Hello tmont3, my answer here is not related to timing constraint as your thread title requests, but it may still help. Apologies if you are already doing this. I am also unclear what is meant by "does not reveal the input signal," can you clarify this?
Since you are describing an input to your FPGA which I am assuming can transition asynchronously from your internal clock, are you double-registering this input signal to remove metastability? If not, it is possible that this could be your problem.
If you are not already doing this, one thing you can try is setting a false path timing constraint on the input, then double-registering the input signal and using the 2nd register stage in your internal logic. For example:
Assume the input pin/port is named "MYINPUT"
.sdc file:
set_false_path -from [get_ports {MYINPUT}]
.vhd file:
signal myinput_q : std_logic_vector (1 downto 0);
input_demet_P : process (CLK, RST_N) begin
if (RST_N = '0') then
myinput_q <= (others => '0');
elsif (rising_edge(CLK)) then
myinput_q <= myinput_q(0) & MYINPUT;
end if;
end process input_demet_P;
...
do something with myinput_q(1) instead of MYINPUT
...