Practice Problems (KEY)

Problem 1

What is aliasing and how can it be prevented?

Aliasing is distortion caused by sampling a signal at a rate that is too low. It can be prevented by selecting a sampling frequency, \(f_s\), that is greater than the Nyquist rate:

\[ f_s > 2f_{\text{High}}. \]

Aliasing can also be prevented by pre-filtering the signal with a low-pass anti-aliasing filter before the ADC, with cutoff frequency \(f_c \le \frac{f_s}{2}\).

Problem 2

What are the three steps for converting an analog signal into digital?

1. Sampling — Taking discrete measurements of the signal in time at the sampling frequency, \(f_s\).

2. Quantizing — Assigning each sampled value to one of a finite number of discrete voltage levels.

3. Encoding — Converting the quantized level into a binary number for storage or transmission.

Problem 3

Given the following spectra, determine the minimum sampling frequency.

\[ f_s = 2f_{\text{High}} = 2(300\ \text{MHz}) = 600\ \text{MHz} \]

\[ f_s = 2f_{\text{High}} = 2(50\ \text{kHz}) = 100\ \text{kHz} \]

Problem 4

Find the number of levels and resolution for an 8-bit ADC with \(V_{\max} = 6\ \text{V}\) and \(V_{\min} = -4\ \text{V}\).

\[ \text{levels} = 2^8 = 256\ \text{levels} \]

\[ \Delta V = \frac{6\ \text{V} - (-4\ \text{V})}{256} = \frac{10\ \text{V}}{256} = 39.0625\ \text{mV/level} \]

Problem 5

Given a cosine input \(v_{in}(t) = 1 + 3\cos\!\left(360^\circ (50\,\text{kHz})t\right)\ \text{V}\):

a. Nyquist sampling frequency:

\[f_s = 2(50\ \text{kHz}) = 100\ \text{kHz}\]

b. Use 120 kHz — the other options (75 kHz, 90 kHz) are below Nyquist and would cause aliasing.

c. Signal range: \(V_{\min} = 1 - 3 = -2\ \text{V}\), \(V_{\max} = 1 + 3 = 4\ \text{V}\)

d. \(V_{\max}=5\ \text{V}\), \(V_{\min}=-3\ \text{V}\) works (covers \(-2\) to \(4\ \text{V}\)). The other two clip the signal.

e. With \(V_{\max}=5\), \(V_{\min}=-3\), \(\Delta V \le 600\ \text{mV}\):

\[2^b = \frac{8\ \text{V}}{0.600\ \text{V/level}} = 13.33 \quad \Rightarrow \quad 2^4 = 16 \ge 14 \quad \Rightarrow \quad b = 4\ \text{bits}\]

f. Actual resolution:

\[\Delta V = \frac{8\ \text{V}}{2^4} = \frac{8}{16} = 500\ \text{mV/level}\]

Problem 6

An input signal \(v_{in}(t) = 6 + 5\cos(360^\circ (6\,\text{kHz})t) + 3\cos(360^\circ (15\,\text{kHz})t)\ \text{V}\) is to be digitized using the ADC below.

Aliasing check: Nyquist requires \(f_s > 2(15\ \text{kHz}) = 30\ \text{kHz}\), but \(f_{s,\text{ADC}} = 20\ \text{kHz}\) — aliasing will occur. Since the 15 kHz component is noise, apply a LPF (cutoff between 6 and 10 kHz) before the ADC to remove it.

Clipping check (after LPF): \(V_{\min} = 6 - 5 = 1\ \text{V}\), \(V_{\max} = 6 + 5 = 11\ \text{V}\) — clipping occurs. Apply a signal conditioning interface (\(K=0.7\), \(B=2.3\ \text{V}\)) to map into the ADC range.

Problem 7

What are the advantages of digital signals?

• Less susceptible to noise

• Easily stored and recovered

• Easier encryption and signal processing

Problem 8

T / F. Improving the resolution is always better.

False. Increasing resolution (more bits) increases memory usage and required bandwidth — it’s an engineering tradeoff.

Problem 9

T / F. Increasing the sampling rate improves resolution.

