LTI Systems and Periodic Excitation

define $H(f)=\int_{-\infty}^{\infty}h(t)e^{-j 2\pi ft}dt$

$$y(t)=T[x(t)]=T[\sum_k x_k e^{j \omega_0 kt}] = \sum_k x_k T[e^{j\omega_0 kt}]$$

$$T[e^{j\omega_0 kt}]=\int_{-\infty}^{\infty} h(\tau)e^{j\omega_0 k(t-\tau)} d\tau =e^{j\omega_0 kt} H(\frac{k}{T})$$

Therefore,

$$y(t)=\sum_k x_k H(\frac{k}{T}) e^{j\omega_0kt}$$

Which is why the frequency response of a system H(f) is so useful for sinusoidal signal inputs

• Parseval

  
  

  $$\int_0^T \vert x(t)\vert^2dt=\sum_k\vert x_k\vert^2$$
  
  

  i.e. energy in a period $\Rightarrow$ add up energy in spectral components

• Fourier Xform

  Just spread period out:
  

  $$X(f)=\int_{-\infty}^{\infty}x(t) e^{-j2\pi ft}dt$$
  
  

  
  

  $$x(t)=\int_{-\infty}^{\infty}X(f) e^{j2\pi ft}df$$
  
  

  Look! no $1/2\pi$ and integrals have f not $\omega=2\pi f$. This is the FT pair - f(t) must satisfy Dirichlet

• Useful stuff

  1.

  $F[\delta(t)]=1$
  therefore

  
  

  $$\delta(t)=\int_{-\infty}^{\infty}e^{j2\pi ft}df$$

  
  

  2.

  x(t) real $\Rightarrow X(f)=X(^*(-f))$     (Hermitian)

  3.

  x(t) odd     X(f) imag

  4.

  x(t) even     X(f) real

  5.

  Linearity

  
  

  F[a x(t) + by(t)]=a X(f)+bX(f)

  
  

  6.

  Duality
  If X(f)=F[x(t)], then
  F[X(-t)]=x(f)
  and F[X(t)]=x(-f)
  notice the absence of $2\pi$ factors! Isn't it lovely!?!?

  7.

  Time shift

  
  

  $$F[x(t-t_0)]=e^{-j2\pi f t_0}X(f)$$

  
  

  8.

  Scaling

  
  

  $$F[x(at)]=\frac{1}{\vert a\vert}X(\frac{f}{a}) \mbox{\qquad} a\ne0 \quad\mbox{and real}$$

  
  

  9.

  Convolution

  
  

  $$F[x(t)\ast y(t)]=X(f)Y(f)$$

  
  

  10.

  Modulation

  
  

  $$F[x(t)e^{j 2\pi f_0 t}]=X(f-f_0)$$

  
  

  11.

  Parseval

  
  

  $$\int_{-\infty}^{\infty}x(t)y^*(t)dt=\int_{-\infty}^{\infty}X(f)Y^*(f)df$$

  
  

  which is the general form. This implies
  

  $$\int_{-\infty}^{\infty}\vert x(t)\vert^2 dt=\int_{-\infty}^{\infty}\vert X(f)\vert^2 df$$

  
  

  which is the more usual (and useful) form.

  12.

  Autocorrelation

  
  

  $$R_x(\tau)=\int_{-\infty}^{\infty}x(t)x^*(t-\tau)dt$$

  
  

  (There are others comparable definition) then

  
  

  $$\underbrace{F}_{w.r.t.\tau}[R_x(\tau]=\vert X(f)\vert^2$$

  
  

  13.

  Differentiation

  
  

  $$x(t)\rightarrow X(f)$$

  
  

  
  

  $$\frac{dx}{dt}\rightarrow(j2\pi f)X(f)$$

  
  

  
  

  $$\frac{d^nx}{dt^n}\rightarrow (j2\pi f)^nX(f)$$

  
  

  Likewise,

  
  

  $$\frac{d^nX(f)}{df^n}\Rightarrow (-j2\pi t)^nx(t)$$

  
  

  14.

  Integration

  
  

  $$\int_{-\infty}^{\tau} x(\tau)d\tau= \frac{1}{j2\pi f} X(f)+\underbrace{\frac{1}{2} X(0)\delta(f)}$$

  
  

  Note that if x(t) has no DC component then this disappears

  15.

  Moments

  (will be useful for probability functions)
  

  $$\int_{-\infty}^{\infty}t^n x(t)dt=\left(\frac{j}{2\pi}\right)^n \left.\frac{d^n X(f)}{df^n}\right\vert _{f=0}$$

  
  

  from differentiation theorem

• Useful Transforms

  (might want to memorize them)
  

  $$\left[ \begin{array}{l @{\rightarrow} r} \delta(t) & 1\\ 1 & \delta(f) \end{array} \right.$$
  
  

  
  

  $$\left[ \begin{array}{l @{\rightarrow} r} \delta(t-t_0) & e^{-j2\pi ft_0}\\ e^{j2\pi ft} & \delta(f-f_0) \end{array} \right.$$
  
  

