Z Transform MCQ [Free PDF] – Objective Question Answer for Z Transform Quiz

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.

We know that according to the periodicity and symmetry property,

If W4100 = Wx200, then what is the value of x is

100/4 = 200/x = >x = 8.

For a rational z-transform X(z) to be zero, the numerator of X(z) is zero and the solutions of the numerator are called ‘zeros’ of X(z).

For a rational z-transform X(z) to be infinity, the denominator of X(z) is zero and the solutions of the denominator are called ‘poles’ of X(z).

If X(z) has M finite zeros and N finite poles, then X(z) can be rewritten as X(z) = z -M+N.X'(z).
So, if N>M then z has positive power. So, it has |N-M| zeros at origin.

If X(z) has M finite zeros and N finite poles, then X(z) can be rewritten as X(z) = z-M+N.X'(z).

So, if N < M then z has a negative power. So, it has |N-M| poles at the origin.