False. Sampling rate and resolution are independent. \(f_s\) divides the signal along the time axis; resolution \(\Delta V\) divides the signal along the voltage axis.

Problem 10

T / F. A smaller resolution is a better resolution.

True (technically) — a smaller \(\Delta V\) means less quantization error. However, achieving it requires more bits, increasing memory and bandwidth requirements, so it involves an engineering tradeoff.

Problem 11

ADC: \(V_{\max} = 0.5\ \text{V}\), \(V_{\min} = -0.5\ \text{V}\), 3-bit. \(v(t) = 200 + 400\cos\!\left(360^\circ (200\,\text{kHz})t\right)\ \text{mV}\)

a. Transducer interface: Signal range: \(v_{\min} = -200\ \text{mV}\), \(v_{\max} = 600\ \text{mV}\). Solve for \(K\) and \(B\):

\[ K = 1.25 \qquad B = -0.25\ \text{V} \]

\[ v_{\text{out}} = 1.25\,v_{\text{in}} - 0.25\ \text{V} \]

b. Sampling rate: \(f_s = 2(200\ \text{kHz}) = 400\ \text{kHz}\)

c. Resolution:

\[ \Delta V = \frac{0.5 - (-0.5)}{2^3} = \frac{1\ \text{V}}{8} = 125\ \text{mV/level} \]

d. Encoded output for 315 mV input:

\[ v_{\text{out}} = 1.25(0.315) - 0.25 = 0.14375\ \text{V} \]

\[ EL = \frac{0.14375 - (-0.5)}{0.125} = 5.15 \quad \Rightarrow \quad QL = 5 \quad \Rightarrow \quad \boxed{101} \]

Problem 12

What is the resolution of a 10-bit ADC with \(V_{\max} = 5\ \text{V}\) and \(V_{\min} = -3\ \text{V}\)?

\[ \Delta V=\frac{5\ \text{V}-(-3\ \text{V})}{2^{10}} =\frac{8\ \text{V}}{1024} = 7.8125\ \text{mV/level} \]

Problem 13

Given \(V(t) = \cos\!\left(360^\circ (8\,\text{kHz})t\right) + 2\cos\!\left(360^\circ (15\,\text{kHz})t\right)\ \text{V}\), create an ADC with 5 bits and determine resolution.

Signal limits: \(V_{\min} = -3\ \text{V}\), \(V_{\max} = 3\ \text{V}\)

Sampling rate: \(f_s = 2(15\ \text{kHz}) = 30\ \text{kHz}\)

\[ \Delta V=\frac{3\ \text{V}-(-3\ \text{V})}{2^5}=\frac{6\ \text{V}}{32}=187.5\ \text{mV/level} \]

b. Encode \(1.62\ \text{V}\):

\[ EL=\frac{1.62-(-3)}{0.1875}=24.64 \quad \Rightarrow \quad QL=24 \quad \Rightarrow \quad \boxed{11000} \]

Problem 14

Given \(v_{in}(t) = 4 + 4\cos\!\left(360^\circ (2\,\text{kHz})t\right)\ \text{V}\), \(\text{max } QE = 200\ \text{mV}\):

Signal limits: \(V_{\min} = 0\ \text{V}\), \(V_{\max} = 8\ \text{V}\)

Sampling rate: \(f_s = 2(2\ \text{kHz}) = 4\ \text{kHz}\)

Bits required:

\[ 2^b = \frac{8\ \text{V}}{0.200\ \text{V/level}} = 40 \quad \Rightarrow \quad 2^6 = 64 \ge 40 \quad \Rightarrow \quad b = 6 \]

Actual resolution:

\[ \Delta V = \frac{8\ \text{V}}{64} = 125\ \text{mV/level} \]

For \(V_{in} = 1.6\ \text{V}\):

\[ EL = \frac{1.6 - 0}{0.125} = 12.8 \quad \Rightarrow \quad QL = 12 \quad \Rightarrow \quad \boxed{001100} \]

\[ QE = 0.125(12.8 - 12) = 100\ \text{mV} \]