Lesson 25 Practice Problems (KEY)

Lesson 25 Practice Problems (KEY)#

Problem 1

An accelerometer outputs \(100\ \text{mV}\) to \(500\ \text{mV}\); frequencies of interest are below \(4500\ \text{Hz}\). Refer to the ADC diagram in Problem 1.

a. Gain and Bias for Signal Conditioning

Map the accelerometer range \([0.1\ \text{V},\ 0.5\ \text{V}]\) to the ADC range \([2\ \text{V},\ 8\ \text{V}]\) using:

\[ v_{\text{out}} = K v_{\text{in}} + B \]

Minimum condition:

\[ (0.1\ \text{V})K + B = 2\ \text{V} \]

Maximum condition:

\[ (0.5\ \text{V})K + B = 8\ \text{V} \]

Solving:

\[ K = 15 \qquad B = 0.5\ \text{V} \]

Answer: \(v_{\text{out}} = 15\,v_{\text{in}} + 0.5\ \text{V}\)

b. Can the accelerometer be used directly with the ADC?

Answer: No. The accelerometer range (\(0.1\) – \(0.5\ \text{V}\)) does not span the ADC range (\(2\) – \(8\ \text{V}\)). Signal conditioning is required.

c. Does the ADC meet the Nyquist rate?

\[ f_s > 2f_{\text{High}} \implies f_s > 2(4500\ \text{Hz}) = 9\ \text{kHz} \]

Answer: Yes, provided \(f_s > 9\ \text{kHz}\) (confirmed from the ADC diagram).

d. Digital output for \(v_{in}(t) = 400\ \text{mV}\)

Conditioned voltage:

\[ v_{\text{out}} = 15(0.4\ \text{V}) + 0.5\ \text{V} = 6.5\ \text{V} \]

Resolution (\(b = 10\) bits):

\[ \Delta V = \frac{8\ \text{V} - 2\ \text{V}}{2^{10}} = \frac{6\ \text{V}}{1024} = 5.859\ \text{mV/level} \]

Expected level:

\[ EL = \frac{6.5\ \text{V} - 2\ \text{V}}{5.859\ \text{mV/level}} = 768 \]

Binary encoding (10-bit):

\(2^9\)

\(2^8\)

\(2^7\)

\(2^6\)

\(2^5\)

\(2^4\)

\(2^3\)

\(2^2\)

\(2^1\)

\(2^0\)

512

256

128

64

32

16

8

4

2

1

1

1

0

0

0

0

0

0

0

0

Answer: Digital output = \(1100000000_2\)

e. Bit rate into the FDR

\[ \text{bit rate} = b f_s = (10\ \text{bits})(10\ \text{kHz}) = 100\ \text{kbps} \]

f. Time to fill the FDR (1 MB card)

\[ \text{Time} = \frac{1\ \text{MB} \cdot 2^{20}\ \text{B/MB} \cdot 8\ \text{bits/B}}{100{,}000\ \text{bits/s}} = 83.9\ \text{s} \]
Problem 2

A low-pass filter (LPF) limits the highest frequency component \(f_{\text{High}}\) of the input signal before it reaches the ADC. To prevent aliasing, the cutoff frequency must satisfy:

\[ f_c \le \frac{f_s}{2} \]

This ensures no frequency components above the Nyquist frequency enter the ADC.

Problem 3 — True / False

T / F: An ADC’s maximum quantization error cannot be greater than its resolution.

Answer: False. The maximum quantization error is \(\frac{\Delta V}{2}\), which is half the resolution — always less than \(\Delta V\), not equal to or greater than it.

