LTI System Analysis MCQ Quiz – Objective Question with Answer for LTI System Analysis

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

The formula for calculating N point DFT is given as

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

From the formula given at every step of computing, we are performing N complex multiplications and N-1 complex additions. So, in a total to perform N-point DFT we perform N2 complex multiplications and N(N-1) complex additions.

If XN represents the N point DFT of the sequence xN in the matrix form, then we know that XN = WN.xN

By pre-multiplying both sides by WN-1, we get
xN=WN-1.XN

But we know that the inverse DFT of XN is defined as
xN=1/N*XN

Thus by comparing the above two equations we get
WN-1=1/N WN*

The first step is to determine the matrix W4. By exploiting the periodicity property of W4 and the symmetry property

\(W_{N}^{k+N/2}=-W_{N^k}\)

The matrix W4 may be expressed as

W4=\(\begin{bmatrix}W_4^0&W_4^0&W_4^0&W_4^1\\W_4^0&W_4^0&W_4^2&W_4^3\\W_4^0&W_4^2&W_4^0&W_4^3\\W_4^4&W_4^6&W_4^6&W_4^9\end{bmatrix}=\begin{bmatrix}W_4^0&W_4^0&W_4^0&W_4^1\\W_4^0&W_4^0&W_4^2&W_4^3\\W_4^0&W_4^2&W_4^0&W_4^3\\W_4^0&W_4^2&W_4^2&W_4^1\end{bmatrix}\)

=\(\begin{bmatrix}1&1&1&1\\1&-j&-1&j\\1&-1&1&-1\\1&j&-1&-j\end{bmatrix}\)

Then X4=W4.x4=\(\begin{bmatrix}6\\-2+2j\\-2\\-2-2j\end{bmatrix}\)

The Fourier series coefficients are given by the expression
ck=\(\frac{1}{N} \sum_{n=0}^{N-1} x(n)e^{-j2πkn/N} = \frac{1}{N}X(k)=> X(k)=Nc_k\)

Given x(n)={0,1,2,3}

We know that the 4-point DFT of the above given sequence is given by the expression
X(k)=\(\sum_{n=0}^{N-1} x(n)e^{-j2πkn/N} \)
In this case N=4
=>X(0)=6, X(1)=-2+2j, X(2)=-2, X(3)=-2-2j.