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:<math>C_1 = 0</math>
:<math>C_1 = 0</math>




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Bisher wurden die Stromkreise von einer Gleichstromquelle, einer Wechselstromquelle und einer exponentiellen Quelle betrieben. Wenn wir den Strom einer Schaltung finden können, die von einer Dirac-Deltafunktion oder einer Stoßspannungsquelle δ erzeugt wird, dann kann das Convolution Integral verwendet werden, um den Strom zu einer bestimmten Spannungsquelle zu finden!

Beispiel Impulsantwort

Der Strom wird durch die Ableitung des durch eine Gleichspannungsquelle gefundenen Stroms ermittelt! Angenommen, das Ziel ist es, den δ-Strom einer LR-Schaltung der Serie zu finden, so dass in Zukunft das Convolution Integral verwendet werden kann, um den Strom einer beliebigen Quelle zu finden.

Wählen Sie eine DC-Quelle von 1 Volt (das reale Vs kann dann davon abweichen). Die besondere homogene Lösung (stationärer Zustand) ist 0, die homogene Lösung zur inhomogenen Gleichung hat die Form:

Angenommen, der Strom im Induktor ist zunächst Null. Die Anfangsspannung wird 1 sein und über dem Induktor liegen (da kein Strom fließt):

::

Wenn der Strom im Induktor zunächst Null ist, dann:

