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

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

Answer 2

Input	LTI System	Output	Reason
$\Phi (\lambda ,t)\,\!$	$\Longrightarrow$	$\Psi (\lambda ,t)\,\!$	Given

Question 3

If a signal x(t) only has frequency components near DC, $\left|X(f)\right|=0$ 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
$v(f)\,\!$	$=\int _{-\infty }^{\infty }x(t)\cos(2\pi f_{0}t)e^{-j2\pi ft}\,dt$	v(f) is x(t) after a low pass filter (cutoff frequency = f_max), multiplied by cos(2\pi f_0 t)
	$=\int _{-\infty }^{\infty }x(t){\frac {e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}}{2}}e^{-j2\pi ft}\,dt$
	$={\frac {1}{2}}\int _{-\infty }^{\infty }x(t)\left(e^{j2\pi (f_{0}-f)t}+e^{-j2\pi (f_{0}+f)t}\right)\,dt$
	$={\frac {1}{2}}\left[X(f_{0}-f)+X(f_{0}+f)\right]$
$w(t)\,\!$	$=v(t)\cdot \cos(2\pi f_{0}t)$	w(t) is v(t) multiplied by cos(2\pi f_0 t)
	$=x(t)\,\cos(2\pi f_{0}t)^{2}$
	$=x(t)\,{\frac {e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}}{2}}\,{\frac {e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}}{2}}$
	$x(t)\left[{\frac {e^{j2\pi (2f_{0})t}}{4}}+{\frac {e^{j2\pi f_{0}t}}{2}}+{\frac {e^{j2\pi (-2f_{0})t}}{4}}\right]$	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).