Signals & Systems Questions and Answers – Fourier Series Analysis using Circuits

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This set of Signals & Systems written test Questions & Answers focuses on “Fourier Series Analysis using Circuits”.

1. Given x (t) = [2 + e-3t] u (t). The final value of x(t) is ___________
a) 2
b) 3
c) e-3t
d) 0
View Answer

Answer: a
Explanation: Given x (t) = [2 + e-3t] u (t)
Applying final value theorem, we get, limt→∞⁡ x(t)= lims→0 sX(s)
= lims→0 \(s[\frac{1}{s} + \frac{1}{s+3}]\)
= lims→0⁡ \(\frac{2s+3}{s+3}\)
= 2.
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2. The Fourier series of the given signal is _______________
signals-systems-written-test-questions-answers-q2
a) -4/π sin x
b) 4/π sin x
c) 4/π cos x
d) -4/π cos x
View Answer

Answer: b
Explanation: We know that the Fourier series is given by a0 + \(∑_{n=1}^∞\) (an cos nx + bn sin⁡ nx)
This can be also written as a0 + a1cos x + b1 sin x + a2 cos 2x + b2 sin 2x ……
Here, c0 = a0 = 0
So, cn = bn = \(2∫_0^t f(t).sin⁡ωt \,dt\)
∴ c1 = b1 = \(2∫_0^t sin⁡ωt \,dt\)
= \(– \frac{2}{ω} [cos⁡ωt]_0^t\)
= \(-\frac{2}{π}\)[-1-1] = 4/π
And a1 = \(2∫_0^t f(t).cos⁡ωt \,dt\)
= \(\frac{2}{ω}[sin⁡ωt]_0^t\)
= 0
So, the Fourier series can be written as 0 + 0.cos x + 4/π sin x
= 4/π sin x.

3. For the circuit given below, the effective inductance of the circuit across the terminal AB, is ___________
signals-systems-written-test-questions-answers-q3
a) 9 H
b) 21 H
c) 11 H
d) 6 H
View Answer

Answer: c
Explanation: LEFF across AB = L1 + L2 + L3 – 2M12 – 2M13 + 2M23
= (4 + 5 + 6 – 2×1 – 2×3 + 2×2)
= (15 – 2 – 6 + 4)
= (15 – 8 + 4)
= (15 – 4)
= 11.

4. For the circuit given below, the inductance measured across the terminals 1 and 2 was 15 H with open terminals 3 and 4. It was 30 H when terminals 3 and 4 were short-circuited. Both the inductors are having inductances 2H. The coefficient of coupling is ______________
signals-systems-written-test-questions-answers-q4
a) 1
b) 0.707
c) 0.5
d) Inderminate due to insufficient data
View Answer

Answer: d
Explanation: When 2 coils are connected in series, then effective inductance,
LEFF = L1 + L2 ± 2M
For this case, LEFF = L1 + L2 – 2M
However, L1 + L2 + 2M or L1 + L2 – 2M cannot be determined.
Hence, data is insufficient to calculate the value of coupling coefficient (k).

5. In the circuit given below, the resonant frequency is ________________
signals-systems-written-test-questions-answers-q5
a) \(\frac{1}{2π\sqrt{3}}\) Hz
b) \(\frac{1}{4π\sqrt{3}}\) Hz
c) \(\frac{1}{4π\sqrt{2}}\) Hz
d) \(\frac{1}{π\sqrt{2}}\) Hz
View Answer

Answer: b
Explanation: LEFF = L1 + L2 + 2M
= 2 + 2 + 2 × 1
Or, LEFF = 6 H
At resonance, ωL = \(\frac{1}{ωC}\)
Or, ω = \(\frac{1}{\sqrt{L_{EFF} C}} = \frac{1}{\sqrt{12}}\)
Or, I = \(\frac{1}{4π\sqrt{3}}\) Hz.
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6. Q meter operator is the principle of __________
a) Series resonance
b) Current resonance
c) Self-inductance
d) Eddy currents
View Answer

Answer: a
Explanation: We know that, Q = \(\frac{ωL}{R}\)
From the above relation, we can say that it works on series resonance.

7. The probability cumulative distribution function must be monotone as well as ___________
a) Increasing
b) Decreasing
c) Non-increasing
d) Non-decreasing
View Answer

Answer: d
Explanation: The probability cumulative distribution function increases to 1 monotonically and there after remains constant. Therefore we can say that the probability cumulative distribution function must be monotone as well as non-decreasing.

8. The impedance seen by the source in the circuit is given by __________
signals-systems-written-test-questions-answers-q8
a) (0.54 + j0.313) Ω
b) (4 – j2) Ω
c) (4.54 – j1.69) Ω
d) (4 + j2) Ω
View Answer

Answer: c
Explanation: Z1 = 10∠30° × \((\frac{1}{4})^2\)
Z1 = (0.54 + j0.31) Ω
Total impedance= (4 – j2) + (0.56 + j0.31)
= (4.54 – j1.69) Ω.

