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The correct choice is (a) Glitch

To EXPLAIN I WOULD say: The application of the input code WORD results in a high-amplitude transient, called a “glitch”. This is especially the case when TWO consecutive code words to the A/D differ by several bits.

The correct CHOICE is (a) H(t)=\(\BEGIN{cases}1,0≤t≤T\\0,otherwise\end{cases}\)

Best explanation: W hen viewed as a linear filter, the S/H has an IMPULSE response:

h(t)=\(\begin{cases}1,0≤t≤T\\0,otherwise\end{cases}\)

The correct answer is (b) By using a S/H circuit

Easy explanation: The usual way to remedy this problem is to USE an S/H circuit designed to serve as a “deglitcher”. Hence the basic task of the S/H is to hold the output of the D/A converter constant at the previous output value until the NEW sample at the output of the D/A reaches steady state, and then it samples and holds the new value in the next sampling interval. Thus the S/H APPROXIMATES the analog signal by a series of RECTANGULAR pulses WHOSE height is equal to the corresponding value of the signal pulse.

The correct ANSWER is (b) Setting TIME

Easy explanation: An important parameter of a D/A CONVERTER is its settling time, which is defined as the time required for the output of the D/A converter to REACH and remain within a given fraction (usually,±1/2 LSB) of the final value, after application of the input code WORD.

The correct option is (a) True

Explanation: D/A CONVERSION is USUALLY performed by combining a D/A converter with a sample-and hold (S/H) and followed by a low pass (smoothing) filter. The D/A converter accepts at its input, ELECTRICAL signals that correspond to a binary word, and produces an output voltage or current that is proportional to the VALUE of the binary word.

Correct option is (a) True

The explanation is: The reconstruction of the signal from its samples as a linear FILTERING PROCESS in which a discrete-time SEQUENCE of SHORT pulses (ideally impulses) with amplitudes EQUAL to the signal samples, excites an analog filter.

The correct OPTION is (a) True

To explain: The ideal reconstruction filter is an ideal low pass filter and its IMPULSE response extends for all time. Hence the filter is noncausal and physically nonrealizable. ALTHOUGH the interpolation filter with impulse response given can be approximated closely with some delay, the resulting function is STILL impractical for most applications where D/A conversion are required.

Correct ANSWER is (C) Linearity error

To explain: We note that practical A/D CONVERTERS, linearity error (the DIFFERENCES between transition values are not all equal or uniformly changing).

The correct answer is (a) \(\sum_{-\infty}^\infty x(NT) \FRAC{sin⁡(π/T) (t-nT)}{(π/T)(t-nT)}\)

To explain: x(t) = \(\sum_{-\infty}^\infty x(nT) \frac{sin⁡(π/T) (t-nT)}{(π/T)(t-nT)}\) where the SAMPLING interval T = 1/Fs=1/2B, Fs is the sampling frequency and B is the bandwidth of the ANALOG SIGNAL.

Right choice is (b) G(t)=\(\frac{sin⁡(πt/T)}{(πt/T)}\)

Easy explanation: The RECONSTRUCTION of the signal x (t) from its samples as an interpolation PROBLEM and have described the function:g(t)=\(\frac{sin⁡(πt/T)}{(πt/T)}\).

Correct answer is (c) H(F)=\(\begin{CASES}T, |F|≤ \frac{1}{2T} = F_s/2\\0,|F| > \frac{1}{2T}\END{cases}\)

The explanation: The analog filter CORRESPONDING to the IDEAL interpolator has a frequency response:

H(F)=\(\begin{cases}T, |F|≤ \frac{1}{2T} = F_s/2\\0,|F| > \frac{1}{2T}\end{cases}\), H(F) is the Fourier transform of the interpolation function g(t).

The correct ANSWER is (b) Offset error

Explanation: We NOTE that practical A/D converters MAY have offset error (the first TRANSITION may not occur at exactly + 1/2 LSB).

The correct choice is (a) Δ = (R)/2^(b+1)

Explanation: The coding process in an A/D CONVERTER assigns a unique binary NUMBER to each quantization level. If we have L levels, we need at least L different binary numbers. With a word length of b + 1 bits we can represent 2^b+1 distinct binary numbers. Hence we should have 2^(b+1) > L or, equivalently, b + 1 > LOG2 L. Then the step size or the RESOLUTION of the A/D converter is given by

Δ = (R)/2^(b+1), where R is the range of the quantizer.

The CORRECT option is (b) xq(N) = Q[x(n)] = \(\hat{x}_k\), if x(n) ∈ Ik

To explain: The possible outputs of the quantizer (i.e., the QUANTIZATION levels) are denoted as \(\hat{x}_1, \hat{x}_2,…\hat{x}_L\). The operation of the quantizer is defined by the relation, xq(n) ≡ Q[x(n)]= \(\hat{x}_k\), if x(n) ∈ Ik.

The correct choice is (b) False

To explain I would say: If the dynamic RANGE of the signal, defined as xmax-xmin, is larger than the range of the QUANTIZER, the samples that exceed the quantizer range are CLIPPED, resulting in a LARGE (greater than \({\Delta}{2}\)) QUANTIZATION error.

