Responses to Impulse Train

Example 1

Consider the steady state response of a first order RC circuit with the parameter $\tau$ to a square impulse train of period $T$:

$$x(t+T)=x(t)=\left\{ \begin{array}{cl}1&0<t<T_0\\ 0&T_0<t<T\end{array}\right.$$ (274)

where $T_0$ is the on-time and $T_1=T-T_0$ is off-time of the impulse train. Consider the response $y(t)=v_C(t)$ during both the on and off periods:

• The on-time response can be found based on the assumed initial condition $y(t)\big\vert _{t=0}=V_0$:

  $$y(t)=1+(V_0-1)e^{-t/\tau},\;\;\;\;\;\;\;\;(0<t<T_0)$$ (275)

  In particular, $y(t)\big\vert _{t=T_0}=y(T_0)=1+(V_0-1)e^{-T_0/\tau}=V_1$
• The off-time response can be found based on the initial condition $y(T_0)=V_1$ to be

  $$y(t)=0+(V_1-0)e^{-(t-T_0)/\tau}=V_1 e^{-(t-T_0)/\tau},\;\;\;\;\;\;(T_0<t<T)$$ (276)

  In particular, $y(T)=y(0)=V_1 e^{-(T-T_0)/\tau}=V_1 e^{-T_1/\tau}=V_0$

The two voltages $V_0$ and $V_1$ can then be found by solving these two simultaneous equations:

$$\left\{\begin{array}{l} V_0=V_1 e^{-T_1/\tau}\\ V_1=1+(V_0-1)e^{-T_0/\tau}\end{array}\right.$$ (277)

Substituting the first equation into the second one we can find $V_1$ and then $V_0$:

$$V_1=\frac{1-e^{-T_0/\tau}}{1-e^{-T/\tau}},\;\;\;\;\;\;\;\; V_0=\frac{e^{-T_1/\tau}-e^{-T/\tau}}{1-e^{-T/\tau}}$$ (278)

Consider some special cases:

• When $T_0\gg\tau$ and $T_1\gg\tau$, all exponential terms become zero and we have $V_0=0$ and $V_1=1$, i.e., the capacitor is fully charged and then fully discharged.
• When $T_0\ll\tau$ and $T_1\ll\tau$, all exponential terms can be approximated by $e^{-x}\approx 1-x$, and we have:

  $$V_1=\frac{1-e^{-T_0/\tau}}{1-e^{-T/\tau}}\approx\frac{T_0}{T}, \;... ...c{e^{-T_1/\tau}-e^{-T/\tau}}{1-e^{-T/\tau}}\approx\frac{T-T_1}{T}=\frac{T_0}{T}$$ (279)

  which is the average value of the impulse train.

• when $T_0=T_1=T/2$, we have

  $$1-e^{-T/\tau}=(1+e^{-T_0/\tau})(1-e^{-T_0/\tau})$$ (280)

  and

  $$V_1=\frac{1-e^{-T_0/\tau}}{1-e^{-T/\tau}}=\frac{1}{1+e^{-T_0/\tau... ...u}}{1-e^{-T/\tau}}=\frac{e^{-T_0/\tau}}{1+e^{-T_0/\tau}}, \;\;\;\;\;\;V_0+V_1=1$$ (281)

Here $T_0=T_1=T/2=1$, $\tau=0.5$.

Example 2

Consider the steady state response of an undamped second order system with parameters $\zeta=0$ and $\omega_n$ to a symmetric square impulse train of period $2T_0$:

$$x(t+2T_0)=x(t)=\left\{ \begin{array}{cl}1&0<t<T_0\\ 0&T_0<t<2T_0\end{array}\right.$$ (282)

where $T_0$ is both the on and off-time of the impulse train, which can also be written as:

$$x(t)=u(t)-u(t-T_0)+u(t-2T_0)-u(t-3T_0)+u(t-4T_0)- \cdots$$ (283)

The response of the 2nd order system can be obtained based on its step response $y(t)=1-\cos(\omega_nt)$ (with some constant coefficient which is neglected):

• $$y(t)=1-\cos\omega_nt,\;\;\;\;\;\;\;\;\;\;\;\;\;\;(0<t<T_0)$$ (284)

• $$y(t)=(1-\cos\omega_nt)-(1-\cos\omega_n(t-T_0)) =-\cos\omega_nt+\cos\omega_n(t-T_0),\;\;\;\;\;\;\;\;\;\;\;\;\;\;(T_0<t<2T_0)$$ (285)

• 
  

  $$y(t) = (1-\cos\omega_nt)-(1-\cos\omega_n(t-T_0))+(1-\cos\omega_n(t-2T_0))$$

  $$= 1-\cos\omega_n(t-2T_0)+\cos\omega_n(t-T_0)-\cos\omega_nt, \;\;\;\;\;\;\;\;\;\;\;\;\;\;(2T_0<t<3T_0)$$ (286)

  
  
• 
  

  $$y(t) = (1-\cos\omega_nt)-(1-\cos\omega_n(t-T_0))+(1-\cos\omega_n(t-2T_0)) -(1-\cos\omega_nt(t-3T_0))$$

  $$= -\cos\omega_nt+\cos\omega_n(t-T_0)-\cos\omega_n(t-2T_0)+\cos\omega_n(t-3T_0), \;\;\;\;\;\;\;\;\;\;\;\;\;\;(3T_0<t<4T_0)$$ (287)

  
  

We now consider two special cases:

• If $T_0=T=2\pi/\omega_n$ is the same as the natural period of the 2nd order system, then $\cos\omega_n(t-kT_0)=\cos\omega_nt$ and we have the following for all integer $k$:

  • $$y(t)=1-\cos\omega_nt,\;\;\;\;\;\;\;2kT_0<t<(2k+1)T_0$$ (288)

  • $$y(t)=0,\;\;\;\;\;\;\;\;\;\;\;\;(2k-1)T_0<t<2kT_0$$ (289)

  We see that as $t\rightarrow\infty$, more terms will be included in $y(t)$, which is zero when an even number of terms are included, but $1-\cos(\omega_nt)$ when an odd number of terms are included.

  

• If $T_0=T/2=\pi/\omega_n$, the period $2T_0$ of the impulse train is the same as the natural period $T=2\pi/\omega_n$ of the 2nd system,

  • $$y(t)=1-\cos\omega_nt,\;\;\;\;\;\;\;\;\;\;\;\;\;\;(0<t<T_0)$$ (290)

  • $$y(t)=\cos\omega_n(t-T/2)-\cos\omega_nt=-2\cos\omega_nt, \;\;\;\;\;\;\;\;\;\;\;\;\;\;(T_0<t<2T_0)$$ (291)

  • $$y(t)=1-\cos\omega_n(t-T)+\cos\omega_n(t-T/2)-\cos\omega_nt =1-3\cos\omega_nt,\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2T_0<t<3T_0)$$ (292)

  • $$y(t)=\cos\omega_n(t-T/2)-\cos\omega_nt+\cos\omega_n(t-3T/2)-\cos\omega_n(t-T)=-4\cos\omega_nt,\;\;\;\;\;\;\;\;\;\;\;\;\;\;(3T_0<t<4T_0)$$ (293)

  This is a sinusoid growing linearly.