A discrete-time signal x[n] = δ [n - 3] + 2δ [n - 5] has a z-transform X(z). If Y(z) = X(-z) is the z-transform of another signal y[n], then

Understanding the Z-transform Relationship: Y(z) = X(-z)

Problem Setup

We are given a discrete-time signal $x[n]$ defined as a sum of shifted unit impulse functions:

$$x[n]=\delta [n-3]+2\delta [n-5]$$

Its z-transform is denoted as $X(z)$. We are also introduced to another signal, $y[n]$, whose z-transform $Y(z)$ is related to $X(z)$ by the transformation:

$$Y(z)=X(-z)$$

The task is to determine the relationship between the signal $y[n]$ and the original signal $x[n]$.

Calculating the Z-transform X(z)

The z-transform $X(z)$ of a discrete-time signal $x[n]$ is fundamentally defined as:

$$X(z)=\sum _{n=-\infty }^{\infty }x[n]z^{-n}$$

A key property of the z-transform is the time-shifting property for the unit impulse function:

- Time-Shifting Property: $Z\{\delta [n-k]\}=z^{-k}$

Applying this property to the given signal $x[n]$:

$$X(z)=Z\{\delta [n-3]+2\delta [n-5]\}$$

Utilizing the linearity property of the z-transform, which states $Z\{a\cdot x[n]+b\cdot y[n]\}=a\cdot X(z)+b\cdot Y(z)$, we can write:

$$X(z)=Z\{\delta [n-3]\}+Z\{2\delta [n-5]\}$$

$$X(z)=z^{-3}+2z^{-5}$$

Deriving the Transformed Z-transform X(-z)

The problem provides the relationship $Y(z)=X(-z)$. To find $Y(z)$, we substitute $-z$ for $z$ in the expression for $X(z)$:

$$X(-z)=(-z{)}^{-3}+2(-z{)}^{-5}$$

Simplifying the terms:

$$X(-z)=\frac{1}{(-z{)}^3}+\frac{2}{(-z{)}^5}$$

Since $(-z{)}^k=(-1{)}^k{z}^k$:

$$X(-z)=\frac{1}{(-1{)}^3{z}^3}+\frac{2}{(-1{)}^5{z}^5}$$

$$X(-z)=\frac{1}{-z^3}+\frac{2}{-z^5}$$

$$X(-z)=-z^{-3}-2z^{-5}$$

Thus, we have $Y(z)=-z^{-3}-2z^{-5}$.

Finding the Signal y[n] using Inverse Z-transform

To find the signal $y[n]$, we perform the inverse z-transform on $Y(z)=-z^{-3}-2z^{-5}$.

We can express $Y(z)$ using the known z-transform pairs:

$$Y(z)=-1\cdot z^{-3}-2\cdot z^{-5}$$

Recalling the property $z^{-k}$ corresponds to $\delta [n-k]$:

$$Y(z)=-1\cdot Z\{\delta [n-3]\}-2\cdot Z\{\delta [n-5]\}$$

Using linearity again for the inverse transform:

$$Y(z)=Z\{-\delta [n-3]\}+Z\{-2\delta [n-5]\}$$

$$Y(z)=Z\{-\delta [n-3]-2\delta [n-5]\}$$

From this, we identify the signal $y[n]$ as:

$$y[n]=-\delta [n-3]-2\delta [n-5]$$

Establishing the Relationship between y[n] and x[n]

We now compare the derived signal $y[n]$ with the original signal $x[n]$.

Original signal:

$$x[n]=\delta [n-3]+2\delta [n-5]$$

Derived signal:

$$y[n]=-\delta [n-3]-2\delta [n-5]$$

By factoring out $-1$ from the expression for $y[n]$:

$$y[n]=-1\cdot (\delta [n-3]+2\delta [n-5])$$

Substituting the definition of $x[n]$:

$$y[n]=-x[n]$$

This relationship indicates that the signal $y[n]$ is the negative of the original signal $x[n]$.