Digital Electronic MCQ Number 00926

Answer: a [Reason:] We know that the ideal low pass filter is non-causal. Hence, a ideal low pass filter cannot be realized in practice.

Answer: d [Reason:] At n=0, the equation for 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.

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

Answer: c [Reason:] If |H(ω)| is square integrable and if the integral 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)|e^jθ(ω) is causal.

Answer: a [Reason:] 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.

Answer: d [Reason:] Given h(n) is causal and h(n)= h_e(n)+h_o(n)
=>h_e(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 h_o(n) for 1≤n≤∞.
=>h(n)= 2h_e(n)u(n)-h_e(0)δ(n) ,n ≥ 0.

Answer: b [Reason:] Given h(n) is causal and h(n)= h_e(n)+h_o(n)
=>h_e(n)=1/2[h(n)+h(-n)]
Now, if h(n) is causal, it is possible to recover h(n) from its even part h_e(n) for 0≤n≤∞ or from its odd component h_o(n) for 1≤n≤∞.
=>h(n)= 2h_o(n)u(n)+h(0)δ(n) ,n ≥ 1
since ho(n)=0 for n=0, we cannot recover h(0) from h_o(n) and hence we must also know h(0).

Answer: c [Reason:] . If h(n) is absolutely summable, i.e., BIBO stable, then the frequency response H(ω) exists and
H(ω)= H_R(ω)+j H_I(ω)
where H_R(ω) and H_I(ω) are the Fourier transforms of he(n) and ho(n) respectively.

Answer: a [Reason:] 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, H_R(ω) and H_I(ω) are interdependent and cannot be specified independently when the system is causal.

Answer: d [Reason:] 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).

Answer: b [Reason:] If the H_I(ω) is uniquely determined from HR(ω) through the integral relationship. This integral is called as discrete Hilbert transform.

Answer: a [Reason:] 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 Gibbs phenomenon, which results from the truncation of h(n) to achieve causality.

Answer: c [Reason:] Pass band edge ripple is the frequency at which the pass band starts transiting to the stop band.

Answer: b [Reason:] If ωP and ωS represents the pass band edge ripple and stop band edge ripple, then the transition width -ω_P+ ω_SS gives the bandwidth of the filter.

Answer: a [Reason:] Since the present output depends on the value of the previous output, the system is called as Recursive system.

Answer: c [Reason:] From the given equation, there are M+N delays, so it requires M+N number of delay elements and it has to perform M+N+1 multiplications, so it require that many number of multipliers.

Answer: d [Reason:] Since the output of the system depend only on the past values of output and the present value of the input, the system is called as “purely recursive” system.

Answer: a [Reason:] If the coefficients of the past values of the output in the difference equation of the system, then the system is said to be FIR system.

Answer: b [Reason:] Since the output of the system depends only on the present value of the input and the past values of the output, the system is a purely recursive system.

Answer: c [Reason:] The equation of the difference equation of any system is defined as

In the given diagram, N=M=2
So, substitute the values of the N and M in the above equation.
We get, y(n)= -a1y(n-1)-a2y(n-2)+ b0x(n)+b1x(n-1)+b2x(n-2)

Answer: d [Reason:] The output of the system according to the direct form given is
y(n)= b0x(n)+b1x(n-1)+b2x(n-2)
Since the output of the system is purely dependent on the present and past values of the input, the system is called as FIR system.

Answer: b [Reason:] For a system to be recursive, the output of the system must be dependent only on the past values of the output. For an FIR system the output of the system must be depending only on the present and past values of the input. So, FIR system is not an recursive system.

Answer: a [Reason:] A normal FIR non-recursive system with the impulse response h(n)= 1/(M+1) is the system which is used to compute the moving average of a signal x(n).

Answer: d [Reason:] The given system equation is
It can be expressed as follows

Answer: b [Reason:] Since the present output depends on the value of the future output, the system is not called as Recursive system.

Answer: a [Reason:] If we know the output of a system y(n) of a system and if we can determine the input x(n) of the system uniquely, then the system is said to be invertible. That is there should be one-to-one correspondence between the input and output signals.

Answer: c [Reason:] . If h(n) is the impulse response of an LTI system T and h1(n) is the impulse response of the inverse system T-1, then we know that h(n)*h1(n)=δ(n)=> [h(n)*h1(n)]*x(n)=x(n).

