Signals And Systems Set 3

Electrical Engineering

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This set of Signals and Systems Multiple Choice Questions & Answers (MCQs) focuses on Signals And Systems Set 3

Q1 | The fourier series coefficients of the signal are carried from –T/2 to T/2.
  • true
  • false
Answer: true
Explaination:
yes, the coefficients evaluation can be done from –t/2 to t/2. it is done for the simplification of the signal.
Q2 | What is the polar form of the fourier series?
  • x(t) = c0 + ∑cncos(nwt+ϕn)
  • x(t) = c0 + ∑cncos(ϕn)
  • x(t) = ∑cncos(nwt+ϕn)
  • x(t) = c0+ ∑cos(nwt+ϕn)
Answer: x(t) = c0 + ∑cncos(nwt+ϕn)
Explaination:
x(t) = c0 + ∑cncos(nwt+ϕn), is the polar form of the fourier series.
Q3 | What is a line spectrum?
  • plot showing magnitudes of waveforms are called line spectrum
  • plot showing each of harmonic amplitudes in the wave is called line spectrum
  • plot showing each of harmonic amplitudes in the wave is called line spectrum
  • plot showing each of harmonic amplitudes called line spectrum
Answer: plot showing each of harmonic amplitudes in the wave is called line spectrum
Answer: plot showing each of harmonic amplitudes in the wave is called line spectrum
Explaination:
the plot showing each of harmonic amplitudes in the wave is called line spectrum. the line rapidly decreases for waves with rapidly convergent series.
Q4 | Fourier series is not true in case of discrete time signals.
  • true
  • false
Answer: false
Explaination:
fourier series is also true in
Q5 | What is the disadvantage of exponential Fourier series?
  • it is tough to calculate
  • it is not easily visualized
  • it cannot be easily visualized as sinusoids
  • it is hard for manipulation
Answer: it cannot be easily visualized as sinusoids
Explaination:
the major disadvantage of exponential fourier series is that it cannot be easily visualized as sinusoids. moreover, it is easier to calculate and easy for manipulation leave aside the disadvantage.
Q6 | Fourier series uses which domain representation of signals?
  • time domain representation
  • frequency domain representation
  • both combined
  • neither depends on the situation
Answer: frequency domain representation
Explaination:
fourier series uses frequency domain representation of signals.
Q7 | How does Fourier series make it easier to represent periodic signals?
  • harmonically related
  • periodically related
  • sinusoidally related
  • exponentially related
Answer: harmonically related
Explaination:
fourier series makes it easier to represent periodic signals as it is a mathematical tool that allows the representation of any periodic signals as the sum of harmonically related sinusoids.
Q8 | Which of the following is correct regarding to impulse signal?
  • x[n]δ[n] = x[0]δ[n]
  • x[n]δ[n] = δ[n]
  • x[n]δ[n] = x[n] d) x[n]δ[n] = x[0]
Answer: x[n]δ[n] = x[0]δ[n]
Explaination:
when the input x[n] is multiplied with an impulse signal, the result will be impulse signal with magnitude of x[n] at that time.
Q9 | Weighted superposition of time-shifted impulse responses is termed as                for discrete-time signals.
  • convolution integral
  • convolution multiple
  • convolution sum
  • convolution
Answer: convolution sum
Explaination:
weighted superposition of time-shifted impulse responses is called
Q10 | Find the convolution sum of sequences x1[n] = (1, 2, 3) and x2[n] = (2, 1, 4).
  • {2, 5, 12, 11, 12}
  • {2, 12, 5, 11, 12}
  • {2, 11, 5, 12, 12}
  • {-2, 5,-12, 11, 12}
Answer: {2, 5, 12, 11, 12}
Explaination:
x1[n] = δ(n)+2δ(n-1)+3δ(n-2) and x2[n] = 2δ(n)+δ(n-1)+4δ(n-2)
Q11 | The Fourier series for the function f (x) = sin2x is                              
  • 0.5 + 0.5 sin 2x
  • 0.5 – 0.5 sin 2x
  • 0.5 + 0.5 cos 2x
  • 0.5 – 0.5 cos 2x
Answer: 0.5 – 0.5 cos 2x
Explaination:
f(x) = sin2x now, f(x) = sin2x =  1−cos2x
Q12 | The type of systems which are characterized by input and the output capable of taking any value in a particular set of values are called as                      
  • analog
  • discrete
  • digital
  • continuous
Answer: continuous
Explaination:
we know that continuous systems have a restriction on the basis of the upper bound and lower bound. however within this set, the input and output can assume any value. hence, there are infinite values attainable in this system.
Q13 | Then, y[k] = x[3k-2] is                              
  • y[k] = 1, for k = 0, 1 and 0 otherwise
  • y[k] = 1, for k = 1 and -1 for k=-1
  • y[k] = 1, for k = 0, 1 and -1 otherwise
