1. Which of the following is the Analysis equation of Fourier Transform?
a) F(ω)=∫∞−∞f(t)ejωtdt
b) F(ω)=∫∞0f(t)e−jωtdt
c) F(ω)=∫∞0f(t)ejωtdt
d) F(ω)=∫∞−∞f(t)e−jωtdt
Answer: d
Explanation: For converting time domain to frequency domain, we use analysis equation. The Analysis equation of Fourier Transform is F(ω)=∫∞−∞f(t)e−jωtdt.

2. Choose the correct synthesis equation.
a) f(t)=12π∫∞−∞F(ω)e−jωtdω
b) f(t)=12π∫∞−∞F(ω)ejωtdω
c) f(t)=12π∫∞0F(ω)e−jωtdω
d) f(t)=12π∫∞0F(ω)ejωtdω
Answer: b
Explanation: Synthesis equation converts from frequency domain to time domain. The synthesis equation of fourier transform is f(t)=12π∫∞−∞F(ω)ejωtdω.

3. Find the fourier transform of an exponential signal f(t) = e^-at u(t), a>0.
a) 1a+jω
b) 1a−jω
c) 1−a+jω
d) 1−a−jω
Answer: a
Explanation: Given f(t)= e^-at u(t)
We know that u(t)={01t<0t>0
Fourier transform,
F(ω)=∫∞−∞f(t)e−jωtdt=∫∞−∞e−atu(t)e−jωtdt=∫∞0e−(a+jω)tdt
F(ω) = 1a+jω, a>0.

4. Find the fourier transform of the function f(t) = e^-a|t|, a>0.
a) 2aa2−ω2
b) 2aa2+ω2
c) 2aω2−a2
d) aa2+ω2
Answer: b
Explanation: The given two-sided exponential function f(t) = e^-a|t|, a>0 can be expressed as
f(t)={e−ateatt≥0t≤0
The Fourier transform is
F(ω)=∫∞−∞f(t)e−jωtdt=∫0−∞f(t)e−jωtdt+∫∞0f(t)e−jωtdt
F(ω)=1a+jω+1a−jω=2aa2+ω2.

5. Gate function is defined as ______________
a) G(t)={10|t|<τ2elsewhere
b) G(t)={10|t|>τ2elsewhere
c) G(t)={10|t|≤τ2elsewhere
d) G(t)={10|t|≥τ2elsewhere
Answer: a
Explanation: A gate function is a rectangular function defined as
G(t)=rect(tτ)={10|t|<τ2elsewhere
Where τ is pulse width.

6. Find the fourier transform of the gate function.
a) 1ωsin(ωτ2)
b) 1ωcos(ωτ2)
c) 2ωsin(ωτ2)
d) 2ωcos(ωτ2)
Answer: c
Explanation: Gate function is defined as
G(t)={10|t|<τ2elsewhere
The fourier transform is F(ω)=∫∞−∞f(t)e−jωtdt=∫τ/2−τ/2e−jωtdt=2ωsin(ωτ2).

7. Choose the wrong option.
a) G(t) = rect(tτ)
b) G(t) = u(t + τ2) – u(t-τ2)
c) G(ω) = τ sa(wτ2)
d) G(f) = τ sinc(f)
Answer: d
Explanation: Fourier transform of gate function, G(ω) = 2ωsin(wτ2)
Multiplying and dividing by τ we get
G(ω)=τsin(wτ2)wτ2=τsin(2πfτ2)2πfτ2=τsin(πτf)πτf=τsinc(τf).

8. Bandwidth of the gate function is __________
a) τ Hz
b) 1τ Hz
c) 2τ Hz
d) 2τ Hz
Answer: b
Explanation: The practical bandwidth of the gate function corresponds to the first zero crossing in the spectrum. Therefore, the bandwidth of the pulse or gate function is 2πτ or 1τ Hz.

9. Which of the following is not a fourier transform pair?
a) u(t)↔πδ(ω)+1jω
b) sgn(t)↔2jω
c) A↔2πδ(ω2)
d) G(t)↔sa(ωτ2)
Answer: d
Explanation: G(t)↔sa(ωτ2) is not a fourier transform pair.
G(t)↔τsa(ωτ2) (or) G(t)↔G(t)τsinc(τf).

10. Find the fourier transform of the unit step function.
a) πδ(ω) + 1ω
b) πδ(ω) + 1jω
c) πδ(ω) – 1jω
d) δ(ω) + 1jω
Answer: b
Explanation: We know that sgn(t) = 2u(t) – 1.
u(t) = 12[sgn(t)+1] Its Fourier transform is F[u(t)] = 12 F[sgn(t)] + 12 F[1]
As the Fourier transforms F[1] = 2πδ(ω) and [sgn(t)] = 2jω, hence
F[u(t)] = πδ(ω) + 1jω.

11. Fourier transform NMR spectrometer allows NMR transitions to be observed simultaneously.
a) True
b) False
Answer: a
Explanation: Fourier transform NMR spectrometer is a type of NMR spectrometer. It allows samples to be observed simultaneously instead of serially.

12. Fourier transform NMR spectrometer has which of the following characteristics?
a) Increased sensitivity, long time to obtain data
b) Decreased sensitivity, long time to obtain data
c) Increased sensitivity, reduced time to obtain data
d) Decreased sensitivity, reduced time to obtain data
Answer: c
Explanation: Fourier transform NMR spectrometer has increased sensitivity. It takes less time to obtain NMR data.

13. Which of the following cannot be done due to the multiplex advantage?
a) Repetitive signals can be summed
b) Repetitive signals can be averaged
c) Increases signal to noise ratio
d) Decreases signal to noise ratio
Answer: d
Explanation: The multiplexing advantage allows improvement of the signal to noise ratio. It also allows repetitive signals to be summed and averaged.

14. Two coils are necessary for Fourier transform NMR spectroscopy.
a) True
b) False
Answer: b
Explanation: Only one coil is necessary for Fourier transform NMR spectroscopy. The coil serves as both antenna for transmitting and receiving RF radiation.

15. Which of the following is the disadvantage of conventional mode of spectrometer or continuous-wave NMR spectrometer?
a) They are unstable
b) Maintenance is difficult
c) High operating cost
d) Excitation is inefficient
Answer: d
Explanation: The disadvantage of conventional mode of a spectrometer or continuous-wave NMR spectrometer is that the excitation is inefficient. Only a narrow band of frequencies are contributing to the signal at a time.

16. Which of the following must be done to improve sensitivity?
a) Frequency sweep mode must be preferred
b) Field sweep mode must be preferred
c) Single channel excitation must be preferred
d) Multichannel excitation must be preferred
Answer: d
Explanation: Sweep techniques have less sensitivity. To improve sensitivity, multichannel excitation must be preferred.

17. Which of the following is the disadvantage of multichannel excitation?
a) Low sensitivity
b) Low resolution
c) Small number of frequencies is present
d) Uneconomical
Answer: d
Explanation: Large number of frequencies is present in multichannel excitation. This is uneconomical since a receiver is required for each channel.

