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In an AM system receiver. the mixer translates the carrier frequency fc to a fixed intermediate frequency of 450 kHz. Given a broadcast signal of frequency 900 kHz. What is the corresponding image frequency when a superhetrodyne receiver is employed with local oscillator frequency greater than the broadcast signal frequency?
A signal containing only two frequency components (3 kHz and 6 kHz) is sampled at the rate of 8 kHz, and then passed through a low pass filter with a cut-off frequency of 8 kHz. The filter output
A signal contains components at 400 Hz and 2400 Hz This signal modulates a carrier of frequency 100 MHz. However, after demodulation, it is found that the 400 Hz signal component is present. The channel bandwidth is 15 kHz. What is the reason for the higher frequency signal not to be detected properly?
What is the bandwidth of the signal if the upper sideband frequency and the lower sideband frequency are 7350 KHz and 7210 KHz respectively?
Consider the following statements regarding modulation : 1. The communication systems, where a weak signal is transmitted by the use of a modulation is essential in carrier signal. 2. When the frequency of the high frequency signal is varied in accordance with the intensity of the low-frequency weak signal, the modulation is said to be frequency modulation. 3. The process of recovering original baseband signal from the modulated signal is called phase modulation. Which of the above statements are correct ? 
What is the upper sideband frequency if the carrier frequency and the modulating signal frequency is 30 KHz and 80 Hz respectively?
If in a broadcasting studio, a 1000 kHz carrier is modulated by an audio signal of frequency range 100-5000 kHz, the width of channel is __________ kHz.
Let x(t) = 2cos(800πt) + cos(1400πt).x(t) is sampled with the rectangular pulse train shown in the figure. The only spectral components (in kHz) present in the sampled signal in the frequency range 2.5 kHz to 3.5 kHz are
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A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(t). The resulting signal is passed through an ideal low pass filter with bandwidth 1 kHz. The output of the low pass filter would be :
A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(t) where
$$g\left( t \right) = \sum\limits_{k = - \infty }^\infty {{{\left( { - 1} \right)}^k}\delta \left( {t - 0.5 \times {{10}^{ - 4}}k} \right)} $$
The resulting signal is then passed through an ideal low pass filter with bandwidth 1 kHz. The output of the low pass filter would be