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QUESTION: 1

The ideal low pass filter cannot be realized in practice.

Solution:

Explanation: We know that the ideal low pass filter is non-causal. Hence, a ideal low pass filter cannot be realized in practice.

QUESTION: 2

The following diagram represents the unit sample response of which of the following filters?

Solution:

Explanation: At n=0, the equation for ideal low pass filter is given as h(n)=ω/π.

From the given figure, h(0)=0.25=>ω=π/4.

Thus the given figure represents the unit sample response of an ideal low pass filter at ω=π/4.

QUESTION: 3

If h(n) has finite energy and h(n)=0 for n<0, then which of the following are true?

Solution:

Explanation: If h(n) has finite energy and h(n)=0 for n<0, then according to the Paley-Wiener theorem, we have

QUESTION: 4

If |H(ω)| is square integrable and if the integralis finite, then the filter with the frequency response

Solution:

Explanation: If |H(ω)| is square integrable and if the integral is finite, then we can associate with |H(ω)| and a phase response θ(ω), so that the resulting filter with the frequency response H(ω)=|H(ω)|e^{jθ(ω)} is causal.

QUESTION: 5

The magnitude function |H(ω)| can be zero at some frequencies, but it cannot be zero over any finite band of frequencies.

Solution:

Explanation: One important conclusion that we made from the Paley-Wiener theorem is that the magnitude function |H(ω)| can be zero at some frequencies, but it cannot be zero over any finite band of frequencies, since the integral then becomes infinite. Consequently, any ideal filter is non-causal.

QUESTION: 6

If h(n) is causal and h(n)=he(n)+ho(n),then what is the expression for h(n) in terms of only h_{e}(n)?

Solution:

Explanation: Given h(n) is causal and h(n)= h_{e}(n)+h_{o}(n)

=>h_{e}(n)=1/2[h(n)+h(-n)] Now, if h(n) is causal, it is possible to recover h(n) from its even part he(n) for 0≤n≤∞ or from its odd component h_{o}(n) for 1≤n≤∞.

=>h(n)= 2h_{e}(n)u(n)-h_{e}(0)δ(n) ,n ≥ 0.

QUESTION: 7

If h(n) is causal and h(n)=he(n)+ho(n),then what is the expression for h(n) in terms of only h_{o}(n)?

Solution:

Explanation: Given h(n) is causal and h(n)= h_{e}(n)+h_{o}(n)

=>h_{e}(n)=1/2[h(n)+h(-n)] Now, if h(n) is causal, it is possible to recover h(n) from its even part h_{e}(n) for 0≤n≤∞ or from its odd component h_{o}(n) for 1≤n≤∞.

=>h(n)= 2h_{o}(n)u(n)+h(0)δ(n) ,n ≥ 1

since ho(n)=0 for n=0, we cannot recover h(0) from h_{o}(n) and hence we must also know h(0).

QUESTION: 8

If h(n) is absolutely summable, i.e., BIBO stable, then the equation for the frequency response H(ω) is given as?

Solution:

Explanation: . If h(n) is absolutely summable, i.e., BIBO stable, then the frequency response H(ω) exists and

H(ω)= H_{R}(ω)+j H_{I}(ω)

where H_{R}(ω) and H_{I}(ω) are the Fourier transforms of he(n) and ho(n) respectively.

QUESTION: 9

H_{R}(ω) and H_{I}(ω) are interdependent and cannot be specified independently when the system is causal.

Solution:

Explanation: Since h(n) is completely specified by he(n), it follows that H(ω) is completely determined if we know HR(ω). Alternatively, H(ω) is completely determined from HI(ω) and h(0). In short, H_{R}(ω) and H_{I}(ω) are interdependent and cannot be specified independently when the system is causal.

QUESTION: 10

The frequency ωP is called as:

Solution:

Explanation: Pass band edge ripple is the frequency at which the pass band starts transiting to the stop band.

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