Frequency Response of Amplifiers Questions and Answers

Answer: A

Holding the product constant:

\[ \begin{aligned} GB &= 100 \times 200\mathrm{k} = 20 \ \mathrm{MHz} \\ B &= \tfrac{20\mathrm{M}}{10} = 2 \ \mathrm{MHz} \end{aligned} \]

giving about 2 MHz. Twenty megahertz is the product itself rather than the new bandwidth. Reducing the gain by ten buys ten times the bandwidth, which is the trade every feedback amplifier makes and the reason a high open-loop gain is worth having.

Answer: B

A single pole falls at a constant number of decibels per octave, so on logarithmic axes it becomes a straight line and the whole response can be sketched from its break points and slopes alone. Phase is plotted separately either way, linear axes can represent any number, and gain is meaningful in any units. A coordinate system that turns the behaviour of interest into straight lines is worth adopting for that reason by itself, and it is why Bode's construction survived the arrival of computers.

Answer: C

Each pole contributes six decibels per octave, so two together give twelve decibels per octave, equivalently forty per decade. Six per octave is one pole, three decibels is the drop at a break rather than a slope, and twenty per decade is one pole expressed in the other unit. Slopes add when poles are cascaded, which is why counting poles is the quickest way to predict a response's steepness and its phase shift together.

Answer: D

Each pole approaches ninety degrees, so two together approach one hundred and eighty degrees - which is exactly the condition that turns negative feedback into positive and makes an amplifier oscillate. That is why a feedback amplifier's stability depends on how many poles lie below the frequency at which the loop gain falls to unity. The phase count and the slope count are the same count, which is what makes a Bode plot a stability tool rather than merely a description.

Answer: A

Two identical stages each 3 dB down at some frequency are 6 dB down together, so the combined 3 dB point sits at a lower frequency than either stage's own. Coupling capacitors set low-frequency poles rather than high-frequency ones, loading reduces gain rather than bandwidth directly, and the product applies per stage. Roll-offs multiply as ratios and add as decibels, and that compounding is what shrinks a cascade's bandwidth even when nothing is loading anything.

Answer: B

For n identical single-pole stages the combined 3 dB point shrinks by a known factor:

\[ B_{n} = B \sqrt{2^{1/n}-1} = 10^{6}(0.64) \]

giving about 640 kHz. Half a megahertz would be the answer if bandwidths simply halved, which they do not. The shrinkage is modest for two stages and becomes severe for five or six, which is why a long cascade of narrow stages is a poor way to obtain both gain and bandwidth.

Answer: C

Exactly as at the high end, each stage's low-frequency roll-off adds to the others, so the combined 3 dB point moves upward and the useful band narrows from both sides. Total capacitance is not a meaningful sum here, and the limit is certainly not set by one stage alone. The symmetry between the two ends is worth noticing: compounding narrows the passband at whichever end the roll-offs occur.

Answer: D

For a single-pole response the rise time multiplied by the bandwidth is about 0.35, so a wider bandwidth means a faster edge and the two are simply two views of one limitation. They are inversely related rather than proportional, and gain does not enter. Being able to move between the frequency and time descriptions of the same limit is what lets an oscilloscope measurement and a swept response be compared directly.

Answer: A

Applying the constant product:

\[ t_{r} = \frac{0.35}{3.5 \times 10^{6}} = 100 \ \mathrm{ns} \]

giving about 100 nanoseconds. The other options misplace a decade or use the numerator directly. Because the product is a fixed number for a given response shape, either quantity can be measured and the other inferred, which is often the quicker route in a laboratory.

Answer: B

A response that rises before it falls emphasises the components near the band edge, and a step contains exactly those components, so the edge overshoots and rings. A slow roll-off gives a slow edge without overshoot, the low-frequency break affects droop rather than overshoot, and phase alone does not produce it. The time and frequency descriptions are two views of the same response, so a defect visible in one is always visible in the other.