Right answer is (c) –\(\frac{\Delta}{2}\) < eq(N) ≤ \(\frac{\Delta}{2}\)

Easy EXPLANATION: The quantization error eq(n) is always in the RANGE – \(\frac{\Delta}{2}\) < eq(n) ≤ \(\frac{\Delta}{2}\), where Δ is QUANTIZER step size.

Right OPTION is (B) FSR

The explanation: The term Full-scale range (FSR) is USED to describe the range of an A/D CONVERTER for bipolar signals (i.e., signals with both positive and negative amplitudes).

The correct answer is (c) a(t) is the Envelope of X(t), θ(t) is CALLED the Phase of x(t)

The EXPLANATION: In the equation x(t) = a(t) COS[2πFct+θ(t)], the signal a(t) is called the Envelope of x(t), and θ(t) is called the phase of x(t).

The correct CHOICE is (b) False

The best explanation: The basic task of the A/D converter is to convert a CONTINUOUS range of input amplitude into a discrete set of DIGITAL code words. This CONVERSION involves the processes of Quantization and Coding.

The correct option is (a) Midrise TYPE

For EXPLANATION: If a zero is ASSIGNED a decision level, the QUANTIZER is of the midrise type.

Right OPTION is (b) x(t) = a(t) cos[2πFct + θ(t)]

To explain: x(t) = Re\([x_l (t) E^{j2πF_c t}]\)

= Re\([a(t) e^{J[2πF_c t + θ(t)]}]\)

= \(a(t) \,cos⁡ [2πF_c t+θ(t)]\)

Hence PROVED.

The correct CHOICE is (b) Mid tread type

Easiest EXPLANATION: If a zero is assigned a QUANTIZATION level, the quantizer is of the mid treat type.

The correct choice is (b) Complex envelope

To explain I would say: In the equation X(t) = Re[xl(t)E^(j2πFct)], Re denotes the real part of the complex valued quantity in the BRACKETS following. The lowpass signal xl (t) is usually called the Complex envelope of the real signal x(t), and is basically the equivalent LOW PASS signal.

Right choice is (b) Quadrature components

Easy explanation: The low frequency signal components uc(t) and US(t) can be viewed as amplitude MODULATIONS IMPRESSED on the carrier components cos2πFct and sin2πFct, respectively. Since these carrier components are in phase quadrature, uc(t) and us(t) are called the Quadrature components of the BANDPASS signal x (t).

The correct answer is (a) a(t) = \(\SQRT{u_c^2 (t)+u_s^2 (t)}\) and θ(t)=\(TAN^{-1}\frac{u_s (t)}{u_c (t)}\)

The best EXPLANATION: A THIRD possible representation of a band pass SIGNAL is obtained by expressing \(x_l (t)=a(t)e^{jθ(t)}\) where a(t) = \(\sqrt{u_c^2 (t)+u_s^2 (t)}\) and θ(t)=\(tan^{-1}\frac{u_s (t)}{u_c (t)}\).

Correct answer is (b) X(t)=\(u_c (t) \,cos⁡2π \,F_c \,t-u_s (t) \,sin⁡2π \,F_c \,t\); ẋ(t)=\(u_s (t) \,cos⁡2π \,F_c t+u_c (t) \,sin⁡2π \,F_c \,t\)

BEST explanation: If we SUBSTITUTE the given EQUATION in other, then we get the REQUIRED resu

The correct option is (a) X+(F+Fc)

To elaborate: The ANALYTIC signal x+(t) is a BANDPASS signal. We OBTAIN an equivalent lowpass representation by performing a frequency translation of X+(F).

The CORRECT answer is (b) Hilbert transformer

Easy explanation: The signal ẋ(t) can be viewed as the output of the filter with IMPULSE RESPONSE h(t) = 1/πt,

 -∞ < t < ∞ when excited by the input signal x(t) then such a filter is CALLED as Hilbert transformer.

Correct choice is (a) \(\begin{cases}&-j(F>0) \\ & 0(F=0)\\ & j(F<0)\end{cases}\)

For explanation I would say: H(F) =\(\int_{-∞}^∞ h(t)E^{-j2πFt} dt\)

=\(\frac{1}{π} \int_{-∞}^∞ 1/t e^{-2πFt} dt\)

=\(\left\{\begin{MATRIX}-j& (F>0)\\0&(F=0) \\ j& (F<0)\end{matrix}\right.\)

We Observe that │H (F)│=1 and the PHASE response &odot;(F) = -1/2π for F > 0 and &odot;(F) = 1/2π for F < 0.

Correct ANSWER is (B) \(\frac{1}{π} \int_{-\INFTY}^\infty \frac{x(t)}{t-τ} dτ\)

Easy EXPLANATION: \(x_+ (t)=[δ(t)+j/πt]*x(t)\)

\(x_+ (t)=x(t)+[j/πt]*x(t)\)

\(ẋ(t)=[j/πt]*x(t)\)

=\(\frac{1}{π} \int_{-\infty}^\infty \frac{x(t)}{t-τ} dτ\)

Hence proved.

