MHB Compute Discrete Time Fourier Transform

[nacho-man]

Hi bros,

so I feel like I am very close, but cannot find out how to go further.

Q.1 Compute the DTFT of the following signals, either directly or using its properties (below a is a fixed constant |a| < 1):

for $x_n = a^n \cos(\lambda_0 n)u_n$ where $\lambda_0 \in (0, \pi)$ and
$u_n$ is the step function, i.e $u_n = 1 $ for $n\ge 0 $ and $0$ otherwise.so,

$X(e^{i \lambda}) = \sum_{n=0}^{+\infty} x_n e^{-i\lambda n}$$X(e^{i \lambda}) = \sum_{n=0}^{+\infty} a^n \cos(\lambda_0 n) e^{-i\lambda n}$

using euler's formula: $\cos(\lambda_0 n)$ = $\frac{e^{i\lambda_0 n}+ e^{-i\lambda_0 n}}{2} $

so $X(e^{i \lambda}) = \frac{1}{2} \sum_{n=0}^{+\infty} (a e^{-i \lambda}( e^{i\lambda_0}+ e^{-i\lambda_0}))^n $

which gives

$\frac{1}{2} \sum_{n=0}^{+\infty} a^n(e^{i \lambda_0 - i\lambda} + e^{-i \lambda_0 - i\lambda})^n)$

and now i am stuck
.
i think i have it right upto this point, but i do not know how to proceed.

also in our notes, he has said to use the property called modulation, which meanswhere we have $x^n e^{i \lambda_0 n}$ the DTFT will be of the form $X(e^{i(\lambda-\lambda_0)})$ANY HELP IS APPRECIATED! thank you!

 

[I like Serena]

Hi nacho! :)

$X(e^{i \lambda}) = \sum_{n=0}^{+\infty} a^n \cos(\lambda_0 n) e^{-i\lambda n}$

using euler's formula: $\cos(\lambda_0 n)$ = $\frac{e^{i\lambda_0 n}+ e^{-i\lambda_0 n}}{2} $

so $X(e^{i \lambda}) = \frac{1}{2} \sum_{n=0}^{+\infty} (a e^{-i \lambda}( e^{i\lambda_0}+ e^{-i\lambda_0}))^n $

Not so fast.
Let's first do only the substitution.

$$X(e^{i \lambda}) = \sum_{n=0}^{+\infty} a^n \cdot \frac{1}{2} (e^{i\lambda_0 n}+ e^{-i\lambda_0 n}) \cdot e^{-i \lambda n} $$

This is different from what you have.

which gives

$\frac{1}{2} \sum_{n=0}^{+\infty} a^n(e^{i \lambda_0 - i\lambda} + e^{-i \lambda_0 - i\lambda})^n)$

and now i am stuck
.
i think i have it right upto this point, but i do not know how to proceed.

Let's redo that and simplify to:
$$X(e^{i \lambda})
= \frac{1}{2} \sum_{n=0}^{+\infty} a^n (e^{-i(\lambda - \lambda_0) n}+ e^{-i(\lambda + \lambda_0) n})
= \frac{1}{2} \sum_{n=0}^{+\infty} a^n e^{-i(\lambda - \lambda_0) n} + \frac{1}{2} \sum_{n=0}^{+\infty} a^n e^{-i(\lambda + \lambda_0) n}
$$

also in our notes, he has said to use the property called modulation, which meanswhere we have $x^n e^{i \lambda_0 n}$ the DTFT will be of the form $X(e^{i(\lambda-\lambda_0)})$

Can you apply this now? (Wondering)