This operator shifts the first operand the specified number of bits to the right. Excess bits shifted off to the right are discarded. Copies of the leftmost bit are shifted in from the left. Since the new leftmost bit has the same value as the previous leftmost bit, the sign bit (the leftmost bit) does not change. Hence the name "sign-propagating".
Consider the 32-bit binary representations of the decimal (base 10) numbers 9 and -9:
9 (base 10): 00000000000000000000000000001001 (base 2)
-9 (base 10): 11111111111111111111111111110111 (base 2)
Notice that the binary representation of the negative decimal (base 10) number -9 is the two's complement of the binary representation of the positive decimal (base 10) number 9. That is, it's calculated by inverting all the bits of 00000000000000000000000000001001 and adding 1.
In both cases, the sign of the binary number is given by its leftmost bit: for the positive decimal number 9, the leftmost bit of the binary representation is 0, and for the negative decimal number -9, the leftmost bit of the binary representation is 1.
Given those binary representations of the decimal (base 10) numbers 9, and -9:
9 >> 2 yields 2:
9 (base 10): 00000000000000000000000000001001 (base 2)
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9 >> 2 (base 10): 00000000000000000000000000000010 (base 2) = 2 (base 10)
Notice how two rightmost bits, 01, have been shifted off, and two copies of the leftmost bit, 0 have been shifted in from the left.
-9 >> 2 yields -3:
-9 (base 10): 11111111111111111111111111110111 (base 2)
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-9 >> 2 (base 10): 11111111111111111111111111111101 (base 2) = -3 (base 10)
Notice how two rightmost bits, 11, have been shifted off. But as far as the leftmost bits: in this case, the leftmost bit is 1. So two copies of that leftmost 1 bit have been shifted in from the left — which preserves the negative sign.
The binary representation 11111111111111111111111111111101 is equal to the negative decimal (base 10) number -3, because all negative integers are stored as two's complements, and this one can be calculated by inverting all the bits of the binary representation of the positive decimal (base 10) number 3, which is 00000000000000000000000000000011, and then adding one.
The left operand will be converted to a 32-bit integer, which means floating point numbers will be truncated, and number not within the 32-bit bounds will over-/underflow.
The right operand will be converted to an unsigned 32-bit integer and then taken modulo 32, so the actual shift offset will always be a positive integer between 0 and 31, inclusive. For example, 100 >> 32 is the same as 100 >> 0 (and produces 100) because 32 modulo 32 is 0.