What does the symbol $W_t$ mean in the SHA-256 specification? I'm really confused about the computation about SHA 256.

$$W_t = \begin{cases}
M_t^{(i)} &amp; 0 \leq t \leq 15 \\
&amp;\\
\sigma_1^{256}(W_{t-2}) + W_{t-7} + \sigma_0^{256}(W_{t-15}) + W_{t-16} &amp; 16 \leq t \leq 63 
\end{cases}$$

The $W_t$ variable: What's the value of $W_t$? How do I get that value?

I'm still confused because of the explanation using the English language. I'm a beginner guys, any help from you will be a plus point for me about hashing.
It is in SHA-256 message schedule (NIST-FIPS 180-4);

The message $M$ with length $l$ is first padded as the usual way;


append 1 to the end of the message,
then, add $k$ zero bits such that $$l+1+k \equiv 448 \mod 512$$
finally, add the length of the message in 64-bit. Now, the total padded length is divisible by 512.


After padding, the padded message parsed into $M^{1},\ldots,M^{N}$ where each has size 512-bit.

The sub index $t$ represents 32-bits in 512 bits. Thus, The $M_t^{(i)}$ is the $t$-th 32-bit in the $M^{(i)}$ for $0 \leq t \leq 15$ 

The $W_t$ is defined as your equation;

$$W_t = \begin{cases}
M_t^{(i)} &amp; 0 \leq t \leq 15 \\
&amp;\\
\sigma_1^{256}(W_{t-2}) + W_{t-7} + \sigma_0^{256}(W_{t-15}) + W_{t-16} &amp; 16 \leq t \leq 63 
\end{cases}$$

Where
$$\sigma_1^{256}(x) = \operatorname{ROTR}^{17}(x) \oplus  \operatorname{ROTR}^{19}(x) \oplus  \operatorname{SHR^{10}(x)}$$
$$\sigma_0^{256}(x) = \operatorname{ROTR}^{7}(x) \oplus  \operatorname{ROTR}^{18}(x) \oplus  \operatorname{SHR^{3}(x)}$$

$\sigma_1^{256}(x)$ and $\sigma_0^{256}(x)$ operate on 32 bits and produce 32 bits.

Note: the 256 above the $\sigma$ represents the 256 in SHA-256. Similarly, there is $\sigma_0^{512}(x)$ and $\sigma_1^{512}(x)$ for SHA-512.