MATH52810/26/18

We use Laplace transforms to investigate how (electrical) circuits response to perturbations.

1Heaviside functions, $t$-shifting

Definition. The Heaviside (step) function $u(t-a)$is defined as

The Laplace transform is

Theorem. (Frequency-shifting) If $F=ℒ\left(f\right)$, then the frequency-shifted function $F(s-a)$ is the Laplace transform of $\stackrel{\sim }{f}\left(t\right)=e^{at}f(t-a)$

Theorem. (Time-shifting) If $F=ℒ\left(f\right)$, then the time-shifted function $\stackrel{\sim }{f}\left(t\right)=f(t-a)u(t-a)$ has Laplace transform

2Circuit response to perturbations

2.1Switch circuit on over time interval

A perturbation of amplitude $V_0$ applied for time interval $t\in [a,b]$ (e.g., voltage for electrical circuits) is

Figure 1. Single rectangular wave

2.1.1$R-C$ (resistance-storage) circuit

• Model

  ${\displaystyle Ry\left(t\right)+\frac{q\left(t\right)}{K}=Ry\left(t\right)+\frac{1}{K}\underset{0}{\overset{t}{\int }}y\left(\tau \right)d\tau =v\left(t\right)=V_0[u(t-a)-u(t-b)]}$

  In[28]:=

  tmodel = R y[t] + 1/K Integrate[y[tau],{tau,0,t}] == v[t,V0,a,b]

  ${\displaystyle \frac{\underset{0}{\overset{t}{\int }}y\left(\tau \right)d\tau }{K}+Ry\left(t\right)=\text{V0}(\theta (t-a)-\theta (t-b))}$

  In[29]:=

• Laplace transform

  ${\displaystyle RY\left(s\right)+\frac{Y\left(s\right)}{sK}=\frac{V_0}{s}[e^{-as}-e^{-bs}]}$

  In[29]:=

  smodel = Simplify[LaplaceTransform[tmodel,t,s],a>0 && b>0] /. LaplaceTransform[y[t], t, s] -> Y[s]

  ${\displaystyle \frac{\text{V0}(e^{-bs}-e^{-as})+Y\left(s\right)(\frac{1}{K}+Rs)}{s}=0}$

  In[30]:=

• Solution in frequency space

  ${\displaystyle I\left(s\right)=\frac{V_0/R}{s+{(RK)}^{-1}}[e^{-as}-e^{-bs}]=F\left(s\right)[e^{-as}-e^{-bs}]}$

  In[32]:=

  sol[s_]=Simplify[Y[s] /. Solve[smodel,Y[s]][[1,1]]]

  ${\displaystyle \frac{K\text{V0}(e^{-as}-e^{-bs})}{KRs+1}}$

  In[33]:=

  InputForm[sol[s]]

  ${\displaystyle \text{((E^∧(-(a*s)) - E^∧(-(b*s)))*K*V0)/(1 + K*R*s)}}$

  In[34]:=

• Response function in time

  ${\displaystyle f\left(t\right)={ℒ}^{-1}\left(F\right)\left(t\right)=\frac{V_0}{R}{e}^{-t/(RC)}}$

  In[34]:=

  F[s_]=(K*V0)/(1 + K*R*s)

  ${\displaystyle \frac{K\text{V0}}{KRs+1}}$

  In[35]:=

  f[t_]=InverseLaplaceTransform[F[s],s,t]

  ${\displaystyle \frac{\text{V0}{e}^{-\frac{t}{KR}}}{R}}$

  In[36]:=

• Response to single rectangular wave from frequency shifting

  ${\displaystyle i\left(t\right)=f(t-a)u(t-a)-f(t-b)u(t-b)}$

  In[42]:=

  y[t_,V0_,R_,K_,a_,b_] = f[t-a] UnitStep[t-a] - f[t-b] UnitStep[t-b]

  ${\displaystyle \frac{\text{V0}\theta (t-a)e^{-\frac{t-a}{KR}}}{R}-\frac{\text{V0}\theta (t-b)e^{-\frac{t-b}{KR}}}{R}}$

  In[43]:=

  yi[t_,V0_,R_,K_,a_,b_]=InverseLaplaceTransform[sol[s],s,t]

  ${\displaystyle \frac{\text{V0}{e}^{-\frac{t}{KR}}(\theta (t-a)e^{\frac{a}{KR}}-\theta (t-b)e^{\frac{b}{KR}})}{R}}$

  In[44]:=

  params = {V0->1,R->1,K->1,a->1,b->2}; solplt = Plot[{y[t,V0,R,K,a,b],yi[t,V0,R,K,a,b]} /. params,{t,0,4},Axes->False,Frame->True,FrameLabel->{"t","v(t)"}]; Export["/home/student/courses/MATH528/solplt.png",solplt]

  ${\displaystyle \text{/home/student/courses/MATH528/solplt.png}}$

  In[45]:=

Figure 2. Circuit response to single rectangular wave