C square so this term actually gives you the cutoff frequency so this term is responsible for to give us a cutoff frequency so let us so the cutoff frequency in this case is 4 radians per second so I have received a cut-off as for radians per second from the given data now we have the cutoff frequency of analog filter we know what cutoff frequency does digital filter demands from there I can easily calculate my T so how to calculate T for that we have a relation of Omega that relation says capital Omega is equal to 2 by T tan off ok Omega by 2 now we know this Omega we know this Omega also so I can easily calculate my T so let us rearrange this equation I want T so I will put T over there then 2 by Omega tan of Omega by 2 so here I will use Omega C and Omega R so that will give me 2 upon 4 tan off your it is PI by 2 so I am going to get PI by 4 right because the given Omega R is PI by 2 so I am going to get PI by 4 so that T is 1 by so I got capital T as 1 by 2 seconds so we got capital T now we have to just replace our H of s and get a job Z so I can write as H of Z equals to H a of s if s is equal to 2 by T 1 minus Z inverse upon 1 plus Z inverse okay so H of Z will be equal to our H a of s was s plus so I will write first here s plus point 1 divided by s plus point 1 the whole square plus 16 and I have to substitute s as 2 by T is 1 by 2 so it will be 1 by 2 virtual for ultimately so one minus Z inverse upon one plus Z inverse okay now let us do them so replace all SS with 4 into 1 minus Z inverse upon 1 plus Z inverse plus 0.1 divided by here I will also have 4 in the capital bracket 4 into 1 minus Z inverse upon 1 plus Z inverse plus 0.1 and this whole raise to 2 plus 16 okay let us solve this now to solve this problem I will get this as a 4 in the numerator I will make it as 1 minus Z inverse plus 0.1 into 1 plus Z inverse the whole divided by 1 plus Z inverse ok divided by this will be again I’ll keep as 4 into 1 minus Z inverse plus 0.1 into 1 minus 1 plus Z inverse ok the whole squared divided by let us make it the whole divided by because I have 16 over here so I will put plus 16 into 1 plus Z inverse the whole square and I will write it as 1 plus Z inverse the whole square what I did was I have brought this numerator denominator multiplied over here there will be a whole denominator I have divided the squares and have multiplied this over here now I will cancel 1 square and 1 1 plus Z inverse and then this will go in the numerator so it will be now 4 into 1 minus Z inverse plus 0.1 into 1 plus Z inverse the whole multiplied by 1 plus Z inverse divided by Y or either you can open it up now or you can keep it as it is so this will be 4 into 1 minus Z inverse plus 0.1 1 plus Z inverse the whole square plus 16 into 1 plus Z inverse the whole square and the final answer of this equation will be 0.12 8 z is 2 Z square plus 0.006 z minus 0.12 – ok the whole divided by Z square plus 0.1 – 3 x of 0 into 6z plus 0.975 which is my H of Z if in exams you do not have enough time to solve this problem you can leave the problem at this step there’s no issues or you can go one step further and keep everything in terms of 0 is 2 minus 2 and leave the problem there’s no harm in that ok so or else you can go on solving if you have a lot of time you can go on solving and reach till this step from here to here there are many steps in between which I have skipped but the readers are encouraged to solve the steps and find this answer as a solution thank you

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