# The Response of LTI Systems to Complex Exponentials

This is a continuation from the previous tutorial - ** singularity functions**.

As we indicated in previous tutorials, it is advantageous in the study of LTI systems to represent signals as linear combinations of basic signals that possess the following two properties:

- The set of basic signals can be used to construct a broad and useful class of signals.
- The response of an LTI system to each signal should be simple enough in structure to provide us with a convenient representation for the response of the system to any signal constructed as a linear combination of the basic signals.

Much of the importance of Fourier analysis results from the fact that both of these properties are provided by the set of complex exponential signals in continuous and discrete time—i.e., signals of the form \(e^{st}\) in continuous time and \(z^n\) in discrete time, where \(s\) and \(z\) are complex numbers.

In subsequent tutorials, we will examine the first property in some detail. In this tutorial, we focus on the second property and, in this way, provide motivation for the use of Fourier series and transforms in the analysis of LTI systems.

The importance of complex exponentials in the study of LTI systems stems from the fact that the response of an LTI system to a complex exponential input is the same complex exponential with only a change in amplitude; that is,

\[\tag{3.1}\text{continuous time:}\qquad{e^{st}}\rightarrow{H(s)}e^{st}\]

\[\tag{3.2}\text{discrete time:}\qquad{z^n}\rightarrow{H(z)}z^n\]

where the complex amplitude factor \(H(s)\) or \(H(z)\) will in general be a function of the complex variable \(s\) or \(z\).

A signal for which the system output is a (possibly complex) constant times the input is referred to as an ** eigenfunction** of the system, and the amplitude factor is referred to as the system's

**.**

*eigenvalue*To show that complex exponentials are indeed eigenfunctions of LTI systems, let us consider a continuous-time LTI system with impulse response \(h(t)\). For an input \(x(t)\), we can determine the output through the use of the convolution integral, so that with \(x(t)=e^{st}\)

\[\tag{3.3}\begin{align}y(t)&=\int\limits_{-\infty}^{+\infty}h(\tau)x(t-\tau)\text{d}\tau\\&=\int\limits_{-\infty}^{+\infty}h(\tau)e^{s(t-\tau)}\text{d}\tau\end{align}\]

Expressing \(e^{s(t-\tau)}\) as \(e^{st}e^{-s\tau}\), and noting that \(e^{st}\) can be moved outside the integral, we see that eq. (3.3) becomes

\[\tag{3.4}y(t)=e^{st}\int\limits_{-\infty}^{+\infty}h(\tau)e^{-s\tau}\text{d}\tau\]

Assuming that the integral on the right-hand side of eq. (3.4) converges, the response to \(e^{st}\) is of the form

\[\tag{3.5}y(t)=H(s)e^{st}\]

where \(H(s)\) is a complex constant whose value depends on \(s\) and which is related to the system impulse response by

\[\tag{3.6}H(s)=\int\limits_{-\infty}^{+\infty}h(\tau)e^{-s\tau}\text{d}\tau\]

Hence, we have shown that complex exponentials are eigenfunctions of LTI systems. The constant \(H(s)\) for a specific value of \(s\) is then the eigenvalue associated with the eigenfunction \(e^{st}\).

In an exactly parallel manner, we can show that complex exponential sequences are eigenfunctions of discrete-time LTI systems. That is, suppose that an LTI system with impulse response \(h[n]\) has as its input the sequence

\[\tag{3.7}x[n]=z^n\]

where \(z\) is a complex number. Then the output of the system can be determined from the convolution sum as

\[\tag{3.8}\begin{align}y[n]&=\sum_{k=-\infty}^{+\infty}h[k]x[n-k]\\&=\sum_{k=-\infty}^{+\infty}h[k]z^{n-k}=z^n\sum_{k=-\infty}^{+\infty}h[k]z^{-k}\end{align}\]

From this expression, we see that if the input \(x[n]\) is the complex exponential given by eq. (3.7), then, assuming that the summation on the right-hand side of eq. (3.8) converges, the output is the same complex exponential multiplied by a constant that depends on the value of \(z\). That is

\[\tag{3.9}y[n]=H(z)z^n\]

where

\[\tag{3.10}H(z)=\sum_{k=-\infty}^{+\infty}h[k]z^{-k}\]

Consequently, as in the continuous-time case, complex exponentials are eigenfunctions of discrete-time LTI systems. The constant \(H(z)\) for a specified value of \(z\) is the eigenvalue associated with the eigenfunction \(z^n\).

For the analysis of LTI systems, the usefulness of decomposing more general signals in terms of eigenfunctions can be seen from an example. Let \(x(t)\) correspond to a linear combination of three complex exponentials; that is,

\[\tag{3.11}x(t)=a_1e^{s_1t}+a_2e^{s_2t}+a_3e^{s_3t}\]

From the eigenfunction property, the response to each separately is

\[a_1e^{s_1t}\rightarrow{a_1}H(s_1)e^{s_1t}\]

\[a_2e^{s_2t}\rightarrow{a_2}H(s_2)e^{s_2t}\]

\[a_3e^{s_3t}\rightarrow{a_3}H(s_3)e^{s_3t}\]

and from the superposition property the response to the sum is the sum of the responses, so that

\[\tag{3.12}y(t)=a_1H(s_1)e^{s_1t}+a_2H(s_2)e^{s_2t}+a_3H(s_3)e^{s_3t}\]

More generally, in continuous time, eq. (3.5), together with the superposition property, implies that the representation of signals as a linear combination of complex exponentials leads to a convenient expression for the response of an LTI system.

