diff --git a/ToegepasteAnalogeElektronica/SmithKaartOefening2.ipynb b/ToegepasteAnalogeElektronica/SmithKaartOefening2.ipynb
index f95c452a..985f9d48 100644
--- a/ToegepasteAnalogeElektronica/SmithKaartOefening2.ipynb
+++ b/ToegepasteAnalogeElektronica/SmithKaartOefening2.ipynb
@@ -3,6 +3,11 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
     "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-05T23:37:50.949025Z",
@@ -12,7 +17,7 @@
      "shell.execute_reply.started": "2020-12-05T23:37:50.948029Z"
     },
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": [
      "remove_cell4reveal"
@@ -63,9 +68,14 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
     "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -78,9 +88,14 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
     "editable": true,
     "slideshow": {
-     "slide_type": ""
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -90,25 +105,16 @@
     "We plaatsen $z= \\frac{Z_L}{Z_0}=1.6$ op de Smith kaart (rode dotje) en we zoeken waar we de r=1 cirkel snijden in het bovenste deel van de Smith kaart (groene dotje)."
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
-    },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
-   },
-   "source": [
-    "###### Smith kaart van de eerste oplossing."
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 12,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "Smith kaart van de eerste oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T19:48:14.450583Z",
      "iopub.status.busy": "2020-12-06T19:48:14.450583Z",
@@ -125,6 +131,9 @@
       "name": "smith2s1"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -159,8 +168,14 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -182,9 +197,12 @@
    "cell_type": "code",
    "execution_count": null,
    "metadata": {
-    "jupyter": {
-     "source_hidden": true
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
     },
+    "editable": true,
     "slideshow": {
      "slide_type": "skip"
     },
@@ -199,9 +217,12 @@
    "cell_type": "code",
    "execution_count": null,
    "metadata": {
-    "jupyter": {
-     "source_hidden": true
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
     },
+    "editable": true,
     "slideshow": {
      "slide_type": "skip"
     },
@@ -216,6 +237,12 @@
    "cell_type": "code",
    "execution_count": 6,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T19:40:36.450044Z",
      "iopub.status.busy": "2020-12-06T19:40:36.449046Z",
@@ -258,6 +285,12 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
      "slide_type": "skip"
     },
@@ -270,13 +303,23 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-11-29T09:50:19.955878Z",
      "iopub.status.busy": "2020-11-29T09:50:19.954881Z",
      "iopub.status.idle": "2020-11-29T09:50:19.960865Z",
      "shell.execute_reply": "2020-11-29T09:50:19.959867Z",
      "shell.execute_reply.started": "2020-11-29T09:50:19.955878Z"
-    }
+    },
+    "slideshow": {
+     "slide_type": "skip"
+    },
+    "tags": []
    },
    "source": [
     "<p class=\"alert alert-success\">\n",
@@ -287,8 +330,14 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -296,25 +345,16 @@
     "Het schema dat we nodig hebben om deze eerst oplossing te realiseren wordt dus:"
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
-    },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
-   },
-   "source": [
-    "###### circuit van de eerste oplossing"
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 13,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "circuit van de eerste oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2024-11-18T13:29:00.271302Z",
      "iopub.status.busy": "2024-11-18T13:29:00.270916Z",
@@ -328,6 +368,9 @@
       "name": "smith2f1"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -1111,8 +1154,14 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -1122,25 +1171,16 @@
     "We plaatsen $\\frac{Z_L}{Z_0}=1.6$ op de Smith kaart en we zoeken waar we de r=1 cirkel snijden in het onderste deel van de kaart."
