Using a 16-bit 2,s Complement normalised floating-point representation; 10-bit fractional mantissa and a 6-bit integer exponent: express 2.171875
I am lost with the given algorithm for a normalized representation.

The most significant bits should be kept, which is not what I see in your example.


1.111010010 x 2^4 = .1111010010 x 2^5 (This keeps the dot leftmost, and
one can possibly truncate bits to the right - or even round 0xxx down, and 1xxx up for positive numbers). Having a 1 immediately after the dot, allows the highest precision, compare with .00001111010010,
sign 1 bit = 0 (both ones complement, twos complement)
mantissa '1' + 8 bits = 111101001
exponent 6 bits two-complements = 000101
result: 0 11110101001 000101


This is just one possible variant: a ones complement mantissa with extra sign bit (but almost a two's complement mantissa). Here it shifted till a .1xxx
for a nonzero number. (Hence that 1 is almost redundant.)

Now your recipe:


1.111010010 x 2^4 = 1111.010010 x 2^1
sign 1 bit = 0 (both ones complement, twos complement)
mantissa 9 bits = 1111 01001
exponent 6 bits two-complements = 000001
result: 0 11110101001 000001


It only differs in the implicit +4.

For the .1 method:


2.171875 = number &lt; 1, times 2^exp = 0,54296875 x 2^2
0,54296875 = 0.1xxx
etcetera