21. What is the Fourier transform X(ω) of a finite energy discrete time signal x(n)?
A. \(\sum_{n = -∞}^∞x(n)e^{-jωn}\)
B. \(\sum_{n = 0}^∞x(n)e^{-jωn}\)
C. \(\sum_{n = 0}^{N-1}x(n)e^{-jωn}\)
D. None of the mentioned
Show Explanation Answer: A
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) = \(\sum_{n = -∞}^∞x(n)e^{-jωn}\)
22. What is the period of the Fourier transform X(ω) of the signal x(n)?
A. π
Show Explanation Answer: D
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 a period 2π.
23. What is the synthesis equation of the discrete-time signal x(n), whose Fourier transform is X(ω)?
A. \(2π\int_0^2π X(ω) e^jωn dω\)
B. \(\frac{1}{π} \int_0^{2π} X(ω) e^jωn dω\)
C. \(\frac{1}{2π} \int_0^{2π} X(ω) e^jωn dω\)
D. None of the mentioned
Show Explanation Answer: C
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 the synthesis equation or inverse transform equation.
24. What is the value of discrete-time signal x(n) at n = 0 whose Fourier transform is represented as below?
A. ωc.π
Show Explanation Answer: C
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.
25. What is the value of discrete time signal x(n) at n≠0 whose Fourier transform is represented as below?
A. \(\frac{ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}\)
B. \(\frac{-ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}\)
C. \(ω_c.\pi \frac{sin ω_c.n}{ω_c.n}\)
D. None of the mentioned
Show Explanation Answer: A
We know that
x(n) = \(\frac{1}{2\pi} \int_{-\pi}^{\pi}X(\omegA.e^{j\omega n} dω\)
= \(\frac{1}{2\pi} \int_{-ω_c}^{ω_c}1.e^{j\omega n} dω = \frac{sin ω_c.n}{ω_c.n}\)
= \(\frac{ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}\)
26. The oscillatory behavior of the approximation of XN(ω) to the function X(ω) at a point of discontinuity of X(ω) is known as the Gibbs phenomenon.
A. True
Show Explanation Answer: A
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 of X(ω) is known as the Gibbs phenomenon.
27. What is the energy of a discrete time signal in terms of X(ω)?
A. \(2π\int_{-π}^π |X(ω)|^2 dω\)
B. \(\frac{1}{2π} \int_{-π}^π |X(ω)|^2 dω\)
C. \(\frac{1}{2π} \int_0^π |X(ω)|^2 dω\)
D. None of the mentioned
Show Explanation Answer: B
We know that,
Ex = \(\sum_{n = -∞}^∞ |x(n)|^2\)
= \(\sum_{n = -∞}^∞ x(n).x^*(n)\)
= \(\sum_{n = -∞}^∞ x(n)\frac{1}{2π} \int_{-π}^π X^*(ω) e^{-jωn} dω\)
= \(\frac{1}{2π} \int_{-π}^π|X(ω)|^2 dω\)
28. Which of the following relation is true if the signal x(n) is real?
A. X*(ω) = X(ω)
B. X*(ω) = X(-ω)
C. X*(ω) = -X(ω)
D. None of the mentioned
Show Explanation Answer: B
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).
29. For a signal x(n) to exhibit even symmetry, it should satisfy the condition |X(-ω)| = | X(ω)|.
A. True
B. False
Show Explanation Answer: A
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 sides should be equal.
So, |X*(ω)| = |X(-ω)| = > |X(-ω)| = |X(ω)|.
30. What is the energy density spectrum Sxx(ω) of the signal x(n) = anu(n), |a|<1?
A. \(\frac{1}{1+2acosω+a^2}\)
B. \(\frac{1}{1+2asinω+a^2}\)
C. \(\frac{1}{1-2asinω+a^2}\)
D. \(\frac{1}{1-2acosω+a^2}\)
Show Explanation Answer: D
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) = \(\frac{1}{(1-ae^{-jω})(1-ae^{jω})} = \frac{1}{1-2acosω+a^2}\).
