Continuous Time Signal Analysis MCQ Quiz – Objective Question with Answer for Continuous Time Signal Analysis

Answer: A

If the given signal is x(t) and F0 is the reciprocal of the time period of the signal and ck is the Fourier coefficient then the Fourier series representation of x(t) is given as \(\sum_{k=-\infty}^{\infty}c_k e^{j2πkF_0 t}\).

Answer: C

When we apply integration to the definition of Fourier series representation, we get

ckTp=\(\int_{t_0}^{t_0+T_p} x(t)e^{-j2πkF_0 t} dt\)
=>ck=\(\frac{1}{T_p} \int_{t_0}^{t_0+T_p} x(t)e^{-j2πkF_0 t} dt\)

Answer: D

For any signal x(t) to be represented as Fourier series, it should satisfy the Dirichlet conditions which are x(t) has a finite number of discontinuities in any period, x(t) has a finite number of maxima, and minima during any period and x(t) is absolutely integrable in any period.

Answer: B
Since we are synthesizing the Fourier series of the signal x(t), we call it a synthesis equation, whereas the equation giving the definition of Fourier series coefficients is known as the analysis equation.

Answer: B

In general, Fourier coefficients ck are complex valued. Moreover, it is easily shown that if the periodic signal is real, ck and c-k are complex conjugates. As a result
ck=|ck|ejθkand ck=|ck|e-jθk
Consequently, we obtain the Fourier series as

x(t)=\(c_0+2\sum_{k=1}^{\infty}|c_k|cos(2πkF_0 t+θ_k)\)

Answer: A

cos(2πkF0 t+θk) = cos2πkF0 t.cosθk-sin2πkF0 t.sinθk
θk is a constant for a given signal.
So, the other form of Fourier series representation of the signal x(t) is

\(a_0+\sum_{k=1}^∞(a_k cos2πkF_0 t – b_k sin2πkF_0 t)\).

Answer: D

The average power of a periodic signal x(t) is given as
The average power of a periodic signal x(t) is given as

\(\frac{1}{T_p} \int_{t_0}^{t_0+T_p}|x(t)|^2 dt\)

=\(\frac{1}{T_p} \int_{t_0}^{t_0+T_p}x(t).x^* (t) dt\)

=\(\frac{1}{T_p} \int_{t_0}^{t_0+T_p}x(t).\sum_{k=-\infty}^{\infty} c_k * e^{-j2πkF_0 t} dt\)

By interchanging the positions of integral and summation and by applying the integration, we get

=\(\sum_{k=-\infty}^{\infty}|c_k |^2\)

Answer: C

When we plot a graph of |ck|2 as a function of frequencies kF0, k=0,±1,±2… the following spectrum is obtained which is known as the Power density spectrum.

Answer: A

We know that Fourier series coefficients are complex-valued, so we can represent ck in the following way.
ck=|ck|ejθk
When we plot |ck| as a function of frequency, the spectrum thus obtained is known as the Magnitude voltage spectrum.

Answer: B

We know that, for an periodic signal, the Fourier series coefficient is

ck=\(\frac{1}{T_p} \int_{-T_p/2}^{T_p/2} x(t)e^{-j2πkF_0 t} dt\)

If we consider a signal x(t) as non-periodic, it is true that x(t)=0 for |t|>Tp/2. Consequently, the limits on the integral in the above equation can be replaced by -∞ to ∞. Hence,

ck=\(\frac{1}{T_p} \int_{-\infty}^∞ x(t)e^{-j2πkF_0 t} dt\)

Answer: C

Let us consider a signal x(t) whose Fourier transform X(F) is given as

X(F)=\(\int_{-∞}^∞ x(t)e^{-j2πF_0 t}dt\)

and the Fourier series coefficient is given as

ck=\(\frac{1}{T_p} \int_{-∞}^∞ x(t)e^{-j2πkF_0 t}dt\)

