Complete HW2 and HW3

homework2/homework2.md

@@ -10,7 +10,7 @@
 | `0001`| s.0.3 | 0.125 |
 | `1000`| s.4.0 | -8 | 
 | `1110`| u.1.3 | 1.75 |
-| `1110`| s.1.2 | x |
-| `0x7fff`| s.0.15 | x |
-| `0x8000`| x | -1 |
-| `0x3fff`| s.5.10 | x |
+| `1110`| s.1.2 | -1.5 |
+| `0x7fff`| s.0.15 | 0.999969482421875 |
+| `0x8000`| s.0.15 | -1 |
+| `0x3fff`| s.5.10 | 15.9990234375 |

homework3/homework3.md

@@ -0,0 +1,38 @@
+# Hardware Description Language
+
+# Homework 3
+
+## Blizzard Finnegan
+
+Author's note:
+All answers here are using the SI standard prefixes (that is, multiples of 1000).  For more information, please see the [linked article](https://en.wikipedia.org/wiki/Binary_prefix).
+
+1. An average audio CD can contain 650MB of data. Given that the audio is stored *in stereo* 16-bit samples at a rate of 44100Hz, how many minutes of audio can a CD contain?
+
+$$
+stereo=\frac{2\times bits}{sample}\\
+\frac{32 bit}{1sample}\times\frac{44100sample}{1sec}\times\frac{650MB}{1 disc}\times\frac{1byte}{8bit}\\
+\frac{1sample}{4B}\times\frac{1sec}{44100sample}\times\frac{650MB}{1 disc}\times\frac{1000kB}{1MB}\times\frac{1000B}{1kB}=\frac{3684sec}{disc}\\
+3684sec \approx1h, 1min, 24sec
+$$
+
+2. It takes 10 bits to encode an **nxm** memory, where m is 8 bits. What is the size of the memory?
+   
+   10 bit address size: $2^{10}=1024$addresses
+   
+   Each address contains 8 bits (1byte).
+   
+   Total memory contains 8096 bits, or 1024 bytes. This is 1.024kB or 1KiB.
+   
+   
+
+3. How many bits does it take to encode a memory that contains 1 second of a mono, 16-bit audio snippet sampled at 8000 Hz?
+   
+   $$
+   \frac{16bit}{sample}\times\frac{8000sample}{sec}\times\frac{1sec}{1}=128000bit\\
+128kbit\ or\ 16kByte
+   $$
+   
+   
+
+