From fdd6ccef311895ed1f0ea5de7213cf9f141da715 Mon Sep 17 00:00:00 2001 From: Jan Genoe Date: Sat, 14 Dec 2024 17:55:37 +0100 Subject: [PATCH] pptx titels --- AnalogeElektronica2/Klasse-B-vb4.ipynb | 52 +- .../Semiconductor industry.ipynb | 228 +++- .../SmithKaartOefening1.ipynb | 123 +- .../SmithKaartOefening4.ipynb | 1015 ++++++++++++----- .../SmithKaartOefening5.ipynb | 944 +++++++++------ .../SmithKaartOefening6.ipynb | 57 +- .../SmithKaartOefening7.ipynb | 36 +- ToegepasteAnalogeElektronica/reflecties.ipynb | 6 +- 8 files changed, 1773 insertions(+), 688 deletions(-) diff --git a/AnalogeElektronica2/Klasse-B-vb4.ipynb b/AnalogeElektronica2/Klasse-B-vb4.ipynb index 7b5944794..03d9a614f 100644 --- a/AnalogeElektronica2/Klasse-B-vb4.ipynb +++ b/AnalogeElektronica2/Klasse-B-vb4.ipynb @@ -23,8 +23,8 @@ "slide_type": "skip" }, "tags": [ - "hide-input", - "remove_cell4reveal" + "remove_cell4reveal", + "remove_cell4pptx" ] }, "source": [ @@ -58,10 +58,10 @@ "slide_type": "skip" }, "tags": [ - "hide-input", "remove_cell4reveal", - "remove-input", - "remove_cell" + "remove-cell4PDF", + "remove-cell4BOOK", + "remove_cell4pptx" ] }, "outputs": [], @@ -75,7 +75,7 @@ "KULeuvenSlides": { "slide_code": "normal", "slide_ref": "", - "slide_title": " {cite}`jiangFundamentalsAudioClass2017`" + "slide_title": "" }, "citation-manager": { "citations": { @@ -91,9 +91,7 @@ "slideshow": { "slide_type": "skip" }, - "tags": [ - "hide-input" - ] + "tags": [] }, "source": [ "In {cite}`jiangFundamentalsAudioClass2017`\n", @@ -167,23 +165,6 @@ "slide_ref": "", "slide_title": "" }, - "editable": true, - "slideshow": { - "slide_type": "slide" - }, - "tags": [ - "remove_cell", - "slide_overzicht" - ] - }, - "source": [ - "###### Overzicht\n", - " \n" - ] - }, - { - "cell_type": "markdown", - "metadata": { "citation-manager": { "citations": { "vks4r": [ @@ -194,12 +175,11 @@ ] } }, + "editable": true, "slideshow": { "slide_type": "skip" }, - "tags": [ - "hide-input" - ] + "tags": [] }, "source": [ "Wanneer we de spanning aan de gate en de source invullen krijgen we:\n", @@ -218,6 +198,12 @@ "cell_type": "code", "execution_count": 1, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "mystnb": { "figure": { "caption": "Transfer curve en stroom van de beide transistors.", @@ -227,9 +213,7 @@ "slideshow": { "slide_type": "slide" }, - "tags": [ - "remove-input" - ] + "tags": [] }, "outputs": [ { @@ -340,9 +324,7 @@ "slideshow": { "slide_type": "slide" }, - "tags": [ - "remove-input" - ] + "tags": [] }, "outputs": [ { diff --git a/MicroEnNanoTechnologie/Semiconductor industry.ipynb b/MicroEnNanoTechnologie/Semiconductor industry.ipynb index 1c8033f36..fb99da5e4 100644 --- a/MicroEnNanoTechnologie/Semiconductor industry.ipynb +++ b/MicroEnNanoTechnologie/Semiconductor industry.ipynb @@ -3,11 +3,19 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, "editable": true, "slideshow": { - "slide_type": "" + "slide_type": "skip" }, - "tags": [] + "tags": [ + "remove_cell4reveal", + "remove_cell4pptx" + ] }, "source": [ "# Semiconductor industry" @@ -15,7 +23,21 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [ + "remove_cell4pptx", + "remove_cell4reveal" + ] + }, "source": [ "Deze pagina overloopt alle fabs in de wereld waar er momenteel chips worden gemaakt, op voorwaarde dat deze fab gekend is op [wikipedia](https://en.wikipedia.org/wiki/List_of_semiconductor_fabrication_plants).\n" ] @@ -24,7 +46,27 @@ "cell_type": "code", "execution_count": null, "metadata": { - "tags": [] + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "" + }, + "tags": [ + "remove_cell4reveal", + "remove-cell4PDF", + "remove-cell4BOOK", + "remove_cell4pptx" + ] }, "outputs": [], "source": [ @@ -36,6 +78,11 @@ "cell_type": "code", "execution_count": 6, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, "editable": true, "execution": { "iopub.execute_input": "2023-02-19T14:58:17.467046Z", @@ -44,10 +91,21 @@ "shell.execute_reply": "2023-02-19T14:58:17.471877Z", "shell.execute_reply.started": "2023-02-19T14:58:17.467046Z" }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "" }, - "tags": [] + "tags": [ + "remove_cell4reveal", + "remove-cell4PDF", + "remove_cell4pptx", + "remove-cell4BOOK" + ] }, "outputs": [ { @@ -66,6 +124,12 @@ "cell_type": "code", "execution_count": 22, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2021-01-06T16:37:28.067495Z", "iopub.status.busy": "2021-01-06T16:37:28.067495Z", @@ -73,8 +137,20 @@ "shell.execute_reply": "2021-01-06T16:37:28.072481Z", "shell.execute_reply.started": "2021-01-06T16:37:28.067495Z" }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "" + }, "tags": [ - "remove_cell" + "remove_cell4pptx", + "remove-cell4BOOK", + "remove_cell4reveal", + "remove-cell4PDF" ] }, "outputs": [ @@ -105,6 +181,12 @@ "cell_type": "code", "execution_count": 13, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2021-01-06T16:33:50.030919Z", "iopub.status.busy": "2021-01-06T16:33:50.030919Z", @@ -112,8 +194,20 @@ "shell.execute_reply": "2021-01-06T16:33:50.039860Z", "shell.execute_reply.started": "2021-01-06T16:33:50.030919Z" }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "skip" + }, "tags": [ - "remove_cell" + "remove_cell4reveal", + "remove-cell4PDF", + "remove_cell4pptx", + "remove-cell4BOOK" ] }, "outputs": [ @@ -225,6 +319,12 @@ "cell_type": "code", "execution_count": 45, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2021-01-06T16:56:48.385537Z", "iopub.status.busy": "2021-01-06T16:56:48.385537Z", @@ -232,8 +332,20 @@ "shell.execute_reply": "2021-01-06T16:56:48.390524Z", "shell.execute_reply.started": "2021-01-06T16:56:48.385537Z" }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "" + }, "tags": [ - "remove_cell" + "remove_cell4reveal", + "remove-cell4PDF", + "remove-cell4BOOK", + "remove_cell4pptx" ] }, "outputs": [], @@ -250,6 +362,12 @@ "cell_type": "code", "execution_count": 46, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2021-01-06T16:56:51.471634Z", "iopub.status.busy": "2021-01-06T16:56:51.470636Z", @@ -257,8 +375,20 @@ "shell.execute_reply": "2021-01-06T16:56:51.478615Z", "shell.execute_reply.started": "2021-01-06T16:56:51.471634Z" }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "skip" + }, "tags": [ - "remove_cell" + "remove_cell4reveal", + "remove-cell4PDF", + "remove_cell4pptx", + "remove-cell4BOOK" ] }, "outputs": [ @@ -392,6 +522,12 @@ "cell_type": "code", "execution_count": 44, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2021-01-06T16:56:30.748153Z", "iopub.status.busy": "2021-01-06T16:56:30.747156Z", @@ -399,8 +535,20 @@ "shell.execute_reply": "2021-01-06T16:56:30.754134Z", "shell.execute_reply.started": "2021-01-06T16:56:30.748153Z" }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "" + }, "tags": [ - "remove_cell" + "remove-cell4BOOK", + "remove-cell4PDF", + "remove_cell4reveal", + "remove_cell4pptx" ] }, "outputs": [ @@ -423,6 +571,12 @@ "cell_type": "code", "execution_count": 31, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2021-01-06T16:51:31.472070Z", "iopub.status.busy": "2021-01-06T16:51:31.472070Z", @@ -430,8 +584,20 @@ "shell.execute_reply": "2021-01-06T16:51:31.497999Z", "shell.execute_reply.started": "2021-01-06T16:51:31.472070Z" }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "" + }, "tags": [ - "remove_cell" + "remove_cell4reveal", + "remove-cell4PDF", + "remove-cell4BOOK", + "remove_cell4pptx" ] }, "outputs": [ @@ -978,6 +1144,12 @@ "cell_type": "code", "execution_count": 28, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2021-01-06T16:40:54.787594Z", "iopub.status.busy": "2021-01-06T16:40:54.786596Z", @@ -985,8 +1157,20 @@ "shell.execute_reply": "2021-01-06T16:40:54.792580Z", "shell.execute_reply.started": "2021-01-06T16:40:54.787594Z" }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "" + }, "tags": [ - "remove_cell" + "remove-cell4PDF", + "remove-cell4BOOK", + "remove_cell4pptx", + "remove_cell4reveal" ] }, "outputs": [ @@ -1018,6 +1202,12 @@ "cell_type": "code", "execution_count": 27, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2021-01-06T16:38:57.924453Z", "iopub.status.busy": "2021-01-06T16:38:57.924453Z", @@ -1025,8 +1215,20 @@ "shell.execute_reply": "2021-01-06T16:38:57.935390Z", "shell.execute_reply.started": "2021-01-06T16:38:57.924453Z" }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "" + }, "tags": [ - "remove_cell" + "remove-cell4PDF", + "remove-cell4BOOK", + "remove_cell4pptx", + "remove_cell4reveal" ] }, "outputs": [ diff --git a/ToegepasteAnalogeElektronica/SmithKaartOefening1.ipynb b/ToegepasteAnalogeElektronica/SmithKaartOefening1.ipynb index 4decdb39b..229a864ec 100644 --- a/ToegepasteAnalogeElektronica/SmithKaartOefening1.ipynb +++ b/ToegepasteAnalogeElektronica/SmithKaartOefening1.ipynb @@ -17,10 +17,11 @@ "shell.execute_reply.started": "2020-12-05T23:37:50.948029Z" }, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [ - "remove_cell4reveal" + "remove_cell4reveal", + "remove_cell4pptx" ] }, "source": [ @@ -56,7 +57,8 @@ "tags": [ "remove-cell4BOOK", "remove_cell4pptx", - "remove-cell4PDF" + "remove-cell4PDF", + "remove_cell4reveal" ] }, "outputs": [], @@ -94,7 +96,7 @@ }, "editable": true, "slideshow": { - "slide_type": "" + "slide_type": "skip" }, "tags": [] }, @@ -130,7 +132,7 @@ } }, "slideshow": { - "slide_type": "" + "slide_type": "slide" }, "tags": [] }, @@ -208,7 +210,7 @@ } }, "slideshow": { - "slide_type": "" + "slide_type": "slide" }, "tags": [] }, @@ -237,7 +239,7 @@ "KULeuvenSlides": { "slide_code": "normal", "slide_ref": "", - "slide_title": "" + "slide_title": "Smith kaart van de admittantie" }, "editable": true, "execution": { @@ -254,7 +256,7 @@ } }, "slideshow": { - "slide_type": "" + "slide_type": "slide" }, "tags": [] }, @@ -279,6 +281,11 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, "editable": true, "slideshow": { "slide_type": "skip" @@ -295,6 +302,12 @@ "cell_type": "code", "execution_count": 16, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2021-12-05T21:17:12.924626Z", "iopub.status.busy": "2021-12-05T21:17:12.924626Z", @@ -302,8 +315,11 @@ "shell.execute_reply": "2021-12-05T21:17:12.967514Z", "shell.execute_reply.started": "2021-12-05T21:17:12.924626Z" }, - "jupyter": { - "source_hidden": true + "mystnb": { + "figure": { + "caption": "", + "name": "" + } }, "slideshow": { "slide_type": "slide" @@ -344,8 +360,17 @@ "cell_type": "code", "execution_count": null, "metadata": { - "jupyter": { - "source_hidden": true + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } }, "slideshow": { "slide_type": "skip" @@ -378,8 +403,17 @@ "cell_type": "code", "execution_count": null, "metadata": { - "jupyter": { - "source_hidden": true + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } }, "slideshow": { "slide_type": "skip" @@ -394,6 +428,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "slide" }, @@ -419,7 +459,7 @@ }, "editable": true, "slideshow": { - "slide_type": "" + "slide_type": "skip" }, "tags": [] }, @@ -451,7 +491,7 @@ } }, "slideshow": { - "slide_type": "" + "slide_type": "slide" }, "tags": [] }, @@ -1225,8 +1265,14 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [] }, @@ -1261,8 +1307,14 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [] }, @@ -1280,8 +1332,17 @@ "cell_type": "code", "execution_count": null, "metadata": { - "jupyter": { - "source_hidden": true + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } }, "slideshow": { "slide_type": "skip" @@ -1295,7 +1356,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "
\n", "Dit laat ons toe vast te stellen dat de spanningsamplitude als gevolg van de aanpassing gaat stijgen van 5 V (normaal op de lijn) naar 8V ter hoogte van de aanpassing.\n", @@ -1313,8 +1385,11 @@ "slide_title": "" }, "editable": true, - "jupyter": { - "source_hidden": true + "mystnb": { + "figure": { + "caption": "", + "name": "" + } }, "slideshow": { "slide_type": "skip" @@ -1350,7 +1425,7 @@ } }, "slideshow": { - "slide_type": "" + "slide_type": "slide" }, "tags": [] }, diff --git a/ToegepasteAnalogeElektronica/SmithKaartOefening4.ipynb b/ToegepasteAnalogeElektronica/SmithKaartOefening4.ipynb index 4137ad5e2..35743eadf 100644 --- a/ToegepasteAnalogeElektronica/SmithKaartOefening4.ipynb +++ b/ToegepasteAnalogeElektronica/SmithKaartOefening4.ipynb @@ -3,12 +3,18 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [ - "remove_cell4reveal" + "remove_cell4reveal", + "remove_cell4pptx" ] }, "source": [ @@ -56,6 +62,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "slide" }, @@ -69,7 +81,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "## Oplossing\n", "\n", @@ -81,9 +104,14 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, "editable": true, "slideshow": { - "slide_type": "" + "slide_type": "skip" }, "tags": [] }, @@ -93,25 +121,16 @@ "We plaatsen $z= \\frac{Z_L}{Z_0}=3$ 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": 6, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de eerste oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T09:51:17.585980Z", "iopub.status.busy": "2020-12-04T09:51:17.585980Z", @@ -128,6 +147,9 @@ "name": "smith4s1" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -161,7 +183,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "Als we van de aanpassing (groene dot) naar de belasting gaan (rode dot), draaien we in tegenwijzerzin over een afstand van $300^o$. Dit geeft een afstand van $\\frac{300}{360}\\frac{\\lambda}{2}$=0.416 $\\lambda$ = 10.41cm. ($\\lambda$= 25 cm is gegeven)\n", "\n", @@ -180,8 +213,20 @@ "cell_type": "code", "execution_count": null, "metadata": { - "jupyter": { - "source_hidden": true + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "" }, "tags": [] }, @@ -194,6 +239,12 @@ "cell_type": "code", "execution_count": 70, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T23:16:27.311508Z", "iopub.status.busy": "2020-12-04T23:16:27.311508Z", @@ -204,6 +255,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -241,7 +298,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen impedantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] @@ -263,32 +331,34 @@ "

" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Het schema dat we nodig hebben om deze eerst oplossing te realiseren wordt dus:" - ] - }, { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "" }, - "tags": [ - "remove_cell", - "slide_title" - ] + "tags": [] }, "source": [ - "###### circuit van de eerste oplossing." + "Het schema dat we nodig hebben om deze eerst oplossing te realiseren wordt dus:" ] }, { "cell_type": "code", "execution_count": 58, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de eerste oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T22:40:20.061846Z", "iopub.status.busy": "2020-12-04T22:40:20.061846Z", @@ -305,6 +375,9 @@ "name": "smith4f1" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -985,6 +1058,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "slide" }, @@ -996,25 +1075,16 @@ "We plaatsen $\\frac{Z_L}{Z_0}=3$ 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": 8, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de tweede oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T09:52:34.330153Z", "iopub.status.busy": "2020-12-04T09:52:34.330153Z", @@ -1031,6 +1101,9 @@ "name": "smith4s2" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -1058,7 +1131,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "We zoomen vervolgens in op het deel van de cirkel die de eenheidscirkel snijdt." ] @@ -1067,6 +1151,12 @@ "cell_type": "code", "execution_count": 73, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T23:26:16.506237Z", "iopub.status.busy": "2020-12-04T23:26:16.506237Z", @@ -1077,6 +1167,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -1109,7 +1205,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen impedantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] @@ -1155,32 +1262,34 @@ "57.73502691896258/2/np.pi/6e8" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Het schema dat we nodig hebben om deze tweede oplossing te realiseren wordt dus:" - ] - }, { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "" }, - "tags": [ - "remove_cell", - "slide_title" - ] + "tags": [] }, "source": [ - "###### circuit van de tweede oplossing." + "Het schema dat we nodig hebben om deze tweede oplossing te realiseren wordt dus:" ] }, { "cell_type": "code", "execution_count": 71, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de tweede oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T23:23:30.911006Z", "iopub.status.busy": "2020-12-04T23:23:30.911006Z", @@ -1197,6 +1306,9 @@ "name": "smith4f2" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -1847,41 +1959,54 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

