Question:
A signal \( x(t) \) is band-limited between 100 Hz and 200 Hz. A signal \( y(t) \) is related to \( x(t) \) as follows:
y(t) = x(2t - 5)
The statement that is always true is_________
A signal \( x(t) \) is band-limited between 100 Hz and 200 Hz. A signal \( y(t) \) is related to \( x(t) \) as follows:
y(t) = x(2t - 5)
The statement that is always true is_________
y(t) = x(2t - 5)
The statement that is always true is_________
Show Hint
When a signal is scaled in time, its frequency range is inversely scaled by the same factor. A time scaling by a factor of 2 doubles the frequency range.
Updated On: Nov 25, 2025
- ( y(t) \) is band-limited between 50 Hz and 100 Hz
- ( y(t) \) is band-limited between 100 Hz and 200 Hz
- ( y(t) \) is band-limited between 200 Hz and 400 Hz
- ( y(t) \) is not band-limited
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understanding the effect of time scaling.
When a signal \( x(t) \) is scaled in time, i.e., replaced by \( x(at + b) \), the frequency spectrum of the signal is scaled by a factor of \( 1/|a| \). In this case, the signal \( y(t) = x(2t - 5) \) corresponds to a time scaling by a factor of 2. This scaling halves the period of the signal, which doubles its frequency range.
Step 2: Analyzing the options.
- (A) Incorrect, time scaling by a factor of 2 causes the bandwidth to increase, not decrease.
- (B) Incorrect, the frequency range will not remain between 100 Hz and 200 Hz.
- (C) Correct, scaling by 2 increases the frequency range from 100 Hz-200 Hz to 200 Hz-400 Hz.
- (D) Incorrect, \( y(t) \) is still band-limited, just with a different frequency range. Step 3: Conclusion.
The correct answer is (C) \( y(t) \) is band-limited between 200 Hz and 400 Hz.
When a signal \( x(t) \) is scaled in time, i.e., replaced by \( x(at + b) \), the frequency spectrum of the signal is scaled by a factor of \( 1/|a| \). In this case, the signal \( y(t) = x(2t - 5) \) corresponds to a time scaling by a factor of 2. This scaling halves the period of the signal, which doubles its frequency range.
Step 2: Analyzing the options.
- (A) Incorrect, time scaling by a factor of 2 causes the bandwidth to increase, not decrease.
- (B) Incorrect, the frequency range will not remain between 100 Hz and 200 Hz.
- (C) Correct, scaling by 2 increases the frequency range from 100 Hz-200 Hz to 200 Hz-400 Hz.
- (D) Incorrect, \( y(t) \) is still band-limited, just with a different frequency range. Step 3: Conclusion.
The correct answer is (C) \( y(t) \) is band-limited between 200 Hz and 400 Hz.
Was this answer helpful?
0
0
Top Questions on DFT and FFT
Questions Asked in GATE IN exam
- Consider the function \( f(z) = \frac{1}{(z+1)(z+2)(z+3)} \). The residue of \( f(z) \) at \( z = -1 \) is _________.
- GATE IN - 2022
- Residue Theorem
- In the circuit shown, the load is driven by a sinusoidal ac voltage source \( V_1 = 100 \angle 0^\circ \, \text{V} \) at 50 Hz. Given \( R_1 = 20 \, \Omega \), \( C_1 = \left(\frac{1000}{\pi}\right) \, \mu\text{F} \), \( L_1 = \left(\frac{20}{\pi}\right) \, \text{mH} \), and \( R_2 = 4 \, \Omega \), the power factor is _________ (round off to one decimal place)
- GATE IN - 2022
- Transient analysis of RLC circuits with ac excitation
- In the circuit shown, the capacitance \( C_0 = 10 \, \mu\text{F} \) and inductance \( L_0 = 1 \, \text{mH} \) and the diode is ideal. The capacitor is initially charged to 10 V and the current in the inductor is initially zero. If the switch is closed at \( t = 0 \, \text{s} \), the voltage \( V_C(t) \) (in volts) across the capacitor at \( t = 0.5 \, \text{s} \) is _________ (round off to one decimal place)
- GATE IN - 2022
- Diodes: Characteristics and Applications
- A 1 kHz sine-wave generator having an internal resistance of 50Ω generates an open-circuit voltage of 10 Vp-p. When a capacitor is connected across the output terminals, the voltage drops to 8 Vp-p. The capacitance of the capacitor (in microfarads) is _________.
- GATE IN - 2022
- Characteristics of ideal and practical operational amplifiers
- In the Wheatstone bridge circuit shown, \( R_1 = 1.5 \, k\Omega \) and \( R_2 = R_3 = R_4 = 1 \, k\Omega \). The switch is initially open and the voltage between the points C and D is \( V_{CD} \). Upon closing the switch at \( t = 0 \), the resistance in the arm AD changes by an amount \( \delta R_1 \), and the voltage between C and D changes by \( \delta V_{CD} \). The sensitivity of the bridge in volt/kilohm, defined as \( \left| \frac{\delta V_{CD}}{\delta R_1} \right| \), is _________.
- GATE IN - 2022
- Bridges and Potentiometers
View More Questions