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Right choice is (b) –\(\sum_{n=-∞}^∞ X(n)sin⁡(ωn)\)

The best EXPLANATION: If the signal x(n) is real, then xI(n)=0

We KNOW that,

XI(ω)=\(\sum_{n=-∞}^∞ x_R (n)sinωn-x_I (n)cosωn\)

Now substitute xI(n)=0 in the above equation=>xR(n)=x(n)

=> XI(ω)=-\(\sum_{n=-∞}^∞ x(n)sin⁡(ωn)\).

Correct CHOICE is (a) Radio BROADCAST

To elaborate: Radio broadcast SIGNAL is an ELECTROMAGNETIC signal which has a frequency range of 30kHz-3MHz.

The correct option is (C) 0-100

The EXPLANATION: Electroencephalogram(EEG) signal has a FREQUENCY range of 0-100 HZ.

Right answer is (a) True

Easy explanation: 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.

Right OPTION is (d) (F1-F2)⋘\(\frac{F_1+F_2}{2}\)(factor of 10 or more)

The BEST I can explain: If the DIFFERENCE in the cutoff frequencies is much LESS than the mean frequency, the such a band PASS signal is known as narrow band signal.

Right answer is (a) Low frequency SIGNAL

Explanation: We know that, for a low frequency signal, the POWER signal has its power DENSITY spectrum CONCENTRATED about zero frequency.

Right choice is (d) \(\frac{1}{2π} \int_{-π}^π \theta(ω) E^{jωn} dω\)

The EXPLANATION is: 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)|e^jθ(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)=\(\frac{1}{2π} \int_{-π}^πθ(ω) e^{jωn} dω\)

Correct CHOICE is (d) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω-π)/2}\)

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

The correct choice is (B) False

For EXPLANATION I would say: The given x(n) do not have Z-transform. But the sequence have finite energy. So, the given sequence x(n) has a FOURIER transform.

Correct answer is (a) |z|=1

For EXPLANATION: 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.e^jω

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

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

Correct answer is (d) \(\frac{1}{1-2acosω+a^2}\)

Explanation: Since |a|<1, the sequence x(n) is ABSOLUTELY SUMMABLE, as can be verified by applying the geometric summation FORMULA.

\(\sum_{n=-∞}^∞|x(n)| = \sum_{n=-∞}^∞ |a|^n = \frac{1}{1-|a|} \LT ∞\)

Hence the Fourier transform of x(n) exists and is obtained as

X(ω) = \(\sum_{n=-∞}^∞ a^n e^{-jωn}=\sum_{n=-∞}^∞ (ae^{-jω})^n\)

Since |ae^-jω|=|a|<1, use of the geometric summation formula again yields

X(ω)=\(\frac{1}{1-ae^{-jω}}\)

The energy density spectrum is given by

Sxx(ω)=|X(ω)|^2= X(ω).X*(ω)=\(\frac{1}{(1-ae^{-jω})(1-ae^{jω})} = \frac{1}{1-2acosω+a^2}\).

Correct choice is (a) True

For explanation: We know that, if a signal x(N) is real, then

X*(ω)=X(-ω)

If the signal is EVEN SYMMETRIC, then the MAGNITUDE on both the sides should be equal.

So, |X*(ω)|=|X(-ω)| => |X(-ω)|=|X(ω)|.

Right choice is (b) X*(ω)=X(-ω)

The explanation: We know that,

X(ω)=\(\sum_{N=-∞}^∞ x(n)e^{-jωn}\)

=> X*(ω)=\([\sum_{n=-∞}^∞ x(n)e^{-jωn}]^*\)

Given the signal x(n) is REAL. THEREFORE,

X*(ω)=\(\sum_{n=-∞}^∞ x(n)e^{jωn}\)

=> X*(ω)=X(-ω).

The correct choice is (b) \(\FRAC{1}{2π} \int_{-π}^π |X(ω)|^2 dω\)

For explanation I WOULD SAY: We know that, EX=\(\sum_{n=-∞}^∞ |x(n)|^2\)

=\(\sum_{n=-∞}^∞ x(n).x^*(n)\)

=\(\sum_{n=-∞}^∞ x(n)\frac{1}{2π} \int_{-π}^π X^*(ω) e^{-jωn} dω\)

=\(\frac{1}{2π} \int_{-π}^π|X(ω)|^2 dω\)

The CORRECT choice is (a) True

Best explanation: We note that there is a SIGNIFICANT oscillatory overshoot at ω=ωc, independent of the VALUE of N. As N increases, the oscillations become more rapid, but the size of the RIPPLE remains the same. One can show that as N→∞, the oscillations converge to the point of the discontinuity at ω=ωc. The oscillatory behavior of the approximation of XN(ω) to the function X(ω) at a point of discontinuity ofX(ω) is known as Gibbs phenomenon.