\n", "De optimale inductantiewaarde in serie wordt dus 15 nH en de nodige lengte van de coax tussen de antenne en de aanpassing is 2.1 cm\n", "

" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "### Oplossing 3: Bijplaatsen van een capaciteit in parallel\n", - "\n", - "We plaatsen $\\frac{Z_L}{Z_0}=3$ 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": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, - "tags": [ - "remove_cell", - "slide_title" - ] + "tags": [] }, "source": [ - "###### Smith kaart van de derde oplossing." + "### Oplossing 3: Bijplaatsen van een capaciteit in parallel\n", + "\n", + "We plaatsen $\\frac{Z_L}{Z_0}=3$ op de Smith kaart en we zoeken waar we de y=1 cirkel snijden in het bovenste deel van de kaart." ] }, { "cell_type": "code", "execution_count": 17, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de derde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T10:45:30.794279Z", "iopub.status.busy": "2020-12-04T10:45:30.794279Z", @@ -1898,6 +2023,9 @@ "name": "smith4s3" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -1923,32 +2051,34 @@ "smitplot(zlijn,Z_0,tt='y')" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "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": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, - "tags": [ - "remove_cell", - "slide_title" - ] + "tags": [] }, "source": [ - "###### Smith kaart van de derde oplossing als admitantie" + "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": 20, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de derde oplossing als admitantie" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T10:47:00.558289Z", "iopub.status.busy": "2020-12-04T10:47:00.558289Z", @@ -1961,10 +2091,13 @@ }, "mystnb": { "figure": { - "caption": "cSmith kaart van de derde oplossing als admitantie.", + "caption": "Smith kaart van de derde oplossing als admitantie.", "name": "smith4s3a" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -1995,6 +2128,12 @@ "cell_type": "code", "execution_count": 74, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T23:27:08.715498Z", "iopub.status.busy": "2020-12-04T23:27:08.715498Z", @@ -2005,6 +2144,15 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -2040,7 +2188,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen admitantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] @@ -2061,7 +2220,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

