Encrypting text using Caeser cipher in CFB mode I am interested in learning and understanding cryptography and its application in every day life. One thing I can't seem to get my head around is encrypting plaintext using a block mode such as CFB. I want to use a Caesar block cipher in CFB mode to encrypt the following plaintext: lucozade with a key: 3. 

So in this case, I get that using the Caeser block cipher the ciphertext for lucozade would be oxfrcdgh as the letters have shifted three places to the right. What I can't do is this encryption using CFB mode. How do I go about doing that? Like would I have to use the ciphertext oxfrcdgh as the IV?

Any help would be great
Caesar cipher

You are confusing Caesar cipher with block ciphers.

In Caesar cipher the ciphertext is calculated as

$$c_i = p_i + 3 \bmod 26$$

There is no security in Caesar Cipher. A ciphertext only attack possible, as worst scenario.

CFB Caesar cipher

$c_i = c_{i-1} + 3 + p_i \bmod 26 $, with $c_0 = IV$

And, the decryption;

$p_i = c_i - 3 -c_{i-1} \bmod 26 $, with $c_0 = IV$

Example : $P=lucozade = \{11,20,2,14,25,0,3,4\}$

Let choose $IV = f = 5$

\begin{align}
  c_1 &amp;\equiv IV + 3 + p_1 \bmod 26 \equiv 5 + 3 + 11 \bmod 26 \equiv 19 \text { that is } T\\
  c_2 &amp;\equiv c_1 + 3 + p_2 \bmod 26 \equiv 19 + 3 + 20 \bmod 26 \equiv 16 \text { that is } Q\\
  c_3 &amp;\equiv c_2 + 3 + p_3 \bmod 26 \equiv 16 + 3 + 2 \bmod 26 \equiv 19 \text { that is } T\\
  c_4 &amp;\equiv c_3 + 3 + p_4 \bmod 26 \equiv 19 + 3 + 14 \bmod 26 \equiv 10 \text { that is } K\\
\end{align}