diff --git a/ME_2046/HW3/simplifying.py b/ME_2046/HW3/simplifying.py
new file mode 100644
index 0000000..5d92933
--- /dev/null
+++ b/ME_2046/HW3/simplifying.py
@@ -0,0 +1,103 @@
+import sympy as sm
+import numpy as np
+sm.init_printing()
+
+s = sm.symbols('s')
+z = sm.symbols('z')
+T = sm.symbols('T')
+
+#########################################################
+print('PROBLEM 2:')
+print('Part a:')
+
+ZOH = (1-sm.exp(-s*T))/(s*T)
+
+G_1_s = 1/s*ZOH
+
+bilinear_s = 2/T *(z-1)/(z+1)
+
+G_1_k = G_1_s.subs({s:bilinear_s}).simplify()
+
+print("G_1_k = ")
+sm.pprint(G_1_k)
+
+print('Part b:')
+
+theta_s_r_s = ZOH*1/s**2
+theta_k_r_k = theta_s_r_s.subs({s:bilinear_s}).simplify()
+print("theta_k_r_k = ")
+sm.pprint(theta_k_r_k)
+
+
+#########################################################
+print('PROBLEM 3:')
+A = sm.Matrix([[0, 1, 0, 0],
+               [0, 0, 1, 0],
+               [0, 0, 0, 1],
+               [1, 0, 0, 0]])
+B = sm.Matrix([0, 0, 0, 1])
+C = sm.Matrix([1, 0, 0, 0]).transpose()
+D = sm.Matrix([1])
+
+print('System Matricies A, B, C, D')
+sm.pprint(A)
+sm.pprint(B)
+sm.pprint(C)
+sm.pprint(D)
+
+#Recursive Solution
+k = sm.symbols('k', integer = True, real = True, positive = True)
+
+def y(k,u):
+    term_1 = C * A**k * x_0
+
+    term_2 = sm.Matrix([0,0,0,0])
+    for j in range(np.size(u)):
+        term_2 = term_2 + A**(k-j-1) * B * u[j]
+    term_2 = C*term_2
+
+    term_3 = D*u[-1]
+
+    return term_1+term_2+term_3
+
+print('Part c:')
+x_0 = sm.Matrix([2, 1, 3, 0])
+u = sm.Matrix([0])
+
+output = y(k, u)
+output = output[0].expand().simplify()
+print('y(k) = ')
+sm.pprint(output)
+
+
+
+print('Part d:')
+x_0 = sm.Matrix([0, 0, 0, 0])
+u = sm.Matrix([2, 1, 3, 0])
+output = y(k, u)
+output = output[0].expand().simplify()
+print('y(k) = ')
+sm.pprint(output)
+
+print('Part e:')
+print('These are the same exact response between parts C and D. This makes sense, becasue we defined our states as just being delays in a chain. The result is that the input at timestep k trickles down through each state in k+1, k+2, and k+3. This means that our states save our input in a way, s.t. loading this initial state mathematically produces an identical result as loading those inputs in one time step at a time. \n \n This is reflected by the two algebraeic expressions being the same.')
+
+#########################################################
+print('PROBLEM 4:')
+print('Part a:')
+
+"""
+x(k+2) - x(k+1) + 0.25 x(k) = u(k+2)
+
+Applying Z transform:
+z^2 X - z X + 0.25 X = z^2 U
+X (z^2 - z + 0.25) = z^2 U
+X/U = z^2 / (z^2 - z + 0.25)
+"""
+# Use SymPy to do partial frac
+z = sm.symbols('z')
+
+X_U = z**2/(z**2 - z + 0.25)
+sm.pprint(X_U.apart())
+
+