2.2d LTI Frequency Response

LTI Frequency Response

The $2\pi$-periodic continuous frequency response of a discrete-time LTI system can be expressed by convolving its filter $h(n)$ with an eigenfunction input $x(n)=e^{j\omega n}$, such that

y(n) &= e^{j\omega n}\sum^{\infty}_{k=-\infty} h(k)e^{-j\omega k} \\

&=H(\omega)e^{j \omega n},
\end{align}$$

where $H(\omega )={\sum }_{k=-\infty }^{\infty }h(k)e^{-j\omega k}$ is the frequency response of the system, while $\omega$ ($\text{rad}/\text{sample}$) is the fundamental period of the eigenfunction.

Similarly, in continuous-time, the non-periodic continuous frequency response of an LTI system can be expressed by convolving its filter $h(t)$ with an eigenfunction input $x(t)=e^{j\omega t}$, such that

y(t) &= e^{j\omega t}\int^{\infty}_{-\infty}h(\tau)e^{-j\omega \tau}d\tau \\ &= H(\omega)e^{j \omega t},

\end{align}$$

where $H(\omega )={\int }_{-\infty }^{\infty }h(\tau )e^{-j\omega \tau }d\tau$ is the frequency response of the system.

Note (New Frequencies) $y$ of a system contains frequencies that are not in its input $x$, then that system cannot be an LTI system; no new frequencies can be created in the output of an LTI system.

If the output

Intuition

9/30 lecture around 40 min to 1:10

Properties

(Standard Conversion) $h(t)=\alpha e^{-\alpha t}⟺H(\omega )=\frac{\beta }{\alpha +j\omega }$.

(D-T Eigenfunction Periodicity) $e^{j\omega n}$ is $2\pi$-periodic in $\omega$ and is only periodic in $n$ if $\omega$ is a rational multiple of $\pi$.

We need to decompose this problem into two sub-problems:

1. We begin by equating $e^{j(\omega +W)n}=e^{j\omega n}$, which simplifies into $e^{jWn}=1$ and into $Wn=2\pi$. This means that $n=\frac{2\pi }{W}=\omega$.

(D-T $2\pi$-periodicity) $H(\omega +2\pi )=H(\omega )$

We can prove by taking the following:
$H(\omega +2\pi )={\sum }_{k=-\infty }^{\infty }h(k)e^{-j\omega k}{e}^{-j2\pi k}.$
Since for any integer $k$, $e^{j2k\pi }=e^{j2\pi }=1$ due to $2\pi$-periodicity of complex exponentials,

