Update AC_01_Basics_Multirate.ipynb

Author: Profeta

AC_01_Basics_Multirate.ipynb

@@ -283,7 +283,7 @@
     "Here we get $c=e^{j\\frac {2 \\pi}{N}n}$, and the sum can hence be computed in a closed form:\n",
     "\n",
     "$$ \\large\n",
-    " \\sum _{k = 0}^{N - 1} {e^{j \\frac{2 π} { N} \\cdot n \\cdot k} } =\\frac{e^{j \\frac{2  π} { N} \\cdot n \\cdot N} - 1}  {e^{j \\frac{2  π} {N }\\cdot n} - 1}\n",
+    " \\sum _{k = 0}^{N - 1} {e^{j \\frac{2 \\pi} { N} \\cdot n \\cdot k} } =\\frac{e^{j \\frac{2  \\pi} { N} \\cdot n \\cdot N} - 1}  {e^{j \\frac{2  π} {N }\\cdot n} - 1}\n",
     "$$\n",
     "\n",
     "Now we have to prove that this is indeed true (since we only applied a guess). To do that, we can look at the sum. It is a geometric sum (a sum over a constant with the summation index in the exponent):\n",
@@ -301,7 +301,7 @@
     "x^d \\left ( n \\right )  =x \\left ( n \\right )  \\cdot \\Delta_N \\left ( n \\right ) $$\n",
     "\n",
     "$$\\large\n",
-    "=x  ( n  ) \\cdot \\frac{1}{N}  \\sum _ {k = 0} ^{N - 1} e^{j \\frac{2 π} {N} \\cdot k \\cdot n}\n",
+    "=x  ( n  ) \\cdot \\frac{1}{N}  \\sum _ {k = 0} ^{N - 1} e^{j \\frac{2 \\pi} {N} \\cdot k \\cdot n}\n",
     "$$\n",
     "\n",
     "Taking its Discrete Time Fourier transform now results in\n",
@@ -1035,13 +1035,13 @@
     "This can be seen in following pictures,\n",
     "\n",
     "<center>\n",
-    "    <img src='./images/lecture6-6.png' width='700'>\n",
+    "    <img src='./images/Lecture6-6.png' width='700'>\n",
     "    Figure: The magnitude spectrum of a signal. The 2 boxes symbolize the passband of an ideal bandpass, here a high pass.\n",
-    "    <img src='./images/lecture6-8.png' width='700'>\n",
+    "    <img src='./images/Lecture6-8.png' width='700'>\n",
     "    Figure: The signal spectrum after passing through the high pass. \n",
-    "    <img src='./images/lecture6-7.png' width='700' >\n",
+    "    <img src='./images/Lecture6-7.png' width='700' >\n",
     "    Figure: Signal spectrum after multiplication with the unit pulse train, for N=2, hence setting every second value to zero (the zeros still in the sequence). Observe that we shift and add the signal by multiples of $2\\pi /2=\\pi$, and in effect we obtain „mirrored“ images of the high frequencies to the low frequencies (since we assume a real valued signal). Observe that the mirrored spectra and the original spectrum don't overlap, which makes reconstruction easy.\n",
-    "    <img src='./images/lecture6-9.png' width='700'>\n",
+    "    <img src='./images/Lecture6-9.png' width='700'>\n",
     "    Figure: Signal spectrum after downsampling (removing the zeros) by N (2 in this example). Observe the stretching of the spectrum by a factor of 2. \n",
     "</center>\n",
     "\n",
@@ -1129,7 +1129,7 @@
     "\n",
     "<center>\n",
     "    <br>\n",
-    "    <img src='./images/lecture6-10.png' width='700'>\n",
+    "    <img src='./images/Lecture6-10.png' width='700'>\n",
     "</center>\n",
     "\n",
     "Observe that we **violate the conventional Nyquist** criterium, because our high pass passes the high frequencies. But then the sampling **mirrors** those frequencies **to the lower range**, such that we can apply the traditional Nyquist sampling theorem. This method is also known as bandpass Nyquist. This is an important principle for filter banks and wavelets. It says that we can perfectly reconstruct a bandpass signal in a filter bank, if we sample with twice the rate as the **bandwidth** of our bandpass signal (assuming ideal filters, to avoid spectral overlap of aliasing components)."
@@ -1203,7 +1203,7 @@
     "\n",
     "Using (see: Lecture ADSP, Slides 06)\n",
     "$$\\large\n",
-    "\\Delta_N(n)= \\frac{1} {N } \\sum_{k = 0} ^ {N - 1} e^{j \\frac{2 π} {N} ∙ k ∙ n}\n",
+    "\\Delta_N(n)= \\frac{1} {N } \\sum_{k = 0} ^ {N - 1} e^{j \\frac{2 \\pi} {N} \\cdot k \\cdot n}\n",
     "$$\n",
     "\n",
     "this becomes\n",
@@ -1506,7 +1506,6 @@
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@@ -1863,7 +1862,7 @@
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