[<< wikibooks] Circuit Theory/Convolution Integral/Examples/example49/VL
Given that the source voltage is (2t-3t2), find voltage across the resistor.
This is the VL solution. 
Outline:


=== Transfer Function ===

  
    
      
        H
        (
        s
        )
        =
        
          
            
              V
              
                L
              
            
            
              V
              
                S
              
            
          
        
        =
        
          
            s
            
              4
              +
              s
              +
              
                
                  1
                  
                    0.25
                    s
                  
                
              
            
          
        
      
    
    {\displaystyle H(s)={\frac {V_{L}}{V_{S}}}={\frac {s}{4+s+{\frac {1}{0.25s}}}}}
  simplify(s/(4 + s + 1/(0.25*s)))

  
    
      
        H
        (
        s
        )
        =
        
          
            
              s
              
                2
              
            
            
              
                s
                
                  2
                
              
              +
              4
              s
              +
              4
            
          
        
      
    
    {\displaystyle H(s)={\frac {s^{2}}{s^{2}+4s+4}}}
  


=== Homogeneous Solution ===
solve(s^2 + 4.0*s + 4.0,s)
There are two equal roots at s = -2, so the solution has the form:

  
    
      
        
          V
          
            
              L
              
                h
              
            
          
        
        (
        t
        )
        =
        A
        
          e
          
            −
            2
            t
          
        
        +
        B
        t
        
          e
          
            −
            2
            t
          
        
        +
        
          C
          
            1
          
        
      
    
    {\displaystyle V_{L_{h}}(t)=Ae^{-2t}+Bte^{-2t}+C_{1}}
  


=== Particular Solution ===
After a long time attached to a unit step function source, the inductor has shorted and the capacitor has opened. All the drop is across the capacitor.

  
    
      
        
          V
          
            
              L
              
                p
              
            
          
        
        =
        0
      
    
    {\displaystyle V_{L_{p}}=0}
  This also means that C1 has to be zero.


=== Initial Conditions ===
So far the full equation is:

  
    
      
        
          V
          
            L
          
        
        (
        t
        )
        =
        A
        
          e
          
            −
            2
            t
          
        
        +
        B
        t
        
          e
          
            −
            2
            t
          
        
      
    
    {\displaystyle V_{L}(t)=Ae^{-2t}+Bte^{-2t}}
  Initial voltage is all across the inductor. 

  
    
      
        
          V
          
            L
          
        
        (
        0
        )
        =
        1
        =
        A
      
    
    {\displaystyle V_{L}(0)=1=A}
  

  
    
      
        A
        =
        1
      
    
    {\displaystyle A=1}
  At this point will have to do integral .. to get to the current. There is no other way to use the known initial conditions: current (initially zero), and VC (initially zero). Will have to introduce integration constant and then evaluate that. More chance of mistakes, more complex, so start over with something else.

  
    
      
        i
        (
        t
        )
        =
        
          
            1
            L
          
        
        
          ∫
          
            0
          
          
            t
          
        
        
          V
          
            L
          
        
        (
        t
        )
        d
        t
        =
        
          ∫
          
            0
          
          
            t
          
        
        (
        
          e
          
            −
            2
            x
          
        
        −
        B
        (
        x
        )
        
          e
          
            −
            2
            x
          
        
        )
        d
        x
      
    
    {\displaystyle i(t)={\frac {1}{L}}\int _{0}^{t}V_{L}(t)dt=\int _{0}^{t}(e^{-2x}-B(x)e^{-2x})dx}
  f := (exp(-2*x) - B*x*exp(-2*x));
S :=int(f,x=0..t)

  
    
      
        i
        (
        t
        )
        =
        B
        ∗
        (
        
          
            
              
                e
                
                  −
                  2
                  t
                
              
              (
              2
              t
              +
              1
              )
            
            4
          
        
        −
        
          
            1
            4
          
        
        )
        −
        
          
            
              e
              
                −
                2
                t
              
            
            2
          
        
        +
        
          
            1
            2
          
        
        +
        
          C
          
            1
          
        
      
    
    {\displaystyle i(t)=B*({\frac {e^{-2t}(2t+1)}{4}}-{\frac {1}{4}})-{\frac {e^{-2t}}{2}}+{\frac {1}{2}}+C_{1}}
  

  
    
      
        i
        (
        0
        )
        =
        0
        =
        
          C
          
            1
          
        
      
    
    {\displaystyle i(0)=0=C_{1}}
  Ok so C1 is zero. Now need to find B.
Find B by doing another integral to get VC:

  
    
      
        
          V
          
            C
          
        
        (
        t
        )
        =
        
          
            1
            C
          
        
        
          ∫
          
            0
          
          
            t
          
        
        i
        (
        t
        )
        d
        t
        =
        4
        ∗
        
          ∫
          
            0
          
          
            t
          
        
        (
        B
        ∗
        (
        
          
            
              
                e
                
                  −
                  2
                  t
                
              
              (
              2
              t
              +
              1
              )
            
