HW 08

Question 1

If the sound track of a movie was played into a high fidelity playback system at twice the correct speed, what happens to a sine wave's frequency, amplitude and phase, relative to what happens at the correct speed? Explain your answers.

Answer 1

Frequency: The frequency is doubled

Amplitude: The amplitude remains the same

Phase: Remains the same

Question 2

Suppose $x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\mathrm{\Omega }(\beta )\mathrm{\Phi }(\beta ,t)d\beta$ and $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{\Phi }}^{*}(\beta ,t)\mathrm{\Phi }(\lambda ,t)dt=\delta (\beta -\lambda )$ where $x(t)$ is any real function of t. If we have a linear time invariant system where an input of $\mathrm{\Phi }(\lambda ,t)$ produces an output of $\mathrm{\Psi }(\lambda ,t)$.

• How do you find $\mathrm{\Omega }(\beta )$ if you are given $x(t)$?
• What is the output due to $cos(2\pi ft)$?

Answer 2

Input	LTI System	Output	Reason
$\mathrm{\Phi }(\lambda ,t)$	$⟹$	$\mathrm{\Psi }(\lambda ,t)$	Given

Question 3

If a signal x(t) only has frequency components near DC, $|X(f)|=0$ for $|f|>f_{max}$, then x(t) is known as a baseband signal. When x(t) is a baseband signal, $x(t)\cos (2\pi f_{0}t)$ is known as a double sideband (DSB) signal. Sometimes a double sideband signal is used to send information over a radio frequency communications link. The transmitter and receiver are shown below.

• Find the Fourier Transform of the DSB signal, $v(t)=x(t)\cos (2\pi f_{0}t)$.
• What is the lowest $f_0$ that can be used and still have the communications system work?
• How does the bandwidth of v(t) compare to the bandwidth of x(t)?
• What does the spectrum of w(t) look like and how does it compare to that of x(t)? A graph would be appropriate showing the spectrum of x(t) and that of w(t).

Answer 3

$v(t)$	$=x(t)\cos (2\pi f_{0}t)$	x(t) is the original signal
$v(f)$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)\cos (2\pi f_{0}t)e^{-j2\pi ft}dt$	v(f) is x(t) after a low pass filter (cutoff frequency = f_max), multiplied by cos(2\pi f_0 t)
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)\frac{e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}}{2}{e}^{-j2\pi ft}dt$
	$=\frac{1}{2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)(e^{j2\pi (f_0-f)t}+e^{-j2\pi (f_0+f)t})dt$
	$=\frac{1}{2}[X(f_0-f)+X(f_0+f)]$
$w(t)$	$=v(t)\cdot \cos (2\pi f_{0}t)$	w(t) is v(t) multiplied by cos(2\pi f_0 t)
	$=x(t)\cos (2\pi f_{0}t{)}^2$
	$=x(t)\frac{e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}}{2}\frac{e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}}{2}$
	$x(t)[\frac{e^{j2\pi (2f_0)t}}{4}+\frac{e^{j2\pi f_{0}t}}{2}+\frac{e^{j2\pi (-2f_0)t}}{4}]$	Need help seeing the math

• The bandwidth of v(t) is 1/2 that of x(t). The cosine splits the (amplitude? of the) signal up in half and moves it up and down by f_0.
• The spectrum of w(t) is 1/2 of x(t) at the center frequency. At +/- 2*f_0, 1/4th of x(t).