Fehler beim Parsen (Konvertierungsfehler. 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data-semantic-children="6,5" data-semantic-content="4" data-semantic-parent="8"><mrow data-semantic-type="prefixop" data-semantic-role="negative" data-semantic-annotation="clearspeak:simple" data-semantic-id="6" data-semantic-children="3" data-semantic-content="2" data-semantic-parent="7"><mo data-semantic-type="operator" data-semantic-role="subtraction" data-semantic-id="2" data-semantic-parent="6" data-semantic-operator="prefixop,−">−<!-- − --></mo><mn data-semantic-type="number" data-semantic-role="integer" data-semantic-font="normal" data-semantic-annotation="clearspeak:simple" data-semantic-id="3" data-semantic-parent="6">1</mn></mrow><mrow class="MJX-TeXAtom-ORD"><mo data-semantic-type="operator" data-semantic-role="division" data-semantic-id="4" data-semantic-parent="7" data-semantic-operator="infixop,/">/</mo></mrow><mi data-semantic-type="identifier" data-semantic-role="latinletter" data-semantic-font="italic" data-semantic-annotation="clearspeak:simple" data-semantic-id="5" 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annotation="clearspeak:simple" id="6">−<content><operator role="subtraction" id="2">−</operator></content><children><number role="integer" font="normal" annotation="clearspeak:simple" id="3">1</number></children></prefixop><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="5">R</identifier></children></infixop></children></relseq></stree>', 'success' => true, 'log' => 'success', 'mathoidStyle' => 'vertical-align: -0.838ex; width:10.739ex; height:2.843ex;', 'sanetex' => '{\\displaystyle A=-1/R}', 'speech' => 'upper A equals negative 1 divided by upper R', ), )“): {\displaystyle i(0) = 0 = A + C}
Das impliziert:
Fehler beim Parsen (Konvertierungsfehler. Der Server („cli“) hat berichtet: „array ( 'nohash' => array ( ), 'success' => true, '956d5beec9b0b11bb3230fe8a567b85c' => (object) array( 'speakText' => 'upper A equals negative 1 divided by upper R', 'mml' => '<math xmlns="http://www.w3.org/1998/Math/MathML" display="block" alttext="upper A equals negative 1 divided by upper R"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0" data-semantic-type="relseq" data-semantic-role="equality" data-semantic-id="8" data-semantic-children="0,7" data-semantic-content="1"><mi data-semantic-type="identifier" data-semantic-role="latinletter" data-semantic-font="italic" data-semantic-annotation="clearspeak:simple" data-semantic-id="0" data-semantic-parent="8">A</mi><mo data-semantic-type="relation" data-semantic-role="equality" data-semantic-id="1" data-semantic-parent="8" data-semantic-operator="relseq,=">=</mo><mrow data-semantic-type="infixop" data-semantic-role="division" data-semantic-id="7" data-semantic-children="6,5" data-semantic-content="4" data-semantic-parent="8"><mrow data-semantic-type="prefixop" data-semantic-role="negative" data-semantic-annotation="clearspeak:simple" data-semantic-id="6" data-semantic-children="3" data-semantic-content="2" data-semantic-parent="7"><mo data-semantic-type="operator" data-semantic-role="subtraction" data-semantic-id="2" data-semantic-parent="6" data-semantic-operator="prefixop,−">−<!-- − --></mo><mn data-semantic-type="number" data-semantic-role="integer" data-semantic-font="normal" data-semantic-annotation="clearspeak:simple" data-semantic-id="3" data-semantic-parent="6">1</mn></mrow><mrow class="MJX-TeXAtom-ORD"><mo data-semantic-type="operator" data-semantic-role="division" data-semantic-id="4" data-semantic-parent="7" data-semantic-operator="infixop,/">/</mo></mrow><mi data-semantic-type="identifier" data-semantic-role="latinletter" data-semantic-font="italic" data-semantic-annotation="clearspeak:simple" data-semantic-id="5" 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xlink:href="#E1-MJMAIN-3D" x="1028" y="0"></use> <use xlink:href="#E1-MJMAIN-2212" x="2084" y="0"></use> <use xlink:href="#E1-MJMAIN-31" x="2863" y="0"></use> <use xlink:href="#E1-MJMAIN-2F" x="3363" y="0"></use> <use xlink:href="#E1-MJMATHI-52" x="3864" y="0"></use></g></svg>', 'width' => '10.739ex', 'height' => '2.843ex', 'style' => 'vertical-align: -0.838ex;', 'streeJson' => (object) array( 'stree' => (object) array( 'type' => 'relseq', 'role' => 'equality', 'id' => '8', '$t' => '=', 'children' => array ( 0 => (object) array( 'type' => 'identifier', 'role' => 'latinletter', 'font' => 'italic', 'annotation' => 'clearspeak:simple', 'id' => '0', '$t' => 'A', ), 1 => (object) array( 'type' => 'infixop', 'role' => 'division', 'id' => '7', '$t' => '/', 'children' => array ( 0 => (object) array( 'type' => 'prefixop', 'role' => 'negative', 'annotation' => 'clearspeak:simple', 'id' => '6', '$t' => '−', 'children' => array ( 0 => (object) array( 'type' => 'number', 'role' => 'integer', 'font' 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annotation="clearspeak:simple" id="6">−<content><operator role="subtraction" id="2">−</operator></content><children><number role="integer" font="normal" annotation="clearspeak:simple" id="3">1</number></children></prefixop><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="5">R</identifier></children></infixop></children></relseq></stree>', 'success' => true, 'log' => 'success', 'mathoidStyle' => 'vertical-align: -0.838ex; width:10.739ex; height:2.843ex;', 'sanetex' => '{\\displaystyle A=-1/R}', 'speech' => 'upper A equals negative 1 divided by upper R', ), )“): {\displaystyle C = -A = 1/R}
Die Antwort auf das Einschalten einer Gleichspannungsquelle bei t=0 bis ein Volt (die so genannte Unit Response μ) lautet also:
Fehler beim Parsen (Konvertierungsfehler. Der Server („cli“) hat berichtet: „array ( 'nohash' => array ( ), 'success' => true, '956d5beec9b0b11bb3230fe8a567b85c' => (object) array( 'speakText' => 'upper A equals negative 1 divided by upper R', 'mml' => '<math xmlns="http://www.w3.org/1998/Math/MathML" display="block" alttext="upper A equals negative 1 divided by upper R"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0" data-semantic-type="relseq" data-semantic-role="equality" data-semantic-id="8" data-semantic-children="0,7" data-semantic-content="1"><mi data-semantic-type="identifier" data-semantic-role="latinletter" data-semantic-font="italic" data-semantic-annotation="clearspeak:simple" data-semantic-id="0" data-semantic-parent="8">A</mi><mo data-semantic-type="relation" data-semantic-role="equality" data-semantic-id="1" data-semantic-parent="8" data-semantic-operator="relseq,=">=</mo><mrow data-semantic-type="infixop" data-semantic-role="division" data-semantic-id="7" 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data-semantic-parent="7">R</mi></mrow></mstyle></mrow><annotation encoding="application/x-tex">{\\displaystyle A=-1/R}</annotation></semantics></math>', 'svg' => '<svg xmlns:xlink="http://www.w3.org/1999/xlink" width="10.739ex" height="2.843ex" style="vertical-align: -0.838ex;" viewBox="0 -863.1 4623.6 1223.9" role="img" focusable="false" xmlns="http://www.w3.org/2000/svg" aria-labelledby="MathJax-SVG-1-Title"><title id="MathJax-SVG-1-Title">upper A equals negative 1 divided by upper R</title><defs aria-hidden="true"><path stroke-width="1" id="E1-MJMATHI-41" d="M208 74Q208 50 254 46Q272 46 272 35Q272 34 270 22Q267 8 264 4T251 0Q249 0 239 0T205 1T141 2Q70 2 50 0H42Q35 7 35 11Q37 38 48 46H62Q132 49 164 96Q170 102 345 401T523 704Q530 716 547 716H555H572Q578 707 578 706L606 383Q634 60 636 57Q641 46 701 46Q726 46 726 36Q726 34 723 22Q720 7 718 4T704 0Q701 0 690 0T651 1T578 2Q484 2 455 0H443Q437 6 437 9T439 27Q443 40 445 43L449 46H469Q523 49 533 63L521 213H283L249 155Q208 86 208 74ZM516 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Wenn man die Ableitung daraus zieht, erhält man den Impuls (δ) Strom ist:

