FIR Differentiator Design MCQ [Free PDF] – Objective Question Answer for FIR Differentiator Design Quiz

In the design of FIR differentiators based on the Chebyshev approximation criterion, the weighting function W(ω) is specified in the program as
W(ω)=1/ω
in order that the relative ripple in the pass band is a constant.

We know that the weighting function is
W(ω)=1/ω
in order that the relative ripple in the passband is a constant. Thus, the absolute error between the desired response ω and the approximation Hr(ω) increases as ω varies from 0 to 2πfp.

The important parameters in a differentiator are its length, its bandwidth, and the peak relative error of the approximation. The interrelationship among these three parameters can be easily displayed parametrically.

In view of the fact that the ideal differentiator has an anti-symmetric unit sample response, we shall confine our attention to FIR designs in which h(n)=-h(M-1-n).

Designs based on M odd are particularly poor if the bandwidth exceeds 0.45. The problem is basically the zero in the frequency response at ω=π(f=1/2). When fp < 0.45, good designs are obtained for M odd.

An ideal Hilbert transformer is an all-pass filter.

An ideal Hilbert transformer is an all-pass filter that imparts a 90° phase shift on the signal at its input.

The frequency response of an ideal Hilbert transform is given as
H(ω) = -j ;0 ≤ ω ≤ π
H(ω) = j ;-π ≤ ω ≤ 0

Hilbert transforms are frequently used in communication systems and signal processing, as, for example, in the generation of SSB modulated signals, radar signal processing, and speech signal processing.

We know that the frequency response of an ideal Hilbert transformer is given as
H(ω)= -j ;0 < ω < π
j ;-π < ω < 0

Thus the unit sample response of an ideal Hilbert transform is obtained as

h(n)=\(\frac{2}{\pi} \frac{(sin(\frac{πn}{2}))^{2}}{n}\); n≠0

h(n)=0; n=0

We know that the unit sample response of the Hilbert transform is given as

h(n)=\(\frac{2}{\pi} \frac{(sin(\frac{πn}{2}))^{2}}{n}\); n≠0

h(n)=0; n=0

it sample response of an ideal Hilbert transform is infinite in duration and non-causal.

We know that the unit sample response of the Hilbert transform is given as

h(n)=\(\frac{2}{\pi} \frac{(sin(\frac{πn}{2}))^{2}}{n}\); n≠0

h(n)=0; n=0

Thus from the above equation, we can tell that h(n)=-h(-n). Thus the unit sample response of the Hilbert transform is anti-symmetric in nature.

In view of the fact that the ideal Hilbert transformer has an anti-symmetric unit sample response, we shall confine our attention to FIR designs in which h(n)=-h(M-1-n).

Our choice of an anti-symmetric unit sample response is consistent with having a purely imaginary frequency response characteristic.

We know that when h(n) is anti-symmetric, the real-valued frequency response characteristic is zero at ω=0 for both M odd and even and at ω=π when M is odd. Clearly, then, it is impossible to design an all-pass digital Hilbert transformer.

The bandwidth of the Hilbert transformer need only cover the bandwidth of the signal to be phase shifted. Consequently, we specify the desired real-valued frequency response of a Hilbert transformer filter is
H(ω)=1; 2π fl < ω < 2πfu
where fl and fu are the cutoff frequencies.

The unit sample response of the Hilbert transformer is given as

h(n)=\(\frac{2}{\pi} \frac{(sin(\frac{πn}{2}))^{2}}{n}\); n≠0

h(n)=0; n=0
From the above equation, it is clear that h(n) becomes zero for even values of ‘n’.