18. How can the need for an array of narrow-band filters and detectors be eliminated?
a) By using multi-channel excitation
b) By using a detector for each frequency
c) By reducing the number of detectors
d) By using Fourier transform
Answer: d
Explanation: Fourier transform plays the role of a multichannel receiver. Hence, the need for array of narrow-band filters can be eliminated

19. Fourier transform can be accomplished by using which of the following components?
a) Spin decoder
b) Detector
c) Spectrum analyser
d) Oscilloscope
Answer: c
Explanation: Fourier transform can be accomplished by using a spectrum analyser. Any complex waveform can be converted to frequencies using Fourier transform.

20. A 15.4 MHz crystal generates the _________ resonance frequency.
a) Hydrogen
b) Deuterium
c) Tritium
d) Helium
Answer: b
Explanation: A 15.4 MHz crystal generates the deuterium resonance frequency. This resonance signal is used to lock the magnetic signal to clock frequency.

21. If x(n)=x_R(n)+jx_I(n) is a complex sequence whose Fourier transform is given as X(ω)=X_R(ω)+jX_I(ω), then what is the value of X_R(ω)?
a) ∑∞n=0x_R (n)cosωn-x_I (n)sinωn
b) ∑∞n=0x_R (n)cosωn+x_I (n)sinωn
c) ∑∞n=−∞x_R (n)cosωn+x_I (n)sinωn
d) ∑∞n=−∞x_R (n)cosωn-x_I (n)sinωn
Answer: c
Explanation: We know that X(ω)=∑∞n=−∞ x(n)e^-jωn
By substituting e^-jω = cosω – jsinω in the above equation and separating the real and imaginary parts we get
X_R(ω)=∑∞n=−∞x_R (n)cosωn+x_I (n)sinωn

22. If x(n)=x_R(n)+jx_I(n) is a complex sequence whose Fourier transform is given as X(ω)=X_R(ω)+jX_I(ω), then what is the value of x_I(n)?
a) 12π∫2π0[X_R(ω) sinωn+ X_I(ω) cosωn] dω
b) ∫2π0[X_R(ω) sinωn+ X_I(ω) cosωn] dω
c) 12π∫2π0[X_R(ω) sinωn – X_I(ω) cosωn] dω
d) None of the mentioned
Answer: a
Explanation: We know that the inverse transform or the synthesis equation of a signal x(n) is given as
x(n)=12π∫2π0 X(ω)e^jωn dω
By substituting e^jω = cosω + jsinω in the above equation and separating the real and imaginary parts we get
x_I(n)=12π∫2π0[X_R(ω) sinωn+ X_I(ω) cosωn] dω

23. If x(n) is a real sequence, then what is the value of X_I(ω)?
a) ∑∞n=−∞x(n)sin(ωn)
b) –∑∞n=−∞x(n)sin(ωn)
c) ∑∞n=−∞x(n)cos(ωn)
d) –∑∞n=−∞x(n)cos(ωn)
Answer: b
Explanation: If the signal x(n) is real, then x_I(n)=0
We know that,
X_I(ω)=∑∞n=−∞xR(n)sinωn−xI(n)cosωn
Now substitute x_I(n)=0 in the above equation=>x_R(n)=x(n)
=> X_I(ω)=-∑∞n=−∞x(n)sin(ωn).

24. 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
Explanation: We know that, if x(n) is a real sequence
X_R(ω)=∑∞n=−∞ x(n)cosωn=>X_R(-ω)= X_R(ω)
X_I(ω)=-∑∞n=−∞ x(n)sin⁡(ωn)=>X_I(-ω)=-X_I(ω)
If we combine the above two equations, we get
X*(ω)=X(-ω)

25. If x(n) is a real signal, then x(n)=1π∫π0[X_R(ω) cosωn- X_I(ω) sinωn] dω.
a) True
b) False
Answer: a
Explanation: We know that if x(n) is a real signal, then x_I(n)=0 and x_R(n)=x(n)
We know that, x_R(n)=x(n)=12π∫2π0[X_R(ω) cosωn- X_I(ω) sinωn] dω
Since both X_R(ω) cosωn and X_I(ω) sinωn are even, x(n) is also even
=> x(n)=1π∫π0[X_R(ω) cosωn- X_I(ω) sinωn] dω

26. If x(n) is a real and odd sequence, then what is the expression for x(n)?
a) 1π∫π0[X_I(ω) sinωn] dω
b) –1π∫π0[X_I(ω) sinωn] dω
c) 1π∫π0[X_I(ω) cosωn] dω
d) –1π∫π0[X_I(ω) cosωn] dω
Answer: b
Explanation: If x(n) is real and odd then, x(n)cosωn is odd and x(n) sinωn is even. Consequently
X_R(ω)=0
X_I(ω)=−2∑∞n=1x(n)sinωn
=>x(n)=-1π∫π0[X_I(ω) sinωn] dω

27. What is the value of X_R(ω) given X(ω)=11−ae−jω,|a|<1?
a) asinω1−2acosω+a2
b) 1+acosω1−2acosω+a2
c) 1−acosω1−2acosω+a2
d) −asinω1−2acosω+a2
Answer: c
Explanation: Given, X(ω)=11−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)=1−aejω(1−ae−jω)(1−aejω)=1−acosω−jasinω1−2acosω+a2
This expression can be subdivided into real and imaginary parts, thus we obtain
X_R(ω)=1−acosω1−2acosω+a2.

28. What is the value of X_I(ω) given 11−ae−jω, |a|<1?
a) asinω1−2acosω+a2
b) 1+acosω1−2acosω+a2
c) 1−acosω1−2acosω+a2
d) −asinω1−2acosω+a2
Answer: d
Explanation: Given, X(ω)=11−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)=1−aejω(1−ae−jω)(1−aejω)=1−acosω−jasinω1−2acosω+a2
This expression can be subdivided into real and imaginary parts, thus we obtain
X_I(ω)=−asinω1−2acosω+a2.

29. What is the value of |X(ω)| given X(ω)=1/(1-ae^-jω), |a|<1?
a) 11−2acosω+a2√
b) 11+2acosω+a2√
c) 11−2acosω+a2
d) 11+2acosω+a2
Answer: a
Explanation: For the given X(ω)=1/(1-ae^-jω), |a|<1 we obtain
X_I(ω)=(-asinω)/(1-2acosω+a²) and X_R(ω)=(1-acosω)/(1-2acosω+a²)
We know that |X(ω)|=XR(ω)2+XI(ω)2−−−−−−−−−−−−−−√
Thus on calculating, we obtain
|X(ω)| = 11−2acosω+a2√.

30. If x(n)=A, -M<n<M,; x(n)=0, elsewhere. Then what is the Fourier transform of the signal?
a) Asin(M−12)ωsin(ω2)
b) A²sin(M+12)ωsin(ω2)
c) Asin(M+12)ωsin(ω2)
d) sin(M−12)ωsin(ω2)
Answer: c
Explanation: Clearly, x(n)=x(-n). Thus the signal x(n) is real and even signal. So, we know that
X(ω)=XR(ω)=A(1+2∑∞n=1cosωn)
On simplifying the above equation, we obtain
X(ω)=Asin(M+12)ωsin(ω2).