\n", "De optimale condensatorwaarde in parallel wordt dus 6.12 pF en de nodige lengte van de coax tussen de antenne en de aanpassing is 8.3 cm \n", @@ -2070,39 +2240,52 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "$$ j \\omega C= 0.023 j $$\n", "\n", "$$ C =\\frac{0.023}{2 \\pi \\cdot 600 \\times 10^6}= 6.12 pF $$" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Het schema dat we nodig hebben om deze derde oplossing te realiseren wordt dus:" - ] - }, { "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 derde oplossing." + "Het schema dat we nodig hebben om deze derde oplossing te realiseren wordt dus:" ] }, { "cell_type": "code", "execution_count": 59, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de derde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T22:43:11.353254Z", "iopub.status.busy": "2020-12-04T22:43:11.353254Z", @@ -2119,6 +2302,9 @@ "name": "smith4f3" } }, + "slideshow": { + "slide_type": "" + }, "tags": [] }, "outputs": [ @@ -2871,34 +3057,36 @@ "d.draw()" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "### Oplossing 4: Bijplaatsen van een spoel in parallel\n", - "\n", - "We plaatsen $\\frac{Z_0}{Z_L}=\\frac{1}{3}$ 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": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, - "tags": [ - "remove_cell", - "slide_title" - ] + "tags": [] }, "source": [ - "###### Smith kaart van de vierde oplossing." + "### Oplossing 4: Bijplaatsen van een spoel in parallel\n", + "\n", + "We plaatsen $\\frac{Z_0}{Z_L}=\\frac{1}{3}$ op de Smith kaart en we zoeken waar we de y=1 cirkel snijden in het bovenste deel van de kaart." ] }, { "cell_type": "code", "execution_count": 23, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de vierde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T10:48:36.208812Z", "iopub.status.busy": "2020-12-04T10:48:36.208812Z", @@ -2915,6 +3103,9 @@ "name": "smith4s4" } }, + "slideshow": { + "slide_type": "" + }, "tags": [] }, "outputs": [ @@ -2945,6 +3136,12 @@ "cell_type": "code", "execution_count": 75, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T23:28:32.074815Z", "iopub.status.busy": "2020-12-04T23:28:32.074815Z", @@ -2955,6 +3152,15 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -2984,14 +3190,36 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen admitantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "$$ \\frac{1}{j \\omega L}= - 0.023 j $$\n", "\n", @@ -3023,32 +3251,34 @@ "

" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Het schema dat we nodig hebben om deze vierde oplossing te realiseren wordt dus:" - ] - }, { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "" }, - "tags": [ - "remove_cell", - "slide_title" - ] + "tags": [] }, "source": [ - "###### circuit van de vierde oplossing." + "Het schema dat we nodig hebben om deze vierde oplossing te realiseren wordt dus:" ] }, { "cell_type": "code", "execution_count": 109, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de vierde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T23:16:30.790937Z", "iopub.status.busy": "2020-12-05T23:16:30.790937Z", @@ -3065,6 +3295,9 @@ "name": "smith4f4" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -3756,34 +3989,36 @@ "d.draw()" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "### Oplossing 5: Oplossing 3 waarbij de condensator vervangen is door een open transmissielijn\n", - "\n", - "De y = 1 - j 1.15 compenseren we door +j 1.15 vertrekkende vanuit g=0." - ] - }, { "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": [ - "###### Smith kaart van de vijfde oplossing." + "### Oplossing 5: Oplossing 3 waarbij de condensator vervangen is door een open transmissielijn\n", + "\n", + "De y = 1 - j 1.15 compenseren we door +j 1.15 vertrekkende vanuit g=0." ] }, { "cell_type": "code", "execution_count": 50, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de vijfde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T12:21:29.581057Z", "iopub.status.busy": "2020-12-04T12:21:29.581057Z", @@ -3800,6 +4035,9 @@ "name": "smith4s5" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -3842,6 +4080,12 @@ "cell_type": "code", "execution_count": 85, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T23:46:49.440764Z", "iopub.status.busy": "2020-12-04T23:46:49.439732Z", @@ -3852,6 +4096,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -3877,7 +4127,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen admitantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] @@ -3886,6 +4147,12 @@ "cell_type": "code", "execution_count": 101, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:02:32.720826Z", "iopub.status.busy": "2020-12-05T00:02:32.720826Z", @@ -3896,6 +4163,15 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -3926,7 +4202,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen admitantie van de open transmissielijn als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] @@ -3947,32 +4234,34 @@ "

" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Het schema dat we nodig hebben om deze vijfde oplossing te realiseren wordt dus:" - ] - }, { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "" }, - "tags": [ - "remove_cell", - "slide_title" - ] + "tags": [] }, "source": [ - "###### circuit van de vijfde oplossing." + "Het schema dat we nodig hebben om deze vijfde oplossing te realiseren wordt dus:" ] }, { "cell_type": "code", "execution_count": 65, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de vijfde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T22:53:34.623109Z", "iopub.status.busy": "2020-12-04T22:53:34.623109Z", @@ -3989,6 +4278,9 @@ "name": "smith4f5" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -4713,34 +5005,36 @@ "d.draw()" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "### Oplossing 6: Oplossing 3 waarbij de condensator vervangen is door een kortgesloten transmissielijn\n", - "\n", - "De y = 1 - j 1.15 compenseren we door +j 1.15 vertrekkende vanuit g=$\\infty$" - ] - }, { "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": [ - "###### Smith kaart van de zesde oplossing." + "### Oplossing 6: Oplossing 3 waarbij de condensator vervangen is door een kortgesloten transmissielijn\n", + "\n", + "De y = 1 - j 1.15 compenseren we door +j 1.15 vertrekkende vanuit g=$\\infty$" ] }, { "cell_type": "code", "execution_count": 53, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de zesde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T12:32:21.030137Z", "iopub.status.busy": "2020-12-04T12:32:21.030137Z", @@ -4757,6 +5051,9 @@ "name": "smith4s6" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -4781,6 +5078,12 @@ "cell_type": "code", "execution_count": 82, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T23:46:31.706843Z", "iopub.status.busy": "2020-12-04T23:46:31.706843Z", @@ -4791,6 +5094,15 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -4813,7 +5125,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen admitantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] @@ -4822,6 +5145,12 @@ "cell_type": "code", "execution_count": 102, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:02:50.736014Z", "iopub.status.busy": "2020-12-05T00:02:50.736014Z", @@ -4832,6 +5161,15 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -4856,7 +5194,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

