Impulse response: Difference between revisions

Latest revision as of 07:59, 6 December 2021

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So far circuits have been driven by a DC source, an AC source and an exponential source. If we can find the current of a circuit generated by a Dirac delta function or impulse voltage source δ, then the convolution integral can be used to find the current to any given voltage source!

Example Impulse Response

The current is found by taking the derivative of the current found due to a DC voltage source! Say the goal is to find the δ current of a series LR circuit ... so that in the future the convolution integral can be used to find the current given any arbitrary source.

Choose a DC source of 1 volt (the real Vs then can scale off this). The particular homogeneous solution (steady state) is 0. The homogeneous solution to the non-homogeneous equation has the form:

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

Assume the current initially in the inductor is zero. The initial voltage is going to be 1 and is going to be across the inductor (since no current is flowing):

v(t)=L{di(t) \over dt}

:

v(0)=1=L*(-{\frac {AR}{L}})

:

A=-1/R

If the current in the inductor is initially zero, then:

i(0)=0=A+C

Which implies that:

C=-A=1/R

So the response to a DC voltage source turning on at t=0 to one volt (called the unit response μ) is:

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

Taking the derivative of this, get the impulse (δ) current is:

i_{\delta }(t)={\frac {e^{-{\frac {t}{\frac {L}{R}}}}}{L}}

Now the current due to any arbitrary V_S(t) can be found using the convolution integral:

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}

Don't think i_δ as current. It is really ${d \over dt}{\frac {current}{1volt}}$ . V_S(τ) turns into a multiplier.

LRC Example

Find the time domain expression for i_o given that I_s = cos(t + π/2)μ(t) amp.

Earlier the step response for this problem was found:

i_{o_{\mu }}={\frac {1}{2}}(1-e^{-t}(\cos t+\sin t))

The impulse response is going to be the derivative of this:

i_{o_{\delta }}={di_{o_{\mu }} \over dt}=0+{\frac {1}{2}}e^{-t}(\cos t+\sin t)-{\frac {1}{2}}e^{-t}(-\sin t+\cos t)

i_{o_{\delta }}={\frac {1}{2}}e^{-t}(\cos t+\sin t+\sin t-\cos t)=e^{-t}\sin t

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The Mupad code to solve the integral (substituting x for τ) is:

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

Finding the integration constant

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This implies:

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id="19">sin</function><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="21">t</identifier></children></appl><appl role="prefix function" annotation="clearspeak:simple" id="43"><content><punctuation role="application" id="42">⁡</punctuation><function role="prefix function" font="normal" id="23">sin</function></content><children><function role="prefix function" font="normal" id="23">sin</function><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="25">t</identifier></children></appl></children></infixop><appl role="prefix function" annotation="clearspeak:simple" id="41"><content><punctuation role="application" id="40">⁡</punctuation><function role="prefix function" font="normal" id="27">cos</function></content><children><function role="prefix function" font="normal" id="27">cos</function><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="29">t</identifier></children></appl></children></infixop></children></fenced></children></infixop><infixop role="implicit" annotation="clearspeak:unit" id="57">⁢<content><operator role="multiplication" id="56">⁢</operator></content><children><superscript role="latinletter" id="36"><children><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="32">e</identifier><prefixop role="negative" annotation="clearspeak:simple" id="35">−<content><operator role="subtraction" id="33">−</operator></content><children><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="34">t</identifier></children></prefixop></children></superscript><appl role="prefix function" annotation="clearspeak:simple" id="52"><content><punctuation role="application" id="51">⁡</punctuation><function role="prefix function" font="normal" id="37">sin</function></content><children><function role="prefix function" font="normal" id="37">sin</function><identifier role="latinletter" font="italic" annotation="clearspeak:simple" id="39">t</identifier></children></appl></children></infixop></children></relseq></stree>', 'success' => true, 'log' => 'success', 'mathoidStyle' => 'vertical-align: -1.838ex; width:48.81ex; height:5.176ex;', 'sanetex' => '{\\displaystyle i_{o_{\\delta }}={\\frac {1}{2}}e^{-t}(\\cos t+\\sin t+\\sin t-\\cos t)=e^{-t}\\sin t}', 'speech' => 'i Subscript o Sub Subscript delta Baseline equals one half e Superscript negative t Baseline left parenthesis cosine t plus sine t plus sine t minus cosine t right parenthesis equals e Superscript negative t Baseline sine t', ), )"): {\displaystyle C_1 = 0}

Tour 3 Documentation System ‹ Operating instructionsnext: Categorized product images

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 So far circuits have been driven by a DC source, an AC source and an exponential source. If we can find the current of a circuit generated by a Dirac delta function or impulse voltage source δ, then the convolution integral can be used to find the current to any given voltage source!
 ==Example Impulse Response==
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 :<math>C_1 = 0</math>
-[[en:{{FULLPAGENAME}}|en:Impulse Response]][[de:Impulsantwort]]
+[[en:{{FULLPAGENAME}}|Impulse Response]][[de:Impulsantwort]]
+{{ContentNav|homelink=Documentation System|hometext=3|pagePrevious=Electric Pallet Truck Alligator{{!}}Operating instructions|pageNext=:Category:Bike{{!}}Categorized product images}}