Signals and Systems Test 6

Concept:

If \(x\left[ n \right]\mathop \leftrightarrow \limits^z X\left( z \right)\) 

Then \(x\left( {n - {n_0}} \right)\mathop \leftrightarrow \limits^z X\left( z \right){z^{ - {n_0}}}\)

Application:

For the given X₁(z), X₂(z), and X₃(z), we can write:

X₁(z) = z¹ X₂ (z) = z² X₃(z)

or z^-2 X₁(z) = z^-1 X₂ (z) = X₃(z)

Taking the inverse z-transform, we get:

x₁(n – 2) = x₂ (n - 1) = x₃(n)

In case of linear phase FIR filter, to have a one zero, z = 0 as z₀ other zeros are.

\({z_0},\frac{1}{{{z_0}}},\left[ {z_0^*} \right],{\left[ {\frac{1}{{{z_0}}}} \right]^*}\)

Z₀ = 3 + 4j

\(\left[ {z_0^*} \right] = 3 - 4j\)

\(\frac{1}{{{z_0}}} = \frac{1}{{3 + 4j}} = \frac{{3 - j4}}{{25}} = \frac{3}{{25}} - \frac{{j4}}{{25}}\)

\({\left[ {\frac{1}{{{z_0}}}} \right]^*} = \frac{3}{{2j}} + j\frac{4}{{25}}\)

For H(z) to be stable the poles of H(z) must be inside the unit circle.

For the inverse of H(z) to be stable the poles inverse of H(z) must be inside the unit circle.

The poles of inverse of H(z) are the zeros of H(z)

Concept:

The even part of a signal x(n) is defined as:

\({x_e}\left( n \right) = \frac{1}{2}\left[ {x\left( n \right) + x\left( { - n} \right)} \right]\)     ---(1)

Property:

If \(x\left( n \right)\mathop \leftrightarrow \limits^x X\left( z \right);\left| z \right| > a\)

Then \(x\left( { - n} \right)\mathop \leftrightarrow \limits^x X\left( {\frac{1}{z}} \right);\left| {\frac{1}{z}} \right| > a\)

Application:

From Equation (1), the z-transform of the even part of the signal will be:

\({X_e}\left( z \right) = \frac{1}{2}\left[ {X\left( z \right) + X\left( {\frac{1}{z}} \right)} \right]\) 

Given \(X\left( z \right) = \frac{z}{{2 - 0.4}}\), the above expression becomes:

\({X_e}\left( z \right) = \underbrace {\frac{1}{2}\left( {\frac{z}{{z - 0.4}}} \right)}_I + \underbrace {\frac{1}{2}\left( {\frac{{\frac{1}{z}}}{{\frac{1}{z} - 0.4}}} \right)}_{II}\) 

ROC for the I term will be |z| > 0.4

ROC for the II term will be:

\(\left| z \right| < \frac{1}{{0.4}},\;i.e.\;\left| z \right| < 2.5\) 

Now the ROC of X_e(z) will be the intersection of two, i.e.

0.4 < |z| < 2.5

x(n) = (0.5)ⁿ u(n)

y(n) = δ(n) - 2δ(n – 1)

Taking the z-transform of input and output, we get:

\(X\left( z \right) = \frac{z}{{z - 0.5}}\) 

Y(z) = 1 – 2z^-1

\(Y\left( z \right) = \frac{{z - 2}}{z}\) 

Transfer function of the filter will be:

\(H\left( z \right) = \frac{{Y\left( z \right)}}{{X\left( z \right)}}\) 

\(= \left( {\frac{{z - 2}}{z}} \right)\left( {\frac{{z - 0.5}}{z}} \right)\) 

\(= \frac{{{z^2} - 2.5\;z + 1}}{{{z^2}}}\) 

= 1 – 2.5 z^-1 + z^-2

Taking the inverse z-transform, we get:

h(n) = δ[n] – 2.5 δ[n – 1] + δ[n – 2]

∴ h[1] = - 2.5