Discrete Time Convolution

Discrete-Time signal as Sum of Weighted impulses We know the property of impulse function $latex \displaystyle x(n) \delta(n) = x(0) \delta(n) &s=2 &bg=ffffff$ $latex \displaystyle x(n) \delta(n-k) = x(k) \delta(n-k) &s=2 &bg=ffffff$ Now apply summation on both side of above equation $latex \displaystyle \sum_{k=-\infty}^{\infty} x(n) \delta(n-k) = \sum_{k=-\infty}^{\infty} x(k) \delta(n-k) &s=3 &bg=ffffff$ $latex \displaystyle x(n) …

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Properties of Z transform

Linearity $latex \displaystyle x_1(n) \rightleftharpoons X_1(z) &s=2 &bg=ffffff$ ROC : $latex \displaystyle a_1 < |z| < b_1 &s=2 &bg=ffffff$ $latex \displaystyle x_2(n) \rightleftharpoons X_2(z) &s=2 &bg=ffffff$ ROC : $latex \displaystyle a_2 <|z| < b_2 &s=2 &bg=ffffff$ According to property of Linearity $latex \displaystyle a_1 x_1(n) + a_2 x_2(n) \rightleftharpoons a_1 X_1(z) + a_2 X_2(z) &s=2 …

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 …

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 …

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 …

Z transform of [u(n-1)/n^2]

We have seen the Z transform of $latex \displaystyle \frac{u(n-1)}{n} &s=3 &bg=ffffff$in previous post. $latex \displaystyle \frac{u(n-1)}{n} \rightleftharpoons ln(\frac{z}{z-1}) &s=3 &bg=ffffff$ $latex \displaystyle x(n) \rightleftharpoons X(z) &s=3 &bg=ffffff$ $latex \displaystyle nx(n) \rightleftharpoons -z\frac{dX(z)}{dz} &s=3 &bg=ffffff$ $latex \displaystyle \frac{u(n-1)}{n^2} \rightleftharpoons Y(z) &s=3 &bg=ffffff$ $latex \displaystyle n(\frac{u(n-1)}{n^2}) \rightleftharpoons -z\frac{dY(z)}{dz} &s=3 &bg=ffffff$ $latex \displaystyle -z\frac{dY(z)}{dz} = ln(\frac{z}{z-1}) &s=3 …

Z transform of [z / (z+2)^3]

$latex \displaystyle X(z) = \frac{z}{(z+2)^3} &s=3 &bg=ffffff$ $latex \displaystyle X(z) = \frac{z^{-2}}{(1+\frac{2}{z})^3} &s=3 &bg=ffffff$ Now we know the Binomial Expansion $latex \displaystyle (1-x)^{-n} = \sum_{k=0}^{\infty}\binom{n+k-1}{k}x^k &s=3 &bg=ffffff$ $latex \displaystyle (1-x)^{-n} = 1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3 +...... &s=2 &bg=ffffff$ Only when$latex \displaystyle|x|<1 &s=2 &bg=ffffff$ $latex \displaystyle X(z) = \frac{z^{-2}}{(1+\frac{2}{z})^3} &s=3 &bg=ffffff$ Compare with equation of standard Bionomial expansion $latex …

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 …

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 …