Problem 4

ADC for a music signal. Refer to the diagram in Problem 4 (\(f_s = 25\ \text{kHz}\), \(b = 5\ \text{bits}\), \(V_{\min} = -10\ \text{V}\), \(V_{\max} = 10\ \text{V}\)).

a. Maximum input frequency to prevent aliasing

\[ f_{\text{High}} \le \frac{f_s}{2} = \frac{25\ \text{kHz}}{2} = 12.5\ \text{kHz} \]

Answer: The LPF must limit the input to \(12.5\ \text{kHz}\) or lower.

b. Resolution

\[ \Delta V = \frac{V_{\max} - V_{\min}}{2^b} = \frac{10\ \text{V} - (-10\ \text{V})}{2^5} = \frac{20\ \text{V}}{32} = 625\ \text{mV/level} \]

c. Digital bandwidth (bit rate)

\[ \text{bit rate} = b f_s = (5\ \text{bits})(25\ \text{kHz}) = 125\ \text{kbps} \]
Problem 5

Two ADCs: ADC A (\(V_{\min}=0\), \(V_{\max}=5\ \text{V}\), \(b=6\), \(f_s=12\ \text{kHz}\)) and ADC B (\(V_{\min}=-2\ \text{V}\), \(V_{\max}=10\ \text{V}\), \(b=7\), \(f_s=10\ \text{kHz}\)).

a. Resolution

\[ \Delta V_A = \frac{5\ \text{V} - 0}{2^6} = \frac{5\ \text{V}}{64} = 78.1\ \text{mV/level} \]
\[ \Delta V_B = \frac{10\ \text{V} - (-2\ \text{V})}{2^7} = \frac{12\ \text{V}}{128} = 93.8\ \text{mV/level} \]

b. Better (smaller) resolution

Since \(\Delta V_A < \Delta V_B\), ADC A has the better resolution.

c. More samples in 10 seconds

\(f_{s,A} = 12\ \text{kHz} > f_{s,B} = 10\ \text{kHz}\), so ADC A takes more samples.

d. Which fills a 2 Gbit drive faster?

\[ \text{bit rate}_A = (12\ \text{kHz})(6) = 72\ \text{kbps} \]
\[ \text{bit rate}_B = (10\ \text{kHz})(7) = 70\ \text{kbps} \]

Answer: ADC A fills the drive faster because its bit rate is higher.

Problem 6

Accelerometer: \(-10\ \text{mV}\) at \(-2\ \text{G}\), \(+20\ \text{mV}\) at \(+8\ \text{G}\). ADC: \(V_{\min} = -2\ \text{V}\), \(V_{\max} = 12\ \text{V}\), \(b = 4\ \text{bits}\), \(f_s = 10\ \text{kHz}\).

a. Transducer interface (gain \(K\) and bias \(B\))

Map \([-10\ \text{mV},\ 20\ \text{mV}]\) to \([-2\ \text{V},\ 12\ \text{V}]\):

Minimum condition:

\[ (-0.010\ \text{V})K + B = -2\ \text{V} \]

Maximum condition:

\[ (0.020\ \text{V})K + B = 12\ \text{V} \]

Solving:

\[ K = 466.7 \qquad B = 2.67\ \text{V} \]

To prevent aliasing, add an LPF with:

\[ f_c = \frac{f_s}{2} = \frac{10\ \text{kHz}}{2} = 5\ \text{kHz} \]

b. Resolution

\[ \Delta V = \frac{12\ \text{V} - (-2\ \text{V})}{2^4} = \frac{14\ \text{V}}{16} = 875\ \text{mV/level} \]
Problem 7

ADC system: \(b = 4\ \text{bits}\), \(f_s = 10\ \text{kHz}\). FDR capacity: \(275\ \text{MB}\).

Solve: Compute the bit rate:

\[ \text{bit rate} = b f_s = (4\ \text{bits})(10\ \text{kHz}) = 40\ \text{kbps} \]

Convert storage to bits, then divide by bit rate:

\[ \text{Time} = \frac{275\ \text{MB} \cdot 2^{20}\ \text{B/MB} \cdot 8\ \text{bits/B}}{40{,}000\ \text{bits/s}} \cdot \frac{1\ \text{min}}{60\ \text{s}} = 959\ \text{min} \approx 16\ \text{hr} \]

Answer: The FDR can record approximately 16 hours of data.