Digital Electronic MCQ Number 00924

Answer: a [Reason:] The memory of the system, describes, in some case, the ‘state’ of the system, the output of the system is called as ‘zero-state response’.

Answer: b [Reason:] The zero-state response depends on the nature of the system and the input signal. Since this output is a response forced upon it by the input signal, it is also known as ‘Forced response’.

Answer: a [Reason:] For a zero-input response, the input is zero and the output of the system is independent of the input of the system. So, the response if such system is also known as Natural or Free response.

Answer: d [Reason:] By making the input x(n)=0 we will get a homogenous difference equation and the solution of that difference equation is known as Homogenous or Complementary solution.

Answer: c [Reason:] The assumed solution obtained by assigning x(n)=0 is

Answer: b [Reason:] Given difference equation is y(n)-3y(n-1)-4y(n-2)=0—-(1)
Let y(n)=λn
Substituting y(n) in the given equation
=> λⁿ – 3λ^n-1 – 4λ^n-2 = 0
=> λ^n-2(λ² – 3λ – 4) = 0
the roots of the above equation are λ=-1,4
Therefore, general form of the solution of the homogenous equation is

The zero-input response of the system can be calculated from the homogenous solution by evaluating the constants in the above equation, given the initial conditions y(-1) and y(-2).
From the given equation (1)
y(0)=3y(-1)+4y(-2)
y(1)=3y(0)+4y(-1)
=3[3y(-1)+4y(-2)]+4y(-1)
=13y(-1)+12y(-2)
From the equation (2)
y(0)=C1+C2 and
y(1)=C1(-1)+C2(4)=-C1+4C2
By equating these two set of relations, we have
C1+C2=3y(-1)+4y(-2)=15
-C1+4C2=13y(-1)+12y(-2)=65
On solving the above two equations we get C1=-1 and C2=16
Therefore the zero-input response is Y_zi(n) = (-1)ⁿ⁺¹ + (4)ⁿ⁺².

Answer: a [Reason:] The assumed solution of the difference equation to the forcing equation x(n), called the particular solution of the difference equation is
y_p(n)=Kx(n)=Ku(n) (where K is a scale factor)
Substitute the above equation in the given equation
=>Ku(n)+aKu(n-1)=u(n)
To determine K we must evaluate the above equation for any n>=1, so that no term vanishes.
=> K+aK=1
=>K=1/(1+a)
Therefore the particular solution is y_p(n)= 1/(1+a) u(n).

Answer: c [Reason:] The assumed solution of the difference equation to the forcing equation x(n), called the particular solution of the difference equation is
y_p(n)=Kx(n)=K2ⁿu(n) (where K is a scale factor)
Upon substituting y_p(n) into the difference equation, we obtain
K2ⁿu(n)=5/6K2^n-1u(n-1)-1/6 K2^n-2u(n-2)+2ⁿu(n)
To determine K we must evaluate the above equation for any n>=2, so that no term vanishes.
=> 4K= 5/6(2K)-1/6 (K)+4
=> K= 8/5
=> y_p(n)= (8/5) 2ⁿ.

Answer: b [Reason:] The linearity property of the linear constant coefficient difference equation allows us to add the homogenous and particular solution in order to obtain the total solution.

Answer: d [Reason:] The homogenous solution of the given equation is y_h(n)=C1(-1)ⁿ+C2(4)ⁿ—-(1)
To find the impulse response, x(n)=δ(n)
now, for n=0 and n=1 we get
y(0)=1 and
y(1)=3+2=5
From equation (1) we get
y(0)=C1+C2 and
y(1)=-C1+4C2
On solving the above two set of equations we get
C1=- 1/5 and C2= 6/5
=>h(n)= [-1/5 (-1)ⁿ + 6/5 (4)ⁿ]u(n).

Answer: a [Reason:] The formula for calculating N point DFT is given as

From the formula given at every step of computing we are performing N complex multiplications and N-1 complex additions. So, in a total to perform N-point DFT we perform N² complex multiplications and N(N-1) complex additions.

Answer: d [Reason:] The formula for calculating N point DFT is given as

From the formula given at every step of computing we are performing N complex multiplications and N-1 complex additions. So, it requires 4N real multiplications and 4N-2 real additions for any value of ‘k’ to compute DFT of the sequence.

Answer: b [Reason:] According to the symmetry property, we get W_Nk+N/2=-W_N^k.