Correct choice is (a) \(\frac{ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}\)

The best I can explain: We know that, X(n)=\(\frac{1}{2\pi} \int_{-\pi}^{\pi}X(\omega)E^{j\omega n} dω\)

=\(\frac{1}{2\pi} \int_{-ω_c}^{ω_c}1.e^{j\omega n} dω=\frac{sin ω_c.n}{ω_c.n}\)

=\(\frac{ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}\)

Correct option is (C) ωc/π

For explanation: We know that, x(n)=\(\frac{1}{2\pi} \int_{-\pi}^{\pi}X(\omega)E^{j\omega n} dω\)

=\(\frac{1}{2\pi} \int_{-ω_c}^{ω_c}1.e^{j\omega n} dω\)

At n=0,

x(n)=x(0)=\(\int_{-ω_c}^{ω_c}1 dω=\frac{1}{2\pi}(2 ω_c)=\frac{ω_c}{\pi_ω}\)

THEREFORE, the value of the signal x(n) at n=0 is ωc/π.

Right answer is (c) \(\frac{1}{2π} \int_0^{2π} X(ω) e^jωn dω\)

For explanation I would say: We KNOW that the Fourier transform of the discrete time SIGNAL x(n) is

X(ω)=\(\sum_{n=-∞}^∞ x(n)e^{-jωn}\)

By calculating the inverse Fourier transform of the above equation, we get

x(n)=\(\frac{1}{2π} \int_0^{2π} X(ω) e^{jωn} dω\)

The above equation is KNOWN as SYNTHESIS equation or inverse transform equation.

The correct choice is (d) 2π

The best I can explain: Let X(ω) be the FOURIER transform of a discrete time signal x(N) which is GIVEN as

X(ω)=\(\sum_{n=-∞}^∞x(n)e^{-jωn}\)

Now X(ω+2πk)=\(\sum_{n=-∞}^∞ x(n)e^{-j(ω+2πk)n}\)

=\(\sum_{n=-∞}^∞ x(n)e^{-jωn}e^{-j2πkn}\)

=\(\sum_{n=-∞}^∞ x(n)e^{-jωn}= X(ω)\)

So, the Fourier transform of a discrete time FINITE energy signal is periodic with period 2π.

Correct option is (a) \(\sum_{n=-∞}^∞x(n)e^{-jωn}\)

Easy explanation: If we consider a SIGNAL x(n) which is discrete in NATURE and has finite energy, then the Fourier TRANSFORM of that signal is given as

X(ω)=\(\sum_{n=-∞}^∞x(n)e^{-jωn}\)

The correct OPTION is (d) \(\FRAC{1}{N} \sum_{n=0}^{N-1}|x(n)|^2 \)

Explanation: Let us CONSIDER a discrete time periodic signal x(n) with PERIOD N.

The average power of that signal is given as

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

Correct choice is (B) \(\sum_{k=-\infty}^{\infty}|c_k|\)

EXPLANATION: 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

=\(\sum_{k=-∞}^∞|c_k |^2\)

The correct option is (a) c1=c2=c3=c4=0,c1=c5=1/2

Easiest EXPLANATION: 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)

Therefore, 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.

The CORRECT OPTION is (b) False

For explanation I would say: 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.

Right option is (a) E^j2πkn/N

For explanation: We know that,

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

In the above equation, ck REPRESENTS the amplitude and e^j2πkn/N represents the phase associated with the FREQUENCY component of DTFS.

Right answer is (d) All of the mentioned

Easiest explanation: 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

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

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

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

Correct answer is (d) \(\frac{1}{N} \sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}\)

To EXPLAIN: 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=\(\frac{1}{N} \sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}\)

Correct choice is (c) ck= 1/TP(X(kF0))

The best I can explain: 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=\(\frac{1}{T_p} X(kF_0)\)

The CORRECT choice is (c) Power density spectrum

The BEST explanation: 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 Power density spectrum.

The correct CHOICE is (a) Magnitude voltage spectrum

The explanation: We know that, Fourier SERIES coefficients are complex valued, so we can REPRESENT ck in the following way.

ck=|ck|e^jθk

When we plot |ck| as a function of frequency, the spectrum thus OBTAINED is KNOWN as Magnitude voltage spectrum.

Correct option is (b) \(\frac{1}{T_p} \int_{-\INFTY}^∞ x(t)e^{-j2πkF_0 t} dt\)

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

Right answer is (d) \(\sum_{k=-\infty}^{\infty}|c_k|^2\)

The EXPLANATION: 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

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

Correct ANSWER is (a) True

The BEST I can explain: 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

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

Correct option is (b) False

Explanation: SINCE we are synthesizing the FOURIER series of the SIGNAL x(t), we call it as synthesis equation, where as the equation giving the definition of Fourier series coefficients is KNOWN as analysis equation.

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

To elaborate: 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|e^jθkand ck=|ck|e^-jθk

Consequently, we obtain the Fourier series as x(t)=\(c_0+2\sum_{k=1}^{\infty}|c_k|cos(2πkF_0 t+θ_k)\)

Right OPTION is (d) all of the mentioned

Easiest explanation: 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 finite number of maxima and minima during any period and x(t) is absolutely INTEGRABLE in any period.

Correct choice is (C) \(\frac{1}{T_p} \int_{t_0}^{t_0+T_p} x(t)e^{-j2πkF_0 t} dt\)

For explanation I would say: 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=\(\frac{1}{T_p} \int_{t_0}^{t_0+T_p} x(t)e^{-j2πkF_0 t} dt\)

The CORRECT answer is (a) \(\sum_{k=-\infty}^{\infty}c_k e^{j2πkF_0 t}\)

Easy explanation: 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 \(\sum_{k=-\infty}^{\infty}c_k e^{j2πkF_0 t}\).