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
-    },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
-   },
-   "source": [
-    "###### Smith kaart van de tweede oplossing"
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 13,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "Smith kaart van de tweede oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T19:48:53.726890Z",
      "iopub.status.busy": "2020-12-06T19:48:53.726890Z",
@@ -1157,6 +1197,9 @@
       "name": "smith2s2"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -1179,6 +1222,12 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
      "slide_type": "skip"
     },
@@ -1192,6 +1241,12 @@
    "cell_type": "code",
    "execution_count": 14,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T19:49:43.921568Z",
      "iopub.status.busy": "2020-12-06T19:49:43.920571Z",
@@ -1228,14 +1283,36 @@
   },
   {
    "cell_type": "markdown",
-   "metadata": {},
+   "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
+    "slideshow": {
+     "slide_type": ""
+    },
+    "tags": []
+   },
    "source": [
     "<p class=tabcaption> Overzicht van de bekomen impedantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax</p>"
    ]
   },
   {
    "cell_type": "markdown",
-   "metadata": {},
+   "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
+    "slideshow": {
+     "slide_type": "skip"
+    },
+    "tags": []
+   },
    "source": [
     "De nieuwe z = 1 - j 0.47. Daaruit volgt dat Z = 50 Ohm - j 23.68 Ohm. Het complexe deel van deze impedantie kunnen we compenseren door een spoel met impedantie van ongeveer  j 23.68 Ohm  toe te voegen. Als we het helemaal juist willen hebben kunnen we in de array van zlijn juist gaan kijken waar het reele deel 50 Ohm wordt en wat we dan als complex deel over houden.\n",
     "\n",
@@ -1266,8 +1343,14 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -1275,25 +1358,16 @@
     "Het schema dat we nodig hebben om deze tweede oplossing te realiseren wordt dus:"
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
-    },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
-   },
-   "source": [
-    "###### circuit van de tweede oplossing."
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 14,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "circuit van de tweede oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2024-11-18T13:29:34.949520Z",
      "iopub.status.busy": "2024-11-18T13:29:34.949286Z",
@@ -1307,6 +1381,9 @@
       "name": "smith2f2"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -2070,7 +2147,18 @@
   },
   {
    "cell_type": "markdown",
-   "metadata": {},
+   "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
+    "slideshow": {
+     "slide_type": "skip"
+    },
+    "tags": []
+   },
    "source": [
     "<p class=\"alert alert-success\">\n",
     "De optimale inductantiewaarde in serie wordt dus  12.56 nH en de nodige lengte van de coax tussen de antenne en de aanpassing is 5.3 cm\n",
@@ -2080,8 +2168,14 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -2091,25 +2185,16 @@
     "We plaatsen $\\frac{Z_L}{Z_0}=1.6$ op de Smith kaart en we zoeken waar we de y=1 cirkel snijden in het bovenste deel van de kaart."
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
-    },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
-   },
-   "source": [
-    "###### Smith kaart van de derde oplossing."
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 18,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "Smith kaart van de derde oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T19:55:05.595155Z",
      "iopub.status.busy": "2020-12-06T19:55:05.595155Z",
@@ -2126,6 +2211,9 @@
       "name": "smith2s3"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -2148,34 +2236,31 @@
   {
    "cell_type": "markdown",
    "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
     },
-    "tags": []
-   },
-   "source": [
-    "Omdat een Smith kaart in admitantie het spiegelbeeld is van een Smith kaart in impedentie, kunnen we ook de kaart spiegelen. We moeten dan wel aan de andere kant van de kaart vertrekken. Dit is weergegeven in figuur 6."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
+    "tags": []
    },
    "source": [
-    "###### Smith kaart van de derde oplossing als admittantie."
+    "Omdat een Smith kaart in admitantie het spiegelbeeld is van een Smith kaart in impedentie, kunnen we ook de kaart spiegelen. We moeten dan wel aan de andere kant van de kaart vertrekken. Dit is weergegeven in figuur 6."
    ]
   },
   {
    "cell_type": "code",
    "execution_count": 19,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "Smith kaart van de derde oplossing als admittantie."
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T19:55:53.034219Z",
      "iopub.status.busy": "2020-12-06T19:55:53.033221Z",
@@ -2192,6 +2277,9 @@
       "name": "smith2s3a"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -2216,6 +2304,12 @@
    "cell_type": "code",
    "execution_count": 21,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T19:56:50.281520Z",
      "iopub.status.busy": "2020-12-06T19:56:50.281520Z",
@@ -2227,7 +2321,7 @@
      "source_hidden": true
     },
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -2305,6 +2399,12 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
      "slide_type": "slide"
     },
@@ -2314,25 +2414,16 @@
     "Het schema dat we nodig hebben om deze derde oplossing te realiseren wordt dus:"
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
-    },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
-   },
-   "source": [
-    "###### circuit van de derde oplossing."