31. What is the Fourier transform of the signal x(n)=a^|n|, |a|<1?
a) 1+a21−2acosω+a2
b) 1−a21−2acosω+a2
c) 2a1−2acosω+a2
d) None of the mentioned
Answer: b
Explanation: First we observe x(n) can be expressed as
x(n)=x₁(n)+x₂(n)
where x₁(n)= aⁿ, n>0
=0, elsewhere
x₂(n)=a^-n, n<0 =0, elsewhere Now applying Fourier transform for the above two signals, we get X₁(ω)=11−ae−jω and X₂(ω)=aejω1−aejω
Now, X(ω)=X₁(ω)+ X₂(ω)=11−ae−jω+aejω1−aejω=1−a21−2acosω+a2.

32. If X(ω) is the Fourier transform of the signal x(n), then what is the Fourier transform of the signal x(n-k)?
a) e^jωk. X(-ω)
b) e^jωk. X(ω)
c) e^-jωk. X(-ω)
d) e^-jωk. X(ω)
Answer: d
Explanation: Given
F{x(n)}= X(ω)=∑∞n=−∞x(n)e−jωn
=>F{x(n-k)}=∑∞n=−∞x(n−k)e−jωn=e−jωk.∑∞n=−∞x(n−k)e−jω(n−k)
=>F{x(n-k)}= e^-jωk. X(ω)

33. What is the convolution of the sequences of x₁(n)=x₂(n)={1,1,1}?
a) {1,2,3,2,1}
b) {1,2,3,2,1}
c) {1,1,1,1,1}
d) {1,1,1,1,1}
Answer: a
Explanation: Given x₁(n)=x₂(n)={1,1,1}
By calculating the Fourier transform of the above two signals, we get
X₁(ω)= X₂(ω)=1+ e^jω + e^-jω = 1+2cosω
From the convolution property of Fourier transform we have,
X(ω)= X₁(ω). X₂(ω)=(1+2cosω)²=3+4cosω+2cos2ω
By applying the inverse Fourier transform of the above signal, we get
x₁(n)*x₂(n)={1,2,3,2,1}

34. What is the energy density spectrum of the signal x(n)=aⁿu(n), |a|<1?
a) 11+2acosω+a2
b) 11−2acosω+a2
c) 11−2acosω−a2
d) 11+2acosω−a2
Answer: b
Explanation: Given x(n)= aⁿu(n), |a|<1
The auto correlation of the above signal is
r_xx(l)=11−a2 a^|l|, -∞< l <∞
According to Wiener-Khintchine Theorem,
S_xx(ω)=F{r_xx(l)}=[11−a2].F{a^|l|} = 11−2acosω+a2

35. How do we represent a pairing of a periodic signal with its fourier series coefficients in case of continuous time fourier series?
a) x(t) ↔ X_n
b) x(t) ↔ X_n+1
c) x(t) ↔ X
d) x(n) ↔ X_n
Answer: a
Explanation: In case of continuous time fourier series, for simplicity, we represent a pairing of a periodic signal with its fourier series coefficients as,
x(t) ↔ X_n
here, x(t) is the signal and Xn is the fourier series coefficient.

36. What are the properties of continuous time fourier series?
a) Linearity, time shifting
b) Linearity, time shifting, frequency shifting
c) Linearity, time shifting, frequency shifting, time reversal, time scaling, periodic convolution
d) Linearity, time shifting, frequency shifting, time reversal, time scaling, periodic convolution, multiplication, differentiation
Answer: d
Explanation: Linearity, time shifting, frequency shifting, time reversal, time scaling, periodic convolution, multiplication, differentiation are some of the properties followed by continuous time fourier series. Integration and conjugation are also followed by continuous time fourier series.

37. Integration and conjugation are also followed by continuous time fourier series?
a) True
b) False
Answer: a
Explanation: Linearity, time shifting, frequency shifting, time reversal, time scaling, periodic convolution, multiplication, differentiation are some of the properties followed by continuous time fourier series. Integration and conjugation are also followed by continuous time fourier series.

38. If x(t) and y(t) are two periodic signals with coefficients X_n and Y_n then the linearity is represented as?
a) ax(t) + by(t) = aX_n + bY_n
b) ax (t) + by(t) = X_n + bY_n
c) ax(t) + by(t) = aX_n + Y_n
d) ax(t) + by(t) = X_n + Y_n
Answer: a
Explanation: ax(t) + by(t) = aX_n + bY_n, x(t) and y(t) are two periodic signals with coefficients X_n and Y_n.

39. How is time shifting represented in case of periodic signal?
a) If x(t) is shifted to t₀, X_n is shifted to t₀
b) x(t-t₀), Y_n = X_n e^-njwt0
c) X_n = x(t-t₀), Y_n = X_n e^-njwt0
d) X_n = x(-t₀), Y_n = X_n e^-njwt0
Answer: c
Explanation: If x(t) and y(t) are two periodic signals with coefficients X_n and Y_n, then if a signal is shifted to t₀, then the property says,
X_n = x(t-t₀), Y_n = X_ne^-njwt0

40. What is the frequency shifting property of continuous time fourier series?
a) Multiplication in the time domain by a real sinusoid
b) Multiplication in the time domain by a complex sinusoid
c) Multiplication in the time domain by a sinusoid
d) Addition in the time domain by a complex sinusoid
Answer: b
Explanation: If x(t) and y(t) are two periodic signals with coefficients Xn and Yn,
Then y(t)= e^jmwtx(t)↔Y_n=X_n-m.
Hence, we can see that a frequency shift corresponds to multiplication in the time domain by complex sinusoid whose frequency is equal to the time shift.

41. What is the time reversal property of fourier series coefficients?
a) Time reversal of the corresponding sequence of fourier series
b) Time reversal of the last term of fourier series
c) Time reversal of the corresponding term of fourier series
d) Time reversal of the corresponding sequence
Answer: a
Explanation: x(t)↔ X_n
Y(t) = x(-t)↔Y_n=X_-n.
That is the time reversal property of fourier series coefficients is time reversal of the corresponding sequence of fourier series.

42. It does not depend whether the signal is odd or even, it is always reversal of the corresponding sequence of fourier series.
a) True
b) False
Answer: b
Explanation: It does depend whether the signal is odd or even.
If the signal is even, the reversal is positive and if the signal is odd, the reversal is negative.

43. Why does the signal change while time scaling?
a) Because the frequency changes
b) Time changes
c) Length changes
d) Both frequency and time changes
Answer: a
Explanation: x(t)↔X_n
Y(t) = x(at)↔Y_n = X_n
Hence, the fourier coefficients have not changed but the representation has changed because of changes in fundamental frequency.

44. What is the period of the signal when it is time shifted?
a) Changes according to the situation
b) Different in different situation
c) Remains the same
d) Takes the shifted value
Answer: c
Explanation: The period of the periodic signal does not change even if it is time shifted.
If x(t) and y(t) are two periodic signals with coefficients X_n and Y_n, then if a signal is shifted to t₀, then the property says,
X_n = x(t-t₀), Y_n = X_ne^-njwt0.