Fehler beim Parsen (Konvertierungsfehler. 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Nun der Strom aufgrund einer beliebigen Anzahl VS(t) kann über das Convolution Integral gefunden werden:

Fehler beim Parsen (Konvertierungsfehler. 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annotation="clearspeak:simple" id="6">−<content><operator role="subtraction" id="2">−</operator></content><children><number role="integer" font="normal" annotation="clearspeak:simple" id="3">1</number></children></prefixop><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="5">R</identifier></children></infixop></children></relseq></stree>', 'success' => true, 'log' => 'success', 'mathoidStyle' => 'vertical-align: -0.838ex; width:10.739ex; height:2.843ex;', 'sanetex' => '{\\displaystyle A=-1/R}', 'speech' => 'upper A equals negative 1 divided by upper R', ), )“): {\displaystyle i(t) = \int_0^t i_\delta (t-\tau) V_s(\tau) d\tau = \int_0^t f(t-\tau)g(\tau)d\tau + C_1}

Sie sollten iδ nicht als aktuell betrachten. Es ist wirklich Fehler beim Parsen (Konvertierungsfehler. 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data-semantic-children="6,5" data-semantic-content="4" data-semantic-parent="8"><mrow data-semantic-type="prefixop" data-semantic-role="negative" data-semantic-annotation="clearspeak:simple" data-semantic-id="6" data-semantic-children="3" data-semantic-content="2" data-semantic-parent="7"><mo data-semantic-type="operator" data-semantic-role="subtraction" data-semantic-id="2" data-semantic-parent="6" data-semantic-operator="prefixop,−">−<!-- − --></mo><mn data-semantic-type="number" data-semantic-role="integer" data-semantic-font="normal" data-semantic-annotation="clearspeak:simple" data-semantic-id="3" data-semantic-parent="6">1</mn></mrow><mrow class="MJX-TeXAtom-ORD"><mo data-semantic-type="operator" data-semantic-role="division" data-semantic-id="4" data-semantic-parent="7" data-semantic-operator="infixop,/">/</mo></mrow><mi data-semantic-type="identifier" data-semantic-role="latinletter" data-semantic-font="italic" data-semantic-annotation="clearspeak:simple" data-semantic-id="5" 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annotation="clearspeak:simple" id="6">−<content><operator role="subtraction" id="2">−</operator></content><children><number role="integer" font="normal" annotation="clearspeak:simple" id="3">1</number></children></prefixop><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="5">R</identifier></children></infixop></children></relseq></stree>', 'success' => true, 'log' => 'success', 'mathoidStyle' => 'vertical-align: -0.838ex; width:10.739ex; height:2.843ex;', 'sanetex' => '{\\displaystyle A=-1/R}', 'speech' => 'upper A equals negative 1 divided by upper R', ), )“): {\displaystyle {d \over dt}\frac{current}{1 volt}} . VS(τ) wird zu einem Multiplikator.

