This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Applications of Signals on Circuits”.

1. A circuit tuned to a frequency of 1.5 MHz and having an effective capacitance of 150 pF. In this circuit, the current falls to 70.7 % of its resonant value. The frequency deviates from the resonant frequency by 5 kHz. Q factor is?

a) 50

b) 100

c) 150

d) 200

View Answer

Explanation: \(Q = \frac{ω}{ω1 – ω2} = \frac{f}{f2-f1} \)

Here, f = 1.5 × 10

^{6}Hz

f1 = (1.5 × 10

^{6}– 5 × 10

^{3})

f2 = (1.5 × 10

^{6}+ 5 × 10

^{3})

So, f2 – f1 = 10 × 10

^{3}Hz

\(∴ Q = \frac{1.5 × 10^6}{10 × 10^3}\) = 150.

2. A 440 V, 50 HZ AC source supplies a series LCR circuit with a capacitor and a coil. If the coil has 100 mΩ resistance and 15 mH inductance, then at a resonance frequency of 50 Hz the half power frequencies of the circuit are ______________

a) 50.53 Hz, 49.57 Hz

b) 52.12 HZ, 49.8 Hz

c) 55.02 Hz, 48.95 Hz

d) 50 HZ, 49 Hz

View Answer

Explanation: Bandwidth, BW = \(\frac{f_o}{Q} = \frac{50}{47.115}\) = 1.061 Hz

f

_{2}, higher half power frequency = f

_{0}+ \(\frac{BW}{2}\)

∴ \(f_2 = 50 + \frac{1.061}{2}\) = 50.53 Hz

f

_{1}, lower half power frequency = f

_{0}– \(\frac{BW}{2}\)

∴ f

_{1}= 100 – \(\frac{1.59}{2}\) = 49.47 Hz.

3. The even component of the signal Y (t) = a^{jt} is _________________

a) Sin t

b) Cos t

c) Sinh t

d) Cosh t

View Answer

Explanation: Let Y

_{a}(t) represents the even component of Y (t)

Now, Y

_{a}(t) = \(\frac{1}{2}\)[Y (t) + Y (-t)]

= \(\frac{1}{2}\)[a

^{jt}+ a

^{-jt}]

= cos t.

4. The odd component of the signal Y (t) = a^{jt} is _______________

a) Sin t

b) Cos t

c) Sinh t

d) Cosh t

View Answer

Explanation: Let Y

_{o}(t) represents the odd component of Y (t)

Now, Y

_{o}(t) = \(\frac{1}{2}\)[Y (t) – Y (-t)]

= \(\frac{1}{2}\)[a

^{jt}+ a

^{-jt}]

= sin t.

5. The signal Y (t) = e^{-2t} u (t) is _______________

a) Power signal with P_{∞} = \(\frac{1}{4}\)

b) Power signal with P_{∞} = \(\frac{1}{2}\)

c) Energy signal with E_{∞} = \(\frac{1}{4}\)

d) Energy signal with E_{∞} = 0

View Answer

Explanation: If a signal has E∞ as ∞ and P∞ as a finite value, then the signal is a power signal. If a signal has E∞ as a finite value and P

_{∞}as ∞, then the signal is an energy signal.

|Y (t)| < ∞, E

_{∞}= \(\int_{-∞}^∞ |y(t)|^2 \,dt\)

= \(\int_∞^∞ e^{-2t} u(t) \,dt \)

= \(\in_∞^∞ e^{-2t} \,dt = \frac{1}{2}\)

So, this is not a power signal but an energy signal.

\(P_∞ = lim_{T→∞} \frac{1}{2T} \int_{-T}^T |y(t)|^2 \,dt = ∞.\)

6. The signal X (t) = \(e^{j(2t + \frac{π}{4})}\) is ________________

a) Energy signal with E_{∞} = 2

b) Power signal with P_{∞} = 2

c) Power signal with P_{∞} = 1

d) Energy signal with E_{∞} = 1

View Answer

Explanation: If a signal has E

_{∞}as ∞ and P

_{∞}as a finite value, then the signal is a power signal. If a signal has E

_{∞}as a finite value and P

_{∞}as ∞, then the signal is an energy signal.

|x (t)| = 1, E

_{∞}= \(\int_{-∞}^∞ |x(t)|^2 \,dt = ∞\)

So, this is a power signal not an energy signal.

\(P_∞ = lim_{T→∞} \frac{1}{2T} \int_{-T}^T |x(t)|^2 \,dt = 1.\).

7. Given the signal

Y (t) = cos t, if t>0

Sin t, if t≥0

The correct statement among the following is?

a) Periodic with fundamental period 2π

b) Non-periodic and discontinuous

c) Periodic but with no fundamental period

d) Non-periodic but continuous

View Answer

Explanation: From the graphs of cos and sin, we can infer that at t=0, the function becomes discontinuous.

Since, cos 0 = 1, but sin 0 = 0

As 1 ≠ 0, so, the function X (t) is discontinuous and therefore Non-periodic.

8. Two series resonant filters are shown below. Let the cut-off bandwidth of filter 1 be B_{1} and that of filter 2 be B_{2}. The value of \(\frac{B_1}{B_2}\) is ____________

a) 0.25

b) 1

c) 0.5

d) 0.75

View Answer

Explanation: For series resonant circuit, 3dB bandwidth is \(\frac{R}{L}\)

B

_{1}= \(\frac{R}{L_1}\)

B

_{2}= \(\frac{R}{L_2} = \frac{4R}{L_1}\)

Hence, \(\frac{B_1}{B_2}\) = 0.25.

9. In a series RLC circuit for lower frequency and for higher frequency, power factors are respectively ______________

a) Leading, Lagging

b) Lagging, Leading

c) Independent of Frequency

d) Same in both cases

View Answer

Explanation: A Leading power factor means that the current in the circuit leads the applied voltage. This condition occurs in capacitive circuits. On the other hand, a lagging power factor indicates that current lags the voltage and this condition happens in an inductive circuit.

10. A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 50. If R and L are doubled and C is kept same, the new Q of the circuit is ____________

a) 25.52

b) 35.35

c) 45.45

d) 20.02

View Answer

Explanation: Quality factor Q of the series RLC circuit is given by, \(Q = \frac{1}{R} \sqrt{\frac{L}{C}}\)

Given that Q = 50

Q

_{new}= \(\frac{1}{2R} \sqrt{\frac{2L}{C}}\)

= \(\frac{1}{2} × \frac{1}{R} \sqrt{\frac{2L}{C}}\)

= \(\frac{1}{2} × \sqrt{2} × Q \)

= \(\frac{1}{2} × \sqrt{2}\) × 50 = 35.35.

**Sanfoundry Global Education & Learning Series – Signals & Systems.**

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