Your Answer : (Not Answered)

Correct Answer : C

Only the polarity of the difference signal is encoded.

In DPCM the transmitter quantises and codes the difference between each sample and a prediction of it. Delta modulation is the simplest case: it uses a one-bit quantiser, so it sends only the sign of the difference between the input and its staircase approximation. A 1 means the signal is above the approximation, which steps up; a 0 means it is below, which steps down.

The receiver integrates the bits to rebuild the signal, but integration and summation are operations on the decoded bits, not what is encoded, and differentiation is not used.

Your Answer : (Not Answered)

Correct Answer : D

Slope overload is reduced by increasing the step size.

A delta modulator can change its staircase approximation by only one step \(\Delta\) per sample, so the steepest slope it can follow is \(\Delta f_{s}\). Slope overload occurs when the signal changes faster than that:

\[ \left| \frac{dm(t)}{dt} \right|_{max} \gt \Delta f_{s} \]

Raising \(\Delta\) (or the sampling rate) lets the staircase keep up. Decreasing the sample rate or the step size makes the overload worse; a smaller step reduces granular noise instead, which is why adaptive delta modulation varies the step. 'Differential gain' is not the controlling quantity.

Your Answer : (Not Answered)

Correct Answer : D

DPCM exploits the correlation between samples to decrease the encoding rate.

Successive samples of signals such as speech and video are strongly correlated, so each sample can be predicted well from earlier ones. DPCM transmits only the difference between the actual sample and its prediction. That difference has a smaller range than the sample itself, so fewer bits are needed and the bit rate falls.

The options with 'increase' get the purpose backwards. 'Autocorrelation' names the same statistical dependence, but the standard description, and the key, speak of the correlation between samples.

Your Answer : (Not Answered)

Correct Answer : B

The symbol rate is 50,000 two-bit symbols per second.

Each symbol carries 2 bits, so the symbol (baud) rate is half the bit rate:

\[ \begin{aligned} R_{s} &= \frac{100000\ \text{bits/s}}{2\ \text{bits/symbol}} \\ &= 50000\ \text{symbols/s} \end{aligned} \]

The carrier frequency of 150 kHz and the particular data sequence do not affect the symbol rate; they are distractors. 1,00,000 is the bit rate itself, 25,000 would be for four-bit symbols, and 30,000 does not follow from the data.

Your Answer : (Not Answered)

Correct Answer : C

Only statement (d) is true, so the answer is (d) only.

When the FSK signal is made by switching between two independent oscillators, the two sine waves are not in step at the instant of switching, so the combined waveform jumps in value there: it has a discontinuity in phase, and in general in amplitude too. That makes (a) false, since it says such a signal cannot have amplitude discontinuities, and also (b) false, which claims no phase discontinuities.

When one oscillator is used and only its frequency is changed (a voltage-controlled oscillator), the phase carries on from where it was, so the waveform is continuous: no discontinuity in amplitude or phase. That makes (d) true and (c) false.

Every option other than (d) only contains (a), (b) or (c), which is why options A, B and D fail.

Your Answer : (Not Answered)

Correct Answer : C

The blanks are filled by time base, bit rate: the oscilloscope's time base is synchronized to the bit rate.

An eye diagram is made by feeding a received digital signal to the vertical input of an oscilloscope and triggering the horizontal sweep from the symbol (bit) clock. Every bit period is then drawn on top of the previous ones, and the overlaid traces form the open "eye". If the time base were not locked to the bit rate, the traces would drift across the screen and no eye would form.

The reversed option D would mean the bit rate is locked to the scope sweep, which gets the roles the wrong way round; the bit rate belongs to the signal and the scope follows it. Options A and B use an "intensity base", which is not a term for any oscilloscope timing control (the intensity control sets brightness).

Your Answer : (Not Answered)

Correct Answer : C

The data rate is 691.2 Mb/s.

64 intensity levels need \(\log_{2}64 = 6\) bits per pixel.

\[ \begin{aligned} \text{bits/frame} &= 800 \times 400 \times 6 \\ &= 1{,}920{,}000 \\ \text{rate} &= 1.92 \times 10^{6} \times 360 \\ &= 691.2\ \text{Mb/s} \end{aligned} \]

1382.4 Mb/s is double the answer, as if 12 bits per pixel were used, and 3686 Mb/s and 7372 Mb/s come from using 32 or 64 per pixel instead of the 6 bits.

Your Answer : (Not Answered)

Correct Answer : B

The Fourier transform of the output is \(e^{-2\pi f^{2}}\).

The Gaussian pulse is its own Fourier transform: \(x(t) = e^{-\pi t^{2}}\) has \(X(f) = e^{-\pi f^{2}}\). A filter matched to \(x(t)\) has \(h(t) = x(-t)\) (taking the delay as zero), so \(H(f) = X^{*}(f) = e^{-\pi f^{2}}\), real and even here. Then

\[ \begin{aligned} Y(f) &= X(f)H(f) = e^{-\pi f^{2}} \cdot e^{-\pi f^{2}} \\ &= e^{-2\pi f^{2}} \end{aligned} \]

\(e^{-\pi f^{2}}\) is the input spectrum alone, as if the filter passed everything. \(e^{-\pi f^{2}/2}\) would be a wider spectrum than the input, which filtering by a Gaussian cannot produce, and \(e^{-\pi|f|}\) is the transform of a Cauchy-shaped pulse, not of anything Gaussian.

Your Answer : (Not Answered)

Correct Answer : B

The most suitable choice is QPSK.

With Nyquist signalling, a channel of bandwidth B can carry at most 2B symbols per second in theory, and in practice, with filtering, about B symbols per second. The 3.2 kHz channel therefore supports about 3.2 kbaud. Each scheme needs:

• BPSK, 1 bit per symbol: 4.8 kbaud, more than the channel carries;
• QPSK, 2 bits per symbol: 2.4 kbaud, which fits with margin;
• 8-PSK, 3 bits: 1.6 kbaud; 16-PSK, 4 bits: 1.2 kbaud.

Higher-order PSK would also fit, but its constellation points are closer together and it needs a much higher signal to noise ratio for the same error rate. QPSK is the simplest scheme whose symbol rate fits the band, which makes it the most suitable.

Your Answer : (Not Answered)

Correct Answer : A

The rates are 10 kbaud and 60 kb/s.

The baud rate is the number of symbols sent per second: 10,000 symbols per second is 10 kbaud. Each symbol has 64 possible states, so it carries

\[ \begin{aligned} \log_{2} 64 &= 6\ \text{bits per symbol} \\ R_{b} &= 6 \times 10\,000 = 60\ \text{kb/s} \end{aligned} \]

30 kb/s would be 3 bits per symbol (8 states), and 10 kb/s treats each symbol as one bit. The option 0.64 Mbaud wrongly multiplies the number of states by the symbol rate.

Your Answer : (Not Answered)

Correct Answer : A

The maximum dynamic range is 98.08 dB.

For an n-bit linear quantizer with a full-scale sinusoidal input, the signal to quantization noise ratio, which sets the usable dynamic range, is

\[ \begin{aligned} \text{SQNR} &= 1.76 + 6.02n\ \text{dB} \\ &= 1.76 + 6.02 \times 16 \\ &= 1.76 + 96.32 = 98.08\ \text{dB} \end{aligned} \]

Each extra bit adds about 6 dB. The value 49.04 dB is half the answer, 32 dB is twice the number of bits, and 16 dB is the number of bits itself, none of which is a decibel ratio for 16 bits.