Answer: b [Reason:] Given impulse response is h(n)=(1/2)nu(n)
The system function corresponding to h(n) is
H(z)=1/(1-1/2 z^-1 ) ROC:|z|>1/2
This system is both stable and causal. Since H(z) is all pole system, its inverse is FIR and is given by the system function
HI(z)= 1- 1/2 z^-1
Hence its impulse response is δ(n)-1/2 δ(n-1).

Answer: c [Reason:] The system function of given system is H(z)= 1- 1/2 z^-1
The inverse of the system has a system function as H(z)= 1/(1-1/2 z^-1 )
Thus it has a zero at origin and a pole at z=1/2.So, two possible cases are |z|>1/2 and |z|<1/2
So, h(n)= (1/2)ⁿu(n) for causal and stable(|z|>1/2)
and h(n)= -(1/2)ⁿu(-n-1) for anti causal and unstable for |z|<1/2.

Answer: d [Reason:] Given h(n)= δ(n)-aδ(n-1)
Since h(0)=1, h(1)=-a and h(n)=0 for n≥a, we have
hI(0)=1/h(0)=1.
and
hI(n)=-ahI(n-1) for n≥1
Consequently, hI(1)=a, hI(2)=a²,….hI(n)=aⁿ
Which corresponds to a causal IIR system as expected.

Answer: a [Reason:] Given H(z)=6+z^-1-z^-2
By factoring the system function we find the zeros for the system.
The zeros of the given system are at z=-1/2,1/3
So, the system is minimum phase.

Answer: b [Reason:] Given H(z)= 1-z^-1-z^-2
By factoring the system function we find the zeros for the system.
The zeros of the given system are at z=-2,3
So, the system is maximum phase.

Answer: c [Reason:] Given H(z)= 1-5/2z^-1-3/2z^-2
By factoring the system function we find the zeros for the system.
The zeros of the given system are at z=-1/2, 3
So, the system is mixed phase.

Answer: a [Reason:] For an IIR filter whose system function is defined as H(z)=(B(z))/(A(z)) to be said a minimum phase,
then both the poles and zeros of the system should fall inside the unit circle.

Answer: d [Reason:] For an IIR filter whose system function is defined as H(z)=(B(z))/(A(z)) to be said a mixed phase and if the system is stable and causal, then the poles are inside the unit circle and some, but not all of the zeros are outside the unit circle.

Answer: b [Reason:] The system function is easily determined by taking the z-transforms of x(n) and y(n). Thus we have
H(z)=(Y(z))/(X(z)) = (1+0.7z^-1)/(1-0.7z^-1+0.1z^-2 ) = (1+0.7z^-1)/((1-0.2z^-1)(1-0.5z^-1))
Upon applying partial fractions and applying the inverse z-transform, we get
[4(0.5)ⁿ-3(0.2)ⁿ]u(n).

Answer: d [Reason:] All the methods mentioned above can be used to calculate the inverse z-transform of the given signal.

Answer: a [Reason:] Since the ROC is the exterior circle, we expect x(n) to be a causal signal. Thus we seek a power series expansion in negative powers of ‘z’. By dividing the numerator of X(z) by its denominator, we obtain the power series

So, we obtain x(n)= {1,3/2,7/4,15/8,31/16,….}.

Answer: b [Reason:] In this case the ROC is the interior of a circle. Consequently, the signal x(n) is anti causal. To obtain a power series expansion in positive powers of z, we perform the long division in the following way:

Thus
In this case x(n)=0 for n≥0.Thus we obtain x(n)= {…..62,30,14,6,2,0,0}

Answer: c [Reason:] Using the power series expansion for log(1+x), with |x|<1, we have

Answer: a [Reason:] First, we note that we should reduce the numerator so that the terms z-2 and z -3 are eliminated. Thus we should carry out the long division with these two polynomials written in the reverse order. We stop the division when the order of the remainder becomes z -1. Then we obtain
X(z)= 1+2z -1+(1/6 z^-1)/(1+5/6 z^-1+1/6 z^-2 ).

Answer: d [Reason:] First we eliminate the negative powers of z by multiplying both numerator and denominator by z2.
Thus we obtain X(z)= z²/(z²-1.5z+0.5)
The poles of X(z) are p1=1 and p2=0.5. Consequently, the expansion will be
(X(z))/z = z/((z-1)(z-0.5)) = 2/((z-1) ) – 1/((z-0.5) )( obtained by applying partial fractions)
=>X(z)= 2z/(z-1)-z/(z-0.5).