  • y[k] = 1, for k = 0, 1 and 0 otherwise
Answer: y[k] = 1, for k = 0, 1 and 0 otherwise
Answer: y[k] = 1, for k = 0, 1 and 0 otherwise
Explaination:
y[k] = x [3k-2] now, y [0] = x [-2] = 1
Q14 | A Discrete signal is said to be even or symmetric if X(-n) is equal to                      
  • x(n)
  • 0
  • –x(n)
  • –x(-n)
Answer: x(n)
Explaination:
we know that any signal be it discrete or continuous is said to be even or
Q15 | Impulse response is the output of               system due to impulse input applied at time=0?
  • linear
  • time varying
  • time invariant
  • linear and time invariant
Answer: linear and time invariant
Explaination:
impulse response is the output of lti system due to impulse input applied at time = 0 or n=0. behaviour of an lti system is characterised by the impulse response.
Q16 | For a causal L.T.I. system, the impulse response is 0 for                    
  • t<0
  • t=0
  • t>0
  • always
Answer: t<0
Explaination:
since the system is causal, the impulse response won’t exist for t<0. this is because the output of the system should not depend on future inputs. thus only option t<0 is correct.
Q17 | The convolution of a discrete signal with itself is                    
  • squaring the signal
  • doubling the signal
  • adding two signals
  • is not possible
Answer: squaring the signal
Explaination:
this is proved by the fact that since discrete signals can be thought of as a one variable polynomial with the coefficients, along with the order, representing the amplitude, at an instant equal to the order of the variable, of the signal- they are simply multiplied during convolution.
Q18 | The convolution of a function with an impulse function delayed to an instant 3 in time results in                          
  • an advance in the function by 3 units in time
  • the function itself
  • a delay in the function by 3 units in time
  • cannot be determined
Answer: a delay in the function by 3 units in time
Explaination:
the convolution of an impulse function with a function results in the function itself. but if the impulse function is delayed, the output will also get delayed by
Q19 | A continuous signal can be represented as the product of an impulse function and the signal itself.
  • true
  • false
Answer: true
Explaination:
the continuous signal can be represented as an integral of impulses. this representation buries it down to the form of convolution of two signal where one signal is the impulse function while the other is the continuous signal. hence, the above statement is true.
Q20 | What is the Scope value if the signal generator has a frequency of 2 Hz only?
  • an attenuated signal of same frequency
  • the entire signal at the same frequency
  • the entire signal at reduced frequency
  • an attenuated signal at reduced frequency
Answer: an attenuated signal of same frequency
Explaination:
the above transfer function is that of a high pass filter. the cut-off frequency for allowing signals is 3hz but the given signal frequency is that of 2hz only. if the frequency of the signal generator was more than 2hz- the output would’ve been entire signal at the same frequency but for
Q21 | If two LTI systems with impulse response h1 (t) and h2 (t) and are connected in parallel then output is given by              
  • y(t) = x(t) *(h1(t) + h2(t))
  • y(t) = x(t) + (h1(t) + h2(t))
  • y(t) = x(t) * (h1(t) h2(t))
  • y(t) = (x(t) * h1(t)) + h2(t)
Answer: y(t) = x(t) *(h1(t) + h2(t))
Explaination:
the equivalent impulse response of two systems connected in parallel is the sum of individual impulse responses. it is represented as
Q22 | When two LTI systems with impulse responses ha (t) and hb (t) are cascaded then equivalent response is given by              
  • h(t) = ha(t) + hb(t)
  • h(t) = ha(t) – hb(t)
  • h(t) = ha(t) hb(t)
  • h(t) = ha(t) * hb(t)
Answer: h(t) = ha(t) * hb(t)
Explaination:
the equivalent impulse response of two systems connected in series (cascaded) is given by convolution of individual impulse responses.
Q23 | The condition for memory-less system is given by            
  • h[k] = cδ[k]
  • h[k] = cδ[n-k]
  • h[k] = ch[k]δ[k]
  • h[k] = ch[n-k]δ[k]
Answer: h[k] = cδ[k]
Explaination:
the lti discrete-time system is
Q24 | The causal continuous system with impulse response should satisfy          equation.
  • h(t)=0,t<0
  • h(t)=0,t>0
  • h(t)≠0,t<0
  • h(t)≠0,t≤0
Answer: h(t)=0,t<0
Explaination:
to the continuous system to be causal, the impulse response should satisfy the equation h(t)=0,t<0 and convolution
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