Impulsantwort

Tour 3 Dokumentationssystem ‹ Betriebsanweisungweiter: Produktbilder

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:

$i(t)=Ae^{-{\frac {t}{\frac {L}{R}}}}+C$

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

$v(t)=L{di(t) \over dt}$:$v(0)=1=L*(-{\frac {AR}{L}})$:$A=-1/R$

Wenn der Strom im Induktor zunächst Null ist, dann:

$i(0)=0=A+C$

Das impliziert:

$C=-A=1/R$

Die Antwort auf das Einschalten einer Gleichspannungsquelle bei t=0 bis ein Volt (die so genannte Unit Response μ) lautet also:

$i_{\mu }(t)={\frac {1}{R}}(1-e^{-{\frac {t}{\frac {L}{R}}}})$

Wenn man die Ableitung daraus zieht, erhält man den Impuls (δ) Strom ist:

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width:21.465ex; height:7.009ex;', 'sanetex' => '{\\displaystyle i_{\\mu }(t)={\\frac {1}{R}}(1-e^{-{\\frac {t}{\\frac {L}{R}}}})}', 'speech' => 'i Subscript mu Baseline left parenthesis t right parenthesis equals StartFraction 1 Over upper R EndFraction left parenthesis 1 minus e Superscript minus StartStartFraction t OverOver StartFraction upper L Over upper R EndFraction EndEndFraction Baseline right parenthesis', ), )“): {\displaystyle i_\delta (t) = \frac{e^{-\frac{t}{\frac{L}{R}}}}{L}}

Nun der Strom aufgrund einer beliebigen Anzahl V_S(t) kann über das Convolution Integral gefunden werden:

Fehler beim Parsen (Konvertierungsfehler. 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width:21.465ex; height:7.009ex;', 'sanetex' => '{\\displaystyle i_{\\mu }(t)={\\frac {1}{R}}(1-e^{-{\\frac {t}{\\frac {L}{R}}}})}', 'speech' => 'i Subscript mu Baseline left parenthesis t right parenthesis equals StartFraction 1 Over upper R EndFraction left parenthesis 1 minus e Superscript minus StartStartFraction t OverOver StartFraction upper L Over upper R EndFraction EndEndFraction Baseline right parenthesis', ), )“): {\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 i_δ nicht als aktuell betrachten. Es ist wirklich Fehler beim Parsen (Konvertierungsfehler. Der Server („cli“) hat berichtet: „array ( 'nohash' => array ( ), 'success' => true, 'cfc90e08666fa1fc92f3ed8adca9843a' => (object) array( 'speakText' => 'i Subscript mu Baseline left parenthesis t right parenthesis equals StartFraction 1 Over upper R EndFraction left parenthesis 1 minus e Superscript minus StartStartFraction t OverOver StartFraction upper L Over upper R EndFraction EndEndFraction Baseline right parenthesis', 'mml' => '<math xmlns="http://www.w3.org/1998/Math/MathML" display="block" alttext="i Subscript mu Baseline left parenthesis t right parenthesis equals StartFraction 1 Over upper R EndFraction left parenthesis 1 minus e Superscript minus StartStartFraction t OverOver StartFraction upper L Over upper R EndFraction EndEndFraction Baseline right parenthesis"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0" data-semantic-type="relseq" data-semantic-role="equality" data-semantic-id="30" data-semantic-children="27,29" 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width:21.465ex; height:7.009ex;', 'sanetex' => '{\\displaystyle i_{\\mu }(t)={\\frac {1}{R}}(1-e^{-{\\frac {t}{\\frac {L}{R}}}})}', 'speech' => 'i Subscript mu Baseline left parenthesis t right parenthesis equals StartFraction 1 Over upper R EndFraction left parenthesis 1 minus e Superscript minus StartStartFraction t OverOver StartFraction upper L Over upper R EndFraction EndEndFraction Baseline right parenthesis', ), )“): {\displaystyle {d \over dt}\frac{current}{1 volt}} . V_S(τ) wird zu einem Multiplikator.