Answer: c [Reason:] For a complex valued sequence x(n) of N points, the DFT may be expressed as

Answer: d [Reason:] The expression for XR(k) is given as
So, from the equation we can tell that the computation of XR(k) requires 2N² evaluations of trigonometric functions, 4N² real multiplications and 4N(N-1) real additions.

Answer: a [Reason:] T he development of computationally efficient algorithms for the DFT is made possible if we adopt a divide-and-conquer approach. This approach is based on the decomposition of an N-point DFT into successively smaller DFTs. This basic approach leads to a family of computationally efficient algorithms known collectively as FFT algorithms.

Answer: b [Reason:] If we consider the mapping n=Ml+m, then it leads to an arrangement in which the first row consists of the first M elements of x(n), the second row consists of the next M elements of x(n), and so on.

Answer: a [Reason:] We know that if N=LM, then W_N^mqL= W_N/L^{mq= W_Mmq.}

Answer: d [Reason:] The expression for N point DFT is given as
The first step involves L DFTs, each of M points. Hence this step requires LM2 complex multiplications, second require LM and finally third requires ML2. So, Total complex multiplications= N(L+M+1).

Answer: b [Reason:] The expression for N point DFT is given as
The first step involves L DFTs, each of M points. Hence this step requires LM(M-1) complex additions, second step do not require any additions and finally third step requires ML(L-1) complex additions. So, Total number of complex additions= N(L+M-2).

Answer: c [Reason:] According to one of the algorithm describing the divide and conquer method, if we store the signal in column wise, then compute the M-point DFT of each row and multiply the resulting array by the phase factors WNlq and then compute the L-point DFT of each column and read the result row wise.

Answer: a [Reason:] According to the second algorithm of divide and conquer approach, if the input signal is stored in row wise, then the result must be read column wise.

Answer: d [Reason:] According to the second algorithm of divide and conquer approach, if the input signal is stored in row wise, then we calculate the L point DFT of each column and we multiply the resulting array by the factor W_N^pm.

Answer: d [Reason:] Lattice filters are used extensively in digital signal processing and in the implementation of adaptive filters.

Answer: b [Reason:] Consider a sequence of FIR filer with system function Hm(z)=Am(z), m=0,1,2…M-1
where, by definition, Am(z) is the polynomial

Answer: c [Reason:] We know that Hm(z)=Am(z) and Am(z) is a polynomial whose equation is given as

Answer: a [Reason:] The FIR structure shown in the above figure is intimately related with the topic of linear prediction. Thus the top filter structure shown in the above figure is called a prediction error filter.

Answer: b [Reason:] The single stage lattice filter is as shown below.

Here both the inputs are excited and output is selected from the top branch.
Thus the output of the single stage lattice filter is given by y(n)= x(n)+Kx(n-1).

Answer: d [Reason:] When two single stage lattice filters are cascaded, then the output from the first filter is given by the equation
f₁(n)= x(n)+K₁x(n-1)
g₁(n)=K₁x(n)+x(n-1)
The output from the second filter is obtained as
f₂(n)=f₁(n)+K₂g₁(n-1)
=x(n)+K₂[K₁x(n-1)+x(n-2)]+ K₁x(n-1)
= x(n)+K₁(1+K₂)x(n-1)+K₂x(n-2).

Answer: c [Reason:] The equation for the output of an FIR filter represented in the direct form structure is given as
y(n)=x(n)+ α₂(1)x(n-1)+ α₂(2)x(n-2)
The output from the double stage lattice structure is given by the equation,
f₂(n)= x(n)+K₂(1+K2)x(n-1)+K₂x(n-2)
By comparing the coefficients of both the equations, we get
α₂(1)= K₁(1+K₂).

Answer: a [Reason:] The equation of the output from the second stage lattice filter is given by
f₂(n)= x(n)+K₁(1+K₂)x(n-1)+K₂x(n-2)
In the above equation, the constants K₁ and K₂ are called as reflection coefficients.

Answer: d [Reason:] We get the output from the third stage lattice filter as
A3(z)=1+(13/24)z^-1+(5/8)z^-2+(1/3)z^-3.
Thus the FIR filter coefficients for the direct form structure are (1,13/24,5/8,1/3).

Answer: c [Reason:] Given the system function of the FIR filter is
H(z)= 1+(13/24)z^-1+(5/8)z^-2+(1/3)z^-3
Thus the lattice coefficients corresponding to the given filter is (1/4,1/2,1/3).

Answer: a [Reason:] FIR least square filters are also called as Wiener filters.