45. How is the discrete time impulse function defined in terms of the step function?
a) d[n] = u[n+1] – u[n].
b) d[n] = u[n] – u[n-2].
c) d[n] = u[n] – u[n-1].
d) d[n] = u[n+1] – u[n-1].
Answer: c
Explanation: Using the definition of the Heaviside function, we can come to this conclusion.

46. What is the definition of the delta function in time space intuitively?
a) Defines that there is a point 1 at t=0, and zero everywhere else
b) Defines that there is a point 0 at t=0, and 1 everywhere else
c) Defines 1 for all t > 0, and 0 else
d) Defines an impulse of area 1 at t=0, zero everywhere else
Answer: d
Explanation: Arises from the definition of the delta function. There is a clear difference between just the functional value and the impulse area of the delta function.

47. Is it practically possible for us to provide a perfect impulse to a system?
a) Certainly possible
b) Impossible
c) Possible
d) None of the mentioned
answer: b
Explanation: The spread of the impulse can never be restricted to a single point in time, and thus, we cannot achieve a perfect impulse.

48. The convolution of a discrete time system with a delta function gives
a) the square of the system
b) the system itself
c) the derivative of the system
d) the integral of the system
Answer: b
Explanation: The integral reduces to the the integral calculated at a single point, determined by the centre of the delta function.

49. Find the value of 2sgn(0)d[0] + d[1] + d[45], where sgn(x) is the signum function.
a) 2
b) -2
c) 1
d) 0
Answer: d
Explanation: sgn(0)=0, and d[n] = 0 for all n not equal to zero. Hence the sum reduces to zero.

50. Where h*x denotes h convolved with x, x[n]*d[n-90] reduces to
a) x[n-89].
b) x[n-91].
c) x[n=90].
d) x[n].
Answer: c
Explanation: The function gets shifted by the center of the delta function during convolution.

51. Where h*x denotes h convolved with x, find the value of d[n]*d[n-1].
a) d[n].
b) d[n-1].
c) d²[n].
d) d²[n-1].
Answer: b
Explanation: Using the corollary, if we take d[n] to be the ‘x’ function, it will be shifted by -1 when convolved with d[n-1], thus rendering d[n-1].

52. How is the continuous time impulse function defined in terms of the step function?
a) u(t) = d(d(t))/dt
b) u(t) = d(t)
c) d(t) = du/dt
d) d(t) = u²(t)
View AnswerAnswer: c
Explanation: Using the definition of the Heaviside function, we can come to this conclusion.

53. In which of the following useful signals, is the bilateral Laplace Transform different from the unilateral Laplace Transform?
a) d(t)
b) s(t)
c) u(t)
d) all of the mentioned
Answer: c
Explanation: The bilateral LT is different from the aspect that the integral is applied for the entire time axis, but the unilateral LT is applied only for the positive time axis. Hence, the u(t) [unit step function] differs in that aspect and hence can be used to differentiate the same.

54. What is the relation between the unit impulse function and the unit ramp function?
a) r = dd(t)/dt
b) d = dr/dt
c) d = d²(r)/dt²
d) r = d²(d)/dt²
Answer: c
Explanation: Now, d = du/dt and u = dr/dt. Hence, we obtain the above answer.

55. The term heterodyning refers to
a) Frequency conversion
b) Frequency mixing
c) Frequency conversion & mixing
d) None of the mentioned
Answer: c
Explanation: The method heterodyning means frequency conversion and mixing and this results in a spectral shift.

56. The causes for error performance degradation in communication systems are
a) Interference
b) Electrical noise
c) Effect of filtering
d) All of the mentioned
Answer: d
Explanation: The main causes of error performance degradation are interference electrical noise effect of filtering and also due to the surroundings.

57. Thermal noise in the communication system due to thermal electrons
a) Can be eliminated
b) Cannot be eliminated
c) Can be avoided upto some extent
d) None of the mentioned
Answer: b
Explanation: Thermal noise which cannot be eliminated is caused by the motion of thermal electrons causes degradation in system.

58. White noise has _______ power spectral density.
a) Constant
b) Variable
c) Constant & Variable
d) None of the mentioned
Answer: a
Explanation: The AWGN has constant power spectral density.

59. Which are called as hard decisions?
a) Estimates of message symbol with error correcting codes
b) Estimates of message symbol without error correcting codes
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: If error correcting codes are not present, the detector output consists of estimates of the message symbol which is also called as hard decisions.

60. The filter which is used to recover the pulse with less ISI is called as
a) Matched filter
b) Correlator
c) Matched filter & Correlator
d) None of the mentioned
Answer: b
Explanation: The optimum filter used to recover the pulse with best possible signal to noise ratio and less or no ISI is called as correlator or matched filter.

61. The composite equalizing filter is the combination of
a) Receiving and equalizing filter
b) Transmitting and equalizing filter
c) Amplifier and equalizing filter
d) None of the mentioned
Answer: a
Explanation: The functions of both receiving and equalizing filter can be performed by only the equalizing filter alone. Thus equalizing filter is the combination of equalizing and receiving filter.

62. The sample from the demodulation process consists of sample which is _______ to energy of the received symbol and _____ to noise.
a) Directly and inversely proportional
b) Inversely and directly proportional
c) Both directly proportional
d) Both inversely proportional
Answer: a
Explanation: The output symbol of the sampler consists of sample which is directly proportional to the energy of the received signal and inversely proportional to the noise.

63. The average noise power of white noise is
a) 0
b) Infinity
c) 1
d) None of the mentioned
Answer: b
Explanation: White noise is a idealized process with two sided spectral density equal to constant N0/2 and frequencies varying from minus infinity to plus infinity. Thus the average noise power is infinity.

64. The channel may be affected by
a) Thermal noise
b) Interference from other signals
c) Thermal noise & Interference from other signals
d) None of the mentioned
Answer: c
Explanation: A channel can be modelled as a linear filter with additional noise. The noise comes from thermal noise source and also from interference from other signals.

65. Channels display multi-path due to
a) Scattering
b) Time delayed reflections
c) Diffraction
d) All of the mentioned
Answer: d
Explanation: Wireless wide-band channels display multi-path due to time delayed reflections, diffraction and also scattering.

1. Carrier signal in modulation technique is _______ signal.
a) High frequency
b) Low frequency
c) High amplitude
d) Low amplitude
Answer: a
Explanation: Carrier signal in modulation technique is a high frequency signal. In amplitude modulation, the amplitude of a high frequency carrier signal is varied in accordance to the instantaneous amplitude of the modulating signal.

2. Modulation index of an AM signal is ratio of __________ to the _______
a) Peak carrier amplitude, Peak message signal amplitude
b) Peak message signal amplitude, Peak carrier amplitude
c) Carrier signal frequency, Message signal frequency
d) Message signal frequency, Carrier signal frequency
Answer: b
Explanation: The modulation index k of an AM signal is defined as the ratio of the peak message signal amplitude to the peak carrier amplitude. The modulation index is often expressed as a percentage. It is also called percentage modulation.