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 15,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "circuit van de derde oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2024-11-18T13:29:48.793242Z",
      "iopub.status.busy": "2024-11-18T13:29:48.793027Z",
@@ -2346,6 +2437,9 @@
       "name": "smith2f3"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -3030,8 +3124,14 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -3041,25 +3141,16 @@
     "We plaatsen $\\frac{Z_L}{Z_0}=\\frac{1}{1.6}$ op de Smith kaart en we zoeken waar we de y=1 cirkel snijden in het bovenste deel van de kaart."
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
-    },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
-   },
-   "source": [
-    "###### Smith kaart van de vierde oplossing."
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 24,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "Smith kaart van de vierde oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T20:01:43.005620Z",
      "iopub.status.busy": "2020-12-06T20:01:43.004655Z",
@@ -3076,6 +3167,9 @@
       "name": "smith2s4"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -3100,6 +3194,12 @@
    "cell_type": "code",
    "execution_count": 25,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T20:02:49.439668Z",
      "iopub.status.busy": "2020-12-06T20:02:49.439668Z",
@@ -3107,11 +3207,8 @@
      "shell.execute_reply": "2020-12-06T20:02:49.492527Z",
      "shell.execute_reply.started": "2020-12-06T20:02:49.439668Z"
     },
-    "jupyter": {
-     "source_hidden": true
-    },
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -3181,6 +3278,12 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
      "slide_type": "slide"
     },
@@ -3190,25 +3293,16 @@
     "Het schema dat we nodig hebben om deze vierde oplossing te realiseren wordt dus:"
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
-    },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
-   },
-   "source": [
-    "###### circuit van de vierde oplossing."
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 16,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "circuit van de vierde oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2024-11-18T13:29:55.913452Z",
      "iopub.status.busy": "2024-11-18T13:29:55.913242Z",
@@ -3222,6 +3316,9 @@
       "name": "smith2f4"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -3972,8 +4069,14 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -3983,25 +4086,16 @@
     "De y = 1 - j 0.47 compenseren we door +j 0.47 vertrekkende vanuit g=0."
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
-    },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
-   },
-   "source": [
-    "###### Smith kaart van de vijfde oplossing."
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 35,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "Smith kaart van de vijfde oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T20:15:33.196126Z",
      "iopub.status.busy": "2020-12-06T20:15:33.195131Z",
@@ -4018,6 +4112,9 @@
       "name": "smith2s5"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -4054,6 +4151,12 @@
    "cell_type": "code",
    "execution_count": 29,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T20:11:46.715525Z",
      "iopub.status.busy": "2020-12-06T20:11:46.715525Z",
@@ -4065,7 +4168,7 @@
      "source_hidden": true
     },
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -4090,7 +4193,18 @@
   },
   {
    "cell_type": "markdown",
-   "metadata": {},
+   "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
+    "slideshow": {
+     "slide_type": "skip"
+    },
+    "tags": []
+   },
    "source": [
     "<p class=tabcaption> Overzicht van de bekomen admitantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax</p>"
    ]
@@ -4099,6 +4213,12 @@
    "cell_type": "code",
    "execution_count": 33,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T20:13:32.014798Z",
      "iopub.status.busy": "2020-12-06T20:13:32.013800Z",
@@ -4109,6 +4229,9 @@
     "jupyter": {
      "source_hidden": true
     },
+    "slideshow": {
+     "slide_type": "skip"
+    },
     "tags": []
    },
    "outputs": [
@@ -4139,6 +4262,12 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
      "slide_type": "skip"
     },
@@ -4150,7 +4279,18 @@
   },
   {
    "cell_type": "markdown",
-   "metadata": {},
+   "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
+    "slideshow": {
+     "slide_type": "skip"
+    },
+    "tags": []
+   },
    "source": [
     "Hieruit blijkt dat een open transmissielijn met de lengte van $\\frac{51}{360} \\frac{\\lambda}{2}$ de beste aanpassing geeft."
    ]
@@ -4167,8 +4307,14 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "fragment"
     },
     "tags": []
    },
@@ -4176,25 +4322,16 @@
     "Het schema dat we nodig hebben om deze vijfde oplossing te realiseren wordt dus:"
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
-    },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
-   },
-   "source": [
-    "###### circuit van de vijfde oplossing."
