Question

Find the Transfer Function H(z) for the LTI system with the following Difference Equation: y[n]+1/4y[n−1]−3/8y[n−2]=2x[n]+2x[n−2]

H(z)= Az²+Bz+C/z²+Dz+E
​
[Enter numeric values for A,B,C,… using 2 decimal places, like 0.14,−8.76 ]

Answer 1

In this case, the transfer function H(z) is
`H(z) = (2z^2 + 2)/(z^2 + (1/4)z - (3/8))`

Step-by-step explanation:

Let's denote Y(z) as the z-transform of the output signal y[n], and X(z) as the z-transform of the input signal x[n]. The z-transform of the delayed signal y[n−1] is
`z^(^-^1^)Y(z)`, and the z-transform of the delayed signal
`y[n-2]` is
`z^(^-^2^)Y(z).`

Using these notations, we can rewrite the given difference equation in the z-domain:

`Y(z) + 1/4z^(^-^1^)Y(z) - 3/8z^(^-^2^)Y(z) = 2X(z) + 2z^(^-^2^)X(z)`

Next, we can factor out Y(z) and X(z) on the left side of the equation:

To find the transfer function H(z) for the given LTI system with the difference equation:

`y[n] + 1/4y[n-1] - 3/8y[n-2] = 2x[n] + 2x[n-2]`

We can rewrite the equation in terms of z-transforms:

`Y(z) + (1/4)z^(^-^1^)Y(z) - (3/8)z^(^-^2^)Y(z) = 2X(z) + 2z^(^-^2^)X(z)`

Now, we can factor out Y(z) and X(z):

`Y(z) * (1 + (1/4)z^(^-^1^) - (3/8)z^(^-^2^)) = X(z) * (2 + 2z^(^-^2^))`

Dividing both sides by the expression in parentheses gives us:

`H(z) = Y(z)/X(z) = (2 + 2z^(^-^2^))/(1 + (1/4)z^(^-^1^) - (3/8)z^(^-^2^))`

Now, we need to simplify this expression. Multiply the numerator and denominator by
`z^2` to eliminate the negative powers of z:

`H(z) = (2z^2 + 2)/(z^2 + (1/4)z - (3/8))`

The transfer function H(z) is given by:

`H(z) = (2z^2 + 2)/(z^2 + (1/4)z - (3/8))`