Signal and System Frequency Analysis MCQ Quiz – Objective Question with Answer for Signal and System Frequency Analysis

The Fourier transform of this signal is

X(ω) = \(\sum_{n = 0}^{L-1} Ae^{-jωn}\)

= A.\(\frac{1-e^{-jωL}}{1-e^{-jω}}\)

= \(Ae^{-j(ω/2)(L-1)}\frac{sin⁡(\frac{ωL}{2})}{sin⁡(\frac{ω}{2})}\)

Let us consider the signal to be x(n)

Z{x(n)} = \(\sum_{n = -∞}^∞ x(n)z^{-n} and X(ω) = \sum_{n = -∞}^∞ x(n)e^{-jωn}\)

Now, represent the ‘z’ in the polar form

= > z = r.ejω

= >Z{x(n)} = \(\sum_{n = -∞}^∞ x(n)r^{-n} e^{-jωn}\)

Now Z{x(n)} = X(ω) only when r = 1 = >|z| = 1.

The given x(n) does not have Z-transform. But the sequence has finite energy. So, the given sequence x(n) has a Fourier transform.

Let us consider a sequence x(n) having a z-transform X(z). We assume that x(n) is a stable sequence so that X(z) converges on to the unit circle. The complex cepstrum of the signal x(n) is defined as the sequence cx(n), which is the inverse z-transform of Cx(z), where Cx(z) = ln X(z)
= > cx(z) = X-1(ln X(z))

We know that,

cx(n) = \(\frac{1}{2π} \int_{-π}^π ln⁡(X(ω))e^{jωn} dω\)

If we express X(ω) in terms of its magnitude and phase, say

X(ω) = |X(ω)|ejθ(ω)

Then ln X(ω) = ln |X(ω)|+jθ(ω)

= > cx(n) = \(\frac{1}{2π} \int_{-π}^π[ln|X(ω)|+jθ(ω)]e^{jωn} dω\) = > cx(n) = cm(n)+jcθ(n)(say)

= > cθ(n) = \(\frac{1}{2π} \int_{-π}^πθ(ω) e^{jωn} dω\)

Given x(n) = u(n)

We know that the z-transform of the given signal is

X(z) = \(\frac{1}{1-z^{-1}}\) ROC:|z|>1

X(z) has a pole p = 1 on the unit circle but converges for |z|>1.

If we evaluate X(z) on the unit circle except at z = 1, we obtain

X(ω) = \(\frac{e^{jω/2}}{2jsin(ω/2)} = \frac{1}{2sin⁡(ω/2)} e^{j(ω-π)/2}\)

We know that, for a low-frequency signal, the power signal has its power density spectrum concentrated at about zero frequency.

The anti-aliasing filter is an analog filter that has a twofold purpose. First, it ensures that the bandwidth of the signal to be sampled is limited to the desired frequency range. Using an anti-aliasing filter is to limit the additive noise spectrum and other interference, which often corrupts the desired signal. Usually, additive noise is wideband and exceeds the bandwidth of the desired signal.

Analog signal processing operations cannot be done very precisely either, since electronic components in analog systems have tolerances and introduce noise during their operation. In general, a digital system designer has better control of tolerances in a digital signal processing system than an analog system designer who is designing an equivalent analog system.

In addition to the relatively broad frequency domain classification of signals, it is often desirable to express quantitatively the range of frequencies over which the power or energy density spectrum is concentrated. This quantitative measure is called the ‘bandwidth’ of a signal.