Solution of Exercise 6

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diff --git a/exercise06/helpers/ex_6_4.py b/exercise06/helpers/ex_6_4.py
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+#!/usr/bin/env python3
+
+# SPDX-License-Identifier: BSD-3-Clause
+#
+# Copyright 2020 Philipp Le
+#
+# Redistribution and use in source and binary forms, with or without
+# modification, are permitted provided that the following conditions
+# are met:
+#
+# 1. Redistributions of source code must retain the above copyright
+#    notice, this list of conditions and the following disclaimer.
+#
+# 2. Redistributions in binary form must reproduce the above copyright
+#    notice, this list of conditions and the following disclaimer in the
+#    documentation and/or other materials provided with the distribution.
+#
+# 3. Neither the name of the copyright holder nor the names of its
+#    contributors may be used to endorse or promote products derived from
+#    this software without specific prior written permission.
+#
+# THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
+# AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
+# IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
+# ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE
+# LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
+# CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF
+# SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
+# INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN
+# CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
+# ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF
+# THE POSSIBILITY OF SUCH DAMAGE.
+
+import numpy
+
+x = numpy.array([-0.5, 1, -2, 2])
+
+print("TEST ================================")
+
+print(numpy.fft.fft(x))
+
+print("a) ==================================")
+
+for k in range(4):
+    eq_str = "\\underline{X}[%d] = " % (k)
+    elems = []
+    res = 0
+    for n in range(4):
+        #elems.append("\\underline{x}[%d] \\cdot e^{-j \\frac{2\\pi}{N} %d \\cdot %d}" % (n, k, n))
+        elems.append("\\underline{x}[%d] \\cdot e^{-j \\frac{\\pi}{2} %d \\cdot %d}" % (n, k, n))
+        res += x[n] * numpy.exp(-1j*(numpy.pi/2)*n*k)
+    eq_str += " + ".join(elems)
+    eq_str += " = " + str(res)
+    print (eq_str)
+
+print("d) ==================================")
+
+E = numpy.zeros(2)
+O = numpy.zeros(2)
+for k in range(2):
+    eq_str = "\\underline{E}[%d] = \\underline{x}[0] + \\underline{w}_2^{%d} \\cdot \\underline{x}[2]" % (k, k)
+    E[k] = x[0] + numpy.exp(-1j*(numpy.pi)*k) * x[2]
+    eq_str += " = " + str(E[k])
+    print (eq_str)
+    
+    eq_str = "\\underline{O}[%d] = \\underline{x}[1] + \\underline{w}_2^{%d} \\cdot \\underline{x}[3]" % (k, k)
+    O[k] = x[1] + numpy.exp(-1j*(numpy.pi)*k) * x[3]
+    eq_str += " = " + str(O[k])
+    print (eq_str)
+
+for k in range(4):
+    eq_str = "\\underline{X}[%d] = \\underline{E}[%d] + \\underline{w}_2^{%d} \\cdot \\underline{O}[%d]" % (k, (k%2), k, (k%2))
+    res = E[k%2] + numpy.exp(-1j*(numpy.pi/2)*k) * O[k%2]
+    eq_str += " = " + str(res)
+    print (eq_str)