Specifically, if the input to a continuous-time LTI system is represented as a linear combination of complex exponentials, that is, if

\[\tag{3.13}x(t)=\sum_{k}a_ke^{s_kt}\]

then the output will be

\[\tag{3.14}y(t)=\sum_{k}a_kH(s_k)e^{s_kt}\]

In an exactly analogous manner, if the input to a discrete-time LTI system is represented as a linear combination of complex exponentials, that is, if

\[\tag{3.15}x[n]=\sum_{k}a_kz_k^n\]

then the output will be

\[\tag{3.16}y[n]=\sum_{k}a_kH(z_k)z_k^n\]

In other words, for both continuous time and discrete time, if the input to an LTI system is represented as a linear combination of complex exponentials, then the output can also be represented as a linear combination of the same complex exponential signals.

Each coefficient in this representation of the output is obtained as the product of the corresponding coefficient \(a_k\) of the input and the system's eigenvalue \(H(s_k)\) or \(H(z_k)\) associated with the eigenfunction \(e^{s_kt}\) or \(z_k^n\), respectively.

It was precisely this fact that Euler discovered for the problem of the vibrating string, that Gauss and others used in the analysis of time series, and that motivated Fourier and others after him to consider the question of how broad a class of signals could be represented as a linear combination of complex exponentials.

In the next few tutorials we examine this question for periodic signals, first in continuous time and then in discrete time, and in later tutorials we consider the extension of these representations to aperiodic signals.

Although in general, the variables \(s\) and \(z\) in eqs. (3.1)—(3.16) may be arbitrary complex numbers, Fourier analysis involves restricting our attention to particular forms for these variables.

In particular, in continuous time we focus on purely imaginary values of \(s\)—i.e., \(s=j\omega\)—and thus, we consider only complex exponentials of the form \(e^{j\omega{t}}\).

Similarly, in discrete time we restrict the range of values of \(z\) to those of unit magnitude—i.e., \(z=e^{j\omega}\)—so that we focus on complex exponentials of the form \(e^{j\omega{n}}\).

**Example 3.1**

As an illustration of eqs. (3.5) and (3.6), consider an LTI system for which the input \(x(t)\) and output \(y(t)\) are related by a time shift of 3, i.e.,

\[\tag{3.17}y(t)=x(t-3)\]

If the input to this system is the complex exponential signal \(x(t)=e^{j2t}\), then, from eq. (3.17),

\[\tag{3.18}y(t)=e^{j2(t-3)}=e^{-j6}e^{j2t}\]

Equation (3.18) is in the form of eq. (3.5), as we would expect, since \(e^{j2t}\) is an eigenfunction. The associated eigenvalue is \(H(j2)=e^{-j6}\).

It is straightforward to confirm eq. (3.6) for this example. Specifically, from eq. (3.17), the impulse response of the system is \(h(t)=\delta(t-3)\). Substituting into eq. (3.6), we obtain

\[H(s)=\int\limits_{-\infty}^{+\infty}\delta(\tau-3)e^{-s\tau}\text{d}\tau=e^{-3s}\]

so that \(H(j2)=e^{-j6}\).

As a second example, in this case illustrating eqs. (3.11) and (3.12), consider the input signal \(x(t)=\cos(4t)+\cos(7t)\). From eq. (3.17), \(y(t)\) will of course be

\[\tag{3.19}y(t)=\cos(4(t-3))+\cos(7(t-3))\]

To see that this will also result from eq. (3.12), we first expand \(x(t)\) using Euler's relation:

\[\tag{3.20}x(t)=\frac{1}{2}e^{j4t}+\frac{1}{2}e^{-j4t}+\frac{1}{2}e^{j7t}+\frac{1}{2}e^{-j7t}\]

From eqs. (3.11) and (3.12),

\[y(t)=\frac{1}{2}e^{-j12}e^{j4t}+\frac{1}{2}e^{j12}e^{-j4t}+\frac{1}{2}e^{-j21}e^{j7t}+\frac{1}{2}e^{j21}e^{-j7t}\]

or

\[\begin{align}y(t)&=\frac{1}{2}e^{j4(t-3)}+\frac{1}{2}e^{-j4(t-3)}+\frac{1}{2}e^{j7(t-3)}+\frac{1}{2}e^{-j7(t-3)}\\&=\cos(4(t-3))+\cos(7(t-3))\end{align}\]

For this simple example, multiplication of each periodic exponential component of \(x(t)\)—for example, \(\frac{1}{2}e^{j4t}\)—by the corresponding eigenvalue—e.g., \(H(j4)=e^{-j12}\)—effectively causes the input component to shift in time by 3.

Obviously, in this case we can determine \(y(t)\) in eq. (3.19) by inspection rather than by employing eqs. (3.11) and (3.12). However, as we will see, the general property embodied in eqs. (3.11) and (3.12) not only allows us to calculate the responses of more complex LTI systems, but also provides the basis for the frequency domain representation and analysis of LTI systems.

The next tutorial discusses about ** Fourier series representation of continuous-time periodic signals**.