# Causality and Stability in terms of Z- Transform

Causality $latex \Rightarrow &s=2 &bg=ffffff$   The condition for LTI system to be causal is given as, $latex h(n) = 0 &s=2 &bg=ffffff$    for  $latex n < 0 &s=2 &bg=ffffff$ $latex \Rightarrow &s=2 &bg=ffffff$   Here  $latex h(n) &s=2 &bg=ffffff$   is the unit sample response of the LTI system LTI system is causal if and only if the ROC of the … Continue reading Causality and Stability in terms of Z- Transform

# Unilateral Z Transform

The Unilateral z-transform is also called as one-sided z- transform. It is defined for $latex \displaystyle n >= 0 &s=2 &bg=ffffff$  i.e. Causal sequences. The unilateral z- transform is used to solve difference equations with initial conditions. $latex \displaystyle X(z) = \sum_{n=0}^{\infty} x(n) z^{-n} &s=3 &bg=ffffff$ Unilateral and bilateral transforms are same for causal signals. Properties of Unilateral Z … Continue reading Unilateral Z Transform

# Basics of Z Transform

$latex \displaystyle \Rightarrow &s=2 &bg=ffffff$ Z -transform is used for the analysis of stable, unstable and Marginally stable signals or systems. $latex \displaystyle x(n) \rightleftharpoons X(z) &s=2 &bg=ffffff$ z $latex \displaystyle \rightarrow &s=2 &bg=ffffff$ complex variable $latex \displaystyle z = r e^{j \omega} &s=2 &bg=ffffff$ r $latex \displaystyle \rightarrow &s=2 &bg=ffffff$ Damping factor  $latex \displaystyle \omega \rightarrow &s=2 &bg=ffffff$ Oscillating frequency … Continue reading Basics of Z Transform

# Fibonacci Sequence

Fibonacci sequence is $latex \displaystyle 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89......... &s=3 &bg=ffffff$ We can observe that a recurrence relation exist $latex \displaystyle y(n+2) = y(n)+y(n+1) &s=3 &bg=ffffff$ or $latex \displaystyle y(n) = y(n-1)+y(n-2) &s=3 &bg=ffffff$ with initial conditions $latex \displaystyle y(0) = 0; y(1)=1 &s=3 &bg=ffffff$ Now we … Continue reading Fibonacci Sequence

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# Z transform of u(n-1)/n

We know that Z transform is defined as : $latex \displaystyle X(z) = \sum_{n=-\infty}^{\infty}x(n)z^{-n} &s=3 &bg=ffffff$ $latex \displaystyle u(n) \rightleftharpoons \frac{z}{z-1} &s=3 &bg=ffffff$ By the use of time shifting property $latex \displaystyle x(n-1) \rightleftharpoons z^{-1} X(z) &s=3 &bg=ffffff$ $latex \displaystyle u(n-1) \rightleftharpoons \frac{1}{z-1} &s=3 &bg=ffffff$ Differentiate the General formula of X(z) with respect to z … Continue reading Z transform of u(n-1)/n