LRC Beispiel

Finden Sie den Zeitbereichsausdruck für i_o, da I_s = 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, 'cfc90e08666fa1fc92f3ed8adca9843a' => (object) array( 'speakText' => 'i Subscript mu Baseline left parenthesis t right parenthesis equals StartFraction 1 Over upper R EndFraction left parenthesis 1 minus e Superscript minus StartStartFraction t OverOver StartFraction upper L Over upper R EndFraction EndEndFraction Baseline right parenthesis', 'mml' => '<math xmlns="http://www.w3.org/1998/Math/MathML" display="block" alttext="i Subscript mu Baseline left parenthesis t right parenthesis equals StartFraction 1 Over upper R EndFraction left parenthesis 1 minus e Superscript minus StartStartFraction t OverOver StartFraction upper L Over upper R EndFraction EndEndFraction Baseline right parenthesis"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0" data-semantic-type="relseq" data-semantic-role="equality" data-semantic-id="30" data-semantic-children="27,29" 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role="integer" font="normal" annotation="clearspeak:simple" id="11">1</number><superscript role="latinletter" id="21"><children><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="13">e</identifier><prefixop role="negative" id="20">−<content><operator role="subtraction" id="14">−</operator></content><children><fraction role="division" id="19"><children><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="15">t</identifier><fraction role="division" id="18"><children><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="16">L</identifier><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="17">R</identifier></children></fraction></children></fraction></children></prefixop></children></superscript></children></infixop></children></fenced></children></infixop></children></relseq></stree>', 'success' => true, 'log' => 'success', 'mathoidStyle' => 'vertical-align: -2.005ex; width:21.465ex; height:7.009ex;', 'sanetex' => '{\\displaystyle i_{\\mu }(t)={\\frac {1}{R}}(1-e^{-{\\frac {t}{\\frac {L}{R}}}})}', 'speech' => 'i Subscript mu Baseline left parenthesis t right parenthesis equals StartFraction 1 Over upper R EndFraction left parenthesis 1 minus e Superscript minus StartStartFraction t OverOver StartFraction upper L Over upper R EndFraction EndEndFraction Baseline right parenthesis', ), )“): {\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. 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role="integer" font="normal" annotation="clearspeak:simple" id="11">1</number><superscript role="latinletter" id="21"><children><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="13">e</identifier><prefixop role="negative" id="20">−<content><operator role="subtraction" id="14">−</operator></content><children><fraction role="division" id="19"><children><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="15">t</identifier><fraction role="division" id="18"><children><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="16">L</identifier><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="17">R</identifier></children></fraction></children></fraction></children></prefixop></children></superscript></children></infixop></children></fenced></children></infixop></children></relseq></stree>', 'success' => true, 'log' => 'success', 'mathoidStyle' => 'vertical-align: -2.005ex; width:21.465ex; height:7.009ex;', 'sanetex' => '{\\displaystyle i_{\\mu }(t)={\\frac {1}{R}}(1-e^{-{\\frac {t}{\\frac {L}{R}}}})}', 'speech' => 'i Subscript mu Baseline left parenthesis t right parenthesis equals StartFraction 1 Over upper R EndFraction left parenthesis 1 minus e Superscript minus StartStartFraction t OverOver StartFraction upper L Over upper R EndFraction EndEndFraction Baseline right parenthesis', ), )“): {\displaystyle i_{o_\delta} = {d i_{o_\mu} \over dt} = 0 + \frac{1}{2}e^{-t}(\cos t + \sin t) - \frac{1}{2}e^{-t}(-\sin t + \cos t)}

Fehler beim Parsen (Konvertierungsfehler. 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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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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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Das impliziert:

Fehler beim Parsen (Konvertierungsfehler. 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Tour 3 Dokumentationssystem ‹ Betriebsanweisungweiter: Produktbilder