11. 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.
12. 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.
13. 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)=\(\frac{1}{2\pi} \int_{-\pi}^{\pi}X(\omegA.e^{j\omega n} dω\)
=\(\frac{1}{2\pi} \int_{-ω_c}^{ω_c}1.e^{j\omega n} dω=\frac{sin ω_c.n}{ω_c.n}\)
=\(\frac{ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}\)
14. 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 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.
15. 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=\(\sum_{n=-∞}^∞ |x(n)|^2\)
=\(\sum_{n=-∞}^∞ x(n).x^*(n)\)
=\(\sum_{n=-∞}^∞ x(n)\frac{1}{2π} \int_{-π}^π X^*(ω) e^{-jωn} dω\)
=\(\frac{1}{2π} \int_{-π}^π|X(ω)|^2 dω\)
16. 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).
17. 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)|.
18. 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)=\(\frac{1}{(1-ae^{-jω})(1-ae^{jω})} = \frac{1}{1-2acosω+a^2}\).
19. What is the Fourier transform of the signal x(n) which is defined as shown in the graph below?
A. Ae-j(ω/2)(L)\(\frac{sin(\frac{ωL}{2})}{sin(\frac{ω}{2})}\)
B. Aej(ω/2)(L-1)\(\frac{sin(\frac{ωL}{2})}{sin(\frac{ω}{2})}\)
C. Ae-j(ω/2)(L-1)\(\frac{sin(\frac{ωL}{2})}{sin(\frac{ω}{2})}\)
D. None of the mentioned
Show Explanation Answer: C
The Fourier transform of this signal is
X(ω)=\(\sum_{n=0}^{L-1} Ae^{-jωn}\)
=A.\(\frac{1-e^{-jωL}}{1-e^{-jω}}\)
=\(Ae^{-j(ω/2)(L-1)}\frac{sin(\frac{ωL}{2})}{sin(\frac{ω}{2})}\)
20. Which of the following condition is to be satisfied for the Fourier transform of a sequence to be equal as the Z-transform of the same sequence?
A. |z|=1
Show Explanation Answer: A
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.