            4
          
        
        −
        
          
            1
            4
          
        
        )
        −
        
          
            
              e
              
                −
                2
                t
              
            
            2
          
        
        +
        
          
            1
            2
          
        
        )
        d
        x
      
    
    {\displaystyle V_{C}(t)={\frac {1}{C}}\int _{0}^{t}i(t)dt=4*\int _{0}^{t}(B*({\frac {e^{-2t}(2t+1)}{4}}-{\frac {1}{4}})-{\frac {e^{-2t}}{2}}+{\frac {1}{2}})dx}
  f := (4*(B*(exp(-2*x)*(2*x+1)/4 -1/4) - exp(-2*x)/2 + 1/2));
S :=int(f,x=0..t)

  
    
      
        
          V
          
            C
          
        
        (
        t
        )
        =
        B
        +
        2
        t
        +
        
          e
          
            −
            2
            t
          
        
        −
        B
        t
        −
        B
        
          e
          
            −
            2
            t
          
        
        −
        B
        t
        
          e
          
            −
            2
            t
          
        
        −
        1
        +
        
          C
          
            1
          
        
      
    
    {\displaystyle V_{C}(t)=B+2t+e^{-2t}-Bt-Be^{-2t}-Bte^{-2t}-1+C_{1}}
  

  
    
      
        
          V
          
            C
          
        
        (
        0
        )
        =
        0
        =
        B
        +
        1
        −
        B
        −
        1
        +
        
          C
          
            1
          
        
      
    
    {\displaystyle V_{C}(0)=0=B+1-B-1+C_{1}}
  

  
    
      
        
          C
          
            1
          
        
        =
        0
      
    
    {\displaystyle C_{1}=0}
  Still doesn't help us find B. Guess B = 2 since (2t-Bt) has to equal zero if VC is going to converge on 1. Then see if VC(∞) = 1:

  
    
      
        
          V
          
            C
          
        
        (
        t
        )
        =
        2
        +
        
          e
          
            −
            2
            t
          
        
        −
        2
        
          e
          
            −
            2
            t
          
        
        −
        2
        t
        
          e
          
            −
            2
            t
          
        
        −
        1
      
    
    {\displaystyle V_{C}(t)=2+e^{-2t}-2e^{-2t}-2te^{-2t}-1}
  At t = ∞ what is the 2te-2t term's value?

limit(B*t*exp(-t),t = infinity)
Mupad says 0.

  
    
      
        
          V
          
            C
          
        
        (
        ∞
        )
        =
        2
        −
        1
        =
        1
      
    
    {\displaystyle V_{C}(\infty )=2-1=1}
  Yes! B = 2 works ... looks like the only thing that works .... So:

  
    
      
        i
        (
        t
        )
        =
        2
        ∗
        (
        
          
            
              
                e
                
                  −
                  2
                  t
                
              
              (
              2
              t
              +
              1
              )
            
            4
          
        
        −
        
          
            1
            4
          
        
        )
        −
        
          
            
              e
              
                −
                2
                t
              
            
            2
          
        
        +
        
          
            1
            2
          
        
        =
        t
        
          e
          
            −
            2
            t
          
        
      
    
    {\displaystyle i(t)=2*({\frac {e^{-2t}(2t+1)}{4}}-{\frac {1}{4}})-{\frac {e^{-2t}}{2}}+{\frac {1}{2}}=te^{-2t}}
  simplify(2*(exp(-2*t)(2*t+1)/4-1/4) - exp(-2*t)/2 + 1/2)
This means that VR is:

  
    
      
        
          V
          
            R
          
        
        (
        t
        )
        =
        4
        ∗
        i
        (
        t
        )
        =
        4
        t
        
          e
          
            −
            2
            t
          
        
      
    
    {\displaystyle V_{R}(t)=4*i(t)=4te^{-2t}}
  


=== Impulse Solution ===
Taking the derivative of the above get:

  
    
      
        
          V
          
            R
          
        
        δ
        (
        t
        )
        =
        4
        
          e
          
            −
            2
            t
          
        
        −
        8
        t
        
          e
          
            −
            2
            t
          
        
      
    
    {\displaystyle V_{R}\delta (t)=4e^{-2t}-8te^{-2t}}
  


=== Convolution Integral ===

  
    
      
        
          V
          
            R
          
        
        (
        t
        )
        =
        
          ∫
          
            0
          
          
            t
          
        
        (
        4
        
          e
          
            −
            2
            (
            t
            −
            x
            )
          
        
        −
        8
        (
        t
        −
        x
        )
        
          e
          
            −
            2
            (
            t
            −
            x
            )
          
        
        )
        (
        2
        x
        −
        3
        
          x
          
            2
          
        
        )
        d
        x
      
    
    {\displaystyle V_{R}(t)=\int _{0}^{t}(4e^{-2(t-x)}-8(t-x)e^{-2(t-x)})(2x-3x^{2})dx}
  f := (4*exp(-2*(t-x)) - 8*(t-x)exp(-2*(t-x)))*(2*x-3*x^2);
S :=int(f,x=0..t)

  
    
      
        
          V
          
            R
          
        
        (
        t
        )
        =
        8
        −
        8
        
          e
          
            −
            2
            t
          
        
        −
        10
        t
        
          e
          
            −
            2
            t
          
        
        −
        6
        t
      
    
    {\displaystyle V_{R}(t)=8-8e^{-2t}-10te^{-2t}-6t}
  There will not be any constant since again, V_R(t) = 0 after a long time ... and the capacitor opens.