Difference between revisions of "Impulse Response"

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Impulse Response[edit | edit source]

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[edit | edit source]

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{d i(t) \over dt}
:v(0) = 1 = L * (-\frac{A R}{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 VS(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 iδ as current. It is really {d \over dt}\frac{current}{1 volt}. VS(τ) turns into a multiplier.

LRC Example[edit | edit source]

Find the time domain expression for io given that Is = 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} = {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)
i_{o_\delta} = \frac{1}{2}e^{-t}(\cos t + \sin t + \sin t - \cos t) = e^{-t}\sin t
:I_s = 1 + \cos t
i_o(t) = \int_0^t i_{o_\delta} (t-\tau) I_s(\tau) d\tau + C_1
i_o(t) = \int_0^t e^{-(t-\tau)}\sin (t-\tau) (1 + \cos \tau) d\tau + C_1
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

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[edit | edit source]

i_o(0_+) = 0 = \frac{1}{5}  - \frac{7}{10} + \frac{1}{2} + C_1

This implies:

C_1 = 0
==Impulse Response==
            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:
        

        : <math>i(t) = Ae^{-\frac{t}{\frac{L}{R}}} + C </math>
        
        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):
        

        : <math>v(t) = L{d i(t) \over dt}</math><br />:<math>v(0) = 1 = L * (-\frac{A R}{L})</math><br />:<math>A = -1/R</math>
        
        If the current in the inductor is initially zero, then:
        

        : <math>i(0) = 0 = A + C</math><br />Which implies that:
        
        : <math>C = -A = 1/R</math><br />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>
        
        Taking the derivative of this, get the impulse (δ) current is:
        
        : <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:
        
        : <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>
        
        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.
        

        ==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.
        

        Earlier the step response for this problem was found:
        

        : <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:
        

        : <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><br />:<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>
        
        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>
        

        ==Finding the integration constant==
        
        : <math>i_o(0_+) = 0 = \frac{1}{5}  - \frac{7}{10} + \frac{1}{2} + C_1</math>
        
        This implies:
        
        : <math>C_1 = 0</math>
            

            [[en:{{FULLPAGENAME}}]][[de:Impulsantwort]]
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==Impulse Response==
 
 
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!
 
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==
 
==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.
+
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).
 
Choose a DC source of 1 volt (the real Vs then can scale off this).
Line 45: Line 44:
 
This implies:
 
This implies:
 
: <math>C_1 = 0</math>
 
: <math>C_1 = 0</math>
 +
 +
[[en:{{FULLPAGENAME}}]][[de:Impulsantwort]]

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