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

41. If F1 and F2 are the lower and upper cutoff frequencies of a bandpass signal, then what is the condition to be satisfied to call such a bandpass signal a narrow band signal?

A. (F1-F2)>\(\frac{F_1+F_2}{2}\)(factor of 3 or less)

B. (F1-F2)⋙\(\frac{F_1+F_2}{2}\)(factor of 10 or more)

C. (F1-F2)<\(\frac{F_1+F_2}{2}\)(factor of 3 or less)

D. (F1-F2)⋘\(\frac{F_1+F_2}{2}\)(factor of 10 or more)

Answer: D

If the difference in the cutoff frequencies is much less than the mean frequency, such a bandpass signal is known as a narrowband signal.

42. What is the frequency range(in Hz) of Electroencephalogram(EEG)?

A. 10-40
B. 1000-2000
C. 0-100
D. None of the mentioned

Answer: C

The electroencephalogram (EEG) signal has a frequency range of 0-100 Hz.

43. Which of the following electromagnetic signals has a frequency range of 30kHz-3MHz?

A. Radio broadcast
B. Shortwave radio signal
C. RADAR
D. Infrared signal

Answer: A

A radio broadcast signal is an electromagnetic signal which has a frequency range of 30kHz-3MHz.

44. If x(n) = xR(n)+jxI(n) is a complex sequence whose Fourier transform is given as X(ω) = XR(ω)+jXI(ω), then what is the value of XR(ω)?

A. \(\sum_{n = 0}^∞\)xR (n)cosωn-xI (n)sinωn

B. \(\sum_{n = 0}^∞\)xR (n)cosωn+xI (n)sinωn

C. \(\sum_{n = -∞}^∞\)xR (n)cosωn+xI (n)sinωn

D. \(\sum_{n = -∞}^∞\)xR (n)cosωn-xI (n)sinωn

Answer: C

We know that X(ω) = \(\sum_{n = -∞}^∞\) x(n)e-jωn

By substituting e-jω = cosω – jsinω in the above equation and separating the real and imaginary parts we get

XR(ω) = \(\sum_{n = -∞}^∞\)xR (n)cosωn+xI (n)sinωn

45. If x(n) = xR(n)+jxI(n) is a complex sequence whose Fourier transform is given as X(ω) = XR(ω)+jXI(ω), then what is the value of xI(n)?

A. \(\frac{1}{2π} \int_0^{2π}\)[XR(ω) sinωn+ XI(ω) cosωn] dω

B. \(\int_0^{2π}\)[XR(ω) sinωn+ XI(ω) cosωn] dω

C. \(\frac{1}{2π} \int_0^{2π}\)[XR(ω) sinωn – XI(ω) cosωn] dω

D. None of the mentioned

Answer: A

We know that the inverse transform or the synthesis equation of a signal x(n) is given as

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

By substituting ejω = cosω + jsinω in the above equation and separating the real and imaginary parts we get

xI(n) = \(\frac{1}{2π} \int_0^{2π}\)[XR(ω) sinωn+ XI(ω) cosωn] dω

46. If x(n) is a real sequence, then what is the value of XI(ω)?

A. \(\sum_{n = -∞}^∞ x(n)sin⁡(ωn)\)

B. –\(\sum_{n = -∞}^∞ x(n)sin⁡(ωn)\)

C. \(\sum_{n = -∞}^∞ x(n)cos⁡(ωn)\)

D. –\(\sum_{n = -∞}^∞ x(n)cos⁡(ωn)\)

Answer: B

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

47. Which of the following relations are true if x(n) is real?

A. X(ω) = X(-ω)
B. X(ω) = -X(-ω)
C. X*(ω) = X(ω)
D. X*(ω) = X(-ω)

Answer: D

We know that, if x(n) is a real sequence

XR(ω) = \(\sum_{n = -∞}^∞\) x(n)cosωn = >XR(-ω) = XR(ω)

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

If we combine the above two equations, we get

X*(ω) = X(-ω)

48. If x(n) is a real signal, then x(n) = \(\frac{1}{π}\int_0^π\)[XR(ω) cosωn- XI(ω) sinωn] dω.

A. True
B. False

Answer: A

We know that if x(n) is a real signal, then xI(n) = 0 and xR(n) = x(n)

We know that,

xR(n) = x(n) = \(\frac{1}{2π}\int_0^{2π}\)[XR(ω) cosωn- XI(ω) sinωn] dω

Since both XR(ω) cosωn and XI(ω) sinωn are even, x(n) is also even

= > x(n) = \(\frac{1}{π} \int_0^π\)[XR(ω) cosωn- XI(ω) sinωn] dω

49. If x(n) is a real and odd sequence, then what is the expression for x(n)?

A. \(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω

B. –\(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω

C. \(\frac{1}{π} \int_0^π\)[XI(ω) cosωn] dω

D. –\(\frac{1}{π} \int_0^π\)[XI(ω) cosωn] dω

Answer: B

If x(n) is real and odd then, x(n)cosωn is odd and x(n) sinωn is even. Consequently

XR(ω) = 0

XI(ω) = \(-2\sum_{n = 1}^∞ x(n) sin⁡ωn\)

= >x(n) = -\(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω

50. What is the value of XR(ω) given X(ω) = \(\frac{1}{1-ae^{-jω}}\),|a|<1?

A. \(\frac{asinω}{1-2acosω+a^2}\)

B. \(\frac{1+acosω}{1-2acosω+a^2}\)

C. \(\frac{1-acosω}{1-2acosω+a^2}\)

D. \(\frac{-asinω}{1-2acosω+a^2}\)

Answer: C

Given, X(ω) = \(\frac{1}{1-ae^{-jω}}\), |a|<1

By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain

X(ω) = \(\frac{1-ae^{jω}}{(1-ae^{-jω})(1-ae^{jω})} = \frac{1-acosω-jasinω}{1-2acosω+a^2}\)

This expression can be subdivided into real and imaginary parts, thus we obtain

XR(ω) = \(\frac{1-acosω}{1-2acosω+a^2}\).

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