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We use y{‘}(nT)=-ay(nT)+bx(nT) to substitute for the derivative in y(nT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (nT-T)]+y(nT-T)\) and thus obtain a difference equation for the equivalent discrete-time system. With y(n) = y(nT) and x(n) = x(nT), we obtain the result as of the following?(a) \((1+\frac{aT}{2})Y(z)-(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} [x(n)+x(n-1)]\)(b) \((1+\frac{aT}{n})Y(z)-(1-\frac{aT}{n})y(n-1)=\frac{bT}{n} [x(n)+x(n-1)]\)(c) \((1+\frac{aT}{2})Y(z)+(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} (x(n)-x(n-1))\)(d) \((1+\frac{aT}{2})Y(z)+(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} (x(n)+x(n+1))\)This question was addressed to me by my school principal while I was bunking the class.Question is from IIR Filter Design by the Bilinear Transformation in chapter Discrete Time Systems Implementation of Digital Signal Processing

Answer»

Correct option is (a) \((1+\frac{aT}{2})Y(Z)-(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} [X(n)+x(n-1)]\)

For EXPLANATION: When we substitute the given EQUATION in the derivative of other we get the resultant REQUIRED equation.



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