Question

Problem 5. (10 points) An example of the frequency-shifting property comes from modeling the response of a physical system as an oscillating signal whose amplitude is dropping. The precise form is an exponential times a sinusoid, $x[n] = a^n \cos(\omega_0 n) u[n]$. (a) Use the inverse Euler formula to show that we can express $x[n]$ as $x[n] = \frac{1}{2} e^{j\omega_0 n} a^n u[n] + \frac{1}{2} e^{-j\omega_0 n} a^n u[n]$ (b) By using the DTFT pair in (7.6), applying the frequency-shift property (7.25) twice, and finding a common denominator for the resulting two fractions, show that we can write $X(e^{j\omega})$ as $X(e^{j\omega}) = \frac{1 - a \cos(\omega_0) e^{-j\omega}}{(1 - a e^{j\omega_0} e^{-j\omega})(1 - a e^{-j\omega_0} e^{-j\omega})} = \frac{1 - a \cos(\omega_0) e^{-j\omega}}{1 - 2a \cos(\omega_0) e^{-j\omega} + a^2 e^{-j2\omega}}$

          Problem 5. (10 points)
An example of the frequency-shifting property comes from modeling the response of a physical
system as an oscillating signal whose amplitude is dropping. The precise form is an exponential
times a sinusoid, $x[n] = a^n \cos(\omega_0 n) u[n]$.
(a) Use the inverse Euler formula to show that we can express $x[n]$ as
$x[n] = \frac{1}{2} e^{j\omega_0 n} a^n u[n] + \frac{1}{2} e^{-j\omega_0 n} a^n u[n]$
(b) By using the DTFT pair in (7.6), applying the frequency-shift property (7.25) twice, and
finding a common denominator for the resulting two fractions, show that we can write $X(e^{j\omega})$ as
$X(e^{j\omega}) = \frac{1 - a \cos(\omega_0) e^{-j\omega}}{(1 - a e^{j\omega_0} e^{-j\omega})(1 - a e^{-j\omega_0} e^{-j\omega})} = \frac{1 - a \cos(\omega_0) e^{-j\omega}}{1 - 2a \cos(\omega_0) e^{-j\omega} + a^2 e^{-j2\omega}}$

$Problem 5. (10 points) An example of the frequency-shifting property comes from modeling the response of a physical system as an oscillating signal whose amplitude is dropping. The precise form is an exponential times a sinusoid, x[n] = a^n cos(ω0 n) u[n]. (a) Use the inverse Euler formula to show that we can express x[n] as x[n] = (1)/(2) e^jω0 n a^n u[n] + (1)/(2) e^-jω0 n a^n u[n] (b) By using the DTFT pair in (7.6), applying the frequency-shift property (7.25) twice, and finding a common denominator for the resulting two fractions, show that we can write X(e^jω) as X(e^jω) = (1 - a cos(ω0) e^-jω)/((1 - a e^jω0 e^-jω)(1 - a e^-jω0 e^-jω)) = (1 - a cos(ω0) e^-jω)/(1 - 2a cos(ω0) e^-jω + a^2 e^-j2ω)$

Added by John L.

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