02. Digital_Engineering_HW2

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Digital_Engineering_HW2

문제

1. Program in C language to print as follows: --- What is the level of Quantization? 16 Type the values of 5 sampled points: 0 -1.2 1.5 -4.4 2.5 The level of the input samples are 8 6 10 0 12 Digitized signal after quantization and coding is 1000 0110 1010 0000 1100 Type a binary number for float point number: -1001010.11101 The converted floating number is 11000010100101011101000000000000 --- 2. Cosider followings - The analog input voltage level is from - 5V to 5V - With 16 level, -5V is mapped to Quantization level 0 and 0V is mapped to Quantization level 8. - Map 5V to the last quantization level. - Use Single Precision floating point number

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정답

// Homework #2 - 디지털 신호 양자화와 부동소주점 표현 #define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #include <string.h> #include <stdint.h> #include <math.h> // 필요한 비트 수 계산 int count_bit(int level) { return (int)ceil(log2(level)); // 밑이 2인 로그를 활용하여 2^n에서 n을 찾아냄 } // 양자화 함수 int quantization_lvl(int x, double y) { double min = -5.0; double max = 5.0; double space = (max - min) / x; int quantization = (y - min) / space; if (quantization >= x) { quantization = x - 1; } return quantization; } // 양자화된 값을 지정된 비트 수로 출력 void print_bits(int value, int bit_count) { for (int i = bit_count - 1; i >= 0; i--) { if ((value >> i) & 1) { // ex) 1 = 0001(2) => (0001 >> 3) = 0(0000), (0001 >> 2) = 0(0000), (0001 >> 1) = 0(0000), (0001 >> 1) = 1(0001) printf("1"); } else { printf("0"); }; } printf(" "); } // S-E-F void print_float_bits(float x) { // 공용체 선언 - 과제하면서 알게된 기능 union { float f; uint32_t i; // unint(부호없는) == unsigned int, 32(32비트), t(type의 약어) : 부호없는 32비트 정수형 => Use Single Precision floating point number } u; u.f = x; for (int i = 31; i >= 0; i--) { if ((u.i >> i) & 1) { // 실수값을 부호없는 정수형으로 접근하여 오른쪽으로 i 만큼 옮기는 비트연산자. 0인지 또는 1을 출력 printf("1"); } else { printf("0"); } } printf("\n"); } // 2진수 -> 10진수 double binary_to_double(char * x) { int negative = 0; if (x[0] == '-') { negative = 1; x++; } char * ptr = strchr(x, '.'); double rst = 0; if (ptr) *ptr = '\0'; for (int i = 0; x[i]; i++) { if (x[i] == '1') { rst += pow(2, strlen(x) - 1 - i); } } if (ptr) { ptr++; for (int i = 0; ptr[i]; i++) { if (ptr[i] == '1') { rst += pow(2, -(i+1)); } } } if (negative) { return -rst; } else { return rst; } } // 메인 함수 int main() { double sampled_points[5]; int quantization_lvl_list[5]; int qtz_lvl = 0; // What is the level of Quantization? - 입력 printf("What is the level of Quantization? "); scanf("%d", &qtz_lvl); if (qtz_lvl <= 1) { printf("양자화 레벨은 2 이상이어야 합니다.\n"); return 1; } // Type the values of 5 sampled points: - 출력 printf("Type the values of 5 sampled points: "); for (int i = 0; i < 5; i++) { scanf("%lf", &sampled_points[i]); } // The level of the input samples are - 출력 printf("The level of the input samples are "); for (int i = 0; i < 5; i++) { int q = quantization_lvl(qtz_lvl, sampled_points[i]); printf("%d ", q); quantization_lvl_list[i] = q; } printf("\n"); // Digitized signal after quantization and coding is - 출력 int bit_count = count_bit(qtz_lvl); printf("Digitized signal after quantization and coding is "); for (int i = 0; i < 5; i++) { print_bits(quantization_lvl_list[i], bit_count); } printf("\n"); // Type a binary number for float point number - 입력 char binary_num[100]; printf("Type a binary number for float point number: "); scanf("%s", binary_num); // The converted floating number is - 출력 float a = (float)binary_to_double(binary_num); printf("The converted floating number is "); print_float_bits(a); return 0; }

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결과

What is the level of Quantization? 16 Type the values of 5 sampled points: 0 -1.2 1.5 -4.4 2.5 The level of the input samples are 8 6 10 0 12 Digitized signal after quantization and coding is 1000 0110 1010 0000 1100 Type a binary number for float point number: -1001010.11101 The converted floating number is 11000010100101011101000000000000

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