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 17,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "circuit van de vijfde oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2024-11-18T13:30:04.150664Z",
      "iopub.status.busy": "2024-11-18T13:30:04.150287Z",
@@ -4208,6 +4345,9 @@
       "name": "smith2f5"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -4967,8 +5107,14 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -4978,25 +5124,16 @@
     "De y = 1 - j 0.47 compenseren we door +j 0.47 vertrekkende vanuit g=$\\infty$"
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
-    },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
-   },
-   "source": [
-    "###### Smith kaart van de zesde oplossing."
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 37,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "Smith kaart van de zesde oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T20:24:53.214401Z",
      "iopub.status.busy": "2020-12-06T20:24:53.213404Z",
@@ -5013,6 +5150,9 @@
       "name": "smith2s6"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -5037,7 +5177,13 @@
    "cell_type": "code",
    "execution_count": 40,
    "metadata": {
-    "execution": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
+    "execution": {
      "iopub.execute_input": "2020-12-06T20:26:41.703112Z",
      "iopub.status.busy": "2020-12-06T20:26:41.703112Z",
      "iopub.status.idle": "2020-12-06T20:26:41.750949Z",
@@ -5048,7 +5194,7 @@
      "source_hidden": true
     },
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -5074,6 +5220,12 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
      "slide_type": "skip"
     },
@@ -5087,6 +5239,12 @@
    "cell_type": "code",
    "execution_count": 39,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T20:26:14.542029Z",
      "iopub.status.busy": "2020-12-06T20:26:14.542029Z",
@@ -5094,8 +5252,8 @@
      "shell.execute_reply": "2020-12-06T20:26:14.591895Z",
      "shell.execute_reply.started": "2020-12-06T20:26:14.542029Z"
     },
-    "jupyter": {
-     "source_hidden": true
+    "slideshow": {
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -5121,6 +5279,12 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
      "slide_type": "skip"
     },
@@ -5130,46 +5294,54 @@
     "<p class=tabcaption> Overzicht van de bekomen admitantie van de kortgesloten transmissielijn als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax</p>"
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "<p class=\"alert alert-success\">\n",
-    "Het stukje kortgesloten coax dat we moeten voorzien ter vervanging van de condensator is dus: 16 cm\n",
-    "</p>"
-   ]
-  },
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
    "source": [
-    "Het schema dat we nodig hebben om deze zesde oplossing te realiseren wordt dus:"
+    "<p class=\"alert alert-success\">\n",
+    "Het stukje kortgesloten coax dat we moeten voorzien ter vervanging van de condensator is dus: 16 cm\n",
+    "</p>"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
+    "tags": []
    },
    "source": [
-    "###### circuit van de zesde oplossing."
+    "Het schema dat we nodig hebben om deze zesde oplossing te realiseren wordt dus:"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": 22,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "circuit van de zesde oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2024-11-18T13:31:45.493830Z",
      "iopub.status.busy": "2024-11-18T13:31:45.493445Z",
@@ -5183,6 +5355,9 @@
       "name": "smith2f6"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -5933,8 +6108,14 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -5944,25 +6125,16 @@
     "De y = 1 + j 0.47 compenseren we door -j 0.47 vertrekkende vanuit g=0."
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
-    },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
-   },
-   "source": [
-    "###### Smith kaart van de zevende oplossing."
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 51,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "Smith kaart van de zevende oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T20:34:30.456257Z",
      "iopub.status.busy": "2020-12-06T20:34:30.455259Z",
@@ -5979,6 +6151,9 @@
       "name": "smith2s7"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -6003,6 +6178,12 @@
    "cell_type": "code",
    "execution_count": 43,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T20:30:14.466373Z",
      "iopub.status.busy": "2020-12-06T20:30:14.466373Z",
@@ -6014,7 +6195,7 @@
      "source_hidden": true
     },
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -6041,6 +6222,12 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
      "slide_type": "skip"
     },
@@ -6054,6 +6241,12 @@
    "cell_type": "code",
    "execution_count": 49,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T20:32:42.530119Z",
      "iopub.status.busy": "2020-12-06T20:32:42.530119Z",
@@ -6065,7 +6258,7 @@
      "source_hidden": true
     },
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -6101,46 +6294,54 @@
     "<p class=tabcaption> Overzicht van de bekomen admitantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax</p>"
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "<p class=\"alert alert-success\">\n",
-    "Het stukje open coax dat we moeten voorzien ter vervanging van het spoel is dus: 21.46 cm\n",
-    "</p>"
-   ]
-  },
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
    "source": [
-    "Het schema dat we nodig hebben om deze zevende oplossing te realiseren wordt dus:"
+    "<p class=\"alert alert-success\">\n",
+    "Het stukje open coax dat we moeten voorzien ter vervanging van het spoel is dus: 21.46 cm\n",
+    "</p>"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
+    "tags": []
    },
    "source": [
-    "###### circuit van de zevende oplossing."
