Skip to main content
Engineering LibreTexts

4.7: Frequency-Response Function from Transfer Function

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    For frequency response of a general LTI SISO stable system, we define the input to be a time-varying cosine, with amplitude \(U\) and circular frequency \(\omega\),

    \[u(t)=U \cos \omega t=\frac{U}{2}\left(e^{j \omega t}+e^{-j \omega t}\right)\label{eqn:4.25} \]

    in which we apply the complex exponential form for the cosine that is derived from Euler’s equation (Problem 2.1). The Laplace transform of input Equation \(\ref{eqn:4.25}\) is

    \[L[u(t)]=\frac{U}{2}\left(\frac{1}{s-j \omega}+\frac{1}{s+j \omega}\right)\label{eqn:4.26} \]

    Substituting Equation 4.6.3 and Equation \(\ref{eqn:4.26}\) into Equation 4.6.4 gives

    \[L[x(t)]_{I C_S=0}=\left(\frac{b_{1} s^{m}+b_{2} s^{m-1}+\ldots+b_{m+1}}{a_{1} s^{n}+a_{2} s^{n-1}+\ldots+a_{n+1}}\right) \frac{U}{2}\left(\frac{1}{s-j \omega}+\frac{1}{s+j \omega}\right)\label{eqn:4.27} \]

    By expanding into partial fractions, we will usually be able to cast Equation \(\ref{eqn:4.27}\) into the form

    \[L[x(t)]_{I C_S=0}=\frac{A}{s-j \omega}+\frac{B}{s+j \omega}+\sum_{k=1}^{n} \frac{C_{k}}{s-p_{k}}\label{eqn:4.28} \]

    in which \(A\), \(B\), \(C_k\), and \(\rho_k\) [the poles of \(TF(s)\)] all are constants. Taking the inverse transform of Equation \(\ref{eqn:4.28}\) gives

    \[x(t)=A e^{j \omega t}+B e^{-j \omega t}+\sum_{k=1}^{n} C_{k} e^{p_{k} t}\label{eqn:4.29} \]

    The first two right-hand-side terms of Equation \(\ref{eqn:4.29}\) are associated with steady-state forced sinusoidal response, and the third term is associated with response bounded by real exponential functions. The nature of system stability is determined by the poles \(p_k\), in particular, by their real parts. If \(\operatorname{Re}\left[p_{k}\right]<0\) for all \(k\) = 1, 2, \(\dots\), \(n\), then each of the \(e^{p_{k} t}\) terms is bounded by a decaying exponential, that is, \(\rightarrow\) 0 as \(t \rightarrow \infty\). A system for which for all \(k\) is said to be stable. Therefore, for steady-state sinusoidal response (after all exponentially bounded transients have decayed) of a stable system, only the first two right-hand-side terms of Equation \(\ref{eqn:4.29}\) remain,

    \[x_{s s}(t)=A e^{j \omega t}+B e^{-j \omega t}\label{eqn:4.30} \]

    Our objective now is to find the complex constants \(A\) and \(B\) in Equation \(\ref{eqn:4.30}\). The first step is to recognize that \(x_{s s}(t)\) must be a real function mathematically; that is, in order to represent actual physical behavior, Equation \(\ref{eqn:4.30}\) cannot have any imaginary component. From Problem 2.5, we conclude that \(B\) must be the complex conjugate of \(A\), i.e., \(B=\bar{A}\). Therefore, Equation \(\ref{eqn:4.30}\) becomes

    \[x_{s s}(t)=A e^{j \omega t}+\bar{A} e^{-j \omega t}\label{eqn:4.31} \]

    The Laplace transform of Equation \(\ref{eqn:4.31}\) is

    \[L\left[x_{s s}(t)\right]=\frac{A}{s-j \omega}+\frac{\bar{A}}{s+j \omega}\label{eqn:4.32} \]

    Recognizing that we seek only constant \(A\) for the steady-state sinusoidal response (not any of the \(C_k\) constants associated with transient response), we now combine Equation \(\ref{eqn:4.32}\) and Equation \(\ref{eqn:4.26}\) into Equation 4.6.4,

    \[L\left[x_{s s}(t)\right]=\frac{A}{s-j \omega}+\frac{\bar{A}}{s+j \omega}=T F(s) \frac{U}{2}\left(\frac{1}{s-j \omega}+\frac{1}{s+j \omega}\right)\label{eqn:4.33} \]

    To find complex constant \(A\) and its conjugate, we use the labor-saving method for partial-fraction expansion from Chapter 2,

    \[A=\left[(s-j \omega) T F(s) \frac{U}{2}\left(\frac{1}{s-j \omega}+\frac{1}{s+j \omega}\right)\right]_{s=j \omega}=\frac{U}{2} T F(j \omega)\label{eqn:4.34} \]

    \[\bar{A}=\left[(s+j \omega) T F(s) \frac{U}{2}\left(\frac{1}{s-j \omega}+\frac{1}{s+j \omega}\right)\right]_{s=-j \omega}=\frac{U}{2} T F(-j \omega)\label{eqn:4.35} \]

    Comparing Equation \(\ref{eqn:4.34}\) and Equation \(\ref{eqn:4.35}\) shows that

    \[T F(-j \omega)=\overline{T F(j \omega)}\label{eqn:4.36} \]

    So now we can denote the transfer function terms in general polar form:

    \[T F(j \omega)=|T F(j \omega)| e^{j \phi(\omega)} \label{eqn:4.37} \]

    where the phase is \(\phi(\omega)=\angle[T F(j \omega)]\).

    \[T F(-j \omega)=|T F(j \omega)| e^{-j \phi(\omega)}\label{eqn:4.38} \]

    Substituting Equations \(\ref{eqn:4.37}\), \(\ref{eqn:4.38}\), \(\ref{eqn:4.34}\), and \(\ref{eqn:4.35}\) back into Equation \(\ref{eqn:4.31}\) gives

    \[x_{s s}(t)=A e^{j \omega t}+\bar{A} e^{-j \omega t}=\frac{U}{2}|T F(j \omega)|\left( e^{j(\omega t+\phi(\omega))}+e^{-j(\omega t+\phi(\omega))}\right)\label{eqn:4.39} \]

    Applying again the formula from Euler’s equation that relates the cosine to complex exponentials gives the desired final result,

    \[\begin{align} x_{s s}(t) &=U|T F(j \omega)| \cos (\omega t+\phi(\omega)) \\[4pt] & \equiv X(\omega) \cos (\omega t+\phi(\omega))\label{eqn:4.40} \end{align} \]

    in which \(X(\omega)\) and \(\phi(\omega)\) are, respectively, the amplitude (magnitude) and phase of the steady-state sinusoidal response. Note that \(|T F(j \omega)|=X(\omega) / U\), the magnitude ratio.

    Therefore, we define the complex frequency-response function \(F R F(\omega)\) to be \(T F(j \omega)\), Equation \(\ref{eqn:4.37}\). Expressing \(F R F(\omega)\) in polar form gives us the FRF magnitude ratio and phase directly, and relatively easily (without all the work of finding the particular solution of the ODE by the method of undetermined coefficients, or of finding the complete time response by forward and inverse Laplace transformation, etc.):

    \[F R F(\omega) \equiv T F(j \omega)=\frac{X(\omega)}{U} e^{j \phi(\omega)}\label{eqn:4.41} \]

    This result is general for LTI SISO systems, it is valid for all of the systems considered in this book, and it is widely used in engineering practice.

    This page titled 4.7: Frequency-Response Function from Transfer Function is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by William L. Hallauer Jr. (Virginia Tech Libraries' Open Education Initiative) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.