e) $$ x(n)=n T, \quad $$ para $$ n=0,1,2, \ldots \quad \quad f(z)=\frac{T z}{(z-1)^{2}} \quad $$ si $$ |z|1 $$

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e) $$ x(n)=n T, \quad $$ para $$ n=0,1,2, \ldots \quad \quad f(z)=\frac{T z}{(z-1)^{2}} \quad $$ si $$ |z|>1 $$

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The inverse Z-transform of $$ f(z) $$ is $$ x(n) = Tn $$. To solve this problem, we need to find the inverse Z-transform of $$ f(z) $$ and then use it to find the sequence $$ x(n) $$. Given: $$ x(n) = nT \quad \text{for} \quad n = 0, 1, 2, \ldots $$ $$ f(z) = \frac{Tz}{(z-1)^2} \quad \text{for} \quad |z| > 1 $$ We want to find the inverse Z-transform of $$ f(z) $$, which is $$ X(z) $$. Since $$ x(n) = nT $$, we can use the Z-transform pair for a first-order difference equation: $$ x(n) = nT $$ $$ X(z) = \frac{Tz}{(z-1)^2} $$ Now, we need to find the inverse Z-transform of $$ X(z) $$. The inverse Z-transform of $$ \frac{z}{(z-1)^2} $$ is $$ n $$, and since we have an additional $$ T $$ in the numerator, the inverse Z-transform of $$ \frac{Tz}{(z-1)^2} $$ is $$ Tn $$. Therefore, the inverse Z-transform of $$ f(z) $$ is: $$ x(n) = Tn $$ This is the sequence that corresponds to the given $$ f(z) $$ for $$ |z| > 1 $$.

e) \( x(n)=n T, \quad \) para \( n=0,1,2, \ldots \quad \quad f(z)=\frac{T z}{(z-1)^{2}} \quad \) si \( |z|>1 \)

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e) \( x(n)=n T, \quad \) para \( n=0,1,2, \ldots \quad \quad f(z)=\frac{T z}{(z-1)^{2}} \quad \) si \( |z|>1 \)

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