# Fourier Series and LTI Systems

This is a continuation from the previous tutorial - ** Fourier series representation of discrete-time periodic signals**.

In the preceding few tutorials, we have seen that the Fourier series representation can be used to construct any periodic signal in discrete time and essentially all periodic continuous-time signals of practical importance.

In addition, in *the response of LTI systems to complex exponentials tutorial* we saw that the response of an LTI system to a linear combination of complex exponentials takes a particularly simple form.

Specifically, in continuous time, if \(x(t)=e^{st}\) is the input to a continuous-time LTI system, then the output is given by \(y(t)=H(s)e^{st}\), where, from eq. (3.6) [refer to *the response of LTI systems to complex exponentials tutorial*],

\[\tag{3.119}H(s)=\int_{-\infty}^{+\infty}h(\tau)e^{-s\tau}\text{d}\tau\]

in which \(h(t)\) is the impulse response of the LTI system.

Similarly, if \(x[n]=z^n\) is the input to a discrete-time LTI system, then the output is given by \(y[n]=H(z)z^n\), where, from eq. (3.10) [refer to *the response of LTI systems to complex exponentials tutorial*],

\[\tag{3.120}H(z)=\sum_{k=-\infty}^{+\infty}h[k]z^{-k}\]

in which \(h[n]\) is the impulse response of the LTI system.

When \(s\) or \(z\) are general complex numbers, \(H(s)\) and \(H(z)\) are referred to as the ** system functions** of the corresponding systems.

For continuous-time signals and systems in this tutorial and the following few tutorials, we focus on the specific case in which \(\mathcal{Re}\{s\}=0\), so that \(s=j\omega\), and consequently, \(e^{st}\) is of the form \(e^{j\omega{t}}\). This input is a complex exponential at frequency \(\omega\).

The system function of the form \(s=j\omega\)—i.e., \(H(j\omega)\) viewed as a function of \(\omega\)—is referred to as the ** frequency response** of the system and is given by

\[\tag{3.121}H(j\omega)=\int_{-\infty}^{+\infty}h(t)e^{-j\omega{t}}\text{d}t\]

Similarly, for discrete-time signals and systems, we focus in this tutorial and in the next few tutorials on values of \(z\) for which \(|z|=1\), so that \(z=e^{j\omega}\) and \(z^n\) is of the form \(e^{j\omega{n}}\). Then the system function \(H(z)\) for \(z\) restricted to the form \(z=e^{j\omega}\) is referred to as the frequency response of the system and is given by

\[\tag{3.122}H(e^{j\omega})=\sum_{n=-\infty}^{+\infty}h[n]e^{-j\omega{n}}\]

The response of an LTI system to a complex exponential signal of the form \(e^{j\omega{t}}\) (in continuous time) or \(e^{j\omega{n}}\) (in discrete time) is particularly simple to express in terms of the frequency response of the system.

Furthermore, as a result of the superposition property for LTI systems, we can express the response of an LTI system to a linear combination of complex exponentials with equal ease.

In the next few tutorials, we will see how we can use these ideas together with continuous-time and discrete-time Fourier transforms to analyze the response of LTI systems to aperiodic signals.

In this tutorial, as a first look at this important set of concepts and results, we focus on interpreting and understanding this notion in the context of periodic signals.

Consider first the continuous-time case, and let \(x(t)\) be a periodic signal with a Fourier series representation given by

\[\tag{3.123}x(t)=\sum_{k=-\infty}^{+\infty}a_ke^{jk\omega_0t}\]

Suppose that we apply this signal as the input to an LTI system with impulse response \(h(t)\). Then, since each of the complex exponentials in eq. (3.123) is an eigenfunction of the system, as in eq. (3.13) [refer to *the response of LTI systems to complex exponentials tutorial*] with \(s_k=jk\omega_0\), it follows that the output is

\[\tag{3.124}y(t)=\sum_{k=-\infty}^{+\infty}a_kH(jk\omega_0)e^{jk\omega_0t}\]

Thus, \(y(t)\) is also periodic with the same fundamental frequency as \(x(t)\). Furthermore, if \(\{a_k\}\) is the set of Fourier series coefficients for the input \(x(t)\), then \(\{a_kH(jk\omega_0)\}\) is the set of coefficients for the output \(y(t)\). That is, the effect of the LTI system is to modify individually each of the Fourier coefficients of the input through multiplication by the value of the frequency response at the corresponding frequency.