3. If the peak message signal amplitude is half the peak amplitude of the carrier signal, the signal is _____ modulated.
a) 100%
b) 2%
c) 50%
d) 70%
Answer: c
Explanation: The modulation is also expressed in percentage. It is also called percentage modulation. The signal is said to be 50% modulated if the peak message signal amplitude is half the peak amplitude of the carrier signal.

4. A percentage of modulation greater than ___________ will distort the message signal.
a) 10%
b) 25%
c) 50%
d) 100%
Answer: d
Explanation: A percentage of modulation greater than 100% will distort the message signal if detected by an envelope detector. In this case the lower excursion of the signal will drive the carrier amplitude below zero, making it negative (and hence changing its phase).

5. The RF bandwidth of AM is ____________ the maximum frequency contained in the modulating message signal.
a) Equal
b) Two times
c) Four times
d) Ten times
Answer: b
Explanation: The RF bandwidth of an AM signal is equal to B_AM=2f_m. It is double the maximum frequency contained in the modulating message signal. AM spectrum consists of an impulse at the carrier frequency and two sidebands which replicate the message spectrum.

6. Single sideband AM systems occupy same bandwidth as of conventional AM systems.
a) True
b) False
Answer: b
Explanation: Single sideband (SSB) AM systems transmit only one of the sidebands (either upper or lower) about the carrier. Hence, they occupy only half the bandwidth of conventional AM systems.

7. How is the performance of SSB AM systems in fading channels?
a) Poor
b) Best
c) Good
d) Average
Answer: a
Explanation: SSB systems have the advantage of being very bandwidth efficient. But their performance in fading channels is very poor. For proper detection, the frequency of the oscillator at the product detector mixer in the receiver must be same as that of the incoming carrier frequency.

8. Which of the following is a disadvantage of tone-in-band SSB system?
a) High bandwidth
b) Bad adjacent channel protection
c) Effects of multipath
d) Generation and reception of signal is complicated
Answer: d
Explanation: Tone-in-band SSB systems has the advantage of maintaining the low bandwidth property of the SSB signals, while at the same time providing good adjacent channel protection. The tone in band system employs feedforward automatic gain and frequency control to mitigate the effects of multipath induced fading.

9. FFSR in AM systems stands for ________
a) Feedforward signal regeneration
b) Feedbackward signal regeneration
c) Feedbackward system restoration
d) Feedforward system restoration
Answer: a
Explanation: FFSR stands for Feedforward signal regeneration. If the pilot tone and the information bearing signal undergo correlated fading, it is possible at the receiver to counteract the effects of fading through signal processing based on tracking of pilot tone. This process is called FFSR.

10. AM demodulation technique can be divided into _____ and _____ demodulation.
a) Direct, indirect
b) Slope detector, zero crossing
c) Coherent, noncoherent
d) Quadrature detection, coherent detection
Answer: c
Explanation: AM demodulation techniques may be broadly divide into two main categories. They are called coherent and noncoherent demodulation. They are differentiated by the knowledge of transmitted carrier frequency and phase at the receiver.

11. Non coherent detection requires the knowledge of transmitted carrier frequency and phase at the receiver.
a) True
b) False
Answer: b
Explanation: Non coherent detection does require the knowledge of phase information. However, coherent detection requires knowledge of the transmitted carrier frequency and phase at the receiver.

12. A product detector in AM systems is also called ___________
a) Envelope detector
b) Differentiator
c) Integrator
d) Phase detector
Answer: d
Explanation: A product detector is also called a phase detector. It forms a coherent demodulator for AM signals. It is a down converter circuit which converts the input bandpass signal to a baseband signal.

13. AM system use only product detector for demodulation. They never use envelope detectors.
a) True
b) False
Answer: b
Explanation: AM systems can use either product detector or envelope detector for demodulation. As a rule, envelope detectors are useful when input signal power is at least 10dB greater than noise power, whereas product detectors are able to process the AM signals with input signal to noise ratios well below 0 dB.

14. LCD uses ________
a) sematic crystals
b) twisted nematic crystals
c) nematic crystals
d) cholesteric crystals
Answer: b
Explanation: LCD uses liquid crystal display. It uses twisted nematic crystals which are a type of liquid crystal, consisting of a substance called the nematic. The nematic liquid crystal is placed between two plates of polarized glass.

15. Which of the following stage is present in FM receiver but not in AM receiver?
a) Amplitude limiter
b) Demodulator
c) AM amplifier
d) Mixer
Answer: a
Explanation: Amplitude Limiter circuit is used in FM receiver to remove the noise or any variation in amplitude present in the received signal. Thus, the output of the amplitude limiter has a constant amplitude. So it is only used in frequency modulation and not in amplitude modulation.

16. Function of duplexer in a RADAR is to permit the use of same antenna for transmission and reception.
a) True
b) False
Answer: a
Explanation: A duplexer is being an electronic unit, allows bi-directional communication over the same path. The transmitter and receiver can communicate simultaneously. In radar, the duplexer isolates the receiver from the transmitter while allowing them to share a common antenna.

17. Single Sideband Modulation (SSB) is generally reserved for point-to-point communication.
a) True
b) False
Answer: a
Explanation: A point-to-point communication refers to bidirectional communication between only one transmitter and one receiver. In SSB-SC modulation technique, the carrier is suppressed and only one of the two side-bands are transmitted. Thus, it reduces power consumption and lessens bandwidth. Thus, it is preferred for point-to-point communication.

18. For an AM transmitter, class C amplifier can be used after the modulation stage.
a) True
b) False
Answer: b
Explanation: In an AM transmitter, the required transmission power is obtained from class C amplifier, as it is a power amplifier, for low-level or high-level modulation. So it is not used after the modulation stage.

19. For which of the modulated system, the linear amplified modulated stage is used?
a) low level amplitude modulated system
b) high level amplitude modulated system
c) high level frequency modulated system
d) low level frequency modulated system
Answer: a
Explanation: In low-level modulation, the generation of amplitude modulated signal takes place at low power levels. The generated AM signal is then amplified using a chain of linear amplifiers, which are required to avoid waveform distortion. Thus, linear amplified modulated stage is used in low level amplitude modulated system.

20. When noise is passed through a narrow band filter, the output of filter should be?
a) triangular
b) square
c) parabolic
d) sinusoidal
Answer: d
Explanation: Narrow band filter is used to isolate a narrow band of frequencies from a wider bandwidth signal. It is a combination of band pass and band reject filter. When noise gets passed through it, the output of it should be sinusoidal.

21. A narrow band noise can exist in _________
a) AM only
b) PCM only
c) FM only
d) AM and FM both
Answer: d
Explanation: Narrow band filter is used to isolate a narrow band of frequencies from a wider bandwidth signal. It is a combination of band pass and band reject filter. So it can be used in both AM and FM to pass a band of frequencies or to attenuate a band of frequencies.

22. The upper and lower sideband frequencies for 5KHz amplitude modulation with a 30KHz carrier frequency will be?
a) 35KHz and 25KHz
b) 34KHz and 24KHz
c) 25KHz and 35KHz
d) 0.35KHz and 0.25KHz
Answer: a
Explanation: Upper sideband frequency will be (30 + 5) = 35 KHz and Lower sideband frequency will be (30 – 5) = 25 KHz.

