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Next: Evaluation of Fourier Transform Up: Laplace_Transform Previous: Representation of LTI Systems

LTI Systems Characterized by LCCDEs

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

\begin{displaymath}
\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}
\end{displaymath}

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

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

and its transfer function is rational

\begin{displaymath}
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)}
\end{displaymath}

where $K=b_M/a_N$ is a constant and $s_{z_k}, (k=1,2, \cdots, M)$ are the zeros of $H(s)$ (roots of the numerator polynomial $Y(s)$) and $s_{p_k}, (k=1,2, \cdots, N)$ are the poles of $H(s)$ (roots of the denominator polynomial $X(s)$). The LCCDE alone does not completely specify the relationship between $x(t)$ and $y(t)$, as additional information such as the initial conditions is needed. Similarly, 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 anti-causal).

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

\begin{displaymath}v(t)=v_R(t)+v_L(t)=R\;i(t)+L\frac{d}{dt} i(t) \end{displaymath}

Taking Laplace transform of this equation, we get

\begin{displaymath}V(s)=V_R(s)+V_L(s)=RI(s)+LsI(s)=(R+sL) I(s) \end{displaymath}

As the ROCs of both $H_R(s)$ and $H_L(s)$ are the same half plane to the right of the only pole $s_p=-1/\tau$ on the negative side of the real axis, the $j\omega$-axis is contained in ROC and the corresponding frequency response function exists:

\begin{displaymath}H_R(j\omega)=H_R(s)\bigg\vert _{s=j\omega}=\frac{1/\tau}{j\om...
...)=H_R(s)\bigg\vert _{s=j\omega}=\frac{j\omega}{j\omega+1/\tau} \end{displaymath}

Example 2: A voltage $v(t)$ is applied as the input to a resistor $R$, a capacitor $C$ and an inductor $L$ connected in series. According to Kirchhoff's voltage law, the, the system can be described by a differential equation in time domain:

\begin{displaymath}
v(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
\end{displaymath}

or an algebraic equation in s-domain:

\begin{displaymath}V(s)=V_L(s)+V_R(s)+V_C(s)=[sL+R+\frac{1}{sC}]I(s) \end{displaymath}

If the current $i(t)$ through the circuit is treated as the output, then the transfer function of the system is

\begin{displaymath}H(s)=\frac{V(s)}{I(s)}=sL+R+\frac{1}{sC}=Z(s) \end{displaymath}

which is the overall impedance of the circuit composed of the individual impedance of the three elements

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

  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: If we define

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

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

\begin{displaymath}s^2+(R/L)s+(1/LC)=s^2+2\zeta\omega_n s+\omega_n^2=(s-p_1)(s-p_2) \end{displaymath}

with two roots

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

and the transfer functions above can be written in standard forms: 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: System identification: find $h(t)$ and $H(s)$ of an LTI, based on the given input $x(t)$ and output $y(t)$:

\begin{displaymath}\left\{ \begin{array}{l}
x(t)=e^{-3t}u(t) \\
y(t)=h(t)*x(t)=(e^{-t}-e^{-2t})u(t)\end{array} \right. \end{displaymath}

In s-domain, input and output signals are

\begin{displaymath}X(s)=\frac{1}{s+3},\;\;\;\;\;\;R_X:\;\;Re[s]>-3 \end{displaymath}


\begin{displaymath}Y(s)=\frac{1}{s+1}-\frac{1}{s+2}=\frac{1}{(s+1)(s+2)},\;\;\;\;\;R_Y:\;Re[s]>-1\end{displaymath}

The transfer function can therefore be obtained

\begin{displaymath}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} \end{displaymath}

This system $H(s)$ has two poles $p_1=-1$ and $p_2=-2$ and therefore there are three possible ROCs: We need to determine which of these ROCs is true for $H(s)$. As the ROC of a product is the intersection of the ROCs of the factors (without zero-pole cancellation):

\begin{displaymath}Y(s)=H(s)X(s),\;\;\;\;\;\;\;\;\;R_Y=R_H \; \bigcap \; R_X \end{displaymath}

ROC of $H(s)$ must be the third one above, and we have:

\begin{displaymath}h(t)={\cal L}^{-1}\left[\frac{2}{s+1}-\frac{1}{s+2}\right]
=(2 e^{-t}-e^{-2t}) u(t) \end{displaymath}

The equation for $H(s)$ above can be written as:

\begin{displaymath}Y(s)(s^2+3s+2)=X(s)(s+3) \end{displaymath}

Its inverse Laplace transform is the LCCDE of the system:

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


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Next: Evaluation of Fourier Transform Up: Laplace_Transform Previous: Representation of LTI Systems
Ruye Wang 2012-01-28