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 positive frequencies where Y(f) has spectral peaks are

This question was previously asked in

ISRO Scientist EC 2017 (May) Official Paper

Option 2 : 2 KHz and 24 KHz

ISRO Scientist ECE: 2020 Official Paper

414

80 Questions
240 Marks
90 Mins

**Concept:**

\(x\left( t \right)\cos {\omega _c}t\;\;\mathop \leftrightarrow \limits^{\;\;F.T\;\;} \;\frac{1}{2}\left[ {x\left( {F - {F_c}} \right) + x\left( {F + {F_c}} \right)} \right]\;\)

OR

\(\frac{1}{2}\left[ {x\left( {\omega - {\omega _c}} \right) + x\left( {\omega + {\omega _c}} \right)} \right]\)

**Calculation:**

Let,

x_{1}(t) = x(t) ⋅ cos (2π × 10 × 10^{3} t)

\({x_1}\left( F \right) = \frac{1}{2}\left[ {x\left( {F - 10} \right) + x\left( {F + 10} \right)} \right]\)

Now, again

y(t) = x_{2}(t) ⋅ cos [2π 13 × 10^{3} t]

\(y\left( f \right) = \frac{1}{2}\left[ {{x_2}\left( {F - {F_c}} \right) + {x_2}\left( {F + {F_c}} \right)} \right]\)

Y(F) has spectral peaks at positive frequency 2 kHz and 24 kHz.