Digital Signal Processing

General Considerations for Design of Digital Filters

Question 1
Marks : +2 | -2
Pass Ratio : 100%
If |H(ω)| is square integrable and if the integral \\(\\int_{-\\pi}^\\pi |ln⁡|H(ω)||dω\\) is finite, then the filter with the frequency response H(ω)=|H(ω)|ejθ(ω) is?
Anti-causal
Constant
Causal
None of the mentioned
Explanation:
If |H(ω)| is square integrable and if the integral \\(\\int_{-\\pi}^\\pi |ln⁡|H(ω)||dω\\) 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)|ejθ(ω) is causal.
Question 2
Marks : +2 | -2
Pass Ratio : 100%
What is the Fourier transform of the unit step function U(ω)?
πδ(ω)-0.5-j0.5cot(ω/2)
πδ(ω)-0.5+j0.5cot(ω/2)
πδ(ω)+0.5+j0.5cot(ω/2)
πδ(ω)+0.5-j0.5cot(ω/2)
Explanation:
Since the unit step function is not absolutely summable, it has a Fourier transform which is given by the equation
Question 3
Marks : +2 | -2
Pass Ratio : 100%
The following diagram represents the unit sample response of which of the following filters?
Ideal high pass filter
Ideal low pass filter
Ideal high pass filter at ω=π/4
Ideal low pass filter at ω=π/4
Explanation:
At n=0, the equation for ideal low pass filter is given as h(n)=ω/π.
Question 4
Marks : +2 | -2
Pass Ratio : 100%
If h(n) is causal and h(n)=he(n)+ho(n),then what is the expression for h(n) in terms of only ho(n)?
h(n)=2ho(n)u(n)+h(0)δ(n), n ≥ 0
h(n)=2ho(n)u(n)+h(0)δ(n), n ≥ 1
h(n)=2ho(n)u(n)-h(0)δ(n), n ≥ 1
h(n)=2ho(n)u(n)-h(0)δ(n), n ≥ 0
Explanation:
Given h(n) is causal and h(n)= he(n)+ho(n)
Question 5
Marks : +2 | -2
Pass Ratio : 100%
If h(n) is absolutely summable, i.e., BIBO stable, then the equation for the frequency response H(ω) is given as?
HI(ω)-j HR(ω)
HR(ω)-j HI(ω)
HR(ω)+j HI(ω)
HI(ω)+j HR(ω)
Explanation:
If h(n) is absolutely summable, i.e., BIBO stable, then the frequency response H(ω) exists and
Question 6
Marks : +2 | -2
Pass Ratio : 100%
The ideal low pass filter cannot be realized in practice.
True
False
Explanation:
We know that the ideal low pass filter is non-causal. Hence, a ideal low pass filter cannot be realized in practice.
Question 7
Marks : +2 | -2
Pass Ratio : 100%
HR(ω) and HI(ω) are interdependent and cannot be specified independently when the system is causal.
True
False
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, HR(ω) and HI(ω) are interdependent and cannot be specified independently when the system is causal.
Question 8
Marks : +2 | -2
Pass Ratio : 100%
The magnitude function |H(ω)| can be zero at some frequencies, but it cannot be zero over any finite band of frequencies.
True
False
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 9
Marks : +2 | -2
Pass Ratio : 100%
If h(n) has finite energy and h(n)=0 for n<0, then which of the following are true?
\\(\\int_{-π}^π|ln⁡ |H(ω)||dω \\gt -\\infty\\)
\\(\\int_{-π}^π|ln⁡ |H(ω)||dω \\lt \\infty\\)
\\(\\int_{-π}^π|ln⁡|H(ω)||dω = \\infty\\)
None of the mentioned
Explanation:
If h(n) has finite energy and h(n)=0 for n<0, then according to the Paley-Wiener theorem, we have
Question 10
Marks : +2 | -2
Pass Ratio : 100%
If h(n) is causal and h(n)=he(n)+ho(n),then what is the expression for h(n) in terms of only he(n)?
h(n)=2he(n)u(n)+he(0)δ(n), n ≥ 0
h(n)=2he(n)u(n)+he(0)δ(n), n ≥ 1
h(n)=2he(n)u(n)-he(0)δ(n), n ≥ 1
h(n)=2he(n)u(n)-he(0)δ(n), n ≥ 0
Explanation:
Given h(n) is causal and h(n)= he(n)+ho(n)