**Example 3.16**

Suppose that the periodic signal \(x(t)\) discussed in Example 3.2 [refer to the *Fourier series representation of continuous-time periodic signals tutorial*] is the input signal to an LTI system with impulse response

\[h(t)=e^{-t}u(t)\]

To calculate the Fourier series coefficients of the output \(y(t)\), we first compute the frequency response:

\[\tag{3.125}\begin{align}H(j\omega)&=\int_0^{\infty}e^{-\tau}e^{-j\omega\tau}\text{d}\tau\\&=\left.-\frac{1}{1+j\omega}e^{-\tau}e^{-j\omega\tau}\right|_0^{\infty}\\&=\frac{1}{1+j\omega}\end{align}\]

Therefore, using eqs. (3.124) and (3.125), together with the fact that \(\omega_0=2\pi\) in this example, we obtain

\[\tag{3.126}y(t)=\sum_{k=-3}^{+3}b_ke^{jk2\pi{t}}\]

with \(b_k=a_kH(jk2\pi)\), so that

\[\tag{3.127}\begin{align}b_0&=1\\b_1&=\frac{1}{4}\left(\frac{1}{1+j2\pi}\right),{\qquad}b_{-1}=\frac{1}{4}\left(\frac{1}{1-j2\pi}\right)\\b_2&=\frac{1}{2}\left(\frac{1}{1+j4\pi}\right),{\qquad}b_{-2}=\frac{1}{2}\left(\frac{1}{1-j4\pi}\right)\\b_3&=\frac{1}{3}\left(\frac{1}{1+j6\pi}\right),{\qquad}b_{-3}=\frac{1}{3}\left(\frac{1}{1-j6\pi}\right)\\\end{align}\]

Note that \(y(t)\) must be a real-valued signal, since it is the convolution of \(x(t)\) and \(h(t)\), which are both real. This can be verified by examining eq. (3.127) and observing that \(b_k^*=b_{-k}\). Therefore, \(y(t)\) can also be expressed in either of the forms given in eqs. (3.31) and (3.32) [refer to the *Fourier series representation of continuous-time periodic signals tutorial*]; that is,

\[\tag{3.128}y(t)=1+2\sum_{k=1}^3D_k\cos(2\pi{k}t+\theta_k)\]

or

\[\tag{3.129}y(t)=1+2\sum_{k=1}^3[E_k\cos2\pi{kt}-F_k\sin2\pi{kt}]\]

where

\[\tag{3.130}b_k=D_ke^{j\theta_k}=E_k+jF_k,\qquad{k=1,2,3}\]

These coefficients can be evaluated directly from eq. (3.127). For example,

\[\begin{align}D_1&=|b_1|=\frac{1}{4\sqrt{1+4\pi^2}},\qquad\qquad\theta_1=\measuredangle{b_1}=-\tan^{-1}(2\pi)\\E_1&=\mathcal{Re}\{b_1\}=\frac{1}{4(1+4\pi^2)},\qquad{F_1}=\mathcal{Im}\{b_1\}=-\frac{\pi}{2(1+4\pi^2)}\end{align}\]

In discrete time, the relationship between the Fourier series coefficients of the input and output of an LTI system exactly parallels eqs. (3.123) and (3.124). Specifically, let \(x[n]\) be a periodic signal with Fourier series representation given by

\[x[n]=\sum_{k=\langle{N}\rangle}a_ke^{jk(2\pi/N)n}\]

If we apply this signal as the input to an LTI system with impulse response \(h[n]\), then, as in eq. (3.16) [refer to *the response of LTI systems to complex exponentials tutorial*] with \(z_k=e^{jk(2\pi/N)}\), the output is

\[\tag{3.131}y[n]=\sum_{k=\langle{N}\rangle}a_kH(e^{j2\pi{k/N}})e^{jk(2\pi/N)n}\]

Thus, \(y[n]\) is also periodic with the same period as \(x[n]\), and the \(k\)th Fourier coefficient of \(y[n]\) is the product of the \(k\)th Fourier coefficient of the input and the value of the frequency response of the LTI system, \(H(e^{j2\pi{k}/N})\), at the corresponding frequency.

