Consider a sampled signal y(t)=5 × 10-6 x(t) Σn=-∞∞ δ(t-nTs)...

 

Question 35

Consider a sampled signal

y(t)=5 × 10^-6 x(t) Σ_n=-∞^∞ δ(t-nT_s)

where

x(t)=10cos(8π×10³t)

and

T_s=100 µs

When y(t) is passed through an ideal low-pass filter having a cutoff frequency of 5 kHz, the output is

1. 5 × 10^-6 cos(8π×10³t)
2. 5 × 10^-5 cos(8π×10³t)
3. 5 × 10^-1 cos(8π×10³t)
4. 10 cos(8π×10³t)

Solution

Step 1: Find the signal frequency

Since

2πf = 8π×10³

Therefore,

f = (8π×10³)/(2π) =4 kHz

Step 2: Find the sampling frequency

f_s=1/T_s

=1/(100×10^-6) =10000 Hz =10 kHz

Step 3: Check Nyquist condition

Nyquist Frequency=f_s/2=5 kHz

Since

4 kHz < 5 kHz

there is no aliasing.

Step 4: Reconstruction amplitude

The sampled signal is

y(t)=5×10^-6x(t) Σδ(t-nT_s)

The impulse train contributes a factor

1/T_s=10000

Hence,

5×10^-6×10000 =5×10^-2

Original signal amplitude = 10

10×5×10^-2 =0.5 =5×10^-1

Final Output

y(t)=5×10^-1 cos(8π×10³t)

Correct Answer: (c)

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