Digital Filters Design MCQ [Free PDF] – Objective Question Answer for Digital Filters Design Quiz

When the input sequence x(n) to the filter becomes zero, the output of the filter then, after a number of iterations, enters into the limit cycle.

The output remains in the limit cycle until another input of sufficient size is applied that drives the system out of the limit cycle. Similarly, zero input limit cycles occur from non-zero initial conditions with the input x(n)=0.

We note that when the response of the single-pole filter is in the limit cycle, the actual non-linear system acts as an equivalent linear system with a pole at z=1 when the pole is positive and z=-1 when the pole is negative.

Answer: C

Since the quantization product av(n-1) is obtained by rounding, it follows that the quantization error is bounded as

|Qr[av(n-1)]-av(n-1)| ≤ \((\frac{1}{2}).2^{-b}\)

Where ‘b’ is the number of bits (exclusive of sign) used in the representation of the pole ‘a’ and v(n). Consequently, we get

|v(n-1)|-|av(n-1)| ≤ \((\frac{1}{2}).2^{-b}\)

and hence

|v(n-1)| ≤ \(\frac{(\frac{1}{2}).2^{-b}}{1-|a|}\)

We know that

|v(n-1)| ≤ \(\frac{(\frac{1}{2}).2^{-b}}{1-|a|}\)

Given |a|=1/2 and b=4 => |v(n-1)| ≤ 1/16=>

The dead band of a single-pole filter with a pole at 1/2 and represented by 4 bits is (-1/16,1/16).

There is a possible limit cycle mode with zero input, which occurs as a result of rounding the multiplications, corresponding to an equivalent second-order system with poles at z=±1.

In this case, the two-pole filter exhibits oscillations with an amplitude that falls in the dead band bounded by 2-b/(1-|a1|-a2).

It can be easily shown that a necessary and sufficient condition for ensuring that no zero-input overflow limit cycles occur is |a1|+|a2|<1
which is extremely restrictive and hence an unreasonable constraint to impose on any second-order section.

An effective remedy for curing the problem of overflow oscillations is to modify the adder characteristic so that it performs saturation arithmetic. Thus when an overflow is sensed, the output of the adder will be the full-scale value of ±1.

We know that

|v(n-1)| ≤ \(\frac{(\frac{1}{2}).2^{-b}}{1-|a|}\)

Given |a|=3/4 and b=4 => |v(n-1)| ≤ 1/8=>

The dead band is (-1/8,1/8).

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

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.

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

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

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.

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.

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.

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).

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.

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.

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

U(ω)= πδ(ω)+0.5-j0.5cot(ω/2).

If the HI(ω) is uniquely determined from HR(ω) through the integral relationship. This integral is called a discrete Hilbert transform.

Answer: A

Causality has very important implications in the design of frequency-selective filters. One among them is the magnitude |H(ω)| cannot be constant in any finite range of frequencies and the transition from pass-band to stop-band cannot be infinitely sharp.

This is a consequence of the Gibbs phenomenon, which results from the truncation of h(n) to achieve causality.

Passband edge ripple is the frequency at which the passband starts transiting to the stopband.

If ωP and ωS represent the passband edge ripple and stopband edge ripple, then the transition width -ωP+ ωS gives the bandwidth of the filter.

An FIR filter of length M with input x(n) and output y(n) is described by the difference equation

y(n)=\(\sum_{k=0}^{M-1} b_k x(n-k)\)

where

{bk} is the set of filter coefficients.

We can express the output sequence as the convolution of the unit sample response h(n) of the system with the input signal. The lower and upper limits on the convolution sum reflect the causality and finite duration characteristics of the filter.

An FIR filter has an linear phase if its unit sample response satisfies the condition

h(n)= ±h(M-1-n) n=0,1,2…M-1.