23. Phase array radar can track many targets together.
a) True
b) False
Answer: a
Explanation: A phased array radar is an array of radiating elements, with each having a phase-shifter. The phase of the signal being emitted from the radiating element is changed to produce beams, thereby producing constructive or destructive interference for steering the beams in the required direction. Thus, it can track many targets together.

24. A duplex arrangement use separate frequencies for transmission.
a) True
b) False
Answer: a
Explanation: In duplex communication, two-way interaction is favourable simultaneously. Thus, a cordless telephone is duplex which uses separate frequencies for transmission in base and portable units.

25. VSB modulation is used in televisions because it avoids phase distortion at low frequencies.
a) True
b) False
Answer: b
Explanation: Vestigial Sideband Modulation (VSB) is a type of amplitude modulation in which the carrier and only one sideband is completely transmitted and the other sideband is partly transmitted. Thus, television production is done using VSB modulation as it reduces bandwidth to half.

26. A cordless telephone that uses separate frequencies for transmission in base and portable units is called _________
a) half duplex
b) duplex
c) simplex
d) one-way communication
Answer: b
Explanation: In duplex communication, two-way interaction is favourable simultaneously. Thus, a cordless telephone is duplex which uses separate frequencies for transmission in base and portable units.

27. Which polarization is used to reduce the depolarization effect on received waves?
a) Circular polarization
b) Linear polarization
c) Atomic polarization
d) Dipolar polarization
Answer: a
Explanation: In circular polarization at each point the electric field of electromagnetic wave has a constant magnitude but its direction changes as it rotates with time at a steady rate, in a plane perpendicular to the direction of propagation of wave. It is used to reduce depolarization effect on received waves.

28. Circular polarization involves critical alignment between transmitting and receiving antenna.
a) True
b) False
Answer: b
Explanation: In circular polarization at each point the electric field of the electromagnetic wave has a constant magnitude but its direction changes as it rotates with time at a steady rate, in a plane perpendicular to the direction of propagation of the wave. It is used to reduce depolarization effect on received waves. It does not involve alignment between transmitting and receiving antenna.

29. It is only the reflected color that decided the color of an object.
a) True
b) False
Answer: b
Explanation: Color of any object is decided by the reflected color for opaque object and wavelength transmitted through it for transparent object, while both reflector color and wavelength transmitted are considered for a translucent object.

30. What do you understand by the term “carrier” in modulation?
a) voltage to be transmitted
b) resultant wave
c) voltage for which amplitude, phase or frequency can be varied
d) voltage for which amplitude, phase or frequency remains constant
Answer: c
Explanation: Carrier wave is the wave with frequency higher than the message signal, whose certain characteristics like amplitude, phase or frequency are varied with respect to the instantaneous amplitude of the message signal. Thus forming the modulated wave which is the wave to be transmitted.

31. Carrier wave in modulation is a resultant wave.
a) True
b) False
Answer: b
Explanation: Carrier wave is the wave with frequency higher than the message signal, whose certain characteristics like amplitude, phase or frequency are varied with respect to the instantaneous amplitude of the message signal. Thus forming the modulated wave which is the wave to be transmitted.

32. For a low level AM system, amplifier modulated stage must have _________
a) harmonic devices
b) linear devices
c) non-linear devices
d) class A amplifiers
Answer: b
Explanation: In low-level modulation, the generation of amplitude modulated signal takes place at low power levels. The generated AM signal is then amplified using a chain of linear amplifiers, which are required to avoid waveform distortion. Thus, linear devices are used in low level amplitude modulated system.

33. Demodulation is done in ________
a) Channel
b) Receiver
c) Receiving antenna
d) Transducer
Answer: b
Explanation: Demodulation refers to extracting the original message signal from a transmitted modulated wave. The extraction of the message signal is generally carried out in the receiver. Channel is the medium through which the modulated message signal is transferred and Antenna receives the transmitted signal. Transducer converts the electrical signal to sound waves and vice-versa.

34. What is Fidelity?
a) Equally amplifies all the signal frequencies at receiver
b) Ability of receiver to select wanted signal from various incoming signal
c) Minimum magnitude of input signal required to produced a specified output
d) Ability to amplify weak signals
Answer: a
Explanation: Fidelity is the ability of the receiver to reproduce all modulating signals, equally, without any distortion. The ability of receiver to select wanted signal from various incoming signals is called Selectivity while Sensitivity is the minimum magnitude of input signal required to produce a specified output. It is the ability to amplify weak signals.

35. In a receiver, noise is usually developed at ________
a) Audio stage
b) Receiving antenna
c) RF stage
d) IF stage
Answer: c
Explanation: Ability of receiver to selected only wanted signal, and reject other frequencies, out of the various incoming signals, helps the receiver to operate more efficiently. However, at times, the RF amplifier allows a frequency lying close to the desired frequency, to pass to the next stage. This other frequency is undesired and later on is responsible for production of image frequency. Thus, noise is usually developed at RF stage.

36. Which oscillator is used as a local oscillator in radio receiver?
a) Wien-bridge
b) Hartley
c) Crystal
d) Phase Shift
Answer: b
Explanation: Oscillator which is used as a local oscillator in radio receiver is generally a tuned circuit. This tuned circuit consists of inductors and capacitors to determine the resonant frequency, therefore it is an LC tuned circuit. Out of the four options, only Hartley Oscillator has an LC resonant tank circuit.

37. Process of recovering information signal from received carrier is known as ________
a) Sensitivity
b) Selectivity
c) Demodulation
d) Fidelity
Answer: c
Explanation: Demodulation means extracting information or message signal from the transmitted modulated wave, while minimum magnitude of input signal required to produced a specified output is known as Sensitivity. The ability of receiver to select wanted signal from various incoming signals is called Selectivity. Fidelity means reproducing all modulating frequencies equally, without any distortion.

37. What is the use of a varactor diode in radio receiver?
a) Demodulation
b) Mixing
c) Multiplexing
d) Tuning
Answer: d
Explanation: Varactor diode is a diode working in the reverse-bias because of which no current flows through it. It has variable capacitance which varies with applied voltage. Varactor diodes are mainly used in Voltage Controlled Oscillators (VCOs) and RF Filters for tuning the receiver to the incoming signal or different stations.

38. What is the function of radio receiver?
a) to detect and amplify information signal from the carrier
b) to modulate a message signal
c) to produce radio waves
d) to convert one form of energy into other
Answer: a
Explanation: Receiver is to detect and amplify information signal from the carrier. Transmitter is used to modulate message signal and produce radio waves. Transducer is used to convert one signal form of energy into another form.

39. Figure of merit is ________
a) Ratio of output signal to noise ratio to input signal to noise ratio
b) Ratio of input signal to noise ratio to output signal to noise ratio
c) Ratio of output signal to input signal to a system
d) Ratio of input signal to output signal to a system
Answer: a
Explanation: Figure of merit is a numerical quantity based on the characteristic of system that represents a measure of efficiency or effectiveness. It is defined as the ratio of output signal to noise ratio to input signal to noise ratio.

