A superheterodyne receiver has intermediate frequency 15 MHz and local oscillator frequency 3.5 GHz. The received signal is above the oscillator frequency. The image frequency in MHz is ________.
Both a signal above and a signal below the oscillator by the same intermediate frequency produce the same mixer difference frequency.
The desired signal is 3500+15=3515 MHz. Its image lies on the other side, at 3500−15=**3485 MHz**. It is separated from the desired signal by twice the intermediate frequency, 30 MHz.
More Analog Communications questions
GATE 2026 EC Q57– Consider a real, narrowband signal x(t)=A(t)[2 f ct+(t)] where the maximum frequency…
GATE 2025 EC Q14– Consider a frequency-modulated (FM) signal f(t)=A c(2 f ct+3(2 f 1t)+4(6 f 1t)), where A c…
GATE 2025 EC Q28– Consider a message signal m(t) which is bandlimited to [-W,W], where W is in Hz. Consider…
GATE 2024 EC Q28– An amplitude modulator has output (in Volts) s(t)=A(400 t)+B(360 t)+B(440 t). The carrier…
GATE 2023 EC Q27– Let m(t) be a strictly band-limited signal with bandwidth B and energy E. Assuming 0=10B,…
GATE 2023 EC Q45– Let a frequency modulated (FM) signal x(t)=A( ct+k f -tm()d), where m(t) is a message…