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

" ] @@ -4870,32 +5219,34 @@ "

" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Het schema dat we nodig hebben om deze zesde oplossing te realiseren wordt dus:" - ] - }, { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "" }, - "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": 66, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de zesde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T22:56:44.540456Z", "iopub.status.busy": "2020-12-04T22:56:44.540456Z", @@ -4912,6 +5263,9 @@ "name": "smith4f6" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -5673,34 +6027,36 @@ "d.draw()" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "### Oplossing 7: Oplossing 4 waarbij het spoel vervangen is door een open transmissielijn\n", - "\n", - "De y = 1 + j 1.15 compenseren we door -j 1.15 vertrekkende vanuit g=0." - ] - }, { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "" }, - "tags": [ - "remove_cell", - "slide_title" - ] + "tags": [] }, "source": [ - "###### Smith kaart van de zevende oplossing." + "### Oplossing 7: Oplossing 4 waarbij het spoel vervangen is door een open transmissielijn\n", + "\n", + "De y = 1 + j 1.15 compenseren we door -j 1.15 vertrekkende vanuit g=0." ] }, { "cell_type": "code", "execution_count": 93, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de zevende oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T23:51:40.854545Z", "iopub.status.busy": "2020-12-04T23:51:40.854545Z", @@ -5717,6 +6073,9 @@ "name": "smith4s7" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -5748,6 +6107,12 @@ "cell_type": "code", "execution_count": 87, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T23:48:20.145258Z", "iopub.status.busy": "2020-12-04T23:48:20.145258Z", @@ -5758,6 +6123,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -5790,7 +6161,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen admitantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] @@ -5799,6 +6181,12 @@ "cell_type": "code", "execution_count": 103, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:03:01.617832Z", "iopub.status.busy": "2020-12-05T00:03:01.617832Z", @@ -5809,6 +6197,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -5835,46 +6229,70 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen admitantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "" + }, + "tags": [] + }, "source": [ "Het schema dat we nodig hebben om deze zevende oplossing te realiseren wordt dus:" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

\n", - "Het stukje open coax dat we moeten voorzien ter vervanging van het spoel is dus: 9.1 cm\n", - "

" - ] - }, { "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." + "

\n", + "Het stukje open coax dat we moeten voorzien ter vervanging van het spoel is dus: 9.1 cm\n", + "

" ] }, { "cell_type": "code", "execution_count": 76, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de zevende oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T23:41:05.435656Z", "iopub.status.busy": "2020-12-04T23:41:05.435656Z", @@ -5891,6 +6309,9 @@ "name": "smith4f7" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -6596,35 +7017,37 @@ "d.draw()" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "### Oplossing 8: Oplossing 4 waarbij het spoel vervangen is door een kortgesloten transmissielijn\n", - "\n", - "\n", - "De y = 1 + j 1.15 compenseren we door -j 1.15 vertrekkende vanuit g=$\\infty$. " - ] - }, { "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": [ - "###### Smith kaart van de achtste oplossing." + "### Oplossing 8: Oplossing 4 waarbij het spoel vervangen is door een kortgesloten transmissielijn\n", + "\n", + "\n", + "De y = 1 + j 1.15 compenseren we door -j 1.15 vertrekkende vanuit g=$\\infty$. " ] }, { "cell_type": "code", "execution_count": 110, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de achtste oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T23:27:43.604829Z", "iopub.status.busy": "2020-12-05T23:27:43.604829Z", @@ -6641,6 +7064,9 @@ "name": "smith4s8" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -6672,6 +7098,12 @@ "cell_type": "code", "execution_count": 87, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T23:48:20.145258Z", "iopub.status.busy": "2020-12-04T23:48:20.145258Z", @@ -6682,6 +7114,15 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -6711,7 +7152,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen admitantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] @@ -6720,6 +7172,12 @@ "cell_type": "code", "execution_count": 104, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:03:10.560363Z", "iopub.status.busy": "2020-12-05T00:03:10.560363Z", @@ -6730,6 +7188,15 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -6760,39 +7227,52 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

\n", "Het stukje kortgesloten coax dat we moeten voorzien ter vervanging van het spoel is dus: 2.8 cm\n", "

" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Het schema dat we nodig hebben om deze achtste oplossing te realiseren wordt dus:" - ] - }, { "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": 77, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de achtste oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-04T23:42:12.067520Z", "iopub.status.busy": "2020-12-04T23:42:12.067520Z", @@ -6809,6 +7289,9 @@ "name": "smith4f8" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ diff --git a/ToegepasteAnalogeElektronica/SmithKaartOefening5.ipynb b/ToegepasteAnalogeElektronica/SmithKaartOefening5.ipynb index 18e9a584d..3e605ac1e 100644 --- a/ToegepasteAnalogeElektronica/SmithKaartOefening5.ipynb +++ b/ToegepasteAnalogeElektronica/SmithKaartOefening5.ipynb @@ -3,12 +3,18 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [ - "remove_cell4reveal" + "remove_cell4reveal", + "remove_cell4pptx" ] }, "source": [ @@ -56,8 +62,14 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [] }, @@ -70,8 +82,14 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [] }, @@ -84,22 +102,12 @@ { "cell_type": "markdown", "metadata": { - "editable": true, - "slideshow": { - "slide_type": "slide" + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" }, - "tags": [ - "slide_overzicht" - ] - }, - "source": [ - "###### Overzicht\n", - " \n" - ] - }, - { - "cell_type": "markdown", - "metadata": { + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -112,9 +120,14 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [] }, @@ -126,25 +139,16 @@ "We plaatsen $\\frac{Z_L}{Z_0}=2.2$ op de Smith kaart (rode dot) en we zoeken waar we de r=1 cirkel snijden in het bovenste deel van de Smith kaart (groene dot)." ] }, - { - "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": 4, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de eerste oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:08:04.584032Z", "iopub.status.busy": "2020-12-05T00:08:04.584032Z", @@ -158,6 +162,9 @@ "name": "smith51" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -192,6 +199,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -215,8 +228,14 @@ "cell_type": "code", "execution_count": null, "metadata": { - "jupyter": { - "source_hidden": true + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "" }, "tags": [] }, @@ -225,25 +244,16 @@ "1/57.73502691896258/6e8/2/np.pi" ] }, - { - "cell_type": "markdown", - "metadata": { - "slideshow": { - "slide_type": "slide" - }, - "tags": [ - "remove_cell", - "slide_title" - ] - }, - "source": [ - "###### Berekening impedantie voor verschillende lengtes" - ] - }, { "cell_type": "code", "execution_count": 3, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Berekening impedantie voor verschillende lengtes" + }, + "editable": true, "execution": { "iopub.execute_input": "2022-08-26T08:25:14.769362Z", "iopub.status.busy": "2022-08-26T08:25:14.769362Z", @@ -254,6 +264,15 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -287,6 +306,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -299,6 +324,12 @@ { "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", @@ -307,7 +338,7 @@ "shell.execute_reply.started": "2020-11-29T09:50:19.955878Z" }, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [] }, @@ -320,6 +351,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -329,25 +366,16 @@ "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": 6, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de eerste oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2023-11-12T22:53:29.114764Z", "iopub.status.busy": "2023-11-12T22:53:29.114764Z", @@ -361,6 +389,9 @@ "name": "smith51cir" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -1049,6 +1080,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "slide" }, @@ -1060,25 +1097,16 @@ "We plaatsen $\\frac{Z_L}{Z_0}=2.2$ 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": 9, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de tweede oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:12:48.382739Z", "iopub.status.busy": "2020-12-05T00:12:48.382739Z", @@ -1095,6 +1123,9 @@ "name": "smith52" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -1131,6 +1162,12 @@ "cell_type": "code", "execution_count": 10, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:13:15.149505Z", "iopub.status.busy": "2020-12-05T00:13:15.149505Z", @@ -1141,6 +1178,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -1168,14 +1211,36 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen impedantie als functie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] }, { "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 0.81. Daaruit volgt dat Z = 50 Ohm - j 40.5 Ohm. Het complexe deel van deze impedantie kunnen we compenseren door een spoel met impedantie van ongeveer j 40.5 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", @@ -1190,8 +1255,20 @@ "cell_type": "code", "execution_count": null, "metadata": { - "jupyter": { - "source_hidden": true + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "skip" }, "tags": [] }, @@ -1200,32 +1277,34 @@ "57.73502691896258/2/np.pi/6e8" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Het schema dat we nodig hebben om deze tweede oplossing te realiseren wordt dus:" - ] - }, { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "" }, - "tags": [ - "remove_cell", - "slide_title" - ] + "tags": [] }, "source": [ - "###### circuit van de tweede oplossing." + "Het schema dat we nodig hebben om deze tweede oplossing te realiseren wordt dus:" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de tweede oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2023-11-12T22:55:31.447372Z", "iopub.status.busy": "2023-11-12T22:55:31.447372Z", @@ -1239,6 +1318,9 @@ "name": "smith52cir" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -1937,7 +2019,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