39. Superheterodyne principle provides selectivity at ________
a) RF stage
b) IF stage
c) Demodulating Stage
d) Audio Stage
Answer: b
Explanation: A superheterodyne receiver uses frequency mixing to convert the received high frequency signal to a fixed lower intermediate frequency (IF), which can be processed more conveniently than original received frequency. Thus, the principle of selectivity is applied at the IF stage as it consists of very efficient filters to only select a wanted signal and pass it to the Demodulating Stage.

40. A heterodyne frequency changer is ________
a) Mixer
b) Demodulator
c) Modulator
d) Local Oscillator
Answer: a
Explanation: A mixer is a nonlinear electrical circuit that multiplies two signal frequencies applied to it, and produces a new frequency. Mixers are widely used to shift signals from one frequency range to other, which is known as heterodyning process. Generally, Local Oscillator generates a frequency to be applied at one of the input terminals of the mixer. Demodulator decodes the message signal from modulated signal, while modulator encodes message signal for transmission.

41. What is the bandwidth required in SSB signal?
a) f_m
b) 2f_m
c) > 2f_m
d) < 2f_m
Answer: a
Explanation: In an AM modulated system, total bandwidth required is from f_c + f_m to f_c – f_m i.e. bandwidth is equal to 2f_m. In SSB-SC transmission, the carrier and one of the sideband gets suppressed, so the bandwidth becomes f_m only.

42. One of the advantage of using a high frequency carrier wave is that it dissipates very small power.
a) True
b) False
Answer: a
Explanation: The main advantage of using high frequency signals is that the signal gets transmitted over very long distances and thus dissipates very less power. The antenna height required for transmission also gets reduced at high frequencies. And also it allows less noise interference and enables multiplexing. This is the reason for sending the audio signals at high frequency carrier signals for communication purpose.

43. What is the function of RF mixer?
a) Addition of two signals
b) Multiplication of two signals
c) Subtraction of two signals
d) To reduce the amount of noise
Answer: b
Explanation: RF mixer translates the frequencies of the two incoming signals by multiplying them and bringing them to a suitable band which can be processed.

44. The antenna current is 10A. Find the percentage of modulation when the antenna current increases to 10.4A?
a) 50%
b) 30%
c) 28.5%
d) 23%
Answer: c
Explanation:
 which gives m = 0.285 or 28.5%.

45. Find the total power, if the carrier of an AM transmitter is 800W and it is modulated to 50%?
a) 100W
b) 800W
c) 500W
d) 900W
Answer: d
Explanation: P_T=P_C (1+^u2⁄₂), according to the problem P_C = 800W and m = 0.5. On substituting values in the equation we get P_T=800(1+ ^0.52⁄₂) = 900W.

46. Aliasing refers to ________
a) Sampling of signals less than at Nyquist rate
b) Sampling of signals at Nyquist rate
c) Sampling of signals greater than at Nyquist rate
d) Unsampled the original signal
Answer: a
Explanation: Aliasing refers to the sampling of signals less than at Nyquist rate. Nyquist rate states that the rate of sampling of signals should be greater than or equal to twice the bandwidth of modulating signal. It gets reduced if sampling is done at a higher rate than nyquist rate of sampling. Aliasing can be avoided by using anti-aliasing filters.

47. Baseband compression produces ________
a) a small range of frequencies from low to high
b) a small range of different phases
c) a small range of angles
d) a small range of amplitude
Answer: d
Explanation: A signal compression method in a wireless network provides efficient transfer of compressed signal samples over serial data links in the system. Baseband compression produces a small range of amplitude.

48. Automatic Level Control (ALC) is used to keep the modulation index close to 100%.
a) True
b) False
Answer: a
Explanation: ALC stands for Automatic Level Control. It is a technology which is used for automatic control of output power. It helps in maintaining the output when there are varying changes in the input.

49. A signal that ________ must have linear power amplifier.
a) is complex
b) has variable frequency
c) is linear
d) has variable amplitude
Answer: d
Explanation: If any signal has variable amplitude then its amplifier must be linear. Others may or may not have the same but can possess non-linear amplifiers.

50. Transmitters are designed usually to derive a load impedance of ________
a) 50 ohms resistive
b) 150 ohms resistive
c) 250 ohms resistive
d) 500 ohms resistive
Answer: a
Explanation: Transmitter is an electronic device that produces radio waves with an antenna. The transmitter itself generates a radio frequency alternating current, which is applied to the antenna. Transmitters are usually designed to derive a load impedance of 50 ohms resistive.

51. What we called a resistor if a transmitter is connected to a resistor instead of an antenna?
a) a test load
b) a temporary load
c) a dummy load
d) a heavy load
Answer: c
Explanation: If a transmitter is connected to resistor not antenna than it is called dummy load. Such a load is used for testing purposes to set the parameters of the transmitter, as it would have behaved in presence of an actual antenna.

52. A class D amplifiers is very efficient than other amplifiers.
a) True
b) False
Answer: a
Explanation: Class D amplifier is also known as a switching amplifier. It is operate as electronic switches, and not an electric gain device which is commonly used in most amplifiers. It also has high power conversion efficiency unlike other amplifiers.

53. The carrier is suppressed in ________
a) a mixer
b) a frequency multiplier
c) a transducer
d) a balance modulator
Answer: d
Explanation: A mixer is the one which mixes the audio frequency with the carrier frequency. A transducer converts a signal from one form to another. Balance modulator suppresses the carrier and leaves only the sidebands.

54. What is the full form of AFC?
a) Amplitude to frequency conversion
b) Automatic frequency conversion
c) Automatic frequency control
d) Audio frequency control
Answer: c
Explanation: AFC stands for Automatic frequency control. It is a method to automatically keep a resonant circuit tuned to a frequency of an incoming radio signal. It is used in receivers to tune to the desired frequency.

55. Mixing is used in communication to ________
a) raise the carrier frequency
b) lower the carrier frequency
c) to altered the deviation
d) to change the carrier frequency to any required value
Answer: d
Explanation: Mixing is used to change the frequency of carrier by mixing it with a radio frequency signal or audio signal. The frequency can be changed to any required value in communication.

56. On which factor the bandwidth required for a modulated carrier depends?
a) baseband frequency range
b) signal to noise ratio
c) carrier frequency
d) amplitude of carrier frequency
Answer: a
Explanation: Bandwidth can be seen as a range of frequencies within a band, that is used for transmitting a signal. A signal bandwidth depends on the baseband frequency range.

57. Super heterodyne receivers needs an extra circuitry for frequency conversion.
a) True
b) False
Answer: a
Explanation: Super heterodyne receiver mixes the incoming signal frequency with the locally generated signal frequency, using a mixer, in order to convert the incoming RF signal to a low frequency signal which can be processed easily. It has better sensitivity and selectivity but it needs an extra circuitry for frequency conversion.

58. Neper is ________ decibel.
a) smaller than
b) larger than
c) equal to
d) exactly twice of
Answer: b
Explanation: Neper is a unit used to express ratios, such as gain, loss and relative values. One neper is equal to 8.685 dB. Hence it is greater than dB.

