Math Problem Statement
f[n] = \left{\begin{matrix}
1, & 0 \leq n \leq N \
0, & n \geq N
\end{matrix}\right. расчитать дискретное z - образование сигнала
Solution
To calculate the discrete -transform of the given signal , we first need to carefully define the signal.
The signal is a piecewise function:
\begin{cases} 1, & 0 \leq n \leq N \\ 0, & n > N \end{cases}$$ This means that $$f[n] = 1$$ for $$n$$ from 0 to $$N$$, and $$f[n] = 0$$ for $$n > N$$. ### Discrete $$Z$$-Transform Definition The discrete-time $$Z$$-transform of a sequence $$f[n]$$ is defined as: $$F(z) = \sum_{n=0}^{\infty} f[n] z^{-n}$$ ### Apply the Signal to the $$Z$$-Transform Formula Since $$f[n] = 0$$ for $$n > N$$, the sum simplifies to a finite sum over the range $$0 \leq n \leq N$$: $$F(z) = \sum_{n=0}^{N} f[n] z^{-n}$$ Substituting $$f[n] = 1$$ for $$0 \leq n \leq N$$, we get: $$F(z) = \sum_{n=0}^{N} z^{-n}$$ This is a geometric series. The sum of a geometric series with the first term $$a = 1$$ and the common ratio $$r = z^{-1}$$ is given by the formula: $$S_N = \frac{1 - r^{N+1}}{1 - r}$$ Substitute $$r = z^{-1}$$ into the formula: $$F(z) = \frac{1 - z^{-(N+1)}}{1 - z^{-1}} = \frac{1 - z^{-(N+1)}}{z^{-1}(z - 1)} = \frac{z^{N+1} - 1}{z^{N}(z - 1)}$$ Thus, the $$Z$$-transform of the signal $$f[n]$$ is: $$F(z) = \frac{z^{N+1} - 1}{z^N(z - 1)}$$ ### Summary The discrete $$Z$$-transform of the signal $$f[n]$$, where $$f[n] = 1$$ for $$0 \leq n \leq N$$ and $$f[n] = 0$$ for $$n > N$$, is: $$F(z) = \frac{z^{N+1} - 1}{z^N(z - 1)}$$ --- Would you like more details on this process or further clarification? Here are some related questions for you: 1. What are the properties of the $$Z$$-transform? 2. How do you compute the inverse $$Z$$-transform? 3. What is the region of convergence (ROC) for this $$Z$$-transform? 4. How does the length of the signal (value of $$N$$) affect the $$Z$$-transform? 5. How can we extend this to more complex signals with different forms? **Tip**: When calculating the $$Z$$-transform of a signal, look for any simplifications based on the range of the signal. If the signal is finite (nonzero only for a limited range of $$n$$), the sum can often be simplified as a finite series.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Discrete Mathematics
Z-Transform
Geometric Series
Formulas
F(z) = \sum_{n=0}^{\infty} f[n] z^{-n}
S_N = \frac{1 - r^{N+1}}{1 - r}
F(z) = \frac{z^{N+1} - 1}{z^N(z - 1)}
Theorems
Geometric Series Summation Formula
Suitable Grade Level
Undergraduate Level
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