Z Transform Properties MCQ Quiz – Objective Question with Answer for Z Transform Properties

According to the linearity property of z-transform, if X(z) and Y(z) are the z-transforms of x(n) and y(n) respectively then, the z-transform of x(n)+y(n) is X(z)+Y(z).

Let us divide the given x(n) into x1(n)=3(2n)u(n) and x2(n)= 4(3n)u(n)
and x(n)=x1(n)-x2(n)

From the definition of z-transform

X1(z)=\(\frac{3}{1-2z^{-1}}\) and X2(z)=\(\frac{4}{1-3z^{-1}}\)

So, from the linearity property of z-transform

X(z)=X1(z)-X2(z)

=> X(z)=\(\frac{3}{1-2z^{-1}}-\frac{4}{1-3z^{-1}}\).

By Euler’s identity, the given signal x(n) can be written as

x(n) = sin(jω0n)u(n)=\(\frac{1}{2j}[e^{jω_0n} u(n)-e^{-jω_0n} u(n)]\)

Thus X(z)=\(\frac{1}{2j}[\frac{1}{1-e^{jω_0} z^{-1}}-\frac{1}{1-e^{-jω_0} z^{-1}}]\)

On simplification, we obtain

=> \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\).

According to the definition of Z-transform

X(z)=\(\sum_{n=-\infty}^{\infty} x(n) z^{-n}\)
=>Z{x(n-k)}=\(X^1(z)=\sum_{n=-\infty}^{\infty} x(n-k) z^{-n}\)

Let n-k=l

=> X1(z)=\(\sum_{l=-\infty}^{\infty} x(l) z^{-l-k}=z^{-k}.\sum_{l=-\infty}^{\infty} x(l) z^{-l}= z^{-k}X(z)\)

We know that \(Z{u(n)}=\frac{1}{1-z^{-1}}\)

And by the time shifting property, we have Z{x(n-k)}=z-k.Z{x(n)}

=>Z{u(n-N)}=\(z^{-N}.\frac{1}{1-z^{-1}}\)

=>Z{u(n)-u(n-N)}=\(\frac{1-z^{-N}}{1-z^{-1}}\).

We know that from the definition of z-transform

Z{anx(n)}=\(\sum_{n=-\infty}^{\infty}a^n x(n) z^{-n}=\sum_{n=-\infty}^{\infty}x(n)(a^{-1}z)^{-n}=X(a^{-1}z)\).

Given ROC of X(z) is r1<|z|<r2
Then ROC of X(a-1z) will be given by r1<|a-1z |<r2=|a|r1<|z|<|a|r2

we know that by the linearity property of z-transform of sin(ω0n)u(n) is

X(z)=\(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\)

Now by the scaling in the z-domain property, we have z-transform of an (sin(ω0n))u(n) as

X(az)=\(\frac{az^{-1} sin\omega_0}{1-2az^{-1} cosω_0+a^2 z^{-2}}\)

From the definition of z-transform, we have

Z{x(-n)}=\(\sum_{n=-\infty}^{\infty} x(-n) z^{-n}=\sum_{n=-\infty}^{\infty} x(-n) (z^{-1})^{-(-n)}=X(z^{-1})\)

From the definition of z-transform, we have

X(z)=\(\sum_{n=-\infty}^{\infty} x(n) z^{-n}\)

On differentiating both sides, we have

\(\frac{dX(z)}{dz}=\sum_{n=-\infty}^{\infty} (-n) x(n) z^{-n-1}=-z^{-1} \sum_{n=-\infty}^{\infty}nx(n) z^{-n}=-z^{-1}Z\{nx(n)\}\)

Therefore, we get \(-z\frac{dX(z)}{dz}\) = Z{nx(n)}.

We know that Z{anu(n)}=\(\frac{1}{1-az^{-1}}\)=X(z) (say)
Now the z-transform of nanu(n)=\(-z\frac{dX(z)}{dz} = \frac{az^{-1}}{(1-az^{-1})^2}\)

A sampling rate conversion by the rational factor I/D is accomplished by cascading an interpolator with a decimator.

We emphasize that the importance of performing the interpolation first and decimation second is to preserve the desired spectral characteristics of x(n).

In the diagram given, an interpolator is in a cascade with a decimator which together performs the action of sampling rate conversion by a factor I/D.

We know that the Discrete Fourier transform of a signal x(n) is given as

X(k)=\(\sum_{n=0}^{N-1} x(n)e^{-j2πkn/N}=\sum_{n=0}^{N-1} x(n) W_N^{kn}\)

Thus we get Nth rot of unity WN= e-j2π/N