59. The coupling used in stoneman transmission bridge is ________
a) Inductive
b) Resistive
c) Capacitive
d) Combination of inductive and capacitive
Answer: c
Explanation: Stoneman transmission bridge is used in telephone communication to separate voice path from signal path and battery. Capacitive coupling is used in it.

60. Which statement is true about Square Law modulators?
a) it is used for frequency modulation
b) it is used for pulse width modulation
c) it is used for amplitude modulation
d) it is used for phase modulation
Answer: c
Explanation: In amplitude modulation, the amplitude of the carrier wave is varied with respect to the instantaneous amplitude of the wave to be transmitted. Square Law modulators are generally used for generation of amplitude modulation. They have nonlinear current-voltage characteristics. They are highly nonlinear in low voltage region.

61. Which statement is true about ring modulator?
a) it is used for DSB-SC generation
b) it is used for SSB-SC generation
c) it is a summation modulator
d) it is used for AM generation

Answer: a
Explanation: Ring Modulator is used for generating DSB-SC waves. It is a product modulator having four diodes connected in the form of a ring, which suppresses the carrier and allows only the two sidebands to be passed.

62. In ergodic process, ensemble and time averages are identical.
a) False
b) True
Answer: b
Explanation: In ergodic process, ensemble and time averages are identical to each other.

63. What is the amount of information in continuous signal?
a) zero
b) infinite
c) 2 bits
d) 4 bits
Answer: b
Explanation: In any continuous signal, the amount of information that comes out is always infinite.

64. Audio frequency range lies between _________
a) 2 MHz to 20 MHz
b) 20 Hz to 20 KHz
c) 20 KHz to 200 KHz
d) 20 MHz to 200 MHz
Answer: b
Explanation: As per standards audio frequency range lies in between 20 Hz to 20 KHz. This is the human audible range.

65. The image frequency of a superheterodyne receiver ___________
a) is created within the receiver itself
b) is due to insufficient adjacent channel rejection
c) is not rejected by the IF tuned circuits
d) is independent of the frequency to which the receiver is tuned
Answer: c
Explanation: To eliminate the image frequency of a superheterodyne receiver, adequate attenuation on the incoming signal by the RF amplifier filter is needed. It is not rejected by the IF tuned circuits and thus, it produces image frequency.

66. Which one of the following circuits could not demodulate SSB?
a) Phase Shift Method
b) Filter Method
c) Weaver Demodulator
d) Phase discriminator
Answer: d
Explanation: Phase discriminator cannot be used to demodulate SSB. It is used to discriminate between different phases.

67. Which of the following statement is true about a radio detector?
a) the linearity is worse than in phase discriminator
b) there is stabilization against signal strength variations
c) the final output is twice of that obtainable from a similar phase discriminator
d) the circuit is same as in a discriminator, except that the diodes are reversed.
Answer: a
Explanation: Detector, also known as a demodulator, in a radio receiver extracts the original information that is contained in a modulated radio wave. The linearity of a radio detector is worse than in phase discriminator. However, the radio detector does not get affected by amplitude variations and minimizes the limiting required.

68. To prevent overloading of the IF amplifier in a receiver, one should use _________
a) squelch
b) variable selectivity
c) double conversion
d) variable sensitivity
Answer: a
Explanation: Squelch is a circuit that is preferably used to prevent overloading of the IF amplifier in a receiver.

69. In a broadcast superheterodyne receiver, the ____________
a) Local oscillator always operates below the frequency of transmitted signal
b) Input of mixer must be tuned to the signal frequency
c) Local oscillator frequency should be double than that of IF
d) RF amplifier normally works at 455 KHz above the carrier frequency
Answer: b
Explanation: For a superheterodyne receiver, one of the inputs of the mixer must be tuned to the signal frequency and another input is of the local oscillator frequency. Tuning means that the circuit works for that one signal frequency only.

70. Which of the following statement is false regarding Armstrong modulation system?
a) The system is basically phase modulated and not frequency modulated
b) Automatic Frequency Control (AFC) is not needed as a crystal oscillator is used
c) Frequency multiplication must be used
d) Equalization is necessary
Answer: d
Explanation: The Armstrong method generates a double sideband carrier signal, phase shifts this signal and then reinserts the carrier to produce a frequency modulated signal. In Armstrong modulation system equalization is not necessary.

71. If the target cross section is changing, the best system for accurate tracking is __________
a) sequential switching
b) lobe switching
c) conical scanning
d) monopulse radar
Answer: d
Explanation: Monopulse radar compares the received signal from a single radar pulse against itself in order to compare the signal in various aspects. If the target cross section is changing then it is the best system used for accurate tracking.

72. After a target has been acquired, the best scanning system for tracking is _________
a) circular
b) spiral
c) conical
d) helical
Answer: c
Explanation: Conical is the best scanning system for tracking. But it is only preferred for tracking after a target has been acquired.

73. Which layer is used for detection of bad frames in wide area network?
a) frame layer
b) link layer
c) physical layer
d) session layer
Answer: b
Explanation: Wide area network covers a long geographical area and thus it also has least connected devices. Link layer is used in WAN for detection of bad frames.

74. TEM stands for Transmitted Electromagnetic wave.
a) True
b) False
Answer: b
Explanation: TEM stands for Transverse Electromagnetic wave. Since both electric and magnetic field are transverse to the direction of wave propagation.

75. What is the reason of attenuation in free space?
a) decrease in energy per square meter due to expansion
b) decrease in energy per square meter due to compression
c) losses in upper atmosphere due to expansion
d) losses in upper atmosphere due to compression
Answer: a
Explanation: In free space, the basic reason of attenuation is the decrease in energy per square meter. It generally happens due to expansion.

76. What is the range of UHF?
a) below the microwave range
b) above the microwave range
c) same as the microwave range
d) it does not include the microwave range in any way
Answer: a
Explanation: Ultra high frequency is the ITU designation for radio frequencies that lies under 300MHz to 3 GHz i.e. its range is below the microwave range.

77. Striplines and microstrips are used to __________
a) couple sections of waveguide
b) couple waveguides to antennas
c) couple components on a circuit board
d) couple electromagnetic waves in an order
Answer: c
Explanation: The main use of Striplines and microstrips in any circuit is to couple components on a circuit board.

78. What is the another term used for a single microwave link?
a) section
b) hop
c) link
d) skip
Answer: b
Explanation: Microwave link is a communication system that uses a beam of radio waves in the microwave frequency range. Hop is also used in places of single microwave link.

79. In digital microwave systems, additional repeaters increase the ________
a) selectivity
b) noise level
c) jitter
d) sensitivity
Answer: c
Explanation: In telecommunication, jitter can be seen as a deviation from true periodicity of a presumably periodic signal. Higher level of jitter are more likely to occur on either slow or heavily congested link. In any circuit, adding additional repeaters finally increase the jitter only.

80. What is the full form of SDLC?
a) Synchronous Data Link Character
b) Synchronous Data Line Control
c) Synchronous Data Line Character
d) Synchronous Data Link Control
Answer: d
Explanation: SDLC stands for Synchronous Data Link Control. It is a layer 2 protocol for IBM’s Systems Network Architecture. It supports multipoint links as well as error-correction. It helps in the proper flow of data to different levels.