**Example 3.17**

Consider an LTI system with impulse response \(h[n]=\alpha^nu[n]\), \(-1\lt\alpha\lt1\), and with the input

\[\tag{3.132}x[n]=\cos\left(\frac{2\pi{n}}{N}\right)\]

As in Example 3.10 [refer to the *Fourier series representation of discrete-time periodic signals tutorial*], \(x[n]\) can be written in Fourier series form as

\[x[n]=\frac{1}{2}e^{j(2\pi/N)n}+\frac{1}{2}e^{-j(2\pi/N)n}\]

Also, from eq. (3.122),

\[\tag{3.133}H(e^{j\omega})=\sum_{n=0}^{\infty}\alpha^ne^{-j\omega{n}}=\sum_{n=0}^{\infty}\left(\alpha{e}^{-j\omega}\right)^n\]

This geometric series can be evaluated, yielding

\[\tag{3.134}H(e^{j\omega})=\frac{1}{1-\alpha{e}^{-j\omega}}\]

Using eq. (3.131), we then obtain the Fourier series for the output:

\[\tag{3.135}\begin{align}y[n]&=\frac{1}{2}H\left(e^{j2\pi/N}\right)e^{j(2\pi/N)n}+\frac{1}{2}H\left(e^{-j2\pi/N}\right)e^{-j(2\pi/N)n}\\&=\frac{1}{2}\left(\frac{1}{1-\alpha{e}^{-j2\pi/N}}\right)e^{j(2\pi/N)n}+\frac{1}{2}\left(\frac{1}{1-\alpha{e}^{j2\pi/N}}\right)e^{-j(2\pi/N)n}\end{align}\]

If we write

\[\frac{1}{1-\alpha{e}^{-j2\pi/N}}=re^{j\theta}\]

then eq. (3.135) reduces to

\[\tag{3.136}y[n]=r\cos\left(\frac{2\pi}{N}n+\theta\right)\]

For example, if \(N=4\),

\[\frac{1}{1-\alpha{e}^{-j2\pi/4}}=\frac{1}{1+\alpha{j}}=\frac{1}{\sqrt{1+\alpha^2}}e^{j(-\tan^{-1}(\alpha))}\]

and thus,

\[y[n]=\frac{1}{\sqrt{1+\alpha^2}}\cos\left(\frac{\pi{n}}{2}-\tan^{-1}(\alpha)\right)\]

We note that for expressions such as eqs. (3.124) and (3.131) to make sense, the frequency responses \(H(j\omega)\) and \(H(e^{j\omega})\) in eqs. (3.121) and (3.122) must be well defined and finite. As we will see in later tutorials, this will be the case if the LTI systems under consideration are stable.

For example, the LTI system in Example 3.16, with impulse response \(h(t)=e^{-t}u(t)\), is stable and has a well-defined frequency response given by eq. (3.125). On the other hand, an LTI system with impulse response \(h(t)=e^tu(t)\) is unstable, and it is easy to check that the integral in eq. (3.121) for \(H(j\omega)\) diverges for any value of \(\omega\).

Similarly, the LTI system in Example 3.17, with impulse response \(h[n]=\alpha^nu[n]\), is stable for \(|\alpha|\lt1\) and has frequency response given by eq. (3.134). However, if \(|\alpha|\gt1\), the system is unstable, and then the summation in eq. (3.133) diverges.

The next tutorial discusses about ** filtering**.