Topic: Communication Systems
Consider a system shown in the figure. Let $X(f)$ and $Y(f)$ denote the Fourier transforms of $x(t)$ and $y(t)$ respectively. The ideal HPF has the cut-off frequency 10 kHz.
(The figure shows $x(t)$ passing through a balanced modulator with 10 kHz carrier, then a HPF with 10 kHz cut-off, then another balanced modulator with 13 kHz carrier to produce $y(t)$). The positive frequencies where $Y(f)$ has spectral peaks are
- A. 1 kHz and 24 kHz
- B. 2 kHz and 24 kHz
- C. 1 kHz and 14 kHz
- D. 2 kHz and 14 kHz
Answer: B. 2 kHz and 24 kHz
Explanation: Peaks of $X(f)$ are at 1 kHz and -1 kHz initially.
Output of the first balanced modulator (DSB-SC with $f_{c1}=10\text{ kHz}$) has peaks at $f_{c1} \pm 1\text{ kHz} = 11\text{ kHz}$ and $9\text{ kHz}$.
The ideal HPF with cut-off 10 kHz will pass only the 11 kHz component.
The second balanced modulator (DSB-SC with $f_{c2}=13\text{ kHz}$) will shift this 11 kHz component to $f_{c2} \pm 11\text{ kHz}$.
Positive frequencies = $13\text{ kHz} + 11\text{ kHz} = 24\text{ kHz}$ and $13\text{ kHz} - 11\text{ kHz} = 2\text{ kHz}$.