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| {{ContentNav|homelink=Documentation System|hometext=3|pagePrevious=Electric Pallet Truck Alligator{{!}}Operating instructions|pageNext=:Category:Bike{{!}}Categorized product images}} | | <includeonly><templatestyles src="Languages/styles.css" /><div class="zeroheight"> |
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| | {{#vardefine:cl|{{CONTENTLANG}}}} |
| | {{#vardefine:l1|de}} |
| | {{#vardefine:l2|fr}} |
| | {{#vardefine:l3|nl}} |
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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!
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| ==Example Impulse Response==
| | {{#vardefine:clswitcher|[[{{NAMESPACE}}:{{PAGENAME}}|{{#var:cl}}]]{{#ifexist:{{NAMESPACE}}:{{PAGENAME}}/{{#var:l1}}|[[{{NAMESPACE}}:{{PAGENAME}}/{{#var:l1}}{{!}}{{#var:l1}}]]|}}{{#ifexist:{{NAMESPACE}}:{{PAGENAME}}/{{#var:l2}}|[[{{NAMESPACE}}:{{PAGENAME}}/{{#var:l2}}{{!}}{{#var:l2}}]]|}}{{#ifexist:{{NAMESPACE}}:{{PAGENAME}}/{{#var:l3}}|[[{{NAMESPACE}}:{{PAGENAME}}/{{#var:l3}}{{!}}{{#var:l3}}]]|}}}} |
| 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.
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| 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:
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| :<math>i(t) = Ae^{-\frac{t}{\frac{L}{R}}} + C </math> | | </div> |
| | | <div id="langbar-hz">{{DISPLAYTITLE:{{{1|}}}}}<div class="lang-button" role="button" aria-label="selectlanguage"><span class="bi bi-globe2"></span> |
| 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):
| | {{#switch: {{SUBPAGENAME}} |
| | | |{{#var:l1}}={{#var:switcher}} |
| :<math>v(t) = L{d i(t) \over dt}</math>:<math>v(0) = 1 = L * (-\frac{A R}{L})</math>:<math>A = -1/R</math> | | |{{#var:l2}}={{#var:switcher}} |
| | | |{{#var:l3}}={{#var:switcher}} |
| If the current in the inductor is initially zero, then:
| | |{{#var:clswitcher}}}}</div></div></includeonly> |
| | | <noinclude> |
| :<math>i(0) = 0 = A + C</math>Which implies that: | | <templatedata> |
| :<math>C = -A = 1/R</math>So the response to a DC voltage source turning on at t=0 to one volt (called the unit response μ) is: | | { |
| :<math>i_\mu (t) = \frac{1}{R}(1 - e^{-\frac{t}{\frac{L}{R}}})</math> | | "params": { |
| | | "1": { |
| Taking the derivative of this, get the impulse (δ) current is:
| | "label": "Displaytitle", |
| | | "suggested": true |
| :<math>i_\delta (t) = \frac{e^{-\frac{t}{\frac{L}{R}}}}{L}</math> | | } |
| | | }, |
| Now the current due to any arbitrary V<sub>S</sub>(t) can be found using the convolution integral:
| | "description": { |
| | | "en": "Language switcher", |
| :<math>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</math> | | "de": "Sprachwechsler", |
| | | "fr": "Sélecteur de langue", |
| Don't think i<sub>δ</sub> as current. It is really <math>{d \over dt}\frac{current}{1 volt}</math>. V<sub>S</sub>(τ) turns into a multiplier.
| | "nl": "Taal verandering" |
| ==LRC Example==
| | } |
| Find the time domain expression for i<sub>o</sub> given that I<sub>s</sub> = cos(t + π/2)μ(t) amp.
| | } |
| | | </templatedata> |
| Earlier the step response for this problem was found:
| | </noinclude> |
| | |
| :<math> i_{o_\mu} = \frac{1}{2}(1 - e^{-t}(\cos t + \sin t))</math> | |
| | |
| The impulse response is going to be the derivative of this:
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| | |
| :<math>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)</math> | |
| :<math>i_{o_\delta} = \frac{1}{2}e^{-t}(\cos t + \sin t + \sin t - \cos t) = e^{-t}\sin t</math>:<math>I_s = 1 + \cos t</math> | |
| :<math>i_o(t) = \int_0^t i_{o_\delta} (t-\tau) I_s(\tau) d\tau + C_1</math> | |
| :<math>i_o(t) = \int_0^t e^{-(t-\tau)}\sin (t-\tau) (1 + \cos \tau) d\tau + C_1</math>
| |
| :<math>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</math> | |
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| The Mupad code to solve the integral (substituting x for τ) is:<pre>f := exp(-(t-x)) *sin(t-x) *(1 + cos(x));<br>S := int(f,x = 0..t)</pre>
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| ==Finding the integration constant==
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| :<math>i_o(0_+) = 0 = \frac{1}{5} - \frac{7}{10} + \frac{1}{2} + C_1</math> | |
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| This implies:
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| :<math>C_1 = 0</math>
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| [[en:{{FULLPAGENAME}}|Impulse Response]][[de:Impulsantwort]]
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