H(\omega + 2\pi) &= \sum^{\infty}_{k=-\infty}h(k)e^{-j\omega k}e^{-j2\pi k} \\ &= \sum^{\infty}_{k=-\infty}h(k)e^{-j\omega k} \times 1 \\ &= H(\omega). \end{align}$$ #### (Real Conjugate) If $h(n)=h\textasteriskcentered(n)$ or $h(t) = h\textasteriskcentered(t)$, then $H\textasteriskcentered(\omega) = H(-\omega)$. Saying that a function is equal to its complex conjugate simply means that it is real. We proceed with a simple proof, where $$\begin{align} H\textasteriskcentered(\omega) &= \sum^{\infty}_{k=-\infty}h\textasteriskcentered(k)e^{-j\omega k} \\ &= \sum^{\infty}_{k=-\infty}h(k)e^{j(-\omega) k} \\ &= H(-\omega). \end{align}$$ #### (Even Magnitude) If $H\textasteriskcentered(\omega) = H(-\omega)$, then $|H(\omega)| = |H(-\omega)|$. The LHS implies that $H$ is real. #### (Odd Phase) If $H\textasteriskcentered(\omega) = H(-\omega)$, then $\angle H(\omega) = -\angle H(-\omega)$. The LHS implies that $H$ is real. # Examples >[!example] Does $1 + e^{j\frac{\pi}{4} n} \xrightarrow{H} 2$ describe an LTI system? > >We can quickly express the input as >$$x(n) = e^{j0n} + e^{j\frac{\pi}{4}n},$$ >and the output as >$$y(n) = 2e^{j0n}.$$ >We know that for $x(n)=e^{j \omega n}$ the output will be $y(n) = H(\omega)x(n) = H(\omega)e^{j\omega n}$. Therefore, >$$y(n) = H(0)e^{j0n} + H(\frac{\pi}{4})e^{j\frac{\pi}{4}n}.$$ >Since $H(\omega)$ only depends on $\omega$ (so we can't set $H(\frac{\pi}{4})=e^{-j \frac{\pi}{4} n}$ as it depends on $n$), there is only one valid solution at >$$H(0) = 2, \enspace H(\frac{\pi}{4})=0.$$ >While this result is consistent with the eigenfunction behavior of a D-T LTI system, it does not mean that $H$ is a discrete-time LTI system. Therefore, we cannot determine the LTI properties of $H$ solely based on the eigenfunction property. >[!example] Does $1 + e^{j\frac{\pi}{4} n} \xrightarrow{H} (-1)^n$ describe an LTI system? > >Once again, we express the input as >$$x(n) = e^{j0n} + e^{j\frac{\pi}{4}n},$$ >and this time we express the output as >$$y(n) = e^{j \pi n}.$$ >We can see that >$$y(n) = e^{j \pi n} = H(0)e^{j0n} + H(\frac{\pi}{4})e^{j\frac{\pi}{4}n}.$$ >This is not an LTI system, as the output has different resonant frequency $w_0 = \pi$ than the input. > [!example] Find the frequency response of a $2$-point moving average. > > We know that the $2$-point moving average can be expressed in terms of Kronecker deltas, such that > $$h(n) = \frac{1}{2}\delta(n) + \frac{1}{2}\delta(n-1).$$ > Therefore, > $$y(n) = e^{j \omega n}\frac{1}{2}\sum^{\infty}_{k=-\infty}[\delta(k) + \delta(k-1)]e^{-j\omega k}.$$ > By inspection, we see that > $$\begin{align} > H(\omega) &= \frac{1}{2}\sum^{\infty}_{k=-\infty}[\delta(k) + \delta(k-1)]e^{-j\omega k} \\ > &= \frac{1}{2} + \frac{1}{2}e^{-j\omega} \\ > &= e^{-j\frac{\omega}{2}}\left(\frac{e^{j\frac{\omega}{2}} + e^{-j\frac{\omega}{2}}}{2} \right) \\ > &= \cos(\frac{\omega}{2})e^{-j\frac{\omega}{2}}. >\end{align}$$ >Then, the magnitude of $H(\omega)$ can be expressed as >$$\begin{align} >|H(\omega)| &= |\cos(\frac{\omega}{2})| \cdot |e^{-j\frac{\omega}{2}}| \\ >&= |\cos(\frac{\omega}{2})| \cdot 1 \\ >&= |\cos(\frac{\omega}{2})|, >\end{align}$$ >where $|e^{-j\frac{\omega}{2}}| = 1$ by the laws of the unit circle. > >Note that we can express $H(\omega)$ as a [[2.2j.2 Phasor|phasor]], such that $H(\omega) = |\cos(\frac{\omega}{2})| \angle -\frac{\omega}{2}$. We can also plot the frequency response on a bode plot, such that it consists of its magnitude and phase plots, >[!example] Find the output of $x(n)=\cos(\omega_0 n)$. > >We can start by expressing $x(n)$ as a sum of complex exponentials, such that >$$x(n)=\cos(\omega_0 n) = \frac{1}{2}e^{j\omega_0 n} + \frac{1}{2}e^{-j\omega_0 n}.$$ >With this in mind, we can use the frequency response formula to formulate $y(n)$, such that >$$\begin{align} >y(n) = \frac{1}{2}H(\omega_0)e^{j\omega_0 n} + \frac{1}{2}H(-\omega_0)e^{-j\omega_0 n} >\end{align}$$ >The second term on the RHS is $H(-\omega)$, since $h(n)$ is real and therefore by the real conjugate property, $h(n) = h\textasteriskcentered(n)$. > >We can do some analysis here. If $h(n)$ was complex valued, then >$$\begin{align} >y(n) &= \frac{1}{2}H(\omega_0)e^{j\omega_0 n} + \frac{1}{2}H\textasteriskcentered(\omega_0)e^{-j\omega_0 n} \\ >&= \frac{1}{2}x + \frac{1}{2}x\textasteriskcentered \\ >&= \frac{1}{2}2\textrm{Re}\{x\} \\ >&= \textrm{Re}\{ x\} \\ >\end{align}$$ >where $x \in \mathbb{C}$. Note here that $x = H(\omega_0)e^{j\omega_0 n}$ is different from $x(n)=\cos(\omega_0 n)$. > >Here, we remember that the frequency response can be represented as a phasor, such that $H(\omega_0) = |H(\omega_0)|e^{j\angle H(\omega_0)}$. Then, >$$\begin{align} >y(n) &= \textrm{Re}\{|H(\omega_0)|e^{j\angle H(\omega_0)}e^{j\omega_0n}\} \\ >&= \textrm{Re}\{|H(\omega_0)|e^{j(\omega_0n + \angle H(\omega_0)}\} \\ >&= |H(\omega_0)|cos(\omega_0n + \angle H(\omega_0)). >\end{align}$$ >**FOR EXAMPLE**, if we were to use the FIR two-point moving average filter from the example above, our output would simplify to >$$y(n)=|cos\frac{\omega_0}{2}|cos(\omega_0n - \frac{\omega}{2}).$$ >Given an input $x(n)=\cos\left(\frac{\pi}{2}n\right)$, where $\omega_0 = \frac{\pi}{2}$, our output will be equal to >$$\begin{align} >y(n) &= |\cos\frac{\frac{\pi}{2}}{2}|\cos(\frac{\pi}{2}n - \frac{\frac{\pi}{2}}{2}) \\ >&= \frac{\sqrt{2}}{2}\cos(\frac{\pi}{2}n - \frac{\pi}{4}). >\end{align}$$ >If, for example, we add a phase to our input, such that $x(n) = \cos(\omega n + \phi)$, then >$$y(n) = |H(\omega_0)|$$