Q:

What's the bit pattern for minimal value in double 64 bit

I assumed that the minimal positive value that can be put into double floating point is this
0 0000000000 00000000000000000000000000000000000000000000000000000

And in scientific form it's this:
1 x 2^{-1023}

However, this article states that:

As mentioned above, zero is not directly representable in the straight
  format, due to the assumption of a leading 1 (we'd need to specify a
  true zero mantissa to yield a value of zero).  Zero is a special value
  denoted with an exponent field of all zero bits, and a fraction field
  of all zero bits. 

So what's the pattern for minival positive value? 

A:

When the exponent part is all zero, the numbers are subnormals or denormals, where the implicit digit is 0 instead of 1.  Therefore
0 00000000000 0000000000000000000000000000000000000000000000000000

is zero, and the next number
0 00000000000 0000000000000000000000000000000000000000000000000001

is the smallest positive number, being equal to
0.00000000000000000000000000000000000000000000000000012×2-1022 = 2-1074 &approx 5.0×10-324
UPDATE: so why is the exponent -1022, when the bias is 1023?  The subnormals have the same exponent as the first binade of normal floats (so 1-1023 = 1022).  This is so that the spread is continuous, i.e. the largest subnormal is
0.11111111111111111111111111111111111111111111111111112×2-1022
and the next floating point number (the smallest normal):
1.00000000000000000000000000000000000000000000000000002×2-1022

