Solving Z Transform of Difference Equation: x(n+2)-0.3679x(n+1)+0.3679x(n)

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In summary, the problem is to determine the output x(n) for a difference equation with a given input function u(n). The initial conditions for u(n) are specified, but there are no initial conditions given for x(n). To solve the problem, the input function needs to be analyzed and an expression for it needs to be found in order to determine the output.
  • #1
angel23
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Homework Statement


consider the difference equation
x(n+2)-0.3679x(n+1)+0.3679x(n)=0.3679u(n+1)+u(n)
where u(n) is the input and given by
u(n)=0 n<0
u(0)=1
u(1)=0.2142
u(2)=-0.2142
u(n)=0 n=3,4,5...
determine the output x(n)





The Attempt at a Solution


let A=0.3679 and B=0.3642
i took z transform for both sides so it became.
z^2 x(z)-z^2x(0)-zx(1)-A[zx(z)-zx(0)]+Ax(z)=A[zu(z)-zu(0)]+Bu(z)
x(z)[z^2-Az+A]+x(0)[Az-z^2]-zx(1)=A[zu(z)-zu(0)]+Bu(z)
x(z)[z^2-Az+A]=A[zu(z)-zu(0)]+Bu(z)-x(0)[Az-z^2]+zx(1)

then what can i do?
also how can i make use of u(1)=0.2142
u(2)=-0.2142
i am too confused and need help.
 
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  • #2
To solve the problem, you need to specify some initial conditions. As it stands, it's unsolvable.
 
  • #3
this is the full question given to me to solve and i am sure it is complete.
 
  • #4
late reply ,, have you known the correct solution ?

the key that the input is not a step function , u have to use the u(0) , u(1) ... etc to draw the input and find expression to it .
 

1. What is the purpose of solving the Z transform of a difference equation?

The Z transform of a difference equation is used to convert a discrete-time signal into a frequency domain representation. This allows for analysis and manipulation of the signal using techniques from complex analysis and system theory.

2. How do you solve the Z transform of a difference equation?

To solve the Z transform of a difference equation, you need to first find the transfer function by taking the Z transform of each term in the equation. Then, you can use techniques such as partial fraction decomposition or the inverse Z transform to find the solution in the time domain.

3. What does the term "x(n+2)" represent in this equation?

The term "x(n+2)" represents the value of the signal x at time n+2. This is commonly used in difference equations to represent a delayed version of the signal.

4. How does the coefficient 0.3679 affect the solution of the equation?

The coefficient 0.3679 is a damping factor that affects the behavior of the signal. It determines the rate at which the signal decays or grows over time, and can have a significant impact on the shape and characteristics of the signal in the time and frequency domains.

5. Is solving the Z transform of a difference equation necessary for all signals?

No, solving the Z transform is not necessary for all signals. It is primarily used for analyzing and manipulating discrete-time signals, and may not be applicable for continuous-time signals. Additionally, some signals may have a simple or known Z transform, making it unnecessary to solve for it.

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