LTI Systems Characterized by LCCDEs

If an LTI system can be described by an LCCDE in time domain

$$\sum_{k=0}^N a_k \frac{d^k y(t)}{dt^k} =\sum_{k=0}^M b_k \frac{d^k x(t)}{dt^k}$$

then after taking Laplace transform of the LCCDE, it can be represented as an algebraic equation in the $s$ domain

$$Y(s)[\sum_{k=0}^N a_k s^k]=X(s)[\sum_{k=0}^M b_k s^k]$$

and its transfer function is rational

$$H(s)=\frac{Y(s)}{X(s)}=\frac{\sum_{k=0}^M b_k s^k}{\sum_{k=0... ...-s_{z_k})}{\prod_{k=1}^N (s-s_{z_k})} =K \; \frac{N(s)}{D(s)}$$

where $K=b_M/a_N$ is a constant and $s_{z_k}, (k=1,2, \cdots, M)$ are the roots of the numerator polynomial and $s_{p_k}, (k=1,2, \cdots, N)$ are the roots of the denominator polynomial. Note that just as the LCCDE alone does not completely specify the relationship between $x(t)$ and $y(t)$ (additional information such as the initial conditions is needed), the transfer function $H(s)$ does not completely specify the system. For example, the same $H(s)$ with different ROCs will represent different systems (e.g., causal or anticausal).

Example 1: A circuit consisting an inductor $L$ and a resistor $R$ with input voltage $x(t)$ applied to the two element in series can be described by an LCCDE:

$$x(t)=v_R(t)+v_L(t)=R\;i(t)+L\frac{d}{dt} i(t)$$

Taking Laplace transform of this equation, we get

$$X(s)=V_R(s)+V_L(s)=RI(s)+LsI(s)=(R+sL) I(s)$$

The ratio between the voltage $X(s)$ and current $I(s)$ is defined as the impedance of the circuit:

$$Z(s)\stackrel{\triangle}{=}\frac{X(s)}{I(s)}=R+sL$$

If the output is the voltage across the resistor $v_R(t)$, then the transfer function of the system (a voltage divider) is

$$H_R(s)=\frac{V_R(s)}{X(s)}=\frac{R}{R+sL}=\frac{R/L}{R/L+s}=\frac{1/\tau}{s+1/\tau}$$

where $\tau\stackrel{\triangle}{=}L/R > 0$. In time domain, the impulse response of the system is

$$h(t)=\frac{1}{\tau} e^{-t/\tau} u(t)$$

If the output is the voltage across the inductor $v_L(t)$, then the transfer function is

$$H_L(s)=\frac{V_L(s)}{X(s)}=\frac{sL}{R+sL}=\frac{s}{1/\tau+s} =1-\frac{1/\tau}{s+1/\tau}=1-H_R(s)$$

with impulse response in time domain:

$$h(t)=\delta(t)-\frac{1}{\tau} e^{-t/\tau} u(t)$$

As the ROC of both $H_R(s)$ and $H_L(s)$ is the half plane to the right of the only pole $s_p=-1/\tau$ on the negative side the real axis, the $j\omega$-axis is contained in ROC and the corresponding Fourier transforms exist:

$$H_R(j\omega)=\frac{1/\tau}{j\omega+1/\tau},\;\;\;\;\mbox{and}\;\;\;\; H_R(j\omega)=\frac{j\omega}{j\omega+1/\tau}$$

Example 2: A voltage $x(t)$ is applied as the input to a resistor $R$, a capacitor $C$ and an inductor $L$ connected in series. The system can be described by a differential equation in time domain:

$$x(t)=v_L(t)+v_R(t)+v_C(t)=L\frac{d}{dt}\;i(t)+R\;i(t) +\frac{1}{C}\int_{-\infty}^t i(\tau)d\tau$$

or an algebraic equation in s-domain:

$$X(s)=V_L(s)+V_R(s)+V_C(s)=[sL+R+\frac{1}{sC}]I(s)$$

and the overall impedance of the circuit is defined as the ratio between voltage $V(s)$ and the current $I(s)$

$$Z(s)\stackrel{\triangle}{=}\frac{X(s)}{I(s)}=sL+R+\frac{1}{sC}$$

which is composed of the individual impedance of the three elements

$$Z_L(s)=sL,\;\;\;\;Z_R(s)=R,\;\;\;\;\;Z_C(s)=\frac{1}{sC}$$

	resistor $R$	capacitor $C$	inductor $L$
time domain	$i=\frac{v}{R}$	$i=\frac{1}{C}\frac{dv}{dt}$	$v=\frac{1}{L}\frac{di}{dt}$
s-domain	$V_R=IR$	$V_C=I/Cs$	$V_L=IsL$
impedance $Z=V/I$	$R$	$1/sC$	$sL$