LRC Beispiel

Finden Sie den Zeitbereichsausdruck für io, da Is = cos(t + π/2)μ(t) amp.

Früher wurde die Step-Response für dieses Problem gefunden:

Fehler beim Parsen (Konvertierungsfehler. Der Server („cli“) hat berichtet: „array ( 'nohash' => array ( ), 'success' => true, '956d5beec9b0b11bb3230fe8a567b85c' => (object) array( 'speakText' => 'upper A equals negative 1 divided by upper R', 'mml' => '<math xmlns="http://www.w3.org/1998/Math/MathML" display="block" alttext="upper A equals negative 1 divided by upper R"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0" data-semantic-type="relseq" data-semantic-role="equality" data-semantic-id="8" data-semantic-children="0,7" data-semantic-content="1"><mi data-semantic-type="identifier" data-semantic-role="latinletter" data-semantic-font="italic" data-semantic-annotation="clearspeak:simple" data-semantic-id="0" data-semantic-parent="8">A</mi><mo data-semantic-type="relation" data-semantic-role="equality" data-semantic-id="1" data-semantic-parent="8" data-semantic-operator="relseq,=">=</mo><mrow data-semantic-type="infixop" data-semantic-role="division" data-semantic-id="7" data-semantic-children="6,5" data-semantic-content="4" data-semantic-parent="8"><mrow data-semantic-type="prefixop" data-semantic-role="negative" data-semantic-annotation="clearspeak:simple" data-semantic-id="6" data-semantic-children="3" data-semantic-content="2" data-semantic-parent="7"><mo data-semantic-type="operator" data-semantic-role="subtraction" data-semantic-id="2" data-semantic-parent="6" data-semantic-operator="prefixop,−">−<!-- − --></mo><mn data-semantic-type="number" data-semantic-role="integer" data-semantic-font="normal" data-semantic-annotation="clearspeak:simple" data-semantic-id="3" data-semantic-parent="6">1</mn></mrow><mrow class="MJX-TeXAtom-ORD"><mo data-semantic-type="operator" data-semantic-role="division" data-semantic-id="4" data-semantic-parent="7" data-semantic-operator="infixop,/">/</mo></mrow><mi data-semantic-type="identifier" data-semantic-role="latinletter" data-semantic-font="italic" data-semantic-annotation="clearspeak:simple" data-semantic-id="5" data-semantic-parent="7">R</mi></mrow></mstyle></mrow><annotation encoding="application/x-tex">{\\displaystyle A=-1/R}</annotation></semantics></math>', 'svg' => '<svg xmlns:xlink="http://www.w3.org/1999/xlink" width="10.739ex" height="2.843ex" style="vertical-align: -0.838ex;" viewBox="0 -863.1 4623.6 1223.9" role="img" focusable="false" xmlns="http://www.w3.org/2000/svg" aria-labelledby="MathJax-SVG-1-Title"><title id="MathJax-SVG-1-Title">upper A equals negative 1 divided by upper R</title><defs aria-hidden="true"><path stroke-width="1" id="E1-MJMATHI-41" d="M208 74Q208 50 254 46Q272 46 272 35Q272 34 270 22Q267 8 264 4T251 0Q249 0 239 0T205 1T141 2Q70 2 50 0H42Q35 7 35 11Q37 38 48 46H62Q132 49 164 96Q170 102 345 401T523 704Q530 716 547 716H555H572Q578 707 578 706L606 383Q634 60 636 57Q641 46 701 46Q726 46 726 36Q726 34 723 22Q720 7 718 4T704 0Q701 0 690 0T651 1T578 2Q484 2 455 0H443Q437 6 437 9T439 27Q443 40 445 43L449 46H469Q523 49 533 63L521 213H283L249 155Q208 86 208 74ZM516 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649Q193 676 204 682Q206 683 378 683Q550 682 564 680Q620 672 658 652T712 606T733 563T739 529Q739 484 710 445T643 385T576 351T538 338L545 333Q612 295 612 223Q612 212 607 162T602 80V71Q602 53 603 43T614 25T640 16Q668 16 686 38T712 85Q717 99 720 102T735 105Q755 105 755 93Q755 75 731 36Q693 -21 641 -21H632Q571 -21 531 4T487 82Q487 109 502 166T517 239Q517 290 474 313Q459 320 449 321T378 323H309L277 193Q244 61 244 59Q244 55 245 54T252 50T269 48T302 46H333Q339 38 339 37T336 19Q332 6 326 0H311Q275 2 180 2Q146 2 117 2T71 2T50 1Q33 1 33 10Q33 12 36 24Q41 43 46 45Q50 46 61 46H67Q94 46 127 49Q141 52 146 61Q149 65 218 339T287 628Q287 635 230 637ZM630 554Q630 586 609 608T523 636Q521 636 500 636T462 637H440Q393 637 386 627Q385 624 352 494T319 361Q319 360 388 360Q466 361 492 367Q556 377 592 426Q608 449 619 486T630 554Z"></path></defs><g stroke="currentColor" fill="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)" aria-hidden="true"> <use xlink:href="#E1-MJMATHI-41" x="0" y="0"></use> <use 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annotation="clearspeak:simple" id="6">−<content><operator role="subtraction" id="2">−</operator></content><children><number role="integer" font="normal" annotation="clearspeak:simple" id="3">1</number></children></prefixop><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="5">R</identifier></children></infixop></children></relseq></stree>', 'success' => true, 'log' => 'success', 'mathoidStyle' => 'vertical-align: -0.838ex; width:10.739ex; height:2.843ex;', 'sanetex' => '{\\displaystyle A=-1/R}', 'speech' => 'upper A equals negative 1 divided by upper R', ), )“): {\displaystyle i_{o_\mu} = \frac{1}{2}(1 - e^{-t}(\cos t + \sin t))}

