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# Answer to Question #195202 in Electrical Engineering for Hridoy JJ

Question #195202

A system is known whose input - output relationship is determined by the following difference equation :

y(n)-1/2y(n-1)=x(n)+1/2x(n-1)

Find the system function H ( z ) and plot the pole - zero plot

1
2021-05-19T06:14:34-0400

In the general case, the representation of a linear system using a linear difference equation with constant coefficients

"\\sum_{i=0}^N a_i y[n-i]=\\sum_{j=0}^M b_j x[n-j]"

Taking the Z-transform of the equation (using linearity and time-shifting laws) yields

"Y(z)\\sum_{i=0}^N a_i z^{-i}=X(z)\\sum_{j=0}^M b_j z^{-j}"

reordering the result gives a transfer function

"H(z)=\\frac{Y(z)}{X(z)}=\\frac{\\sum_{j=0}^M b_j z^{-j}}{\\sum_{i=0}^N a_i z^{-i}}=\\frac {b_0 +b_1 z^{-1}+...+b_M z^{-M}}{a_0 +a_1 z^{-1}+...+a_Nz^{-N}}"

where the numerator has M roots (corresponding to zeros of H) and the denominator has N roots (corresponding to poles of H)

Zeros and poles are usually complex, and in order to plot them on the complex plane (pole-zero plot) we rewrite the transfer function in terms of zeros and poles

"H(z)=\\frac {b_0 +b_1 z^{-1}+...+b_M z^{-M}}{a_0 +a_1 z^{-1}+...+a_Nz^{-N}}=\\frac{(1-q_1 z^{-1})(1-q_2 z^{-1})...(1-q_M z^{-1})}{(1-p_1 z^{-1})(1-p_2 z^{-1})...(1-p_N z^{-1})}"

where qk is the k-th zero and pk is the k-th pole.

So, if the linear difference equation

"y[n]-\\frac 1 2 y[n-1]=x[n]+\\frac 1 2 x[n-1]"

where

"M=N=1,\\space a_0=b_0=1, \\space a_1 =-\\frac 1 2, \\space b_1=\\frac 1 2"

Then the transfer function is

"H(z)=\\frac{Y(z)}{X(z)}=\\frac{b_0+b_1 z^{-1}}{a_0+a_1 z^{-1}}=\\frac {1+\\frac 1 2 z^{-1}}{1-\\frac 1 2 z^{-1}}"

the numerator has root (corresponding to zero of H)

"q_1 =-\\frac 1 2"

the denominator has root (corresponding to pole of H)

"p_1=\\frac 1 2"

The DT LTI system

The pole-zero plot

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