1. The ideal low pass filter cannot be realized in practice.

A. True
B. False

Answer: A

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

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

A. Ideal high pass filter
B. Ideal low pass filter
C. Ideal high pass filter at ω=π/4
D. Ideal low pass filter at ω=π/4

Answer: D

At n=0, the equation for the 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.

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

A. \(\int_{-π}^π|ln |H(ω)||dω \gt -\infty\)
B. \(\int_{-π}^π|ln |H(ω)||dω \lt \infty\)
C. \(\int_{-π}^π|ln|H(ω)||dω = \infty\)
D. None of the mentioned

Answer: B

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

\(\int_{-π}^π|ln |H(ω)||dω \lt \infty\)

4. 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?

A. Anti-causal
B. Constant
C. Causal
D. None of the mentioned

Answer: C

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.

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

A. True
B. False

Answer: A

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.

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 he(n)?

A. h(n)=2he(n)u(n)+he(0)δ(n), n ≥ 0
B. h(n)=2he(n)u(n)+he(0)δ(n), n ≥ 1
C. h(n)=2he(n)u(n)-he(0)δ(n), n ≥ 1
D. h(n)=2he(n)u(n)-he(0)δ(n), n ≥ 0

Answer: D

Given h(n) is causal and h(n)= he(n)+ho(n)

=>he(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 ho(n) for 1≤n≤∞.

=>h(n)= 2he(n)u(n)-he(0)δ(n), n ≥ 0.

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 ho(n)?

A. h(n)=2ho(n)u(n)+h(0)δ(n), n ≥ 0
B. h(n)=2ho(n)u(n)+h(0)δ(n), n ≥ 1
C. h(n)=2ho(n)u(n)-h(0)δ(n), n ≥ 1
D. h(n)=2ho(n)u(n)-h(0)δ(n), n ≥ 0

Answer: B

Given h(n) is causal and h(n)= he(n)+ho(n)

=>he(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 ho(n) for 1≤n≤∞.

=>h(n)= 2ho(n)u(n)+h(0)δ(n), n ≥ 1

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

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

A. HI(ω)-j HR(ω)
B. HR(ω)-j HI(ω)
C. HR(ω)+j HI(ω)
D. HI(ω)+j HR(ω)

Answer: C

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

H(ω)= HR(ω)+j HI(ω)

where HR(ω) and HI(ω) are the Fourier transforms of he(n) and ho(n) respectively.

9. HR(ω) and HI(ω) are interdependent and cannot be specified independently when the system is causal.

A. True
B. False

Answer: A

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.

10. What is the Fourier transform of the unit step function U(ω)?

A. πδ(ω)-0.5-j0.5cot(ω/2)
B. πδ(ω)-0.5+j0.5cot(ω/2)
C. πδ(ω)+0.5+j0.5cot(ω/2)
D. πδ(ω)+0.5-j0.5cot(ω/2)

Answer: D

Since the unit step function is not absolutely summable, it has a Fourier transform which is given by the equation