Signals & System Lecture 17

Main note

To find the Z-transform of the given signal:

$$y(n)=3^{n}u(n)\cdot (-10{)}^{n}u(-n-1)$$

we need to analyze the components separately and understand their contributions in the Z-domain.

Step 1: Analyzing the Signal Components

1. First Term: (3^n u(n))

   This is a right-sided sequence where (u(n)) is the unit step function that is 1 for (n \geq 0) and 0 otherwise. Thus, (3^n u(n)) exists only for (n \geq 0).

   The Z-transform of (3^n u(n)) is given by:

   $$Z\{3^{n}u(n)\}=\frac{1}{1-3z^{-1}},|z|>3$$

2. Second Term: ((-10)^n u(-n-1))

   This is a left-sided sequence where (u(-n-1)) is a step function that is 1 for (n \leq -1) and 0 otherwise. Thus, ((-10)^n u(-n-1)) exists only for (n \leq -1).

   The Z-transform of ((-10)^n u(-n-1)) is:

   $$Z\{(-10{)}^{n}u(-n-1)\}=\frac{z}{z+10},|z|<10$$

Step 2: Finding the Z-Transform of (y(n))

Since $(y(n)=3^{n}u(n)\cdot (-10{)}^{n}u(-n-1)$, we notice that:

• $(3^{n}u(n)isnon-zeroonlyforn\ge 0$,
• $(-10{)}^{n}u(-n-1)$ is non-zero only for $n\le -1$.

Thus, their product (y(n)) is zero for all (n) (since they do not overlap over any value of (n)).

Conclusion

Since (y(n) = 0) for all (n), the Z-transform of (y(n)) is:

$$Y(z)=0$$

Initial Value and Final Value Theorem for Z-Transform

1. Initial Value Theorem:
   The initial value theorem provides the value of a discrete-time signal at $n=0$ directly from its Z-transform.

   $x(0)={\lim }_{z\to \infty }X(z)$

   where $X(z)$ is the Z-transform of the signal $x(n)$.

2. Final Value Theorem:
   The final value theorem gives the steady-state (long-term) value of a discrete-time signal, assuming the limit exists.

   ${\lim }_{n\to \infty }x(n)={\lim }_{z\to 1}(z-1)X(z)$

   The final value theorem is valid only if all poles of $(z-1)X(z)$ lie inside the unit circle, except for a simple pole at $z=1$.

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Solving $X(z)=\frac{1}{1-z^2}$

Given: $X(z)=\frac{1}{1-z^2}$

Step 1: Perform Partial Fraction Expansion

Rewrite $X(z)$ as: $X(z)=\frac{1}{(1-z)(1+z)}$

Using partial fraction decomposition: $X(z)=\frac{A}{1-z}+\frac{B}{1+z}$

Multiplying through by $(1-z)(1+z)$: $1=A(1+z)+B(1-z)$

Expanding and collecting terms: $1=A+Az+B-Bz$ $1=(A+B)+(A-B)z$

Equating coefficients:

1. For the constant term: $A+B=1$
2. For the $z$ term: $A-B=0$

Solving these equations:

• From $A-B=0$, we get $A=B$
• Substitute $A=B$ into $A+B=1$: $2A=1\Rightarrow A=\frac{1}{2}$

Thus, $A=\frac{1}{2}$ and $B=\frac{1}{2}$.

Now, we can rewrite $X(z)$ as: $X(z)=\frac{1}{2}\cdot \frac{1}{1-z}+\frac{1}{2}\cdot \frac{1}{1+z}$

Step 2: Find the Inverse Z-Transform

Using standard Z-transform pairs:

• The inverse Z-transform of $\frac{1}{1-z}$ is $u(n)$
• The inverse Z-transform of $\frac{1}{1+z}$ is $(-1{)}^{n}u(n)$

Therefore: $x(n)=\frac{1}{2}u(n)+\frac{1}{2}(-1{)}^{n}u(n)$

So, the time-domain sequence $x(n)$ is: $x(n)=\frac{1}{2}+\frac{1}{2}(-1{)}^n$

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Applying Initial and Final Value Theorems

1. Initial Value:
   Using the initial value theorem: $x(0)={\lim }_{z\to \infty }X(z)$ Substituting $X(z)=\frac{1}{1-z^2}$: $x(0)={\lim }_{z\to \infty }\frac{1}{1-z^2}=0$

2. Final Value:
   Using the final value theorem: ${\lim }_{n\to \infty }x(n)={\lim }_{z\to 1}(z-1)X(z)$ Substituting $X(z)=\frac{1}{1-z^2}$: ${\lim }_{n\to \infty }x(n)={\lim }_{z\to 1}(z-1)\cdot \frac{1}{1-z^2}$ Simplifying $\frac{1}{1-z^2}$ as $\frac{1}{(1-z)(1+z)}$: ${\lim }_{n\to \infty }x(n)={\lim }_{z\to 1}\frac{z-1}{(1-z)(1+z)}={\lim }_{z\to 1}\frac{1}{1+z}=\frac{1}{2}$

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Summary of Results

• Initial Value: $x(0)=0$
• Final Value: ${\lim }_{n\to \infty }x(n)=\frac{1}{2}$
• Time-Domain Expression: $x(n)=\frac{1}{2}+\frac{1}{2}(-1{)}^n$

References

Information

• date: 2024.11.05
• time: 10:21