VA 24-0296 2018-2024 free printable template

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Let's take a look at the direct form one flow graph implementation of a system function here's a specific example of what I mean I have a system function H of Z second-order in this case and this is the flow graph implementation of that system function these numerical coefficients somehow get involved in the specific numbers that are showing up on our flow graph let's see if we can better understand how to get those numbers to begin with let's write the system function as a difference equation H of Z can be expanded as Y of Z divided by X of Z Y of Z is the Z transform of the output of our system and X of Z is the Z transform of the input to our system the system function is the ratio of two polynomials in Z inverse I'll write the numerator and the standard form using B for the coefficients these are arranged as B 0 plus B 1 z inverse plus B 2 Z inverse squared and so forth the denominator polynomial is similar but now using a for the coefficients now we can write this more compactly using summation notation in the numerator we start at K equals zero and go up to m capital M write this as B sub K times Z inverse to the K power similarly in the denominator to keep things general let me write this as n instead of M we start at K equals zero we're actually dealing with the sub K coefficients now and multiply that by Z inverse to the K power let me consider the case of N equals two just to keep our subsequent development a little more specific let me rewrite things by moving the pieces of the polynomial round to each side of the equation than I can apply the distributive property for Y of Z and X of Z now the inverse z-transform of each of these terms looks like Y of n in minus 1 this is y of n minus 2 so Z inverse raised to the various powers introduces a delay and from this we have the difference equation corresponding to our system function now I'd like to rewrite the difference equation in a better form for implementation as a flow graph I'd like to do is concentrate on how do we calculate the next output value Y of n this will be a function of both previous outputs and a current and present or current and previous input values let me begin by setting a 0 equal to 1 also I'm going to redefine the eight coefficients to use negative signs rather than positive signs this way I can take all of these terms and push them off to the right side of the equation, and now I have Y of n as a function of both previous outputs and the current input and previous inputs more generally I can again use summation notation to write this more compactly we see that the eight coefficients begin at K equals 1 now run out to n the B coefficient still start at 0 though, and they run out to capital M alright let me put this on the other side of the equation and let's move everything over to the z-transform domain again Y of Z is something that I can pull out front same thing for X of Z, and now I see why if Z is common we rewrite this as 1 minus the summation at this...