The correct option is (C) X+(F)=2V(F)X(F)

To explain: In a real valued signal x(t), has a frequency content concentrated in a narrow band of frequencies in the VICINITY of a frequency Fc. Such a signal which has only positive frequencies can be EXPRESSED as X+(F)=2V(F)X(F)

Where X+(F) is a Fourier transform of x(t) and V(F) is UNIT step function.

The CORRECT choice is (d) F^(-1)[2V(F)]*F^(-1)[X(F)]

To EXPLAIN I would say: Given EXPRESSION, X+(F)=2V(F)X(F).It can be calculated as follows

\(x_+ (t)=\int_{-∞}^∞ X_+ (F)e^{j2πFt} dF\)

=\(F^{-1} [2V(F)]*F^{-1} [X(F)]\)

Right ANSWER is (d) Both ANALYTIC SIGNAL & Pre-envelope of x(t)

To explain I would SAY: From the given expression, \(x_+ (t)=F^{-1} [2V(F)] * F^{-1}[X(F)]\).

The correct option is (d) X (F) = \(\frac{1}{2} [X_l (F-F_c)+X_l^* (-F-F_c)]\)

BEST explanation: X(F) = \(\frac{1}{2} [X_l (F-F_c)+X_l^* (-F-F_c)]\), where Xl(F) is the Fourier transform of xl(t). This is the basic RELATIONSHIP between the spectrum o f the real band PASS signal x(t) and the spectrum of the EQUIVALENT low pass signal xl(t).

Correct answer is (a) \(\sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(π/(2T^{‘}))(t-2nT^{‘}+T^{‘})}\)

The explanation: The low PASS signal components US(t) can be EXPRESSED in terms of samples of the

band pass signal as follows:

\(u_s (t) = \sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(π/(2T^{‘}))(t-2nT^{‘}+T^{‘})}\)

Right answer is (b) \(\sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \FRAC{sin⁡(π/(2T^{‘})) (t-2nT^{‘})}{(π/(2T^{‘}))(t-2nT^{‘})}\)

Explanation: The LOW pass signal COMPONENTS uc(t) can be expressed in TERMS of samples of the

band pass signal as follows:

\(u_c (t) = \sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘})}{(π/(2T^{‘}))(t-2nT^{‘})}\).

Right ANSWER is (b) \(\sum_{m=-∞}^∞ u_s (mT_1-\FRAC{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1+\frac{T_1}{2})}{(π/T_1)(t-mT_1+\frac{T_1}{2})}\)

The best I can explain: To RECONSTRUCT the equivalent low pass signals. Thus, ACCORDING to the sampling theorem for low pass signals with T1=1/B .

\(u_s (t)=\sum_{m=-∞}^∞ u_s (mT_1-T_1/2) \frac{sin⁡(π/T_1) (t-mT_1+T_1/2)}{(π/T_1)(t-mT_1+T_1/2)}\)

The correct option is (a) \(\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(π/T_1)(t-mT_1)}\)

Easiest EXPLANATION: To reconstruct the equivalent low PASS SIGNALS. Thus, according to the SAMPLING theorem for low pass signals with T1=1/B.

\(u_c (t)=\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(π/T_1)(t-mT_1)}\).

The CORRECT option is (b) Fc‘= Fc+B/2-B’/2

Explanation: A NEW centre frequency for the increased bandwidth SIGNAL is Fc‘ = Fc+B/2-B’/2

The correct choice is (b) US – lowpass SIGNAL component

The best I can explain: With the odd-numbered SAMPLES of x(t), which occur at the rate of B samples per second, produce samples of the LOW pass signal component us.

Right ANSWER is (a) uc-lowpass SIGNAL component

The best explanation: With the even-numbered samples of x(t), which OCCUR at the RATE of B samples per second, PRODUCE samples of the low pass signal component uc.

Right choice is (a) = \(u_c (t) cos⁡2π F_c t-u_s (t) sin⁡2π F_c t\) by the quadrature carriers cos[2πFct] and sin[2πFct] and lowpass FILTERING the products to eliminate the signal components of 2Fc.

To explain: It is CERTAINLY advantageous to perform a frequency shift of the band pass signal by and sampling the equivalent low pass signal. Such a frequency shift can be ACHIEVED by multiplying the band pass signal as given in the above EQUATION by the quadrature carriers cos[2πFct] and sin[2πFct] and low pass filtering the products to eliminate the signal components at 2Fc. Clearly, the multiplication and the SUBSEQUENT filtering are first performed in the analog domain and then the outputs of the filters are sampled.

Right option is (d) All of the mentioned

To EXPLAIN I would say: One way to reduce these two types of distortion is to use an integrator in front of the DM. This has two effects. First, it emphasizes the low frequencies of x (t) and INCREASES the CORRELATION of the signal into the DM input. Second, it simplifies the DM decoder because the differentiator (INVERSE system) required at the decoder is canceled by the DM integrator.