By comparing the above two equations, we get

ck=\(\frac{1}{T_p} X(kF_0)\)

Answer: D

Let x(t) be any finite energy signal with Fourier transform X(F). Its energy is

Ex=\(\int_{-∞}^∞|x(t)|^2 dt\)

which in turn, can be expressed in terms of X(F) as follows

Ex=\(\int_{-∞}^∞ x^* (t).x(t)\) dt

=\(\int_{-∞}^∞ x(t) dt[\int_{-∞}^∞X^* (F)e^{-j2πF_0 t} dt]\)

=\(\int_{-∞}^∞ X^* (F) dt[\int_{-∞}^∞ x(t)e^{-j2πF_0 t} dt] \)

\(=\int_{-∞}^∞ |X(F)|^2 dt = \int_{-∞}^∞|X^* (F)|^2 dt = \int_{-∞}^∞X(F).X^* (F) dt\)

Answer: B

Here, the frequency F0 of a continuous-time signal is divided into 2π/N intervals.
So, the Fourier series representation of a discrete-time signal with period N is given as

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

where ck is the Fourier series coefficient

Answer: D

We know that, the Fourier series representation of a discrete signal x(n) is given as

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

Now multiply both sides by the exponential e-j2πln/N and summing the product from n=0 to n=N-1. Thus,

\(\sum_{n=0}^{N-1} x(n)e^{-j2πln/N}=\sum_{n=0}^{N-1}\sum_{k=0}^{N-1}c_k e^{j2π(k-l)n/N}\)

If we perform summation over n first in the right hand side of above equation, we get

\(\sum_{n=0}^{N-1} e^{-j2πkn/N}\) = N, for k-l=0,±N,±2N…
= 0, otherwise

Therefore, the right hand side reduces to Nck
So, we obtain ck=\(\frac{1}{N} \sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}\)

Answer: A

We know that,

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

In the above equation, ck represents the amplitude and ej2πkn/N represents the phase associated with the frequency component of DTFS.

Answer: B

For ω0=√2π, we have f0=1/√2. Since f0 is not a rational number, the signal is not periodic. Consequently, this signal cannot be expanded in a Fourier series.

Answer: A

In this case, f0=1/6 and hence x(n) is periodic with fundamental period N=6.
Given signal is x(n)=cosπn/3=cos2πn/6

=\(\frac{1}{2} e^{j2πn/6}+\frac{1}{2} e^{-j2πn/6}\)

We know that -2π/6=2π-2π/6=10π/6=5(2π/6)

∴ x(n)=\(\frac{1}{2} e^{j2πn/6}+\frac{1}{2} e^{j2π(5)n/6}\)

Compare the above equation with

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

So, we get c1 = c2 = c3 = c4 = 0 and c1 = c5 = 1/2.

Answer: B

The average power of a periodic signal x(t) is given as

\(\frac{1}{T_p}\int_{t_0}^{t_0+T_p}|x(t)|^2 dt\)

=\(\frac{1}{T_p}\int_{t_0}^{t_0+T_p} x(t).x^* (t) dt\)

=\(\frac{1}{T_p}\int_{t_0}^{t_0+T_p}x(t).\sum_{k=-∞}^∞ c_k^* e^{-j2πkF_0 t} dt\)

By interchanging the positions of integral and summation and by applying the integration, we get

=\(\sum_{k=-∞}^∞|c_k |^2\)

Answer: D

Let us consider a discrete-time periodic signal x(n) with period N.
The average power of that signal is given as

Px=\(\frac{1}{N} \sum_{n=0}^{N-1}|x(n)|^2\)

Answer: B

We know that

Px=\(\frac{1}{N} \sum_{n=0}^{N-1}|x(n)|^2\)

=\(\frac{1}{N} \sum_{n=0}^{N-1}x(n).x^*(n)\)

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

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

=\(\sum_{k=0}^{N-1}|c_k |^2\)