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The ideal low pass filter cannot be realized in practice.
Explanation: We know that the ideal low pass filter is noncausal. Hence, a ideal low pass filter cannot be realized in practice.
The following diagram represents the unit sample response of which of the following filters?
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.
If h(n) has finite energy and h(n)=0 for n<0, then which of the following are true?
Explanation: If h(n) has finite energy and h(n)=0 for n<0, then according to the PaleyWiener theorem, we have
If H(ω) is square integrable and if the integralis finite, then the filter with the frequency response
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.
The magnitude function H(ω) can be zero at some frequencies, but it cannot be zero over any finite band of frequencies.
Explanation: One important conclusion that we made from the PaleyWiener 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 noncausal.
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)?
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.
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)?
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).
If h(n) is absolutely summable, i.e., BIBO stable, then the equation for the frequency response H(ω) is given as?
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.
H_{R}(ω) and H_{I}(ω) are interdependent and cannot be specified independently when the system is causal.
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.
Explanation: Pass band edge ripple is the frequency at which the pass band starts transiting to the stop band.
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