  
  

  $$\cos(2\pi f_0 t) \rightarrow \frac{\delta(f-f_0)+\delta(f+f_0)}{2}$$
  
  

  
  

  $$\sin(2\pi f_0 t) \rightarrow \left(\frac{-\delta(f+f_0)+\delta(f-f_0)}{2j}\right)$$
  
  

  
  

  $$rect(t) \rightarrow sinc(f)=\frac{\sin \pi f}{\pi f}$$
  
  

  
  

  $$sinc(t) \rightarrow rect(f)$$
  
  

  
  

  $$e^{-\alpha t} u(t) \rightarrow \frac{1}{\alpha+j2\pi f} \;\;\;\; \alpha>0$$
  
  

  
  

  $$sgn(t) \rightarrow \frac{1}{j\pi f}$$
  
  

  
  

  $$u(t) \rightarrow \frac{1}{j2\pi f}+\frac{1}{2}\delta(f)$$
  
  

  (from integration theorem)

• Periodic signals

  Qualitative $\Rightarrow$ energy concentrated at harmonics
  So x'(t)=x(t+T)

  
  

  $$X(f)=X \left(\frac{k}{T} \right)= \sum_{k=-\infty}^{\infty} x_k \delta(f-\frac{k}{T})$$
  
  

  
  

  $$x_k=\frac{1}{T}\int_{-\infty}^{\infty}x(t) e^{-j\frac{2\pi}{T}kt} dt$$
  
  

  i.e. Fourier series coeff.

• FT and LTI systems

  
  

  H(f)= F[h(t)]
  
  

  
  

  $$Y(f)=F[\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau]=H(f)Y(f)$$
  
  

  as we used before

  If $x(t)=e^{j2\pi f_0 t}$

  
  

  $$Y(f)=\delta(f-f_0)H(f)\Rightarrow y(t)=\underbrace{H(f_0)}_{\mbox{const}} \underbrace{e^{j2\pi f_0 t}}_{\mbox{x(t)}}$$
  
  

  $e^{j2\pi f_0 t}$ is an EIGENFUNCTION of LTI systems
  $\rightarrow$ that's why we decompose signals onto $e^{j\omega t}$

• Signal energy

  $voltage^2 \Rightarrow \propto \mbox{units of watts=joules/sec}$

  
  

  $$E_x=\int_{-\infty}^{\infty}\vert x(t)\vert^2dt=\mbox{joules}$$
  
  

  
  

  $$P_x=\lim_{T\rightarrow\infty} \frac{1}{T}\int_{-T/2}^{T/2}\vert x(t)\vert^2dt=\mbox{watts}$$
  
  

  energy signal $E_x<\infty$
  power signal $0<P_x<\infty$
  or OTHER signals (don't really care for this designation, but it's analytically useful sometimes)

• E-type
  
  $$\begin{eqnarray*}R_x(\tau) &\equiv& x(\tau)\ast x^*(-\tau)\\ &=&\int_{-\infty}^... ...au)x^*(t)dt\\ & \stackrel{F[]}{\Rightarrow}& \vert X(f)\vert^2 \end{eqnarray*}$$
  
  Clearly,

  
  

  $$E_x=R_x(0)\mbox{\qquad and can use Parseval as well}$$
  
  

  
  

  $$y(t)=x(t)\ast h(t)\Rightarrow R_x(\tau)\ast R_h(\tau)=R_y(\tau)$$
  
  

  since

  
  

  $$R_x(\tau)\rightarrow \vert X(f)\vert^2\Rightarrow \mbox{notation} G_x(f)=\mbox{ESD of x(t)}$$
  
  

  
  

  $$\vert X(f)\vert^2\equiv\mbox{energy spectral density}$$
  
  
  --------------------
  
  
  G_y(f)=|G_x(f)||H(f)|²
  
  
  --------------------

  rough units: F[joules(t)] $\Rightarrow$ joules.sec=joules/(1/sec)=joules/Hz

• Power signals

  
  

  $$R_x(\tau)\equiv \lim_{T\rightarrow \infty}\frac{1}{T}\int_{-T/2}^{T/2} x(t)x^*(t-\tau)dt$$
  
  

  
  

  $$P_x=R_x(0)\mbox{\qquad (parseval too)}$$
  
  

  define
  

  $$\begin{eqnarray*}S_X(f)&\equiv& PSD (\mbox{\qquad power spectral density})\\ &=&F[R_x(\tau)] \end{eqnarray*}$$
  
  

  as with energy signals

  
  

  $$y(t)=x(t)\ast h(t)\Rightarrow R_y(\tau)=R_x(\tau)\ast \underbrace{h(\tau)\ast h^*(-\tau)}_{\mbox{ESD of system}}$$
  
  

  so

  
  

  S_y(f)=S_x(f)|H(f)|²
  
  

• aside

  Periodic signals are power signals x(t) periodic, $y(t)=x(t)\ast h(t)$

  
  

  $$P_y=\sum_n\vert x_n\vert^2\vert H\left(\frac{n}{T}\right)\vert^2 \mbox{\qquad obtained via Parseval}$$