+    "Het schema dat we nodig hebben om deze zevende oplossing te realiseren wordt dus:"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": 23,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "circuit van de zevende oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2024-11-18T13:31:56.686456Z",
      "iopub.status.busy": "2024-11-18T13:31:56.686244Z",
@@ -6154,6 +6355,9 @@
       "name": "smith2f7"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -6936,8 +7140,14 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -6948,25 +7158,16 @@
     "De y = 1 + j 0.47 compenseren we door -j 0.47 vertrekkende vanuit g=$\\infty$. "
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "slideshow": {
-     "slide_type": "slide"
-    },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
-   },
-   "source": [
-    "###### Smith kaart van de achtste oplossing."
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 52,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "Smith kaart van de achtste oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T20:36:24.363514Z",
      "iopub.status.busy": "2020-12-06T20:36:24.363514Z",
@@ -6983,6 +7184,9 @@
       "name": "smith2s8"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [
@@ -7007,6 +7211,12 @@
    "cell_type": "code",
    "execution_count": 53,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T20:36:59.284509Z",
      "iopub.status.busy": "2020-12-06T20:36:59.283477Z",
@@ -7018,7 +7228,7 @@
      "source_hidden": true
     },
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -7044,7 +7254,18 @@
   },
   {
    "cell_type": "markdown",
-   "metadata": {},
+   "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
+    "slideshow": {
+     "slide_type": "skip"
+    },
+    "tags": []
+   },
    "source": [
     "<p class=tabcaption> Overzicht van de bekomen admitantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax</p>"
    ]
@@ -7053,6 +7274,12 @@
    "cell_type": "code",
    "execution_count": 57,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2020-12-06T20:39:29.499118Z",
      "iopub.status.busy": "2020-12-06T20:39:29.499118Z",
@@ -7064,7 +7291,7 @@
      "source_hidden": true
     },
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
@@ -7091,6 +7318,12 @@
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
      "slide_type": "skip"
     },
@@ -7100,46 +7333,54 @@
     "<p class=tabcaption> Overzicht van de bekomen admitantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax</p>"
    ]
   },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "<p class=\"alert alert-success\">\n",
-    "Het stukje kortgesloten coax dat we moeten voorzien ter vervanging van het spoel is dus: 9 cm\n",
-    "</p>"
-   ]
-  },
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
     "tags": []
    },
    "source": [
-    "Het schema dat we nodig hebben om deze achtste oplossing te realiseren wordt dus:"
+    "<p class=\"alert alert-success\">\n",
+    "Het stukje kortgesloten coax dat we moeten voorzien ter vervanging van het spoel is dus: 9 cm\n",
+    "</p>"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": ""
+    },
+    "editable": true,
     "slideshow": {
-     "slide_type": "slide"
+     "slide_type": "skip"
     },
-    "tags": [
-     "remove_cell",
-     "slide_title"
-    ]
+    "tags": []
    },
    "source": [
-    "###### circuit van de achtste oplossing."
+    "Het schema dat we nodig hebben om deze achtste oplossing te realiseren wordt dus:"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": 24,
    "metadata": {
+    "KULeuvenSlides": {
+     "slide_code": "normal",
+     "slide_ref": "",
+     "slide_title": "circuit van de achtste oplossing"
+    },
+    "editable": true,
     "execution": {
      "iopub.execute_input": "2024-11-18T13:32:07.620594Z",
      "iopub.status.busy": "2024-11-18T13:32:07.620111Z",
@@ -7147,12 +7388,18 @@
      "shell.execute_reply": "2024-11-18T13:32:07.731304Z",
      "shell.execute_reply.started": "2024-11-18T13:32:07.620574Z"
     },
+    "jupyter": {
+     "source_hidden": true
+    },
     "mystnb": {
      "figure": {
       "caption": "circuit van de achtste oplossing.",
       "name": "smith2f8"
      }
     },
+    "slideshow": {
+     "slide_type": "slide"
+    },
     "tags": []
    },
    "outputs": [