# Signals & Systems Questions and Answers – Applications of Signals on Circuits

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

Explanation: $$Q = \frac{ω}{ω1 – ω2} = \frac{f}{f2-f1}$$
Here, f = 1.5 × 106 Hz
f1 = (1.5 × 106 – 5 × 103)
f2 = (1.5 × 106 + 5 × 103)
So, f2 – f1 = 10 × 103 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

Explanation: Bandwidth, BW = $$\frac{f_o}{Q} = \frac{50}{47.115}$$ = 1.061 Hz
f2, higher half power frequency = f0 + $$\frac{BW}{2}$$
∴ $$f_2 = 50 + \frac{1.061}{2}$$ = 50.53 Hz
f1, lower half power frequency = f0 – $$\frac{BW}{2}$$
∴ f1 = 100 – $$\frac{1.59}{2}$$ = 49.47 Hz.

3. The even component of the signal Y (t) = ajt is _________________
a) Sin t
b) Cos t
c) Sinh t
d) Cosh t

Explanation: Let Ya (t) represents the even component of Y (t)
Now, Ya (t) = $$\frac{1}{2}$$[Y (t) + Y (-t)]
= $$\frac{1}{2}$$[ajt + a-jt]
= cos t.

4. The odd component of the signal Y (t) = ajt is _______________
a) Sin t
b) Cos t
c) Sinh t
d) Cosh t

Explanation: Let Yo (t) represents the odd component of Y (t)
Now, Yo (t) = $$\frac{1}{2}$$[Y (t) – Y (-t)]
= $$\frac{1}{2}$$[ajt + 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

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

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

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 B1 and that of filter 2 be B2. The value of $$\frac{B_1}{B_2}$$ is ____________

a) 0.25
b) 1
c) 0.5
d) 0.75

Explanation: For series resonant circuit, 3dB bandwidth is $$\frac{R}{L}$$
B1 = $$\frac{R}{L_1}$$
B2 = $$\frac{R}{L_2} = \frac{4R}{L_1}$$
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 ______________
c) Independent of Frequency
d) Same in both cases

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

Explanation: Quality factor Q of the series RLC circuit is given by, $$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$$
Given that Q = 50
Qnew = $$\frac{1}{2R} \sqrt{\frac{2L}{C}}$$
= $$\frac{1}{2} × \frac{1}{R} \sqrt{\frac{2L}{C}}$$
= $$\frac{1}{2} × \sqrt{2} × Q$$
= $$\frac{1}{2} × \sqrt{2}$$ × 50 = 35.35.

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