If the output is the voltage across one of the three elements ($V_L$, $V_R$, or $V_C$), the transfer function $H(s)$ can be easily obtained by treating the series circuit as a voltage divider:

• Output is voltage across the capacitor $v_C(t)$
  
  $$H_C(s)=\frac{1/sC}{Ls+R+1/sC}=\frac{1/LC}{s^2+(R/L)s+(1/LC)}$$
  
  

• Output is voltage across the resistor $v_R(t)$
  
  $$H_R(s)=\frac{R}{Ls+R+1/sC}=\frac{(R/L)s}{s^2+(R/L)s+(1/LC)}$$
  
  

• Output is voltage across the inductor $v_L(t)$
  
  $$H_L(s)=\frac{sL}{Ls+R+1/sC}=\frac{s^2}{s^2+(R/L)s+(1/LC)}$$
  
  

If we define

$$\zeta \stackrel{\triangle}{=}\frac{R}{2}\sqrt{\frac{C}{L}},\;\;\;\;\; \omega_n \stackrel{\triangle}{=}\frac{1}{\sqrt{LC}}$$

the common denominator of the transfer functions can be written in standard (canonical) form

$$s^2+(R/L)s+(1/LC)=s^2+2\zeta\omega_n s+\omega_n^2=(s-p_1)(s-p_2)$$

with two roots

$$p_{1,2}=(-\zeta \pm \sqrt{\zeta^2-1})\omega_n =(-\zeta \pm j \sqrt{1-\zeta^2})\omega_n$$

and the transfer functions above can be written in standard forms:

• Output across C:
  
  $$H_C(s)=\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$$
  
  
  with two poles $p_1, p_2$ and no zeros.
• Output across R:
  
  $$H_R(s)=\frac{2\zeta\omega_n s}{s^2+2\zeta\omega_n s+\omega_n^2}$$
  
  
  with two poles $p_1, p_2$ and one zero at the origin.
• Output across L:
  
  $$H_L(s)=\frac{s^2}{s^2+2\zeta\omega_n s+\omega_n^2}$$
  
  
  with two poles $p_1, p_2$ and two repeated zeros at the origin.

As to be discussed later, the magnitude and phase of the corresponding frequency response function $H(j\omega)$ can be qualitatively determined in the s-plane, and it turns out that the three transfer functions behave like low-pass, band-pass and high-pass filter, respectively. Moreover, when the common real part $-\zeta \omega_n$ of the two complex conjugate poles is small (i.e., $0<\zeta < 0.5$), there will be a narrow pass-band around $\omega=\omega_n$ in all three cases.

Example 3:

The input to an LTI is $x(t)=e^{-3t}u(t)$ and the output is

$$y(t)=h(t)*x(t)=(e^{-t}-e^{-2t})u(t)$$

We want to identify the system by finding $h(t)$ and $H(s)$. In s-domain, input and output signals are

$$X(s)=\frac{1}{s+3},\;\;\;\;Re[s]>-3$$

and

$$Y(s)=H(s)X(s)=\frac{1}{s+1}-\frac{1}{s+2}=\frac{1}{(s+1)(s+2)},\;\;\;\;Re[s]>-1$$

The transfer function can therefore be obtained

$$H(s)=\frac{Y(s)}{X(s)}=\frac{s+3}{(s+1)(s+2)}=\frac{s+3}{s^2+3s+2} =\frac{2}{s+1}-\frac{1}{s+2}$$

This system $H(s)$ has two poles $p_1=-1$ and $p_2=-2$ and therefore three possible ROCs:

• $Re[s]<-2$, $h(t)$ is left sided (anticausal);
• $-2 < Re[s] < -1$, $h(t)$ is two sided (noncausal);
• $Re[s]>-1$, $h(t)$ is right sided (causal).

To determine which of these ROCs the system has, recall that the ROC of a product $Y(s)=H(s)X(s)$ should be no less than the intersection of the ROCs of $H(s)$ and $X(s)$, i.e., the ROC of $H(s)$ must be $Re[s]>-1$, i.e., the system is causal and stable:

$$h(t)=(2 e^{-t}-e^{-2t}) u(t)$$

The inverse Laplace transform of $Y(s)=H(s)X(s)$ is the LCCDE of the system:

$$\frac{d^2}{dt^2}y(t)+ 3\frac{d}{dt}y(t)+ 2y(t)=\frac{d}{dt}x(t)+ 3x(t)$$