Test: Backward Difference Method - Electrical Engineering (EE) MCQ

Explanation: The analog filter in the time domain is governed by the following difference equation,

Explanation: A simple approximation to the first order derivative is given by the first backward difference. The first backward difference is defined by
[y(n)-y(n-1)]/T.

Explanation: We know that s=(1-z^-1)/T=> z=1/(1-sT).

Explanation: Letting s=σ+jΩ in the equation z=1/(1-sT) and by letting σ=0, we get
|z-0.5|=0.5
Thus the image of the jΩ axis of the s-domain is a circle with centre at z=0.5 in z-domain

Explanation: Letting s=σ+jΩ in the equation z=1/(1-sT) and by letting σ=0, we get
|z-0.5|=0.5
Thus the image of the jΩ axis of the s-domain is a circle of radius 0.5 centered at z=0.5 in z-domain.

Explanation: Since jΩ axis is not mapped to the circle |z|=1, we can expect that the frequency response H(ω) will be considerably distorted with respect to H(jΩ).

Explanation: The left half of the s-plane is mapped inside the circle of |z-0.5|=0.5 in the z-plane, which completely lies in the right half z-plane.

Explanation: An analog high pass filter cannot be mapped to a digital high pass filter because the poles of the digital filter cannot lie in the correct region, which is the left-half of the z-plane(z < 0) in this case.

Explanation: We know that z=1/(1-sT)=> s=(1-z^-1)/T.

Explanation: The first backward difference of y(n) is given by the equation
[y(n)-y(n-1)]/T
Thus the z-transform of the first backward difference of y(n) is given as