To solve for the Z-transform of $k \sin a n$, where $k \geq 0$, we follow these steps:

General Formula for Z-transform:

The Z-transform of a discrete-time signal $x[n]$ is defined as: [ Z[x[n]] = \sum_{n=0}^{\infty} x[n] z^{-n}. ]

Given Signal:

Here, $x[n] = k \sin(a n)$. Let's compute its Z-transform step by step.

1. Write the signal explicitly: $Z[k \sin(a n)] = \sum_{n=0}^\infty k \sin(a n) z^{-n}.$

   Since $k$ is a constant, it can be factored out: $Z[k \sin(a n)] = k \sum_{n=0}^\infty \sin(a n) z^{-n}.$

2. Use the standard formula for $\sin(a n)$: Recall that $\sin(a n) = \frac{e^{j a n} - e^{-j a n}}{2j}$. Substituting this: $\sum_{n=0}^\infty \sin(a n) z^{-n} = \frac{1}{2j} \sum_{n=0}^\infty \left( e^{j a n} z^{-n} - e^{-j a n} z^{-n} \right).$

3. Separate the two terms: Let’s split the summation into two parts: $\sum_{n=0}^\infty \sin(a n) z^{-n} = \frac{1}{2j} \left( \sum_{n=0}^\infty (z e^{j a})^{-n} - \sum_{n=0}^\infty (z e^{-j a})^{-n} \right).$

4. Apply the geometric series formula: For a geometric series $\sum_{n=0}^\infty r^n = \frac{1}{1-r}$, valid when $|r| < 1$:

   • For the first term: $r = z^{-1} e^{-j a}$,
   • For the second term: $r = z^{-1} e^{j a}$.

   Using this, the summations become: $\sum_{n=0}^\infty \sin(a n) z^{-n} = \frac{1}{2j} \left( \frac{1}{1 - z^{-1} e^{j a}} - \frac{1}{1 - z^{-1} e^{-j a}} \right).$

5. Simplify the expressions: Combine the fractions: $\sum_{n=0}^\infty \sin(a n) z^{-n} = \frac{1}{2j} \cdot \frac{(1 - z^{-1} e^{-j a}) - (1 - z^{-1} e^{j a})}{(1 - z^{-1} e^{j a})(1 - z^{-1} e^{-j a})}.$

   Simplify the numerator: $(1 - z^{-1} e^{-j a}) - (1 - z^{-1} e^{j a}) = z^{-1} (e^{j a} - e^{-j a}).$

   Using $e^{j a} - e^{-j a} = 2j \sin(a)$, the numerator becomes: $z^{-1} \cdot 2j \sin(a).$

   The denominator simplifies to: $(1 - z^{-1} e^{j a})(1 - z^{-1} e^{-j a}) = 1 - 2z^{-1} \cos(a) + z^{-2}.$

6. Final Expression: Substitute these back into the fraction: $Z[\sin(a n)] = \frac{\sin(a) z^{-1}}{1 - 2z^{-1} \cos(a) + z^{-2}}.$

   Therefore: $Z[k \sin(a n)] = k \cdot \frac{\sin(a) z^{-1}}{1 - 2z^{-1} \cos(a) + z^{-2}}.$

This is the Z-transform of $k \sin(a n)$, valid for $k \geq 0$.

Would you like a deeper explanation of any specific step or further examples?

Related Questions:

1. What is the Z-transform of $\cos(a n)$?
2. How does the region of convergence (ROC) affect this Z-transform?
3. Can we derive the inverse Z-transform from this result?
4. How does $k$ being a constant influence the transform?
5. What happens if $k$ is not constant but varies with $n$?

Tip:

Always ensure the region of convergence (ROC) is determined for practical applications of the Z