## Signed binary 0111 0101 1100 0100_{(2)} to an integer in decimal system (in base 10) = ?

### 1. Is this a positive or a negative number?

#### In a signed binary, first bit (the leftmost) is reserved for the sign, 1 = negative, 0 = positive. This bit does not count when calculating the absolute value.

#### 0111 0101 1100 0100 is the binary representation of a positive integer, on 16 bits (2 Bytes).

### 2. Construct the unsigned binary number, exclude the first bit (the leftmost), that is reserved for the sign:

#### 0111 0101 1100 0100 = 111 0101 1100 0100

### 3. Map the unsigned binary number's digits versus the corresponding powers of 2 that their place value represent:

2^{14}

1 2^{13}

1 2^{12}

1 2^{11}

0 2^{10}

1 2^{9}

0 2^{8}

1 2^{7}

1 2^{6}

1 2^{5}

0 2^{4}

0 2^{3}

0 2^{2}

1 2^{1}

0 2^{0}

0

### 4. Multiply each bit by its corresponding power of 2 and add all the terms up:

#### 111 0101 1100 0100_{(2)} =

#### (1 × 2^{14} + 1 × 2^{13} + 1 × 2^{12} + 0 × 2^{11} + 1 × 2^{10} + 0 × 2^{9} + 1 × 2^{8} + 1 × 2^{7} + 1 × 2^{6} + 0 × 2^{5} + 0 × 2^{4} + 0 × 2^{3} + 1 × 2^{2} + 0 × 2^{1} + 0 × 2^{0})_{(10)} =

#### (16 384 + 8 192 + 4 096 + 0 + 1 024 + 0 + 256 + 128 + 64 + 0 + 0 + 0 + 4 + 0 + 0)_{(10)} =

#### (16 384 + 8 192 + 4 096 + 1 024 + 256 + 128 + 64 + 4)_{(10)} =

#### 30 148_{(10)}

### 5. If needed, adjust the sign of the integer number by the first digit (leftmost) of the signed binary:

#### 0111 0101 1100 0100_{(2)} = 30 148_{(10)}

## Number 0111 0101 1100 0100_{(2)} converted from signed binary to an integer in decimal system (in base 10):

0111 0101 1100 0100_{(2)} = 30 148_{(10)}

#### Spaces used to group digits: for binary, by 4; for decimal, by 3.

### More operations of this kind:

## Convert signed binary numbers to integers in decimal system (base 10)

#### First bit (the leftmost) is reserved for the sign, 1 = negative, 0 = positive. This bit does not count when calculating the absolute value.

#### Entered binary number length must be: 2, 4, 8, 16, 32, or 64 - otherwise extra bits on 0 will be added in front (to the left).