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GATE EC 2017 Official Paper: Shift 2

Option 3 : band-pass filter

__ ____Bandpass ____Filter__:

This filter passes a certain band of frequencies and blocks low and high frequencies.

It has two corner frequencies, smaller (fc1) and larger (fc2).

It attenuates signals with frequencies below fc1 and above fc2.

**Calculation:**

x(n) = 5δ (n) – 7δ (n - 1) + 7δ (n - 3) – 5δ (n - 4)

Since δ(n) ↔ 1

δ(n - to) ↔ e-jωto

Taking the Fourier transform of x[n], we get:

H(e^{jω }) = 5 – 7e^{-jω } + 7e^{-3jω } – 5e^{-4jω}

= 5(1 – e^{-4jω}) + 7(e^{-j3ω} – e^{-jω})

= 5 e^{-j2ω} (e^{j2ω} – e^{-j2ω}) – 7e^{-j2ω} (e^{jω} – e^{-jω })

= 5e^{-j2ω} (2j sin (2ω)) – 7e^{-j2ω} (2j sin ω )

= e^{-j2ω} [10j sin 2ω – 14j sin ω]

When ω = 0; |H(e^{jω})| = 0 (Low frequency)

When \({\rm{\omega }} = \frac{\pi }{2},\left| {H\left( {{e^{j{\rm{\omega }}}}} \right)} \right| = 14\) (medium frequency)

When \({\rm{\omega }} = \pi ;\left| {H\left( {{e^{j{\rm{\omega }}}}} \right)} \right| = 0\) (High frequency)

So, the given filter is aCT 1: Ratio and Proportion

3941

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