9. In the circuit given below, the capacitor is initially having a charge of 10 C. 1 second after the switch is closed, the current in the circuit is ________
signals-systems-written-test-questions-answers-q9
a) 14.7 A
b) 18.5 A
c) 40.0 A
d) 50.0 A
View Answer

Answer: a
Explanation: Using KVL, 100 = \(R\frac{dq}{dt} + \frac{q}{C}\)
100 C = RC\(\frac{dq}{dt}\) + q
Or, \(\int_{q_o}^q \frac{dq}{100C-q} = \frac{1}{RC} \int_0^t \,dt\)
100C – q = (100C – qo)e-t/RC
I = \(\frac{dq}{dt} = \frac{100C – q_o}{RC} e^{-1/1}\)
∴ e-t/RC = 40e-1 = 14.7 A.
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10. Convolution is used to find ____________
a) Impulse of an LTI system
b) Frequency response of an LTI system
c) Time response of an LTI system
d) Phase response of an LTI system
View Answer

Answer: c
Explanation: We know that, the time response of an LTI system is given by
Y (n) = \(∑_{k=-∞}^∞ x(k)h(n-k)\)
This can also be written as y (n) = x (n) * h (n), which is the convolution of two systems. Therefore convolution is used to find the time response of an LTI system.

11. The Fourier transform of a rectangular function is ___________
a) Another rectangular pulse
b) Triangular pulse
c) Sinc function
d) Impulse
View Answer

Answer: c
Explanation: Substituting the square pulse function f (t) in the below equation as shown,
F (jω) = \(\int_{-∞}^∞ f(t) e^{jωt} \,dt\)
This gives us the Sinc function.

12. The property of Fourier Transform which states that the compression in time domain is equivalent to expansion in the frequency domain is ___________
a) Duality
b) Scaling
c) Time scaling
d) Frequency shifting
View Answer

Answer: b
Explanation: We know that the definition of Fourier Transform states that Fourier Transform is a function derived from a given function and representing it by a series of sinusoidal functions. The property of Fourier Transform which states that the compression in the time domain is equivalent to expansion in the frequency domain is Scaling can be found by putting the value of pulse function in the definition of Fourier Transform.

13. A coil of inductance 10 H, resistance 40 Ω is connected as shown in the figure. After the switch S has been in connection with point 1 for a very long time, it is moved to point 2 at t=0. For the value of R obtained, the time taken for 95 % of the stored energy dissipated is close to?
signals-systems-written-test-questions-answers-q13
a) 0.10 s
b) 0.15 s
c) 0.50 s
d) 1.0 s
View Answer

Answer: c
Explanation: For source free circuit,
I (t) = Io \(e^{-\frac{R}{T} \,t}\)
∴ I (t) = 0.05 = 2 × \(e^{-\frac{60}{10} \,t}\)
Or, t = 0.61 ≈ 0.5 s.
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14. The function which has its Fourier transform, Laplace transform, and Z transform unity is __________
a) Gaussian
b) Impulse
c) Sinc
d) Pulse
View Answer

Answer: b
Explanation: The definition of Fourier transform is \(\int_{-∞}^∞ f(x) e^{2πjxf} \,dx\)
The definition of Laplace transform is \(\int_0^∞ f(t) e^{-st} \,dt\)
The definition of Z-transform is \(∑_{n=-∞}^∞ x[n]z^{-n}\)
Substituting f (t) = δ (t) in the definitions of Fourier Laplace and Z-transform, we get the transforms in each case as 1. Hence an impulse function has unity Fourier Laplace and Z-transform.

15. N(t) = 2, 0≤t<4;
t2, t≥4;
The Laplace transform of M (t) is ___________
a) \(\frac{2}{s} – e^{-4s} (\frac{2}{s^3} – \frac{8}{s^2} – \frac{14}{s})\)
b) \(\frac{2}{s} + e^{-4s} (\frac{2}{s^3} – \frac{8}{s^2} – \frac{14}{s})\)
c) \(\frac{2}{s} – e^{-4s} (\frac{2}{s^3} + \frac{8}{s^2} + \frac{14}{s})\)
d) \(\frac{2}{s} + e^{-4s} (\frac{2}{s^3} + \frac{8}{s^2} + \frac{14}{s})\)
View Answer

Answer: d
Explanation: N (t) = 2 + u (t) (t2 – 2)
L {2 + u (t) (t2-2)} = \(\frac{2}{s}\) + L {u (t) (t2-2)}
= \(\frac{2}{s} + e^{-4s}\) L {(t+4)2 – 2}
= \(\frac{2}{s} + e^{-4s}\) L {t2 + 8t + 14}
= \(\frac{2}{s} + e^{-4s} (\frac{2}{s^3} + \frac{8}{s^2} + \frac{14}{s})\).

Sanfoundry Global Education & Learning Series – Signals & Systems.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn