HW 07

Question 1

Show that a periodic signal, $x(t)=x(t+T)\,\!$, always has a Fourier transform which is a series of impulse functions spaced out every 1/T Hz. What do the areas (both magnitude and phase) under each impulse function represent?

Answer 1

If a function is periodic, $x(t)=x(t+T)\,\!$, and it meets the Dirichlet conditions, then we can write it as $x(t)=\sum _{n=-\infty }^{\infty }\alpha _{n}e^{j\,2\,\pi \,n\,t/T}$

The inner product of two vectors requires a complex conjugate. ${\vec {a}}\cdot {\vec {b}}=\sum _{i=1}^{n}a_{i}\cdot b_{i}^{*}$

$x(t)\cdot e^{j\,2\,\pi \,f\,t/T}$	$=\int _{-\infty }^{\infty }x(t)e^{-j\,2\,\pi \,f\,t}\,dt$
	$=\int _{-\infty }^{\infty }\left(\sum _{n=-\infty }^{\infty }\alpha _{n}e^{j\,2\,\pi \,n\,t/T}\right)e^{-j\,2\pi \,f\,t}\,dt$
	$=\sum _{n=-\infty }^{\infty }\alpha _{n}\int _{-\infty }^{\infty }e^{j\,2\,\pi \,f\,t}e^{-j\,2\,\pi \,n\,t/T}\,dt$
	$=\sum _{n=-\infty }^{\infty }\alpha _{n}\int _{-\infty }^{\infty }e^{j\,2\pi \,\left({\frac {n}{T}}-f\right)\,t}\,dt$
	$=\sum _{n=-\infty }^{\infty }\alpha _{n}\delta \left({\frac {n}{T}}-f\right)$

• N is the independant variable which makes the impulse functions spaced out every 1/T units. Since the delta function is an even function, you can also rewrite it as $\delta \left(f-{\frac {n}{T}}\right)$

• Remember from Relation to the Fourier Series that $2\,\left|\alpha _{n}\right|$ is the magnitude of the cosine/sine wave and $\theta _{n}\,\!$ is the phase shift.

Question 2

A few summers ago I found myself on Guam trying to figure out which bends in the transmission line leading to the antennas were causing the msot reflections. Ideally, I needed a time domain reflectometer (TDR), which excites the tranmission line with something close to an impulse funciton and measures the signals reflected back as a function of time using a high speed oscilloscope. For example, if there is a big bend at 150 meters from the end of the line the instrument is connected to, there will be a blip on the oscilloscope 1.0 microseconds after the incident pulse was applied. (It takes 0.5 microseconds for the pulse to travel down to the bend, where part of it is reflected back toward the soruce, and 0.5 microseconds for the reflection to travel back).

Unfortunately I didn't have a TDR available. However, there was a very fancy network analyzer that would apply a sinusoidal wave, $\sin(2\pi \,f\,t)$, to the transmission line and measure the steady-state voltage reflected back (both magnitude and phase). Furthermore, the network analyzer would make this measurement as a function of frequency for practically any frequency range you could dream up. If this reflection as a function of frequncy was measured as $R(f)\left(\sin(2\pi \,f\,t+\psi (f)\right)$ by the network analyzer, write an expression for the function that would be seen on the TDR for the same transmission line. Hint: This is kind of like our "favorite game"

Answer 2

• Note that you can't simply multiply the input by a variable. You can mulitply it by a constant however.
• The idea is to build back to an impulse function.

$\sin(2\,\pi \,f\,t)$	$\longrightarrow$	$R(f)\,\sin(2\,\pi \,f\,t+\psi (f))$	Given
$\cos(2\,\pi \,f\,t)$	$\longrightarrow$	$R(f)\,\sin(2\,\pi \,f\,t+\psi (f))$	Time Shift
$\cos(2\,\pi \,f\,t)+j\,\sin(2\,\pi \,f\,t)$	$\longrightarrow$	$R(f)\left[\cos(2\,\pi \,f\,t+\psi (f))+j\,\sin(2\,\pi \,f\,t+\psi (f))\right]$	Linearity
$e^{j\,2\,\pi \,f\,t}$		$R(f)e^{j\,2\,\pi \,f\,t+\psi (f)}$
$\int _{-\infty }^{\infty }1\cdot e^{j\,2\,\pi \,f\,t}\,df$	$\longrightarrow$	$\int _{-\infty }^{\infty }R(f)e^{j\left(\,2\,\pi \,f\,t+\psi (f)\right)}\,df$	Superposition
$F^{-1}\left[1\right]$		$F^{-1}\left[R(f)\,e^{\psi (f)}\right]$
$\delta (t)\,\!$

Question 3

By investigating its impulse response, argue that an ideal "brick wall" low pass filter having $H(f)=[u(f+f_{0})-u(f-f_{0})]\,e^{-j\,2\,\pi \,f\,t_{0}}$ is always going to be an unrealizeable filter, meaing that we won't ever be able to build a system with this exact response.

Answer 3

• We're interested in taking it back to the time domain

$H(f)\,\!$	$=[u(f+f_{0})-u(f-f_{0})]\,e^{-j\,2\,\pi \,f\,t_{0}}$
$H(t)\,\!$	$=\int _{-\infty }^{\infty }[u(f+f_{0})-u(f-f_{0})]\,e^{-j\,2\,\pi \,f\,t_{0}}e^{j\,2\,\pi \,f\,t}\,df$
	$=\int _{-f_{0}}^{f_{0}}\,e^{j\,2\,\pi \,f\,\left(t-t_{0}\right)}\,df$
	$=\left.{\frac {\,e^{j\,2\,\pi \,f\,(t-t_{0})}}{j\,2\,\pi \,(t-t_{0})}}\right|_{-f_{0}}^{f_{0}}$
	$={\frac {\,e^{j\,2\,\pi \,f_{0}\,(t-t_{0})}-e^{-j\,2\,\pi \,f_{0}\,(t-t_{0})}}{j\,2\,\pi \,(t-t_{0})}}$
	$={\frac {\sin(2\,\pi \,f_{0}\,(t-t_{0}))}{\pi \,(t-t_{0})}}$
	$={\frac {2\,f_{0}}{2\,f_{0}}}{\frac {\sin(2\,\pi \,f_{0}\,(t-t_{0}))}{\pi \,(t-t_{0})}}$
	$=2\,f_{0}\,\mathrm {sinc} (2\,\pi \,f_{0}\,(t-t_{0}))$

• The impulse response is a sinc function. The sinc function was "bouncing" around before the impulse function even came along. In other words, the impulse response knew when the impulse would occur, even though it hadn't happened yet (and it knew this, at t=-\infty). This is a non-causal system and is thus unrealizeable. If it had a u[t] stuck on the end of it, then it would become a causal system and would be perfectly legit.