Logic analyzer for signal debugging in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of signal capture buffer code

What is the output of the following embedded C code that simulates capturing 8 digital signal samples into a buffer?

Embedded C

#include <stdio.h>
#include <stdint.h>

int main() {
    uint8_t signal_samples[8] = {1, 0, 1, 1, 0, 0, 1, 0};
    uint8_t buffer = 0;
    for (int i = 0; i < 8; i++) {
        buffer |= (signal_samples[i] << i);
    }
    printf("0x%02X\n", buffer);
    return 0;
}

A 0x4D

B 0x5B

C 0xA5

D 0x6B

💡 Hint

Remember that each bit is shifted by its index and ORed into the buffer.

The buffer accumulates bits from the signal_samples array, each shifted by its index. The bits are: 1 (bit0), 0 (bit1), 1 (bit2), 1 (bit3), 0 (bit4), 0 (bit5), 1 (bit6), 0 (bit7). This forms binary 01001101 which is 0x4D in hex.

🧠 Conceptual

intermediate

Understanding signal sampling rate importance

Why is it important to sample a digital signal at a rate higher than twice its highest frequency component when using a logic analyzer?

A To reduce the amount of data stored in the buffer

B To avoid aliasing and accurately reconstruct the signal transitions

C To increase the voltage level of the signal

D To synchronize the signal with the microcontroller clock

💡 Hint

Think about how sampling relates to signal frequency and reconstruction.

Sampling at more than twice the highest frequency (Nyquist rate) prevents aliasing, which causes distortion and incorrect signal representation.

🔧 Debug

advanced

Debugging incorrect signal capture in ISR

Given the following ISR code snippet for capturing a signal edge, what is the main bug causing incorrect signal data capture?

Embedded C

volatile uint8_t signal_buffer = 0;
volatile uint8_t bit_index = 0;

void ISR_signal_edge() {
    uint8_t signal_value = READ_SIGNAL_PIN();
    signal_buffer |= (signal_value << bit_index);
    bit_index++;
    if (bit_index >= 8) {
        bit_index = 0;
        PROCESS_SIGNAL(signal_buffer);
        signal_buffer = 0;
    }
}

A The signal_buffer is reset before processing

B The bit_index variable is not incremented

C The signal_value is not read from the correct pin

D No synchronization for shared variables between ISR and main code

💡 Hint

Consider what happens if main code reads signal_buffer while ISR modifies it.

Since signal_buffer and bit_index are shared between ISR and main code, they need atomic access or synchronization to avoid race conditions causing corrupted data.

📝 Syntax

advanced

Identify syntax error in signal capture loop

Which option contains the correct syntax for capturing 8 bits from a digital input pin into a byte variable?

Embedded C

uint8_t capture_signal() {
    uint8_t result = 0;
    for (int i = 0; i < 8; i++) {
        result |= (READ_PIN() << i)
    }
    return result;
}

A Add a semicolon after the bitwise OR operation line

B Change the for loop to while loop without braces

C Remove the shift operator <<

D Declare result as int instead of uint8_t

💡 Hint

Check each line for missing punctuation.

The line 'result |= (READ_PIN() << i)' is missing a semicolon at the end, causing a syntax error.

🚀 Application

expert

Calculate number of captured signal edges

A logic analyzer captures signal edges by storing each edge event in a 16-bit unsigned integer array of size 64. Each element stores the timestamp of an edge in microseconds. If the analyzer runs for 10 milliseconds and captures edges every 150 microseconds on average, how many edges will be stored in the array after the run?

A 66 edges

B 67 edges

C 64 edges

D 60 edges

💡 Hint

Calculate how many 150 microsecond intervals fit into 10 milliseconds, then consider array size.

10 ms = 10,000 us. 10,000 / 150 ≈ 66.66 edges, but array size is 64, so only 64 edges can be stored.