Test: Optimum Equiripple Filter Design - 2 - Electronics and Communication Engineering (ECE) MCQ

Explanation: If the filter has a symmetric unit sample response, then we know that
h(n)=h(M-1-n)
and for M odd in this case, Q(ω)=1.

Explanation: If the filter has a anti-symmetric unit sample response, then we know that
h(n)= -h(M-1-n)
and for M odd in this case, Q(ω)=sin(ω).

Explanation: The real valued desired frequency response Hdr(ω) is simply defined to be unity in the pass band and zero in the stop band.

Explanation: According to Alternation theorem, a necessary and sufficient condition for Pω) to be unique, best weighted chebyshev approximation, is that the error function E(ω) must exhibit at least L+2 extremal frequencies in S.

Explanation: In general, the filter designs that contain maximum number of alternations or ripples are called as maximal ripple filters.

Explanation: Initially, we neither know the set of external frequencies nor the parameters. To solve for the parameters, we use an iterative algorithm called the Remez exchange algorithm, in which we begin by guessing at the set of extremal frequencies.

Explanation: |E(ω)|≥δ for some frequencies on the dense set, then a new set of frequencies corresponding to the L+2 largest peaks of |E(ω)| are selected and computation is repeated. Since the new set of L+2 extremal frequencies are selected to increase in each iteration until it converges to the upper bound, this implies that when |E(ω)|≤δ for all frequencies on the dense set, the optimal solution has been found in terms of the polynomial H(ω).

Explanation: FX denotes an array of maximum size 10 that specifies the weight function in each band.

Explanation: The chebyshev approximation problem is viewed as an optimum design criterion on the sense that the weighted approximation error between the desired frequency response and the actual frequency response is spread evenly across the pass band and evenly across the stop band of the filter minimizing the maximum error. The resulting filter designs have ripples in both pass band and stop band.

Explanation: If the filter has a symmetric unit sample response, then we know that
h(n)=h(M-1-n)
and for M even in this case, Q(ω)= cos(ω/2).