Answer: b [Reason:] To eliminate the negative powers of z, we multiply both numerator and denominator by z2. Thus,
X(z)=(z(z+1))/(z^-2-z+0.5)
The poles of X(z) are complex conjugates p1=0.5+0.5j and p2=0.5-0.5j
Consequently the expansion will be
X(z)= (z(0.5-1.5j))/(z-0.5-0.5j) + (z(0.5+1.5j))/(z-0.5+0.5j).

Answer: c [Reason:] First we express X(z) in terms of positive powers of z, in the form X(z)=z³/((z+1)〖(z-1)〗² )
X(z) has a simple pole at z=-1 and a double pole at z=1. In such a case the approximate partial fraction expansion is
(X(z))/z = z²/((z+1)〖(z-1)〗² ) =A/(z+1) + B/(z-1) + C/〖(z-1)〗²
On simplifying, we get the values of A, B and C as 1/4, 3/4 and 1/2 respectively.
Therefore, we get X(z)= z/(4(z+1)) + 3z/(4(z-1)) + z/(2〖(z-1)〗² ) .

Answer: a [Reason:] The partial fraction expansion for the given X(z) is
X(z)= 2z/(z-1)-z/(z-0.5)
In case when ROC is |z|>1, the signal x(n) is causal and both the terms in the above equation are causal terms. Thus, when we apply inverse z-transform to the above equation, we get
x(n)=2(1)ⁿu(n)-(0.5)ⁿu(n)=(2-0.5ⁿ)u(n).

Answer: c [Reason:] The partial fraction expansion for the given X(z) is
X(z)= 2z/(z-1)-z/(z-0.5)
In case when ROC is |z|<0.5,the signal is anti causal. Thus both the terms in the above equation are anti causal terms. So, if we apply inverse z-transform to the above equation we get x(n)= [-2+0.5ⁿ]u(-n-1).

Answer: b [Reason:] The partial fraction expansion of the given X(z) is
X(z)= 2z/(z-1)-z/(z-0.5)
In this case ROC is 0.5<|z|<1 is a ring, which implies that the signal is two sided. Thus one of the signal corresponds to a causal signal and the other corresponds to an anti causal signal. Obviously, the ROC given is the overlapping of the regions |z|>0.5 and |z|<1. Hence the pole p2=0.5 provides the causal part and the pole p1=1 provides the anti causal part. SO, if we apply the inverse z-transform we get x(n)= -2u(-n-1)-(0.5)ⁿu(n).

Answer: d [Reason:] The partial fraction expansion of X(z) is X(z)= z/(4(z+1)) + 3z/(4(z-1)) + z/(2[(z-1)]²)
When we apply the inverse z-transform for the above equation, we get
x(n)=[1/4(-1)ⁿ+3/4+n/2]u(n).

Answer: b [Reason:] Let us assume that h_d(n) is specified for n > 0, and the digital filter is an all-pole filter.

Answer: d [Reason:] In this method, we consider the cascade connection of the desired filter H_d(z) with the reciprocal, all zero filter 1/H(z).

Answer: a [Reason:] We excite the cascade configuration by the unit sample sequence δ(n). Thus the input to the inverse system 1/H(z) is h_d(n) and the output is y(n). Ideally, y_d(n)= δ(n).

Answer: c [Reason:] The condition that y_d(0)= y(0)=1 is satisfied by selecting b₀=h_d(0).

Answer: a [Reason:] For n > 0, y(n) represents the error between the desired output y_d(n)=0 and the actual output.

Answer: b [Reason:] The parameters {a_k} are selected to minimize the sum of squares of the error sequence.

Answer: b [Reason:] By differentiating the square of the error sequence with respect to the parameters {a_k}, it is easily established that we obtain the set of linear equations.

Answer: d [Reason:] In a practical design problem, the desired impulse response h_d(n) is specified for a finite set of points, say 0 < n <L where L>> N. In such a case, the correlation sequence can be computed from the finite sequence h_d(n).

Answer: a [Reason:] We know that we can perform pole-zero approximation for H_d(z) by using the least squares method.

Answer: c [Reason:] Nevertheless, we can use the desired response h_d(n) for n < M to construct an estimate of h_d(n).

Answer: a [Reason:] The parameters {b_k} are selected to determine the zeros of the filter that can be obtained where h(n)=h_d(n).

Answer: a [Reason:] We find the coefficients {b_k} by pade approximation and find the coefficients {a_k} by least squares method. Thus the foregoing approach for determining the poles and zeros of H(z) is sometimes called as Prony’s method.