Die Impulsantwort wird die Ableitung davon sein:

Fehler beim Parsen (Konvertierungsfehler. Der Server („cli“) hat berichtet: „array ( 'nohash' => array ( ), 'success' => true, '956d5beec9b0b11bb3230fe8a567b85c' => (object) array( 'speakText' => 'upper A equals negative 1 divided by upper R', 'mml' => '<math xmlns="http://www.w3.org/1998/Math/MathML" display="block" alttext="upper A equals negative 1 divided by upper R"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0" data-semantic-type="relseq" data-semantic-role="equality" data-semantic-id="8" data-semantic-children="0,7" data-semantic-content="1"><mi data-semantic-type="identifier" data-semantic-role="latinletter" data-semantic-font="italic" data-semantic-annotation="clearspeak:simple" data-semantic-id="0" data-semantic-parent="8">A</mi><mo data-semantic-type="relation" data-semantic-role="equality" data-semantic-id="1" data-semantic-parent="8" data-semantic-operator="relseq,=">=</mo><mrow data-semantic-type="infixop" data-semantic-role="division" data-semantic-id="7" 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Fehler beim Parsen (Konvertierungsfehler. 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Der Server („cli“) hat berichtet: „array ( 'nohash' => array ( ), 'success' => true, '956d5beec9b0b11bb3230fe8a567b85c' => (object) array( 'speakText' => 'upper A equals negative 1 divided by upper R', 'mml' => '<math xmlns="http://www.w3.org/1998/Math/MathML" display="block" alttext="upper A equals negative 1 divided by upper R"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0" data-semantic-type="relseq" data-semantic-role="equality" data-semantic-id="8" data-semantic-children="0,7" data-semantic-content="1"><mi data-semantic-type="identifier" data-semantic-role="latinletter" data-semantic-font="italic" data-semantic-annotation="clearspeak:simple" data-semantic-id="0" data-semantic-parent="8">A</mi><mo data-semantic-type="relation" data-semantic-role="equality" data-semantic-id="1" data-semantic-parent="8" data-semantic-operator="relseq,=">=</mo><mrow data-semantic-type="infixop" data-semantic-role="division" data-semantic-id="7" 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Fehler beim Parsen (Konvertierungsfehler. 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Fehler beim Parsen (Konvertierungsfehler. 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Fehler beim Parsen (Konvertierungsfehler. 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annotation="clearspeak:simple" id="6">−<content><operator role="subtraction" id="2">−</operator></content><children><number role="integer" font="normal" annotation="clearspeak:simple" id="3">1</number></children></prefixop><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="5">R</identifier></children></infixop></children></relseq></stree>', 'success' => true, 'log' => 'success', 'mathoidStyle' => 'vertical-align: -0.838ex; width:10.739ex; height:2.843ex;', 'sanetex' => '{\\displaystyle A=-1/R}', 'speech' => 'upper A equals negative 1 divided by upper R', ), )“): {\displaystyle i_o(t) = \frac{\cos t}{5} + \frac{2 \sin t}{5} - \frac{7 e^{-t}\cos t}{10} - \frac{11 e^{-t}\sin t}{10} + \frac{1}{2} + C_1}

Der Mupad-Code zur Lösung des Integrals (ersetzt x durch τ) ist:

f := exp(-(t-x)) *sin(t-x) *(1 + cos(x));<br>S := int(f,x = 0..t)

Auffinden der Integrationskonstante

Fehler beim Parsen (Konvertierungsfehler. 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data-semantic-children="6,5" data-semantic-content="4" data-semantic-parent="8"><mrow data-semantic-type="prefixop" data-semantic-role="negative" data-semantic-annotation="clearspeak:simple" data-semantic-id="6" data-semantic-children="3" data-semantic-content="2" data-semantic-parent="7"><mo data-semantic-type="operator" data-semantic-role="subtraction" data-semantic-id="2" data-semantic-parent="6" data-semantic-operator="prefixop,−">−<!-- − --></mo><mn data-semantic-type="number" data-semantic-role="integer" data-semantic-font="normal" data-semantic-annotation="clearspeak:simple" data-semantic-id="3" data-semantic-parent="6">1</mn></mrow><mrow class="MJX-TeXAtom-ORD"><mo data-semantic-type="operator" data-semantic-role="division" data-semantic-id="4" data-semantic-parent="7" data-semantic-operator="infixop,/">/</mo></mrow><mi data-semantic-type="identifier" data-semantic-role="latinletter" data-semantic-font="italic" data-semantic-annotation="clearspeak:simple" data-semantic-id="5" 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annotation="clearspeak:simple" id="6">−<content><operator role="subtraction" id="2">−</operator></content><children><number role="integer" font="normal" annotation="clearspeak:simple" id="3">1</number></children></prefixop><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="5">R</identifier></children></infixop></children></relseq></stree>', 'success' => true, 'log' => 'success', 'mathoidStyle' => 'vertical-align: -0.838ex; width:10.739ex; height:2.843ex;', 'sanetex' => '{\\displaystyle A=-1/R}', 'speech' => 'upper A equals negative 1 divided by upper R', ), )“): {\displaystyle i_o(0_+) = 0 = \frac{1}{5} - \frac{7}{10} + \frac{1}{2} + C_1}

Das impliziert:

Fehler beim Parsen (Konvertierungsfehler. 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annotation="clearspeak:simple" id="6">−<content><operator role="subtraction" id="2">−</operator></content><children><number role="integer" font="normal" annotation="clearspeak:simple" id="3">1</number></children></prefixop><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="5">R</identifier></children></infixop></children></relseq></stree>', 'success' => true, 'log' => 'success', 'mathoidStyle' => 'vertical-align: -0.838ex; width:10.739ex; height:2.843ex;', 'sanetex' => '{\\displaystyle A=-1/R}', 'speech' => 'upper A equals negative 1 divided by upper R', ), )“): {\displaystyle C_1 = 0}


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