Answer: d [Reason:] The inverse to a linear time invariant system with impulse response h(n) is defined as the system whose impulse response is h_I(n), satisfy the following condition h(n)*h_I(n)=δ(n).

Answer: c [Reason:] The inverse to a linear time invariant system with impulse response h(n) and system function H(z) is defined as the system whose impulse response is h_I(n) and system function H_I(z), satisfy the following condition
H(z).H_I(z)=1.

Answer: b [Reason:] In most of the practical applications, it is desirable to restrict the inverse filter to be an FIR filter.

Answer: a [Reason:] In many practical applications, it is desirable to restrict the inverse filter to be FIR. One of the simple method to get this requirement is to truncate h_I(n).

Answer: c [Reason:] In the process of truncating, we incur a total squared approximation error where M+1 is the length of the truncated filter.

Answer: b [Reason:] We can use the least squares error criterion to optimize the M+1 coefficients of the FIR filter.

Answer: d [Reason:] Since from the block diagram, the coefficients of the FIR filter coefficients are optimized by the least squares error criterion, it belongs to the least squares FIR inverse filter or wiener filter.

Answer: a [Reason:] When ε is minimized with respect to the filter coefficients, we obtain the set of linear equations which are dependent on the auto correlation sequence of the signal h(n).

Answer: d [Reason:] When ε is minimized with respect to the filter coefficients, we obtain the set of linear equations

and we know that r_dh(l) depends on the desired output d(n).

Answer: a [Reason:] The optimum, in the least square sense, FIR filter that satisfies the linear equations in is called the wiener filter.

Answer: b [Reason:] If the optimum least squares FIR filter is to be an approximate inverse filter, the desired response is
d(n)=δ(n).

Answer: c [Reason:] We observe that the matrix is not only symmetric but it also has the special property that all the elements along any diagonal are equal. Such a matrix is called a Toeplitz matrix and lends itself to efficient inversion by means of an algorithm.

Answer: b [Reason:] The Levinson-Durbin algorithm is the algorithm which is used for the efficient inversion of Toeplitz matrix which requires a number of computations proportional to M² instead of the usual M³.

Answer: a [Reason:] If the given signal is x(t) and F0 is the reciprocal of the time period of the signal and ck is the Fourier coefficient then the Fourier series representation of x(t) is given as .

Answer: c [Reason:] When we apply integration to the definition of Fourier series representation, we get

Answer: d [Reason:] For any signal x(t) to be represented as Fourier series, it should satisfy the Dirichlet conditions which are x(t) has a finite number of discontinuities in any period, x(t) has finite number of maxima and minima during any period and x(t) is absolutely integrable in any period.

Answer: b [Reason:] Since we are synthesizing the Fourier series of the signal x(t), we call it as synthesis equation, where as the equation giving the definition of Fourier series coefficients is known as analysis equation.

Answer: b [Reason:] In general, Fourier coefficients ck are complex valued. Moreover, it is easily shown that if the periodic signal is real, ck and c-k are complex conjugates. As a result
c_k=|c_k|e^jθkand c_k=|c_k|e^-jθk
Consequently, we obtain the Fourier series as

Answer: a [Reason:] cos(2πkF₀ t+θ_k)= cos2πkF₀ t.cosθ_k-sin2πkF₀ t.sinθ_k
θ_k is a constant for a given signal.
So, the other form of Fourier series representation of the signal x(t) is

Answer: d [Reason:] The average power of a periodic signal x(t) is given as

By interchanging the positions of integral and summation and by applying the integration, we get

Answer: c [Reason:] When we plot a graph of |c_k |² as a function of frequencies kF₀, k=0,±1,±2… the following spectrum is obtained which is known as Power density spectrum.

Answer: a [Reason:] We know that, Fourier series coefficients are complex valued, so we can represent ck in the following way.
c_k=|c_k|e^jθk
When we plot |c_k| as a function of frequency, the spectrum thus obtained is known as Magnitude voltage spectrum.

Answer: b [Reason:] We know that, for an periodic signal, the Fourier series coefficient is

If we consider a signal x(t) as non-periodic, it is true that x(t)=0 for |t|>Tp/2. Consequently, the limits on the integral in the above equation can be replaced by -∞ to ∞. Hence,

Answer: c [Reason:] Let us consider a signal x(t) whose Fourier transform X(F) is given as

Answer: d [Reason:] Let x(t) be any finite energy signal with Fourier transform X(F). Its energy is