Final answer:
The given autocorrelation function is not valid because it does not meet the conditions for a wide-sense stationary random process.
Step-by-step explanation:
A valid autocorrelation function must satisfy certain conditions such as being an even function and non-negative for all values of τ.
In this case, the function RX(t,τ) is not an even function and can take negative values depending on the value of sinτ.
Therefore, the given autocorrelation function RX(t,τ) = sinτ / t is not a valid autocorrelation function for a wide-sense stationary random process.
An example of a valid autocorrelation function for a wide-sense stationary random process would be RX(t,τ) = 2A / t2. This function satisfies the conditions of being an even function and non-negative for all values of τ.
Therefore, the given autocorrelation function RX(t,τ)=sinτ / t is not a valid autocorrelation function for a wide-sense stationary random process.