\n", "De optimale inductantiewaarde in serie wordt dus 15 nH en de nodige lengte van de coax tussen de antenne en de aanpassing is 2.1 cm\n", @@ -1947,8 +2040,14 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [] }, @@ -1958,25 +2057,16 @@ "We plaatsen $\\frac{Z_L}{Z_0}=2.2$ 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": 12, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de derde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:16:11.809879Z", "iopub.status.busy": "2020-12-05T00:16:11.809879Z", @@ -1993,6 +2083,9 @@ "name": "smith53" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -2016,6 +2109,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -2025,25 +2124,16 @@ "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": { - "slideshow": { - "slide_type": "slide" - }, - "tags": [ - "remove_cell", - "slide_title" - ] - }, - "source": [ - "###### Smith kaart van de derde oplossing als admitantie" - ] - }, { "cell_type": "code", "execution_count": 13, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de derde oplossing als admitantie" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:16:42.029099Z", "iopub.status.busy": "2020-12-05T00:16:42.029099Z", @@ -2060,6 +2150,9 @@ "name": "smith53b" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -2081,25 +2174,16 @@ "smitplot(ylijn,Y_0)" ] }, - { - "cell_type": "markdown", - "metadata": { - "slideshow": { - "slide_type": "slide" - }, - "tags": [ - "remove_cell", - "slide_title" - ] - }, - "source": [ - "###### Berekening admittantie voor verschillende lengtes" - ] - }, { "cell_type": "code", "execution_count": 4, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Berekening admittantie voor verschillende lengtes" + }, + "editable": true, "execution": { "iopub.execute_input": "2022-08-26T08:25:50.808947Z", "iopub.status.busy": "2022-08-26T08:25:50.806937Z", @@ -2110,6 +2194,15 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -2139,6 +2232,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -2152,8 +2251,17 @@ "cell_type": "code", "execution_count": null, "metadata": { - "jupyter": { - "source_hidden": true + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } }, "slideshow": { "slide_type": "skip" @@ -2168,8 +2276,14 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "fragment" + "slide_type": "skip" }, "tags": [] }, @@ -2182,6 +2296,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -2191,25 +2311,16 @@ "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": 9, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de derde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2023-11-12T22:57:25.181091Z", "iopub.status.busy": "2023-11-12T22:57:25.181091Z", @@ -2223,6 +2334,9 @@ "name": "smith53cir" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -2979,6 +3093,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "slide" }, @@ -2990,25 +3110,16 @@ "We plaatsen $\\frac{Z_0}{Z_L}=\\frac{1}{2.2}$ 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": 17, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de vierde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:19:49.313758Z", "iopub.status.busy": "2020-12-05T00:19:49.313758Z", @@ -3025,6 +3136,9 @@ "name": "smith54" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -3050,6 +3164,12 @@ "cell_type": "code", "execution_count": 6, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2022-08-26T08:59:57.624278Z", "iopub.status.busy": "2022-08-26T08:59:57.623296Z", @@ -3060,6 +3180,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -3130,6 +3256,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -3139,25 +3271,16 @@ "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": 10, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de vierde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2023-11-12T23:05:11.032823Z", "iopub.status.busy": "2023-11-12T23:05:11.032823Z", @@ -3171,6 +3294,9 @@ "name": "smith54cir" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -3941,6 +4067,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "slide" }, @@ -3952,25 +4084,16 @@ "De y = 1 - j 0.85 compenseren we door +j 0.85 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": 27, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de vijfde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:28:20.304502Z", "iopub.status.busy": "2020-12-05T00:28:20.304502Z", @@ -3987,6 +4110,9 @@ "name": "smith55" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -4024,6 +4150,12 @@ "cell_type": "code", "execution_count": 7, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2022-08-26T09:00:13.375085Z", "iopub.status.busy": "2022-08-26T09:00:13.374122Z", @@ -4034,6 +4166,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -4061,6 +4199,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -4074,6 +4218,12 @@ "cell_type": "code", "execution_count": 5, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2022-08-26T08:58:31.517019Z", "iopub.status.busy": "2022-08-26T08:58:31.515017Z", @@ -4084,6 +4234,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -4117,6 +4273,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -4129,6 +4291,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -4141,8 +4309,14 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [] }, @@ -4155,6 +4329,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -4164,25 +4344,16 @@ "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": 11, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de vijfde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2023-11-12T23:05:22.904862Z", "iopub.status.busy": "2023-11-12T23:05:22.904862Z", @@ -4196,6 +4367,9 @@ "name": "smith55cir" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -4942,6 +5116,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "slide" }, @@ -4953,25 +5133,16 @@ "De y = 1 - j 0.85 compenseren we door +j 0.85 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": 29, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de zesde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:30:34.630325Z", "iopub.status.busy": "2020-12-05T00:30:34.614671Z", @@ -4988,6 +5159,9 @@ "name": "smith56" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -5012,6 +5186,12 @@ "cell_type": "code", "execution_count": 31, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:32:23.142842Z", "iopub.status.busy": "2020-12-05T00:32:23.142842Z", @@ -5019,8 +5199,11 @@ "shell.execute_reply": "2020-12-05T00:32:23.181196Z", "shell.execute_reply.started": "2020-12-05T00:32:23.142842Z" }, - "jupyter": { - "source_hidden": true + "mystnb": { + "figure": { + "caption": "", + "name": "" + } }, "slideshow": { "slide_type": "slide" @@ -5047,7 +5230,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen admitantie als funktie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] @@ -5056,6 +5250,12 @@ "cell_type": "code", "execution_count": 32, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T00:32:32.322101Z", "iopub.status.busy": "2020-12-05T00:32:32.322101Z", @@ -5066,6 +5266,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -5092,7 +5298,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen admittantie van de kortgesloten transmissielijn als functie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] @@ -5114,6 +5331,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -5123,25 +5346,16 @@ "Het schema dat we nodig hebben om deze zesde oplossing te realiseren wordt dus:" ] }, - { - "cell_type": "markdown", - "metadata": { - "slideshow": { - "slide_type": "slide" - }, - "tags": [ - "remove_cell", - "slide_title" - ] - }, - "source": [ - "###### circuit van de zesde oplossing." - ] - }, { "cell_type": "code", "execution_count": 13, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de zesde oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2023-11-12T23:05:51.129356Z", "iopub.status.busy": "2023-11-12T23:05:51.128358Z", @@ -5155,6 +5369,9 @@ "name": "smith56cir" } }, + "slideshow": { + "slide_type": "" + }, "tags": [] }, "outputs": [ @@ -5913,8 +6130,14 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [] }, @@ -5924,25 +6147,16 @@ "De y = 1 + j 0.85 compenseren we door -j 0.85 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": 42, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Smith kaart van de zevende oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T11:05:48.315693Z", "iopub.status.busy": "2020-12-05T11:05:48.315693Z", @@ -5959,6 +6173,9 @@ "name": "smith57" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -5985,6 +6202,12 @@ "cell_type": "code", "execution_count": 38, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T11:04:00.429831Z", "iopub.status.busy": "2020-12-05T11:04:00.429831Z", @@ -5995,6 +6218,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -6037,6 +6266,12 @@ "cell_type": "code", "execution_count": 41, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T11:05:13.251850Z", "iopub.status.busy": "2020-12-05T11:05:13.251850Z", @@ -6047,6 +6282,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -6074,6 +6315,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -6086,6 +6333,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -6098,8 +6351,14 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [] }, @@ -6109,25 +6368,16 @@ "

" ] }, - { - "cell_type": "markdown", - "metadata": { - "slideshow": { - "slide_type": "slide" - }, - "tags": [ - "remove_cell", - "slide_title" - ] - }, - "source": [ - "###### circuit van de zevende oplossing." - ] - }, { "cell_type": "code", "execution_count": 15, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de zevende oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2023-11-12T23:06:08.120447Z", "iopub.status.busy": "2023-11-12T23:06:08.120447Z", @@ -6141,6 +6391,9 @@ "name": "smith57cir" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -6886,6 +7139,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "slide" }, @@ -6897,25 +7156,16 @@ "De y = 1 + j 0.85 compenseren we door -j 0.85 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": "2020-12-05T11:10:48.661998Z", "iopub.status.busy": "2020-12-05T11:10:48.660999Z", @@ -6932,6 +7182,9 @@ "name": "smith58" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -6963,6 +7216,12 @@ "cell_type": "code", "execution_count": 45, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T11:08:06.299549Z", "iopub.status.busy": "2020-12-05T11:08:06.299549Z", @@ -6973,6 +7232,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -7010,6 +7275,12 @@ "cell_type": "code", "execution_count": 51, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "execution": { "iopub.execute_input": "2020-12-05T11:10:34.919250Z", "iopub.status.busy": "2020-12-05T11:10:34.919250Z", @@ -7020,6 +7291,12 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, "slideshow": { "slide_type": "slide" }, @@ -7046,7 +7323,18 @@ }, { "cell_type": "markdown", - "metadata": {}, + "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "slideshow": { + "slide_type": "skip" + }, + "tags": [] + }, "source": [ "

Overzicht van de bekomen admittantie als functie van de hoek op de Smith kaart. De laatste kolom geeft de nodige lengte van de coax

" ] @@ -7054,8 +7342,14 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { - "slide_type": "slide" + "slide_type": "skip" }, "tags": [] }, @@ -7068,6 +7362,12 @@ { "cell_type": "markdown", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -7077,25 +7377,16 @@ "Het schema dat we nodig hebben om deze achtste oplossing te realiseren wordt dus:" ] }, - { - "cell_type": "markdown", - "metadata": { - "slideshow": { - "slide_type": "slide" - }, - "tags": [ - "remove_cell", - "slide_title" - ] - }, - "source": [ - "###### circuit van de achtste oplossing." - ] - }, { "cell_type": "code", "execution_count": 16, "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "circuit van de achtste oplossing" + }, + "editable": true, "execution": { "iopub.execute_input": "2023-11-12T23:06:17.147795Z", "iopub.status.busy": "2023-11-12T23:06:17.146465Z", @@ -7109,6 +7400,9 @@ "name": "smith58cir" } }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -7838,7 +8132,7 @@ "metadata": { "KULeuvenSlides": { "authors": "Jan Genoe", - "date": "December 2022", + "date": "December 2024", "kuleuven_presentation_style": "1425", "subtitle": "Oefening 5", "title": "Smith Kaart" diff --git a/ToegepasteAnalogeElektronica/SmithKaartOefening6.ipynb b/ToegepasteAnalogeElektronica/SmithKaartOefening6.ipynb index 25b496930..e7caa4b0f 100644 --- a/ToegepasteAnalogeElektronica/SmithKaartOefening6.ipynb +++ b/ToegepasteAnalogeElektronica/SmithKaartOefening6.ipynb @@ -222,7 +222,7 @@ } }, "slideshow": { - "slide_type": "" + "slide_type": "slide" }, "tags": [] }, @@ -248,6 +248,12 @@ "cell_type": "markdown", "id": "89fbbb08-a96f-4370-9975-eb35ad515c82", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, "slideshow": { "slide_type": "skip" }, @@ -259,27 +265,17 @@ "Op de kaart lezen we af dat dit al ongeveer 30 na graden is. Dat is 1/12 van de volledige omtrek van de cirkel Als een gevolg gaat de lengte van de transmissielijn 1/24 deel van de golflengte zijn. We hebben dus een stukje van 2.8 mm nodig. In de tabel hieronder zien we dit in meer detail uitgewerkt:\n" ] }, - { - "cell_type": "markdown", - "id": "875e5ddd", - "metadata": { - "editable": true, - "slideshow": { - "slide_type": "slide" - }, - "tags": [ - "slide_title" - ] - }, - "source": [ - "###### Berekening admittantie voor verschillende lengtes" - ] - }, { "cell_type": "code", "execution_count": 3, "id": "59c0370e-1a81-4043-af1d-c144d4581966", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Berekening admittantie voor verschillende lengtes" + }, + "editable": true, "execution": { "iopub.execute_input": "2022-08-26T08:17:52.624869Z", "iopub.status.busy": "2022-08-26T08:17:52.624869Z", @@ -290,6 +286,15 @@ "jupyter": { "source_hidden": true }, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "slide" + }, "tags": [] }, "outputs": [ @@ -319,8 +324,20 @@ "execution_count": null, "id": "9ed1108a-0baa-418c-b76b-8fbc8be0c2d8", "metadata": { - "jupyter": { - "source_hidden": true + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, + "editable": true, + "mystnb": { + "figure": { + "caption": "", + "name": "" + } + }, + "slideshow": { + "slide_type": "" }, "tags": [] }, @@ -514,7 +531,7 @@ "KULeuvenSlides": { "slide_code": "normal", "slide_ref": "", - "slide_title": "circuit van de aangepaste antenne." + "slide_title": "Circuit van de aangepaste antenne" }, "editable": true, "execution": { diff --git a/ToegepasteAnalogeElektronica/SmithKaartOefening7.ipynb b/ToegepasteAnalogeElektronica/SmithKaartOefening7.ipynb index 03769d77d..d0ddf051b 100644 --- a/ToegepasteAnalogeElektronica/SmithKaartOefening7.ipynb +++ b/ToegepasteAnalogeElektronica/SmithKaartOefening7.ipynb @@ -4,12 +4,18 @@ "cell_type": "markdown", "id": "f3cd07a6-3b50-4a6b-b871-abf08b437a01", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, "editable": true, "slideshow": { "slide_type": "slide" }, "tags": [ - "remove_cell4reveal" + "remove_cell4reveal", + "remove_cell4pptx" ] }, "source": [ @@ -59,6 +65,11 @@ "cell_type": "markdown", "id": "34460840-0dcd-4421-a0f6-69c94390a310", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, "editable": true, "slideshow": { "slide_type": "skip" @@ -80,6 +91,11 @@ "cell_type": "markdown", "id": "f9d50dfb-90af-48d6-9a64-d540bde0b804", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, "citation-manager": { "citations": { "btvbi": [ @@ -111,7 +127,9 @@ "id": "c3c9e8b1-2166-4770-aa58-6be4755ea380", "metadata": { "KULeuvenSlides": { - "slide_title": "circuit van de aangepaste antenne" + "slide_code": "normal", + "slide_ref": "", + "slide_title": "Circuit van de aangepaste antenne" }, "editable": true, "execution": { @@ -450,6 +468,11 @@ "cell_type": "markdown", "id": "d3759f7e-94b5-47df-9722-69c0a998f648", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, "citation-manager": { "citations": { "btvbi": [ @@ -475,6 +498,8 @@ "id": "f818d4f4-e36a-46e2-a996-0985b5d50330", "metadata": { "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", "slide_title": "Initiele waarde van de admittantie" }, "editable": true, @@ -678,6 +703,11 @@ "cell_type": "markdown", "id": "a1619bd3-596f-45ac-a52c-a8ea4279b89d", "metadata": { + "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", + "slide_title": "" + }, "editable": true, "slideshow": { "slide_type": "skip" @@ -695,6 +725,8 @@ "id": "b393270d-bf09-4d03-b7ba-aba56cc3c95d", "metadata": { "KULeuvenSlides": { + "slide_code": "normal", + "slide_ref": "", "slide_title": "admittantie: aanpassing+open transmissielijn" }, "editable": true, diff --git a/ToegepasteAnalogeElektronica/reflecties.ipynb b/ToegepasteAnalogeElektronica/reflecties.ipynb index 0f0bcc460..42c612319 100644 --- a/ToegepasteAnalogeElektronica/reflecties.ipynb +++ b/ToegepasteAnalogeElektronica/reflecties.ipynb @@ -374,10 +374,10 @@ "metadata": { "KULeuvenSlides": { "authors": "Jan Genoe", - "date": "December 2022", + "date": "December 2024", "kuleuven_presentation_style": "1425", - "subtitle": "subtitle", - "title": "titel" + "subtitle": "", + "title": "Reflecties op transmissielijnen" }, "citation-manager": { "items": {}